SOLUTIONS MANUAL for Engineering Mechanics Statics 6e Anthony Bedford, Wallace Fowler by StudyGuide

Chapter 1 Problem 1.1 In 1967, the International Committee of Weights and Measures defined one second to be the time required for 9,192,631,770 cycles of the transition between two quantum states of the cesium-133 atom. Express the number of cycles in two seconds to four significant digits.

Solution:

Problem 1.2 The base of natural logarithms is e = 2.71828183... . (a) Express e to three significant digits. (b) Determine the value of e 2 to three significant digits. (c) Use the value of e you obtained in part (a) to determine the value of e 2 to three significant digits.

Solution:

The number of cycles in two seconds is 2(9,192,631, 770) = 18,385, 263, 540. Expressed to four significant digits, this is 18,390, 000, 000 or 1.839E10 cycles. 18,390,000,000 or 1.839E10 cycles.

(a) The rounded-off value is e = 2.72. (b) e 2 = 7.38905610..., so to three significant digits it is e 2 = 7.39. (c) Squaring the three-digit number we obtained in part (a) and expressing it to three significant digits, we obtain e 2 = 7.40. (a) e = 2.72. (b) e 2 = 7.39. (c) e 2 = 7.40.

[Comparing the answers of parts (b) and (c) demonstrates the hazard of using rounded-off values in calculations.]

Problem 1.3 The base of natural logarithms (see Problem 1.2) is given by the infinite series 1 1 1 e = 2+ + + + . 2! 3! 4! Its value can be approximated by summing the first few terms of the series. How many terms are needed for the approximate value rounded off to five digits to be equal to the exact value rounded off to five digits?

Solution: The exact value rounded off to five significant digits is e = 2.7183. Let N be the number of terms summed. We obtain the results N

Sum

2.5

2.666...

2.708333...

2.71666...

2.7180555...

2.7182539...

We see that summing seven terms gives the rounded-off value 2.7183. Seven.

Problem 1.4 The opening in the soccer goal is 24 ft wide and 8 ft high, so its area is 24 ft × 8 ft = 192 ft 2 . What is its area in m 2 to three significant digits? Solution: A = 192 ft 2

1m ( 3.281 ) = 17.8 m ft

A = 17.8 m 2

Problem 1.4

Problem 1.5 In 2020, teams from China and Nepal, based on their independent measurements using GPS satellites, determined that the height of Mount Everest is 8848.86 meters. Determine the height of the mountain to three significant digits (a) in kilometers; (b) in miles.

Solution: (a) The height of the mountain in kilometers to three significant digits is 8848.86 m = 8848.86 m

1 km ( 1000 ) = 8.85 km. m

(b) Its height in miles to three significant digits is 8848.86 m = 8848.86 m

ft 1 mi ( 3.281 )( 5280 ) = 5.50 mi. 1m ft

(a) 8.85 km. (b) 5.50 mi.

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Problem 1.6 The distance D = 6 in (inches). The magnitude of the moment of the force F = 12 lb (pounds) about point P is defined to be the product M P = FD. What is the value of M P in N-m (newton-meters)?

Solution: The value of M P is M P = FD = (12 lb)(6 in) = 72 lb-in. Converting units, the value of M P in N-m is

72 lb-in = 72 lb-in

1N ( 0.2248 )( 1 m ) = 8.14 N-m. lb 39.37 in 8.14 N-m.

P D

Problem 1.6

Problem 1.7 The length of this Boeing 737 is 110 ft 4 in and its wingspan is 117 ft 5 in. Its maximum takeoff weight is 154,500 lb. Its maximum range is 3365 nautical miles. Express each of these quantities in SI units to three significant digits.

Solution: Because 1 ft = 12 in, the length in meters is m (110 + 124 ft )( 0.3048 ) = 33.6 m. 1 ft In the same way, the wingspan in meters is m (117 + 125 ft )( 0.3048 ) = 35.8 m. 1 ft The weight in newtons is N ( 4.448 ) = 687,000 N. 1 lb

( 154, 500 lb )

One nautical mile is 1852 meters. Therefore, the range in meters is (3365 nautical miles)

1852 m ( 1 nautical ) = 6.23E6 m. mile

Length = 33.6 m, wingspan = 35.8 m,

Problem 1.7

weight = 687 kN, range = 6230 km.

Problem 1.8 The maglev (magnetic levitation) train from Shanghai to the airport at Pudong reaches a speed of 430 km/h. Determine its speed (a) in mi/h; (b) in ft/s. Solution: (a)

(

)

km 0.6214 mi = 267mi/h . h 1 km v = 267 mi/h

v = 430

(b)

(

km 1000 m h 1 km = 392 ft/s.

v = 430

1 ft )( 0.3048 )( 1 h ) m 3600 s

v = 392 ft/s Source: Courtesy of Qilai Shen/EPA/Shutterstock.

Problem 1.8

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Problem 1.9 In the 2006 Winter Olympics, the men’s 15-km cross-country skiing race was won by Andrus Veerpalu of Estonia in a time of 38 minutes, 1.3 seconds. Determine his average speed (the distance traveled divided by the time required) to three significant digits (a) in km/h; (b) in mi/h.

Solution: (a)

 60 min  15 km   1.3 min  1 h  38 + 60 = 23.7 km/h.

v =

(

)

v = 23.7 km/h

(b)

 1 mi  v = (23.7 km/h) = 14.7 mi/h.  1.609 km  v = 14.7 mi/h

Problem 1.10 The Porsche’s engine exerts 229 ft-lb (foot-pounds) of torque at 4600 rpm. Determine the value of the torque inN-m (newton-meters). Solution: T = 229 ft-lb

1N ( 0.2248 )( 1 m ) = 310 N-m. lb 3.281 ft

T = 310 N-m

Problem 1.10

Problem 1.11 The kinetic energy of the man in Practice Example 1.1 is defined by 12 mv 2, where m is his mass and υ is his velocity. The man’s mass is 68 kg and he is moving at 6 m/s, so his kinetic energy is 12 (68 kg)(6 m/s) 2 = 1224 kg-m 2 /s 2. What is his kinetic energy in US customary units?

Solution:

Problem 1.12 The acceleration due to gravity at sea level in SI units is g = 9.81 m/s 2 . By converting units, use this value to determine the acceleration due to gravity at sea level in US customary units.

Solution:

Problem 1.13 The value of the universal gravitational constant in SI units is G = 6.67E−11 m 3 /kg-s 2 . Use this value and convert units to determine the value of G in US customary units.

Solution:

2  1 slug  1 ft T = 1224 kg-m 2 /s 2    14.59 kg  0.3048 m

(

)

= 903 slug-ft 2 /s. T = 903 slug-ft 2 /s

Use Table 1.2. The result is: 1ft ( sm )( 0.3048m ) = 32.185...( sft ) = 32.2( sft ).

g = 9.81

Converting units, 6.67E − 11 m 3 /kg-s 2 = 6.67E − 11

m 3 3.281 ft 3 kg-s 2 1m

(

)

  1 kg  0.0685 slug  = 3.44E − 8 ft 3 /slug-s 2 . G = 3.44E − 8 ft 3 /slug-s 2

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Problem 1.14 The density (mass per unit volume) of aluminum is 2700 kg/m 3 . Determine its density in slug/ft 3 .

Solution: Converting units, the density is  0.0685 slug  1 m 3 2700 kg/m 3 = 2700 kg/m 3     3.281 ft 1 kg

(

)

5.24 slug/ft 3 . 5.24 slug/ft 3 .

Problem 1.15 The cross-sectional area of the C12×30 American Standard Channel steel beam is A = 8.81 in 2 . What is its cross-sectional area in mm 2 ?

Solution: A = 8.81 in 2

( 25.41 inmm ) = 5680 mm

Problem 1.15

Problem 1.16 A pressure transducer measures a value of 300 lb/in 2. Determine the value of the pressure in pascals. A pascal (Pa) is one newton per square meter.

Solution: Convert the units using Table 1.2 and the definition of the Pascal unit. The result: 300

N 12 in 1ft ( inlb )( 4.448 )( 1 ft ) ( 0.3048 ) 1 lb m

= 2.0683...(10 6 )

( mN ) = 2.07(10 ) Pa. 2

Problem 1.17 A horsepower is 550 ft-lb/s. A watt is 1 N-m/s. Determine how many watts are generated by the engines of the passenger jet if they are producing 7000 horsepower. Solution:  550 ft-lb/s  1 m P = 7000 hp    1 hp  3.28 1ft

(

1N )( 0.2248 ) lb

= 5.22 × 10 6 W. P = 5.22 × 10 6 W

Problem 1.17

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Problem 1.18 A typical speed of the head of a driver on the PGA Tour is 100 miles per hour. Determine the speed in ft/s, m/s, and km/h. Solution: The speed in ft/s is 100 mi/h = 100 mi/h The speed in m/s is

ft 1h ( 5280 )( 3600 ) = 147 ft/s. 1 mi s

147 ft/s = 147 ft/s The speed in km/h is

m ( 0.3048 ) = 44.7 m/s. 1 ft

100 mi/h = 100 mi/h

km ( 1.609 ) = 161 km/h. 1 mi

147 ft/s, 44.7 m/s, 161 km/h.

Source: Courtesy of Takanakai/123RF.

Problem 1.18

Problem 1.19 In 2000, a carbon nanotube was shown to support a tensile stress of 63 GPa (gigapascals). A pascal is 1 N/m 2 . Determine the value of this tensile stress in lb/in 2 . (This is the amount of force in pounds that could theoretically be supported in tension by a bar of this material with a cross-sectional area of one square inch.)

Solution:

Problem 1.20 In dynamics, the moments of inertia of an object are expressed in units of ( mass ) × ( length ) 2. If one of the moments of inertia of a particular object is 600 slug-ft 2, what is its value in kg-m 2?

Solution:

Converting units, the tensile stress is

(

N 1 lb m 2 4.448 N = 9.14E6 lb/in 2 .

63E9 N/m 2 = 63E9

1m )( 39.37 ) in

9.14E6 lb/in 2 .

Converting units, 2  14.59 kg  1 m 600 slug-ft 2 = 600 slug-ft 2    1 slug  3.281 ft

(

)

= 813 kg-m 2 . 813 kg-m 2

Problem 1.21 defined by

The airplane’s drag coefficient C D is

D , S 12 ρ v 2 where D is the drag force exerted on the airplane by the air, S is the wing area, ρ is the density (mass per unit volume) of the air, and v is the magnitude of the airplane’s velocity. At the instant shown, the drag force D = 5300 N, the wing area is S = 100 m 2 , the air density is ρ = 1.226 kg/m 3 , and the velocity is v = 60 m/s. (a) What is the value of the drag coefficient? (b) Determine the values of D, S, ρ, and v in terms of US customary units and use them to calculate the drag coefficient. CD =

Source: Courtesy of Fasttailwind/Shutterstock.

Solution: (a)

Problem 1.21

The drag coefficient is CD = =

(b)

D 1 S ρv 2 2

D = 5300 N 1 2

5300 N

the reference area is

( 100 m 2 ) ( 1.226 kg/m 3 )( 60 m/s ) 2

= 0.0240.

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Converting units, the drag force is

S = 100 m 2

lb ( 0.2248 ) = 1190 lb, 1N 2

ft ( 3.281 ) = 1080 ft , 1m 2

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1.21 (Continued) the density is  0.0685 slug  1 m 3 ρ = 1.226 kg/m 3     3.281 ft 1 kg

(

)

0.00238 slug/ft 3 ,

and the velocity is v = 60 m/s

ft ( 3.281 ) = 197 ft/s. 1m

Therefore, the drag coefficient is CD = =

D 1 S ρv 2 2 1 2

1190 lb

( 1080 ft 2 ) ( 0.00238 slug/ft 3 )( 197 ft/s ) 2

= 0.0240. (a) C D = 0.0240. (b) D = 1190 lb, S = 1080 ft 2 , ρ = 0.00238 slug/ft 3 , v = 197 ft/s, C D = 0.0240.

Problem 1.22 A person has a mass of 72 kg. (a) What is their weight at sea level in newtons? (b) What is their mass in slugs? (c) What is their weight at sea level in pounds?

(c)

Solution: (a)

W = mg = ( 4.93 slugs )( 32.2 ft/s 2 ) = 159 lb. Or we could use a conversion factor to convert their weight from newtons to pounds, obtaining

From Eq. (1.6), their weight is W = mg

lb ( 0.2248 ) 1N

W = ( 706 N )

= ( 72 kg )( 9.81 m/s 2 ) = 706 N. (b)

Using Eq. (1.6), their weight is

= 159 lb.

Using a unit conversion factor, their mass in slugs is

(a) 706 N. (b) 4.93 slugs. (c) 159 lb.

 0.0685 slug  m = ( 72 kg )    1 kg = 4.93 slugs.

Problem 1.23 The 1 ft × 1 ft × 1 ft cube of iron weighs 490 lb at sea level. Determine the weight in newtons of a 1 m × 1 m × 1 m cube of the same material at sea level.

1 ft

Solution:

The weight density is γ = 490 lb 1 ft 3

The weight of the 1 m 3 cube is: W = γV =

1 ft 3

lb 1 ft 1N ( 490 )(1 m) ( 0.3048 ) ( 0.2248 ) 1 ft m lb 3

1 ft

Problem 1.23

= 77.0 kN

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Problem 1.24 The area of the Pacific Ocean is 64,186,000 square miles and its average depth is 12,925 ft. Assume that the weight per unit volume of ocean water is 64 lb/ft 3. Determine the mass of the Pacific Ocean (a) in slugs; (b) in kilograms.

Solution: The volume of the ocean is V = (64,186, 000 mi 2 )(12, 925 ft)

( 5,1280mi ft )

= 2.312 × 10 19 ft 3 . (a) m = ρV =

64 lb/ft ( 32.2 )(2.312 × 10 ft ) ft/s 3

= 4.60 × 10 19 slugs (b)  14.59 kg  m = (4.60 × 10 19 slugs)   1slug  = 6.71 × 10 20 kg

Problem 1.25 Two oranges weighing 8 oz (ounces) each at sea level are 3 ft apart. (a) What is the magnitude of the gravitational force F they exert on each other? (b) If a stationary object of mass m is subjected to a constant force F and no other forces, the time required for it to reach a speed v is mv t = . F

Solution: (a)

m =

W 0.5 lb = = 0.0155 slug. g 32.2 ft/s 2

In U.S. customary units, the universal gravitational constant is G = 3.44E − 8 ft 3 /slug-s 2 . From Eq. (1.1), the gravitational force the oranges exert on each other is

If you subjected one of the oranges to a constant force equal to the force you determined in part (a) and no other forces, how long would it take to reach a speed of 1 ft/s?

3 ft

One pound equals 10 ounces, so the weight of each orange is W = 0.5 lb, and its mass is

Gm 2 r2 ( 3.44E − 8 ft 3 /slug-s 2 )( 0.0155 slug ) 2 = ( 3 ft ) 2 = 9.22E − 13 lb.

F =

(b)

From the given equation, the time required is mv F ( 0.0155 slug )( 1 ft/s ) = 9.22E − 13 lb = 1.68E10 s (534 years).

t =

Problem 1.25

(a) 9.22E − 13 lb. (b) 1.68E10 s (534 years).

Problem 1.26 A person weighs 180 lb at sea level. The radius of the Earth is 3960 mi. What force is exerted on the person by the gravitational attraction of the Earth if he is in a space station in orbit 200 mi above the surface of the Earth?

Solution: Use Eq. (1.5). 2

( Rr ) =  Wg  g R R+ H  3960 = W ( 3960 + 200 )

W = mg

= (180)(0.90616) = 163 lb.

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Problem 1.27 The acceleration due to gravity on the surface of the Moon is 1.62 m/s 2 . The Moon’s radius is R M = 1738 km. (a) What is the weight in newtons on the surface of the Moon of an object that has a mass of 10 kg? (b) Using the approach described in Practice Example 1.6, determine the force exerted on the object by the gravity of the Moon if the object is located 1738 km above the Moon’s surface.

Solution: (a)

(b)

W = mg M = (10 kg)(1.26 m/s 2 ) = 12.6 N W = 12.6 N Adapting Eq. 1.4, we have a M = g M R M r

( ) . The force is then

F = ma M = (10 kg)(1.62 m/s 2 )

1738 km ( 1738 km + 1738 km )

= 4.05 N F = 4.05 N

Problem 1.28 If an object is near the surface of the Earth, the variation of its weight with distance from the center of the Earth can often be neglected. The acceleration due to gravity at sea level is g = 9.81 m/s 2. The radius of the Earth is 6370 km. The weight of an object at sea level is mg, where m is its mass. At what height above the surface of the Earth does the weight of the object decrease to 0.99 mg ?

Solution:

Problem 1.29 A person has a mass of 72 kg. The radius of the Earth is 6370 km. (a) What is his weight at sea level in newtons and in pounds? (b) If he is in a space station 420 km above the surface of the Earth, what force is exerted on him by the Earth’s gravitational attraction in newtons and in pounds?

(b)

Use a variation of Eq. (1.5).  R E  2 W = mg   = 0.99mg  R E + h  Solve for the radial height, h = RE

= 32.09 ... km = 32,100 m = 32.1 km.

To apply Eq. (1.5), we need his distance from the center of the earth: r = 6370 km + 420 km = 6790 km. Then from Eq. (1.5), the grativational force on him is W = mg

Solution: (a)

1 − 1 ) = (6370)(1.0050378 − 1.0) ( 0.99

From Eq. (1.6), his weight at sea level is

= (72 kg)(9.81 m/s 2 )

W = mg = (72 kg)(9.81 m/s 2 ) = 706 N. Using a conversion factor, his weight at sea level in pounds is

(

0.2248 lb W = 706 N 1N = 159 lb.

RE 2 r2

)

(6,370, 000 m) 2 (6, 790, 000 m) 2

= 622 N. The force on him in pounds is W = 622 N

lb ( 0.2248 ) 1N

= 140 lb. (a) 706 N, 159 lb. (b) 622 N, 140 lb.

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Problem 1.30 If you model the Earth as a homogeneous sphere, at what height above sea level does your weight decrease to one-half of its value at sea level? Determine the answer (a) in kilometers; (b) in miles. (The radius of the Earth is 6370 km. )

Gm E . ( RE + H ) 2 Dividing this equation by Eq. (1) yields

Solution:

Solving for H, we obtain

(a)

0.5g =

0.5 =

Your weight will decrease to 0.5 times its value at sea level at the height at which the earth’s acceleration due to gravity decreases to 0.5 times its value at sea level. From Eq. (1.3), the acceleration due to gravity at a distance r from the center of the earth is Gm E , r2 where m E is the earth’s mass. The acceleration due to gravity at sea level is a =

Gm E , (1) R E2 where R E is the earth’s radius. Let H be the height above sea level at which the acceleration due to gravity equals 0.5g. That is, g =

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R E2 . ( RE + H ) 2

1 − 1) R ( 0.5 1 = ( − 1 ) (6370 km) 0.5

H =

= 2640 km. (b)

In terms of miles, the value of H is H = 2640 km

1 mi ( 1.609 ) km

= 1640 mi. (a) 2640 km. (b) 1640 mi.

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Chapter 2 Problem 2.1 Consider vectors U and V oriented as shown. Their magnitudes are U = 8 and V = 3. Graphically determine the magnitude of the vector 2U − 3V.

Solution: The vector 2U has the same direction as U and magnitude 16. The vector −3V has the direction opposite to V and magnitude 9. We choose a scale and add these vectors to obtain the vector 2U − 3 V: 45o

208 458

2U U V

2U – 3V 20o 3V

By measuring its magnitude, we estimate that 2U − 3V = 21.5. 2U − 3V = 21.5.

Problem 2.2 Suppose that the pylon in Example 2.2 is moved closer to the stadium so that the angle between the forces F AB and F AC is 45 °. Draw a sketch of the structure with the cables in their new orientation. The magnitudes of the forces are F AB = 100 kN and F AC = 60 kN. Graphically determine the magnitude and direction of the sum of the forces exerted on the pylon at A by the two cables.

Solution: FAB

45o

FAB + FAC

100 kN

28o 60 kN

FAC

From the diagram we estimate that F AB + F AC = 148 kN at 48 ° relative to the horizontal. 148 kN at 28 ° from horizontal.

Problem 2.3 Two unit vectors e 1 and e 2 are oriented as shown. Graphically determine the magnitude of the vector U = 2e 1 + 3e 2 .

Solution: The vector 2e 1 has the same direction as e 1 and magnitude 2. The vector 3e 2 has the same direction as e 2 and magnitude 3. We choose a scale and add these vectors to obtain the vector 2e 1 + 3 e 2 :

2e1 1 3e2 3e2

608

2e1 e1

By measuring its magnitude, we estimate that 2e 1 + 3 e 2 = 4.4. U = 4.4.

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Problem 2.4 Two unit vectors e 1 and e 2 are oriented as shown. Graphically determine the magnitude of the vector U = 4 e 1 − 3e 2 .

Solution: The vector 4e 1 has the same direction as e 1 and magnitude 4. The vector −3e 2 has the direction opposite to e 2 and magnitude 3. We choose a scale and add these vectors to obtain the vector 4 e 1 − 3 e 2 : 4e1

608

]3e2

4e1 ] 3e2

By measuring its magnitude, we estimate that 4 e 1 − 3 e 2 = 3.6. U = 3.6.

Problem 2.5 Three forces act on the ring. Their sum is zero: F A + FB + FC = 0. The magnitudes FB = 100 N and FC = 160 N. Graphically determine the magnitude of F A and the angle α.

Solution: If the sum of the three forces equals zero, the three force vectors form a closed triangle when placed head-to-tail. We choose a scale and draw the force triangle: o

FB o

408

FA a 808 By measuring, we estimate that F A = 140 lb, α = 42 °. F A = 140 lb, α = 42 °.

Problem 2.6 Three forces act on the ring. Their sum is zero: F A + FB + FC = 0. The magnitude FC = 200 lb. Graphically determine the value of the angle α for which the magnitude of F A is a minimum. If α has that value, what are the magnitudes of F A and FB ?

Solution: The magnitude and direction of FC are known. The direction of FB is known. If the sum of the three forces equals zero, the three force vectors form a closed triangle when placed head-to-tail. Choosing a scale and drawing the force triangle, we can see that the magnitude of F A is a minimum when it is perpendicular to the line of action of FB . Constructing the triangle in this way,

408

40 FA

808 a

and measuring, we estimate that α = 50 °, F A = 173 lb, FB = 100 lb. α = 50 °, F A = 173 lb, FB = 100 lb.

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Problem 2.7 A T-shaped structural member is suspended from cables. The member is subjected to three forces: the forces F A and FB exerted by the cables and its weight W. The weight of the member is W = 1000 lb. The sum of the forces is zero. Graphically determine the magnitudes of F A and FB .

Solution:

The directions of all three vectors and the magnitude of W are known. If the sum of the three forces equals zero, the three force vectors form a closed triangle when placed head-to-tail. We choose a scale and draw the force triangle:

FB 308 W

408 408

308

408 F A

308

By measuring, we estimate that F A = 530 lb, FB = 690 lb. F A = 530 lb, FB = 690 lb.

Problem 2.8 A T-shaped structural member is suspended from cables. The member is subjected to three forces: the forces F A and FB exerted by the cables and its weight W. The sum of the forces is zero. Suppose that the magnitude of the force exerted by either cable must not exceed 1200 lb. Graphically determine the largest magnitude of the weight W that can be suspended in this way.

Solution:

The directions of all three vectors are known. If the sum of the three forces equals zero, the three force vectors form a closed triangle when placed head-to-tail. We draw the force triangle:

FB 308 W

408

308

The force FB has the larger magnitude of the two cable forces. It must not exceed 1200 lb. Assuming it has that value and measuring, we estimate that the magnitude of W is 1740 lb.

408 F A

308

1740 lb.

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Problem 2.9 The 2700-kg tractor is at rest on the inclined surface. The angle α = 30 ° relative to the horizontal. The forces exerted on the tractor’s tires by the inclined surface are represented by the normal force N, which is perpendicular to the surface, and the friction force f, which is tangential to the surface. The sum of the normal force, the friction force, and the tractor’s weight is zero: N + f + W = 0. Graphically determine the magnitudes of N and f in newtons.

Solution: The tractor’s weight is (2700 kg)(9.81 m/s 2 ) = 26.5 kN. The directions of all three vectors and the magnitude of W are known. If the sum of the three forces equals zero, the three force vectors form a closed triangle when placed head-to-tail. We choose a scale and draw the force triangle: f 308

W N 308

By measuring, we estimate that N = 23.0 kN, f = 13.5 kN. W

N = 23.0 kN, f = 13.5 kN.

Source: Courtesy of Tka4ko/Shutterstock.

Problem 2.10 The 2700-kg tractor is at rest on the inclined surface. The forces exerted on the tractor’s tires by the inclined surface are represented by the normal force N, which is perpendicular to the surface, and the friction force f, which is tangential to the surface. The sum of the normal force, the friction force, and the tractor’s weight is zero: N + f + W = 0. Graphically determine the value of the angle α relative to the horizontal for which the magnitudes of N and f are equal. What is their magnitude?

Solution:

The tractor’s weight is (2700 kg)(9.81 m/s 2 ) = 26.5 kN. If the sum of the three forces equals zero, the three force vectors form a closed triangle when placed head-to-tail. The vector f is normal (perpendicular) to the vector N, so these two vectors can be of equal magnitude only if they have the following configuration:

f 458 W

N 458 We see immediately that α = 45 °. Measuring, we estimate that N = f = 18.8 kN. f a

α = 45 °, N = f = 18.8 kN. W N

Source: Courtesy of Tka4ko/Shutterstock.

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Problem 2.11 A spherical storage tank is suspended from cables. The tank is subjected to three forces: the forces F A and FB exerted by the cables and its weight W. The weight of the tank is W = 800 lb. The vector sum of the forces acting on the tank equals zero. Graphically determine the magnitudes of F A and FB .

Solution:

Measuring a vertical distance of 8 units and drawing two lines at 20 ° to where they intersect, we obtain the diagram

FB 208 208

408

208

Measuring the magnitudes, we estimate that F A = FB = 424 lb. F A = FB = 424 lb. W

Solution:

Problem 2.12 The rope ABC exerts forces FBA and FBC of equal magnitude on the block at B. The magnitude of the total force exerted on the block by the two forces is 200 lb. Graphically determine FBA .

Draw the vectors accurately and then measure the unknown

magnitudes.

|FBA| 5 174 lb |FBC|

FBC

208 208

B |FBA| |R| 5 200 lb FBA

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Problem 2.13 Two snowcats tow an emergency shelter to a new location near McMurdo Station, Antarctica. (The top view is shown. The cables are horizontal.) The total force F A + FB exerted on the shelter is in the direction of the line L and has a magnitude of 300 lb. Graphically determine the magnitudes of F A and FB .

Solution: Measuring a vertical distance of 3 units and drawing two lines at 30 ° and 50 ° as shown to identify where they intersect, we obtain the diagram FA FA 1 FB

508

300 lb

308

508

Measuring the magnitudes, we estimate that F A = 150 lb, FB = 234 lb. F A = 150 lb, FB = 234 lb.

Top View

Problem 2.14 A surveyor determines that the horizontal distance from A to B is 400 m and the horizontal distance from A to C is 600 m. Graphically determine the magnitude of the vector rBC and the angle α.

Solution: Draw the vectors accurately and then measure the unknown magnitude and angle. |rBC| 5 390 m a 5 21.28

North B

A a

|rBC|

rBC C 608 208

East

Problem 2.15 The figure shows three forces acting on a joint of a structure. The magnitude of FC is 60 kN, and F A + FB + FC = 0. Graphically determine the magnitudes of F A and FB .

Solution: Measuring a vertical distance of 6 units and drawing two lines at 40 ° and 15 ° as shown to identify where they intersect, we obtain the diagram

y FC

158

FB 158

408 FA

60 kN FC 408 Measuring the magnitudes, FB = 110 kN.

estimate

that

F A = 138 kN,

F A = 138 kN, FB = 110 kN.

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Problem 2.16 explain why

By drawing sketches of the vectors,

First we add U and V, then add W:

U + ( V + W) = (U + V) + W. Solution:

Consider three arbitrary vectors:

U U1V

U1V V

(U 1 V) 1 W

From this diagram we can also see that U + ( V + W) = (U + V) + W:

U V 1 W

W U 1 (V 1 W)

Problem 2.17 Two forces are given in terms of their components by F A = 60 i − 20 j (lb), FB = 30 i + 40 j (lb). (a) What are the magnitudes of the forces? (b) What is the magnitude of their sum F A + FB?

Solution: (a) FA =

(60 lb) 2 + (−20 lb) 2

= 63.2 lb. FB =

Strategy: The magnitude of a vector in terms of its components is given by Eq. (2.8).

(30 lb) 2 + (40 lb) 2

= 50.0 lb. (b) F A + FB = (60 + 30) i + (−20 + 40) j (lb) = 90 i + 20 j (lb). F A + FB =

(90 lb) 2 + (20 lb) 2

= 92.2 lb. (a) F A = 63.2 lb, FB = 50.0 lb. (b) F A + FB = 92.2 lb.

Problem 2.18 Consider three vectors A = 12 i + 8 j, B = B x i + B y j, and C = C x i + C y j. Their sum is zero, A + B + C = 0, and the components of B and C satisfy the relations B x = −2 B y and C x = 3C y . What are the magnitudes of B and C?

Solving these equations together with the two given equations, we obtain

Solution:

If the sum of the three vectors is zero, that implies that each component of their sum is zero. That yields the two equations

B =

12 + B x + C x = 0,

C = (−16.8) 2 + (−5.6) 2 = 17.7.

B = 4.8i − 2.4 j, C = −16.8i − 5.6 j. Therefore

8 + B y + C y = 0.

(4.8) 2 + (−2.4) 2

= 5.37,

B = 5.37, C = 17.7.

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Problem 2.19 The vector r = xi + yj extends to a point on the curve described by the equation shown. Determine the values of x and y for which the magnitude of the vector is a minimum. What is the minimum magnitude? y

We equate the derivative of this expression to zero, −2 x +

1 3 x = 0, 4

obtaining the roots x = 0 and x = ± 8. Substituting these values into the expression for the magnitude in terms of x, we find that r = 4 at x = 0 and r = 3.46 at x = ± 8. So the minimum magnitude occurs at x = ± 8. At both of these points, y = 2. If we examine a graph of the magnitude as a function of x, 8 7.5

(x, y)

Solution: r =

1 4 + x . 16

We want to find the value of x for which the magnitude of r is a minimum. We can simplify things a bit by seeking the value of x for which the square of the magnitude of r is a minimum: r 2 = 16 − x 2 +

5.5 5

x2 + y2.

Substituting the relation y = 4 − (1/4) x 2 into this relation gives r =

4.5

The magnitude of r is

16 − x 2

Magnitude

6.5 y 5 4 2 1 x2 4

1 4 x . 16

3 –6

–4

–2

0 x

we see that this is an interesting case, x = 0 is a stationary point but is not the minimum magnitude. x = ± 8, y = 2, magnitude = 3.46.

Problem 2.20 The forces F A = 90 i (kN) and FB = 60 j (kN). The force FC = c ( −2 i − j ), where c is a parameter. Determine the value of c so that the magnitude of the total force exerted on the beam by the three forces is a minimum. What is the minimum magnitude? y FB

3.5

FC FA

Solution:

The total force is

F A + FB + FC = (90 − 2c) i + (60 − c) j (kN). Its magnitude is F A + FB + FC =

(90 − 2c) 2 + (60 − c) 2

8100 − 360 c + 4 c 2 + 3600 − 120 c + c 2

11,700 − 480 c + 5c 2 .

We can seek the value of c that makes the square of the magnitude a minimum. We set d (11,700 − 480 c + 5c 2 ) = −480 + 10 c = 0, dc obtaining c = 48 kN. Substituting this value into the expression for the magnitude gives F A + FB + FC = 13.4 kN. c = 48 kN, magnitude = 13.4 kN.

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Problem 2.21 The forces acting on the sailplane are its weight W = −500 j (lb), the lift L, and the drag D. The angle between the lift vector L and the vertical y-axis is 20°, and the drag vector is perpendicular to the lift vector. At the instant shown, the sum of the forces on the sailplane is zero: L + D + W = 0. Determine the magnitudes of L and D.

Solution: y L

208

208 W

x L Let L and D denote the magnitudes of the lift and drag. Then D

L = L sin 20 °i + L cos 20 ° j, D = −D cos 20 °i + D sin 20 ° j. We have

L + D + W = ( L sin 20 ° − D cos 20 °) i + ( L cos 20 ° + D sin 20 ° − 500 lb) j

= 0. Each component must equal zero, resulting in two equations: L sin 20 ° − D cos 20 ° = 0, L cos 20 ° + D sin 20 ° = 500 lb. Solving, we obtain L = 470 lb, D = 171 lb. L = 470 lb, D = 171 lb.

Problem 2.22 Two unit vectors e 1 and e 2 are oriented as shown. Determine the magnitude of the vector U = 2e 1 + 3e 2 . Strategy: Introduce a cartesian coordinate system and express e 1 and e 2 in terms of their components.

Solution:

We introduce the coordinate system shown: y

e2 608 e1 e2

608

The unit vectors have magnitude one, so we can express them as e 1 = i, e 2 = cos60 °i + sin 60 ° j. e1

With these expressions, the vector U = 2e 1 + 3e 2 = (2 + 3cos60 °) i + 3sin 60 ° j. Its magnitude is U = (2 + 3cos60 °) 2 + (3sin 60°) 2 = 4.36. U = 4.36.

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Problem 2.23 Two forces act on the support. Their magnitudes are F A = 300 lb, FB = 200 lb. Determine the magnitude F A + FB of the total force exerted on the support by the two forces. y

Solution:

Using similar triangles to determine the components, the force F A in terms of its components is 11 (300 lb) i + (11) 2 + (5) 2 = 273i + 124 j (lb).

FA =

5 (300 lb) j (11) 2 + (5) 2

The force FB in terms of its components is

FB FA 5

7 (200 lb) i + (7) 2 + (5) 2 = −163i + 116 j (lb).

FB = −

5 (200 lb) j (7) 2 + (5) 2

The total force is x

F A + FB = 110 i + 240 j (lb), and the magnitude of the total force is F A + FB =

(110 lb) 2 + (240 lb) 2

= 265 lb. F A + FB = 265 lb.

Problem 2.24 The person exerts a 30-lb force F to push the crate onto a truck. (a) Express F in terms of components using the coordinate system shown. (b) The weight of the crate is 120 lb. What is the magnitude of the sum of the two forces exerted by the person and the crate’s weight?

Solution: y

F Fy

208 Fx

W F

208

(a) The force F in terms of its components is

F = Fx i + Fy j x

= (30 lb) cos 20 ° i + (30 lb)sin 20 ° j = 28.2 i + 10.3 j (lb). (b) The sum of the forces is F + W = 28.2 i + 10.3 j − 120 j (lb) = 28.2 i − 110 j (lb). Its magnitude is F + W = (28.2 lb) 2 + (−110 lb) 2 = 113 lb. (a) F = 28.2 i − 110 j (lb). (b) 113 lb.

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Problem 2.25 The missile’s engine exerts a 130-kN thrust F. (a) Express F in terms of components using the coordinate system shown. (b) The mass of the missile is 1800 kg. What is the magnitude of the sum of the thrust F and the missile’s weight?

(a) The thrust F in terms of its components is F = Fx i + Fy j 4 3 (130 kN) i + (130 kN) j 5 5 = 104 i + 78 j (kN). =

(b) The missile’s weight is

W = mg = (1800 kg)(9.81 m/s 2 ) = 17.7 kg.

The sum of the forces is

F + W = 104 i + 78 j − 17.7 j (kN) = 104 i + 60.3 j (kN). x

Its magnitude is

Solution:

F + W = (104 kN) 2 + (60.3 kN) 2 = 120 kN.

(a) F = 104 i + 78 j (kN). (b) 120 kN.

F Fy

Fx W

Problem 2.26 The magnitude of the force F A is 8 kN. The magnitude of the vertical force FB is 2 kN. For what value of the angle α in the range 0 ≤ α ≤ 90 ° is the magnitude of the sum of the two forces equal to 9 kN ?

Solution: In terms of the coordinate system shown, the two forces in terms of their components are F A = (8 kN)(cos αi + sin α j), FB = (2 kN) j. The magnitude of their sum is

F A + FB = = FA

68 + 32sin α (kN).

We want to determine the value of α for which F A + FB =

(8cos α) 2 + (8sin α + 2) 2 (kN)

68 + 32sin α (kN) = 9 kN.

Squaring both sides gives the equation 68 + 32sin α = 81. Solving, we obtain α = 24.0 °. α = 24.0 °. y

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Problem 2.27 The x-axis is parallel to the bar AB. The x ′-axis is parallel to the bar BC. The magnitude of the vector F is 200 lb. (a) Express F in terms of its components in terms of the x – y coordinate system. Use these components to determine the m agnitude of F. (b) Express F in terms of its components in terms of the x ′ – y ′ coordinate system. Use these components to determine the magnitude of F. y9

5 ft

x9 x

(a) The angle between bar AB and bar BC is arctan

( 105 ) = 26.6°.

Therefore, the angle between the vector F and the x-axis is 26.6 ° + 40 ° = 66.6 °. The vector F expressed in terms of the x -y coordinate system is F = (200 lb) cos66.6 °i + (200 lb)sin 66.6 ° j = 79.5i + 183.5 j (lb). Using these components to determine the magnitude of F yields

F 408

Solution:

F = (79.5 lb) 2 + (183.5 lb) 2 = 200 lb. (b) The vector F expressed in terms of the x ′-y ′ coordinate system is F = (200 lb) cos 40 °i′ + (200 lb)sin 40 ° j′ = 153.2 i′ + 128.6 j′ (lb).

10 ft

Using these components to determine the magnitude of F yields F = (153.2 lb) 2 + (128.6 lb) 2 = 200 lb. (a) F = 79.5i + 183.5 j (lb), F = 200 lb. (b) F = 153.2 i′ + 128.6 j′ (lb), F = 200 lb.

Problem 2.28 Suppose that the magnitude of the component of the vector F parallel to the x ′-axis is 150 lb. (a) What is the magnitude of F? (b) Express F in terms of its components in terms of the x – y coordinate system. y9

5 ft

F cos 40 ° = 150 lb for the magnitude of F yields F = 196 lb. (b) The angle between bar AB and bar BC is

F 408

x9 x

Solution: (a) The component of F parallel to the x ′-axis is F cos 40 °. Solving

10 ft

arctan

( 105 ) = 26.6°.

Therefore the angle between the vector F and the x-axis is 26.6 ° + 40 ° = 66.6 °. The vector F expressed in terms of the x -y coordinate system is F = (196 lb) cos66.6 °i + (196 lb)sin 66.6 ° j = 77.9 i + 180 j (lb). (a) F = 196 lb. (b) F = 77.9 i + 180 j (lb).

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Problem 2.29 The magnitudes A = 12 and B = 10. (a) Express A in terms of its components in terms of the x ′ – y ′ coordinate system. (b) Express B in terms of its components in terms of the x – y coordinate system. y9

Solution: (a) The angle between the vector A and the x ′-axis is 20 ° + 40 ° = 60 °. The vector A expressed in terms of the x ′-y ′ coordinate system is A = 12 cos60 °i′ + 12sin 60 ° j′ = 6 i′ + 10.4 j′. (b) The angle between the vector B and the x-axis is 15 ° − 40 ° = −25 °. The vector B expressed in terms of the x -y coordinate system is B = 10 cos 25 °i − 10 sin 25 ° j = 9.06 i − 4.23 j.

158

408

(a) A = 6 i′ + 10.4 j′. (b) B = 9.06 i − 4.23 j.

A 208

A = 12

Problem 2.30 The magnitudes B = 10. Determine A + B . y9

and

Solution:

To determine the sum of A and B, we must express them in terms of their components in terms of the same coordinate system. The angle between the vector A and the x ′-axis is 20 ° + 40 ° = 60 °. The vector A expressed in terms of the x ′-y ′ coordinate system is A = 12 cos60 °i′ + 12sin 60 ° j′ = 6 i′ + 10.4 j′.

The vector B expressed in terms of the x ′-y ′ coordinate system is 158

408

B x9

B = 10 cos15 °i′ + 10 sin15 ° j′ = 9.66 i′ + 2.59 j′. The sum of A and B is A + B = (6 i′ + 10.4 j′) + (9.66 i′ + 2.59 j′) = 15.7 i′ + 13.0 j′.

Therefore A 208

A + B = (15.7) 2 + (13.0) 2 = 20.3. A + B = 20.3.

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Problem 2.31 In Practice Example 2.3, the cable AB exerts a 900-N force on the top of the tower. Suppose that the attachment point B is moved in the horizontal direction farther from the tower, and assume that the magnitude of the force F the cable exerts on the top of the tower is proportional to the length of the cable. (a) What is the distance from the tower to point B if the magnitude of the force is 1000 N? (b) Express the 1000-N force F in terms of components using the coordinate system shown.

Solution:

In the new problem assume that point B is located a distance d away from the base. The lengths in the original problem and in the new problem are given by L original = L new =

(40 m) 2 + (80 m) 2 =

8000 m 2

d 2 + (80 m) 2

(a) The force is proportional to the length. Therefore 1000 N = (900 N) d =

d 2 + (80 m) 2 8000 m 2

(8000 m 2 )

N ( 1000 ) − (80 m) = 59.0 m. 900 N 2

d = 59.0 m (b) The force F is then  80 m j  d 2 + (80 m) 2 

 d F = (1000 N)  i −  d 2 + (80 m) 2 = (593i − 805 j) N. F = (593i − 805 j) N

Solution:

Problem 2.32 Determine the position vector r AB in terms of its components if (a) θ = 30 °; (b) θ = 225 °.

(a) r AB = (60) cos(30 °) i + (60)sin(30 °) j, or r AB = 51.96 i + 30 j mm. And (b) r AB = (60) cos(225 °)i + (60)sin (225 °)j or r AB = −42.4 i − 42.4 j mm.

150 mm

60 mm

y rAB B u A

150 mm

60 mm

rBC C

FAB A

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FBC C

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Problem 2.33 In Example 2.4, the coordinates of the fixed point A are (17, 1) ft. The driver lowers the bed of the truck into a new position in which the coordinates of point B are (9, 3) ft. The magnitude of the force F exerted on the bed by the hydraulic cylinder when the bed is in the new position is 4800 lb. Draw a sketch of the new situation. Express F in terms of components.

Solution:

Problem 2.34 Using laser instruments, a surveyor determines that the coordinates of point B are x = 120 m, y = 122 m. He knows that rOA = 125 m and r AB = 62 m. Determine the vector rOA .

Solution:

θ = tan −1

( 82 ftft ) = 14.04°

F = 4800 lb(− cos θ i + sin θ j). F = (−4660 i + 1160 j) lb

a B A

N b x

rAB

We can apply geometry. The angle β = arctan (122 m/120 m) = 45.5 °. The side b = (122 m) 2 + (120 m) 2 = 171 m, and we are told that a = 125 m, c = 62 m. Applying the law of cosines (Appendix A), rOA

c 2 = a 2 + b 2 − 2ab cos α,

Proposed roadway

we obtain cos α = 0.956, so α = 16.3 °. We can now determine the position vector of A: x

rOA = (125 m) cos(α + β ) i + (125 m)sin(α + β ) j = 59.1i + 110 j (m). rOA = 59.1i + 110 j (m).

Problem 2.35 The magnitude of the position vector rBA from point B to point A is 6 m and the magnitude of the position vector rCA from point C to point A is 4 m. What are the components of rBA? Solution:

y 3m B

The coordinates are: A( x A , y A ), B(0, 0), C (3 m, 0) Thus

rBA = ( x A − 0) i + ( y A − 0) j ⇒ (6 m) 2 = x 2A + y 2A rCA = ( x A − 3 m) i + ( y A − 0) j ⇒ (4 m) 2 = ( x A − 3 m) 2 + y 2A Solving these two equations, we find x A = 4.833 m, y A = ±3.555 m. We choose the “-” sign and find

rBA = (4.83i − 3.56 j) m

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Problem 2.36 In Problem 2.35, determine the components of a unit vector e CA that points from point C toward point A. Strategy: Determine the components of rCA and then divide the vector rCA by its magnitude. Solution:

y 3m B

From the previous problem, we have

rCA = (1.83i − 3.56 j) m, rCA =

1.83 2 + 3.56 2 m = 3.56 m.

Thus e CA =

rCA = (0.458i − 0.889 j) rCA

Problem 2.37 The hydraulic cylinder BC exerts a force F on the boom of the crane at C. The force points from B toward C, and its magnitude is F = 300 kN. Determine the components of F.

Solution: y

2.4 m

a C 2.4 m 1m

1.2 m x

1.8 m

The angle α = arctan(2.4/1.2) = 63.4 °, so F = (300 kN)(cos αi + sin α j) = 134 i + 268 j (kN).

1.2 m

Alternatively, the distance from B to C is

7m x

(1.2 m) 2 + (2.4 m) 2 = 2.68 m, so 1.2 m 2.4 m (300 kN) i + (300 kN) j 2.68 m 2.68 m = 134 i + 268 j (kN).

F =

F = 134 i + 268 j (kN).

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Problem 2.38 The length of the bar AB is 0.6 m. Determine the components of a unit vector e AB that points from point A toward point B.

Solution:

We need to find the coordinates of point B( x, y ) We have the two equations (0.3 m + x ) 2 + y 2 = (0.6 m) 2 x 2 + y 2 = (0.4 m) 2 B

Solving we find x = 0.183 m, y = 0.356 m.

e AB =

0.6

0.4

Thus r AB (0.183 m − [−0.3 m]) i + (0.356 m) j = r AB (0.183 m + 0.3 m) 2 + (0.356 m) 2

= (0.806 i + 0.593 j)

Problem 2.39 The link AC exerts a force F AC on the hydraulic cylinder at C. The force points from C toward A, and its magnitude is F AC = 650 N. Determine the components of F AC .

0.3 m

Solution: y

C FAC

1m D

0.6 m a

Hydraulic cylinder

1m 0.6 m

0.15 m x

0.6 m 0.15 m

Scoop

The angle α = arctan(0.15/0.6) = 14.0 °, so x

F AC = (650 N)(sin αi − cos α j) = 158i − 631 j (N). Alternatively, the distance from A to C is (0.15 m) 2 + (0.6 m) 2 = 0.618 m, so 0.15 m 0.6 m (650 N) i − (650 N) j 0.618 m 0.618 m = 158i − 631 j (N).

F AC =

F AC = 158i − 631 j (N).

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Problem 2.40 The link CD exerts a force FCD on the hydraulic cylinder at C. The force points from C toward D, and its magnitude is FCD = 600 N. Determine the components of FCD .

1m D C

Hydraulic cylinder

Solution:

The angle α = arctan(0.4/1.0) = 21.8 °, so

0.6 m

FCD = (600 N)(cos αi + sin α j) = 557 i + 223 j (N).

0.6 m

Alternatively, the distance from C to D is

Scoop

0.15 m

(1 m) 2 + (0.4 m) 2 = 1.077 m, so

1.0 m 0.4 m FCD = (600 N) i + (600 N) j 1.077 m 1.077 m = 557 i + 223 j (N).

FAC

FCD = 557 i + 223 j (N).

0.6 m a

A 0.15 m x

Problem 2.41 A surveyor finds that the length of the line OA is 1500 m and the length of the line OB is 2000 m. (a) Determine the components of the position vector from point A to point B. (b) Determine the components of a unit vector that points from point A toward point B.

Solution:

We need to find the coordinates of points A and B

rOA = 1500 cos60 °i + 1500 sin 60 ° j rOA = 750 i + 1299 j (m) Point A is at (750, 1299) (m) rOB = 2000 cos30 °i + 2000 sin 30 ° j (m) rOB = 1732 i + 1000 j (m) Point B is at (1732, 1000) (m)

(a) The vector from A to B is r AB = ( x B − x A ) i + ( y B − y A ) j

Proposed A bridge

r AB = 982 i − 299 j (m). (b) The unit vector e AB is

e AB =

r AB 982 i − 299 j = 1026.6 r AB

e AB = 0.957 i − 0.291 j. 608 308 O

River x

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Problem 2.42 The magnitudes of the forces exerted by the cables are T1 = 2800 lb, T2 = 3200 lb, T3 = 4000 lb, and T4 = 5000 lb. What is the magnitude of the total force exerted by the four cables? y T4

518

408

T2 298

Solution:

The x-component of the total force is

T x = T1 cos 9 ° + T2 cos 29 ° T3 cos 40 ° + T4 cos 51 ° T x = (2800 lb) cos 9 ° + (3200 lb) cos 29 ° + (4000 lb) cos 40 ° + (5000 lb) cos 51 ° T x = 11,800 lb The y-component of the total force is T y = T1 sin 9 ° + T2 sin 29 ° + T3 sin 40 ° + T4 sin 51 ° T y = (2800 lb)sin 9 ° + (3200 lb)sin 29 ° + (4000 lb)sin 40 ° + (5000 lb)sin 51 ° T y = 8450 lb The magnitude of the total force is T =

T x2 + T y2 =

(11,800 lb) 2 + (8450 lb) 2 = 14,500 lb

T = 14,500 lb

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Problem 2.43 The tensions in the four cables are T1 = T , T2 = 1.5T , T3 = 2.5T , and T4 = 3 T . Determine the value of T so that the four cables exert a total force of 24 kN magnitude on the support. y T4

518

408

T2 298

Solution:

The sum of the x components of the forces is

T cos 9 ° + 1.5T cos 29 ° + 2.5T cos 40 ° + 3T cos 51 ° = 6.10 T . The sum of the y components of the forces is T sin 9 ° + 1.5T sin 29 ° + 2.5T sin 40 ° + 3T sin 51 ° = 4.82T . The magnitude of the total force is (6.10T ) 2 + (4.82T ) 2 = 7.78T . Equating this expression to 24 kN, we obtain T = 3.09 kN.

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Problem 2.44 The rope ABC exerts forces FBA and FBC on the block at B. Their magnitudes are equal: FBA = FBC . The magnitude of the total force exerted on the block at B by the rope is FBA + FBC = 920 N. Determine FBA by expressing the forces FBA and FBC in terms of components.

Solution: FBC = F (cos 20 °i + sin 20 ° j) FBA = F (− j) FBC + FBA = F (cos 20 °i + [sin 20 ° − 1] j) Therefore (920 N) 2 = F 2 (cos 2 20 ° + [sin 20 ° − 1] 2 ) ⇒ F = 802 N

FBC C

208 FBC

B B 208 FBA A

FBA

Problem 2.45 The magnitude of the horizontal force F1 is 5 kN and F1 + F2 + F3 = 0. What are the magnitudes of F2 and F3? Solution:

F3 308 F1

Using components we have

∑ Fx : 5 kN + F2 cos 45 ° − F3 cos30 ° = 0 458

∑ Fy : −F2 sin 45 ° + F3 sin 30 ° = 0 Solving simultaneously yields: ⇒ F2 = 9.66 kN, F3 = 13.66 kN

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Problem 2.46 Four groups engage in a tug-of-war. The magnitudes of the forces exerted by groups B, C, and D are FB = 800 lb, FC = 1000 lb, and FD = 900 lb. If the vector sum of the four forces equals zero, what is the magnitude of F A and the angle α ?

y FB

708

Solution: The strategy is to use the angles and magnitudes to determine the force vector components, to solve for the unknown force F A and then take its magnitude. The force vectors are

FB = 800( i cos110 ° + j sin110 °) = −273.6 i + 751.75 j

308 208 FD

FC = 1000( i cos30 ° + j sin 30 °) = 866 i + 500 j x

FD = 900( i cos( −20°) + j sin(−20°)) = 845.72 i − 307.8 j F A = F A ( i cos(180 + α) + j sin(180 + α)) = F A (−i cos α − j sin α) The sum vanishes: F A + FB + FC + FD = i(1438.1 − F A cosα) + j(944 − F A sin α) = 0 From which F A = 1438.1i + 944 j. The magnitude is FA =

(1438) 2 + (944) 2 = 1720 lb

The angle is: tan α =

944 = 0.6565, or α = 33.3 ° 1438

Problem 2.47 In Example 2.5, suppose that the attachment point of cable A is moved so that the angle between the cable and the wall increases from 40° to 55 °. Draw a sketch showing the forces exerted on the hook by the two cables. If you want the total force F A + FB to have a magnitude of 200 lb and be in the direction perpendicular to the wall, what are the necessary magnitudes of F A and FB?

Solution: Let FA and FB be the magnitudes of FA and FB . The component of the total force parallel to the wall must be zero. And the sum of the components perpendicular to the wall must be 200 lb. FA cos 55 ° − FB cos 20 ° = 0 FA sin 55 ° + FB sin 20 ° = 200 lb Solving we find F A 5 195 lb F B 5 119 lb

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Problem 2.48 The bracket must support the two forces shown, where F1 = F2 = 2 kN. An engineer determines that the bracket will safely support a total force of magnitude 3.5 kN in any direction. Assume that 0 ≤ α ≤ 90 °. What is the safe range of the angle α ?

F2 a

Solution:

∑ Fx : (2 kN) + (2 kN) cos α = (2 kN)(1 + cos α) ∑ Fy : (2 kN)sin α F2

Thus the total force has a magnitude given by F1

F = 2 kN (1 + cos α) 2 + (sin α) 2 = 2 kN 2 + 2 cos α = 3.5 kN

Thus when we are at the limits we have 2 + 2 cos α =

F1 + F2

17 49 ⇒ cos α = ⇒ α = 57.9 ° ( 3.52 kNkN ) = 16 32

In order to be safe we must have 57.9 ° ≤ α ≤ 90 °

Problem 2.49 The figure shows three forces acting on a joint of a structure. The magnitude of FC is 80 kN, and F A + FB + FC = 0. What are the magnitudes of F A and FB?

y FC FB 158

Solution:

Denote the magnitudes of F A and FB by A and B. Writing the vectors in terms of their magnitudes,

408 FA

F A = A cos 40 °i + A sin 40 ° j, FB = −B cos15 °i − B sin15 ° j, FC = −(80 kN) j. We know that F A + FB + FC = ( A cos 40 ° − B cos15 °) i + [ A sin 40 ° − B sin15 ° − (80 kN) ] j = 0 . If a vector equals zero, each of its components equals zero, so we obtain the two equations

A cos 40 ° − B cos15 ° = 0, A sin 40 ° − B sin15 ° − (80 kN) = 0.

Solving them we obtain A = F A = 183 kN, B = FB = 145 kN. 408 F A = 183 kN, FB = 145 kN.

158 FB

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Problem 2.50 Four coplanar forces act on a beam. The forces FB and FC are vertical. The vector sum of the forces is zero. The magnitudes FB = 10 kN and FC = 5 kN. Determine the magnitudes of F A and FD .

308 FA FB

Solution:

Use the angles and magnitudes to determine the vectors, and then solve for the unknowns. The vectors are: F A = F A ( i cos30 ° + j sin 30°) = 0.866 F A i + 0.5 F A j FB = 0 i − 10 j, FC = 0 i + 5 j, FD = − FD i + 0 j.

Take the sum of each component in the x- and y- directions:

∑ Fx = (0.866 FA − FD )i = 0 and

∑ Fy = (0.5 FA − (10 − 5)) j = 0. From the second equation we get F A = 10 kN . Using this value in the first equation, we get FD = 8.7 kN

Problem 2.51 Six forces act on a beam that forms part of a building’s frame. The vector sum of the forces is zero. The magnitudes FB = FE = 20 kN, FC = 16 kN, and FD = 9 kN. Determine the magnitudes of F A and FG .

Solution:

Write each force in terms of its magnitude and direction as

F = F cos θ i + F sin θ j where θ is measured counterclockwise from the +x-axis. Thus, (all forces in kN) F A = F A cos110 °i + FA sin110 ° j (kN) FB = 20 cos 270 °i + 20 sin 270 ° j (kN) FC = 16 cos140 °i + 16sin140° j (kN) FD = 9 cos 40°i + 9sin 40° j (kN) FE = 20 cos 270°i + 20 sin 270° j (kN) FG = FG cos 50°i + FG sin 50° j (kN) We know that the x components and y components of the forces must add separately to zero. Thus

FA 708

FD 408

 FAx + FBx + FCx + FDx + FEx + FGx = 0   FAy + FBy + FCy + FDy + FEy + FGy = 0 

FG 508

408

 F A cos110 ° + 0 − 12.26 + 6.89 + 0 + FG cos 50 ° = 0   F A sin110 ° − 20 + 10.28 + 5.79 − 20 + FG sin 50 ° = 0  Solving, we get F A = 13.0 kN

FG = 15.3 kN y

U x

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Problem 2.52 The total weight of the man and parasail is W = 230 lb. The drag force D is perpendicular to the lift force L. If the vector sum of the three forces is zero, what are the magnitudes of L and D?

Solution: Let L and D be the magnitudes of the lift and drag forces. We can use similar triangles to express the vectors L and D in terms of components. Then the sum of the forces is zero. Breaking into components we have 2 L− 22 + 52

5 D = 0 22 + 52

5 L− 22 + 52

2 D − 230 lb = 0 22 + 52

y L 5

Solving we find

D = 85.4 lb, L = 214 lb

Problem 2.53 The three forces acting on the car are shown. The force T is parallel to the x-axis and the magnitude of the force W is 14 kN. If T + W + N = 0, what are the magnitudes of the forces T and N ?

Solution: ∑ Fx : T − N sin 20 ° = 0 ∑ Fy : N cos 20 ° − 14 kN = 0 Solving we find N = 14.90 N, T = 5.10 N

208

208 N

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Problem 2.54 The sum of the three forces acting on the beam is zero: F A + FB + FC = 0. The magnitude F A = 800 lb. Determine the magnitudes of FC and FB .

Solution: y FA

458

508

458

808 FB

808

Let FA , FB , FC denote the magnitudes of the forces. The sum of the x components of the forces is

ΣFx = −FA cos 45 ° − FB cos80 ° + FC cos 50 ° = 0. The sum of the x components of the forces is ΣFy = FA sin 45 ° − FB sin80 ° + FC sin 50 ° = 0. Solving these two equations, we obtain FB = 1590 lb, FC = 1310 lb. FB = 1590 lb, FC = 1310 lb.

Problem 2.55 The total force exerted on the top of the mast B by the sailboat’s forestay AB and backstay BC is 180 i − 820 j (N). What are the magnitudes of the forces exerted at B by the cables AB and BC ?

Solution:

We first identify the forces:

F AB = T AB

(−4.0 mi − 11.8 mj) (−4.0 m) 2 + (11.8 m) 2

FBC = TBC

(5.0 mi − 12.0 mj) (5.0 m) 2 + (−12.0 m) 2

y B (4, 13) m

Then if we add the force we find ∑ Fx : −

4 T AB + 155.24

5 TBC = 180 N 169

∑ Fy : −

11.8 T AB − 155.24

12 TBC = −820 N 169

Solving simultaneously yields: ⇒ T AB = 226 N, T AC = 657 N

A (0, 1.2) m

C (9, 1) m

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Problem 2.56 The structure shown forms part of a truss designed by an architectural engineer to support the roof of an orchestra shell. The members AB, AC, and AD exert forces F AB , F AC , and F AD on the joint A. The magnitude F AB = 4 kN. If the vector sum of the three forces equals zero, what are the magnitudes of F AC and F AD? Solution:

−2 i+ 22 + 32

−3 j = −0.5547 i − 0.8320 j 22 + 32

e AC =

−4 i+ 4 2 + 12

1 j = −0.9701i + 0.2425 j 4 2 + 12

e AB =

4 i+ 42 + 22

2 j = 0.89443i + 0.4472 j 42 + 22

FAB

FAC

FAD

(4, 2) m x

Determine the unit vectors parallel to each force:

e AD =

(24, 1) m

(22, 23) m

The forces are F AD = F AD e AD , F AC = F AC e AC , C

F AB = F AB e AB = 3.578i + 1.789 j. Since the vector sum of the forces vanishes, the x- and y-components vanish separately:

A D

∑ Fx = (−0.5547 FAD − 0.9701 FAC + 3.578)i = 0, and

∑ Fy = (−0.8320 FAD + 0.2425 FAC + 1.789) j = 0 These simultaneous equations in two unknowns can be solved by any standard procedure. An HP-28S hand held calculator was used here: The results: F AC = 2.108 kN , F AD = 2.764 kN

Problem 2.57 The distance s = 45 in. (a) Determine the unit vector e BA that points from B toward A. (b) Use the unit vector you obtained in (a) to determine the coordinates of the collar C.

A (14, 45) in

Solution: (a) The unit vector is the position vector from B to A divided by its magnitude rBA = ([14 − 75]i + [45 − 12] j)in = (−61i + 33 j)in rBA =

(−61 in) 2 + (33 in) 2 = 69.35 in 1 e BA = (−61i + 33 j) in = (−0.880 i + 0.476 j) 69.35 in

s B (75, 12) in x

e BA = (−0.880 i + 0.476 j) (b) To find the coordinates of point C we will write a vector from the origin to point C. rC = r A + r AC = r A + se BA = (75i + 12 j) in + (45 in)(−0.880 i + 0.476 j) rC = (35.4 i) + 33.4 j) in Thus the coordinates of C are C (35.4, 33.4) in

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Problem 2.58 Determine the x and y coordinates of the collar C as functions of the distance s. y

Solution:

The coordinates of the point C are given by

x C = x B + s(−0.880) and y C = y B + s(0.476). Thus, the coordinates of point C are x C = 75 − 0.880 s in and y C = 12 + 0.476s in. Note from the solution of Problem 2.57 above, 0 ≤ s ≤ 69.4 in.

A (14, 45) in s

B (75, 12) in x

Problem 2.59 The vector U can be expressed in terms of its components in terms of either the x–y coordinate system or the x′–y′ coordinate system:

Solution: y y9

U = U xi + U y j = U x ′ i ′ + U y ′ j′. It can be shown that the component U x′ is given in terms of the components U x and U y and the angle θ by U x ′ = U x cos θ + U y sin θ. Determine the component U y′ in terms of the components U x and U y and the angle θ . Strategy: Draw a diagram showing the components U x , U y , U x′ , and U y′ and apply trigonometry to determine U y′ . y9

Ux u Ux sinu

u Uy sinu

Uy9

From the diagram, U y ' = U y cos θ − U x sin θ. U

U y ' = U y cos θ − U x sin θ.

x9 u

Problem 2.60 Let r be the position vector from point C to the point that is a distance s meters from point A along the straight line between A and B. Express r in terms of components. (Your answer will be in terms of s.)

Solution:

rB /A = ([10 − 3]i + [9 − 4] j) m = (7 i + 5 j) m r B /A = e B /A =

y B

First define the unit vector that points from A to B.

(7 m) 2 + (5 m) 2 = 1 (7 i + 5 j) 74

74 m

Let P be the point that is a distance s along the line from A to B. The coordinates of point P are

(10, 9) m

xp = 3 m + s

s r

( 774 ) = (3 + 0.814s) m

yp = 4 m + s

A (3, 4) m

( 574 ) = (4 + 0.581s) m.

The vector r that points from C to P is then

C (9, 3) m

r = ([3 + 0.814 s − 9]i + [4 + 0.581s − 3] j) m x r = ([0.814 s − 6]i + [0.581s + 1] j) m

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Problem 2.61 A vector U = 3i − 4 j − 12k. What is its magnitude? Strategy: The magnitude of a vector is given in terms of its components by Eq. (2.14).

Solution:

Problem 2.62 Two forces are given in terms of their components by F A = 60 i − 20 j + 30 k (lb), FB = 30 i + 40 j − 10 k (lb). (a) What are the magnitudes of the forces? (b) What is the magnitude of their sum F A + FB? Strategy: The magnitude of a vector in terms of its components is given by Eq. (2.14).

Solution:

U =

Use definition given in Eq. (14). The vector magnitude is

3 2 + (−4) 2 + (−12) 2 = 13

(a) The magnitudes of the vectors are FA =

(60 lb) 2 + (−20 lb) 2 + (30 lb) 2 = 70.0 lb,

FB =

(30 lb) 2 + (40 lb) 2 + (−10 lb) 2 = 51.0 lb.

(b) The sum of the vectors is F A + FB = 90 i + 20 j + 20 k (lb). The magnitude of the sum is F A + FB =

(90 lb) 2 + (20 lb) 2 + (20 lb) 2 = 94.3 lb.

(a) F A = 70.0 lb, FB = 51.0 lb. (b) F A + FB = 94.3 lb.

Problem 2.63 Consider three vectors A = 12 i + 8 j − 16k, B = B x i + B y j + B z k, and C = C x i + C y j + C z k. Their sum is zero, A + B + C = 0, and the components of B and C satisfy the relations B x = −2 B y , B y = −B z , and C x = 3C y . What are the magnitudes of B and C?

Solution:

The sum of the vectors is zero, which yields the three

equations ΣFx = 12 + B x + C x = 0, ΣFy = 8 + B y + C y = 0, ΣFz = −16 + B z + C z = 0. We also have the three given equations B x = −2 B y , B y = −B z , C x = 3C y . Solving these six equations, we obtain B x = 4.8, B y = −2.4, B z = 2.4, C x = −16.8, C y = −5.6, C z = 13.6. The magnitudes of the vectors are B =

Bx 2 + By 2 + Bz 2

= (4.8) 2 + (−2.4 lb) 2 + (2.4) 2 = 5.88, C =

Cx 2 + Cy 2 + Cz 2

= (−16.8) 2 + (−5.6) 2 + (13.6) 2 = 22.3. B = 5.88, C = 22.3.

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Problem 2.64 A vector U = U x i + U y j + U z k. Its magnitude U = 30. Its components are related by the equations U y = −2U x and U z = 4U y . Determine the components.

Solution:

Substitute the relations between the components, determine the magnitude, and solve for the unknowns. Thus U = U x i + (−2U x ) j + (4(−2U x ))k = U x (1i − 2 j − 8k) where U x can be factored out since it is a scalar. Take the magnitude, noting that the absolute value of U x must be taken: 30 = U x

1 2 + 2 2 + 8 2 = U x (8.31).

Solving, we get U x = 3.612, or U x = ±3.61. The two possible vectors are U = +3.61i + (−2(3.61)) j + (4(−2)(3.61))k = 3.61i − 7.22 j − 28.9k U = −3.61i + (−2(−3.61) j + 4(−2)(−3.61)k = −3.61i + 7.22 j + 28.9k

Problem 2.65 An object is acted on by two forces F1 = 20 i + 30 j − 24 k (kN) and F2 = −60 i + 20 j + 40 k (kN). What is the magnitude of the total force acting on the object?

Solution: F1 = (20 i + 30 j − 24 k) kN F2 = (−60 i + 20 j + 40 k) kN F = F1 + F2 = (−40 i + 50 j + 16k) kN Thus F =

Problem 2.66 The position vector r AB extends from point A to point B. (a) Express r AB in terms of components. (b) Determine its magnitude r AB .

(−40 kN) 2 + (50 kN) 2 + (16 kN) 2 = 66 kN

y (22, 6, 3) ft A rAB

Solution:

(a) The components of r AB are r AB = (12 ft − (−2 ft)) i + (−2 ft − 6 ft) j + (8 ft − 3 ft)k = 14 i − 8 j + 5k (ft). (b)

(12, 22, 8) ft z

Its magnitude is r AB =

(14 ft) 2 + (−8 ft) 2 + (5 ft) 2

= 16.9 ft. (a) r AB = 14 i − 8 j + 5k (ft). (b) r AB = 16.9 ft.

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Problem 2.67 The position vector r AB extends from point A to point B. (a) Determine the direction cosines of r AB . (b) Determine the components of a unit vector that has the same direction as r AB .

(b) The direction cosines are the components of the desired unit vector. (Notice that the direction cosines are the components of r AB divided by its magnitude.) Therefore, e = 0.829 i − 0.474 j + 0.296k. (a) cos θ x = 0.829, cos θ y = −0.474, cos θ z = 0.296.

(b) e = 0.829 i − 0.474 j + 0.296k.

(22, 6, 3) ft A rAB x (12, 22, 8) ft B

Solution: (a) The components of r AB are r AB = (12 ft − (−2 ft)) i + (−2 ft − 6 ft) j + (8 ft − 3 ft)k = 14 i − 8 j + 5k (ft), and its magnitude is r AB =

(14 ft) 2 + (−8 ft) 2 + (5 ft) 2

= 16.9 ft. Its direction cosines are 14 ft = 0.829, 16.9 ft −8 ft cos θ y = = −0.474, 16.9 ft 5 ft cos θ z = = 0.296. 16.9 ft cos θ x =

Problem 2.68 A force vector is given in terms of its components by F = 10 i − 20 j − 20 k (N). (a) What are the direction cosines of F? (b) Determine the components of a unit vector e that has the same direction as F.

Solution: F = (10 i − 20 j − 20 k) N F =

(10 N) 2 + (−20 N) 2 + (−20 N) 2 = 30 N

(a)

cos θ x =

(b)

e = (0.333i − 0.667 j − 0.667k)

10 N −20 N = 0.333, cos θ y = = −0.667, 30 N 30 N −20 N cos θ z = = −0.667 30 N

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Problem 2.69 Express the position vector rBD from point B to point D in terms of components. What are its direction cosines?

Solution: y

(4, 3, 1) m

(0, 5, 5) m

(0, 4, 23) m C

rBD

B (4, 3, 1) m

D (0, 5, 5) m

x x A E

z The vector rBD from point B to point D is rBD = (0 − 4 m) i + (5 m − 3 m) j + (5 m − 1 m) = −4 i + 2 j + 4 k (m). Its magnitude is rBD =

(−4 m) 2 + (2 m) 2 + (4 m) 2

= 6 m. Its direction cosines are −4 m = −0.667, 6m 2m cos θ y = = 0.333, 6m 4m cos θ z = = 0.667. 6m cos θ x =

rBD = −4 i + 2 j + 4 k (m), cos θ x = −0.667, cos θ y = 0.333, cos θ z = 0.667.

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Problem 2.70 The cable from B to C exerts a force on the bar AB at B. The force points from B toward C and its magnitude is 80 N. Express the force in terms of components.

Solution: y (0, 4, 23) m C

rBC

(0, 4, 23) m C

(4, 3, 1) m B

B (4, 3, 1) m

D (0, 5, 5) m

x A

z E

The vector rBC from point B to point C is rBC = (0 − 4 m) i + (4 m − 3 m) j + (−3 m − 1 m) = −4 i + j − 4 k (m). Its magnitude is rBC =

(−4 m) 2 + (1 m) 2 + (−4 m) 2

= 5.74 m. By dividing the vector by its magnitude, we obtain a unit vector that points from point B toward point C: −4 i + j − 4 k (m) 5.74 m = −0.696 i + 0.174 j − 0.696k.

e BC =

y (0, 4, 23) m C eBC

(4, 3, 1) m B

z We obtain the force in terms of its components by multiplying the unit vector by 80 N: (80 N)e BC = −0.55.7 i + 13.9 j − 55.7k (N). −0.55.7 i + 13.9 j − 55.7k (N).

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Problem 2.71* The suspended weight E exerts an 800-N downward force on bar AB at B. The cable from B to C exerts a force on the bar at B that points from B toward C, but its magnitude is unknown. The cable from B to D exerts a force on the bar that points from B toward D, and its magnitude is also unknown. If the weight of bar AB is negligible, the sum of the three forces exerted on it at B is a vector that is parallel to the bar. Use this condition to determine the magnitudes of the forces exerted by cables BC and BD.

Let FBC be the force exerted at B by the cable BC. It can be expressed as FBC = FBC e BC = FBC (−0.696 i + 0.174 j − 0.696k). y

(4, 3, 1) m

(0, 5, 5) m rBC

y (0, 4, 23) m C

x B (4, 3, 1) m

D (0, 5, 5) m

x A

z The vector rBD from point B to point D is rBD = (0 − 4 m) i + (5 m − 3 m) j + (5 m − 1 m)k

= −4 i + 2 j + 4 k (m).

Its magnitude is rBD =

Solution:

(−4 m) 2 + (2 m) 2 + (4 m) 2

= 6 m. y

Dividing rBD by its magnitude, we obtain a unit vector that points from point B toward point D:

(0, 4, 23) m

−4 i + 2 j + 4 k (m) 6m = −0.667 i + 0.333 j + 0.667k.

e BD = rBC

(4, 3, 1) m B

Let FBD be the force exerted at B by the cable BD. It can be expressed as x

FBD = FBD e BD = FBD (−0.667 i + 0.333 j + 0.667k). y

z The vector rBC from point B to point C is

(4, 3, 1) m

rBC = (0 − 4 m) i + (4 m − 3 m) j + (−3 m − 1 m) = −4 i + j − 4 k (m). Its magnitude is rBC =

rBA (0, 0, 0)

(−4 m) 2 + (1 m) 2 + (−4 m) 2

= 5.74 m.

By dividing rBC by its magnitude, we obtain a unit vector that points from point B toward point C: −4 i + j − 4 k (m) 5.74 m = −0.696 i + 0.174 j − 0.696k.

e BC =

z The vector rBA from point B to point A is rBA = (0 − 4 m) i + (0 − 3 m) j + (0 − 1 m)k = −4 i − 3 j − k (m).

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2.71 (Continued) Its magnitude is

where F is an unknown constant. Notice that this condition yields three equations. Its i component is

rBA =

−0.696 FBC − 0.667 FBD = −0.784 F ,

(−4 m) 2 + (−3 m) 2 + (−1 m) 2

= 5.10 m.

its j component is

Dividing rBA by its magnitude, we obtain a unit vector that points from point B toward point A:

0.174 FBC + 0.333 FBD − 800 N = −0.588F , and its k component is −0.696 FBC + 0.667 FBD = −0.196 F .

−4 i − 3 j − k (m) 5.10 m = −0.784 i − 0.588 j − 0.196k.

e BA =

Solving these three equations yields FBC = 657 N, FBD = 411 N, and F = 932 N.

We can write the given condition on the forces at B as

BC : 657 N, BD : 411 N.

FBC + FBD − (800 N) j = Fe BA ,

Problem 2.72 Determine the components of the position vector rBD from point B to point D. Use your result to determine the distance from B to D.

Solution:

We have the following B(5, 0, 3) m, C (6, 0, 0) m, D(4, 3, 1) m

coordinates:

A(0, 0, 0),

rBD = (4 m − 5 m) i + (3 m − 0) j + (1 m − 3 m)k

= (−i + 3 j − 2k) m

D (4, 3, 1) m

rBD =

(−1 m) 2 + (3 m) 2 + (−2 m) 2 = 3.74 m

A C (6, 0, 0) m x z

B (5, 0, 3) m

Problem 2.73 What are the direction cosines of the position vector rBD from point B to point D ?

Solution: cos θ x =

y D (4, 3, 1) m

−1 m 3m = −0.267, cos θ y = = 0.802, 3.74 m 3.74 m −2 m cos θ z = = 0.535 3.74 m

A C (6, 0, 0) m x z

B (5, 0, 3) m

Problem 2.74 Determine the components of the unit vector e CD that points from point C toward point D.

Solution:

We have the following B(5, 0, 3) m, C (6, 0, 0) m, D(4, 3, 1) m

= (−2 i + 3 j + 1k)

D (4, 3, 1) m

rCD = A

(−2 m) 2 + (3 m) 2 + (1 m) 2 = 3.74 m

Thus C (6, 0, 0) m x

A(0, 0, 0),

rCD = (4 m − 6 m) i + (3 m − 0) j + (1 m − 0)k

coordinates:

B (5, 0, 3) m

1 (−2 i + 3 j + k) m 3.74 m = (−0.535i + 0.802 j + 0.267k)

e CD =

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Problem 2.75 What are the direction cosines of the unit vector e CD that points from point C toward point D?

Solution:

Using Problem 2.74

cos θ x = −0.535, cos θ y = 0.802, cos θ z = 0.267

y D (4, 3, 1) m

A C (6, 0, 0) m x z

B (5, 0, 3) m

Problem 2.76 In Example 2.7, suppose that the caisson shifts on the ground to a new position. The magnitude of the force F remains 600 lb. In the new position, the angle between the force F and the x-axis is 60° and the angle between F and the z-axis is 70 °. Express F in terms of components. Solution:

We need to find the angle θ y between the force F and the y-axis. We know that cos 2 θ x + cos 2 θ y + cos 2 θ z = 1 cos θ y = ± 1 − cos 2 θ x − cos 2 θ z = ± 1 − cos 2 60 ° − cos 2 70 ° = ±0.7956 θ y = ± cos −1 (0.7956) = 37.3 ° or 142.7°

We will choose θ y = 37.3 ° because the picture shows the force pointing up. Now Fx = (600 lb) cos60 ° = 300 lb Fy = (600 lb) cos 37.3 ° = 477 lb Fz = (600 lb) cos 70 ° = 205 lb Thus F = (300 i + 477 j + 205k) lb

Problem 2.77 Astronauts on the space shuttle use radar to determine the magnitudes and direction cosines of the position vectors of two satellites A and B. The vector r A from the shuttle to satellite A has magnitude 2 km and direction cosines cos θ x = 0.768, cos θ y = 0.384, cos θ z = 0.512. The vector rB from the shuttle to satellite B has magnitude 4 km and direction cosines cos θ x = 0.743, cos θ y = 0.557, cos θ z = −0.371. What is the distance between the satellites? Solution:

rB x

The two position vectors are:

r A = 2(0.768i + 0.384 j + 0.512k) = 1.536 i + 0.768 j + 1.024 k (km)

rB = 4(0.743i + 0.557 j − 0.371k) = 2.972 i + 2.228 j − 1.484 k (km) The distance is the magnitude of the difference: r A − rB =

(1.536 − 2.927) 2 + (0.768 − 2.228) 2 + (1.024 − (−1.484)) 2

= 3.24 (km)

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Problem 2.78 Archaeologists measure a pre-Columbian ceremonial structure and obtain the dimensions shown. Determine (a) the magnitude and (b) the direction cosines of the position vector from point A to point B.

y 4m

10 m

10 m 8m

Solution:

(a) The coordinates are A (0, 16, 14) m and B (10, 8, 4) m. r AB = ([10 − 0]i + [8 − 16] j + [4 − 14]k) m = (10 i − 8 j − 10 k) m r AB =

10 2 + 8 2 + 10 2 m =

z 264 m = 16.2 m

r AB = 16.2 m (b)

10 = 0.615 264 −8 = −0.492 cos θ y = 264 −10 = −0.615 cos θ z = 264 cos θ x =

Problem 2.79 Consider the structure described in Problem 2.78. After returning to the United States, an archaeologist discovers that a graduate student has erased the only data file containing the dimension b. But from recorded GPS data he is able to calculate that the distance from point B to point C is 16.61 m. (a) What is the distance b? (b) Determine the direction cosines of the position vector from B to C. Solution:

We have the coordinates B (10 m, 8 m, 4 m), C (10 m + b, 0 18 m). rBC = (10 m + b − 10 m) i + (0 − 8 m) j + (18 m − 4 m)k rBC = (b) i + (−8 m) j + (14 m)k

(a) We have (16.61 m) 2 = b 2 + (−8 m) 2 + (14 m) 2 ⇒

b = 3.99 m

(b) The direction cosines of rBC are 3.99 m = 0.240 16.61 m −8 m cos θ y = = −0.482 16.61 m 14 m cos θ z = = 0.843 16.61 m cos θ x =

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Problem 2.80 Observers at A and B use theodolites to measure the direction from their positions to a rocket in flight. If the coordinates of the rocket’s position at a given instant are (4, 4, 2) km, determine the direction cosines of the vectors r AR and rBR that the observers would measure at that instant. Solution:

rAR

The vector r AR is given by

rBR A

r AR = 4 i + 4 j + 2k km and the magnitude of r AR is given by r AR =

x z

B (5, 0, 2) km

(4) 2 + (4) 2 + (2) 2 km = 6 km.

The unit vector along AR is given by u AR = r AR r AR . Thus, u AR = 0.667 i + 0.667 j + 0.333k and the direction cosines are cos θ x = 0.667, cos θ y = 0.667, and cos θ z = 0.333. The vector rBR is given by rBR = ( x R − x B ) i + ( y R − y B ) j + ( z R − z B )k km = (4 − 5) i + (4 − 0) j + (2 − 2)k km and the magnitude of rBR is given by rBR =

(1) 2 + (4) 2 + (0) 2 km = 4.12 km.

The unit vector along BR is given by e BR = rBR rBR . Thus, u BR = −0.242 i + 0.970 j + 0 k and the direction cosines are cos θ x = −0.242, cos θ y = 0.970, and cos θ z = 0.0.

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Problem 2.81 Point p lies in the x–z plane. The line from point p to the tip of the vector r is parallel to the y-axis. If r = 12 i + 6 j + 10 k (m), what are the angles φ and ψ in degrees?

Solution:

The coordinates of the tip of the vector r are (12, 6, 10) m, so the coordinates of point p are (12, 0, 10) m. The length of the line from the origin to point p is

c x f

(12 m) 2 + (10 m) 2 = 15.6 m:

tan ψ =

The tangent of ψ is 6m = 0.384, 15.6 m

which yields θ = 21.0 °. The tangent of φ is tan φ =

(12,6,10) m

10 m = 0.833, 12 m

which yields φ = 39.8 °. φ = 39.8 °, ψ = 21.0 °.

x f

10 m

15.6 m 12 m

Problem 2.82 The height of Mount Everest was originally measured by a surveyor in the following way. He first measured the altitudes of two points and the horizontal distance between them. For example, suppose that the points A and B are 3000 m above sea level and are 10,000 m apart. He then used a theodolite to measure the direction cosines of the vector r AP from point A to the top of the mountain P and the vector rBP from point B to P. Suppose that the direction cosines of r AP are cos θ x = 0.5179, cos θ y = 0.6906, and cos θ z = 0.5048, and the direction cosines of rBP are cos θ x = −0.3743, cos θ y = 0.7486, and cos θ z = 0.5472. Using this data, determine the height of Mount Everest above sea level. z

Solution:

We have the following coordinates A(0, 0, 3000) m, B(10,000, 0, 3000) m, P( x, y, z )

Then r AP = xi + yj + ( z − 3000 m)k = r AP (0.5179 i + 0.6906 j + 0.5048k) rBP = ( x − 10,000 m) i + yj + ( z − 3000 m)k = rBP (−0.3743i + 0.7486 j + 0.5472k) Equating components gives us five equations (one redundant) which we can solve for the five unknowns. x = r AP 0.5179 y = r AP 0.6906 z − 3000 m = r AP 0.5048 ⇒ z = 8848 m x − 10000 m = −rBP − 0.7486 y = rBP 0.5472

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Problem 2.83 The distance from point O to point A is 20 ft. The straight line AB is parallel to the y-axis, and point B is in the x–z plane. Express the vector rOA in terms of components. Strategy: You can express rOA as the sum of a vector from O to B and a vector from B to A. You can then express the vector from O to B as the sum of vector components parallel to the x- and z-axes. See Example 2.8. Solution:

A rOA

308 608

See Example 2.8. The length BA is, from the right

triangle OAB, r AB = rOA sin 30 ° = 20(0.5) = 10 ft.

Similarly, the length OB is rOB = rOA cos30 ° = 20(0.866) = 17.32 ft The vector rOB can be resolved into components along the axes by the right triangles OBP and OBQ and the condition that it lies in the x -z plane. Hence,

rOB = rOB ( i cos30 ° + j cos 90 ° + k cos60 °) or

rOB = 15i + 0 j + 8.66k.

A rOA 308

x Q

z P

308

The vector rBA can be resolved into components from the condition that it is parallel to the y-axis. This vector is rBA = rBA ( i cos 90 ° + j cos 0 ° + k cos 90 °) = 0 i + 10 j + 0 k. The vector rOA is given by rOA = rOB + rBA , from which rOA = 15i + 10 j + 8.66k (ft)

Problem 2.84 The magnitudes of the two force vectors are F A = 140 lb and FB = 100 lb. Determine the magnitude of the sum of the forces F A + FB . Solution:

y FB

We have the vectors

F A = 140 lb([ cos 40 ° sin 50 °]i + [sin 40 °] j + [cos 40 ° cos 50 °]k) F A = (82.2 i + 90.0 j + 68.9k) lb

608

FB = 100 lb([− cos60 ° sin 30 °]i + [sin 60 °] j + [cos60 ° cos30 °]k)

308

FB = (−25.0 i + 86.6 j + 43.3k) lb Adding and taking the magnitude we have

408

508

F A + FB = (57.2 i + 176.6 j + 112.2k) lb FA + FB =

(57.2 lb) 2 + (176.6 lb) 2 + (112.2 lb) 2 = 217 lb

F A + FB = 217 lb

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Problem 2.85 Determine the direction cosines of the vectors F A and FB . Solution:

y FB

We have the vectors

F A = 140 lb([ cos 40 ° sin 50 °]i + [sin 40 °] j + [cos 40 ° cos 50 °]k) F A = (82.2 i + 90.0 j + 68.9k) lb

608

FB = 100 lb([ −cos60 ° sin 30 °]i + [sin 60 °] j + [cos60 ° cos30 °]k)

408

308

FB = (−25.0 i + 86.6 j + 43.3k) lb The direction cosines for FA are

508

82.2 lb 90.0 lb cos θ x = = 0.643, = 0.587, cos θ y = 140 lb 140 lb 68.9 lb cos θ z = = 0.492 140 lb The direction cosines for FB are 86.6 lb −25.0 lb = 0.866, = −0.250, cos θ y = 100 lb 100 lb 43.3 lb cos θ z = = 0.433 100 lb cos θ x =

F A : cos θ x = 0.587, cos θ y = 0.643, cos θ z = 0.492 FB : cos θ x = −0.250, cos θ y = 0.866, cos θ z = 0.433

Problem 2.86 The positions of two airplanes A and B relative to a control tower are determined by radar. Their distances from the control tower are r A = 16 km and rB = 12 km. Determine the components of the position vector r AB from plane A to plane B (that is, the vector that specifies the position of plane B relative to plane A). What is the straight-line distance between the two planes?

Solution:

The components of r A are

r A = (16 km)(cos 20 ° cos30 °i + sin 20 ° j − cos 20 ° sin 30 °k) = 13.0 i + 5.47 j − 7.52k (km). The components of are rB = (12 km)(− cos60 ° sin 40 °i + sin 60 ° j + cos60 ° cos 40 °k) = −3.86 i + 10.4 j + 4.60 k (km). y

y B

B 608

rA 208

408 z

rAB

A rA

308

The vector from plane A to plane B is r AB = rB − r A = −16.9 i + 4.92 j + 12.1k (km). The distance between the planes is r AB =

(−16.9 km) 2 + (4.92 km) 2 + (12.1 km) 2

= 21.3 km. r AB = −16.9 i + 4.92 j + 12.1k (km), distance = 21.3 km.

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Problem 2.87 An engineer calculates that the magnitude of the axial force in one of the beams of a geodesic dome is P = 7.65 kN. The cartesian coordinates of the endpoints A and B of the straight beam are (−12.4, 22.0, −18.4) m and (−9.2, 24.4, −15.6) m, respectively. Express the force P in terms of components.

Solution:

The components of the position vector from B to A are

rBA = ( x A − x B ) i + ( y A − y B ) j + ( z A − z B )k = (−12.4 + 9.2) i + (22.0 − 24.4) j + (−18.4 + 15.6)k = −3.2 i − 2.4 j − 2.8k (m). Dividing this vector by its magnitude, we obtain a unit vector that points from B toward A: e BA = −0.655i − 0.492 j − 0.573k.

Therefore P = P e BA

P A

= 7.65 e BA = −5.01i − 3.76 j − 4.39k (kN).

Problem 2.88 The cable BC exerts an 8-kN force F on the bar AB at B. (a) Determine the components of a unit vector that points from point B toward point C. (b) Express F in terms of components.

Solution: (a) e BC = rBC = rBC e BC =

( x C − x B ) i + ( y C − y B ) j + ( z C − z B )k ( x C − x B ) 2 + ( y C − y B ) 2 + (z C − z B ) 2

−2 i − 6 j + 3k 2 6 3 = − i− j+ k 7 7 7 22 + 62 + 32

e BC = −0.286 i − 0.857 j + 0.429k

B (5, 6, 1) m

C (3, 0, 4) m z

(b) F = F e BC = 8e BC = −2.29 i − 6.86 j + 3.43k (kN)

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Problem 2.89 A cable extends from point C to point E. It exerts a 50-lb force T on the plate at C that is directed along the line from C to E. Express T in terms of components.

6 ft A

Solution:

Find the unit vector e CE and multiply it times the magnitude of the force to get the vector in component form, r e CE = CE = rCE

208

C 4 ft

The coordinates of point C are (4, −4 sin 20 °, 4 cos 20 °) or (4, −1.37, 3.76) (ft) The coordinates of point E are (0, 2, 6) (ft) e CE =

(0 − 4) i + (2 − (−1.37)) j + (6 − 3.76)k 4 2 + 3.37 2 + 2.24 2

6 ft

e CE = −0.703i + 0.592 j + 0.394 k

T = 50 e CE (lb)

T = −35.2 i + 29.6 j + 19.7k (lb)

2 ft z

4 ft C

4 ft

The vector rBD from point B to point D is

rBD = (18 in − 8 in) i + (−8 in − 0) j + (7 in − 0)

= 10 i − 8 j + 7k (in).

8 in

14 in

Its magnitude is rBD =

208

Problem 2.90 A cable extends from point B to point D. It exerts a 30-lb force on the bar at B that is directed along the line from B to D. Express the force in terms of components. Solution:

4 ft

2 ft

( x E − x C ) i + ( y E − y C ) j + ( z E − z C )k ( x E − x C ) 2 + ( y E − y C ) 2 + (z E − z C ) 2

(10 in) 2 + (−8 in) 2 + (7 in) 2

= 14.6 in.

(18, 28, 7) in

By dividing this vector by its magnitude, we obtain a unit vector that points from point B toward point D: 10 i − 8 j + 7k (in) 14.6 in = 0.685i − 0.548 j + 0.480 k.

e BD =

(8, 0, 0) in

We obtain the force in terms of its components by multiplying this unit vector by 30 lb:

eBD z

(30 lb)e BD = 20.6 i − 16.4 j + 14.4 k (lb).

(18, 28, 7) in D

20.6 i − 16.4 j + 14.4 k (lb). y

(8, 0, 0) in

B rBD z D

(18, 28, 7) in

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Problem 2.91 The cable AB exerts a 200-lb force F AB at point A that is directed along the line from A to B. Express F AB in terms of components. Solution:

8 ft

The coordinates of B are B(0, 6, 8). The position vector

from A to B is r AB = (0 − 6) i + (6 − 0) j + (8 − 10)k = −6 i + 6 j − 2k The magnitude is r AB =

6 ft B

6 2 + 6 2 + 2 2 = 8.718 ft.

The unit vector is u AB =

−6 6 2 i+ j− k 8.718 8.718 8.718

FAC

FAB

u AB = −0.6882 i + 0.6882 j − 0.2294 k.

A (6, 0, 10) ft

F AB = F AB u AB = 200(−0.6882 i + 0.6882 j − 0.2294 k) The components of the force are F AB = F AB u AB = 200(−0.6882 i + 0.6882 j − 0.2294 k) or F AB = −137.6 i + 137.6 j − 45.9k

Problem 2.92 The cable CD exerts a 6-kN force on the bar at C that is directed from C toward D. Express the force in terms of components.

y C

(0, 2, –1) m

y 1m

rCD

D (1.5, 0, 0) m x

z 2m y A z

C 1.5 m

(0, 2, –1) m eCD

D x

Solution: The vector rCD from point B to point D is rCD = (1.5 m − 0) i + (0 − 2 m) j + (0 − (−1 m))k = 1.5i − 2 j + k (m). Its magnitude is rCD =

D (1.5, 0, 0) m x

(1.5 m) 2 + (−2 m) 2 + (1 m) 2

= 2.69 m. By dividing this vector by its magnitude, we obtain a unit vector that points from point C toward point D: 1.5i − 2 j + k (m) 2.69 m = 0.557 i − 0.743 j + 0.371k.

e CD =

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We obtain the force in terms of its components by multiplying this unit vector by 6 kN: (6 kN)e CD = 3.34 i − 4.46 j + 2.23k (kN). 3.34 i − 4.46 j + 2.23k (kN).

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Problem 2.93 The 70-m-tall tower is supported by three cables that exert forces F AB , F AC , and F AD on it. The magnitude of each force is 2 kN. Express the total force exerted on the tower by the three cables in terms of components.

Solution:

The coordinates of the points B (40, 0, 0), C (−40, 0, 40) D (−60, 0, −60).

are

A (0, 70, 0),

The position vectors corresponding to the cables are: r AD = (−60 − 0) i + (0 − 70) j + (−60 − 0)k r AD = −60 i − 70 k − 60 k r AC = (−40 − 0) i + (0 − 70) j + (40 − 0)k r AC = −40 i − 70 j + 40 k

FAD

r AB = (40 − 0) i + (0 − 70) j + (0 − 0)k A

r AB = 40 i − 70 j + 0 k

FAB

FAC

The unit vectors corresponding to these position vectors are: D

u AD = 60 m

60 m

= −0.5455i − 0.6364 j − 0.5455k B

40 m

C 40 m

r AD −60 70 60 = i− j− k r AD 110 110 110

u AC =

r AC 40 70 40 = − i− j+ k r AC 90 90 90

= −0.4444 i − 0.7778 j + 0.4444 k

40 m z

u AB =

r AB 40 70 = i− j + 0 k = 0.4963i − 0.8685 j + 0 k 80.6 80.6 r AB

The forces are: F AB = F AB u AB = 0.9926 i − 1.737 j + 0 k F AC = F AC u AC = −0.8888i − 1.5556 j + 0.8888 F AD = F AD u AD = −1.0910 i − 1.2728 j − 1.0910 k The resultant force exerted on the tower by the cables is: FR = F AB + F AC + F AD = −0.9875i − 4.5648 j − 0.2020 k kN

Problem 2.94 The magnitudes of the two force vectors are F A = 4.6 kN and FB = 5.2 kN. Determine the magnitude of the sum of the forces F A + FB . (Notice that the angle between the vector FB and the positive x-axis is 180 ° − 57 ° = 123 °. ) Solution:

y FB 408 578 708

448

758

The vectors in terms of their direction cosines are 508

F A = (4.6 kN)(cos θ Ax i + cos θ Ay j + cos θ Az k) = (4.6 kN)(cos 50 °i + cos 44 ° j + cos 75 °k) = 2.96 i + 3.31 j + 1.19k (kN), FB = (5.2 kN)(cos θ Bx i + cos θ By j + cos θ Bz k)

z x

= (5.2 kN)(cos123 °i + cos 40 ° j + cos 70 °k) = −2.83i + 3.98 j + 1.78k (kN). The magnitude of their sum is F A + FB =

(2.96 − 2.83) 2 + (3.31 + 3.98) 2 + (1.19 + 1.78) 2 (kN)

= 7.87 kN. F A + FB = 7.87 kN.

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Problem 2.95 The rope BD is attached to the bar ABC at B. The distance from A to B is 3 m. Determine the components of the position vector rBD from point B to point D. y

Solution:

The vector from A to C is

r AC = 3 i + 4 j − 6 k (m). Dividing this vector by its magnitude gives a unit vector that points from A toward C. We can obtain the vector from A to B by multiplying that unit vector by 3 m: r AB = (3 m)

  3 i + 4 j − 6 k (m) = (3 m)    (3 m) 2 + (4 m) 2 + (−6 m) 2  = 1.15i + 1.54 j − 2.30 k (m).

r AC r AC

B A

The vector from A to D is 4m

3m z

2m x

r AD = 7 i − 4 k (m). Because r AD = r AB + rBD , we obtain rBD = r AD − r AB = 7 i − 4 k − (1.15i + 1.54 j − 2.30 k) (m) = 5.85i − 1.54 j − 1.70 k (m). rBD = 5.85i − 1.54 j − 1.70 k (m).

Problem 2.96 The rope BD is attached to the bar ABC at B. The distance from A to B is 3 m. The rope exerts a 2-kN force on the bar at B that is directed from B toward D. Determine the components of the force. y

Dividing this vector by its magnitude gives a unit vector that points from A toward C. We can obtain the vector from A to B by multiplying that unit vector by 3 m: r AC r AC

  3 i + 4 j − 6 k (m) = (3 m)    (3 m) 2 + (4 m) 2 + (−6 m) 2  = 1.15i + 1.54 j − 2.30 k (m).

4m B

A 3m z

The vector from A to C is

r AC = 3 i + 4 j − 6 k (m).

r AB = (3 m)

C 6m

Solution:

The vector from A to D is 4m

2m x

r AD = 7 i − 4 k (m). Because r AD = r AB + rBD , we obtain rBD = r AD − r AB = 7 i − 4 k − (1.15i + 1.54 j − 2.30 k) (m) = 5.85i − 1.54 j − 1.70k (m). Using this result, the force exerted on the bar at B is F = (2 kN)

rBD rBD

  5.85i − 1.54 j − 1.70k (m) (2 kN)    (5.85 m) 2 + (−1.54 m) 2 + (−1.70 m) 2  = 1.86 i − 0.489j − 0.540k (kN). 1.86 i − 0.489 j − 0.540 k (kN).

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Problem 2.97 The circular bar has a 4-m radius and lies in the x–y plane. Express the position vector from point B to the collar at A in terms of components.

Solution:

From the figure, the point B is at (0, 4, 3) m. The coordinates of point A are determined by the radius of the circular bar and the angle shown in the figure. The vector from the origin to A is rOA = 4 cos(20 °)i + 4 sin(20 °) j m. Thus, the coordinates of point A are (3.76, 1.37, 0) m. The vector from B to A is given by rBA = ( x A − x B ) i + ( y A − y B ) j + ( z A − z B )k = 3.76 i − 2.63 j −3k m. Finally, the scalar components of the vector from B to A are (3.76, −2.63, −3) m.

4m 208

4m z

Problem 2.98 The cable AB exerts a 60-N force T on the collar at A that is directed along the line from A toward B. Express T in terms of components.

Solution:

We know rBA = 3.76 i − 2.63 j − 3k m from Problem 2.97. The unit vector u AB = −rBA rBA . The unit vector is u AB = −0.686 i + 0.480 j + 0.547k. Hence, the force vector T is given by

T = T (−0.686 i + 0.480 j + 0.547k) N = −41.1i + 28.8 j + 32.8k N

4m 208 4m

Problem 2.99 In Practice Example 2.11, suppose that the vector V is changed to V = 4 i − 6 j − 10 k. (a) What is the value of U ⋅ V ? (b) What is the angle between U and V when they are placed tail to tail?

Solution:

From Active Example 2.4 we have the expression for U.

Thus U = 6 i − 5 j − 3k, V = 4 i − 6k − 10 k U ⋅ V = (6)(4) + (−5)(−6) + (−3)(−10) = 84 cos θ =

U⋅V = V V

84 = 0.814 6 2 + (−5) 2 + (−3) 2 4 2 + (−6) 2 + (−10) 2

θ = cos −1 (0.814) = 35.5 ° (a) U ⋅ V = 84, (b) θ = 35.5 °

Problem 2.100 In Example 2.12, suppose that the coordinates of point B are changed to (6, 4, 4) m. What is the angle θ between the lines AB and AC?

Solution:

Using the new coordinates we have

r AB = (2 i + j + 2k) m, r AB = 3 m r AC = (4 i + 5 j + 2k) m, r AC = 6.71 m cos θ =

r AB ⋅ r AC ((2)(4) + (1)(5) + (2)(2)) m 2 = = 0.845 r AB r AC (3 m)(6.71 m)

θ = cos −1 (0.845) = 32.4 ° θ = 32.4 °

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Problem 2.101

Solution:

Consider the vectors

Applying Eq. (2.23), the dot products are

U ⋅ V = (12)(−6) + (4)(8) + (−8)(14)

U = 12 i + 4 j − 8k, V = −6 i + 8 j + 14 k, W = 8i − 8 j + 8k.

= −152, U ⋅ W = (12)(8) + (4)(−8) + (−8)(8)

Determine the values of the dot products U ⋅ V, U ⋅ W, and V ⋅ W. What can you deduce about the three vectors from the results?

= 0, V ⋅ W = (−6)(8) + (8)(−8) + (14)(8) = 0. U ⋅ V = −152, U ⋅ W = 0, V ⋅ W = 0. U and V are not perpendicular. W is perpendicular to both U and V.

Problem 2.102 Consider the position vectors P and Q. (a) Determine the dot product P ⋅ Q by using the definition, Eq. (2.18). (b) Determine the dot product P ⋅ Q by using Eq. (2.23).

Solution: (a) The two vectors in terms of their components are P = 12 i + 10 k (m), Q = 8i (m). Their magnitudes are

P = (12 m) 2 + (10 m) 2 = 15.6 m, Q = 8 m. The angle between the two vectors is

θ = arctan

(8, 0, 0) m

m ( 10 ) = 39.8°. 12 m

From Eq. (2.18),

P · Q = (15.6 m)(8 m) cos39.8 °

= 96 m 2 .

z (12, 0, 10) m

(b)

From Eq. (2.23), P · Q = (12 m)(8 m) + (0)(0) + (10 m)(0) = 96 m 2 . (a), (b) P · Q = 96 m 2 .

Problem 2.103 Two perpendicular vectors are given in terms of their components by U = U x i − 4 j + 6k and V = 3i + 2 j − 3k. Use the dot product to determine the component U x .

Solution:

When the vectors are perpendicular, U ⋅ V ≡ 0.

Thus U ⋅ V = U xV x + U yV y + U zVz = 0 = 3U x + (−4)(2) + (6)(−3) = 0 3U x = 26 U x = 8.67

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Problem 2.104 The vectors P, Q, and F = Fx i + Fy j lie in the x–y plane. The magnitudes of P and Q are 4 m and 5 m, respectively. The dot products P ⋅ F = 50 N-m (newton-meters) and Q ⋅ F = 60 N-m. Determine the components Fx and Fy .

Solution: The components of P and Q are P = (4 m)(cos15 °i + sin15 ° j), Q = (5 m)(sin 30 °i + cos30 ° j). The two given conditions are P ⋅ F = [ (4 m) cos15 ° ] Fx + [ (4 m)sin15 ° ] Fy = 50 N-m,

Q ⋅ F = [ (5 m)sin 30 ° ] Fx + [ (5 m) cos30 ° ] Fy = 60 N-m.

Solving these two equations, we obtain Fx = 10.9 N, Fy = 7.55 N.

F 308 P 158

Problem 2.105 The magnitudes U = 10 and V = 20. (a) Use Eq. (2.18) to determine U ⋅ V. (b) Use Eq. (2.23) to determine U ⋅ V.

Solution: (a) The definition of the dot product (Eq. (2.18)) is U ⋅ V = U V cos θ. Thus U ⋅ V = (10)(20) cos(45 ° − 30 °) = 193.2

(b) The components of U and V are

U = 10( i cos 45 ° + j sin 45 °) = 7.07 i + 7.07 j U

V = 20( i cos30 ° + j sin 30 °) = 17.32 i + 10 j

458

Problem 2.106 OA and OB?

From Eq. (2.23) U ⋅ V = (7.07)(17.32) + (7.07)(10) = 193.2

308 x

What is the angle between the ropes B

Solution: Let e OA be a unit vector that points from the origin O toward A. In terms of the angles shown, its components are

A 608

e OA = cos35 ° cos15 °i + sin 35 ° j − cos35 ° sin15 °k. 408

The components of a unit vector e OB that points from O toward B are e OB = − cos60 ° sin 40 °i + sin 60 ° j + cos60 ° cos 40 °k.

358 158 x

We can use definition of the dot product to determine the angle between OA and OB: cos θ =

e OA ⋅ e OB e OA e OB

(cos35 ° cos15 °)(− cos60 ° sin 40 °) + (sin 35 °)(sin 60 °) + (− cos35 ° sin15 °)(cos60 ° cos 40 °) (1)(1) = 0.161.

We obtain θ = 80.7 °. 80.7 °.

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Problem 2.107 Use the dot product to determine the angle between the forestay (cable AB) and the backstay (cable BC ) of the sailboat.

Solution:

The unit vector from B to A is rBA e BA = = −0.321i − 0.947 j rBA

The unit vector from B to C is y e BC =

B (4, 13) m

rBC = 0.385i − 0.923 j rBC

From the definition of the dot product, e BA ⋅ e BC = 1 ⋅ 1 ⋅ cosθ, where θ is the angle between BA and BC. Thus cos θ = (−0.321)(0.385) + (−0.947)(−0.923) cos θ = 0.750 θ = 41.3 °

C (9, 1) m

A (0, 1.2) m

Problem 2.108 Determine the angle θ between the lines AB and AC (a) by using the law of cosines (see Appendix A); (b) by using the dot product. y

Solution: (a) We have the distances: AB =

4 2 + 3 2 + 12 m =

AC =

BC =

(5 − 4) 2

m =

+ (−1 − 3) 2

26 m 35 m + (3 + 1) 2 m =

33 m

BC 2 = AB 2 + AC 2 − 2( AB)( AC ) cos θ cos θ = x

u z

+ 32

The law of cosines gives

B (4, 3, 21) m

+ 12

(5, 21, 3) m C

AB 2 + AC 2 − BC 2 = 0.464 ⇒ θ = 62.3 ° 2( AB)( AC )

(b) Using the dot product r AB = (4 i + 3 j − k) m,

r AC = (5i − j + 3k) m

r AB ⋅ r AC = (4 m)(5 m) + (3 m)(−1 m) + (−1 m)(3 m) = 14 m 2 r AB ⋅ r AC = ( AB)( AC ) cos θ Therefore cos θ =

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14 m 2 = 0.464 ⇒ θ = 62.3 ° 26 m 35 m

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Problem 2.109 The ship O measures the positions of the ship A and the airplane B and obtains the coordinates shown. What is the angle θ between the lines of sight OA and OB? y

Solution:

From the coordinates, the position vectors are: rOA = 6 i + 0 j + 3k and rOB = 4 i + 4 j − 4 k The dot product:

rOA ⋅ rOB = (6)(4) + (0)(4) + (3)(−4) = 12 The magnitudes: rOA = rOA =

B (4, 4, 24) km

6 2 + 0 2 + 3 2 = 6.71 km and 4 2 + 4 2 + 4 2 = 6.93 km.

rOA ⋅ rOB = 0.2581, from which θ = ±75 °. rOA rOB From the problem and the construction, only the positive angle makes sense, hence θ = 75 ° From Eq. (2.24) cos θ =

u x

O A (6, 0, 3) km

Problem 2.110 Astronauts on the space shuttle use radar to determine the magnitudes and direction cosines of the position vectors of two satellites A and B. The vector r A from the shuttle to satellite A has magnitude 2 km and direction cosines cos θ x = 0.768, cos θ y = 0.384, cos θ z = 0.512. The vector rB from the shuttle to satellite B has magnitude 4 km and direction cosines cos θ x = 0.743, cos θ y = 0.557, cos θ z = −0.371. What is the angle θ between the vectors r A and rB?

Solution:

The direction cosines of the vectors along r A and rB are the components of the unit vectors in these directions (i.e., u A = cos θ x i + cos θ y j + cos θ z k, where the direction cosines are those for r A ). Thus, through the definition of the dot product, we can find an expression for the cosine of the angle between r A and rB . cos θ = cos θ x A cos θ x B + cos θ y A cos θ y B + cos θ z A cos θ z B . Evaluation of the relation yields cos θ = 0.594 ⇒ θ = 53.5 °.

rB x u

rA z

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Problem 2.111 Segment AB of the bar is parallel to the x-axis. Determine the angle θ between the straight segments AB and BC.

A (0, 3, 1) m

Solution:

The coordinates of point B are (5, 3, 1) m. The vector from B to A is rBA = −5i (m). The vector from B to C is

rBC = (2 m − 5 m) i + (0 − 3 m) j + (4 m − 1 m)k = −3i − 3 j + 3k (m).

The magnitudes of these vectors are

(2, 0, 4) m

rBA = 5 m, rBC =

(−3 m) 2 + (−3 m) 2 + (3 m) 2

= 5.20 m. The value of their dot product is rBA · rBC = (−5 m)(−3 m) + (0)(−3 m) + (0)(3 m) = 15 m 2 . Using the definition of the dot product, cos θ =

rBA · rBC rBA rBC

15 m 2 (5 m)(5.20 m) = 0.577. =

This yields θ = 54.7 °.

Problem 2.112 The person exerts a force F = 60 i − 40 j (N) on the handle of the exercise machine. Use Eq. (2.26) to determine the vector component of F that is parallel to the line from the origin O to where the person grips the handle. Solution:

150 mm

The vector r from the O to where the person grips the

handle is

r = (250 i + 200 j − 150 k) mm, r = 354 mm To produce the unit vector that is parallel to this line we divide by the magnitude e =

200 mm

(250 i + 200 j − 150 k) mm r = = (0.707 i + 0.566 j − 0.424 k) r 354 mm

250 mm

Using Eq. (2.26), we find that the vector component parallel to the line is F p = (e ⋅ F)e = [(0.707)(60 N) + (0.566)(−40 N)](0.707 i + 0.566 j − 0.424 k) F p = (14.0 i + 11.2 j + 8.4 k) N

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Problem 2.113 At the instant shown, the Harrier’s thrust vector is T = 4.22 i + 21.6 j − 3.30 k (kip) and its velocity vector is v = 24.4 i + 8.86 j + 2.44 k (ft/s). (a) Determine the components of the thrust vector parallel and normal (perpendicular) to the velocity vector. (b) The term T ⋅ v is the power being transferred to the plane by its engine. What is the power in horsepower?

Solution: (a) We divide the velocity vector by its magnitude to obtain a unit vector with the same direction: e =

v v

T x

24.4 i + 8.86 j + 2.44 k (ft/s) (24.4 ft/s) 2 + (8.86 ft/s) 2 + (2.44 ft/s) 2 = 0.936 i + 0.340 j + 0.0936k. =

The component of T parallel to the velocity vector is Tp = (e ⋅ T)e. The dot product of e and T is e ⋅ T = (0.936)(4.22 kip) + (0.340)(21.6 kip) + (0.0936)(−3.30 kip) = 11.0 kip. The parallel component is Tp = (e ⋅ T)e = (11.0 kip)(0.936 i + 0.340 j + 0.0936k) = 10.3i + 3.73 j + 1.03k (kip). The component of T normal to the velocity vector is Tn = T − Tp = −6.06 i + 17.9 j − 4.33k (kip). (b) The dot product of T and v is T ⋅ v = (4.22 kip)(24.4 ft/s) + (21.6 kip)(8.86 ft/s) + (−3.30 kip)(2.44 ft/s) = 286 kip-ft/s = 286,000 ft-lb/s. Converting this to horsepower, 286,000 ft-lb/s = 286,000 ft-lb/s

horsepower ( 1 550 ft-lb/s )

= 521 horsepower. (a) Tp = 10.3i + 3.73 j + 1.03k (kip), Tn = −6.06 i + 17.9 j − 4.33k (kip). (b) 521 horsepower.

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Problem 2.114 For the three-dimensional truss shown, determine the angle between members AB and AD. Solution:

F A (4, 3, 4) ft

The vector from A to B is

D (6, 0, 0) ft x

r AB = −4 i − 3 j − 4 k (ft). The vector from A to D is r AD = (6 ft − 4 ft) i + (0 − 3 ft) j + (0 − 4 ft)k

C (5, 0, 6) ft

= 2 i − 3 j − 4 k (ft). The magnitudes of these vectors are r AB =

(−4 ft) 2 + (−3 ft) 2 + (−4 ft) 2

= 6.40 ft, r AD =

(2 ft) 2 + (−3 ft) 2 + (−4 ft) 2

= 5.39 ft. The value of their dot product is r AB ⋅ r AD = (−4 ft)(2 ft) + (−3 ft)(−3 ft) + (−4)(−4 ft) = 17 ft 2 . Using the definition of the dot product, cos θ =

rBA ⋅ rBC rBA rBC

17 ft 2 (6.40 ft)(5.39 ft) = 0.493. =

This yields 60.5 °.

Problem 2.115 For the three-dimensional truss shown, determine the angle between members AC and CD. Solution:

F A (4, 3, 4) ft

The vector from C to A is

D (6, 0, 0) ft x

rCA = (4 ft − 5 ft) i + (3 ft − 0) j + (4 ft − 6 ft)k = −i + 3 j − 2k (ft). z

The vector from C to D is

C (5, 0, 6) ft

rCD = (6 ft − 5 ft) i + (0 − 0) j + (0 − 6 ft)k = i − 6k (ft). The magnitudes of these vectors are rCA =

(−1 ft) 2

+ (3 ft) 2

+ (−2 ft) 2

= 3.74 ft, rCD =

Using the definition of the dot product, cos θ =

(1 ft) 2 + (0) 2 + (−6 ft) 2

= 6.08 ft.

= 11 ft 2 .

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11 ft 2 (3.74 ft)(6.08 ft)

= 0.483.

The value of their dot product is rCA ⋅ rCD = (−1 ft)(1 ft) + (3 ft)(0) + (−2)(−6 ft)

rCA ⋅ rCD rCA rCD

This yields 61.1 °.

12/04/23 4:03 PM

Problem 2.116 The three-dimensional truss is subjected to a force F = 25i − 100 j − 30 k (lb) at A. Determine the components of F parallel and normal (perpendicular) to member AD. Solution:

F A (4, 3, 4) ft

D (6, 0, 0) ft x

The vector from A to D is

r AD = (6 ft − 4 ft) i + (0 − 3 ft) j + (0 − 4 ft)k

C (5, 0, 6) ft

= 2 i − 3 j − 4 k (ft). Dividing this vector by its magnitude yields a unit vector that points from A toward D: e AD =

r AD r AD 2 i − 3 j − 4 k (ft) (2 ft) 2 + (−3 ft) 2 + (−4 ft) 2

= 0.371i − 0.557 j − 0.743k. The component of F parallel to member AD is Fp = (e AD ⋅ F)e AD . The dot product of e AD and F is e AD ⋅ F = (0.371)(25 lb) + (−0.557)(−100 lb) + (−0.743)(−30 lb) = 87.3 lb. The parallel component is Fp = (e AD ⋅ F)e AD = (87.3 lb)(0.371i − 0.557 j − 0.743k) = 32.4 i − 48.6 j − 64.8k (lb). The component of F normal to member AD is Fn = F − Fp = −7.41i − 51.4 j + 34.8k (lb). Fp = 32.4 i − 48.6 j − 64.8k (lb), Fn = −7.41i − 51.4 j + 34.8k (lb).

Problem 2.117 The rope AB exerts a 50-N force T on collar A. Determine the vector component of T parallel to the bar CD. Solution:

0.15 m

We have the following vectors

0.4 m

rCD = (−0.2 i − 0.3 j + 0.25k) m e CD =

rCD = (−0.456 i − 0.684 j + 0.570 k) rCD

A 0.5 m

rOB = (0.5 j + 0.15k) m

rOC = (0.4 i + 0.3 j) m

r e AB = AB = (0.674 i + 0.735 j + 0.079k) r AB

x D

rOA = rOC + (0.2 m)e CD = (0.309 i + 0.163 j + 0.114 k) m r AB = rOB − rOA = (−0.309 i + 0.337 j + 0.036k) m

0.2 m 0.3 m

0.25 m

0.2 m

We can now write the force T and determine the vector component parallel to CD. T = (50 N)e AB = (−33.7 i + 36.7 j + 3.93k) N T p = (e CD ⋅ T)e CD = (3.43i + 5.14 j − 4.29k) N T p = (3.43i + 5.14 j − 4.29k) N

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Problem 2.118 In Problem 2.117, determine the vector component of T normal to the bar CD. Solution:

0.15 m

From Problem 2.117 we have

0.4 m

T = (−33.7 i + 36.7 j + 3.93k) N

Tp = (3.43i + 5.14 j − 4.29k) N

The normal component is then

0.2 m 0.3 m

0.5 m

Tn = T − T p

Tn = (−37.1i + 31.6 j + 8.22k) N.

x D

0.2 m

Problem 2.119 The magnitudes of the two force vectors are F A = 200 lb and FB = 160 lb. What is the angle between the two vectors?

Solution:

0.25 m

The components of F A are

F A = (200 lb)(cos 40 ° sin 50 °i + sin 40 ° j + cos 40 ° cos 50 °k) = 117 i + 129 j + 98.5k (lb). The components of FB are

FB = (160 lb)(− cos60 ° sin 30 °i + sin 60 ° j + cos60° cos30°k) = −40.0 i + 139 j + 69.3k (lb).

FB FA 608 308

408

Their dot product is F A ⋅ FB = (117 lb)(−40.0 lb) + (129 lb)(139 lb) + (98.5 lb)(69.3 lb) = 19,900 lb 2 .

508

Using the definition of the dot product, cos θ = =

F A · FB F A FB 19,900 lb 2 (200 lb)(160 lb)

= 0.623. This yields 51.5 °.

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Problem 2.120 The magnitudes of the two force vectors are F A = 200 lb and FB = 160 lb. Determine the components of F A parallel and normal to FB .

Solution:

The components of F A are

F A = (200 lb)(cos 40 ° sin 50 °i + sin 40 ° j + cos 40 ° cos 50 °k) = 117 i + 129 j + 98.5k (lb). The components of FB are

FB = (160 lb)(− cos60 ° sin 30 °i + sin 60 ° j + cos60° cos30°k) = −40.0 i + 139 j + 69.3k (lb).

FB FA 608 308

408

We divide FB by its magnitude to determine the components of a unit vector with the same direction: x

eB =

508

FB FB

−40.0 i + 139 j + 69.3k (lb) (−40.0 lb) 2 + (139 lb) 2 + (69.3 lb) 2 = −0.250 i + 0.866 j + 0.433k. =

The dot product of e B with F A is e B ⋅ F A = (−0.250)(117 lb) + (0.866)(129 lb) + (0.433)(98.5 lb) = 125 lb. The component of F A that is parallel to FB is F Ap = (e B ⋅ F A )e B = (125 lb)(−0.250 i + 0.866 j + 0.433k) = −31.2 i + 108 j + 54.0 k (lb). The component of F A that is normal to FB is F An = F − F Ap = 149 i + 20.6 j + 44.5k (lb). F Ap = −31.2 i + 108 j + 54.0 k (lb), F An = 149 i + 20.6 j + 44.5k (lb).

Problem 2.121 An astronaut in a maneuvering unit approaches the ISS. At the present instant, the station informs him that his position relative to the origin of the station’s coordinate system is rG = 50 i + 80 j + 180 k (m) and his velocity is v = −2.2 j − 3.6k (m/s). The position of an airlock is r A = −12 i + 20 k (m). Determine the angle between his velocity vector and the line from his position to the airlock’s position.

Solution:

Points G and A are located at G: (50, 80, 180) m and A: (−12, 0, 20) m. The vector rGA is rGA = ( x A − x G ) i + ( y A − y G ) j + ( z A − z G )k = (−12 − 50) i + (0 − 80) j + (20 − 180)k m. The dot product between v and rGA is v ⋅ rGA = v rGA cos θ = v x x GA + v y y GA + v z z GA , where θ is the angle between v and rGA . Substituting in the numerical values, we get θ = 19.7 °.

G A z

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Problem 2.122 In Problem 2.121, determine the vector component of the astronaut’s velocity parallel to the line from his position to the airlock’s position.

Solution:

The coordinates are A (−12, 0, 20) m, G (50, 80, 180) m.

Therefore rGA = (−62 i − 80 j − 160 k) m e GA =

rGA = (−0.327 i − 0.423 j − 0.845k) rGA

The velocity is given as v = (−2.2 j − 3.6k) m/s The vector component parallel to the line is now v p = (e GA ⋅ v)e GA = [(−0.423)(−2.2) + (−0.845)(−3.6)]e GA v p = (−1.30 i − 1.68 j − 3.36k) m/s

Problem 2.123 Point P is at longitude 30°W and latitude 45°N on the Atlantic Ocean between Nova Scotia and France. Point Q is at longitude 60°E and latitude 20°N in the Arabian Sea. Use the dot product to determine the shortest distance along the surface of the earth from P to Q in terms of the radius of the earth R E . Strategy: Use the dot product to determine the angle between the lines OP and OQ; then use the definition of an angle in radians to determine the distance along the surface of the earth from P to Q.

P = rOB + rBP = R E ( i cos λ P cos θ P + j sin θ P + k sin λ P cos θ P ). A similar argument for the point Q yields Q = rOC + rCQ = R E ( i cos λ Q cos θQ + j sin θQ + k sin λ Q cos θQ ) Using the identity cos 2 β + sin 2 β = 1, the magnitudes are P = Q = RE The dot product is P ⋅ Q = R E 2 (cos(λ P − λ Q ) cos θ P cos θQ + sin θ P sin θQ )

Substitute:

cos θ =

P⋅Q = cos(λ P − λ Q ) cos θ P cos θQ + sin θ P sin θQ P Q

P Q 458

308

208

Substitute λ P = +30 °, λ Q = −60 °, θ p = +45 °, θQ = +20 °, to obtain cos θ = 0.2418, or θ = 1.326 radians. Thus the distance is d = 1.326 R E

608

G P

Equator

458 x

Solution:

The distance is the product of the angle and the radius of the sphere, d = R E θ, where θ is in radian measure. From Eqs. (2.18) and (2.24), the angular separation of P and Q is given by

RE b 308 x

608

Q 208 c

 P ⋅ Q  . cos θ =   P Q  The strategy is to determine the angle θ in terms of the latitude and longitude of the two points. Drop a vertical line from each point P and Q to b and c on the equatorial plane. The vector position of P is the sum of the two vectors: P = rOB + rBP . The vector rOB = rOB ( i cos λ P + 0 j + k sin λ P ). From geometry, the magnitude is rOB = R E cos θ P . The vector rBP = rBP (0 i + 1 j + 0 k). From geometry, the magnitude is rBP = R E sin θ P . Substitute and reduce to obtain:

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Problem 2.124

Consider the vectors U = 12 i + 4 j − 8k, V = −6 i + 8 j + 14 k.

(a) Determine the components of the vector W = U × V. (b) Evaluate the dot products U ⋅ W and V ⋅ W. What is the explanation of these two results? Solution: (a) The vector W is W = U×V i = 12

j 4

k −8

−6

= [ (4)(14) − (−8)(8) ] i − [ (12)(14) − (−8)(−6) ] j + [ (12)(8) − (4)(−6) ] k = 120 i − 120 j + 120 k. (b) The dot products are U ⋅ W = (12)(120) + (4)(−120) + (−8)(120) = 0, V ⋅ W = (−6)(120) + (8)(−120) + (14)(120) = 0. (a) W = 120 i − 120 j + 120 k. (b) U ⋅ W = 0, V ⋅ W = 0. From the definition of the cross product, W is perpendicular to both U and V.

Problem 2.125 Two vectors U = 3i + 2 j and V = 2 i + 4 j. (a) What is the cross product U × V? (b) What is the cross product V × U?

Solution:

Use Eq. (2.34) and expand into 2 by 2 determinants.

U×V =

i j k 3 2 0 2 4 0

= i((2)(0) − (4)(0)) − j((3)(0) − (2)(0)) + k((3)(4) − (2)(2)) = 8k

V×U =

i j k 2 4 0 3 2 0

= i((4)(0) − (2)(0)) − j((2)(0) − (3)(0)) + k((2)(2) − (3)(4)) = −8k

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Problem 2.126 Consider the position vectors P and Q. (a) Determine the cross product P × Q by using the definition, Eq. (2.28). (b) Determine the dot product P ⋅ Q by using Eq. (2.34).

The angle between the two vectors is θ = arctan

m ( 10 ) = 39.8°. 12 m

From Eq. (2.28), P × Q = (15.6 m)(8 m)sin 39.8 °e = (80 m 2 )e.

The unit vector e is defined to be perpendicular to P and perpendicular to Q. Either j or −j satisfies those criteria. The remaining criterion states that P, Q, e form a right-handed system, which requires that e = j. Therefore,

(8, 0, 0) m

P × Q = 80 j (m 2 ).

(b) From Eq. (2.34),

z P×Q =

(12, 0, 10) m

Solution:

12 m 0 10 m 8m 0 0

= [(0)(0) − (10 m)(0)]i − [(12 m)(0) − (10 m)(8 m)] j

(a) The two vectors in terms of their components are

+ [(12 m)(0) − (0)(8 m)]k

P = 12 i + 10 k (m), Q = 8i (m).

= 80 j (m 2 ).

Their magnitudes are

(a), (b) P × Q = 80 j (m 2 ).

P = (12 m) 2 + (10 m) 2 = 15.6 m, Q = 8 m.

Problem 2.127 Consider the position vectors P and Q. (a) Determine the cross product Q × P by using the definition, Eq. (2.28). (b) Determine the dot product Q ⋅ P by using Eq. (2.34).

Solution: (a) The two vectors in terms of their components are P = 12 i + 10 k (m), Q = 8i (m). Their magnitudes are P = (12 m) 2 + (10 m) 2 = 15.6 m, Q = 8 m.

The angle between the two vectors is θ = arctan Q

m ( 10 ) = 39.8°. 12 m

From Eq. (2.28),

(8, 0, 0) m x

Q × P = (8 m)(15.6 m)sin 39.8 °e = (80 m 2 )e.

The unit vector e is defined to be perpendicular to Q and p erpendicular to P. Either j or −j satisfies those criteria. The remaining criterion states that Q, P, e form a right-handed system, which requires that e = − j. Therefore,

z (12, 0, 10) m

Q × P = −80 j (m 2 ). (b) From Eq. (2.34),

Q×P =

i j k 8m 0 0 12 m 0 10 m

= [ (0)(10 m) − (0)(0) ] i − [ (8 m)(10 m) − (0)(12 m) ] j + [ (8 m)(0) − (0)(12 m) ] k = −80 j (m 2 ). (a), (b) Q × P = −80 j (m 2 ).

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Problem 2.128 Suppose that the cross product of two vectors U and V is U × V = 0. If U ≠ 0, what do you know about the vector V ?

Solution:

Problem 2.129 The cross product of two vectors U and V is U × V = −30 i + 40 k. The vector V = 4 i − 2 j + 3k. The vector U = 4 i + U y j + U z k. Determine U y and U z .

Solution:

From the given information we have

U×V =

i j 4 Uy

k Uz

4 −2

Either V = 0 or V U

= (3U y + 2U z ) i + (4U z − 12) j + (−8 − 4U y )k U × V = (−30i + 40 k) Equating the components we have 3U y + 2U z = −30, 4U z − 12 = 0,

−8 − 4U y = 40.

Solving any two of these three redundant equations gives U y = −12, U z = 3.

Problem 2.130 (a) Determine the components of a unit vector e p that is parallel to the line AB and points from A toward B. (b) Determine the components of a unit vector e n that is normal (perpendicular) to the line AB. (Make sure your result is a unit vector.) Confirm that it is perpendicular to the line by showing that e p ⋅ e n = 0.

Dividing this vector by its magnitude yields the unit vector e p : ep =

r AB r AB

4 i − 18 j − 6k (ft) (4 ft) 2 + (−18 ft) 2 + (−6 ft) 2 = 0.206 i − 0.928 j − 0.309k. =

(b) The cross product of e p with any non-zero vector that is not parallel to the line will yield a vector that is perpendicular to the line, from the definition of the cross product. For example, the vector

y A

e p × i = (0.206 i − 0.928 j − 0.309k) × i = 0.206( i × i) − 0.928( j × i) − 0.309(k × i) = 0.928k − 0.309 j

(22, 12, 2) ft

is perpendicular to e p . However, although it is the cross product of two unit vectors, it is not a unit vector: ep × i =

= 0.978.

To obtain the desired unit vector, we must divide the vector e p × i by its magnitude:

(2, 26, 24) ft

Solution:

en =

y A

B (2, –6, –4) ft

x z

B (2, –6, –4) ft

(a) The position vector from point A to point B is r AB = (2 ft − (−2 ft)) i + (−6 ft − 12 ft) j + (−4 ft − 2 ft)k = 4 i − 18 j − 6k (ft).

ep × i ep × i −0.309 j + 0.928k 0.978

= −0.316 j + 0.949k.

(–2, 12, 2) ft rAB

(–2, 12, 2) ft

(0.928) 2 + (−0.309) 2

The dot product of this vector with e p is zero, confirming that the two vectors are perpendicular: e p ⋅ e n = (0.206)(0) + (−0.928)(−0.316) + (−0.309)(0.949) = 0. (a) e p = 0.206 i − 0.928 j − 0.309k. (b) For example, e n = −0.316 j + 0.949k.

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Problem 2.131 The force F = 10 i − 4 j (N). Determine the cross product r AB × F.

(6, 3, 0) m A rA B x

(6, 0, 4) m B F

Solution:

The position vector is

A(6, 3, 0)

r AB = (6 − 6) i + (0 − 3) j + (4 − 0)k = 0 i − 3 j + 4 k The cross product: r AB × F =

i j k 0 −3 4 10 −4 0

rA B = i(16) − j(−40) + k(30)

x z

= 16 i + 40 j + 30 k (N-m)

Problem 2.132 By evaluating the cross product U × V, prove the identity sin(θ1 − θ 2 ) = sin θ1 cos θ 2 − cos θ1 sin θ 2 .

B(6, 0, 4)

Solution: Assume that both U and V lie in the x -y plane. The strategy is to use the definition of the cross product (Eq. 2.28) and the Eq. (2.34), and equate the two. From Eq. (2.28) U × V = U V sin (θ1 − θ 2 )e. Since the positive z -axis is out of the paper, and e points into the paper, then e = −k. Take the dot product of both sides with e, and note that k ⋅ k = 1. Thus

( (U ×U VV) ⋅ k )

sin (θ1 − θ 2 ) = − The vectors are: U

U = U (i cos θ1 + j sin θ 2 ), and V = V (i cos θ 2 + j sin θ 2 ).

The cross product is

u1 u2

U×V =

i U cos θ1

j U sin θ1

k 0

V cos θ 2

V sin θ 2

= i(0) − j(0) + k( U V )(cos θ1 sin θ 2 − cos θ 2 sin θ1 ) Substitute into the definition to obtain: sin (θ1 − θ 2 ) = sin θ1 cos θ 2 − cos θ1 sin θ 2 . Q.E.D.

y U V

U1 U2

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Problem 2.133 In Example 2.15, what is the minimum distance from point B to the line OA?

Solution: Let θ be the angle between rOA and rOB . Then the minimum distance is d = rOB sin θ Using the cross product, we have rOA × rOB = rOA rOB sin θ = rOA d ⇒ d =

rOA × rOB rOA

We have rOA = (10 i − 2 j + 3k) m rOB = (6 i + 6 j − 3k) m

rOA × rOB =

i j k 10 −2 3 6 6 −3

= (−12 i + 48 j + 72k) m 2

Thus d =

(−12 m 2 ) + (48 m 2 ) 2 + (72 m 2 ) 2 = 8.22 m (10 m) 2 + (−2 m) 2 + (3 m) 2

d = 8.22 m

Problem 2.134 (a) What is the cross product rOA × rOB? (b) Determine a unit vector e that is perpendicular to rOA and rOB . Solution:

y B (4, 4, 24) m

The two radius vectors are rOB

rOB = 4 i + 4 j − 4 k, rOA = 6 i − 2 j + 3k (a) The cross product is rOA × rOB =

i j k 6 −2 3 4 4 −4

O = i(8 − 12) − j(−24 − 12) + k(24 + 8)

= −4 i + 36 j + 32k (m 2 )

x rOA

A (6, 22, 3) m

The magnitude is rOA × rOB =

4 2 + 36 2 + 32 2 = 48.33 m 2

(b) The unit vector is  r × rOB  e = ± OA  = ± (−0.0828i + 0.7448 j + 0.6621k)  rOA × rOB  (Two vectors.)

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Problem 2.135 Use the cross product to determine the length of the shortest straight line from point B to the straight line that passes through points O and A. y

i j k −4 +36 32 −2 3 6

C × rOA =

C = 172 i + 204 j − 208k

B (4, 4, 24) m

The unit vector in the direction of C is eC =

rOB

C = 0.508i + 0.603 j − 0.614 k C

(The magnitude of C is 338.3)

We now want to find the length of the projection, P, of line OB in direction e c . P = rOB ⋅ e C

rOA

= (4 i + 4 j − 4 k) ⋅ e C

A (6, 22, 3) m

P = 6.90 m y

Solution:

B( 4, 4, –4) m

rOA = 6 i − 2 j + 3k (m) rOB = 4 i + 4 j − 4 k (m) rOA × rOB = C

r OB

(C is ⊥ to both rOA and rOB ) C =

i j k 6 −2 3 4 4 −4

= (+8 − 12) i + (12 + 24) j + (24 + 8)k

C = −4 i + 36 j + 32k

x r OA

P A(6, –2, 3) m

C is ⊥ to both rOA and rOB . Any line ⊥ to the plane formed by C and rOA will be parallel to the line BP on the diagram. C × rOA is such a line. We then need to find the component of rOB in this direction and compute its magnitude.

Problem 2.136 The cable BC exerts a 1000-lb force F on the hook at B. Determine r AB × F. Solution:

The coordinates of points A, B, and C are A (16, 0, 12), B (4, 6, 0), C (4, 0, 8). The position vectors are

F 6 ft

rOA = 16 i + 0 j + 12k, rOB = 4 i + 6 j + 0 k, rOC = 4 i + 0 j + 8k.

8 ft

rBC r − rOB r = OC = AB rBC rOC − rOB r AB

Noting rOC − rOB = (4 − 4) i + (0 − 6) j + (8 − 0)k = 0 i − 6 j + 8k rOC − rOB =

rAB x

The force F acts along the unit vector e BC =

C rAC

4 ft 4 ft

e BC = 0 i − 0.6 j + 0.8k, and F = F e BC = 0 i − 600 j + 800 k (lb).

The vector

6 ft

r AB = (4 − 16) i + (6 − 0) j + (0 − 12)k = −12 i + 6 j − 12k Thus the cross product is r AB × F =

8 ft = −2400 i + 9600 j + 7200 k (ft-lb)

BandF_6e_ISM_CH02.indd 73

12 ft

6 2 + 8 2 = 10. Thus

i j k −12 −12 6 −600 800 0

r x

4 ft 4 ft

12 ft

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Problem 2.137* The force F = 200 i + 180 j + 100 k ( lb ). Determine the vector component of F that is normal (perpendicular) to the flat plate ACD. y

The vector from A to C is r AC = −6 i + 10 j (ft). The vector from A to D is r AD = −2 i + 12k (ft). The cross product r AC × r AD is perpendicular to the plate.

(0, 10, 0) ft

i j k −6 10 0 −2 0 12

r AC × r AD =

= 120 i + 72 j + 20 k(ft 2 ). A

(6, 0, 0) ft z

Dividing this vector by its magnitude yields a unit vector that is perpendicular to the plate: e =

(4, 0, 12) ft

r AC × r AD r AC × r AD

= 0.849 i + 0.509 j + 0.141k. From Eq. (2.26), the vector component of F that is parallel to e, and so is perpendicular to the plate, is

Solution: y

Fn = (e ⋅ F)e = [(0.849)(200 lb) + (0.509)(180 lb)

+ (0.141)(100 lb)](0.849 i + 0.509 j + 0.141k)

(0, 10, 0) ft

= 234 i + 140 j + 39k(lb).

rAC

Fn = 234 i + 140 j + 39k(lb). A

rAD z

(6, 0, 0) ft

D (4, 0, 12) ft y

Problem 2.138 The rope AB exerts a 50-N force T on the collar at A. Let rCA be the position vector from point C to point A. Determine the cross product rCA × T.

0.15 m

Solution:

We define the appropriate vectors.

0.4 m C

rCD = (−0.2 i − 0.3 j + 0.25k) m

r rCA = (0.2 m) CD = (−0.091i − 0.137 j + 0.114 k) m rCD

A 0.5 m

rOB = (0.5 j + 0.15k) m

rOC = (0.4 i + 0.3 j) m

x D

r AB = rOB − (rOC + rCA ) = (0.61i − 1.22 j − 0.305k) m T = (50 N)

0.2 m 0.3 m

0.25 m

0.2 m

r AB = (−33.7 i + 36.7 j + 3.93k) N r AB

Now take the cross product rCA × T =

i j l −0.091 −0.137 0.114 36.7 3.93 −33.7

= (−4.72 i − 3.48 j + −7.96k) N-m

rCA × T = (−47.2 i − 3.48 j + −7.96k) N-m

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Problem 2.139 In Example 2.16, suppose that the attachment point E is moved to the location (0.3, 0.3, 0) m and the magnitude of T increases to 600 N. What is the magnitude of the component of T perpendicular to the door?

Solution:

We first develop the force T.

rCE = (0.3i + 0.1 j) m r T = (600 N) CE = (569 i + 190 j) N rCE From Example 2.16 we know that the unit vector perpendicular to the door is e = (0.358i + 0.894 j + 0.268k) The magnitude of the force perpendicular to the door (parallel to e ) is then Tn = T ⋅ e = (569 N)(0.358) + (190 N)(0.894) = 373 N Tn = 373 N

Problem 2.140 The slender bar AB is 14 ft in length and is perpendicular to the flat plate ACD. (a) Determine the components of a unit vector e that is perpendicular to the plate ACD and points in the direction from A toward B. (b) Determine the coordinates ( x B , y B , z B ) of point B.

The cross product r AC × r AD is perpendicular to the plate and points in the direction from A toward B. i r AC × r AD =

−6 10

−2

= 120 i + 72 j + 20 k (ft 2 ). Dividing this vector by its magnitude yields the desired unit vector: (0, 10, 0) ft

B (xB, yB, zB)

x (6, 0, 0) ft D

r AC × r AD r AC × r AD

= 0.849 i + 0.509 j + 0.141k. (b) The bar is 14 ft long, so we obtain the vector from A to B by multiplying e by 14 ft:

e =

r AB = (14 ft)e = 11.9 i + 7.13 j + 1.98k (ft). Adding the coordinates of point A to the components of this vector, we obtain the coordinates of point B:

(4, 0, 12) ft

( x B , y B , z B ) = (17.9, 7.13, 1.98) ft.

Solution:

(a) e = 0.849 i + 0.509 j + 0.141k.

(b) ( x B , y B , z B ) = (17.9, 7.13, 1.98) ft. (0, 10, 0) ft

rAC A

rAB x

rAD z

(6, 0, 0) ft

D (4, 0, 12) ft

(a) The vector from A to C is r AC = −6 i + 10 j (ft). The vector from A to D is r AD = −2 i + 12k (ft).

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Problem 2.141* Determine the minimum distance from point P to the plane defined by the three points A, B, and C.

y B (0, 5, 0) m P (9, 6, 5) m

Solution:

The strategy is to find the unit vector perpendicular to the plane. The projection of this unit vector on the vector OP : rOP ⋅ e is the distance from the origin to P along the perpendicular to the plane. The projection on e of any vector into the plane (rOA ⋅ e, rOB ⋅ e, or rOC ⋅ e) is the distance from the origin to the plane along this same perpendicular. Thus the distance of P from the plane is

A (3, 0, 0) m C

d = rOP ⋅ e − rOA ⋅ e.

rBC × rBA =

i j k 0 −5 4 3 −5 0

P[9, 6, 5]

B[0, 5, 0] = 20 i + 12 j + 15k.

Problem 2.142 Two cartesian coordinate systems are shown. The unit vectors i′ and j′ of the x ′y ′z ′ coordinate system are given in terms of their components in the xyz coordinate system by

A[3, 0, 0] z

C[0, 0, 4]

Solution: (a) The vector k ′ is k′ = i′ × j′

i′ = 0.743i + 0.557 j + 0.371k, j′ = −0.408i + 0.816 j − 0.408k.

(a) Determine the unit vector k′ of the x ′y ′z ′ coordinate system in terms of its components in the xyz coordinate system. (b) The force F = 40 i + 80 j + 20 k (N). Use the dot product to determine the components of F in terms of the x ′y ′z ′ coordinate system.

The magnitude is rBC × rBA = 27.73, thus the unit vector is e = 0.7212 i + 0.4327 j + 0.5409k. The distance of point P from the plane is d = rOP ⋅ e − rOA ⋅ e = 11.792 − 2.164 = 9.63 m. The second term is the distance of the plane from the origin; the vectors rOB , or rOC could have been used instead of rOA .

(0, 0, 4) m

The position vectors are: rOA = 3i, rOB = 5 j, rOC = 4 k and rOP = 9 i + 6 j + 5k. The unit vector perpendicular to the plane is found from the cross product of any two vectors lying in the plane. Noting: rBC = rOC − rOB = −5 j + 4 k, and rBA = rOA − rOB = 3i − 5 j. The cross product:

i j k 0.743 0.557 0.371 −0.408 0.816 −0.408

= −0.530 i + 0.152 j + 0.834 k. (b) The x ′ component of F is Fx ′ = F ⋅ i′ = (40 N)(0.743) + (80 N)(0.557) + (20 N)(0.371) = 81.7 N, the y ′ component is

Fy ′ = F ⋅ j′

= (40 N)(−0.408) + (80 N)(0.816) + (20 N)(−0.408) = 40.8 N,

and the z ′ component is Fz ′ = F ⋅ k ′

Therefore F = 81.7 i′ + 40.8 j′ + 7.61k ′ (N).

z9 z

= (40 N)(−0.530) + (80 N)(0.152) + (20 N)(0.834) = 7.61 N.

(a) k ′ = −0.530 i + 0.152 j + 0.834 k. (b) F = 81.7 i′ + 40.8 j′ + 7.61k ′ (N).

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Problem 2.143 For the vectors U = 6 i + 2 j − 4 k, V = 2 i + 7 j, and W = 3i + 2k, evaluate the following mixed triple products: (a) U ⋅ ( V × W); (b) W ⋅ ( V × U); (c) V ⋅ (W × U).

Solution: (a)

Use Eq. (2.36). 6 2 −4 2 7 0 3 0 2

U ⋅ ( V × W) =

= 6(14) − 2(4) + (−4)(−21) = 160 (b)

3 0 2 2 7 0 6 2 −4

W ⋅ ( V × U) =

= 3(−28) − (0) + 2(4 − 42) = −160 (c)

2 7 0 3 0 2 6 2 −4

V ⋅ (W × U ) =

= 2(−4) − 7(−12 − 12) + (0) = 160

Problem 2.144 Use the mixed triple product to calculate the volume of the parallelepiped.

Solution:

We are given the coordinates of point D. From the geometry, we need to locate points A and C. The key to doing this is to note that the length of side OD is 200 mm and that side OD is the x-axis. Sides OD, AE , and CG are parallel to the x-axis and the coordinates of the point pairs (O and D), ( A and E), and (C and D) differ only by 200 mm in the x coordinate. Thus, the coordinates of point A are (−60, 90, 30) mm and the coordinates of point C are ( −40, 0, 100) mm. Thus, the vectors rOA , rOD , and rOC are rOD = 200 i mm, rOA = −60 i + 90 j + 30 k mm, and rOC = −40 i + 0 j + 100 k mm. The mixed triple product of the three vectors is the volume of the parallelepiped. The volume is

rOA ⋅ (rOC × rOD ) =

(140, 90, 30) mm

(200, 0, 0) mm x (160, 0, 100) mm

z y

(140, 90, 30) mm E

−60 90 30 −40 0 100 200 0 0

F O

= −60(0) + 90(200)(100) + (30)(0) mm 3 = 1,800,000 mm 3

D (200, 0, 0) mm

G C

(160, 0, 100) mm

Problem 2.145 that

By using Eqs. (2.23) and (2.34), show

U ⋅ (V × W) =

Solution: One strategy is to expand the determinant in terms of its components, take the dot product, and then collapse the expansion. Eq. (2.23) is an expansion of the dot product: Eq. (2.23): U ⋅ V = U X V X + U Y VY + U Z VZ . Eq. (2.34) is the determinant representation of the cross product: Eq. (2.34) U × V =

i UX VX

j UY VY

k UZ VZ

For notational convenience, write P = (U × V). Expand the determinant about its first row: P = i

UY VY

UZ VZ

BandF_6e_ISM_CH02.indd 77

UX VX

UZ VZ

+ k

UX VX

UZ VZ

Since the two-by-two determinants are scalars, this can be written in the form: P = iPX + jPY + kPZ where the scalars PX , PY , and PZ are the two-by-two determinants. Apply Eq. (2.23) to the dot product of a vector Q with P. Thus Q ⋅ P = Q X PX + QY PY + Q Z PZ . Substitute PX , PY , and PZ into this dot product Q ⋅ P = QX

UY VY

UZ VZ

− QY

UX VX

UZ VZ

+ QZ

UX VX

UZ VZ

But this expression can be collapsed into a three-by-three determinant directly, thus:

Q ⋅ (U × V ) =

− j

QX UX VX

QY UY VY

QZ UZ VZ

. This completes the demonstration.

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Problem 2.146 The magnitudes of the two force vectors are F A = 200 lb and FB = 160 lb. What is the magnitude of their cross product F A × FB ? Solution:

y FB FA

The components of F A are

F A = (200 lb)(cos 40 ° sin 50 °i + sin 40 ° j + cos 40 ° cos 50 °k)

608

= 117 i + 129 j + 98.5k (lb).

308

The components of FB are FB = (160 lb)(− cos60 ° sin 30 °i + sin 60 ° j + cos60° cos30°k)

408

508

= −40.0 i + 139 j + 69.3k (lb). Their cross product is F A × FB =

i j k 117 lb 129 lb 98.5 lb −40.0 lb 139 lb 69.3 lb

= −4740 i − 12,100 j + 21,400 k (lb 2 ). Its magnitude is F A × FB =

(−4740 lb 2 ) 2 + (−12,100 lb 2 ) 2 + (21,400 lb 2 ) 2

= 25,000 lb 2 . F A × FB = 25,000 lb 2 .

Problem 2.147 The magnitude of F is 8 kN. Express F in terms of scalar components. y

Solution: The unit vector collinear with the force F is developed as follows: The collinear vector is r = (7 − 3) i + (2 − 7) j = 4 i − 5 j The magnitude: r = 4 2 + 5 2 = 6.403 m. The unit vector is r e = = 0.6247 i − 0.7809 j. The force vector is r

(3, 7) m

F = F e = 4.998i − 6.247 j = 5i − 6.25 j (kN)

(7, 2) m x

Problem 2.148 The magnitude of the vertical force W is 600 lb, and the magnitude of the force B is 1500 lb. Given that A + B + W = 0, determine the magnitude of the force A and the angle α. Solution:

The strategy is to use the condition of force balance to determine the unknowns. The weight vector is W = −600 j. The vector B is

W B a

508

B = 1500 ( i cos 50 ° + j sin 50 °) = 964.2 i + 1149.1 j The vector A is A = A ( i cos(180 + α) + j sin (180 + α)) A = A (−i cos α − j sin α). The forces balance, hence A + B + W = 0, or (964.2 − A cos α) i = 0, and (1149.1 − 600 − A sin α) j = 0. Thus A cos α = 964.2, and A sin α = 549.1. Take the ratio of the two equations to obtain tan α = 0.5695, or α = 29.7 °. Substitute this angle to solve: A = 1110 lb

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Problem 2.149 The magnitude of the vertical force vector A is 200 lb. If A + B + C = 0, what are the magnitudes of the force vectors B and C ? Solution:

The strategy is to express the forces in terms of scalar components, and then solve the force balance equations for the unknowns. C = C (−i cos α − j sin α), where tan α =

50 in

50 = 0.7143, or α = 35.5 °. 70

Thus C = C (−0.8137 i − 0.5812 j). Similarly, B = + B i, and A = +200 j. The force balance equation is A + B + C = 0. Substituting, (−0.8137 C + B ) i = 0, and (−0.5812 C + 200) j = 0. Solving, C = 344.1 lb, B = 280 lb

The strategy is to express the force vectors in terms of scalar components, and then solve the force balance equation for the unknowns. The force vectors are: 50 E = E ( i cos β − j sin β ), where tan β = = 0.5, or 100 β = 26.6 °. Thus

Problem 2.150 The magnitude of the horizontal force vector D is 280 lb. If D + E + F = 0, what are the magnitudes of the force vectors E and F ? Solution:

100 in

70 in

100 in

70 in

50 in

C D

E = E (0.8944 i − 0.4472 j) D = −280 i, and F = F j. The force balance equation is D + E + F = 0. Substitute and resolve into two equations: (0.8944 E − 280) i = 0, and (−0.4472 E + F ) j = 0. Solve: E = 313.1 lb, F = 140 lb

Problem 2.151 Solution:

What are the direction cosines of F ?

Use the definition of the direction cosines and the ensu-

A (4, 4, 2) ft

ing discussion. The magnitude of F: F =

20 2 + 10 2 + 10 2 = 24.5.

The direction cosines are cos θ x =

Fx 20 = = 0.8165, 24.5 F

B (8, 1, 22) ft x

Fy 10 = = 0.4082 F 24.5 F −10 = −0.4082 cos θ z = z = F 24.5

cos θ y =

Problem 2.152 Determine the components of a unit vector parallel to line AB that points from A toward B. y A (4, 4, 2) ft

F 5 20i 1 10j 2 10k (lb)

F 5 20i 1 10j 2 10k (lb) u

Solution:

Use the definition of the unit vector, we get

The position vectors are: r A = 4 i + 4 j + 2k, rB = 8i + 1 j − 2k. The vector from A to B is r AB = (8 − 4) i + (1 − 4) j + (−2 − 2)k = 4 i − 3 j − 4 k. The magnitude: r AB = 4 2 + 3 2 + 4 2 = 6.4. The unit vector is e AB =

r AB 4 3 4 = i− j− k = 0.6247 i − 0.4685 j − 0.6247k 6.4 6.4 6.4 r AB

B (8, 1, 22) ft x z

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Problem 2.153 What is the angle θ between the line AB and the force F ? Solution:

F 5 20i 1 10j 2 10k (lb)

A (4, 4, 2) ft

The vector from A to B is

r AB = (8 − 4) i + (1 − 4) j + (−2 − 2)k (ft) B (8, 1, 22) ft x

= 4 i − 3 j − 4 k (ft). The dot product of this vector with F is z

r AB ⋅ F = (4)(20) + (−3)(10) + (−4)(−10) ft-lb = 90 ft-lb.

From the definition of the dot product

r AB ⋅ F = r AB F cos θ, A

We obtain r ⋅F cos θ = AB = r AB F

90 ft-lb (4 ft) 2 + (−3 ft) 2 + (−4 ft) 2 (20 lb) 2 + (10 lb) 2 + (−10 lb) 2

rAB B

= 0.574.

This gives θ = 55.0 °. z

θ = 55.0 °.

Problem 2.154 Determine the vector component of F that is parallel to the line AB. y A (4, 4, 2) ft

F 5 20i 1 10j 2 10k (lb)

Solution: Use the definition in Eq. (2.26): U P = (e ⋅ U)e, where e is parallel to a line L. From Problem 2.152 the unit vector parallel to line AB is e AB = 0.6247 i − 0.4688 j − 0.6247k. The dot product is e ⋅ F = (0.6247)(20) + (−0.4688)(10) + (−0.6247)(−10) = 14.053. The parallel vector is

(e ⋅ F)e = (14.053)e = 8.78i − 6.59 j − 8.78k (lb) B (8, 1, 22) ft x z y

Problem 2.155 Determine the vector component of F that is normal to the line AB. Solution:

A (4, 4, 2) ft

F 5 20i 1 10j 2 10k (lb) u

Use the Eq. (2.27) and the solution to Problem 2.154.

FN = F − FP = (20 − 8.78) i + (10 + 6.59) j + (−10 + 8.78)k

B (8, 1, 22) ft x

= 11.22 i + 16.59 j − 1.22k (lb) z

Problem 2.156 Determine the vector rBA × F, where rBA is the position vector from B to A.

Solution:

Use the definition in Eq. (2.34). Noting rBA = −r AB , from Problem 2.155 rBA = −4 i + 3 j + 4 k. The cross product is

rBA × F =

i j k −4 3 4 20 10 −10

A (4, 4, 2) ft

F 5 20i 1 10j 2 10k (lb) u

= (−30 − 40) i − (40 − 80) j + (−40 − 60) B (8, 1, 22) ft x

= −70 i + 40 j − 100 k (ft-lb) z

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Problem 2.157 (a) Write the position vector r AB from point A to point B in terms of components. (b) A vector R has magnitude R = 200 lb and is parallel to the line from A to B. Write R in terms of components.

y A (4, 4, 2) ft

Solution:

B (8, 1, 22) ft x

(a) r AB = ([8 − 4]i + [1 − 4] j + [−2 − 2]k) ft r AB = (4 i − 3 j − 4 k) ft

F 5 20i 1 10j 2 10k (lb)

r AB = (125i − 93.7 j − 125k) N r AB

(b) R = (200 N)

R = (125i − 96.3 j − 125k) N

Problem 2.158 The rope exerts a force of magnitude F = 200 lb on the top of the pole at B. (a) Determine the vector r AB × F, where r AB is the position vector from A to B. (b) Determine the vector r AC × F, where r AC is the position vector from A to C.

B (5, 6, 1) ft

Solution:

The strategy is to define the unit vector pointing from B to A, express the force in terms of this unit vector, and take the cross product of the position vectors with this force. The position vectors

r AB = 5i + 6 j + 1k, r AC = 3i + 0 j + 4 k, rBC = (3 − 5) i + (0 − 6) j + (4 − 1)k = −2 i − 6 j + 3k. The magnitude rBC = e BC =

x C (3, 0, 4) ft

2 2 + 6 2 + 3 2 = 7. The unit vector is

rBC = −0.2857 i − 0.8571 j + 0.4286k. rBC

The force vector is F = F e BC = 200 e BC = −57.14 i − 171.42 j + 85.72k. The cross products:

r AB × F =

−57.14 −171.42 85.72 = 685.74 i − 485.74 j − 514.26k = 685.7 i − 485.7 j − 514.3k (ft-lb)

r AC × F =

−57.14 −171.42 85.72 = 685.68i − 485.72 j − 514.26k = 685.7 i − 485.7 j − 514.3k (ft-lb)

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Problem 2.159 The pole supporting the sign is parallel to the x -axis and is 6 ft long. Point A is contained in the y–z plane. (a) Express the vector r in terms of components. (b) What are the direction cosines of r ?

Solution:

The vector r is

r = r (sin 45 °i + cos 45 ° sin 60 ° j + cos 45 ° cos60 °k) The length of the pole is the x component of r. Therefore r sin 45 ° = 6 ft ⇒ r =

(a) A

6 ft = 8.49 ft sin 45 °

r = (6.00 i + 5.20 j + 3.00 k) ft

(b) The direction cosines are

Bedford Falls

cos θ x =

cos θ x = 0.707, cos θ y = 0.612, cos θ z = 0.354

r 608

ry r rx = 0.707, cos θ y = = 0.612, cos θ z = z = 0.354 r r r

458 O

Problem 2.160 The z component of the force F is 80 lb. (a) Express F in terms of components. (b) What are the angles θ x , θ y , and θ z between F and the positive coordinate axes? Solution:

We can write the force as

We know that the z component is 80 lb. Therefore

608 A

F cos 20 ° cos60 ° = 80 lb ⇒ F = 170 lb (a)

F = (139 i + 58.2 j + 80 k) lb

(b) The direction cosines can be found:

208

F = F (cos 20 ° sin 60 °i + sin 20 ° j + cos 20 ° cos60 °k)

139 ( 170 ) = 35.5° 58.2 θ = cos ( ) = 70.0° 170 80 θ = cos ( ) = 62.0° 170 θ x = cos −1 −1

−1

θ x = 35.5 °, θ y = 70.0 °, θ z = 62.0 °

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Problem 2.161 The magnitude of the force vector FB is 2 kN. Express it in terms of components.

F D

(4, 3, 1) m

Solution:

The strategy is to determine the unit vector collinear with FB and then express the force in terms of this unit vector.

The radius vector collinear with FB is

rBD = (4 − 5) i + (3 − 0) j + (1 − 3)k or rBD = −1i + 3 j − 2k.

The magnitude is rBD =

1 2 + 3 2 + 2 2 = 3.74.

(5, 0, 3) m

The unit vector is r e BD = BD = −0.2673i + 0.8018 j − 0.5345k rBD

x (6, 0, 0) m

D(4,3,1)

The force is FB = FB e BD = 2e BD (kN) FB = −0.5345i + 1.6036 j − 1.0693k

= −0.53i + 1.60 j − 1.07k (kN)

C(6,0,0)

FB B(5,0,3)

Problem 2.162 The magnitude of the vertical force vector F is 6 kN. Determine the vector components of F parallel and normal to the line from B to D.

y D

Solution:

The projection of the force F onto the line from B to D is FP = (F ⋅ e BD )e BD . The vertical force has the component F = −6 j (kN). From Problem 2.139, the unit vector pointing from B to D is e BD = −0.2673i + 0.8018 j − 0.5345k. The dot product is F ⋅ e BD = −4.813. Thus the component parallel to the line BD is FP = −4.813e BD = +1.29 i − 3.86 j + 2.57k (kN). The component perpendicular to the line is: FN = F − FP . Thus FN = −1.29 i − 2.14 j − 2.57k (kN)

Problem 2.163 The magnitude of the vertical force vector F is 6 kN. Given that F + F A + FB + FC = 0, what are the magnitudes of F A , FB , and FC ?

F (4, 3, 1) m FC

z B

r AD = 4 i + 3 j + 1k, r AD =

(5, 0, 3) m

x (6, 0, 0) m

26 = 5.1,

e AD = 0.7845i + 0.5883 j + 0.1961k. rBD = −1i + 3 j − 2k, rBD =

14 = 3.74,

e BD = −0.2673i + 0.8018 j − 0.5345k.

F D

rCD = −2 i + 3 j + 1k, rCD = (4, 3, 1) m

The forces are:

F A = F A e AD , FB = FB e BD , FC = FC e CD , F = −6 j (kN).

14 = 3.74,

e CD = −0.5345i + 0.8018 j + 0.2673k

Substituting into the force balance equation C

z B

(5, 0, 3) m

x (6, 0, 0) m

F + F A + FB + FC = 0, (0.7843 F A − 0.2674 FB − 0.5348 FC ) i = 0 (0.5882 F A + 0.8021 FB + 0.8021 FC − 6) j = 0(0.1961 F A − 0.5348 FB + 0.2674 FC )k = 0

Solution: The strategy is to expand the forces into scalar components, and then use the force balance equation to solve for the unknowns. The unit vectors are used to expand the forces into scalar components. The position vectors, magnitudes, and unit vectors are:

BandF_6e_ISM_CH02.indd 83

These simple simultaneous equations can be solved a standard method (e.g., Gauss elimination) or, conveniently, by using a commercial package, such as TK Solver®, Mathcad®, or other. An HP-28S hand held calculator was used here: F A = 2.83 (kN), FB = 2.49 (kN), FC = 2.91 (kN)

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Problem 2.164 The magnitude of the vertical force W is 160 N. The direction cosines of the position vector from A to B are cos θ x = 0.500, cos θ y = 0.866, and cos θ z = 0, and the direction cosines of the position vector from B to C are cos θ x = 0.707, cos θ y = 0.619, and cos θ z = −0.342. Point G is the midpoint of the line from B to C. Determine the vector r AG × W, where r AG is the position vector from A to G.

Solution:

Express the position vectors in terms of scalar components, calculate r AG , and take the cross product. The position vectors are: rBG = 0.3(0.707 i + 0.619 j − 0.342k), rBG = 0.2121i + 0.1857 j − 0.1026k. r AG = r AB + rBG = 0.5121i + 0.7053 j − 0.1026k. W = −160 j

r AG × W =

= −16.44 i + 0 j − 81.95k = −16.4 i + 0 j − 82k (Nm)

G W

600 mm

C 600 mm

Problem 2.165 The rope CE exerts a 500-N force T on the hinged door. (a) Express T in terms of components. (b) Determine the vector component of T parallel to the line from point A to point B.

(0.2, 0.4, 20.1) m

Solution:

We have

rCE = (0.2 i + 0.2 j − 0.1k) m r T = (500 N) CE = (333i + 333 j − 167k) N rCE (a) T = (333i + 333 j − 167k) N

(b) We define the unit vector in the direction of AB and then use this vector to find the component parallel to AB. r AB = (−0.15i + 0.2k) m r e AB = AB = (−0.6 i + 0.8k) r AB

A (0.5, 0, 0) m

T p = (e AB ⋅ T)e AB = ([−0.6][333 N] + [0.8][−167 N])(−0.6 i + 0.8k) x

T p = (200 i − 267k) N

B (0.35, 0, 0.2) m

G B

C (0, 0.2, 0) m

i j k 0.5121 0.7053 −0.1026 0 −160 0

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Problem 2.166 In Problem 2.165, let rBC be the position vector from point B to point C. Determine the cross product rBC × T. E

C (0, 0.2, 0) m

The vector rBC is

The cross product is

i j k −0.35 0.2 −0.2 333 333 −137

= (33.3i − 125 j − 183k) Nm

rBC × T = (33.3i − 125 j − 183k) Nm x

B (0.35, 0, 0.2) m

BandF_6e_ISM_CH02.indd 85

T = (333i + 333 j − 167k) N

rBC × T =

A (0.5, 0, 0) m

From Problem 2.165 we know that

rBC = (−035i + 0.2 j − 0.2k) m

(0.2, 0.4, 20.1) m

Solution:

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Chapter 3 Problem 3.1 The ring weighs 5 lb and is in equilibrium. The y-axis is vertical. (a) Determine the angle α so that the magnitude of the force F2 is a minimum. (b) If α has the value determined in part (a), what are the values of F1 and F2? Strategy: To do part (a), draw a vector diagram of the sum of the three forces acting on the ring.

Solution: y

308

F1 308

F1 x

(a) From the diagram of the sum of the forces on the ring it is clear that the magnitude of F2 is a minimum when F2 is perpendicular to F1 , which means that α = 60 °. (b) In terms of the coordinate system shown, the equilibrium equations are ΣFx = F1 cos30 ° − F2 cos α = 0,

ΣFy = F1 sin 30 ° + F2 sin α − 5 lb = 0. Solving these equations with α = 60 ° yields F1 = 2.50 lb, F2 = 4.33 lb. (a) α = 60 °. (b) F1 = 2.50 lb, F2 = 4.33 lb.

Problem 3.2 The ring weighs 5 lb and is in equilibrium. The force F1 = 4.5 lb . Determine the force F2 and the angle α.

The free-body diagram is shown below the drawing. The equilibrium equations are ΣFx : F1 cos30 ° − F2 cos α = 0 ΣFy : F1 sin 30 ° + F2 sin α − 5 lb = 0

Solution:

We can write these equations as F2 sin α = 5 lb − F1 sin 30 °

F1 308

F2 cos α = F1 cos30 ° Dividing these equations and using the known value for F1 we have. tan α =

F2 =

5 lb − (4.5 lb)sin 30° = 0.706 ⇒ α = 35.2° (4.5 lb) cos30°

(4.5 lb) cos30° = 4.77 lb cos α

F2 = 4.77 lb, α = 35.2°

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Problem 3.3 The forces acting on the airplane are its weight W = 28,000 lb, thrust T = 16,200 lb, lift L, and drag D. Assume that the thrust and drag are parallel, and the lift is perpendicular to the drag. The airplane is in equilibrium. Determine the lift and drag. y

Solution:

In terms of the coordinate system shown, the equilibrium equations are ΣFx = T cos30 ° − L sin 30 ° − D cos30 ° = 0, ΣFy = T sin 30 ° + L cos30 ° − D sin 30 ° − W = 0. Setting T = 16,200 lb and W = 28,000 lb and solving yields L = 24,200 lb, D = 2200 lb.

308

D W x

Problem 3.4 The 200-kg engine block is suspended by the cables AB and AC. The angle α = 40 °. The freebody diagram obtained by isolating the part of the system within the dashed line is shown. Determine the forces T AB and T AC .

Solution: α = 40 ° ΣFx : T AC cos α − T AB cos α = 0 ΣFy : T AC sin α + T AB sin α − 1962 N = 0 Solving:

y B

TAB

TAC

T AB = T AC = 1.526 kN

TAB

TAC

C A

1962 N (200 kg) (9.81 m/s2)

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Problem 3.5 A heavy rope used as a mooring line for a cruise ship sags as shown. If the mass of the rope is 90 kg, what are the tensions in the rope at A and B? 558

Solution: The free-body diagram is shown. The equilibrium equations are ΣFx : TB cos 40 ° − T A cos 55 ° = 0 ΣFy : TB sin 40 ° + T A sin 55 ° − 90(9.81) N = 0

Solving: T A = 679 N, TB = 508 N 408

Problem 3.6 A 110-lb dancer is balanced on one foot. Neglecting the weight of the foot, determine the tension FA in the Achilles tendon and the reaction FJ at the ankle joint. FJ

788

Solution:

The equilibrium equations are

ΣFx = FJ cos 78 ° − FA cos 52 ° = 0, ΣFy = −FJ sin 78 ° + FA sin 52 ° + N = 0. Solving with N = 110 lb yields

FA = 52.2 lb, FJ = 154 lb.

528

N x

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Problem 3.7 The two springs are identical, with unstretched lengths 250 mm and spring constants k = 1200 N/m. (a) Draw the free-body diagram of block A. (b) Draw the free-body diagram of block B. (c) What are the masses of the two blocks? Solution:

The tension in the upper spring acts on block A in the positive Y direction. Solve the spring force-deflection equation for the tension in the upper spring. Apply the equilibrium conditions to block A. Repeat the steps for block B. N TUA = 0 i + 1200 (0.3 m − 0.25 m) j = 0 i + 60 j N m Similarly, the tension in the lower spring acts on block A in the negative Y direction N TLA = 0 i − 1200 (0.28 m − 0.25 m) j = 0 i − 36 j N m

(

)

(

)

300 mm

280 mm

The weight is W A = 0 i − W A j The equilibrium conditions are ΣF = ΣFx + ΣFy = 0, ΣF = W A + TUA + TLA = 0 Collect and combine like terms in i, j ΣFy = (− W A + 60 − 36) j = 0

300 mm

Solve

W A = (60 − 36) = 24 N The mass of A is mA =

WL 24 N = = 2.45 kg g 9.81 m/s 2

280 mm

The free body diagram for block B is shown. The tension in the lower spring TLB = 0 i + 36 j The weight: WB = 0 i − WB j

Tension, upper spring

Apply the equilibrium conditions to block B. ΣF = WB + TLB = 0 Collect and combine like terms in i, j: ΣFy = (− WB + 36) j = 0

Tension, lower spring

Solve: WB = 36 N The mass of B is given by m B =

A Weight, mass A

WB 36 N = = 3.67 kg g 9.81 m/s 2

Tension, lower spring

y B x

BandF_6e_ISM_CH03.indd 89

Weight, mass B

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Problem 3.8 By analyzing the forces on the patella, or kneecap, the compressive force on the human patellofemoral joint can be determined. In the figure, T is the tension in the patellar and quadriceps tendons. The weight of the patella is negligible. If T = 120 N, determine the magnitude P of the compressive force and the angle α. Patella

Solution:

The equilibrium equations are

ΣFx = T cos 70 ° − T cos10 ° + P cos α = 0, ΣFy = −T sin 70 ° + T sin10 ° + P sin α = 0. Arranging these equations as P cos α = T cos10 ° − T cos 70 °,

(1)

P sin α = T sin 70 ° − T sin10 °

(2)

and dividing Eq. (2) by Eq. (1) yields tan α =

sin 70 ° − sin10 ° . cos10 ° − cos 70 °

From this equation we obtain α = 50 °. (You can see from the symmetry of the forces that the angles between the compressive force P and the two tensions must be equal.) We can now use either Eq. (1) or Eq. (2) with T = 120 N to obtain P. The result is P = 120 N.

y T

P = 120 N, α = 50 °.

108 708 P T

Problem 3.9 The inclined surface is smooth. (Remember that “smooth” means that friction is negligible.) The two springs are identical, with unstretched lengths of 250 mm and spring constants k = 1200 N/m. What are the masses of blocks A and B?

Solution: F1 = (1200 N/m)(0.3 − 0.25)m = 60 N F2 = (1200 N/m)(0.28 − 0.25)m = 36 N ΣFB : −F2 + m B g sin 30 ° = 0 ΣFA : −F1 + F2 + m A g sin 30 ° = 0

300 mm

Solving: m A = 4.89 kg, m B = 7.34 kg mA g

A 280 mm

F1 B

F2 mB g

F2 308

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Problem 3.10 The forces acting on a person’s head when it is tilted downward are shown. The y-axis is vertical. The mass of the head is 5 kg. The force FA is exerted by the atlantooccipital joint. The force FE is exerted by the neck extensor muscles. The head is in equilibrium. Determine FA and FE . Solution:

FE 308 FA

The equilibrium equations are mg

ΣFx = FE cos30 ° − FA cos60 ° = 0, ΣFy = −FE sin 30 ° + FA

sin 60 ° − (5 kg)(9.81 m/s 2 ) =

608 x

Solving yields FA = 85.0 N, FE = 49.1 N.

Problem 3.11 The inclined surface is smooth. The 100-kg crate is held stationary by a force T applied to the cable. (a) Draw the free-body diagram of the crate. (b) Determine the force T.

Solution: (a) The FBD T 981 Ν

Ν 608

608

(b) ΣF : T − 981 N sin 60 ° = 0 T = 850 N

Problem 3.12 The 1200-kg car is stationary on the sloping road. (a) If α = 20 °, what are the magnitudes of the total normal and friction forces exerted on the car’s tires by the road? (b) The car can remain stationary only if the total friction force necessary for equilibrium is not greater than 0.6 times the total normal force. What is the largest angle α for which the car can remain stationary?

Solution: (a) α = 20 ° ΣF : N − 11.772 kN cos α = 0 ΣF : F − 11.772 kN sin α = 0 N = 11.06 kN, F = 4.03 kN (b) F = 0.6 N ΣF : N − 11.772 kN cos α = 0 ⇒ α = 31.0 ° ΣF : F − 11.772 kN sin α = 0

11.772 kN

F N

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Problem 3.13 The bar BC exerts a force on the cables at B that is directed along the line from C to B. The mass of the suspended load is 180 kg. What is the tension in cable AB?

Solution:

We draw the free-body diagram of the top end of the bar, letting T be the tension in the cable and F the force exerted by the bar: y B

308 T

308 308 A

308

C The equilibrium equations are ΣFx = −T cos30 ° + F sin 30 ° = 0, ΣFy = −T sin 30 ° + F cos30 ° − mg = 0.

Solving, we obtain T = mg = (180 kg)(9.81 m/s 2 ) = 1770 N. 1.77 kN.

Problem 3.14 The 600-lb box is held in place on the smooth bed of the dump truck by the rope AB. (a) If α = 25 °, what is the tension in the rope? (b) If the rope will safely support a tension of 400 lb, what is the maximum allowable value of α?

B A a

Solution:

Isolate the box. Resolve the forces into scalar components, and solve the equilibrium equations.

The external forces are the weight, the tension in the rope, and the normal force exerted by the surface. The angle between the x axis and the weight vector is −(90 − α) (or 270 + α). The weight vector is W = W ( i sin α − j cos α) = (600)( i sin α − j cos α) The projections of the rope tension and the normal force are T = − Tx i + 0 j

N = 0i + N y j

A B

The equilibrium conditions are a

ΣF = W + N + T = 0 Substitute, and collect like terms ΣFx = (600 sin α − Tx ) i = 0 ΣFy = (−600 cos α + N y ) j = 0 Solve for the unknown tension when For α = 25 ° Tx = 600 sin α = 253.6 lb.

y T

For a tension of 400 lb, (600 sin α − 400) = 0. Solve for the unknown angle sin α =

400 = 0.667 or α = 41.84 ° 600

x N W

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Problem 3.15 The 80-lb box is held in place on the smooth inclined surface by the rope AB. Determine the tension in the rope and the normal force exerted on the box by the inclined surface. A

Solution: The equilibrium equations (in terms of a coordinate system with the x axis parallel to the inclined surface) are ΣFx : (80 lb)sin 50 ° − T cos 50 = 0 ΣFx : N − (80 lb) cos 50 ° − T sin 50 = 0 Solving: T = 95.34 lb, N = 124 lb

308

508

Problem 3.16 The wire used in segments AB and BC will safely support a tension no greater than 1.2 kN. Based on this criterion, what is the largest mass m that can be supported in this way? What are the resulting tensions in segments AB and BC ?

Solution:

We draw the free-body diagram of joint B:

TC TA

208

458

208

158

B The equilibrium equations are

158

ΣFx = −T A cos 20 ° + TC cos 45 ° + mg sin15 ° = 0, ΣFy = T A sin 20 ° + TC sin 45 ° − mg cos15 ° = 0. Solving for T A and TC , we obtain T A = 0.996 mg, TC = 0.904 mg. The largest tension is in wire AB. Setting T A = 0.996 m(9.81 m/s 2 ) = 1200 N, the largest mass is m = 128 kg. This gives TC = 1140 N. m

BandF_6e_ISM_CH03.indd 93

m = 128 kg. AB: 1.20 kN, BC: 1.14 kN.

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Problem 3.17 The 50-lb box is held in equilibrium on the smooth inclined surface by the string AB. What is the tension in the string? What is the magnitude of the normal force exerted on the box by the smooth surface?

Solution: T

5 in a

A 8 in

x 8 in The angle α in the diagram is 20 in

( 208 inin ) = 21.8°.

α = arctan

The angle γ is γ = arctan

( 85 inin ) = 58.0°,

and the angle β is β = γ − α = 36.2 °. The equilibrium equations are ΣFx = W sin α − T cos β = 0, ΣFy = −W cos α + T sin β + N = 0. From the first equation, W sin α cos β (50 lb)sin 21.8 ° = cos 36.2 ° = 23.0 lb.

T =

Then from the second equilibrium equation, the normal force is N = W cos α − T sin β = (50 lb) cos 21.8 ° − (23.0 lb)sin 36.2 ° = 32.8 lb. Tension = 23.0 lb, normal force = 32.8 lb.

Problem 3.18 A 10-kg painting is hung with a wire supported by a nail. The length of the wire is 1.3 m. (a) What is the tension in the wire? (b) What is the magnitude of the force exerted on the nail by the wire?

Solution: (a) ΣFy : 98.1 N − 2

5 T = 0 13

T = 128 N (b) Force = 98.1 N

98.1 N 1.2 m

T 12

5 12

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Problem 3.19 The force exerted on the kite surfer by her kite is 55 i + 32 j (lb). (The y-axis is vertical.) Her weight including equipment is 145 lb. If she is in equilibrium, what is the magnitude of the force exerted on her by the water?

Solution:

Let F = Fx i + Fy j denote the force exerted on him by the water. The equilibrium equations are

ΣFx = Fx + 32 lb = 0, ΣFy = Fy + 55 lb − 185 lb = 0. We see that

Fx = −32 lb, Fy = 130 lb.

The magnitude is

(−32 lb) 2 + (130 lb) 2 = 134 lb. y

134 lb.

Source: Courtesy of Arterra Picture Library/ Clement Philippe/Alamy Stock Photo.

Problem 3.20 The 600-lb box is held in place on the smooth bed of the truck by the rope AB. (a) If α = 30 , what is the tension in the rope? (b) The rope will safely support a tension no greater than 400 lb. Based on this criterion, what is the largest allowable value of α?

Solution: TAB 30° y W

B N A

308 a

(a) Let T AB be the tension in the rope and W the weight of the box. The sum of the forces on the box in the direction parallel to the bed is T AB cos30 ° − W sin α = 0. Solving for T AB , T AB =

W sin α . cos30 °

(1)

Setting α = 30 ° and W = 600 lb yields T AB = 346 lb. (b) Setting T AB equal to 400 lb and W = 600 lb in Eq. (1) and solving for α yields α = 35.3 °. (a) Tension = 346 lb. (b) α = 35.3 °.

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Problem 3.21 Assume that the 80-kg climber is in equilibrium. The angles of the segments A and B of his rope relative to the horizontal are shown. What are the tensions in the two segments?

Solution:  ΣFx = T R cos(15 °) − T L cos(14°) = 0   ΣFy = T R sin (15°) + T L sin (14°) − mg = 0  Solving, we get T L = 1.56 kN, T R = 1.57 kN y

TL 148

158 x

408

mg 5 (80) (9.81) N

A B 108

Source: Courtesy of Design Pics Inc/Tom Evans/ Alamy Stock Photo.

Problem 3.22 The mass of each suspended object is 12 kg. What are the tensions in cables AB and BD? A

Solution:

The weight of each suspended object is mg = (12 kg)(9.81 m/s 2 ) = 118 N. Consider the free-body diagram of point B: y

TAB

458 D 458

458

TBD

The equilibrium equations are ΣFx = TBD sin 45 ° − T AB sin 45 ° = 0, ΣFy = TBD cos 45 ° + T AB cos 45 ° − mg = 0. Solving, we obtain T AB = TBD = 83.2 N. AB: 83.2 N, BD: 83.2 N.

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Problem 3.23 The mass of each suspended object is 12 kg. What are the tensions in cables CD and DE ? A

C 458 D 458

458

Solution:

The weight of each suspended object is mg = (12 kg)(9.81 m/s 2 ) = 118 N. Before we can analyze the cables that join at D, we need to determine the tension in cable BD. Consider the free-body diagram of point B: y TAB

TBD

458 458

x The equilibrium equations are ΣFx = TBD sin 45 ° − T AB sin 45 ° = 0, ΣFy = TBD cos 45 ° + T AB cos 45 ° − mg = 0. Solving, we obtain T AB = TBD = 83.2 N. Now consider the free-body diagram of D: y TCD

458

D TDE

458 TBD

x The equilibrium equations are ΣFx = T DE − TCD cos 45 ° − (83.2 N) cos 45 ° = 0, ΣFy = TCD sin 45 ° − (83.2 N)sin 45 ° − mg = 0. Solving yields TCD = 250 N, T DE = 235 N. CD: 250 N, DE: 235 N.

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Problem 3.24 The person wants to cause the 200-lb crate to start sliding toward the right. To achieve this, the horizontal component of the force exerted on the crate by the rope must equal 0.35 times the normal force exerted on the crate by the floor. In Fig. a, the person pulls on the rope in the direction shown. In Fig. b, the person attaches the rope to a support as shown and pulls upward on the rope. What is the magnitude of the force the person must exert on the rope in each case?

Solution:

The friction force F fr is given by

F fr = 0.35 N (a) For equilibrium we have ΣFx : T cos 20 ° − 0.35 N = 0 ΣFy : T sin 20 ° − 200 lb + N = 0 Solving: T = 66.1 lb (b) The person exerts the force F. Using the free-body diagram of the crate and of the point on the rope where the person grabs the rope, we find ΣFx : T L − 0.35 N = 0 ΣFy : N − 200 lb = 0 ΣFx : −T L + T R cos10 ° = 0

208

ΣFy : F − T R sin10 ° = 0 Solving we find F = 12.3 lb

(a)

108

(b)

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Problem 3.25 The boxes are at rest on the smooth surfaces. The rope connecting them is horizontal. The mass of box A is m A = 20 kg. What is the mass of box B? What is the tension in the rope?

Solution:

We draw the free-body diagrams of the individual

boxes: y

T B

T mB g

mA g

308

408

x The equilibrium equations for box A are ΣFx = T − N A sin 40 ° = 0, ΣFy = −m A g + N A cos 40 ° = 0. The equilibrium equations for box B are ΣFx = −T + N B sin 30 ° = 0, ΣFy = −m B g + N B cos30 ° = 0. Solving these four equations with m A = 20 kg, we obtain T = 165 N, m B = 29.1 kg, N A = 256 N, N B = 329 N. m B = 29.1 kg, tension = 165 N.

Problem 3.26 The mass of the suspended object is 300 kg. Points A and C are at the same height. Cable AB is 2 m long and cable BC is 4 m long. Determine the tensions in cables AB and BC. Strategy: Use relations from Appendix A to analyze the geometry.

Solution: y Tab

ac ab

Tbc

5m A

C Let ab = 2 m, bc = 4 m, ac = 5 m. We can use the law of cosines (Appendix A) to determine the two angles, B

ab 2 = bc 2 + ac 2 − 2bc ac cos θ c , bc 2 = ab 2 + ac 2 − 2ab ac cos θ a , obtaining θ a = 49.5 °, θ c = 22.3 °. The equilibrium equations for the joint where the cables join are ΣFx = −Tab cos θ a + Tbc cos θ c = 0, ΣFx = Tab sin θ a + Tbc sin θ c − mg = 0. Solving them, we obtain Tab = 2.87 kN, Tbc = 2.01 kN. AB: 2.87 kN. BC: 2.01 kN.

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Problem 3.27* The mass of the suspended object is 300 kg. Points A and C are at the same height. Cable BC is 4 m long and the length of cable AB is adjustable. If you don’t want the tension in either cable to exceed 2.4 kN, what is the minimum acceptable length of cable AB?

Solution: y Tab

ac ab

Tbc

mg A

Let ab denote the length of cable AB, and let bc = 4 m, ac = 5 m. For a given value of ab, the law of cosines (Appendix A) can be used to determine the two angles: ab 2 = bc 2 + ac 2 − 2bc ac cos θ c , bc 2 = ab 2 + ac 2 − 2ab ac cos θ a , Then the equilibrium equations for the joint where the cables join are ΣFx = −Tab cos θ a + Tbc cos θ c = 0, ΣFx = Tab sin θ a + Tbc sin θ c − mg = 0. The graph shows the values of Tab and Tbc as a function of ab. From it, we estimate that the minimum length of cable AB is 2.9 m. 3 2.8

Tensions, KN

2.6 AB

2.4 2.2 2

1.8 1.6

2.2

2.4

2.6

2.8

3.2

3.4

3.6

3.8

Length of cable AB, meters

2.9 m.

100

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Problem 3.28 The length of the cable segment AB is 10 ft. The angle α = 30 °. What force T is necessary to support the 600-lb crate?

Solution:

From the geometry,

308

10 ft A

10 ft

b 10 ft

T B

the angle β is  10 ft − (10 ft) cos30 °  β = arctan   = 15 °.   (10 ft)sin 30 ° Let T AB be the tension in cable AB. From the free-body diagram, y TAB b

308 600 lb x the equilibrium equations are ΣFx = T sin β − T AB cos30 ° = 0, ΣFy = T cos β + T AB sin 30 ° − 600 lb = 0. Solving, we obtain T = 538 lb and T AB = 161 lb. T = 538 lb.

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Problem 3.29* The length of the cable segment AB is 10 ft. If the 600-lb crate is in equilibrium and T = 500 lb, what is the angle α?

the equilibrium equations are ΣFx = T sin β − T AB cos α = 0, ΣFy = T cos β + T AB sin α − 600 lb = 0. T =

10 ft

600 lb . (cos β + tan α sin β )

Eliminating T AB , we obtain

a T

T =

600 lb . (cos β + tan α sin β )

(2)

If we choose a value of α, we can determine β from Eq. (1) and then determine T from Eq. (2). By drawing a graph of T as a function of α, we can estimate the value of α for which T = 500 lb. From the graph,

540

Solution:

From the geometry,

530

10 ft

520

T, pounds

b 10 ft B the angle β is

510

500

 10 ft − (10 ft) cos α  β = arctan  .   (10 ft)sin α

(1) 490

Let T AB be the tension in cable AB. From the free-body diagram, y TAB b

480

37 38 a, degrees

We estimate that T = 500 lb at α = 38 °. Using software designed to solve nonlinear algebraic equations, we obtain α = 38.0094 °.

600 lb x

102

α = 38.0 °.

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Problem 3.30 An astronaut candidate conducts experiments on an airbearing platform. While she carries out calibrations, the platform is held in place by the horizontal tethers AB, AC , and AD. The forces exerted by the tethers are the only horizontal forces acting on the platform. If the tension in tether AC is 2 N, what are the tensions in the other two tethers?

TOP VIEW D

4.0 m

A 3.5 m B

C 3.0 m

Solution: tan α =

Isolate the platform. The angles α and β are

1.5 m

( 1.5 ) = 0.429, α = 23.2°. 3.5

3.0 m

1.5 m

Also, tan β =

C 3.5 m

( 3.0 ) = 0.857, β = 40.6°. 3.5

4.0 m

The angle between the tether AB and the positive x axis is (180 ° − β ), hence y

T AB = T AB ( i cos(180 ° − β ) + j sin (180 ° − β ))

T AB = T AB (−i cos β + j sin β ).

The angle between the tether AC and the positive x axis is (180 ° + α). The tension is

T AC = T AC ( i cos(180 ° + α) + j sin (180 ° + α)) = T AC (−i cos α − j sin α).

The tether AD is aligned with the positive x axis, T AD = T AD i + 0 j. The equilibrium condition:

Solve:

ΣF = T AD + T AB + T AC = 0.

 sin α  T AB =  T ,  sin β  AC

Substitute and collect like terms, ΣFx = (− T AB cos β − T AC cos α + T AD ) i = 0, ΣFy = ( T AB sin β − T AC sin α) j = 0.

 T sin(α + β )  T AD =  AC .  sin β For T AC = 2 N, α = 23.2 ° and β = 40.6 °, T AB = 1.21 N,

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T AD = 2.76 N

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Problem 3.31 The bucket contains concrete and weighs 5800 lb. What are the tensions in the cables AB and AC ? (5, 34) ft y

(20, 34) ft

Solution:

The angles are

α = tan −1

( 3412 −− 165 ) = 68.7°

β = tan −1

− 16 ( 34 ) = 66.0° 20 − 12

Now from equilibrium we have ΣFx : −T AB cos α + T AC cos β = 0 ΣFy : T AB sin α + T AC sin β − 660 lb = 0 Solving yields T AB = 319 lb, T AC = 470 lb

(12, 16) ft

Problem 3.32 The 30-lb slider A is held in place on the smooth inclined bar by the wire. What is the tension in the wire?

Solution:

The forces on the free-body diagram of the slider are its weight W, the normal force N exerted by the smooth bar, and the force T exerted by the wire. The sum of the forces in the x direction is ΣFx = T cos 40 ° − W cos30 ° = 0. Solving for T, we obtain W cos30 ° cos 40 ° (30 lb) cos30 ° = cos 40 ° = 33.9 lb.

T = A

208

33.9 lb. 608

104

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Problem 3.33 This device is designed to exert a large force on the stationary object at the left. The jack exerts a 400-lb upward force at B. As a result, the link BC exerts a force on the fixed support at C that is directed along the line from B to C, and the link AB exerts a force on the object at A that is directed along the line from B to A. What is the magnitude of the force exerted by link AB on the object at A? What is the magnitude of its horizontal component? C

A B 108

Solution:

From the free-body diagram of the joint B, y FAB

FBC

108

208 400 lb x

the equilibrium equations are ΣFx = FAB cos10 ° − FBC cos 20 ° = 0,

208

ΣFy = −FAB sin10 ° − FBC sin 20 ° + 400 lb = 0. Solving yields FAB = 752 lb, FBC = 788 lb. The horizontal component of FAB is FAB cos10 ° = 740 lb. Force = 752 lb, horizontal component = 740 lb.

Problem 3.34 The structural joint is in equilibrium. If FA = 1000 lb and FD = 5000 lb, what are FB and FC ? FC 808 FB 658

358

Solution:

In terms of the coordinate system shown, the equilibrium equations are ΣFx = −FA − FB cos35 ° − FC cos65 ° + FD = 0,

y FC

ΣFy = FB sin 35 ° − FC sin 65 = 0. Solving with FA = 1000 lb, FD = 5000 lb, we obtain FB = 3680 lb, FC = 2330 lb.

FB = 3680 lb, FC = 2330 lb. FA

358

658 FD

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Problem 3.35 The collar A slides on the smooth vertical bar. The masses m A = 20 kg and m B = 10 kg. When h = 0.1 m, the spring is unstretched. When the system is in equilibrium, h = 0.3 m. Determine the spring constant k.

Solution:

When the spring is unstretched, the length of the rope from

the collar A to the mass B is L 0 = (0.25 m) 2 + (0.1 m) 2 = 0.269 m. When the system is in equilibrium, the length of the rope from the collar A to the mass B is L = (0.25 m) 2 + (0.3 m) 2 = 0.391 m. This tells us that the force exerted on the mass B by the spring when the system is in equilibrium is F = k(L − L 0 )

0.25 m

= k (0.121 m). Consider the free-body diagrams of the collar and the mass:

T T

A N

mA g

mB g

F x

Here the tension in the rope is denoted by T. When the system is in equilibrium, the angle α = arctan(0.3/0.25) = 50.2 °. The equilibrium equation in the vertical direction for the collar A is T sin α − m A g = 0: T sin 50.2 ° − (20 kg)(9.81 m/s 2 ) = 0. The equilibrium equation in the vertical direction for the mass B is T − m B g − F = 0: T − (10 kg)(9.81 m/s 2 ) − k (0.121 m) = 0. The two equilibrium equations can be solved for T and k. We obtain T = 255 N, k = 1300 N/m. k = 1300 N/m.

106

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Problem 3.36* The collar A slides on the smooth vertical bar. The masses m A = 16 kg and m B = 20 kg. When h = 0.1 m, the spring is unstretched. If the spring constant is k = 300 N/m, what is the distance h at which the system is in equilibrium?

Solution:

When the spring is unstretched, the length of the rope from the collar A to the mass B is L 0 = (0.25 m) 2 + (0.1 m) 2 = 0.269 m. The length of the rope from the collar A to the mass B is L = (0.25 m) 2 + h 2 . The force exerted on the mass B by the spring is F = k ( L − L 0 ).

(1)

Consider the free-body diagrams of the collar and the mass: 0.25 m T T h

mBg

A mA g

F x

Here the tension in the rope is denoted by T. The angle α = arctan(h /0.25 m). The equilibrium equation in the vertical direction for the collar A is T sin α − m A g = 0.

(2)

The equilibrium equation in the vertical direction for the mass B is T − m B g − F = 0.

(3)

If we choose a value of h, we can determine L and the angle α. We can then determine the spring force F from Eq. (1) and the tension T from Eq. (2). By doing this for a range of values of h, we can identify the value of h for which the equilibrium Eq. (3) is satisfied. That is, we look for the value of h for which the function f (h) = T − m B g − F is zero. The resulting graph is show below. 250 200

f(h), newtons

150 100 50 0 –50 –100

0.1

0.15

0.2

0.25 0.3 h, meters

0.35

0.4

0.45

We estimate that f (h) = 0 at h = 0.25 m. By using software designed to solve nonlinear algebraic equations, we obtain h = 0.2507 m. h = 0.25 m.

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Problem 3.37 The system of cables suspends a 1000-lb bank of lights above a movie set. Determine the tensions in cables AB, CD, and CE. 20 ft

Isolate juncture C. The angle between the positive x axis and the cable CA is (180 ° − α). The tension is TCA = TCA ( i cos(180 ° + α) + j sin (180 ° + α)), or TCA = TCA (−i cos α − j sin α).

18 ft

The tension in the cable CE is B

TCE = i TCE + 0 j.

C E

458

308 A

The tension in the cable CD is TCD = 0 i + j TCD . The equilibrium conditions are ΣF = 0 = TCA + TCE + TCD = 0 Substitute t and collect like terms, ΣFx = ( TCE − TCA cos α) i = 0, ΣFy = ( TCD − TCA sin α) j = 0. Solve: TCE = TCA cos α,

Solution:

Isolate juncture A, and solve the equilibrium equations. Repeat for the cable juncture C.

TCD = TCA sin α; for TCA = 732 lb and α = 30 °,

The angle between the cable AC and the positive x-axis is α. The tension in AC is T AC = T AC ( i cos α + j sin α)

T AB = 896.6 lb,

The angle between the x axis and AB is (180 ° − β ). The tension is

TCD = 366 lb

T AB = T AB ( i cos(180 − β ) + j sin (180 − β ))

TCE = 634 lb, B

T AB = (−i cos β + j sin β ).

C A

The weight is W = 0 i − W j.

a y

The equilibrium conditions are

ΣF = 0 = W + T AB + T AC = 0. Substitute and collect like terms, ΣFx = ( T AC cos α − T AB cos β ) i = 0

ΣFy = ( T AB sin β + T AC sin α − W ) j = 0. Solving, we get  W cos β   cos α  T AB =  T , and T AC =   sin (α + β )   cos β  AC

108

(

)

T AC T AB

0.866 = (732) = 896.5 lb 0.7071

(

E y

W = 1000 lb, and α = 30 °, β = 45 ° 0.7071 = (1000) = 732.05 lb 0.9659

908

a x

)

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Problem 3.38 The unstretched length of the spring AB is 12 in and the spring constant is k = 50 lb/in. The cable BC is 10 in long. If the system is in equilibrium when the angle α = 25 °, what is the weight W of the suspended object?

22 in

a B

Solution:

22 in A

D a

From the figure showing the geometry, the distance BD is BD = BC sin α = (10 in)sin 25 ° = 4.23 in.

From the free-body diagram, the equilibrium equations are

The distance AD is

ΣFy = T sin α + F sin β − W = 0.

AD = 22 in − BC cos α = 22 in − (10 in) cos 25 ° = 12.9 in.

Substituting the known information, these equations are

The angleβ between the spring and the horizontal is

( BD ) AD 4.23 in = arctan ( ) 12.9 in

β = arctan

ΣFx = T cos α − F cos β = 0,

T cos 25 ° − (80.5 lb) cos18.1 ° = 0, T sin 25 ° + (80.5 lb)sin18.1 ° − W = 0. Solving, we obtain T = 84.4 lb, W = 60.7 lb. W = 60.7 lb.

= 18.1 °. The length of the spring is L =

AD 2 + BD 2

= (12.9 in) 2 + (4.23 in) 2 = 13.6 in. The force exerted by the spring is F = k(L − L 0 ) = (50 lb/in)(13.6 in − 12 in) = 80.5 lb.

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Problem 3.39 Each ring weighs 5 lb. (The positive y-axis points upward.) If the tension in rope B is 10 lb, what is the tension in rope C?

Solution:

Consider the free-body diagram of the left ring:

5 lb

308

208

708

x 10 lb

208

308 408

The equilibrium equations are ΣFx = −T A cos 20 ° + TC cos30° = 0,

ΣFy = T A sin 20 ° + TC sin 30 ° − 5 lb − 10 lb = 0. x

Solving, we obtain T A = 17.0 lb, TC = 18.4 lb. TC = 18.4 lb.

Problem 3.40 Each ring weighs 5 lb. (The positive y-axis points upward.) If the tension in rope B is 10 lb, what is the tension in rope D?

Solution:

Consider the free-body diagram of the left ring: y

5 lb

308

208

708 10 lb

208

308 408

The equilibrium equations are ΣFx = −T A cos 20 ° + TC cos30° = 0, ΣFy = T A sin 20 ° + TC sin 30 ° − 5 lb − 10 lb = 0.

B x

Solving, we obtain T A = 17.0 lb, TC = 18.4 lb. Now consider the free-body diagram of the right ring: y TE 708 x 308 TC

5 lb

408 TD

The equilibrium equations are ΣFx = −TC cos30 ° + T D cos 40 ° + TE cos 70 ° = 0, ΣFy = −TC sin 30 ° − T D sin 40 ° + TE sin 70 ° − 5 lb = 0. Solving these equations with TC = 18.4 lb, we obtain T D = 10.8 lb, TE = 22.5 lb. T D = 10.8 lb.

110

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Problem 3.41 The unstretched length of the spring is 0.25 m, and the spring constant is k = 430 N/m. What is the mass of the suspended object?

Solution:

The length of the spring in the position shown is L = (0.16 m) 2 + (0.32 m) 2 = 0.358 m, so the tension in the spring is T AB = k(0.358 m − 0.25 m)

0.32 m

= (430 N/m)(0.358 m − 0.25 m) = 46.3 N.

0.24 m

Consider the free-body diagram of the suspended object:

0.16 m k

TAD

TAB

x mg

The angles α = arctan(0.16/0.24) = 33.7 °, β = arctan(0.16/0.32) = 26.6 °. The equilibrium equations are ΣFx = −T AB cos β + TBC cos α = 0, ΣFy = T AB sin β + TBC sin α − mg = 0. Solving yields TBC = 49.8 N, mg = 48.4 N. The suspended mass is mg 48.4 N = g 9.81 m/s 2 = 4.93 kg.

m =

4.93 kg.

Problem 3.42 The unstretched length of the spring is 12 in, and the spring constant is k = 3 lb/in. The suspended weight W = 50 lb. What are the tensions in cables AB and AD?

The free-body diagram of the box is TS

TAB

TAD b

12 in

10 in

14 in

y C

6 in B

6 in 50 lb

D 10 in The equilibrium equations are

A W

Solution:

The length of the spring in the position shown is

L = (10 in) 2 + (16 in) 2 = 18.9 in.

ΣFx = −T AB cos α + TS cos β + T AD cos γ = 0, ΣFy = T AB sin α + TS sin β + T AD sin γ − 50 lb = 0. Solving these equations with T AB = 38.6 lb, T AD = 20.3 lb.

TS = 20.6 lb,

obtain

T AB = 38.6 lb, T AD = 20.3 lb.

The tension in the spring is TS = k ( L − L 0 ) = (3 lb/in)(18.9 in − 12 in) = 20.6 lb.

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Problem 3.43* The length of the cable ABC is 1.4 m. The 2-kN force is applied to a small pulley. The system is stationary. What is the tension in the cable?

C 0.75 m

0.25 m

a h

0.75 m

158

Solution:

tan α =

2 kN T

Examine the geometry

h 2 + (0.75 m) 2 +

h 2 + (0.25 m) 2 = 1.4 m

h h , tan β = 0.75 m 0.25 m

⇒ h = 0.458 m, α = 31.39 °, β = 61.35 ° Now draw a FBD and solve for the tension. We can use either of the equilibrium equations ΣFx : −T cos α + T cos β + (2 kN)sin15 ° = 0

2 kN

T = 1.38 kN

158

Problem 3.44 The masses m1 = 12 kg and m 2 = 6 kg are suspended by the cable system shown. The cable BC is horizontal. Determine the angle α and the tensions in the cables AB, BC , and CD.

From the equilibrium equations ΣFx = TCD cos 70 ° − TBC = 0, ΣFy = TCD sin 70° − m 2 g = 0, we obtain TBC = 21.4 N, TCD = 62.6 N. Next we consider joint B: y

TAB

a a

B TBC

B C

708

m1 g

Solution:

The equilibrium equations are

First we consider joint C:

ΣFx = TBC − T AB cos α = 0,

ΣFy = T AB sin α − m1g = 0.

TCD

Writing these equations as C

708

TBC

m2 g

112

T AB cos α = TBC ,

(*)

T AB sin α = m1g and dividing the second equation by the first one, we obtain tan α = m1g /TBc . Because we now know that TBC = 21.4 N, we can use this equation to obtain α = 79.7 °. Then from Eq. (*) we obtain T AB = 120 N. α = 79.7 °. AB: 120 N. BC: 21.4 N. CD: 62.6 N.

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Problem 3.45 The weights W1 and W2 are suspended from the cable system shown. If W2 = 20 lb, what are the tensions in cables AB, BC , and CD ?

30 in

D 16 in

20 in

Solution:

Next we consider joint B:

First we consider joint C:

y TAB

TCD TBC

TBC x

The angles α = arctan(4/30) = 7.59 °, β = arctan(16/30) = 28.1 °. From the equilibrium equations

The angle γ = arctan(20/30) = 33.7 °. From the equilibrium equations

ΣFx = −TBC cos α + TCD cos β = 0,

ΣFx = −T AB cos γ + TBC cos α = 0,

ΣFy = −TBC sin α + TCD sin β = W2 ,

ΣFy = T AB sin γ + TBC sin α = W1 ,

we obtain TBC = 50.4 lb, TCD = 56.7 lb.

we obtain T AB = 60.1 lb, W1 = 40 lb. AB: 60.1 lb. BC: 50.4 lb. CD: 56.7 lb.

Problem 3.46 Assume that W2 = W1 /2. If you don’t want the tension anywhere in the supporting cable to exceed 200 lb, what is the largest acceptable value of W1 ?

TCD TBC

D 16 in

20 in

W2 W2

Solution:

30 in

The angles γ = arctan(20/30) = 33.7 °, α = arctan(4/30) = 7.59 °, β = arctan(16/30) = 28.1 °. From the two free-body diagrams, we have the equilibrium equations

Consider the free-body diagrams of the weights: y

(1)

ΣFx = TCD cos β − TBC cos α = 0.

TAB g

ΣFx = TBC cos α − T AB cos γ = 0,

TBC x

From these equations we obtain the relations T AB = 1.19TBC , TCD = 1.12TBC, which show that the largest tension occurs in cable AB. Setting T AB = 200 lb, from Eq. (1) and the equilibrium equation ΣFy = T AB sin γ + TBC sin α − W1 = 0, we obtain TBC = 168 lb, W1 = 133 lb. W1 = 133 lb.

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Problem 3.47* The objects A, B, and C weigh 30 lb, 35 lb, and 20 lb, respectively. Determine the angles α1 and α 2 . Solution:

The equilibrium equations for the point where the

cables join are

ΣFx = −W A cos α1 + WC cos α 2 = 0,

(1)

ΣFy = W A sin α1 + WC sin α 2 − W B = 0.

(2)

Using the identity sin 2 θ + cos 2 θ WA

1 − sin 2 α1

= WC

= 1, Eq. (1) can be written

1 − sin 2 α

C B

Squaring both sides, this equation can be written W A 2 sin 2 α1 = W A 2 − WC 2 + WC 2 sin 2 α 2 . (3)

Eq. (2) can be written W A sin α1 = W B − WC sin α 2 .

Squaring this equation yields

W A 2 sin 2 α1 = W B 2 − 2W BWC sin α 2 + WC 2 sin 2 α 2 . Substituting this result into Eq. (3) gives the equation

W B 2 − 2W BWC sin α 2 + WC 2 sin 2 α 2 = W A 2 − WC 2 + WC 2 sin 2 α 2 . Solving this equation for sin α 2 yields sin α 2 = =

W B 2 − W A 2 + WC 2 2W BWC (35 lb) 2 − (30 lb) 2 + (20 lb) 2 . 2(35 lb)(20 lb)

From this equation we obtain α 2 = 31.2 °. We can then determine α1 from Eq. (1), obtaining α1 = 55.2 °. α1 = 55.2 °, α 2 = 31.2 °.

Problem 3.48 The 50-lb cylinder rests on two smooth surfaces. (a) Draw the free-body diagram of the cylinder. (b) If α = 30 °, what are the magnitudes of the forces exerted on the cylinder by the left and right surfaces?

NR NL

W x

458

The angle between the positive x-axis and the left hand force is normal (90 − α); the normal force is N L = N L ( i sin α + j cos α). The weight is W = 0 i − W j. The equilibrium conditions are

∑ F = W + N R + N L = 0. Substitute and collect like terms,

Solution:

Isolate the cylinder. (a) The free body diagram of the isolated cylinder is shown. (b) The forces acting are the weight and the normal forces exerted by the surfaces. The angle between the normal force on the right and the x-axis is (90 + β ). The normal force is N R = N R ( i cos(90 + β ) + j sin (90 + β )) N R = N R (−i sin β + j cos β ).

ΣFx = (− N R sin β + N L sin α) i = 0, ΣFy = ( N R cos β + N L cos α − W ) j = 0. Solve:  sin α  N R =  N ,  sin β  L and  W sin β  N L =  .  sin(α + β )  For W = 50 lb, and α = 30 °, β = 45 °, the normal forces are N L = 36.6 lb, N R = 25.9 lb

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Problem 3.49 If the magnitude of the normal force exerted on the 50-lb cylinder by the left surface is 40 lb, what is the angle α?

Solution: y

x 50 lb 458

458 a

We have the equilibrium equations ΣFx = N L sin α − N R cos 45 ° = 0, ΣFy = N L cos α + N R cos 45 ° = 50 lb. Eliminating the term N R results in the equation cos α + sin α = r , where we define r = 50 lb/N L . Using the identity sin 2 α + cos 2 α = 1, we can solve cos α in terms of sin α and obtain the quadratic equation r2 − 1 sin 2 α − r sin α + = 0. 2 Solving, we obtain the roots sin α = 0.294 and sin α = 0.956, yielding the angles α = 17.1 ° and α = 72.9 °. α = 17.1 ° or 72.9 °.

Problem 3.50 The two springs are identical, with unstretched length 0.4 m. When the 50-kg mass is suspended at B, the length of each spring increases to 0.6 m. What is the spring constant k?

Solution: F = k (0.6 m − 0.4 m) ΣFy : 2 F sin 60° − 490.5 N = 0 k = 1416 N/m F

0.6 m A C

608

k kk B

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Problem 3.51 The cable AB is 0.5 m in length. The unstretched length of the spring is 0.4 m. When the 50-kg mass is suspended at B, the length of the spring increases to 0.45 m. What is the spring constant k?

Solution:

The Geometry Law of Cosines and Law of Sines 0.7 2 = 0.5 2 + 0.45 2 − 2(0.5)(0.45) cos β sin φ sin β sin θ = = 0.45 m 0.5 m 0.7 m β = 94.8 °, θ = 39.8 ° φ = 45.4 ° Now do the statics

0.7 m

F = k (0.45 m − 0.4 m) ΣFx : −T AB cos θ + F cos φ = 0

ΣFy : T AB sin θ + F sin φ − 490.5 N = 0 C

Solving: k = 7560 N/m

0.7 m u

B 0.5 m

0.45 m b

TAB

490.5 N

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Problem 3.52 In the left figure, the three springs are unstretched. The spring constant of the center spring is k1 = 200 N/m , and the spring constant of each outer spring is k 2 = 160 N/m. In the right figure, a mass m has been suspended from the springs. If h = 1 m, what is the mass m in kilograms?

0.45 m

k1 k2

0.65 m h

Solution: L 02 =

The unstretched length of the outer springs is

(0.45 m) 2 + (0.65 m) 2 = 0.791 m.

The stretched length of the outer springs is L2 =

where the angle α = arctan(0.45/1.0) = 24.2 °. From the equilibrium equation ΣFy = T1 + 2T2 cos α − mg = 0: 70 N + 2(49.0 N) cos 24.2 ° − m(9.81 m/s 2 ) = 0,

(0.45 m) 2 + (1.0 m) 2 = 1.10 m.

The tensions in the stretched inner and outer springs are T1 = k1 ( L1 − L 01 ) = (200 N/m)(1.0 m − 0.65 m) = 70 N, T2 = k 2 ( L1 − L 01 )

we obtain m = 16.2 kg. m = 16.2 kg.

= (160 N/m)(1.10 m − 0.791 m) = 49.0 N. The free-body diagram of the suspended mass is T2

T1 a

T2 a

mg x

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Problem 3.53* The horizontal force F holds the small slider of weight W in equilibrium on the smooth curved bar. The centerline of the bar is described by the equation shown. Show that F is given in terms of W and the x coordinate of the slider by F = 2Wx.

Solution:

The forces on the slider are its weight W, the force F, and the normal force N exerted by the smooth bar:

dy u

u F

N y 5 x2

F As indicated in the figure, the slope of the bar is the derivative dy/dx of the function describing the centerline of the bar: dy tan θ = . dx The equilibrium equation in the direction parallel to the bar is x x

F cos θ − W sin θ = 0. Solving for F yields sin θ W cos θ = (tan θ)W dy = W. dx

F =

The derivative dy /dx = 2 x, so the force F is F = 2Wx.

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Problem 3.54 The 20-lb slider A is held in place on the smooth inclined bar by the spring. The unstretched length of the spring is 6 in. What is the spring constant k?

Solution:

The length of the spring is

L = (5 in) 2 + (6 in) 2 = 7.81 in and its unstretched length is L 0 = 6 in,

5 in

so the spring force is F = k(L − L 0 ) = k (7.81 in − 6 in) = k (1.81 in).

k 6 in

5 in

308

6 in

308 x W

Let W denote the weight of the slider. The sum of the forces on the slider in the direction parallel to the bar is ΣFx = W sin 30 ° −

5 in F L

= (20 lb)sin 30 ° −

5 in k (1.81 in) = 0. 7.81 in

Solving for k yields (20 lb)(7.81 in)sin 30 ° (5 in)(1.81 in) = 8.63 lb/in.

k =

k = 8.63 lb/in.

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Problem 3.55 The mass of each pulley of the system is m and the mass of the suspended object A is m A . Determine the force T necessary for the system to be in equilibrium.

Solution:

Draw free body diagrams of each pulley and the object A. Each pulley and the object A must be in equilibrium. The weights of the pulleys and object A are W = mg and W A = m A g. The equilibrium equations for the weight A, the lower pulley, second pulley, third pulley, and the top pulley are, respectively, B − W A = 0, 2C − B − W = 0, 2 D − C − W = 0, 2T − D − W = 0, and FS − 2T − W = 0. Begin with the first equation and solve for B, substitute for B in the second equation and solve for C, substitute for C in the third equation and solve for D, and substitute for D in the fourth equation and solve for T, to get T in terms of W and W A . The result is WA W + , 2 2 W W 3W 7W D = A + , and T = A + , 4 4 8 8

B = WA, C =

or in terms of the masses, g T = (m A + 7m). 8 Fs W T

T T

T W

D D

C C B B WA

Problem 3.56 The suspended mass m1 = 50 kg. Neglecting the masses of the pulleys, determine the value of the mass m 2 necessary for the system to be in equilibrium.

Solution: ΣFC : T1 + 2m 2 g − m1g = 0 ΣFB : T1 − 2m 2 g = 0 m2 =

m1 = 12.5 kg 4

A T1

C m2 m1 g

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Problem 3.57 The boy is lifting himself using the block and tackle shown. If the weight of the block and tackle is negligible, and the combined weight of the boy and the beam he is sitting on is 120 lb, what force does he have to exert on the rope to raise himself at a constant rate? (Neglect the deviation of the ropes from the vertical.)

Solution:

A free-body diagram can be obtained by cutting the four ropes between the two pulleys of the block and tackle and the rope the boy is holding. The tension has the same value T in all five of these ropes. So the upward force on the free-body diagram is 5T and the downward force is the 120-lb weight. Therefore the force the boy must exert is T = (120 lb)/5 = 24 lb T = 24 lb

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Problem 3.58 Neglect the weights of the pulleys and the deviations of the rope from the vertical. The object A weighs 200 lb. Determine the force T necessary to support it.

Solution:

The tensions in the rope on either side of a given pulley must be equal; otherwise the pulley would rotate. That means that the tension T is the same in each segment of the rope in this problem. We just need to find a suitable free-body diagram to determine it. If we cut the rope segments between the top two pulleys and the bottom two pulleys and isolate the bottom two, TT

T T

200 lb

we see that 5T = 200 lb. The tension is T = 40 lb.

Problem 3.59 The number of pulleys in the type of system shown could obviously be extended to an arbitrary number N. (a) Neglecting the weights of the pulleys, determine the force T required to support the weight W as a function of the number of pulleys N in the system. (b) Using the result of part (a), determine the force T required to support the weight W for a system with 10 pulleys. T

Solution:

By extrapolation of the previous problem

(a)

T =

W 2N

(b)

T =

W 1024

W (a) One pulley W (b) Two pulleys

W (c) Three pulleys

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Problem 3.60 The 20-lb sphere is supported by the smooth inclined surfaces. (The positive y-axis points upward. The sphere’s weight acts at its center.) What are the magnitudes of the forces exerted on the sphere by the surfaces at A and B?

Solution:

From the free-body diagram of the sphere,

y 308

NB x

W NA B 208 the equilibrium equations are ΣFx = −N A sin 20 ° + N B cos30 ° = 0, ΣFy = N A cos 20 ° − N B sin 30 ° − 20 lb = 0.

308

A 208

Solving yields N A = 26.9 lb, N B = 10.6 lb. A: 26.9 lb, B: 10.6 lb. x

Problem 3.61 The 20-lb sphere is held in equilibrium on the smooth circular surface by the horizontal force. What is the angle α?

Solution:

From the free-body diagram of the sphere,

15 lb

20 lb

N a 15 lb the equilibrium equations are ΣFx = −N sin α + 15 lb = 0, ΣFy = N cos α − 20 lb = 0. Eliminating N gives the relation tan α = 15/20, so we obtain α = 36.9 °. α = 36.9 °.

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Problem 3.62 The piston A is held in equilibrium in the smooth inclined slot by the linear spring. The unstretched length of the spring is 0.18 m, and the spring constant is k = 6 kN/m. The distance d = 0.12 m. What is the mass of the piston?

Solution: x

d 0.18 m F

308

0.18 m A 308

The forces on the free-body diagram of the piston are its weight mg, the normal force N exerted by the slot, and the force F exerted by the spring. The length of the spring is L =

d 2 + (0.18 m) 2

= (0.12 m) 2 + (0.18 m) 2 = 0.216 m and its unstretched length is L 0 = 0.18 m, so the spring force is F = k(L − L 0 ) = (6000 N/m)(0.216 m − 0.180 m) = 218 N. From the free-body diagram, the sum of the forces in the x direction is d ΣFx = F − mg sin 30 ° = 0. L Solving for m yields dF gL sin 30 ° (0.12 m)(218 N) = (9.81 m/s 2 )(0.216 m)sin 30 ° = 24.7 kg.

m =

24.7 kg.

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Problem 3.63 The force F = 800 i (lb) acts where the cables AB, AC , and AD are joined. What are the tensions in the three cables? y

Solution: (0, 6, 0) ft

rD/A

rB/A

(0, 4, 6) ft

(12, 4, 2) ft

rC/A

D (0, 6, 0) ft

(0, 4, 6) ft

y D

(6, 0, 0) ft x

(6, 0, 0) ft

The position vectors

rB /A = (6 ft − 12 ft) i + (0 − 4 ft) j + (0 − 2 ft)k = −6 i − 4 j − 2k (ft), rC /A = (0 − 12 ft) i + (4 ft − 4 ft) j + (6 ft − 2 ft)k = −12 i + 4 k (ft), rD /A = (0 − 12 ft) i + (6 ft − 4 ft) j + (0 − 2 ft)k = −12 i + 2 j − 2k (ft).

y (0, 6, 0) ft

eAD

(12, 4, 2) ft

eAC C (0, 4, 6) ft

A eAB

(6, 0, 0) ft

z Dividing these vectors by their magnitudes gives the unit vectors r B /A = −0.802 i − 0.535 j − 0.267k, r B /A r e AC = C /A = −0.949 i + 0.316k, rC /A r e AD = D /A = −0.973i + 0.162 j − 0.162k. r D /A e AB =

From the equilibrium equation for joint A, T AB e AB + T AC e AC + T AD e AD + F = 0, we obtain the three equations −0.802T AB − 0.949T AC − 0.973T AD + 800 lb = 0, −0.535T AB + 0.162T AD = 0, −0.267T AB + 0.316T AC − 0.162T AD = 0. Solving, we obtain T AB = 125 lb, T AC = 316 lb, T AD = 411 lb. AB: 125 lb. AC: 316 lb. AD: 411 lb.

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Problem 3.64 The force F = 600 i + 200 j + 100 k (lb) Solution: acts where the cables AB, AC , and AD are joined. What are the tensions in the three cables?

(0, 6, 0) ft D

(0, 4, 6) ft

(12, 4, 2) ft B

rB/A

(0, 4, 6) ft

(12, 4, 2) ft A

rC/A

D (0, 6, 0) ft

rD/A

(6, 0, 0) ft x

(6, 0, 0) ft

The position vectors rB /A = (6 ft − 12 ft) i + (0 − 4 ft) j + (0 − 2 ft)k = −6 i − 4 j − 2k (ft), rC /A = (0 − 12 ft) i + (4 ft − 4 ft) j + (6 ft − 2 ft)k = −12 i + 4 k (ft), rD /A = (0 − 12 ft) i + (6 ft − 4 ft) j + (0 − 2 ft)k = −12 i + 2 j − 2k (ft).

y (0, 6, 0) ft D

eAD

(12, 4, 2) ft A

eAC

C (0, 4, 6) ft

eAB

B (6, 0, 0) ft

From the equilibrium equation for joint A, T AB e AB + T AC e AC + T AD e AD + F = 0, we obtain the three equations −0.802T AB − 0.949T AC − 0.973T AD + 600 lb = 0, −0.535T AB + 0.162T AD + 200 lb = 0, −0.267T AB + 0.316T AC − 0.162T AD + 100 lb = 0. Solving, we obtain T AB = 421 lb, T AC = 119 lb, T AD = 154 lb. AB: 421 lb. AC: 119 lb. AD: 154 lb.

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Problem 3.65* Suppose that you want to apply a 1000-lb force F at point A in a direction such that the resulting tensions in cables AB, AC , and AD are equal. Determine the components of F.

F D (0, 6, 0) ft A (12, 4, 2) ft

C B (0, 4, 6) ft

(6, 0, 0) ft

Solution:

We first write the position vectors

r AB = (−6 i − 4 j − 2k) ft r AC = (−12 i + 6k) ft r AD = (−12 i + 2 j − 2k) ft Now we can use these vectors to define the force vectors r AB = T (−0.802 i − 0.535 j − 0.267k) r AB r T AC = T AC = T (−0.949 i + 0.316k) r AC r T AD = T AD = T (−0.973i + 0.162 j − 0.162k) r AD T AB = T

The force F can be written F = ( Fx i + Fy j + Fz k) The equilibrium equations are then ΣFx : −0.802T − 0.949T − 0.973T + Fx = 0 ⇒ Fx = 2.72T ΣFy : −0.535T + 0.162T + Fy = 0 ⇒ Fy = 0.732T ΣFz : −0.267T + 0.316T − 0.162T + Fz = 0 ⇒ Fz = 0.113T We also have the constraint equation ⇒ T = 363 lb

Fx2 + Fy2 + Fz2 = 1000 lb

Solving, we find Fx = 990 lb, Fy = 135 lb, Fz = 41.2 lb

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Problem 3.66 The mass of the suspended object is m = 20 kg. The y-axis is vertical. The weight of bar AB is negligible. Bar AB exerts a force at B that is directed along the line from A to B. What are the tensions in the cables BC and BD?

e AB = 0.693i + 0.693 j + 0.174 k, e BC = −0.743i + 0.557 j + 0.371k, e BD = −0.535i + 0.267 j − 0.802k. y

(0, 6, 25) m

(0, 7, 3) m

Dividing these vectors by their magnitudes, we obtain unit vectors that point from A toward B, from B toward C, and from B toward D:

(0, 7, 3) m eBC

(0, 6, 25) m

B (4, 4, 1) m

eAB

eBD (4, 4, 1) m B

A z

x m

Let F denote the force exerted on the cables at B by the bar AB. Let the tensions in the cables be denoted by TBC and TBD . The equilibrium equation for the top end of the bar is Fe AB + TBC e BC + TBD e BD − mg j = 0.

Solution:

The three components of this equation yield the equations y (0, 7, 3) m C

0.693F − 0.743TBC − 0.535TBD = 0,

(0, 6, 25) m D

0.693F + 0.557TBC + 0.267TBD − mg = 0, 0.174 F + 0.371TBC − 0.802TBD = 0.

rBD

rBC rAB

(4, 4, 1) m B

Solving, we obtain F = 170 N, TBC = 99.7 N, TBD = 83.1 N. BC : 99.7 N, BD : 83.1 N.

A z

The position vectors r AB = 4 i + 4 j + k (m), rBC = (0 − 4) i + (7 − 4) j + (3 − 1)k (m) = − 4 i + 3 j + 2k (m), rBD = (0 − 4) i + (6 − 4) j + (−5 − 1)k (m) = − 4 i + 2 j − 6k (m).

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Problem 3.67 The bulldozer exerts a force F = 2 i (kip) at A. What are the tensions in cables AB, AC , and AD?

y 6 ft C

8 ft 2 ft B

3 ft z

D 4 ft

8 ft x

Solution:

Isolate the cable juncture. Express the tensions in terms of unit vectors. Solve the equilibrium equations. The coordinates of points A, B, C , D are: A(8, 0, 0), B(0, 3, 8), C (0, 2, −6), D(0, −4, 0).

The radius vectors for these points are r A = 8i + 0 j + 0 k,

rB = 0 i + 3 j + 8k,

rC = 0 i + 2 j − 6k,

rD = 0 i + 4 j + 0 k.

By definition, the unit vector parallel to the tension in cable AB is r − rA e AB = B . rB − r A Carrying out the operations for each of the cables, the results are: e AB = −0.6835i + 0.2563 j + 0.6835k, e AC = −0.7845i + 0.1961 j − 0.5883k, e AD = −0.8944 i − 0.4472 j + 0 k. The tensions in the cables are expressed in terms of the unit vectors, T AB = T AB e AB , T AC = T AC e AC , T AD = T AD e AD . The external force acting on the juncture is F = 2000 i + 0 j + 0 k. The equilibrium conditions are

∑ F = 0 = TAB + TAC + TAD + F = 0. Substitute the vectors into the equilibrium conditions: ΣFx = (−0.6835 T AB − 0.7845 T AC − 0.8944 T AD + 2000) i = 0 ΣFy = (0.2563 T AB + 0.1961 T AC − 0.4472 T AD ) j = 0 ΣFz = (0.6835 T AB − 0.5883 T AC + 0 T AD )k = 0 The commercial program TK Solver Plus was used to solve these equations. The results are T AB = 780.31 lb , T AC = 906.49 lb , T AD = 844.74 lb .

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Problem 3.68 The 50-kg mass m is supported by three forces with magnitudes R, S , and T. The direction of each vector is defined in terms of a right triangle whose lower side lies in the x – z plane. Determine the magnitudes R, S , and T.

Solution:

The three force vectors are

R(cos 40 ° cos15 °i + sin 40 ° j + cos 40 ° sin15 °k) = R(0.740 i + 0.643 j + 0.198k), S (− cos 50 ° sin10 °i + sin 50 ° j − cos 50 ° cos10 °k) = S (−0.112 i + 0.766 j − 0.633k), T (− cos60 ° sin 20 °i + sin 60 ° j + cos60 ° cos 20 °k) = T (−0.171i + 0.866 j + 0.470 k). From these expressions, the three components of the equations of equilibrium for the joint where the cables join are

y S T

0.740 R − 0.112S − 0.171T = 0, 0.643 R + 0.766S + 0.866T − mg = 0, 0.198 R − 0.633S + 0.470T = 0.

508 108 608

Solving, we obtain R = 101 N, S = 239 N, T = 280 N. R

R = 101 N, S = 239 N, T = 280 N.

408

208

158

z m

Problem 3.69 The 20-kg mass is suspended by cables attached to three vertical 2-m posts. Point A is at (0, 1.2, 0) m. Determine the tensions in cables AB, AC , and AD.

C B D A

1m 1m 2m

0.3 m

Solution:

Points A, B, C , and D are located at

A(0, 1.2, 0), B(−0.3, 2, 1), C (0, 2, −1), D(2, 2, 0)

Write the unit vectors e AB , e AC , e AD e AB = −0.228i + 0.608 j + 0.760 k

e AC = 0 i + 0.625 j − 0.781k

FAC FAB

FAD

e AD = 0.928i + 0.371 j + 0 k

The forces are

F AB = −0.228FAB i + 0.608FAB j + 0.760 FAB k W

F AC = 0 FAC i + 0.625FAC j − 0.781FAC k F AD = 0.928FAD i + 0.371FAD j + 0 k W = −(20)(9.81) j

(20) (9.81) N

The equations of equilibrium are  ΣF = −0.228F + 0 + 0.928F = 0 x AB AD    ΣFy = 0.608FAB + 0.625FAC + 0.371FAD − 20(9.81) = 0   ΣF = 0.760 F − 0.781F + 0 = 0 z AB AC 

130

We have 3 eqns in 3 unknowns solving, we get FAB = 150.0 N FAC = 146.1 N FAD = 36.9 N

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Problem 3.70 The horizontal rectangular plate is supported by cables. The dimensions are in meters. A strain gage indicates that the tension in cable AB is 1.05 kN. Determine the tensions in cables AC and AD and the mass of the plate.

Dividing these vectors by their magnitudes, we obtain unit vectors that point from A toward B, from A toward C, and from A toward D: e AB = 0.391i − 0.893 j + 0.223k, e AC = 0.179 i − 0.954 j − 0.239k, e AD = −0.527 i − 0.843 j + 0.105k.

y A (0, 1.6, 0) m eAD

A 1.6

eAC eAB

(–1.0, 0, 0.2) m

0.2

0.3

(0.3, 0, –0.4) m

D C

1.0

0.7

0.4 0.4

B (0.7, 0, 0.4) m

Let the tensions in the cables be denoted by T AB , T AC , T AD . The equilibrium equation for the portion of the cables where they are joined is T AB e AB + T AC e AC + T AD e AD + mg j = 0 .

Solution:

The three components of this equation yield the equations 0.391T AB + 0.179T AC − 0.527T AD = 0,

−0.893T AB − 0.954T AC − 0.843T AD + mg = 0, 0.223T AB − 0.239T AC + 0.105T AD = 0.

A (0, 1.6, 0) m rAD

Solving with T AB = 1050 N, we obtain m = 360 kg, T AC = 1560 N, T AD = 1310 N.

rAC

AC : 1.56 kN, AD : 1.31 kN, mass = 360 kg. rAB

(21.0, 0, 0.2) m

(0.3, 0, 20.4) m

D C z

B (0.7, 0, 0.4) m

The position vectors r AB = (0.7 − 0) i + (0 − 1.6) j + (0.4 − 0)k (m) = 0.7 i − 1.6 j + 0.4 k (m), r AC = (0.3 − 0) i + (0 − 1.6) j + (−0.4 − 0)k (m) = 0.3i − 1.6 j − 0.4 k (m), r AD = (−1.0 − 0) i + (0 − 1.6) j + (0.2 − 0)k (m) = −1.0 i − 1.6 j + 0.2k (m).

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Problem 3.71 The suspended object has a mass of 40 kg. Determine the tensions in cables AB, AC , and AD.

Dividing these vectors by their magnitudes, we obtain unit vectors that point from A toward B, from A toward C, and from A toward D: e AB = 0.667 i + 0.667 j − 0.333k, e AC = −0.287 i + 0.575 j − 0.766k,

e AD = −0.512 i + 0.768 j + 0.384 k.

y C

0.3 m

C (20.3, 0, 20.8) m

0.8 m 0.3 m

0.6 m

(20.4, 0, 0.3) m

D B

0.4 m

0.3 m

0.6 m

B eAD

(0.6, 0, 20.3) m

eAC eAB

A (0, 20.6, 0) m

A Let the tensions in the cables be denoted by T AB , T AC , T AD . The equilibrium equation for the portion of the cables where they are joined is T AB e AB + T AC e AC + T AD e AD − mg j = 0 . The three components of this equation yield the equations

Solution:

0.667T AB − 0.287T AC − 0.333T AD = 0, 0.667T AB + 0.575T AC − 0.766T AD − mg = 0,

0.693T AB + 0.768T AC + 0.384T AD = 0.

C (20.3, 0, 20.8) m D

AB : 232 N, AC : 39.4 N, AD : 280 N.

(20.4, 0, 0.3) m z

Solving, we obtain T AB = 232 N, T AC = 39.4 N, T AD = 280 N.

(0.6, 0, 20.3) m

rAC

rAD

rAB

A (0, 20.6, 0) m The position vectors r AB = (0.6 − 0) i + (0 − (−0.6)) j + (−0.3 − 0)k (m) = 0.6 i + 0.6 j − 0.3k (m), r AC = (−0.3 − 0) i + (0 − (−0.6)) j + (−0.8 − 0)k (m) = −0.3i + 0.6 j − 0.8k (m), r AD = (−0.4 − 0) i + (0 − (−0.6)) j + (0.3 − 0)k (m) = −0.4 i + 0.6 j + 0.3k (m).

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Problem 3.72 The 680-kg load suspended from the helicopter is in equilibrium. The aerodynamic drag force on the load is horizontal. The y-axis is vertical, and cable O A lies in the x – y plane. Determine the magnitude of the drag force and the tension in cable O A.

Solution: ∑ Fx = TOA sin10 ° − D = 0, ∑ Fy = TOA cos10 ° − (680)(9.81) = 0. Solving, we obtain D = 1176 N, TOA = 6774 N. y TOA

108 y

108

x (680) (9.81) N

B D

Problem 3.73 The coordinates of the three cable attachment points B, C , and D are (−3.3, − 4.5, 0) m, (1.1, − 5.3, 1) m, and (1.6, − 5.4, −1) m, respectively. What are the tensions in cables O B, OC , and OD?

Solution:

The position vectors from O to pts B, C , and D are

rOB = −3.3i − 4.5 j (m), rOC = 1.1i − 5.3 j + k (m), rOD = 1.6 i − 5.4 j − k (m). Dividing by the magnitudes, we obtain the unit vectors e OB = −0.591i − 0.806 j, e OC = 0.200 i − 0.963 j + 0.182k, e OD = 0.280 i − 0.944 j − 0.175k. Using these unit vectors, we obtain the equilibrium equations

∑ Fx = TOA sin10° − 0.591TOB + 0.200TOC + 0.280TOD = 0,

∑ Fy = TOA cos10 ° − 0.806TOB − 0.963TOC − 0.944TOD = 0,

108

∑ Fz = 0.182TOC − 0.175TOD = 0.

From the solution of Problem 3.72, TOA = 6774 N. Solving these equations, we obtain TOB = 3.60 kN,

TOC = 1.94 kN,

TOD = 2.02 kN.

y TOA

B C

108 x

TOB

BandF_6e_ISM_CH03.indd 133

TOC

TOD

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Problem 3.74 The 100-lb object is held at rest on a smooth plane surface by the cables AB and AC . The unit vector e = 0.743i + 0.557 j + 0.371k is normal to the surface. What are the tensions in cables AB and BC ? What is the magnitude of the normal force exerted on the object by the smooth surface? y

Solution:

The position vectors

r AB = (6 − 10) i + (14 − 6) j + (8 − 4)k (ft) = −4 i + 8 j + 4 k (ft), r AC = (9 − 10) i + (12 − 6) j + (0 − 4)k (ft) = −1.0 i + 6 j − 4 k (ft). Dividing these vectors by their magnitudes, we obtain unit vectors that point from A toward B and from A toward C:

(6, 14, 8) ft B

(9, 12, 0) ft C

e AB = −0.408i + 0.816 j + 0.408k, e AC = −0.137 i + 0.824 j − 0.549k.

A (10, 6, 4) ft

Let the tensions in the cables be denoted by T AB , T AC . The normal force exerted on the object by the surface can be expressed in terms of the given unit vector as Ne. The equilibrium equation for the object is T AB e AB + T AC e AC + Ne − W j = 0,

x z

where W is the weight of the object. The three components of this equation yield the equations −0.408T AB − 0.137T AC + 0.743 N = 0, 0.816T AB + 0.824T AC + 0.557 N − W = 0, 0.408T AB − 0.549T AC + 0.371N = 0. Solving, we obtain T AB = 43.4 lb, T AC = 55.3 lb, N = 34.1 lb. AB: 43.4 lb, AC: 55.3 lb, normal force = 34.1 lb.

Problem 3.75* The 3400-lb car is at rest on the plane surface. The unit vector e n = 0.456 i + 0.570 j + 0.684 k is perpendicular to the surface. Determine the magnitudes of the total normal force N and the total friction force f exerted on the surface by the car’s wheels.

y en

Solution: The forces on the car are its weight, the normal force, and the friction force. The normal force is in the direction of the unit vector, so it can be written x

N = Ne n = N (0.456 i + 0.570 j + 0.684 k) The equilibrium equation is

Ne n + f − (3400 lb) j = 0 The friction force f is perpendicular to N, so we can eliminate the friction force from the equilibrium equation by taking the dot product of the equation with e n . ( Ne n + f − (3400 lb) j) ⋅ e n = N − (3400 lb)( j ⋅ e n ) = 0 N = (3400 lb)(0.57) = 1940 lb Now we can solve for the friction force f. f = (3400 lb) j − Ne n = (3400 lb) j − (1940 lb)(0.456 i + 0.570 j + 0.684 k) f = (−884 i + 2300 j − 1330 k) lb f =

(−884 lb) 2 + (2300 lb) 2 + (−1330 lb) 2 = 2790 lb

N = 1940 lb, f = 2790 lb

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Problem 3.76 The circular ring is suspended from the wires AB, AC , and AD. Each wire is 1 m in length. The mass of the ring is 5 kg. The relative positions of the attachment points of the wires are shown in the top view. What are the tensions in the three wires?

A C

608

208 A

608

D D 1.00 m

Solution:

Let the points B,C,D lie in the x-z plane of the coordinate system shown. The radius of the suspended ring is 0.5 m and the length of each cable segment is 1 m, so the y coordinate of point A (the y-axis point upward) is (1 m) 2 − (0.5 m) 2 = 0.866 m. The coordinates of the points are A : (0, 0.866, 0) m,

B : (−0.5cos 20 °, 0, −0.5sin 20 °) = (−0.470, 0, −0.171) m, C : (0.5cos60 °, 0, −0.5sin 60 °) = (0.250, 0, −0.433) m,

y A

rAB 608

208 A

D : (0.5cos60 °, 0, 0.5sin 60 °) = (0.250, 0, 0.433) m.

rAC

rAB

rAC

rAD

The position vectors D

r AB = (−0.470 − 0) i + (0 − 0.866) j + (−0.171 − 0)k (m) = −0.470 i − 0.866 j − 0.171k (m),

608

r AC = (0.250 − 0) i + (0 − 0.866) j + (−0.433 − 0)k (m) = 0.250 i − 0.866 j − 0.433k (m), r AD = (0.250 − 0) i + (0 − 0.866) j + (0.433 − 0)k (m) = 0.250 i − 0.866 j + 0.433k (m). Because the three cable segments are each of unit length, these position vectors are unit vectors that point from A toward B, from A toward C, and from A toward D. Let the tensions in the wires be denoted by T AB , T AC , T AD . The equilibrium equation for the portion of the cables where they are joined is T AB r AB + T AC r AC + T AD r AD + mg j = 0 . The three components of this equation yield the equations −0.470T AB + 0.250T AC + 0.250T AD = 0, −0.866T AB − 0.866T AC − 0.866T AD + mg = 0, −0.171T AB − 0.433T AC + 0.433T AD = 0. Solving, we obtain T AB = 19.7 N, T AC = 14.6 N, T AD = 22.4 N. AB: 19.7 N, AC: 14.6 N, AD: 22.4 N.

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Problem 3.77 The 100-lb collar A is held in equilibrium on the smooth bar CD by the cable AB. The collar A is at the midpoint of the bar. The y-axis is vertical. What is the tension in cable AB?

Solution:

The coordinates of the collar A are (4, 3, 1.5) ft. Let T be the tension in cable AB, and let r AB be the vector from A to B. The force exerted on the collar by cable AB is T = T

−4 i + j + 2.5k (ft) (−4 ft) 2 + (1 ft) 2 + (2.5 ft) 2 = T (−0.830 i + 0.207 j + 0.518k).

= T (0, 6, 0) ft D

The forces acting on the collar A are its weight W = −(100 lb) j, the force exerted by the cable AB, and the normal force N exerted by the smooth bar. The sum of the forces is zero:

(0, 4, 4) ft B

T + W + N = 0. x C

r AB r AB

(8, 0, 3) ft

Taking the dot product of this equation with a unit vector e that is parallel to the bar CD yields e ⋅ T + e ⋅ W = 0,

(1)

because e ⋅ N = 0. The only unknown in Eq. (1) will be the tension T in cable AB. To apply it, we must determine the unit vector e. The vector from D to C is rDC = 8i − 6 j + 3k (ft). Dividing this vector by its magnitude yields a unit vector parallel to the bar CD: e =

rDC rDC

8i − 6 j + 3k (ft) (8 ft) 2 + (−6 ft) 2 + (3 ft) 2 = 0.766 i − 0.575 j + 0.287k. =

Using this result, Eq. (1) is (0.766 i − 0.575 j + 0.287k) ⋅ [T (−0.830 i + 0.207 j + 0.518k)] + (0.766 i − 0.575 j + 0.287k) ⋅ [−(100 lb) j] = 0. Evaluating the dot products, we obtain T = 94.9 lb. 94.9 lb.

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Problem 3.78 The 200-kg slider at A is held in place on the smooth vertical bar by the cable AB. (a) Determine the tension in the cable. (b) Determine the force exerted on the slider by the bar.

2m B

Solution:

The coordinates of the points A, B are A(2, 2, 0), B(0, 5, 2). The vector positions

r A = 2 i + 2 j + 0 k, rB = 0 i + 5 j + 2k The equilibrium conditions are:

Eliminate the slider bar normal force as follows: The bar is parallel to the y axis, hence the unit vector parallel to the bar is e B = 0 i + 1 j + 0 k. The dot product of the unit vector and the normal force vanishes: e B ⋅ N = 0. Take the dot product of e B with the equilibrium conditions: e B ⋅ N = 0. Σe B ⋅ F = e B ⋅ T + e B ⋅ W = 0.

ΣF = T + N + W = 0.

2m x

2m z

The weight is e B ⋅ W = 1 j ⋅ (− j W ) = − W = −(200)(9.81) = −1962 N. The unit vector parallel to the cable is by definition, r − rA e AB = B . rB − r A

Substitute the vectors and carry out the operation:

e AB = −0.4851i + 0.7278 j + 0.4851k. (a) The tension in the cable is T = T e AB . Substitute into the modified equilibrium condition

Σe B F = (0.7276 T − 1962) = 0. Solve: T = 2696.5 N from which the tension vector is T = T e AB = −1308i + 1962 j + 1308k. (b)

The equilibrium conditions are ΣF = 0 = T + N + W = −1308i + 1308k + N = 0. Solve for the normal force: N = 1308i − 1308k. The magnitude is N = 1850 N.

BandF_6e_ISM_CH03.indd 137

Note: For this specific configuration, the problem can be solved without eliminating the slider bar normal force, since it does not appear in the y-component of the equilibrium equation (the slider bar is parallel to the y-axis). However, in the general case, the slider bar will not be parallel to an axis, and the unknown normal force will be projected onto all components of the equilibrium equations (see Problem 3.79 below). In this general situation, it will be necessary to eliminate the slider bar normal force by some procedure equivalent to that used above. End Note.

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Problem 3.79 In Example 3.6, suppose that the cable AC is replaced by a longer one so that the distance from point B to the slider C increases from 6 ft to 8 ft. Determine the tension in the cable. Solution:

We can use the dot product to eliminate N from the equation

(

rBC = (8 ft) e BD

(

Te CA + N − (100 lb) j = 0

[ Te CA + N − (100 lb) j ] ⋅ e BD = T (e CA ⋅ e BD ) − (100 lb)( j ⋅ e BD ) = 0

The vector from B to C is now 4 7 4 rBC = (8 ft) i − j + k 9 9 9

Using N to stand for the normal force between the bar and the slider, we can write the equilibrium equation:

)

rBC = (3.56 i − 6.22 j + 3.56k) ft

)

4 7 4 T   [−0.495] +  −  [0.867] +   [ 0.0619 ] − (100 lb)(−0.778) = 0 9  9 9 T (−0.867) + (77.8 lb) = 0 ⇒ T = 89.8 lb T = 89.8 lb

We can now find the unit vector form C to A. rCA = rOA − (rOB + rBC ) = [(7 j + 4 k) − {(7 j) + (3.56 i − 6.22 j + 3.56k)}] ft rCA = (−3.56 i + 6.22 j + 0.444 k) ft r e CA = CA = (−0.495i + 0.867 j + 0.0619k) rCA

Problem 3.80 The cable AB keeps the 8-kg collar A in place on the smooth bar CD. The y-axis points upward. What is the tension in the cable? Solution:

0.15 m

We develop the following position vectors and unit

0.4 m

vectors

rCD = (−0.2 i − 0.3 j + 0.25k) m e CD =

rCD = (−0.456 i − 0.684 j + 0.570 k) rCD

rCA = (0.2 m)e CD = (−0.091i − 0.137 j + 0.114 k) m r AB = rOB − (rOC + rCA ) r AB = [(0.5 j + 0.15k) − ({0.4 i + 0.3 j} + {−0.091i − 0.137 j + 0.114 k})] m r AB = (−0.309 i + 0.337 j + 0.036k) m e AB =

0.2 m 0.3 m

0.5 m D

x 0.25 m

0.2 m z

r AB = (−0.674 i + 0.735 j + 0.079k) r AB

We can now write the equilibrium equation for the slider using N to stand for the normal force between the slider and the bar CD. Te AB + N − (8 kg)(9.81 m/s 2 ) j = 0 To eliminate the normal force N we take a dot product with e CD . [ Te AB + N − (8 kg)(9.81 m/s 2 ) j ] ⋅ e CD = 0 T (e AB ⋅ e CD ) − (78.5 N)( j ⋅ e CD ) = 0 T ([−0.674][−0.456] + [0.735][−0.684] + [0.079][0.570]) − (78.5 N)(−0.684) = 0 T (−0.150) + 53.6 N = 0 T = 357 N

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Problem 3.81* Determine the magnitude of the normal force exerted on the collar A by the smooth bar.

Solution:

From Problem 3.81 we have

e AB = (−0.674 i + 0.735 j + 0.079k) T = 357 N

The equilibrium equation is

0.15 m

Te AB + N − (78.5 N) j = 0 0.4 m

We can now solve for the normal force N. N = (78.5 N) j − (357 N)(−0.674 i + 0.735 j + 0.079k)

N = (240 i − 184 j − 28.1k) N

0.5 m

The magnitude of N is

0.2 m 0.3 m

N =

(240 N) 2 + (−184 N) 2 + (−28.1 N) 2

N = 304 N

0.25 m

D 0.2 m z

Problem 3.82* The 20-kg object A is at rest on the rough triangular surface BCD. The y-axis is vertical. What are the magnitudes of the normal and friction forces exerted on the object by the surface? y

The dot product of the weight with e is −mg j i

(0.337 i + 0.842 j + 0.421k) = −0.842 mg.

N = 0.842(20 kg)(9.81 m/s 2 ) = 165 N. B

x (1.0, 0, 0) m D (0, 0, 0.8) m

e = 0.337 i + 0.842 j + 0.421k.

Therefore, from Eq. (1), the magnitude of the normal force is (0, 0.4, 0) m

Dividing this vector by its magnitude yields the unit vector e:

Since N + f = −W, we know that N + f = W = mg. Also, N and f are perpendicular, so the magnitude of their sum is N+ f =

N 2 + f 2.

The magnitude of the friction force is therefore f =

Solution:

Let e be a unit vector that is perpendicular to the surface and points upward. The forces on the object A are its weight W = −mg j, the normal force N = Ne, and the friction force f. Note that f is perpendicular to N. The sum of the forces is zero. W + N + f = 0:

(mg) 2 − N 2

= (20 kg) 2 (9.81 m/s 2 ) 2 − (165 N) 2 = 106 N. Normal force = 165 N, friction force = 106 N.

−mg j + Ne + f = 0 . The dot product of this equation with e is −mg j ⋅ e + Ne ⋅ e + f ⋅ e = 0: −mg j ⋅ e + N = 0,

(1)

where we used the fact that f ⋅ e = 0. We can solve this equation for N if we can determine the components of the unit vector e. The vector from C to D is rCD = −0.4 j + 0.8k (m). The vector from C to B is rCB = i − 0.4 j (m). By definition, the cross product rCD × rCB is a vector that is perpendicular to the surface and points upward: rCD × rCB =

i j k 0 −0.4 0.8 1 −0.4 0

= 0.32 i + 0.8 j + 0.4 k (m 2 ).

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Problem 3.83 The 30-kg crate is held in place on the smooth surface by the rope AB. Determine the tension in the rope and the magnitude of the normal force exerted on the crate by the surface.

Solution: T 458

A 308 458

N N x

The equilibrium equations are ΣFx = −T cos 45 ° + N sin 30 ° = 0, 308

ΣFy = T sin 45 ° + N cos30 ° − mg = 0. Solving, we obtain T = 152 N, N = 215 N. Tension is 152 N, normal force is 215 N.

Problem 3.84 The system shown is called Russell’s traction. If the sum of the downward forces exerted at A and B by the patient’s leg is 32.2 lb, what is the weight W?

Solution:

The force in the cable is W everywhere. The free-body diagram of the leg is shown. The downward force is given, but the horizontal force FH is unknown. The equilibrium equation in the vertical direction is Σ Fy : W sin 25 ° + W sin 60 ° − 32.2 lb = 0

Thus W =

32.2 lb sin 25 ° + sin 60 °

W = 25.0 lb 608

208 258

B A

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Problem 3.85 The engine block is suspended by cables. The free-body diagram obtained by isolating the part of the system within the dashed line is shown. If the tension T AB = 330 lb, what are the tension T AC and the weight W of the engine block? y B

TAB

TAC

508

408

Solution:

The equilibrium equations are

ΣFx = T AC cos 40 ° − T AB cos 50 ° = 0, ΣFy = T AC sin 40 ° + T AB sin 50 ° − W = 0. Using the value T AB = 330 lb, the first equation can be solved for T AC , then the second equation can be solved for W. We obtain T AC = 277 lb, W = 431 lb.

C A

Problem 3.86 The cable AB is horizontal, and the box on the right weighs 100 lb. The surfaces are smooth. (a) What is the tension in the cable? (b) What is the weight of the box on the left?

Solution:

We have the following equilibrium equations

ΣFyB : N B cos 40 ° − 100 lb = 0 ΣFxB : N B sin 40 ° − T = 0 ΣFxA : T − N A sin 20 ° = 0

ΣFyA : N A cos 20 ° − W A = 0

Solving these equations sequentially, we find N B = 131 lb, T = 83.9 lb

208

N A = 245 lb, W A = 230.5 lb 408

Thus we have T = 83.9 lb, W A = 230.5 lb

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Problem 3.87 Assume that the forces exerted on the 170-lb climber by the slanted walls of the “chimney” are perpendicular to the walls. If the climber is in equilibrium and is exerting a 160-lb force on the rope, what are the magnitudes of the forces exerted on the climber by the left and right walls?

108

Solution: The forces in the free-body diagram are in the directions shown on the figure. The equilibrium equations are: ΣFx : −T sin10 ° + N L cos 4 ° − N R cos3 ° = 0 ΣFy : T cos10 ° − (170 lb) + N L sin 40 ° + N R sin 3 ° = 0 where T = 160 lb. Solving we find N L = 114 lb, N R = 85.8 lb Left Wall: 114 lb Right Wall: 85.8 lb

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Problem 3.88 The weights of the pulleys are negligible. The tension T = 50 lb. What is the weight of the suspended object A?

We see that TB = 3T . Continuing this way, we draw the free-body diagram of pulley 2: TC

TB TB TB So TC = 3TB = 9T . We now know enough to draw a free-body diagram of the object A and the three lower pulleys:

T T

Solution: 3

W We conclude that W = 2T + 2TB + 2TC = 26T .

1300 lb.

The key thing to remember is that the tension on either side of a given pulley is the same. Following this rule, we draw the free-body diagram of pulley 1: TB

TTT

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Problem 3.89 The assembly A, including the pulley, weighs 60 lb. What force F is necessary for the system to be in equilibrium?

Solution:

From the free body diagram of the assembly A, we have 3F − 60 = 0, or F = 20 lb

F F F F

F F

60 lb.

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Problem 3.90 The mass of block A is 42 kg, and the mass of block B is 50 kg. The surfaces are smooth. If the blocks are in equilibrium, what is the force F?

B F

458 A 208

Solution:

Isolate the top block. Solve the equilibrium equations. The weight is. The angle between the normal force N 1 and the positive x-axis is. The normal force is. The force N 2 is. The equilibrium conditions are

B N2

ΣF = N 1 + N 2 + W = 0 from which

ΣFx = (0.7071 N 1 − N 2 ) i = 0

ΣFy = (0.7071 N 1 − 490.5) j = 0. Solve:

N 1 = 693.7 N,

N 2 = 490.5 N

y N1

Isolate the bottom block. The weight is W = 0 i − W j = 0 i − (42)(9.81) j = 0 i − 412.02 j (N). The angle between the normal force N 1 and the positive x-axis is (270 ° − 45 °) = 225 °. The normal force:

A b

x N3

N 1 = N 1 ( i cos 225 ° + j sin 225 °) = N 1 (−0.7071i − 0.7071 j). The angle between the normal force N 3 and the positive x-axis is (90 ° − 20 °) = 70 °. The normal force is N 1 = N 3 ( i cos 70 ° + j sin 70 °) = N 3 (0.3420 i + 0.9397 j). The force is . . .F = F i + 0 j. The equilibrium conditions are ΣF = W + N 1 + N 3 + F = 0, from which:

ΣFx = (−0.7071 N 1 + 0.3420 N 3 + F ) i = 0 ΣFy = (−0.7071 N 1 + 0.9397 N 3 − 412) j = 0

For N 1 = 693.7 N from above:

F = 162 N

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Problem 3.91 The climber A is being helped up an icy slope by two friends. His mass is 80 kg, and the direction cosines of the force exerted on him by the slope are cos θ x = −0.286, cos θ y = 0.429, and cos θ z = 0.857. The y-axis is vertical. If the climber is in equilibrium in the position shown, what are the tensions in the ropes AB and AC and the magnitude of the force exerted on him by the slope?

B (2, 2, 0) m

rB = 2 i + 2 j + 0 k,

C A

The vector locations of the points A, B, C are: r A = 3i + 0 j + 4 k,

A (3, 0, 4) m

Solution: Get the unit vectors parallel to the ropes using the coordinates of the end points. Express the tensions in terms of these unit vectors, and solve the equilibrium conditions. The rope tensions, the normal force, and the weight act on the climber. The coordinates of points A, B, C are given by the problem, A(3, 0, 4), B(2, 2, 0), C(5, 2, −1).

C (5, 2, 21) m

rC = 5i + 2 j − 1k.

The unit vector parallel to the tension acting between the points A, B in the direction of B is r − rA e AB = B rB − r A

Substitute and collect like terms,

The unit vectors are

ΣFz = (0.8729 T AB + 0.8704 T AC − 0.857 N )k = 0

e AB = −0.2182 i + 0.4364 j − 0.8729k,

ΣFx = (−0.2182 T AB + 0.3482 T AC − 0.286 N ) i = 0 ΣFy = (0.4364 T AB + 0.3482 T AC + 0.429 N − 784.8) j = 0

We have three linear equations in three unknowns. The solution is:

e AC = 0.3482 i + 0.3482 j − 0.8704 k,

T AB = 100.7 N ,

and e N = −0.286 i + 0.429 j + 0.857k.

T AC = 889.0 N ,

N = 1005.5 N .

where the last was given by the Problem statement. The forces are expressed in terms of the unit vectors, T AB = T AB e AB ,

T AC = T AC e AC ,

N = N eN.

The weight is W = 0 i − W j + 0 k = 0 i − (80)(9.81) j + 0 k − 0 i − 784.8 j + 0 k. The equilibrium conditions are ΣF = 0 = T AB + T AC + N + W = 0.

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Problem 3.92 Consider the climber A being helped by his friends in Problem 3.91. To try to make the tensions in the ropes more equal, the friend at B moves to the position (4, 2, 0) m. What are the new tensions in the ropes AB and AC and the magnitude of the force exerted on the climber by the slope? Solution: Get the unit vectors parallel to the ropes using the coordinates of the end points. Express the tensions in terms of these unit vectors, and solve the equilibrium conditions. The coordinates of points A, B, C are A(3, 0, 4), B(4, 2, 0), C (5, 2, −1). The vector locations of the points A, B, C are: r A = 3i + 0 j + 4 k,

rB = 4 i + 2 j + 0 k,

B (2, 2, 0) m

A (3, 0, 4) m

C (5, 2, 21) m

rC = 5i + 2 j − 1k.

The unit vectors are e AB = +0.2182 i + 0.4364 j − 0.8729k, e AC = +0.3482 i + 0.3482 j − 0.8704 k, e N = −0.286 i + 0.429 j + 0.857k. where the last was given by the problem statement. The forces are expressed in terms of the unit vectors, T AB = T AB e AB ,

T AC = T AC e AC ,

N = N eN.

The weight is W = 0 i − W j + 0 k = 0 i − (80)(9.81) j + 0 k − 0 i − 784.8 j + 0 k. The equilibrium conditions are ΣF = 0 = T AB + T AC + N + W = 0. Substitute and collect like terms, ΣFx = (+0.281 T AB + 0.3482 T AC − 0.286 N ) i = 0 ΣFy = (0.4364 T AB + 0.3482 T AC + 0.429 N − 784.8) j = 0 ΣFz = (0.8729 T AB + 0.8704 T AC − 0.857 N )k = 0 The HP-28S hand held calculator was used to solve these simultaneous equations. The solution is: T AB = 420.5 N ,

T AC = 532.7 N ,

N = 969.3 N .

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Problem 3.93 A climber helps a friend up an icy slope. The friend is hauling a box of supplies. If the mass of the friend is 90 kg and the mass of the supplies is 22 kg, what are the tensions in the ropes AB and CD? Assume that the slope is smooth. That is, only normal forces are exerted on the friend and the box by the slope.

A 208

C 408 758

Solution:

Isolate the box. The weight vector is

W2 = (22)(9.81) j = −215.8 j (N).

608

The angle between the normal force and the positive x-axis is (90 ° − 60 °) = 30 °. The normal force is N B = N B (0.866 i − 0.5 j). The angle between the rope CD and the positive x-axis is (180 ° − 75 °) = 105 °; the tension is: T2 = T2 ( i cos105 ° + j sin105 °) = T2 (−0.2588i + 0.9659 j)

T y

The equilibrium conditions are ΣFx = (0.866 N B + 0.2588 T2 ) i = 0,

ΣFy = (0.5 N B + 0.9659 T2 − 215.8) j = 0. Solve:

N B = 57.8 N,

x W

T2 = 193.5 N.

Isolate the friend. The weight is W = −(90)(9.81) j = −882.9 j (N).

208

The angle between the normal force and the positive x-axis is (90 ° − 40 °) = 50 °. The normal force is: N F = N F (0.6428i + 0.7660 j). The angle between the lower rope and the x-axis is −75 °; the tension is

408 W

758 T2 x

T2 = T2 (0.2588i + 0.9659 j). The angle between the tension in the upper rope and the positive x-axis is (180 ° − 20 °) = 160 °, the tension is T1 = T1 (0.9397 i + 0.3420 j). The equilibrium conditions are ΣF = W + T1 + T2 + N F = 0. From which: ΣFx = (0.6428 N F + 0.2588 T2 − 0.9397 T1 ) i = 0 ΣFy = (−0.7660 N F − 0.9659 T2 + 0.3420 T1 − 882.9) j = 0 Solve, for T2 = 193.5 N. The result: N F = 1051.6 N ,

148

T1 = 772.6 N .

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Problem 3.94 The 1080-kg car is moving at constant speed on a road with the slope shown. The aerodynamic forces on the car are the drag D = 420 N and the lift L = 260 N, which is perpendicular to the road. Determine the magnitudes of the normal and friction forces exerted on the car by the road.

Solution:

The equilibrium equations are

158

ΣFx = f − D − mg sin15 ° = 0, ΣFy = L + N − mg cos15 ° = 0. Solving, we obtain N = 9970 N, f = 3160 N. Normal force is 9.97 kN, friction force is 3.16 kN.

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x f

D N mg

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Problem 3.95 An engineer doing preliminary design studies for a new radio telescope envisions a triangular receiving platform suspended by cables from three equally spaced 40-m towers. The receiving platform has a mass of 20 Mg (megagrams) and is 10 m below the tops of the towers. What tension would the cables be subjected to?

TOP VIEW

20 m 65 m

Solution: Isolate the platform. Choose a coordinate system with the origin at the center of the platform, with the z-axis vertical, and the x, y axes as shown. Express the tensions in terms of unit vectors, and solve the equilibrium conditions. The cable connections at the platform are labeled a, b, c, and the cable connections at the towers are labeled A, B, C . The horizontal distance from the origin (center of the platform) to any tower is given by L =

A c x

65 = 37.5 m. 2sin (60)

The coordinates of points A, B, C are The tensions in the cables are expressed in terms of the unit vectors, A(37.5, 0, 10),

B(37.5cos(120 °), 37.5sin (120 °).10),

C (37.5cos(240 °), 37.5sin (240°), 10),

TbB = TbB e bB ,

TcC = TcC e cC .

The weight is

The vector locations are: r A = 37.5i + 0 j + 10 k,

TaA = TaA e aA ,

W = 0 i − 0 j − (20000)(9.81)k = 0 i + 0 j − 196200 k. rB = 18.764 i + 32.5 j + 10 k,

The equilibrium conditions are

rC = 18.764 i + −32.5 j + 10 k.

ΣF = 0 = TaA + TbB + TcC + W = 0,

The distance from the origin to any cable connection on the platform is

from which:

d =

20 = 11.547 m. 2sin (60 °)

ΣFx = (0.9333 TaA − 0.4666 TbB − 0.4666 TcC ) i = 0

The coordinates of the cable connections are a(11.547, 0, 0),

b(11.547 cos(120 °),

ΣFz = (0.3592 TaA − 0.3592 TbB

11547sin (120 °), 0),

c(11.547 cos(240 °), 11.547sin (240 °), 0). The vector locations of these points are, ra = 11.547 i + 0 j + 0 k,

ΣFy = (0 TaA + 0.8082 TbB − 0.8082 TcC ) j = 0

+ 0.3592 TcC − 196200 )k = 0 The commercial package TK Solver Plus was used to solve these equations. The results:

rb = 5.774 i + 10 j + 0 k,

rc = 5.774 i + 10 j + 0 k. The unit vector parallel to the tension acting between the points A, a in the direction of A is by definition r − ra e aA = A . r A − ra Perform this operation for each of the unit vectors to obtain

TaA = 182.1 kN ,

TbB = 182.1 kN ,

TcC = 182.1 kN .

Check: For this geometry, where from symmetry all cable tensions may be assumed to be the same, only the z-component of the equilibrium equations is required: ΣFz = 3 T sin θ − 196200 = 0, Where θ = tan −1

( 37.5 −1011.547 ) = 21.07°,

from which each tension is T = 182.1 kN.

check.

e aA = +0.9333i + 0 j − 0.3592k e bB = −0.4667 i + 0.8082 j − 0.3592k e cC = −0.4667 i + 0.8082 j + 0.3592k

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Problem 3.96 To support the tent, the tension in the rope AB must be 35 lb. What are the tensions in the ropes AC , AD, and AE?

(0, 5, 0) ft

Solution:

(0, 6, 6) ft We develop the following position vectors

(6, 4, 3) ft A B (8, 4, 3) ft

r AB = (2 i) ft E (3, 0, 3) ft

r AC = (−6 i + j − 3k) ft r AD = (−6 i + 2 j + 3k) ft r AE = (−3i − 4 j) ft

If we divide by the respective magnitudes we can develop the unit vectors that are parallel to these position vectors. e AB = 1.00 i e AC = −0.885i + 0.147 j − 0.442k e AD = −0.857 i + 0.286 j + 0.429k e AE = −6.00 i − 0.800 j The equilibrium equation is T AB e AB + T AC e AC + T AD e AD + T AE e AE = 0. If we break this up into components, we have ΣFx : T AB − 0.885T AC − 0.857T AD − 0.600T AE = 0 ΣFy : 0.147T AC + 0.286T AD − 0.800T AE = 0 ΣFz : −0.442T AC + 0.429T AD = 0 If we set T AB = 35 lb, we cans solve for the other tensions. The result is T AC = 16.7 lb, T AD = 17.2 lb, T AE = 9.21 lb

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Problem 3.97 Cable AB is attached to the top of the vertical 3-m post, and its tension is 50 kN. What are the tensions in cables AO, AC , and AD?

y 5m 5m C

Solution:

Get the unit vectors parallel to the cables using the coordinates of the end points. Express the tensions in terms of these unit vectors, and solve the equilibrium conditions. The coordinates of points A, B, C , D, O are found from the Problem sketch: The coordinates of the points are A(6, 2, 0), B(12, 3, 0), C (0, 8, 5), D(0, 4, −5), O(0, 0, 0).

4m 8m (6, 2, 0) m B

The vector locations of these points are: z r A = 6 i + 2 j + 0 k,

rB = 12 i + 3 j + 0 k,

rD = 0 i + 4 j − 5k,

rO = 0 i + 0 j + 0 k.

12 m

rC = 0 i + 8 j + 5k,

3m x

The unit vector parallel to the tension acting between the points A, B in the direction of B is by definition r − rA e AB = B . rB − r A

Perform this for each of the unit vectors e AB = +0.9864 i + 0.1644 j + 0 k

e AC = −0.6092 i + 0.6092 j + 0.5077k 8m

e AD = −0.7442 i + 0.2481 j − 0.6202k e AO = −0.9487 i − 0.3162 j + 0 k

(6, 2, 0) m A

The tensions in the cables are expressed in terms of the unit vectors, T AB = T AB e AB = 50 e AB , T AD = T AD e AD ,

T AC = T AC e AC ,

T AO = T AO e AO .

The equilibrium conditions are ΣF = 0 = T AB + T AC + T AD + T AO = 0. Substitute and collect like terms, ΣFx = (0.9864(50) − 0.6092 T AC − 0.7422 T AD − 0.9487 T AO ) i = 0 ΣFy = (0.1644(50) + 0.6092 T AC + 0.2481 T AD − 0.3162 T AO ) j = 0 ΣFz = (+0.5077 T AC − 0.6202 T AD )k = 0. This set of simultaneous equations in the unknown forces may be solved using any of several standard algorithms. The results are: T AO = 43.3 kN,

152

T AC = 6.8 kN,

T AD = 5.5 kN.

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Problem 3.98 The 20-kg collar B is held in equilibrium on the smooth bar AC by the cable BD. The distance from A to B is 3 m. The y-axis is vertical. What is the tension in cable BD?

y C 4m

6m B A 4m

3m z

Solution:

C (0, 4, 0) m rAC

= 3i + 4 j − 6k (m). B

Dividing this vector by its magnitude yields a unit vector that points from A toward C:

The vector from A to C is

r AC = (0 − (−3)) i + (4 − 0) j + (0 − 6)k (m)

e AC =

3m A (23, 0, 6) m

r AC r AC

rBD

D (4, 0, 2) m

3i + 4 j − 6k (m) (3 m) 2 + (4 m) 2 + (−6 m) 2

Using this result, the vector from A to B is

The forces acting on the collar B are its weight W = −mg j, the force T exerted by the cable, and the normal force N exerted by the smooth bar. The equilibrium equation is

r AB = (3 m)e AC

T + W + N = 0.

= 0.384 i + 0.512 j − 0.768k.

= (3 m)(0.384 i + 0.512 j − 0.768k)

Taking the dot product of this equation with the unit vector e AC that is parallel to the bar yields

= 1.15i + 1.54 j − 2.30 k (m). Then the coordinates of the collar B are

e AC ⋅ T + e AC ⋅ W = 0,

x B = −3 m + 1.15 m = −1.85 m,

because e AC ⋅ N = 0. The only unknown in Eq. (1) is the tension T in cable BD. Eq. (1) is

y B = 0 + 1.54 m = 1.54 m, z B = 6 m − 2.30 m = 3.70 m. Now that we have these coordinates, the vector from B to D is

(1)

(0.384 i + 0.512 j − 0.768k) ⋅ [T (0.931i − 0.245 j − 0.270 k)] + (0.384 i + 0.512 j − 0.768k) ⋅ [−mg j] = 0: [(0.384)(0.931) + (0.512)(−0.245) + (−0.768)(−0.270)]T − 0.512mg = 0,

rBD = (4 − (−1.85)) i + (0 − 1.54) j + (2 − 3.70)k (m)

0.440T − 0.512mg = 0.

= 5.85i − 1.54 j − 1.70 k (m). Let T be the tension in cable BD. The force exerted on the collar B by the cable is

We obtain T = 1.16 mg = 1.16(20 kg)(9.81 m/s 2 ) = 228 N. 228 N.

r T = T BD rBD = T

5.85i − 1.54 j − 1.70 k (m) (5.85 m) 2 + (−1.54 m) 2 + (−1.70 m) 2

= T (0.931i − 0.245 j − 0.270 k).

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Problem 3.99* The 20-kg collar B is held in equilibrium on the smooth bar AC by the cable BD. The distance from A to B is 3 m. The y-axis is vertical. What is the magnitude of the normal force exerted on the collar by the smooth bar?

y C 4m

6m B A

Solution:

The vector from A to C is

3m z

r AC = (0 − (−3)) i + (4 − 0) j + (0 − 6)k (m)

= 3i + 4 j − 6k (m).

C (0, 4, 0) m

Dividing this vector by its magnitude yields a unit vector that points from A toward C:

rAC

r e AC = AC r AC

3i + 4 j − 6k (m) (3 m) 2 + (4 m) 2 + (−6 m) 2

= 0.384 i + 0.512 j − 0.768k.

3m A (23, 0, 6) m

rBD

D (4, 0, 2) m

Using this result, the vector from A to B is r AB = (3 m)e AC = (3 m)(0.384 i + 0.512 j − 0.768k) = 1.15i + 1.54 j − 2.30 k (m). Then the coordinates of the collar B are x B = −3 m + 1.15 m = −1.85 m,

because e AC ⋅ N = 0. T in cable BD.

The only unknown in Eq. (2) is the tension

Eq. (2) is (0.384 i + 0.512 j − 0.768k) ⋅ [T (0.931i − 0.245 j − 0.270 k)] +(0.384 i + 0.512 j − 0.768k) ⋅ [−mg j] = 0: [ (0.384)(0.931) + (0.512)(− 0.245) + (− 0.768)(− 0.270) ]T

−0.512mg = 0, 0.440T − 0.512mg = 0.

y B = 0 + 1.54 m = 1.54 m, z B = 6 m − 2.30 m = 3.70 m.

We obtain

Now that we have these coordinates, the vector from B to D is

T = 1.16 mg = 1.16(20 kg)(9.81 m/s 2 ) = 228 N.

rBD = (4 − (−1.85)) i + (0 − 1.54) j + (2 − 3.70)k (m)

Now that the vector T is known, we can determine the normal force N from the equilibrium Eq. (1):

= 5.85i − 1.54 j − 1.70 k (m).

N = −T − W Let T be the tension in cable BD. The force exerted on the collar by the cable is

= −T (0.931i − 0.245 j − 0.270 k) + mg j = −0.931T i + (0.245T + mg) j + 0.270T k.

r T = T BD rBD = T

= −213i + 252 j + 61.7k (N).

5.85i − 1.54 j − 1.70 k (m) (5.85 m) 2 + (−1.54 m) 2 + (−1.70 m) 2

= T (0.931i − 0.245 j − 0.270 k).

154

The magnitude of the normal force is N =

(−213) 2 + (252) 2 + (61.7) 2

= 336 N.

The forces acting on the collar B are its weight W = −mg j, the force T exerted by the cable BD, and the normal force N exerted by the smooth bar. The equilibrium equation is

As a check, we observe that

T + W + N = 0.

N ⋅ e AC = (−213)(0.384) + (252)(0.512) + (61.7)(−0.768)

(1)

= 0.

Taking the dot product of this equation with the unit vector e AC that is parallel to the bar yields

The normal force is indeed normal to the bar.

e AC ⋅ T + e AC ⋅ W = 0,

336 N.

(2)

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Chapter 4 Problem 4.1 In Practice Example 4.1, the 40-kN force points 30° above the horizontal. Suppose that the force points 30° below the horizontal instead. Draw a sketch of the beam with the new orientation of the force. What is the moment of the force about point A? Solution:

The perpendicular distance from A to the line of action of the force is unchanged D = (6 m)sin 30 ° = 3 m The magnitude of the moment is therefore unchanged M = (3 m)(40 kN) = 120 kN-m

However, with its new orientation, the force would tend to cause the beam to rotate about A in the clockwise direction. The moment is clockwise M = 120 kN-m clockwise

Problem 4.2 The mass m1 = 20 kg. The magnitude of the total moment about B due to the forces exerted on bar AB by the weights of the two suspended masses is 170 N-m. What is the magnitude of the total moment due to the forces about point A?

0.35 m

Solution:

The total moment about B is

M B = m 2 (9.81 m/s 2 )(0.35 m) + (20 kg)(9.81 m/s 2 )(0.7 m) = 170 N-m Solving, we find m 2 = 9.51 kg The moment about A is then M A = (20 kg)(9.81 m/s 2 )(0.35 m) + (9.51 kg)(9.81 m/s 2 )(0.7 m)

0.35 m

M A = 134 N-m B

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Solution:

Problem 4.3 The wheels of the overhead crane exert downward forces on the horizontal I-beam at B and C . If the force at B is 40 kip and the force at C is 44 kip, determine the sum of the moments of the forces on the beam about (a) point A , (b) point D .

Use 2-dimensional moment strategy: determine normal distance to line of action D; calculate magnitude DF; determine sign. Add moments. (a) The normal distances from A to the lines of action are D AB = 10 ft, and D AC = 35 ft. The moments are clockwise (negative). Hence, ∑ M A = −10(40) − 35(44) = −1940 ft-kip .

10 ft

25 ft

15 ft

(b) The normal distances from D to the lines of action are D DB = 40 ft, and D DC = 15 ft. The actions are positive; hence ∑ M D = +(40)(40) + (15)(44) = 2260 ft-kip

D 10 ft

25 ft

15 ft

D B

Problem 4.4 The magnitude of the vertical force exerted on the airplane’s wheel by the ground is 26.6 kN . What is the moment due to this force about the pin connection P ?

Solution:

The magnitude of the moment is

(0.36 m)(26.6 kN) = 9.58 kN-m. Its direction is counterclockwise (its value is positive). 9.58 kN-m counterclockwise.

0.48 m

0.36 m

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Problem 4.5 Two forces of equal magnitude F are applied to the wrench as shown. If a 50 N-m moment is required to loosen the nut, what is the necessary value of F ? Solution: ∑ M nut center = ( F cos30 °)(0.3 m) + ( F cos 20 °)(0.38 m) = 50 N-m F =

50 N-m = 81.1 N (0.3 m) cos30 ° + (0.38 m) cos 20 °

308 208 F

Problem 4.6 The force F = 8 kN. What is the moment of the force about point P ? Solution:

300 mm 380 mm

(3, 7) m

The angle between the force F and the x-axis is F

Q (8, 5) m

P (3, 2) m

(7, 2) m

α = tan −1 (5/4) = 51.3 ° The force can then be written F = (8 kN)(cos αi − sin α j) = (5.00 i − 6.25 j) kN The line of action of the j component passes through P, so it exerts no moment about P. The moment of the i component about P is clockwise, and its magnitude is M P = (5 m)(5.00 kN) = 25.0 kN-m

M P = 25.0 kN-m clockwise

Problem 4.7 Assume that 0 ≤ α ≤ 90 ° . The moment of F about point P is 70 ft-lb clockwise, and the moment of F about point Q is 85 ft-lb counterclockwise. Determine the magnitude of F and the angle α. y

Solution:

The components of the force are

Fx = F cos α, Fy = F sin α. The clockwise moment about P is (2 ft) Fx + (8 ft) Fy = (2 ft) F cos α + (8 ft) F sin α = 70 ft-lb,

(1)

and the counterclockwise moment about Q is

(4, 12) ft

(6 ft) Fx = (6 ft) F cos α = 85 ft-lb.

(2)

Dividing Eq. (1) by Eq. (2) yields

a (4, 6) ft

2 ft 8 ft 70 ft-lb + tan α = . 6 ft 6 ft 85 ft-lb

P (12, 4) ft x

Solving this equation yields tan α = 0.368, from which we obtain α = 20.2 °. Then from Eq. (2), F = 15.1 lb. F = 15.1 lb, α = 20.2 °. y Q 6 ft

F Fy

a Fx

2 ft

8 ft x

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Problem 4.8 The magnitude of the force exerted on the wrench is 30 lb. What moment does the force exert about point P ?

Solution:

The components of the force are

Fx = F sin 20 °,

Fy = F cos 20 °.

The counterclockwise moment about P is (2.40 in) Fx + (5.20 in) Fy = (2.40 in) F sin 20 ° + (5.20 in) F cos 20 ° = (2.40 in)(30 lb)sin 20 ° + (5.20 in)(30 lb) cos 20 °

= 171 in-lb. 171 in-lb counterclockwise.

2.40 in

y F

x 5.20 in

5.20 in

208

2.40 in

208

Problem 4.9 The length of the bar AP is 650 mm. The radius of the pulley is 120 mm. Equal forces T = 50 N are applied to the ends of the cable. What is the sum of the moments of the forces (a) about A; (b) about P ?

Solution: (a)

∑ M A = (50 N)(0.12 m) − (50 N)(0.12 m) = 0 MA = 0

(b)

∑ M P = (50 N)(0.12 m) −(50 N cos30 °)(0.65 m sin 45 ° + 0.12 m cos30 °)

458

308

−(50 N sin 30 °)(0.65 m cos 45 ° + 0.12 m sin 30 °) T

M P = −31.4 N-m or M P = 31.4 N-m CW

458

158

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Problem 4.10 The force F = 200 lb. Determine the moment about point P . F 158

508

35 in

Solution:

The components of the force are

Fx = F cos15 ° = (200 lb) cos15 ° = 193 lb, Fy = F sin15 ° = (200 lb)sin15 ° = 51.8 lb. The distances x and y are x = (40 in) cos 20 ° + (35 in) cos 50° = 60.1 in,

208 40 in

y = (40 in)sin 20° + (35 in)sin 50 ° = 40.5 in. The counterclockwise moment about P is xFy − yFx = (60.1 in)(51.8 lb) − (40.5 in)(193 lb) = −4710 in-lb. 4710 in-lb clockwise.

158 Fx

508

208 P

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Problem 4.11 The length of bar AB is 350 mm. The moments exerted about points B and C by the vertical force F are M B = −1.75 kN-m and M C = −4.20 kN-m. Determine the force F and the length of bar AC.

Solution:

We have

1.75 kN-m = F (0.35 m)sin 30 ° ⇒ F = 10 kN 4.20 kN-m = F ( L AC ) cos 20 ° ⇒ L AC = 0.447 m In summary F = 10 kN, L AC = 447 mm B

B 308

C 308

C 208

A 208

d1 308 50

0.3 m

0.450 208 d2

160

600 N

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Problem 4.12 The forces acting on a person’s forearm as he holds a suspended mass m are shown. The mass m = 10 kg and the mass of his forearm is m F = 2.2 kg. The vertical force FB is exerted by the biceps muscle. The sum of the moments of the three forces about the elbow joint E is zero. Determine FB .

FB E

mF g

Solution:

The sum of the counterclockwise moments of the three forces about point E is

∑ M B = (30 mm)FB − (130 mm)m F g − (350 mm)mg

30 mm

= (30 mm)FB − (130 mm)(2.2 kg)(9.81 m/s 2 )

220 mm

100 mm

− (350 mm)(10 kg)(9.81 m/s 2 ) = 0. Solving, we obtain FB = 1240 N. FB = 1.24 kN.

Problem 4.13 Two equal and opposite forces act on the beam. Determine the sum of the moments of the two forces (a) about point P; (b) about point Q; (c) about the point with coordinates x = 7 m, y = 5 m.

y 40 N P

308 2m

Solution: (a) (b)

308

40 N

M P = −(40 N cos30 °)(2 m) + (40 N cos30 °)(4 m)

= 69.3 N-m (CCW ) 308

308

M Q = (40 N cos30 °)(2 m) = 69.3 N-m (CCW )

M = (40 N sin 30 °)(5 m) + (40 N cos30 °)(5 m) − (40 N sin 30 °)(5 m) − (40 N cos30 °)(3 m)

(c)

= 69.3 N-m (CCW )

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Problem 4.14 The mass of the ankle weight is 1.2 kg . If you assume equilibrium, what is the moment about the knee joint at P due to the ankle weight?

Solution:

The moment is

(0.4 m)mg cos 58 ° = (0.4 m)(1.2 kg)(9.81 m/s 2 ) cos 58 ° = 2.50 N-m. Its direction is counterclockwise. 2.50 N-m counterclockwise.

P P

0.4 m

328 0.4 m mg

588

328

Problem 4.15 The magnitudes of the forces exerted on the pillar at D by the cables A, B, and C are equal: FA = FB = FC . The magnitude of the total moment about E due to the forces exerted by the three cables at D is 1350 kN-m . What is FA?

D FC

D FA 6m A

Solution:

The angles between the three cables and the pillar are

α A = tan −1 (4/6) = 33.7 ° α B = tan −1 (8/6) = 53.1 °

E 4m

αC = tan −1 (12/6) = 63.4 ° The vertical components of each force at point D exert no moment about E. Noting that FA = FB = FC , the magnitude of the moment about E due to the horizontal components is ∑ M E = FA (sin α A + sin α B + sin αC )(6 m) = 1350 kN-m Solving for FA yields FA = 100 kN

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Problem 4.16 The force F = 120 lb. The sum of the forces on the beam is zero and the sum of the moments about point A is zero. (a) Determine the values of the forces A x , A y , and B . (b) What is the sum of the moments of the forces about point B ?

y F Ax

B B

Solution:

8 ft

(a) If the sum of the forces on the beam is zero, the sum of the x components of the forces and the sum of the y components of the forces are zero:

4 ft x

∑ Fx = A x = 0, ∑ Fy = A y + B − F = 0. The sum of the moments about A is ∑ M point A = (12 ft) B − (8 ft) F = 0. Solving these equations gives A x = 0, A y = 40 lb, B = 80 lb. (b) The sum of the moments about B is ∑ M point B = −(12 ft) A y + (4 ft) F = −(12 ft)(40 lb) + (4 ft)(120 lb) = 0. (a) A x = 0, A y = 40 lb, B = 80 lb. (b) Zero.

Problem 4.17 The forces F1 = 30 N, F2 = 80 N, and F3 = 40 N. What is the sum of the moments of the forces about point A ?

Solution:

The moment about point A due to F1 is zero. Treating counterclockwise moments as positive the sum of the moments is

F3 308

A F1

∑ M A = F3 sin 30 ° (8 m) + F2 cos 45 ° (2 m) ∑ M A = 273 N-m counterclockwise

458 F2 8m

Problem 4.18 The force F1 = 60 N. The sum of the moments of the three forces about point A is 150 N-m . The sum of the moments of the three forces about point B is 50 N-m . Determine F2 and F3 . Solution:

The sum of the moments about A is

∑ M point A = (2 m) F2 cos 45 ° + (8 m) F3 sin 30 ° = 150 N-m. The sum of the moments about B is ∑ M point B = −(2 m) F1 + (2 m) F3 cos30 ° + (8 m) F3 sin 30° = 50 N-m.

F3 308

2m B

458 F2 8m

Solving these two equations with F1 = 60 N, we obtain F2 = 22.2 N, F3 = 29.7 N. F2 = 22.2 N, F3 = 29.7 N.

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Problem 4.19 The forces FA = 30 lb, FB = 40 lb, FC = 20 lb, and FD = 30 lb. What is the sum of the moments of the forces about the origin of the coordinate system?

308 FA

Solution:

The moment about the origin due to FA and FD is zero. Treating counterclockwise moments as positive, the sum of the moments is

∑ M = −FB (6 ft) + FC (10 ft) = −(40 lb)(6 ft) + (20 lb)(10 ft) = −40 ft-lb

6 ft

4 ft

∑ M = 40 ft-1b clockwise

Problem 4.20 The forces FA = 120 N and FB = 100 N. What total moment is exerted on the nut by the wrench? Solution:

The sum of the counterclockwise moments of the two

forces is ∑ M = (380 mm)FA cos30 ° + (300 mm)FB = (380 mm)(120 N) cos30 ° + (300 mm)(100 N) = 69,500 N-mm.

300 mm 380 mm

Its direction is counterclockwise (its value is positive). 69.5 N-m counterclockwise.

308 FA

Problem 4.21 Three forces act on the car. The sum of the forces is zero and the sum of the moments of the forces about point P is zero. (a) Determine the forces A and B. (b) Determine the sum of the moments of the forces about point Q.

Solution: ∑ Fy : A + B − 2800 lb = 0 ∑ M P : − (2800 lb)(6 ft) + A(9 ft) = 0 (a) Solving we find A = 1867 lb, B = 933 lb

y 6 ft

(b) ∑ M Q = (2800 lb)(3 ft) − B(9 ft) = 0

3 ft

MQ = 0

6 ft

2800 lb

3 ft

x A

P 2800 lb B

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Problem 4.22 A sailboat’s masthead fitting connects the forestay and the backstay to the top of the mast. The total force exerted on the mast by the two stays is directed straight downward and is 480 N in magnitude. Determine the moment the two stays exert about point P . 80 mm

100 mm

Solution:

Because the sum of the forces is vertical and equal to 480 N, we know that ∑ Fx = −TF sin 25° + TB sin 35 ° = 0, ∑ Fy = −TF cos 25 ° − TB cos35 ° = −480 N.

Solving, we obtain TF = 318 N, TB = 234 N. The counterclockwise moment about P is (80 mm)TF cos 25 ° − (100 mm)TB cos35 ° = 3860 N-mm. 3.86 N-m counterclockwise.

y 80 mm

358

258

100 mm x P

Problem 4.23 The members AB, BC , CD , . . . of the structure are each 1 m in length. (a) Determine the sum of the moments of the forces about A. (b) Determine the sum of the moments of the forces about B. 400 N

800 N

358

258 TF

Solution: (a) The counterclockwise moment of the forces about point A is ∑ M A = −(0.5 m)(400 N) − (1.5 m)(800 N) + (2 m)(700 N) = 0. (b) The counterclockwise moment of the forces about point B is ∑ M B = −(0.5 m)(500 N) − (1.0 m)(800 N) + (1.5 m)(700 N) = 0.

(a) Zero. (b) Zero.

800 N

400 N B A

C 500 N

700 N

E C

500 N

Problem 4.24 The tension in the cable is the same on both sides of the pulley. The sum of the moments about point A due to the 800-lb force and the forces exerted on the bar by the cable at B and C is zero. What is the tension in the cable? Solution:

Let T be the tension in the cable. The sum of the moments about A is

0.5 m

308

30 in 30 in

700 N

0.5 m

800 lb 30 in

∑ M A : T (30 in) + T sin 30 ° (90 in) − (800 lb)(60 in) = 0 Solving yields T = 640 lb

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Problem 4.25 The 160-N weights of the arms AB and BC of the robotic manipulator act at their midpoints. Determine the sum of the moments of the three weights about A.

150 mm 600

mm 208

Solution:

The strategy is to find the perpendicular distance from the points to the line of action of the forces, and determine the sum of the moments, using the appropriate sign of the action.

40 N

B m

408

The distance from A to the action line of the weight of the arm AB is:

160 N

d AB = (0.300) cos 40 ° = 0.2298 m 160 N

The distance from A to the action line of the weight of the arm BC is d BC = (0.600)(cos 40 °) + (0.300)(cos 20 °) = 0.7415 m. The distance from A to the line of action of the force is d F = (0.600)(cos 40 °) + (0.600)(cos 20 °) + (0.150)(cos 20 °) = 1.1644 m. The sum of the moments about A is ∑ M A = −d AB (160) − d BC (160) − d F (40) = −202 N-m

Problem 4.26 The Space Shuttle’s attitude thrusters exert two forces of magnitude F . They exert a 560 N-m counterclockwise moment about the center of mass G. Determine F . Solution:

The counterclockwise moment due to the aft thruster

2.2 m 2.2 m F

about G is M aft = (2.2 m) F cos6° − (12 m) F sin 6 °.

18 m

12 m

The counterclockwise moment due to the forward thruster about G is M forward = −(2.2 m) F cos 5° + (18 m) F sin 5 °. Setting M aft + M forward = (2.2 m) F cos6 ° − (12 m) F sin 6 ° − (2.2 m) F cos 5° + (18 m) F sin 5° = 560 N-m, we obtain F = 1800 N. 1.80 kN.

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Problem 4.27 The force F exerts a 200 ft-lb counterclockwise moment about A and a 100 ft-lb clockwise moment about B. What are F and θ ?

A (25, 5) ft

F u (4, 3) ft

Solution:

The strategy is to resolve F into x- and y-components, and compute the perpendicular distance to each component from A and B. The components of F are: F = iFX + jFY . The vector from A to the point of application is:

r AF = (4 − (−5)) i + (3 − 5) j = 9 i − 2 j. The perpendicular distances are d AX = 9 ft, and d AY = 2 ft, and the actions are positive. The moment about A is M A = (9) FY + (2) FX = 200 ft-lb. The vector from B to the point of application is rBF = (4 − 3) i + (3 − (−4)) j = 1i + 7 j; the distances d BX = 1 ft and d BY = 7 ft, the action of FY is positive and the action of FX is negative. The moment about B is M B = (1) FY − (7) FX = −100 ft-lb. The two simultaneous equations have solution: FY = 18.46 lb and FX = 16.92 lb. Take the ratio to find the angle:

B (3, 24) ft y F u

F  18.46 θ = tan −1  Y  = tan −1 = tan −1 (1.091) = 47.5 °.  FX  16.92

(

)

(4, 3) ft x

From the Pythagorean theorem F =

FY2 + FX2 =

18.46 2 + 16.92 2 = 25.04 lb B (3, –4) ft

Problem 4.28 The 620-lb combined weight of the person and the upper arm of the “cherry picker” lift acts at G. Determine the moment due to the weight (a) about point A; (b) about point B.

13 ft G

6 ft

Solution:

(a) The counterclockwise moment about point A is (13 ft)(620 lb) = 8060 ft-lb. (b) The counterclockwise moment about point B is (6 ft)(620 lb) = 3720 ft-lb.

9 ft

6 ft

(a) 8060 ft-lb counterclockwise. (b) 3720 ft-lb counterclockwise. B

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Problem 4.29 Five forces act on a model truss built by a civil engineering student as part of a design project. The dimensions are b = 300 mm and h = 400 mm and F = 100 N. The sum of the moments of the forces about the point where A x and A y act is zero. If the weight of the truss is negligible, what is the force B? Solution:

608

h Ax Ay

The x- and y-components of the force F are

F = − F ( i cos60 ° + j sin 60 °) = − F (0.5i + 0.866 j). The distance from A to the x-component is h and the action is positive. The distances to the y-component are 3b and 5b. The distance to B is 6b. The sum of the moments about A is ∑ M A = 2 F (0.5)(h) − 3b F (0.866) − 5b F (0.866) + 6bB = 0. Substitute and solve: B =

1.6784 F = 93.2 N 1.8

Problem 4.30 The dimensions are b = 3 ft and h = 5 ft, and F = 300 lb. The vector sum of the forces acting on the truss is zero, and the sum of the moments of the forces about the point where A x and A y act is zero. (a) Determine the forces A x , A y , and B. (b) Determine the sum of the moments of the forces about the point where the force B acts.

608

h Ax Ay

Solution:

608

(a) We know that ∑ Fx = A x − F cos60 ° − F cos60 ° = 0, ∑ Fy = A y + B − F sin 60 ° − F sin 60 ° = 0, ∑ M point A = (6b) B + hF cos60 ° − (3b) F sin 60 ° + hF cos60 ° − (5b) F sin 60 ° = 0. Solving, we obtain A x = 300 lb, A y = 257 lb, B = 263 lb. (c)

The sum of the moments about B ∑ M point B = −(6b) A y + hF cos60 ° + (3b) F sin 60 ° + hF cos60 ° + bF sin 60 ° = 0. (This is an example of an important general result. If the sum of the forces acting on an object is zero, and the sum of the moments about one point is zero, then the sum of the moments about any point is zero.) (a) A x = 300 lb, A y = 257 lb, B = 263 lb. (b) Sum of the moments is zero.

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Problem 4.31 The mass m = 70 kg. What is the moment about A due to the force exerted on the beam at B by the cable?

458

308

Solution:

The strategy is to resolve the force at B into components parallel to and normal to the beam, and solve for the moment using the normal component of the force. The force at B is to be determined from the equilibrium conditions on the cable juncture O. Angles are measured from the positive x-axis. The forces at the cable juncture are:

FOB = FOB ( i cos150 ° + j sin150 °) = FOB (−0.866 i + 0.5 j) FOC = FOC ( i cos 45 ° + j sin 45 °) = FOC (0.707 i + 0.707 j). W = (70)(9.81)(0 i − 1 j) = −686.7 j (N).

FOB

The equilibrium conditions are: ∑ Fx = (−0.866 FOB + 0.7070 FOC ) i = 0

FOC O

∑ FY = (0.500 FOB + (.707 FOC ) − 686.7) j = 0.

Solve: FOB = 502.70 N. This is used to resolve the cable tension at B: FB = 502.7( i cos330 ° + j sin 330°) = 435.4 i − 251.4 j. The distance from A to the action line of the y-component at B is 3 m, and the action is negative. The x-component at passes through A, so that the action line distance is zero. The moment at A is M A = −3(251.4) = −754.0 N-m

Problem 4.32 The weights W1 = 20 lb and W2 are suspended from the cable system shown. The cable BC is horizontal. Determine the moment about point P due to the force exerted on the vertical post at D by the cable CD. Solution:

508

6 ft

For the joint B we have the equilibrium equations W1

∑ Fx = FBC − FAB cos 50 ° = 0,

∑ Fy = FAB sin 50 ° − W1 = 0.

Solving, we obtain FAB = 26.1 lb, FBC = 16.8 lb. For the joint C we have the equilibrium equations ∑ Fx = FCD cos β − FBC = 0,

(*)

∑ Fy = FCD sin β − W2 = 0. The counterclockwise moment about P due to the force exerted by the cable CD is M P = (6 ft) FCD cos β . From Eq. (*), FCD cos β = FBC , so the moment is M P = (6 ft) FBC = 101 ft-lb. (Notice that we obtain this result without ever knowing FCD or β .)

FAB 508 y

FCD B

C FBC

b FCD 6 ft

FBC

W2 P x

101 ft-lb counterclockwise.

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Problem 4.33 The bar AB exerts a force at B that helps support the vertical retaining wall. The force is parallel to the bar. The civil engineer wants the bar to exert a 38 kN-m moment about O . What is the magnitude of the force the bar must exert?

Solution:

The strategy is to resolve the force at B into components parallel to and normal to the wall, determine the perpendicular distance from O to the line of action, and compute the moment about O in terms of the magnitude of the force exerted by the bar. By inspection, the bar forms a 3, 4, 5 triangle. The angle the bar makes with the horizontal is cos θ = 35 = 0.600, and sin θ = 45 = 0.800. The force at B is FB = FB (−0.600 i + 0.800 j). The perpendicular distance from O to the line of action of the x-component is (4 + 1) = 5 m, and the action is positive. The distance from O to the line of action of the y-component is 1 m, and the action is positive. The moment about O is ∑ M O = 5(0.600) FB + 1(0.800) FB = 3.8 FB = 38 kN, from which FB = 10 kN

4m FB A

B 1m

4m 1m

3m O 1m

Problem 4.34 The tensions in the four cables are T1 = 2.4 kN, T2 = 3.2 kN, T3 = 4.6 kN, and T4 = 6.0 kN. The lines of action of the forces T1, T2 , T3 , and T4 pass through the origin. (a) Determine the moment of each force about point P . Sum the moments to obtain the total moment about P due to the four forces. (b) Let T = T1 + T2 + T3 + T4 . Determine the components of T and calculate the moment of T about point P .

y P

T4 518

T3 408 T2 298

Solution: (a) The counterclockwise moments of the forces are (2 m) T1 cos 9 ° = (2 m)(2.4 kN) cos 9 ° = 4.74 kN-m, (2 m) T2 cos 29 ° = (2 m)(3.2 kN) cos 29 ° = 5.60 kN-m,

98 x

(2 m) T3 cos 40 ° = (2 m)(4.6 kN) cos 40 ° = 7.05 kN-m, (2 m) T4 cos 51 ° = (2 m)(6.0 kN) cos 51 ° = 7.55 kN-m. Their sum is 24.9 kN-m. (b) Summing the x components and the y components of the forces, we obtain T = 12.5i + 9.55 j kN. The counterclockwise moment of this force about P is (2 m)(12.5 kN) = 24.9 kN-m. (a) 4.74 kN-m, 5.60 kN-m, 7.05 kN-m, 7.55 kN-m counterclockwise. Total moment is 24.9 kN-m counterclockwise. (b) T = 12.5i + 9.55 j (kN-m). Moment is 24.9 kN-m counterclockwise.

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Problem 4.35 The bar BC is 6 m in length. The mass of the suspended load is 260 kg . The cable supporting the suspended load exerts a downward force on the bar at B. Determine the moment about point C due to the downward force. B

Solution:

The distance from C to D is (6 m)sin 30 ° = 3 m. The magnitude of the moment due to the weight of the suspended mass about C is (3 m)mg = (3 m)(260 kg)(9.81 m/s 2 ) = 7650 N. Its direction is clockwise. 7.65 kN clockwise.

B 308 A

mg 308

308 C

Problem 4.36 The bar BC is 12 ft in length. The suspended load weighs 800 lb . If the weight of the bar is negligible, the sum of the moments about point C due to the forces exerted by the two cables at B is zero. What is the tension in cable AB? B

Solution:

The distance from C to D is CD = (12 ft)sin 30 ° = 6 ft. The distance from B to D is BD = (12 ft) cos30 ° = 10.4 ft. The moment of the two forces about C is BD(T AB cos30 °) − CD(T AB sin 30°) − CD(800 lb) = 0. Solving yields T AB = 800 lb. 800 lb.

B 308

TAB

800 lb

308 308 A

308 C

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Problem 4.37 The cable AB exerts a 290-kN force on the crane’s boom at B . The cable AC exerts a 148-kN force on the boom at C . Determine the sum of the moments about P due to the forces the cables AB and AC exert on the boom.

Solution: ∑MP = −

∑ M P = 3.36 MN-m CW

C G

Boom

290

38 m

8 14

16 m

40 m

16 m

8 (148 kN)(16 m) 320

= −3.36 MNm

A B

8 (290 kN)(56 m) − 3200

56 m

Problem 4.38 A single cable goes from B through a pulley at A to C. The tension in the cable is 120 kN. The sum of the moments about P due to the forces the pulley exerts at B and C and the weight acting at G is zero. What is the weight at G ?

Boom

∑ M P = −(56 m)T sin α − (16 m)T sin β + (38 m)W = 0. Solving, we obtain W = 47.6 kN.

C G

The angles α = arctan(8 m/56 m) = 8.13 °, β = arctan (8 m/16 m) = 26.6 °. The sum of the moments about P is

47.6 kN.

A B

Solution:

16 m

T G

38 m 56 m

172

P W

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Problem 4.39 The mass of the luggage carrier and the suitcase combined is 12 kg. Their weight acts at A. The sum of the moments about the origin of the coordinate system due to the weight acting at A and the vertical force F applied to the handle of the luggage carrier is zero. Determine the force F (a) if α = 30°; (b) if α = 50°. Solution:

O is the origin of the coordinate system

∑ M O = F(1.2 m cos α) −(12 kg)(9.81 m/s 2 )(0.28cos α − 0.14 sin α)

0.28 m

= 0

0.14 m 1.2 m

Solving we find F =

(12 kg)(9.81 m/s 2 )(0.28cos α − 0.14 sin α) 1.2 m cos α

a C

(a) For α = 30 ° We find F = 19.54 N (b) For α = 50 ° We find F = 11.10 N

Problem 4.40 The system is in equilibrium. Gas pressure exerts a 2500-lb downward force on the top of the stationary piston. The piston is held in equilibrium in the smooth cylinder by the force exerted at C by the connecting rod BC . The force is directed along the line from B to C , and the connecting rod exerts an equal and opposite force at B on the crankshaft AB. What is the moment about A due to the force exerted on the crankshaft at B?

Solution:

Let F denote the force exerted on the piston at C by the connecting rod. The distance from B to D is BD = (2 in)sin 45 °. The angle α is α = arcsin

45 °   = 16.4 °. ( BD ) = arcsin  (2 in)sin BC 5 in 

From the sum of the vertical forces on the piston, ∑ Fy = F cos α − 2500 lb = 0, we see that F = 2610 lb.

The counterclockwise moment about A due to the force exerted by the connecting rod on the crank at B is [(2 in) cos 45 °]F sin α + [(2 in)sin 45 °]F cos α = 4580 in-lb. 4580 in-lb counterclockwise.

5 in

2500 lb

458 B

a 5 in F

y D

458 2 in A

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Problem 4.41 The hydraulic piston AB exerts a 400-lb force on the ladder at B in the direction parallel to the piston. The sum of the moments about C due to the force exerted on the ladder by the piston and the weight W of the ladder is zero. What is the weight of the ladder? Solution:

6 ft W

3 ft

The angle between the piston AB and the horizontal is

α = tan −1 (3/6) = 26.6 °

B C

The sum of the counterclockwise moment about C is

6 ft

3 ft

∑ M C : W (6 ft) − (400 lb) cos α(3 ft) − (400 lb)sin α(3 ft) = 0 Solving yields W = 268 lb

Problem 4.42 The hydraulic cylinder exerts an 8-kN force at B that is parallel to the cylinder and points from C toward B . Determine the moments of the force about points A and D .

1m D C

Hydraulic cylinder

Solution:

Use x, y coords with origin A. We need the unit vector from C to B, e CB . From the geometry,

0.6 m

e CB = 0.780 i − 0.625 j The force FCB is given by

0.6 m

Scoop

0.15 m

FCB = (0.780)8i − (0.625)8 j kN FCB = 6.25i − 5.00 j kN For the moments about A and D, treat the components of FCB as two separate forces. a + M A = (5,00)(0.15) − (0.6)(6.25) kN ⋅ m a + M A = −3.00 kN ⋅ m

5.00 kN 6.25 kN

C (20.15, 10.6)

For the moment about D

a + ∑ M D = (5 kN)(1 m) + (6.25 kN)(0.4 m)

0.6 m

a + M D = 7.5 kN ⋅ m

0.15 m A (0 , 0)

5.0 kN

D 0,4 m

C 6.25 kN

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Problem 4.43 The structure shown in the diagram is one of two identical structures that support the scoop of the excavator. The bar BC exerts a 700-N force at C that points from C toward B . What is the moment of this force about K ?

320 mm Shaft 100 mm

Scoop

260 mm

B 180 260 mm mm J

D 160 mm

K 380 mm

1040 mm 1120 mm

Solution: MK = −

320

320 (700 N)(0.52 m) = −353 Nm 108800

M K = 353 Nm CW 520 mm

700 N

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Problem 4.44 The bar BC exerts a force at C that points from C toward B . The hydraulic cylinder DH exerts a 1550-N force at D that points from D toward H. The sum of the moments of these two forces about K is zero. What is the magnitude of the force that bar BC exerts at C ?

320 mm C

Shaft 100 mm

Scoop

260 mm

B 180 260 mm mm J

D 160 mm

K 380 mm

1040 mm 1120 mm

Solution:

320

320 F (0.52) = 0 108800

260 mm

1120 (1550 N)(0.26 m) − 1264400

∑MK =

Solving we find

1120

F = 796 N

100 260 mm

1550 N

Problem 4.45 A force vector F = Fx i + Fy j with magnitude F = 20 N acts at a point ( x, y, 0). The moment of the force about the origin is M = 40 k (N-m). (a) If you view the coordinate system so that the unit vector k points toward you, is the moment clockwise or counterclockwise? (b) What is the perpendicular distance from the origin to the line of action of F?

Solution: (a) With the coordinate system oriented as described, if you point the thumb of your right hand in the direction of M, it points toward you. The moment is counterclockwise. (b) The perpendicular distance is D =

M 40 N-m = = 2 m. F 20 N

(a) Counterclockwise. (b) 2 m.

Problem 4.46 Use Eq. (4.2) to determine the moment of the 80-N force about the origin O letting r be the vector (a) from O to A; (b) from O to B.

y 80j (N) B

Solution:

(6, 4, 0) m

(a) M O = rOA × F = 6 i × 80 j = 480 k (N-m). (b) M O = rOB × F = (6 i + 4 j) × 80 j = 480 k (N-m).

176

x A (6, 0, 0) m

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Problem 4.47 The magnitude of the force exerted on the javelin is F = 360 N , and the angle of the vector relative to the horizontal is 50 °. The coordinates of the point where the athlete is holding the javelin are x = 0.52 m, y = 0.14 m. Use Eq. (4.2) to determine the moment the force exerts about the shoulder joint at O. y

Solution:

The force in terms of its components is

F = (360 N)(− cos 50 °i + sin 50 ° j) = −231i + 276 j (N). The vector from O to the point where the javelin is held is r = 0.52 i + 0.14 j (m). The moment is MO = r × F =

i j k 0.52 m 0.14 m 0 −231 N 276 N 0

= [(0.52 m)(276 N) − (0.14 m)(−231 N)]k = 176k (N-m).

M O = 176k (N-m).

Problem 4.48 Use Eq. (4.2) to determine the moment of the 100-kN force (a) about A; (b) about B. y

Solution: (a) The coordinates of A are (0, 6, 0). The coordinates of the point of application of the force are (8, 0, 0). The position vector from A to the point of application of the force is r AF = (8 − 0) i + (0 − 6) j = 8i − 6 j. The force is F = 100 j (kN). The cross product is

A r AF × F = 100j (kN)

i j k 8 −6 0 0 100 0

= 800 k (kN-m)

(b) The coordinates of B are (12, 0, 0). The position vector from B to the point of application of the force is rBF = (8 − 12) i = −4 i. The cross product is:

8m 12 m

rBF × F =

Problem 4.49 The cable AB exerts a 200-N force on the support at A that points from A toward B. Use Eq. (4.2) to determine the moment of this force about point P , (a) letting r be the vector from P to A; (b) letting r be the vector from P to B.

P (0.9, 0.8) m

= −400 k (kN-m)

Solution:

First we express the force as a vector. The force points in the same direction as the position vector AB. AB = (1 − 0.3) mi + (0.2 − 0.5) mj = (0.7 i − 0.3 j) m AB F

i j k −4 0 0 0 100 0

(0.7 m) 2 + (0.3 m) 2 =

0.58 m

200 N = (0.7 i − 0.3 j) 0.58

(a) M P = PA × F = (−0.6 mi − 0.3 mj) × Carrying out the cross product we find

200 N (0.7 i − 0.3 j) 0.58

M P = 102.4 N-mk

(0.3, 0.5) m

(b) M P = PB × F = (0.1 mi − 0.6 mj) ×

Carrying out the cross product we find B

200 N (0.7 i − 0.3 j) 0.58

M P = 102.4 N-mk

(1, 0.2) m x

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Problem 4.50 The line of action of F is contained in the x– y plane. The moment of F about O is 190 k (N-m), and the moment of F about A is 330 k (N-m). What are the components of F? y

Solution:

F = Fx i + Fy j . The moment of F about O is MO =

A (0, 7, 0) m

Write the force F as

i j k 5m 3m 0 Fx Fy 0

= [(5 m) Fy − (3 m) Fx ]k.

The moment of F about A is MA =

(5, 3, 0) m x O

i j k 5 m −4 m 0 Fx Fy 0

= [(5 m) Fy − (−4 m) Fx ]k.

Solving the equations (5 m) Fy − (3 m) Fx = 190 N-m, (5 m) Fy − (−4 m) Fx = 330 N-m, we obtain Fx = 20 N, Fy = 50 N. F = 20 i + 50 j (N).

Problem 4.51 Use Eq. (4.2) to determine the sum of the moments of the three forces (a) about A; (b) about B.

y 6 kN

Solution: (a)

3 kN

3 kN B

M A = 0.2 i × (−3 j) + 0.4 i × 6 j + 0.6 i × (−3 j) = O.

(b)

M B = (−0.2 i) × (−3 j) + (−0.4 i) × 6 j + (−0.6 i) × (−3 j)

0.2 m

= O.

Problem 4.52 (a) The coordinates of point A are (0, 3, 0) m. Use Eq. (4.2) to determine the sum of the moments of the two forces about the origin O. (b) P is an arbitrary point with coordinates (in meters) (x, y, z ). Use Eq. (4.2) to determine the sum of the moments of the two forces about P .

Solution: (a) The force acting at O exerts no moment about O. The moment due to the top force is MO = (b)

i j k 0 3m 0 0 0 −20 N

= 60 k (N-m).

In terms of the coordinates of point P, the position vectors rO /P = −xi − yj − zk, r A /P = −xi + (3 m − y ) j − zk.

220i (N)

The sum of the moments of the two forces about A is P

M P = rO /P × (20 N) i + r A /P × (−20 N) i =

O 20i (N)

i j k −x −y −z 20 N 0 0

j k i −x 3 m − y −z −20 N 0 0

= −20 zj + 20 yk + 20 zj + 20(3 m − y )k = 60 k (N-m). (a), (b) 60 k (N-m).

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Problem 4.53 Three forces act on the plate. Use Eq. (4.2) to determine the sum of the moments of the three forces about point P .

y 4 kN 458

Solution: r1 = (−0.12 i + 0.08 j) m, F1 = (4 cos 45 °i + 4 sin 45 ° j) kN

3 kN

r2 = (0.16 i) m, F2 = (3cos30 °i + 3sin 30 ° j) kN

308 0.18 m

r3 = (0.16 i − 0.1 j) m, F3 = (12 cos 20 °i − 12sin 20 ° j) kN

P 0.10 m

M P = r1 × F1 + r2 × F2 + r3 × F3 M P = (0.145 kN-m)k = (145 N-m)k

208

0.12 m

12 kN

0.28 m

Problem 4.54 (a) Determine the magnitude of the moment of the 150-N force about A by calculating the perpendicular distance from A to the line of action of the force. (b) Use Eq. (4.2) to determine the magnitude of the moment of the 150-N force about A.

150k (N)

(0, 6, 0) m

Solution: A

(a) The perpendicular from A to the line of action of the force lies in the x−y plane d =

(6, 0, 0) m

6 2 + 6 2 = 8.485 m

M = dF = (8.485)(150) = 1270 N-m z

(b) M = (−6 i + 6 j) × (150 k) = −900 j + 900 i N-m M =

900 2 + 900 2 = 1270 N-m

Problem 4.55 The forces exerted on the runway by the airplane’s five wheels are F A = 8.22k (kN), FB = 7.94 k (kN), and FC = 1.12k (kN). Determine the sum of the moments of the three forces about the origin. Solution:

The moment is

M = r A × F A + rB × FB + rC × FC i =

0 1.5 m

+ 0 −1.5 m

8.22 kN

+ 1.86 m 0

7.94 kN

0 1.12 kN

= (1.5)(8.22) i + (−1.5 m)(7.94) i − (1.86)(1.12) j (kN-m)

Source: Courtesy of Miglena Pencheva/Shutterstock.

= 0.420 i − 2.08 j (kN-m).

1.5 m

0.420 i − 2.08 j (kN-m).

1.5 m

1.86 m

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Problem 4.56 What is the magnitude of the moment of F about point B? Solution:

F 5 20i 1 10j 2 10k (lb)

A (4, 4, 2) ft

The position vector from B to A is

rBA = [(4 − 8) i + (4 − 1) j + (2 − (−2))k] ft rBA = (−4 i + 3 j + 4 k) ft

B (8, 1, 22) ft x

The moment of F about B is M B = rBA × F =

i j k −4 3 4 20 10 −10

= (−70 i + 40 j − 100 k) ft-lb

Its magnitude is MB =

(−70 ft-lb) 2 + (40 ft-lb) 2 + (−100 ft-lb) 2 = 128 ft-lb

M B = 128 ft-lb

Problem 4.57 The tension in the cable CD is 6 kN . What is the magnitude of the moment about A due to the force exerted on the bar by the cable at C ?

y C (0, 2, 21) m

eCD 1m

D x (1.5, 0, 0) m

We obtain the force in terms of its components by multiplying this unit vector by 6 kN:

F = (6 kN)e CD A z

= 3.34 i − 4.46 j + 2.23k (kN). D

1.5 m

(0, 2, 21) m

Solution:

y C

rAC

(0, 2, 21) m A

rCD

D (1.5, 0, 0) m x

The vector from A to C is

D (1.5, 0, 0) m

r AC = 2 j − k (m).

The vector rCD from point B to point D is

The moment of the force about A is M A = r AC × F

rCD = (1.5 m − 0) i + (0 − 2 m) j + (0 − (−1 m))k = 1.5i − 2 j + k (m). Its magnitude is rCD =

= [(2)(2.23) − (−1)(−4.46)]i − [0 − (−1)(3.34)] j

(1.5 m) 2 + (−2 m) 2 + (1 m) 2

+ [0 − (2)(3.34)]k (kN-m)

= 2.69 m. By dividing this vector by its magnitude, we obtain a unit vector that points from point C toward point D: 1.5i − 2 j + k (m) e CD = 2.69 m = 0.557 i − 0.743 j + 0.371k.

180

i j k −1 m 0 2m 3.34 kN −4.46 kN 2.23 kN

= −3.34 j − 6.69k (kN-m). Its magnitude is MA =

(−3.34 kN-m) 2 + (−3.34 kN-m) 2 = 7.47 kN-m.

M A = 7.47 kN-m.

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Problem 4.58 The magnitude of the moment of F about A is 1200 ft-lb. What is the magnitude of F?

B (5, 6, 1) ft

Solution: We denote the magnitude of the force by F = F . By dividing the position vector of C relative to B by its magnitude to obtain a unit vector in the direction of F, we can express F in terms of its components: F =

rC /B −2 i − 6 j + 3k (ft) F = F = −0.286 Fi − 0.857 Fj + 0.429 Fk. −2 i − 6 j + 3k (ft) rC /B

The vector from A to B is C (3, 0, 4) ft

r = 5i + 6 j + k (ft).

The moment of F about A is MA = r × F =

i j 5 ft 6 ft −0.286 F −0.857 F

k 1 ft 0.429 F

= 3.43Fi − 2.43Fj − 2.57 Fk. Setting M A 2 = (3.43F ) 2 + (−2.43F ) 2 + (−2.57 F ) 2 = (1200 ft-lb) 2 , we obtain F = 244 lb. F = 244 lb.

Problem 4.59 The force F = 30 i + 20 j − 10 k (N). (a) Determine the magnitude of the moment of F about A. (b) Suppose that you can change the direction of F while keeping its magnitude constant, and you want to choose a direction that maximizes the moment of F about A. What is the magnitude of the resulting maximum moment?

Solution:

The vector from A to the point of application of F is

r = 4 i − 1 j − 7k m and r =

4 2 + 1 2 + 7 2 = 8.12 m

(a) The moment of F about A is

MA = r × F =

y F

MA = (8, 2, 24) m

i j k 4 −1 −7 30 20 −10

= 150 i − 170 j + 110 k N-m

150 2 + 170 2 + 110 2 = 252 N-m

(b) The maximum moment occurs when r ⊥ F. In this case M Amax = r F

(4, 3, 3) m x

Hence, we need F . F =

30 2 + 20 2 + 10 2 = 37.4 (N)

Thus, M Amax = (8.12)(37.4) = 304 N-m

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Problem 4.60 The person exerts a force F = −14 i − 12 j − k lb on the lug wrench. The force she exerts effectively acts at the point with coordinates x = 8 in, y = −10 in, and z = 4 in. Determine the moment of F about the lug nut located at the origin. Solution:

y x

The moment of F about the origin is M = r × F,

where r = 8i − 10 j + 4 k in and F = −14 i − 12 j − k lb: F

M = r×F =

i j k 8 −10 4 −14 −12 −1

= 58i − 48 j − 236k (in-lb). 58i − 48 j − 236k (in-lb).

Problem 4.61 The force F exerted on the grip of the exercise machine points in the direction of the unit vector e = 23 i − 23 j + 13 k , and its magnitude is 120 N. Determine the magnitude of the moment of F about the origin O .

Solution:

The vector from O to the point of application of the

force is r = 0.25i + 0.2 j − 0.15k m and the force is F = F e or F = 80 i − 80 j + 40 k N. The moment of F about O is

150 mm

MO = r × F =

i j k 0.25 0.2 −0.15 80 −80 40

N-m

or M O = −4 i − 22 j − 36k N-m

200 mm

and

250 mm

MO = x

182

4 2 + 22 2 + 36 2 N-m

M O = 42.4 N-m

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Problem 4.62 The force F points in the direction of the unit vector e = 23 i − 23 j + 13 k. The support at O will safely support a moment of 560 N-m magnitude. (a) Based on this criterion, what is the largest safe magnitude of F? (b) If the force F may be exerted in any direction, what is its largest safe magnitude?

Solution:

See the figure of Problem 4.61.

The moment in Problem 4.61 can be written as MO =

i j k 0.25 0.2 −0.15 2 − 23 F + 13 F 3F

where F = F

M O = (−0.0333i − 0.1833 j − 0.3k)F And the magnitude of M O is

150 mm

M O = ( 0.0333 2 + 0.1833 2 + 0.3 2 ) F M O = 0.353 F If we set M O = 560 N-m, we can solve for Fmax 560 = 0.353 Fmax F

(b) If F can be in any direction, then the worst case is when r ⊥ F. The moment in this case is M O = r Fworst

200 mm

Fmax = 1586 N

250 mm

r = x

560 = (0.3536) FWORST = 1584 N

Fworst

Problem 4.63 A civil engineer in Boulder, Colorado, estimates that under the severest expected Chinook winds, the total force on the highway sign will be F = 2.8i − 1.8 j (kN). Let M O be the moment due to F about the base O of the cylindrical column supporting the sign. The y component of M O is called the torsion exerted on the cylindrical column at the base, and the component of M O parallel to the x−z plane is called the bending moment. Determine the magnitudes of the torsion and bending moment.

Solution:

0.25 2 + 0.2 2 + 0.15 2 = 0.3536 m

y F

O x

The total moment is

M = (8 j + 8k) m × (2.8i − 1.8 j) kN = (14.4 i + 22.4 j − 22.4 k) kN-m

We now identify Torsion = M y = 22.4 kN-m Bending moment = =

M x2 + M z2 (14.4 kNm) 2 + (22.4 kNm) 2 = 26.6 kN-m

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Problem 4.64 The weights of the arms OA and AB of the robotic manipulator act at their midpoints. The direction cosines of the centerline of arm OA are cos θ x = 0.500, cos θ y = 0.866, and cos θ z = 0, and the direction cosines of the centerline of arm AB are cos θ x = 0.707, cos θ y = 0.619, and cos θ z = −0.342. What is the sum of the moments about O due to the two forces?

600 mm

Solution:

By definition, the direction cosines are the scalar components of the unit vectors. Thus the unit vectors are e 1 = 0.5i + 0.866 j, and e 2 = 0.707 i + 0.619 j − 0.342k. The position vectors of the midpoints of the arms are

160 N

200 N

r1 = 0.3e 1 = 0.3(0.5i + 0.866 j) = 0.15i + 0.2598 j r2 = 0.6e 1 + 0.3e 2 = 0.512 i + 0.7053 j − 0.1026k.

The sum of moments is M = r1 × W1 + r2 × W2 =

i j k 0.15 0.2598 0 −200 0 0

j k i 0.512 0.7053 −0.1026 −160 0 0

= −16.42 i − 111.92k (N-m)

Problem 4.65 The tension in cable AB is 100 lb. If you want the magnitude of the moment due to the forces exerted on the tree by the two ropes about the base O of the tree to be 1500 ft-lb, what is the necessary tension in rope AC?

(0, 8, 0) ft

Solution: F1 =

We have the forces

T AC 100 lb (−8 j + 10 k), F2 = (14 i − 8 j + 14 k) 164 456

Thus the total moment is

M = (8 ft) j × (F1 + F2 ) = (625 ft lb + 5.24 ft T AC ) i − (5.24 ft)T AC K The magnitude squared is then (625 ft lb + 5.24 ft T AC ) 2 + (5.24 ft T AC ) 2 = (1500 ft lb) 2

B z

(0, 0, 10) ft

(14, 0, 14) ft C

Solving we find T AC = 134 lb

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Problem 4.66 A driver exerts two forces on the car’s steering wheel. The force F A = 2.02 i − 2.43 j + 1.42k (N) is applied at point A with coordinates (53.0, 183, −15.0) mm. The force FB = −3.84 i + 2.02 j − 0.844 k (N) is applied at point B with coordinates (328, 360, − 56.0) mm. The center of the wheel P has coordinates (233, 206, 0) mm. Determine the components of the sum of the moments of the two forces about P . Solution:

y B

P FA

The position vectors of points A and B relative to point

P are x

rPA = r A − rP = (53.0 − 233) i + (183 − 206) j + (−15.0 − 0)k (mm) = −180 i − 23.0 j − 15.0 k (mm), rPB = rB − rP

= (328 − 233) i + (360 − 206) j + (−56.0 − 0)k (mm) = 95.0 i + 154 j − 56.0 k (mm).

B rPB

The sum of the moments about P is M = rPA × F A + rPB × FB

rPA

    i j k i j k        =  −180 mm −23.0 mm −15.0 mm  +  95.0 mm 154 mm −56.0 mm   2.02 N  −3.84 N 2.02 N −0.844 N  −2.43 N 1.42 N      = −86.0 i + 521 j + 1270 k (N-mm).

−0.0860 i + 0.521 j + 1.27k (N-m).

Problem 4.67 The force F = 5i (kN) acts on the ring A where the cables AB, AC , and AD are joined. What is the sum of the moments about point D due to the force F and the three forces exerted on the ring by the cables? Strategy: The ring is in equilibrium. Use what you know about the four forces acting on it. Solution:

y D (0, 6, 0) m A (12, 4, 2) m

The vector from D to A is

rDA = 12 i − 2 j + 2k m.

(6, 0, 0) m

(0, 4, 6) m

The sum of the moments about point D is given by ∑ M D = rDA × F AD + rDA × F AC + rDA × F AB + rDA × F

∑ M D = rDA × (F AD + F AC + F AB + F) However, we are given that ring A is in equilibrium and this implies that

FAD A

(F AD + F AC + F AB + F) = O = 0 Thus, ∑ M D = rDA × (O) = 0

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FAB

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Problem 4.68 In Problem 4.67, determine the moment about point D due to the force exerted on the ring A by the cable AB.

D(0, 6, 0) FAD

A(12, 4, 2) m

FAC

F 5 5i (kN)

D (0, 6, 0) m A

C(0, 4, 6) m

B(6, 0, 0) m

(12, 4, 2) m In component form, we get B

(6, 0, 0) m

(0, 4, 6) m z

Solving, we get

Solution: We need to write the forces as magnitudes times the appropriate unit vectors, write the equilibrium equations for A in component form, and then solve the resulting three equations for the three unknown magnitudes. The unit vectors are of the form e AP =

 i: −0.802 FAB − 0.949 FAC − 0.973FAD + 5 = 0   j: −0.535FAB + (0) FAC + 0.162 FAD = 0   k: −0.267 FAB + 0.316 FAC − 0.162 FAD = 0 

( x P − x A ) i + ( y P − y A ) j + ( z P − z A )k r AP

FAB = 779.5 N, FAC = 1976 N FAD = 2569 N The vector from D to A is rDA = 12 i − 2 j + 2k m The force F AB is given by

Where P takes on values B, C , and D

F AB = FAB e AB

Calculating the unit vectors, we get

F AB = −0.625i − 0.417 j − 0.208k (kN)

 e AB = −0.802 i − 0.535 j − 0.267k   e AC = −0.949 i + 0 j + 0.316k   e AD = −0.973i + 0.162 j − 0.162k 

The moment about D is given by

M D = rDA × F AB =

From equilibrium, we have

i j k 12 −2 2 −0.625 −0.417 −0.208

FAB e AB + FAC e AC + FAD e AD + 5i (kN) = 0

M D = 1.25i + 1.25 j − 6.25k (kN-m)

Problem 4.69 The magnitude of the force F A is 80 N . The length of the straight line from the origin to point P is 6 m. Determine the components of the moment of F A about point P .

Solution:

The force

F A = (80 N)(cos 50 ° cos 20 °i + sin 50 ° j + cos 50 ° sin 20 °k). = 48.3i + 61.3 j + 17.6k (N). The vector from the origin to point P is rOP = (6 m)(− cos 45 ° sin15 °i + sin 45 ° j − cos 45 ° cos15 °k)

= −1.10 i + 4.24 j − 4.10 k (m). FB

The moment of F A about P is

M P = −rOP × F A   i j k     1.10 m 4.24 m 4.10 m =  −    48.3 N 61.3 N 17.6 N   = −326 i + 179 j + 272k (N-m).

458 158

608 158

FA O

(Notice that we needed to take the cross product using the vector from P to O.)

508

−326 i + 179 j + 272k (N-m).

208

rOP O z

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Problem 4.70 The magnitude of the force FB is 120 N . The length of the straight line from the origin to point P is 6 m. (a) Determine the magnitude of the moment of FB about point P . (b) What is the length of the shortest straight line from point P to the line of action of FB?

458 158

608

FA O

158

508 208

Solution: y

(a) The force

FB = (120 N)(− cos60° sin15°i + sin 60° j + cos60° cos15 °k). = −15.5i + 104 j + 58.0 k (N). The vector from the origin to point P is

rOP

rOP = (6 m)(− cos 45 ° sin15 °i + sin 45 ° j − cos 45 ° cos15 °k) O

= −1.10 i + 4.24 j − 4.10 k (m). The moment of FB about P is

M P = −rOP × FB   i j k    =  1.10 m −4.24 m 4.10 m   −15.5 N 104 N 58.0 N    = −672 i − 127 j + 48.2k (N-m).

(Notice that we needed to take the cross product using the vector from P to O.) The magnitude of the moment is M P = 685 N-m. (b)

Let θ denote the angle between the vectors rOP and FB : From the figure we can see that the length of the shortest line from P to the line of action of FB is rOP sin θ. Furthermore, we can use the definition of the dot product of the vectors rOP and FB to determine θ. Observe that

r ⋅ FB cos θ = OP rOP FB =

P rOP

FB u

O z

(−1.10 m)(−15.5 N) + (4.24 m)(104 N) + (−4.10 m)(58.0 N) (−1.10 m) 2 + (4.24 m) 2 + (−4.10 m) 2 (−15.5 N) 2 + (104 N) 2 + (58.0 N) 2

= 0.306, which yields θ = 72.2 °. The length of the shortest line is rOP sin θ = (6 m)sin 72.2 ° = 5.71 m. (a) 685 N-m. (b) 5.71 m.

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Problem 4.71 The collar A is at the midpoint of the bar. Assume that the tension in cable AB is 100 lb. Determine the components of the moment about point C due to the force exerted on the collar by the cable.

y (0, 6, 0) ft D

Solution:

(0, 4, 4) ft

Because it is midway between points C and D, the coordinates of the collar A are (4, 3, 1.5) ft. Therefore the vector from C to A is

rCA = (4 − 8) i + (3 − 0) j + (1.5 − 3)k (ft) C

= −4 i + 3 j − 1.5k (ft),

(8, 0, 3) ft

and the vector from A to B is r AB = (0 − 4) i + (4 − 3) j + (4 − 1.5)k (ft) = −4 i + j + 2.5k (ft).

We can express the force exerted on the collar by the cable in terms of its components as T = (100 lb)

r AB r AB

= (100 lb)(−0.830 i + 0.207 j + 0.518k). The moment of the force about C is

M C = rCA × T   i j k   =  −4 ft 3 ft −1.5 ft   −83.0 lb 20.7 lb 51.8 lb    = 187 i + 332 j + 166k (ft-lb).

A rAB rCA

x C

187 i + 332 j + 166k (ft-lb).

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Problem 4.72* The 100-lb collar A is held in equilibrium on the smooth bar by the cable AB. (The positive y-axis is upward.) The collar is at the midpoint of the bar. Determine the magnitude of the moment about point C due to the force exerted on the collar by the cable.

Let us divide the vector rCA by its magnitude to obtain a unit vector parallel to the bar: rCA = rCA

−4 i + 3 j − 1.5k (ft) (−4 ft) 2 + (3 ft) 2 + (−1.5 ft) 2 = −0.766 i + 0.575 j − 0.287k.

e CA =

The dot product of this unit vector with the equilibrium equation yields the equation

y (0, 6, 0) ft

e CA ⋅ T + e CA ⋅ W = 0,

because e CA ⋅ N = 0. The only unknown in this equation is the tension T in the cable.

(0, 4, 4) ft

In terms of components, A

e CA ⋅ T + e CA ⋅ W = 0: (−0.766)(−0.830T ) + (0.575)(0.207T ) + (−0.287)(0.518T )

+ (0.575)(−100 lb) = 0.

C (8, 0, 3) ft

Solving yields T = 94.9 lb. This gives T = −78.7 i + 19.7 j + 49.2k lb.

Solution:

The moment of this force about C is M C = rCA × T

  i j k   =  −4 ft 3 ft −1.5 ft   −78.7 lb 19.7 lb 49.2 lb    = 177 i + 315 j + 157k (ft-lb).

Its magnitude is M C = 394 ft-lb.

A rAB

394 ft-lb.

rCA

x C

z Because it is midway between points C and D, the coordinates of the collar A are (4, 3, 1.5) ft. Therefore the vector from C to A is rCA = (4 − 8) i + (3 − 0) j + (1.5 − 3)k (ft) = −4 i + 3 j − 1.5k (ft), and the vector from A to B is r AB = (0 − 4) i + (4 − 3) j + (4 − 1.5)k (ft) = −4 i + j + 2.5k (ft). The forces exerted on the collar are its weight W = −100 j (lb), the force T exerted by the cable, and the normal force N exerted by the bar: The equilibrium equation for the collar is T + W + N = 0. We can express the force exerted by the cable in terms of its components as T = T

r AB r AB

= T (−0.830 i + 0.207 j + 0.518k).

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Problem 4.73 The tension in the cable BD is 1 kN. As a result, cable BD exerts a 1-kN force on the “ball” at B that points from B toward D . Determine the moment of this force about point A.

Solution: F =

We have the force and position vectors

1 kN (−4 i + 2 j + 4 k), r = AB = (4 i + 3 j + k) m 6

The moment is then M = r × F = (1.667 i − 3.33 j + 3.33k) kN-m

y C

(0, 4, 23) m B (4, 3, 1) m

D (0, 5, 5) m

x A E z

Problem 4.74* Suppose that the mass of the suspended object E is 100 kg and the mass of the bar AB is 20 kg . Assume that the weight of the bar acts at its midpoint. If the sum of the moments about point A due to the weight of the bar and the forces exerted on the “ball” at B by the three cables BC , BD, and BE is zero, determine the tensions in the cables BC and BD.

F1 = −(100 kg)(9.81 m/s 2 ) j, F3 =

F2 =

TBC (−4 i + j − 4 k), 33

TBD (−4 i + 2 j + 4 k) 6

In addition we have the weight of the bar F4 = −(20 kg)(9.81 m/s 2 ) j The moment around point A is + (2 i + 1.5 j + 0.5k) m × F4 = 0

Carrying out the cross products and breaking into components we find (0, 4, 23) m

M x = 1079 − 2.26TBC + 1.667TBD = 0 M y = 2.089TBC − 3.333TBD = 0

B (4, 3, 1) m

(0, 5, 5) m

We have the following forces applied at point B.

M A = (4 i + 3 j + k) m × (F1 + F2 + F3 )

Solution:

M z = −4316 + 2.785TBC + 3.333TBD = 0 Only two of these three equations are independent. Solving we find x

TBC = 886 N, TBD = 555 N

A E z

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Problem 4.75 The 200-kg slider at A is held in place on the smooth vertical bar by the cable AB. Determine the moment about the bottom of the bar (point C with coordinates x = 2 m, y = z = 0 ) due to the force exerted on the slider by the cable.

Solution:

The slider is in equilibrium. The smooth bar exerts no vertical forces on the slider. Hence, the vertical component of F AB supports the weight of the slider. The unit vector from A to B is determined from the coordinates of points A and B A(2, 2, 0), B(0, 5, 2) m Thus, r AB = −2 i + 3 j + 2k m

e AB = −0.485i + 0.728 j + 0.485k

and F AB = FAB e AB

The horizontal force exerted by the bar on the slider is H = H xi + H zk Equilibrium requires H + F AB − mg j = 0 i : H x − 0.485FAB = 0 m = 200 kg

j : 0.728FAB − mg = 0

k : H z + 0.485FAB = 0 2m

g = 9.81 m/s 2

Solving, we get x

FAB = 2697N = 2, 70 kN H x = 1308N = 1.31 kN H z = −1308N = −1.31 kN rCA = 2 j m F AB = FAB e AB F AB = −1308i + 1962 j + 1308k N

Mc =

−1308 1962 1308 M c = 2616 i + 0 j + 2616k N-m M c = 2.62 i + 2.62 i kN-m

FAB

2mg j

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Problem 4.76 To evaluate the adequacy of the design of the vertical steel post, you must determine the moment about the bottom of the post due to the force exerted on the post at B by the cable AB. A calibrated strain gauge mounted on cable AC indicates that the tension in cable AC is 22 kN . What is the moment?

y 5m 5m C

4m 8m (6, 2, 0) m

Solution:

To find the moment, we must find the force in cable AB. In order to do this, we must find the forces in cables AO and AD also. This requires that we solve the equilibrium problem at A.

O z

Our first task is to write unit vectors e AB , e AO , e AC , and e AD . Each will be of the form e Ai =

12 m

( x i − x A ) i + ( y i − y A ) j + ( z i − z A )k ( x i − x A ) 2 + ( y i − y A ) 2 + (z i − z A ) 2

where i takes on the values B, C , D, and O. We get

D(0, 4,25) m

e AB = 0.986 i + 0.164 j + 0 k

C(0, 8, 5) m

e AC = −0.609 i + 0.609 j + 0.508k e AO = −0.949 i − 0.316 j + 0 k

TAO

We now write the forces as T AB = T AB e AB T AC = T AC e AC T AD = T AD e AD T AO = T AO e AO We then sum the forces and write the equilibrium equations in component form. For equilibrium at A, ∑ F A = 0 ∑ F A = T AB + T AC + T AD + T AO = 0. In component form,  T e + T AC e ACx + T AD e ADx + T AO e AOx = 0  AB ABx T e  AB ABy + T AC e ACy + T AD e ADy + T AO e AOy = 0   T AB e ABz + T AC e ACz + T AD e ADz + T AO e AOz = 0 

TAD

TAC

e AD = −0.744 i + 0.248 j − 0.620 k

192

A(6, 2, 0) m

B(12, 3, 0) m

TAB

O(0, 0, 0) m We know T AC = 22 kN. Substituting this in, we have 3 eqns in 3 unknowns. Solving, we get T AB = 163.05 kN, T AD = 18.01 kN T AO = 141.28 kN We now know that T AB is given as T AB = T AB e AB = 160.8i + 26.8 j (kN) and that the force acting at B is (−T AB ). The moment about the bottom of the post is given by M BOTTOM = r × (−T AB ) = 3 j × (−T AB ) Solving, we get M BOTTOM = 482k (kN-m)

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Problem 4.77 The force F = 20 i + 40 j − 10 k (N). Use both of the procedures described in Example 4.7 to determine the moment due to F about the z-axis.

Solution:

First Method: We can use Eqs. (4.5) and (4.6)

r = (8i) m F = (20 i + 40 j − 10 k) N M z − axis = [k ⋅ (r × F)]k M z − axis

= k ⋅ ( r × F) =

= 320 N-m

20 N 40 N − 10 N M z − axis = (320 N-m)k F x (8, 0, 0) m

Second Method: The y-component of the force is perpendicular to the plane containing the z-axis and the position vector r. The perpendicular distance from the z-axis to the y-component of the force is 8 m. Therefore M z-axis = (40 N)(8 m) = 320 N-m Using the right-hand rule we see that the moment is about the +z-axis. Thus

M z-axis = (320 N-m)k

Problem 4.78 Use Eqs. (4.5) and (4.6) to determine the moment of the 20-N force about (a) the x-axis, (b) the y-axis, (c) the z-axis. (First see if you can write down the results without using the equations.)

Solution:

The force is parallel to the z-axis. The perpendicular distance from the x-axis to the line of action of the force is 4 m. The perpendicular distance from the y-axis is 7 m and the perpendicular distance from the z-axis is 4 2 + 7 2 = 65 m. By inspection, the moment about the x-axis is M x = (4)(20) i (N-m)

M x = 80 i (N-m) By inspection, the moment about the y-axis is M y = (7)(20)(− j) N-m

(7, 4, 0) m

M y = −140 j (N-m) By inspection, the moment about the z-axis is zero since F is parallel to the z-axis.

20 k (N) x

M z = 0 (N-m) Now for the calculations using (4.5) and (4.6)

BandF_6e_ISM_CH04.indd 193

M L = [e ⋅ (r × F)]e Mx =

1 0 0 7 4 0 i = 80 i (N-m) 0 0 20

My =

0 1 0 7 4 0 0 0 20

Mz =

0 0 1 7 4 0 k = 0 k (N-m) 0 0 20

j = −140 j (N-m)

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Problem 4.79 Three forces parallel to the y-axis act on the rectangular plate. Use Eqs. (4.5) and (4.6) to determine the sum of the moments of the forces about the x-axis. (First see if you can write down the result without using the equations.)

∑ M x = [(2)(0.6) − (6)(0.6)]i kN

Calculating the result:

3 kN x

6 kN

M 3 kN =

1 0 0 0 0 0 i = 0 i kN 0 −3 0

M 2 kN =

1 0 0 0 0 .6 i = 1.2 i kN 0 −2 0

M 6 kN =

1 0 0 0 0 .6 i = −3.6 i kN 0 6 0

600 mm

900 mm

By inspection, the 3 kN force has no moment about the x-axis since it acts through the x-axis. The perpendicular distances of the other two forces from the x-axis is 0.6 m. The H 2 kN force has a positive moment and the 6 kN force has a negative about the x-axis.

∑ M x = −2.4 i kN

2 kN

Solution:

∑ M x = M 3 kN + M 2 kN + M 6 kN ∑ M x = 0 + 1.2 i − 3.6 i (kN) ∑ M x = −2.4 i (kN)

Problem 4.80 The three forces are parallel to the y-axis. Determine the sum of the moments of the forces (a) about the y-axis; (b) about the z-axis.

Solution: (a) The magnitude of the moments about the y-axis is M = e Y ⋅ (r × F). The position vectors of the three forces are given in the solution to Problem 4.79. The magnitude for each force is: e Y ⋅ ( r × F) =

y 3 kN x

2 kN

6 kN 900 mm

0.9

0 0

−3 0

= 0,

e Y ⋅ ( r × F) =

0 1 0 0.9 0 0.6 0 6 0

= 0,

e Y ⋅ ( r × F) =

0 1 0 0 0 0.6 0 −2 0

= 0

600 mm

Thus the moment about the y-axis is zero, since the magnitude of each moment is zero. (b) The magnitude of each moment about the z-axis is e Z ⋅ (r × F) =

e Z ⋅ (r × F) =

0.9

0 0

−3 0

0 0 1 0.9 0 0.6 0+ 6 0 0

0 1

0 0.6

= −2.7,

= 5.4,

= 0.

0 −2 0 Thus the moment about the z-axis is ∑ M Z = −2.7e Z + 5.4 e Z = 2.7k (kN-m)

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Problem 4.81 The person exerts a force F = 0.2 i − 0.4 j + 1.2k (lb) on the gate at C . Point C lies in the x – y plane. What moment does the person exert about the gate’s hinge axis, which is coincident with the y-axis ? y

Solution: M = [e ⋅ (r × F)] e e = j, r = 2 i ft, F is given  0 1 0    0 0  j = −2.4 j (ft-lb) My =  2    .2 −.4 1.2 

A C 3.5 ft

Problem 4.82 ponents are

2 ft

Four forces act on the plate. Their com-

Solution:

Note that F A acts at the origin so no moment is generated about the origin. For the other forces we have

F A = −2 i + 4 j + 2k (kN), FB = 3 j − 3k (kN),

i j k 3m 0 0 0 3 kN −3 kN

MO =

FC = 2 j + 3k (kN), FD = 2 i + 6 j + 4 k (kN). Determine the sum of the moments of the forces (a) about the x-axis; (b) about the z-axis.

j k i 3m 0 2m 0 2 kN 3 kN

i j k 0 0 2m 2 kN 6 kN 4 kN

M O = (−16 i + 4 j + 15k) kN-m y

Now we find M x = M O ⋅ i = −16 kN-m, M z = M O ⋅ k = 15 kN-m

FB FA FD

x 2m

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Problem 4.83 The force F = 30 i + 20 j − 10 k (lb). (a) What is the moment of F about the y-axis? (b) Suppose that you keep the magnitude of F fixed, but you change its direction so as to make the moment of F about the y-axis as large as possible. What is the magnitude of the resulting moment?

Solution: (a)

M y = j ⋅ [(4 i + 2 j + 2k) ft × (30 i + 20 j − 10 k) lb] 0 1 0 4 ft 2 ft 2 ft 30 lb 20 lb −10 lb

My =

= 100 ft lb

⇒ M y = (100 ft-lb) j y (b)

M ymax = Fd = ( 30 2 + 20 2 + 10 2 lb)( 4 2 + 2 2 ft) = 167.3 ft-lb

Note that d is the distance from the y-axis, not the distance from the origin. F (4, 2, 2) ft x

Problem 4.84 The moment of the force F about the x-axis is −80 i (ft-lb), the moment about the y-axis is zero, and the moment about the z-axis is 160 k (ft-lb). If Fy = 80 lb, what are Fx and Fz ? y

Solution:

The magnitudes of the moments:

e ⋅ (r × F) =

e Z ⋅ (r × F) =

eX rX FX 0 4 FX

eY rY FY

eZ rZ FZ

0 1 2 −2 80 FZ

= 320 − 2 FX = 160

Solve: FX = 80 lb, FZ = 40 lb, from which the force vector is F = 80 i + 80 j + 40 k F (4, 2, 2) ft z

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Problem 4.85 The robotic manipulator is stationary. The weights of the arms AB and BC act at their midpoints. The direction cosines of the centerline of arm AB are cos θ x = 0.500, cos θ y = 0.866, cos θ z = 0, and the direction cosines of the centerline of arm BC are cos θ x = 0.707, cos θ y = 0.619, cos θ z = −0.342. What total moment is exerted about the z-axis by the weights of the arms?

Solution:

The unit vectors along AB and AC are of the form

e = cos θ x i + cos θ y j + cos θ z k. The unit vectors are e AB = 0.500 i + 0.866 j + 0 k and e BC = 0.707 i + 0.619 j − 0.342k. The vector to point G at the center of arm AB is r AG = 300(0.500 i + 0.866 j + 0 k) = 150 i + 259.8 j + 0 k mm, and the vector from A to the point H at the center of arm BC is given by r AH = r AB + rBH = 600 e AB + 300 e BC = 512.1i + 705.3 j − 102.6k mm.

0 60

e • ( r × F) =

160 N

The weight vectors acting at G and H are WG = −200 j N, and WH = −160 j N. The moment vectors of these forces about the z-axis are of the form eX

rX FX

rY FY

rZ FZ

Here, WG and WH take on the role of F, and e = k. 600 mm

Substituting into the form for the moment of the force at G, we get 200 N

e • (r × F ) =

0 0 1 0.150 0.260 0 0 −200 0

= −30 N-m.

Similarly, for the moment of the force at H, we get A

e • ( r × F) =

Problem 4.86 In Problem 4.85, what total moment is exerted about the x-axis by the weights of the arms?

Solution:

The solution is identical to that of Problem 4.85 except that e = i. Substituting into the form for the moment of the force at G, we get e ⋅ (r × F) =

600 mm

160 N

1 0 0 0.150 0.260 0 0 −200 0

= 0 N-m.

Similarly, for the moment of the force at H, we get e ⋅ (r × F) =

= −81.9 N-m.

The total moment about the z-axis is the sum of the two moments. Hence, M z -axis = −111.9 N-m

0 0 1 0.512 0.705 −0.103 0 −160 0

1 0 0 0.512 0.705 −0.103 0 −160 0

= −16.4 N-m.

The total moment about the x-axis is the sum of the two moments. Hence, M x-axis = −16.4 N-m

200 N

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Problem 4.87 In Practice Example 4.6, suppose that the force changes to F = −2 i + 3 j + 6k (kN). Determine the magnitude of the moment of the force about the axis of the bar BC .

Solution:

We have the following vectors

rBA = (4 i + 2 j − 1k) m F = (−2 i + 3 j + 6k) kN rBC = (4 j − 3k) m r e BC = BC = (0.8 j − 0.6k) rBC The moment of F about the axis of the bar is M BC = e BC ⋅ (r × F) =

0 0.8 −0.6 4 2 −1 −2 3 6

= −27.2 kN-m

Thus M BC = (−27.2 kN-m)e BC , M BC = 27.2 kN-m

Problem 4.88 The person exerts a force F = −14 i − 12 j − k lb on the lug wrench. The force she exerts effectively acts at the point with coordinates x = 8 in, y = −10 in, and z = 4 in. Determine the moment of F about the z-axis, which is the moment that tightens the lug nut. y

Solution: Because the origin is a point on the z-axis, we can use the result of part (a) and apply Eq. (4.4). The moment of F about the z-axis is M z -axis = (k ⋅ M)k = [k ⋅ (58i − 48 j − 236k)]k = −236k (in-lb). Notice that we could have obtained this result from the result of part (a) without using Eq. (4.4). The moment of F about the z-axis is the component of M that is parallel to the z-axis, which is −236k (in-lb). −236k (in-lb).

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Problem 4.89 The force F = −10 i + 5 j − 5k (kip). Determine the moment of F about the line AB. Draw a sketch to indicate the direction of the moment. y

y B

(6, 6, 0) ft

x A (6, 0, 0) ft

(6, 6, 0) ft z F A

(6, 0, 0) ft 230j (ft-kip)

Solution:

Direction of moment

The moment of F about pt. A is

M A = −6 i × F =

i j k −6 0 0 −10 5 −5

A z

= −30 j − 30 k (ft-kip). The unit vector j is parallel to line AB, so the moment about AB is M AB = ( j ⋅ M A ) j = −30 j (ft-kip).

Problem 4.90 The force F = −15i + 10 j + 20 k (kN). Determine the magnitude of the moment of F about the line OA.

A F

Solution:

The vector from the origin O to the point of application

(–2, 6, 3) m

of F is r = −2 i + 6 j + 3k (m). z

  i j k   6m 3 m  =  −2 m  −15 kN 10 kN 20 kN    = 90 i − 5 j + 70 k (kN-m).

y A

The components of a unit vector e OA that points from origin toward point A is e OA = cos60 ° sin 70 °i + sin 60° j + cos60° cos 70°k

(22, 6, 3) m

= 0.470 i + 0.866 j + 0.171k.

= 49.9 (kN-m).

eOA

The component of M O in the direction of e OA is e OA ⋅ M O = (0.470)(90 kN-m) + (0.866)(−5 kN-m) + (0.171)(70 kN-m)

708

The moment due to F about O is MO = r × F

608

49.9 kN-m.

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Problem 4.91 The weight of the hinged rectangular door causes an 80-lb tension in the rope AB. Determine the moment about the x-axis due to the force the rope exerts on the door at B.

Solution: y A

(15, 30, 0) in

rBA

15 in

C x (60, 0, 0) in

rCB

z 30 in

The coordinates of point B are x

[60 in, −(30 in)sin 35 °, (30 in) cos35 °] = (60, − 17.2, 24.6) in. The vector from B to A is

358

30 in

rBA = (15 − 60) i + (30 + 17.2) j + (0 − 24.6)k (in) = −45i + 47.2 j − 24.6k (in).

z B

60 in

We can express the force exerted at B by the rope AB as T = (80 lb)

rBA rBA

= (80 lb)(−0.646 i + 0.677 j − 0.353k). The vector from C to B is rCB = (60 − 60) i + (−17.2 − 0) j + (24.6 − 0)k (in) = −17.2 j + 24.6k (in). The moment due to T about C is M C = rCB × T   i j k    = (80 lb)  −17.2 in 24.6 in  0  −0.646 −0.353  0.677   = −846 i − 1270 j − 889k (in-lb). The moment about the x-axis is −846 i (in-lb). −846 i (in-lb).

Problem 4.92 Determine the moment of the force applied at D about the straight line through the hinges A and B. (The line through A and B lies in the y –z plane.)

Solution: From the figure, we see that the unit vector along the line from A toward B is given by e AB = − sin 20 ° j + cos 20 °k. The position vector is r AD = 4 i ft, and the force vector is as shown in the figure. The moment vector of a force about an axis is of the form

20i 2 60j (lb) A

200

ez rZ .

For this case, 0 x

e ⋅ (r × F) =

4 ft

2 ft z

ey rY

FX FY FZ

6 ft

eX e • ( r × F) = r X

B 20

C 4 ft

− sin 20 ° cos 20 °

− 60

= −240 cos 20 ° ft-lb

= −225.5 ft-lb. The negative sign is because the moment is opposite in direction to the unit vector from A to B.

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Problem 4.93 The tension in the cable CE is 160 lb. Determine the moment of the force exerted by the cable on the hatch at C about the straight line through the hinges A and B. y

6 ft

e • ( r × F) =

20i 2 60j (lb) A

4 ft B 20

From the figure, we see that the unit vector along the line from A toward B is given by e AB = − sin 20 ° j + cos 20 °k. The position vector is rBC = 4 i ft. The coordinates of point C are (4, −4 sin 20 °, 4 cos 20 °). The unit vector along CE is −0.703i + 0.592 j + 0.394 k and the force vector is as shown acting at point D. The moment vector is a force about an axis is of the form

2 ft z

Solution:

rX FX

rY FY

rZ FZ

For this case, rCE = −4 i + 3.368 j + 2.242k TCE = 160 e CE = −112.488i + 94.715 j + 63.049k

e ⋅ (r × F) =

4 ft

0 − sin 20° cos 20° 4 0 0 −112.488 94.715 63.049

= −240 cos 20° ft-lb

= 701 ft-lbs.

Problem 4.94 The coordinates of A are (−2.4, 0, −0.6) m, and the coordinates of B are (−2.2, 0.7, −1.2) m. The force exerted at B by the sailboat’s main sheet AB is 130 N. Determine the moment of the force about the centerline of the mast (the y-axis). Draw a sketch to indicate the direction of the moment.

Solution:

The position vectors:

rOA = −2.4 i − 0.6k (m), rOB = −2.2 i + 0.7 j − 1.2k (m), rBA = (−2.4 + 2.2) i + (0 − 0.7) j + (−0.6 + 1.2)k (m) = −0.2 i − 0.7 j + 0.6k (m). The magnitude is rBA = 0.9434 m.

The unit vector parallel to BA is e BA = −0.2120 i − 0.7420 j + 0.6360 k. The tension is TBA = 130 e BA . The moment of TBA about the origin is M O = rOB × TBA = x

i j k −2.2 0.7 −1.2 , −27.56 −96.46 82.68

or M O = −57.88i + 214.97 j + 231.5k.

The magnitude of the moment about the y-axis is

M Y = e Y ⋅ M O = 214.97 N-m. z

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The moment is M Y = e Y (214.97) = 214.97 j N-m.

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Problem 4.95 The radius of the disk is 0.4 m . The spring AB is unstretched when the angle α = 0. The spring constant is k = 200 N/m. Determine the moment about the x-axis due to the force the spring exerts on the disk at B if α = 60 °.

Solution: y

rBA

y A

608

rCB B

x C

a x z

0.4 m The coordinates of point B are ( B x , B y , B z ) = [1 m, (0.4 m) cos α, −0.4 m sin α] = (1 m, 0.2 m, −0.346 m). z

The vector from the center C of the disk to B is rCB = B y j + B z k. The vector from B to A is

1.0 m

rBA = −(1 m) i + [0.4 m − (0.4 m) cos α] j + [(0.4 m)sin α]k = −i + 0.2 j + 0.346k (m). The unstretched length of the spring is 1.0 m, so the magnitude of the force exerted at B by the spring is F = k ( rBA − 1.0 m) = 15.4 N. The force in terms of its components is F = F

rBA rBA

= −14.3i + 2.86 j + 4.96k (N). The moment due to F about C is   i  M = rCB × F =  0   Fx 

 j k   By Bz  .  Fy Fz  

Its x component is M x = ( B y Fz − B z Fy ) i = 1.98i (N-m). 1.98i (N-m).

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Problem 4.96 The total force exerted on the blades of the turbine by the steam nozzle is F = 20 i − 120 j + 100 k (N), and it effectively acts at the point (100, 80, 300) mm. What moment is exerted about the axis of the turbine (the x-axis )?

Solution: MO =

The moment about the origin is

i j k 0.1 0.08 0.3 20 −120 100

= 44.0 i − 4.0 j − 13.6k (N-m).

The moment about the x-axis is

Fixed Rotating

(M O ⋅ i) i = 44.0 i (N-m).

Problem 4.97 The pneumatic support AB holds a trunk lid in place. It exerts a 35-N force on the fixture at B that points in the direction from A toward B . Determine the magnitude of the moment of the force about the hinge axis of the lid, which is the z-axis.

Solution:

The vector from A to B is

r AB = [(60 − 480) i + (100 − (−40)) j + (−30 − 40)k] mm r AB = (−420 i + 140 j − 70 k) mm The 35-N force can be written F = (35 N)

r AB = (−32.8i + 10.9 j − 5.47k) N r AB

The moment about point O is

M O = rOB × F =

i j k 60 100 −30 −32.8 10.9 −5.47

= (−219 i + 1310 j + 3940 k) N-mm The magnitude of the moment about the z-axis is M z = M O ⋅ k = 3940 N-mm = 3.94 N-m

M z = 3.94 N-m B (60, 100, 230) mm

x A (480, 240, 40) mm

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Problem 4.98 The tension in cable AB is 80 lb. What is the moment about the line CD due to the force exerted by the cable on the wall at B? y 8 ft 3 ft

Solution:

The strategy is to find the moment about the point C exerted by the force at B, and then to find the component of that moment acting along the line CD. The coordinates of the points B, C , D are B (8, 6, 0), C (3, 6, 0), D (3, 0, 0). The position vectors are: rOB = 8i + 6 j, rOC = 3i + 6 j, rOD = 3i. The vector parallel to CD is rCD = rOD − rOC = −6 j. The unit vector parallel to CD is e CD = −1 j. The vector from point C to B is rCB = rOB − rOC = 5i. The position vector of A is rOA = 6 i + 10 k. The vector parallel to BA is rBA = rOA − rOB = −2 i − 6 j + 10 k. The magnitude is rBA = 11.832 ft. The unit vector parallel to BA is

B C

e BA = −0.1690 i − 0.5071 j + 0.8452k.

6 ft

The tension acting at B is x

TBA = 80 e BA = −13.52 i − 40.57 j + 67.62k. The magnitude of the moment about CD due to the tension acting at B is

A (6, 0, 10) ft

M CD = e CD ⋅ (rCB × TBA ) =

0 −1 0 5 0 0 −13.52 −40.57 67.62

= 338.1 (ft lb). The moment about CD is M CD = 338.1e CD = −338.1 j (ft lb). The sense of the moment is along the curled fingers of the right hand when the thumb is parallel to CD, pointing toward D.

Problem 4.99 The magnitude of the force F is 0.2 N and its direction cosines are cos θ x = 0.727, cos θ y = −0.364, and cos θ z = 0.582. Determine the magnitude of the moment of F about the axis AB of the spool. y

Solution:

We have

r AB = (0.3i − 0.1 j − 0.4 k) m, r AB =

(0.3) 2 + (0.1 2 ) + (0.4) 2 m =

e AB =

1 (0.3i − 0.1 j − 0.4 k) 0.26

0.26 m

F = 0.2 N(0.727 i − 0.364 j + 0.582k)

B (200, 400, 0) mm

r AP = (0.26 i − 0.025 j − 0.11k) m

(160, 475, 290) mm

Now the magnitude of the moment about the spool axis AB is

A (2100, 500, 400) mm

F x

M AB =

0.2 N 0.26

0.3 −0.1 −0.4 0.26 m −0.025 m −0.11 m 0.727 −0.364 0.582

= 0.0146 N-m

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Problem 4.100 A motorist applies the two forces shown to loosen a lug nut. The direction cosines of F are 3 cos θ x = 134 , cos θ y = 12 13 , and cos θ z = 13 . If the magnitude of the moment about the x-axis must be 32 ft-lb to loosen the nut, what is the magnitude of the forces the motorist must apply? (See Example 4.8.)

Solution: The unit vectors for the forces are the direction cosines. The position vector of the force F is rOF = −1.333k ft. The magnitude of the moment due to F is M OF = e X ⋅ (rOF × F) =

1 0 0.3077 F

0 0 0.9231F

0 −1.333 0.2308F

M OF = 1.230 F ft lb. y

The magnitude of the moment due to −F is M −OF = e X ⋅ (r−OF × −F)

F =

1 0 0 0 0 1.333 −.3077 F −0.9231F −0.2308F

= 1.230 F ft lb.

The total moment about the x-axis is

∑ M X = 1.230 Fi + 1.230 Fi = 2.46 Fi, from which, for a total magnitude of 32 ft lb, the force to be applied is 16 in

16 in

Problem 4.101 The tension in cable AB is 2 kN. What is the magnitude of the moment about the shaft CD due to the force exerted by the cable at A? Draw a sketch to indicate the direction of the moment about the shaft. 2m

F =

32 = 13 lb 2.46

Solution:

The strategy is to determine the moment about C due to A, and determine the component parallel to CD. The moment is determined from the distance CA and the components of the tension, which is to be found from the magnitude of the tension and the unit vector parallel to AB. The coordinates of the points A, B, C , and D are: A (2, 2, 0), B (3, 0, 1), C (0, 2, 0), and D (0, 0, 0). The unit vector parallel to CD is by inspection e CD = −1 j. The position vectors parallel to DC , DA, and DB: rDC = 2 j, rDA = 2 i + 2 j, rDB = 3i + 1k. The vector parallel to CA is rCA = 2 i. The vector parallel to AB is

r AB = rDB − rDA = 1i − 2 j + 1k. D

The magnitude: r AB = 2.4495 m. The unit vector parallel to AB is e AB = 0.4082 i − 0.8165 j + 0.4082k. The tension is T AB = 2e AB = 0.8165i − 1.633 j + 0.8165k. The magnitude of the moment about CD is

M CD = e CD ⋅ (rCA × T AB ) =

0 −1 0 2 0 0 0.8164 −1.633 0.8165

= 1.633 kN-m. The moment about CD is M CD = e CD M CD = −1.633 j (kN-m). The sense is in the direction of the curled fingers of the right hand when the thumb is parallel to DC, pointed toward D.

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Problem 4.102 The axis of the car’s wheel passes through the origin of the coordinate system and its direction cosines are cos θ x = 0.940, cos θ y = 0, and cos θ z = 0.342. The force exerted on the tire by the road effectively acts at the point x = 0, y = −0.36 m, and z = 0 and has components F = −720 i + 3660 j + 1240 k (N). What is the moment of F about the wheel’s axis?

Solution:

We have to determine the moment about the axle where a unit vector along the axle is e = cos θ x i + cos θ y j + cos θ z k e = 0.940 i + 0 j + 0.342k The vector from the origin to the point of contact with the road is r = 0 i − 0.36 j + 0 k m The force exerted at the point of contact is F = −720 i + 3660 j + 1240 k N The moment of the force F about the axle is M AXLE = [e ⋅ (r × F)]e

0.940 0 0.342 0 0 (0.940 i + 0.342k) (N-m) −0.36 −720 +3660 +1240

M AXLE =

M AXLE = (−508.26)(0.940 i + 0.342k) (N-m) M AXLE = −478i − 174 k (N-m)

Problem 4.103 The direction cosines of the centerline OA are cos θ x = 0.500, cos θ y = 0.866, and cos θ z = 0, and the direction cosines of the line AG are cos θ x = 0.707, cos θ y = 0.619, and cos θ z = −0.342. What is the moment about OA due to the 250-N weight? Draw a sketch to indicate the direction of the moment about the shaft.

Solution:

By definition, the direction cosines are the scalar components of the unit vectors. Thus the unit vectors are e 1 = 0.5i + 0.866 j, and e 2 = 0.707 i + 0.619 j − 0.341k. The force is W = 250 j (N). The position vector of the 250 N weight is rW = 0.600 e 1 + 0.750 e 2 = 0.8303i + 0.9839 j − 0.2565k The moment about OA is M OA = e OA (e OA ⋅ (rW × W))

m 0m

250 N

0.5 0.866 0 0.8303 0.9839 −0.2565 e 1 = −32.06e 1 0 0 −250

= −16 i − 27.77 j (N-m) The moment is anti parallel to the unit vector parallel to OA, with the sense of the moment in the direction of the curled fingers when the thumb of the right hand is directed oppositely to the direction of the unit vector.

600 mm

O z

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Problem 4.104 A driver exerts two forces on the car’s steering wheel. The force F A = 7.20 i − 9.35 j + 5.00 k (oz) is applied at point A with coordinates (2.08, 7.20, − 0.600) in. The force FB = −13.0 i + 7.20 j − 4.30 k (oz) is applied at point B with coordinates (12.9, 14.2, − 2.20) in. The center of the wheel P has coordinates (9.20, 8.10, 0) in. The steering wheel’s shaft is parallel to the unit vector 0.342 j + 0.940 k. What is the magnitude of the sum of the moments of the two forces about the shaft?

FB B

P FA

Solution:

The position vectors of points A and B relative to point

P are

rPA = r A − rP = (2.08 − 9.20) i + (7.20 − 8.10) j + (−0.600 − 0)k (in) = −7.12 i − 0.900 j − 0.600 k (in),

rPB

rPB = rB − rP = (12.9 − 9.20) i + (14.2 − 8.10) j + (−2.20 − 0)k (in)

rPA

= 3.70 i + 6.10 j − 2.20 k (in).

The sum of the moments about P is M = rPA × F A + rPB × FB    i j k i    =  −7.12 in −0.900 in −0.600 in  +  3.70 mm  7.20 oz −9.35 oz  −13.0 oz 5.00 oz    

 j k  6.10 in −2.20 in  7.20 oz −4.30 oz  

= −20.5i + 75.8 j + 179k (in-oz). The dot product with the unit vector parallel to the shaft is e ⋅ M = (0)(−20.5) + (0.342)(75.8) + (0.940)(179) in-oz = 194 in-oz (12.1 in-lb). 12.1 in-lb.

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Problem 4.105* The magnitude of the force F is 10 N. Suppose that you want to choose the direction of the force F so that the magnitude of its moment about the line L is a maximum. Determine the components of F and the magnitude of its moment about L. (There are two solutions for F.) y

Solution: The moment of the general force F = Fx i + Fy j + Fz k about the line is developed by e BA =

3i + 6 j − 6k 1 = ( i + 2 j − 2k), 9 3

rBP = (12 i + 2 j − 2k) m, M BA = e BA ⋅ (rBP × F)

A (3, 8, 0) m

This expression simplifies to M BA = −

22 m ( Fy + Fz ) 3

We also have the constraint that (10 N) 2 = Fx2 + Fy2 + Fz2 L

Since Fx does not contribute to the moment we set it equal to zero. Solving the constraint equation for Fz and substituting this into the expression for the moment we find

F B (0, 2, 6) m

(12, 4, 4) m

M BA = − x

22 (F ± 3 y

100 − Fy2 ). ⇒

dM BA = 0 dFy

⇒ Fy = ±5 2 N ⇒ Fz = ±5 2 We thus have two answers: F = (7.07 j + 7.07k) N or

208

F = −(7.07 j + 7.07k)

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Problem 4.106 The weight W causes a tension of 100 lb in cable CD. If d = 2 ft, what is the moment about the z-axis due to the force exerted by the cable CD at point C ?

(12, 10, 0) ft (0, 3, 0) ft

Solution:

The strategy is to use the unit vector parallel to the bar to locate point C relative to the origin, and then use this location to find the unit vector parallel to the cable CD. With the tension resolved into components about the origin, the moment about the origin can be resolved into components along the z-axis. Denote the top of the bar by T and the bottom of the bar by B. The position vectors of the ends of the bar are:

z (3, 0, 10) ft The position vector of point D is rOD = 0 i + 3 j + 0 k.

rOB = 3i + 0 j + 10 k, rOT = 12 i + 10 j + 0 k.

The vector parallel to CD is

The vector from the bottom to the top of the bar is

rCD = rOD − rOC = −4.074 i + 1.807 j − 8.807k.

rBT = rOT − rOB = 9 i + 10 j − 10 k.

The magnitude is

The magnitude:

rCD =

rBT =

The unit vector parallel to CD is

9 2 + 10 2 + 10 2 = 16.763 ft.

4.074 2 + 1.807 2 + 8.807 2 = 9.87 ft.

The unit vector parallel to the bar, pointing toward the top, is

e CD = −0.4127 i + 0.1831 j − 0.8923k.

e BT = 0.5369 i + 0.5965 j − 0.5965k.

The tension is

The position vector of the point C relative to the bottom of the bar is

TCD = 100 e CD = −41.27 i + 18.31 j − 89.23k lb.

rBC = 2e BT = 1.074 i + 1.193 j − 1.193k.

The magnitude of the moment about the z-axis is

The position vector of point C relative to the origin is rOC = rOB + rBC = 4.074 i + 1.193 j + 8.807k.

M O = e Z ⋅ (rOC × TCD ) =

0 0 1 4.074 1.193 8.807 −41.27 18.31 −89.23

= 123.83 ft lb

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Problem 4.107* The y-axis points upward. The weight of the 4-kg rectangular plate acts at the midpoint G of the plate. The sum of the moments about the straight line through the supports A and B due to the weight of the plate and the force exerted on the plate by the cable CD is zero. What is the tension in the cable?

Solution:

Note that the coordinates of point G are (150, 152.5, 195).

We calculate the moment about the line BA due to the two forces as follows. e BA =

0.1i + 0.07 j − 0.36k 0.1445

r1 = (0.2 i − 0.125 j + 0.03k) m,

y A

(100, 500, 700) mm

F1 = TCD (100, 250, 0) mm

(−0.1i + 0.445 j + 0.31k) 0.304125

r2 = (0.15i − 0.0275 j − 0.165k) m,

F2 = −(4 kg)(9.81 m/s 2 ) j M BA = e BA ⋅ (r1 × F1 + r2 × F2 )

G x

The moment reduces to M BA = 3.871 N-m − (0.17793 m)TCD = 0 ⇒ TCD = 21.8 N

(0, 180, 360) mm C

(200, 55, 390) mm

Problem 4.108 In Practice Example 4.9, suppose that the point of application of the force F is moved from (8, 3, 0) m to (8, 8, 0) m. Draw a sketch showing the new position of the force. From your sketch, will the moment due to the couple be clockwise or counterclockwise? Calculate the moment due to the couple. Represent the moment by its magnitude and a circular arrow indicating its direction. Solution:

From Practice Example 4.9 we know that

F = (10 i − 4 j) N From the sketch, it is evident that the moment will be clockwise. The moment due to the couple is the sum of the moments of the two forces about any point. If we determine the sum of the moments about the point of application of one of the forces, the moment due to that force is zero and we only need to determine the moment due to the other force. Let us determine the moment about the point of application of the force F. The vector from the point of application of F to the point of application of the force −F is r = [(6 − 8) i + (6 − 8) j] m = (−2 i − 2 j) m The sum of the moments of the two forces is

M = r × (−F) =

i j k −2 −2 0 −10 4 0

= −28k N-m

The magnitude of the moment is 28 N-m. Pointing the thumb of the right hand into the page, the right-hand rule indicates that the moment is clockwise. M = 28 N-m clockwise

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Problem 4.109 The two forces lie in the x– y plane. (a) Determine the sum of the moments of the two forces about the origin. (b) What is the sum of the moments of the two forces about the point with coordinates x = 20 in, y = 16 in?

Solution: (a)

The counterclockwise moment of the top force about the origin is (12 in)[(50 lb)sin 20 °] + (8 in)[(50 lb) cos 20 °] = 581 in-lb. The counterclockwise moment of the bottom force about the origin is −(12 in)[(50 lb)sin 20 °] + (8 in)[(50 lb) cos 20 °] = 171 in-lb.

50 lb

Their sum is 752 in-lb counterclockwise, or 752k (in-lb).

(b) The counterclockwise moment of the top force about point P is

208

−(8 in)[(50 lb)sin 20 °] − (8 in)[(50 lb) cos 20 °] = −513 in-lb. 12 in

The counterclockwise moment of the bottom force about point P is

8 in

(8 in)[(50 lb)sin 20 °] + (24 in)[(50 lb) cos 20 °] = 1264 in-lb. Their sum is 752 in-lb counterclockwise, or 752k (in-lb).

(a), (b) 752 in-lb counterclockwise. 8 in

208 50 lb

Problem 4.110 The two forces lie in the x– y plane. (a) If α = 20 °, what is the moment of the couple? (b) For what value of α in the range 0 ≤ α ≤ 90 ° is the magnitude of the moment of the couple a maximum? What is the maximum moment?

(5 m)[(100 N)sin 20 °] + (4 m)[(100 N) cos 20 °] = 547 N-m. The moment of the couple is 547 N-m counterclockwise, or 547k (N-m). (b) Because the magnitude of the moment due to a couple is equal to the product of the magnitude of the forces and the perpendicular distance between their lines of action, we can see that the moment is a maximum when the forces are perpendicular to the line joining their points of application:

(0, 4) m

The total counterclockwise moment about point P is

100 N

P (0, 4) m 100 N

a x

(5, 0) m

100 N

100 N a

(5, 0) m

Solution: (a) Because the moment due to the couple is the same about any point, we can determine the sum of the moments of the forces about any point we choose. Consider the point P: y

208

P (0, 4) m

100 N

From the figure we see that α = arctan(5/4) = 51.3 °. The distance between the points of application is D = (4 m) + (5 m) = 6.40 m, so the magnitude of the moment is D(100 N) = 640 N-m. The moment of the couple is 640 N-m counterclockwise, or 640 k (N-m). (a) 547k (N-m). (b) α = 51.3 °, 640 k (N-m).

100 N

208 (5, 0) m

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Problem 4.111 Point P is contained in the x– y plane, F = 100 N, and the moment of the couple is −500 k (N-m). What are the coordinates of P ? y

Solution:

The force is

F = 100( i cos(−30 °) + j sin(−30 °)) = 86.6 i − 50 j. Let r be the distance OP. The vector parallel to OP is r = r ( i cos 70 ° + j sin 70 °) = r (0.3420 i + 0.9397 j).

P 308

The moment is

F 2F

708

M = r×F =

i 0.3420r 86.6

j 0.9397r −50.0

k 0 0

From which, r =

500 = 5.077 m. From above, 98.48

= −98.48rk.

r = 5.077(0.3420 i + 0.9397 j). The coordinates of P are x = 5.077(0.3420) = 1.74 m, y = 5.077(0.9397) = 4.77 m

Problem 4.112 Three forces of equal magnitude are applied parallel to the sides of an equilateral triangle. (a) Show that the sum of the moments of the forces is the same about any point. (b) Determine the magnitude of the sum of the moments. Solution:

L F

(a)

Resolving one of the forces into vector components parallel to the other two forces results in two equal and opposite forces with the same line of action and one couple. Therefore the moment due to the forces is the same about any point. (b) Determine the moment about one of the vertices of the triangle. A vertex lies on the line of action of two of the forces, so the moment due to them is zero. The perpendicular distance to the line of action of the third force is L cos30 °, so the magnitude of the moment due to the three force is M = FL cos30 °

Problem 4.113 In Example 4.10, suppose that the 200 ft-lb couple is counterclockwise instead of clockwise. Draw a sketch of the beam showing the forces and couple acting on it. What are the forces A and B? Solution:

In Example 4.10 we are given that the sum of the forces is zero and the sum of the moments is zero. Thus ∑ Fy = A + B = 0 ∑ M A = B (4 ft) + 200 ft-lb = 0 Solving we find A = 50 lb, B = −50 lb

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Problem 4.114 The moments of two couples are shown. What is the sum of the moments about point P ?

Solution:

The moment of a couple is the same anywhere in the plane. Hence the sum about the point P is ∑ M = −50 k + 10 k = −40 k ft lb

y 50 ft-lb

(24, 0, 0) ft 10 ft-lb

Problem 4.115 Determine the sum of the moments exerted on the plate by the two couples.

Solution:

The moment due to the 30 lb couple, which acts in a clockwise direction is M 30 = −3(30)k = −90 k ft lb.

y 30 lb

The moment due to the 20 lb couple, which acts in a counterclockwise direction, is

3 ft

M 20 = 9(20)k = 180 k ft lb.

30 lb

The sum of the moments is

2 ft x

20 lb

∑ M = −90 k + 180 k = +90 k ft lb. The sum of the moments is the same anywhere on the plate.

20 lb 5 ft

4 ft

Problem 4.116 Determine the sum of the moments exerted about A by the couple and the two forces.

Solution: Let the x-axis point to the right and the y-axis point upward in the plane of the page. The moments of the forces are M 100 = (−3i) × (100 j) = −300 k (ft-lb),

100 lb

400 lb

and M 400 = (7 i) × (−400 j) = −2800 k (ft-lb).

900 ft-lb A

The moment of the couple is M C = 900 k (ft-lb). Summing the moments, we get M Total = −2200 k (ft-lb)

3 ft

4 ft

3 ft

4 ft

Problem 4.117 Determine the sum of the moments exerted about A by the couple and the two forces. 100 N

∑ M A = (0.2 i) × (−200 j) + (0.4 i + 0.2 j) × (86.7 i + 50 j) + 300 k (N-m)

308 200 N

Solution:

∑ M A = −40 k + 2.66k + 300 k (N-m) 0.2 m

∑ M A = 262.7k (N-m) 263k (N-m)

A 300 N-m 0.2 m

0.2 m

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Problem 4.118 The sum of the moments about point A due to the forces and couples acting on the bar is zero. (a) What is the magnitude of the couple C ? (b) Determine the sum of the moments about point B due to the forces and couples acting on the bar. B 4 kN

(a)

∑ M A = 20 kN-m − (2 kN)(5 m) − (4 kN)(3 m) −(3 kN)(8) + C = 0 C = 26 kN-m

(b) ∑ M B = −(3 kN)(3 m) − (4 kN)(3 m) − (5 kN)(5 m) + 20 kN-m + 26 kN-m = 0

20 kN-m 4 kN

Solution:

2 kN

5 kN 5m

3 kN

Problem 4.119 In Example 4.11, suppose that instead of acting in the positive z-axis direction, the upper 20-N force acts in the positive x-axis direction. Instead of acting in the negative z-axis direction, let the lower 20-N force act in the negative x-axis direction. Draw a sketch of the pipe showing the forces acting on it. Determine the sum of the moments exerted on the pipe by the two couples. Solution:

The magnitude of the moment of the 20-N couple is

unchanged, (2 m)(20 N) = 40 N-m. The direction of the moment vector is perpendicular to eh x – y plane, and the right-hand rule indicates that it points in the negative z-axis direction. The moment of the 20-N couple is (−40 N-m) k. The sum of the moments exerted on the pipe by the two couples is ∑ M = (−40 N-m)k + (30 N) cos60 ° (4 m) j − (30 N)sin 60 °(4 m)k ∑ M = (60 j − 144 k) N-m

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Solution:

Problem 4.120 The magnitude of the force F is 80 lb. The length of the straight line from the origin to point P is 4 ft . Determine the moment of the couple.

y P

–F

rOP

The force F is F = (80 lb)(− cos 50° sin 70°i + sin 50° j + cos 50° cos 70 °k).

408

508

208

The vector from the origin to point P is rOP = (4 ft)(− cos 40 ° sin 20 °i + sin 40 ° j − cos 40 ° cos 20 °k).

708

The moment of the couple is M = rOP × (−F)

i x

= (4 ft)(80 lb) − cos 40 ° sin 20 ° sin 40 ° − cos 40 ° cos 20 ° cos 50 ° sin 70 ° − sin 50 ° − cos 50 ° cos 70 ° = −222 i − 158 j − 60.0 k (ft-lb). −222 i − 158 j − 60.0 k (ft-lb).

Problem 4.121 The magnitude of the force F is 200 N. The length of the straight line from the origin to point P is 3 m . (a) What is the magnitude of the moment of the couple? (b) What is the perpendicular distance between the lines of action of the two forces?

Solution: y P F

rOP

O z

(a)

The force F is F = (200 N)(− cos 50 ° sin 70 °i + sin 50 ° j + cos 50 ° cos 70 °k).

508

708

The vector from the origin to point P is

408

rOP = (3 m)(− cos 40 ° sin 20 °i + sin 40 ° j − cos 40 ° cos 20 °k).

208

The moment of the couple is

M = rOP × (−F) i

= (3 m)(200 N) − cos 40 ° sin 20 ° sin 40 ° − cos 40 ° cos 20 °

cos 50 ° sin 70 ° − sin 50 ° − cos 50 ° cos 70 ° x

= −416 i − 295 j − 113k (N-m). Its magnitude is M = 522 N-m. (b) Let D denote the perpendicular distance between the lines of action. We know that D F = M: D(200 N) = 522 N-m. We obtain D = 2.61 m. (a) 522 N-m. (b) 2.61 m.

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Problem 4.122 What is the magnitude of the sum of the moments exerted on the T-shaped structure by the two couples?

3 ft

50 i 1 20j 2 10k (lb)

3 ft 50j (lb)

3 ft

z 250j (lb)

Solution:

3 ft

The moment of the 50 lb couple can be determined by

inspection: 250 i 220 j 1 10 k (lb)

M 1 = −(50)(3)k = −150 k ft lb. The vector separating the other two force is r = 6k. The moment is M2 = r × F =

= −120 i + 300 j.

3 ft

3 ft 50j (lb)

50 20 −10 The sum of the moments is

3 ft x

∑ M = −120 i + 300 j − 150 k.

–F

The magnitude is M =

–50j (lb)

3 ft

120 2 + 300 2 + 150 2 = 356.23 ft lb

Problem 4.123 A person exerts a couple on the lug wrench. The lower side of the right triangle indicating the direction of F lies in the x–z plane. (a) If the person wants to exert a 20 N-m moment about the x-axis (the moment that loosens the lug nut), what is the necessary magnitude of F? (b) If F has the magnitude determined in (a), what is the resulting moment about the y-axis?

Solution: y

2F 708 308

(a)

0.2 m

z 0.2 m

Let F denote the magnitude of F. The vector F is F = F (cos 70 ° cos30 °i + sin 70 ° j − cos 70 ° sin 30 °k). We can determine the moment of the couple by evaluating the moment about the point of application P of −F. The vector from P to the point of application of the right force is r = −(0.4 m)k. The moment of the couple is

M = r×F

= F

i j k 0 0 −0.4 m cos 70 ° cos30 ° sin 70 ° − cos 70 ° sin 30 °

= F (0.376 i − 0.118 j). Equating the moment about the x-axis to 20 N-m, 0.376 F = 20 N-m, we obtain F = 53.2 N. (b) Using the value of F we obtained in part (a), the moment about the y-axis is F (−0.118 j) = (53.2)(−0.118 j) = −6.30 j N-m. (a) F = 53.2 N. (b) − 6.30 j N-m.

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Problem 4.124 The cables AB and CD exert a couple on the vertical pipe. The tension in each cable is 8 kN . Determine the magnitude of the moment the cables exert on the pipe.

Solution: F AB = 8kN

(1.4 i − 0.6 j + 1.0 k) , rDB = (3.2 i − 2.2 j + 2.4 k) m 3.32

(21.6, 2.2, 21.2) m

C (20.2, 1.6, 20.2) m

M = rBD × F AB = (−3.34 i + 0.702 j + 5.09k) kN-m A

⇒ M = 6.13 kN-m

(0.2, 0.6, 0.2) m

Problem 4.125 A driver exerts two forces on the car’s steering wheel. The force F = 3.84 i − 2.02 j + 0.844 k (N) is applied at point A with coordinates (53.0, 183, −15.0) mm. The force −F is applied at point B with coordinates (328, 360, − 56.0) mm. The steering wheel’s shaft is parallel to the unit vector 0.342 j + 0.940 k. What is the magnitude of the moment the two forces exert about the shaft?

(1.6, 0, 1.2) m

Solution: –F B

rAB A F

2F x

Because the moment of a couple is the same about any point, we are free to choose any convenient point to determine it. The position vector from point A to point B is r AB = rB − r A

A F

= (328 − 53.0) i + (360 − 183) j + (−56.0 + 15.0)k (mm) = 275i + 177 j − 41.0 k (mm).

The moment of the two forces about A is M = rBA × (−F) x

  i j k    =  275 mm 177 mm − 41.0 mm     −3.84 N 2.02 N − 0.844 N    = −66.6 i + 390 j + 1240 k (N-mm). The dot product of the moment with the unit vector parallel to the shaft is (0.342 j + 0.940 k) ⋅ M = (0)(−66.6) + (0.342)(390) + (0.940)(1240) N-mm = 1290 N-mm. 1.29 N-m.

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Problem 4.126 The forces FB = 2 i + 6 j + 3k (kN),

y FB

FC = i − 2 j + 2k (kN),

and the couple M C = M Cy j + M Cz k (kN-m).

Determine the values of M Cy and M Cz so that the sum of the moments of the two forces and the couple about A is zero.

MC C x

Solution:

The sum of the moments of the two forces and the couple about A is r AB × FB + r AC × FC + M C =

i j k 1m 0 0 2 kN 6 kN 3 kN

i j k 2m 0 0 1 kN −2 kN 2 kN

+ M Cy j + M Cz k

= −(1 m)(3 kN) j + (1 m)(6 kN)k − (2 m)(2 kN) j + (2 m)(−2 kN)k + M Cy j + M Cz k = (−7 kN-m + M Cy ) j + (2 kN-m + M Cz )k. This is zero if M Cy = 7 kN-m, M Cz = −2 kN-m. M Cy = 7 kN-m, M Cz = −2 kN-m.

Problem 4.127 Two wrenches are used to tighten an elbow fitting. The force F = 10 k (lb) on the right wrench is applied at (6, −5, −3) in, and the force −F on the left wrench is applied at (4, −5, 3) in. (a) Determine the moment about the x-axis due to the force exerted on the right wrench. (b) Determine the moment of the couple formed by the forces exerted on the two wrenches. (c) Based on the results of (a) and (b), explain why two wrenches are used.

Solution:

The position vector of the force on the right wrench is rR = 6 i − 5 j − 3k. The magnitude of the moment about the x-axis is M R = e X ⋅ ( r R × F) =

1 0 0 6 −5 −3 0 0 10

= −50 in 1b

(a) The moment about the x-axis is

from which MXL = 50i in lb, which is opposite in direction and equal in magnitude to the moment exerted on the x-axis by the right wrench. The left wrench force is applied 2 in nearer the origin than the right wrench force, hence the moment must be absorbed by the space between, where it is wanted.

M R = M R e X = −50 i (in 1b). (b) The moment of the couple is M C = ( r R − r L ) × FR =

i j k 2 0 −6 0 0 10

= −20 j in 1b

(c) The objective is to apply a moment to the elbow relative to connecting pipe, and zero resultant moment to the pipe itself. A resultant moment about the x-axis will affect the joint at the origin. However the use of two wrenches results in a net zero moment about the x-axis the moment is absorbed at the juncture of the elbow and the pipe. This is demonstrated by calculating the moment about the x-axis due to the left wrench: M X = e X ⋅ ( r L × FL ) =

218

1 0 0 4 −5 3 0 0 −10

= 50 in lb

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Problem 4.128 Two equivalent systems of forces act on the beam. Determine the force F and the distance D .

System 1

100 N 50 N x

Solution:

In System 1 the sum of the vertical forces is 150 N. In System 2 it is F. Therefore F = 150 N. In System 1 the moment about the left end of the beam is 200 N-m counterclockwise. In System 2 it is DF counterclockwise. Therefore D =

200 N-m = 1.33 m. 150 N

2m System 2 y F

F = 150 N, D = 1.33 m.

x D

Problem 4.129 Two systems of forces and moments act on the beam. Are they equivalent?

System 1

Solution:

The sums of the forces are:

20 lb

50 ft-lb

10 lb

∑ FX = 0 (both systems) ∑ FY 1 = 10 j − 20 j = −10 j (lb)

2 ft

∑ FY 2 = −20 j + 10 j = −10 j (lb)

2 ft System 2

Thus the sums of the forces are equal. The sums of the moments about the left end are:

∑ M 1 = (−20)(4)k + 50 k = −30 k (ft lb) ∑ M 2 = (+10(2))k − 30 k = −10 k (ft lb)

20 lb

30 ft-lb

10 lb

The sums of the moments are not equal, hence the systems are not equivalent. No 2 ft

x 2 ft

Problem 4.130 Four systems of forces and moments act on an 8-m beam. Which systems are equivalent?

Solution:

For equivalence, the sum of the forces and the sum of the moments about some point (the left end will be used) must be the same. System 1

System 2

System 3

∑ F (kN)

10 j

System 4 10 j

∑ M L (kN-m)

80 k

160 k

80 k

Systems 1, 2, and 4 are equivalent. y x

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Problem 4.131 Two equivalent systems of forces act on the beam. Determine the forces F1 and F2 . System 2

System 1 30 lb

10 lb 50 ft-lb

20 ft-lb

20 lb

x 2 ft

Solution:

20 lb

2 ft

Equate the sums of the vertical forces in the two

systems: −10 lb + 30 lb + F1 = −20 lb + F2 + 20 lb. Equate the sums of the moments about the left end of the beam in the two systems: 50 ft-lb − (2 ft)(10 lb) + (4 ft)(30 lb) + (6 ft) F1 = 20 ft-lb + (4 ft) F2 + (6 ft)(20 lb). Solving these two equations yields F1 = 35 lb, F2 = 55 lb. F1 = 35 lb, F2 = 55 lb.

Problem 4.132 Two equivalent systems of forces and moments act on the structural member. Determine the force F and couple C . System 1

System 2 C

50 N 0.34 m 80 N-m 50 N 60 N

40 N

0.6 m

120 N

0.3 m

Solution:

The sums of the horizontal forces in the two systems are equal. We equate the sums of the vertical forces in the two systems, 40 N + 60 N = −F + 120 N

and the sums of the moments about the left end of the structure in the two systems, 80 N-m + (0.34 m)(50 N) + (0.9 m)(60 N) = C − (0.6 m) F + (0.9 m)(120 N). Solving these two equations yields F = 20 N, C = 55 N-m. F = 20 N, C = 55 N-m.

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Problem 4.133 In system 1, the cable AB exerts a 2.4-kN force on the log at B. In system 2, the cables CD and CE exert forces on the log at D and E . The vector sum of the forces exerted on the log by the cables in system 2 is equal to the force exerted on the log by the cable in system 1. (a) Determine the tensions in cables CD and CE . (b) Is the system of forces exerted on the log by the cables in system 2 equivalent to the force exerted on the log by the cable in system 1? Explain. Solution: (a) The angle between cable AB and the horizontal is arctan

m ( 12 ) = 36.9°. 16 m

and the angle between cable CE and the horizontal is arctan

m ( 12 ) = 24.8°. 26 m

The vector sum of the forces in system 2 is equal to the force in system 1 if TCD cos 63.4 ° + TCE cos 24.8 ° = (2.4 kN) cos36.9 °, TCD sin 63.4 ° + TCE sin 24.8 ° = (2.4 kN)sin 36.9 °. Solving these equations yields TCD = 0.805 kN, TCE = 1.72 kN. (b) The sum of the forces in system 1 is equal to the sum of the forces in system 2. Also, the sum of the moments about point A in system 1 is equal to the sum of the forces about point C (the same point) in system 2. The systems are equivalent. (a) TCD = 0.805 kN, TCE = 1.72 kN. (b) Yes.

The angle between cable CD and the horizontal is arctan

( 126 mm ) = 63.4°, System 1

System 2

12 m

12 m B

16 m

Problem 4.134 Systems 1 and 2 each consist of a couple. If they are equivalent, what is F ? For couples, the sum of the forces vanish for both systems. For System 1, the two forces are located at r11 = 4 i, and r12 = +5 j. The forces are F1 = 200( i cos30 ° + j sin 30 °) = 173.21i + 100 j. The moment due to the couple in System 1 is i j k 4 −5 0 173.21 100 0

= 1266.05k (N-m).

20 m

System 1

System 2 y

y 200 N

Solution:

M 1 = (r11 − r12 ) × F1 =

308

208

200 N 308 4m

(5, 4, 0) m

2m x F

208

For System 2, the positions of the forces are r21 = 2 i, and r22 = 5i + 4 j. The forces are F2 = F ( i cos(−20 °) + j sin(−20 °)) = F (0.9397 i − 0.3420 j). The moment of the couple in System 2 is M 2 = (r21 − r22 ) × F2 = F

i j k −3 −4 0 0.9397 −0.3420 0

= 4.7848Fk,

from which, if the systems are to be equivalent, F =

1266 = 264.6 N 4.7848

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Problem 4.135 Two equivalent systems of forces and moments act on the L-shaped bar. Determine the forces FA and FB and the couple M.

System 1

System 2

120 N-m 3m

60 N

FB 40 N

50 N

Solution:

The sums of the forces for System 1 are

∑ FX = 50, and ∑ FY = −FA + 60. The sums of the forces for System 2 are ∑ FX = FB , and ∑ FY = 40. For equivalent systems: FB = 50 N, and FA = 60 − 40 = 20 N. The sum of the moments about the left end for System 1 is ∑ M 1 = −(3) FA − 120 = −180 N-m. The sum of the moments about the left end for System 2 is ∑ M 2 = −(3) FB + M = −150 + M N-m. Equating the sums of the moments, M = 150 − 180 = −30 N-m

Problem 4.136 Two equivalent systems of forces and moments act on the plate. Determine the force F and the couple M.

System 1 30 lb

System 2 10 lb

8 in The sums of the forces for System 1 are

50 lb

100 in-lb 5 in

5 in

Solution:

30 lb

8 in

30 lb

∑ FX = 30 lb, ∑ FY = 50 − 10 = 40 lb. The sums of the forces for System 2 are ∑ FX = 30 lb ∑ FY = F − 30 lb. For equivalent forces, F = 30 + 40 = 70 lb. The sum of the moments about the lower left corner for System 1 is ∑ M 1 = −(5)(30) − (8)(10) + M = −230 + M in lb. The sum of the moments about the lower left corner for System 2 is ∑ M 2 = −100 in lb. Equating the sum of moments, M = 230 − 100 = 130 in lb

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Problem 4.137 In system 1, a beam is subjected to two forces acting at points A and C . Suppose that you want systems 1 and 2 to be equivalent, and you want system 2 to consist of the force F shown and a second force acting at point B. What is the magnitude of F and what force must act at point B?

Solution:

The sum of the forces in System 1 is zero. The system exerts only a couple on the beam. Therefore the force exerted at B in System 2 must be a downward force of magnitude F: System 1 exerts a counterclockwise couple of magnitude (4 m)[(60 N)sin 30 °] = 120 N-m. System 2 exerts a counterclockwise couple of magnitude (2 m)F , so the systems are equivalent if F = 60 N.

System 1

F = 60 N, a 60-N downward force must act at B. 60 N

60 N

308

F A

B F

System 2

C 2m

F A

Problem 4.138 Three forces and a couple are applied to a beam (system 1). (a) If you represent system 1 by a force applied at A and a couple (system 2), what are F and M ? (b) If you represent system 1 by the force F (system 3), what is the distance D ?

Solution:

The sum of the forces in System 1 is

∑ FX = 0 i, ∑ FY = (−20 + 40 − 30) j = −10 j 1b. The sum of the moments about the left end for System 1 is ∑ M 1 = (2(40) − 4(30) + 30)k = −10 k ft lb. (a) For System 2, the force at A is F = −10 j 1b The moment at A is M 2 = −10 k ft lb (b) For System 3 the force at D is F = −10 j lb. The distance D is the ratio of the magnitude of the moment to the magnitude of the force, where the magnitudes are those in System 1: D =

10 = 1 ft 10

System 1

System 2

y 30 lb

40 lb

20 lb

30 ft-lb

A 2 ft

M x

2 ft System 3 y F x

A D

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Problem 4.139 The cable BC exerts a force on the beam at B. The vector sum of the forces on the beam is zero, and the sum of the moments about the left end of the beam is zero. (a) Determine the forces A x and A y and the tension in cable BC . (b) What is the sum of the moments about the right end of the beam? (c) If you represent the system of forces and moments acting on the beam by an equivalent system consisting of a force F acting at the origin and a couple M, what are F and M ?

y 10 kN 110 kN-m

C B

308

Ay 3m

Solution: (a) Let T denote the magnitude of the force exerted by the cable at B. The sum of the x components of the forces on the beam is zero: ∑ Fx = − A x + T cos30 ° = 0. The sum of the y components of zero: ∑ Fy = − A y − 10 kN + T sin 30 ° = 0. The sum of the moments about the left end of the beam is zero: ∑ M left end = −110 kN-m − (3 m)(10 kN) + (7 m)T sin 30 ° = 0. Solving these equations yields A x = 34.6 kN, A y = 10 kN, T = 40 kN. (b) The sum of the moments about the right end is ∑ M right end = −110 kN-m + (7 m) A y + (4 m)(10 kN) = 0. (c) The sum of the forces acting on the beam is zero, so F = 0. The sum of the moments about the left end of the beam is zero, so M = 0. (a) A x = 34.6 kN, A y = 10.0 kN, tension = 40 kN. (b) Zero. (c) F = 0, M = 0.

Problem 4.140 The bracket is subjected to three forces and a couple. If you represent this system by a force F, what is F and where does its line of action intersect the x-axis?

400 N 180 N

0.4 m

Solution:

We locate a single equivalent force along the x-axis a distance d to the right of the origin. We must satisfy the following three equations:

140 N-m

200 N 0.2 m

0.65 m

∑ Fx = 400 N − 200 N = R x ∑ Fy = 180 N = R y ∑ M O = −(400 N)(0.6 m) + (200 N)(0.2 m) + (180 N)(0.65 m) + 140 Nm = R y d Solving we find R x = 200 N, R y = 180 N, d = 0.317 m

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Problem 4.141 Three forces and a couple are applied to a beam (system 1). (a) If you represent system 1 by a force applied at A and a couple (system 2), what are F and M ? (b) If you represent system 1 by the force F (system 3), what is the distance D ?

The moment about A in the two systems must be equal, so M = −10 kN-m − (2 m)(4 kN) + (5 m)(6 kN) = 12 kN-m. (b) The sum of the forces in Systems 2 and 3 are the same. The moment about A in the two systems must be equal, so

Solution:

M = 12 kN-m = D(4 kN).

(a) The force F must equal the sum of the forces in System 1:

The distance D = 3 m.

F = (2 kN) j − (4 kN) j + (6 kN) j = (4 kN) j.

(a) F = 4 j (kN), M = 12 kN-m. (b) D = 3 m.

System 1 y

2 kN

6 kN

10 kN-m

x 2m

System 2

4 kN

F x

System 3

F A

x D

Solution:

Problem 4.142 The forces FA and FB support the 50-lb weight. (a) Determine the magnitudes FA and FB . (b) If you represent the forces FA and FB by an equivalent force F, what is F, and where does its line of action intersect the x-axis?

(a) The angles α = arctan(6/8) = 36.9 ° and β = arctan(6/10) = 31.0 °. The equilibrium equations are ∑ Fx = −FA cos α + FB cos β = 0, ∑ Fy = FA sin α + FB sin β − 50 lb = 0. Solving, we obtain FA = 46.3 lb, FB = 43.2 lb. (b) If the force F is equivalent to the two forces, it must be equal to their sum:

8 in

10 in

F = (−FA cos α + FB cos β ) i + ( FA sin α + FB sin β ) j

= 50 j (lb). x

If we assume that this force acts somewhere on the x-axis, where does it need to be placed to be equivalent to the forces FA and FB? We equate the sum of the moments about the origin due to the forces FA and FB to the moment about the origin due to F:

6 in

(18 in) FB sin β = x (50 lb),

50 lb

obtaining x = 8 in. (An easier solution is to consider the point where the cables join. The moment about that point due to the forces FA and FB is zero. If we place the force F at that point, or anywhere on the same line of action, it is equivalent to the forces FA and FB . (a) FA = 46.3 lb, FB = 43.2 lb. (b) F = 50 j (lb), x = 8 in.

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Problem 4.143 The distributed force exerted on part of a building foundation by the soil is represented by five forces. If you represent them by a force F, what is F, and where does its line of action intersect the x-axis?

Solution:

The equivalent force must equal the sum of the forces exerted by the soil:

F = (80 + 35 + 30 + 40 + 85) j = 270 j kN

80 kN 3m

The sum of the moments about any point must be equal for the two systems. The sum of the moments are

35 kN

30 kN

40 kN

85 kN

∑ M = 3(35) + 6(30) + 9(40) + 12(85) = 1665 kN-m. Equating the moments for the two systems FD = 1665 kN-m from which D =

1665 kN-m = 6.167 m. 270 kN

Thus the action line intersects the x-axis at a distance D = 6.167 m to the right of the origin.

Problem 4.144 At a particular instant, aerodynamic forces distributed over the airplane’s surface exert the 88-kN and 16-kN vertical forces and the 22 kN-m counterclockwise couple shown. If you represent these forces and couple by a system consisting of a force F acting at the center of mass G and a couple M , what are F and M ?

y 88 kN

16 kN x

G 5m

22 kN-m

5.7 m

Solution:

∑ Fy = 88 kN + 16 kN = R y ∑ M G = −(88 kN)(0.7 m) + (16 kN)(3.3 m) + 22 kN-m = M Solving we find R y = 104 kN, M = 13.2 kN-m

Problem 4.145 If you represent the two forces and couple acting on the airplane by a force F, what is F, and where does its line of action intersect the x-axis?

y 88 kN

16 kN

Solution: 5m

∑ M Origin = (88 kN)(5 m) + (16 kN)(9 m) + 22 kN-m = R y x

F = R y j = 104 kNj, x = 5.83 m

Solution:

The sum of the forces acting at A is in opposition to the weight, or F = W j = 100 j lb. The moment about point A is zero.

22 kN-m

5.7 m

Solving we find

Problem 4.146 The system is in equilibrium. If you represent the forces F AB and F AC by a force F acting at A and a couple M, what are F and M?

∑ Fy = 88 kN + 16 kN = R y

y B

608 408

FAB

100 lb

FAC

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Problem 4.147 Three forces act on the beam. (a) Represent the system by a force F acting at the origin O and a couple M . (b) Represent the system by a single force. Where does the line of action of the force intersect the x-axis?

30 N 5m

Solution: O

(a) The sum of the forces is ∑ FX = 30 i N, and

x 6m

30 N

∑ FY = (30 + 50) j = 80 j N.

50 N

The equivalent at O is F = 30 i + 80 j (N). The sum of the moments about O: ∑ M = (−5(30) + 10(50)) = 350 N-m (b) The solution of Part (a) is the single force. The intersection is the moment divided by the y-component of force: 350 D = = 4.375 m 80

Problem 4.148 Forces exerted on the Boeing 777 by its engines are shown. Assume that the vectors are parallel to the x-axis and lie in the x– y plane. Represent the two forces by an equivalent system consisting of a force F acting at the origin and a couple M . What are F and M ?

Solution:

We want to represent the system y 50,000 lb 30 ft

30 ft 56,000 lb y

by the system

y 50,000 lb

30 ft

30 ft 56,000 lb

The total force in the two systems must be equal, F = 50,000 i + 56,000 i (lb) = 106,000 i (lb), and the moment about the origin must be equal, M = (30 ft)(56,000 lb) − (30 ft)(50,000 lb) = 180,000 ft-lb counterclockwise. F = 106,000 i (lb), M = 180,000 ft-lb counterclockwise.

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Problem 4.149 The forces exerted on the Boeing 777 by its engines are shown. Assume that the vectors are parallel to the x-axis and lie in the x– y plane. Represent the two forces by an equivalent system consisting of a force F. What is F and where does its line of action intersect the y-axis?

Solution:

We want to represent the system

y 50,000 lb 30 ft

30 ft 56,000 lb

by the system

y y

50,000 lb

30 ft

x x

30 ft 56,000 lb

The total force in the two systems must be equal, F = 50,000 i + 56,000 i (lb) = 106,000 i (lb), and the moment about the origin must be equal, −y(106,000 lb) = (30 ft)(56,000 lb) − (30 ft)(50,000 lb). We see that F = 106,000 i (lb) and y = −1.70 ft. F = 106,000 i (lb), y = −1.70 ft.

Problem 4.150 If you represent the three forces acting on the beam cross section by a force F, what is F, and where does its line of action intersect the x-axis? Solution:

The sum of the forces is

∑ FX = (500 − 500) i = 0.

y 500 lb

6 in

800 lb

∑ FY = 800 j. Thus a force and a couple with moment M = 500 k ft lb act on the cross section. The equivalent force is F = 800 j which acts at a pos500 itive x-axis location of D = = 0.625 ft = 7.5 in to the right 800 of the origin.

228

x 6 in

500 lb

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Problem 4.151 In Practice Example 4.12, suppose that the force FB is changed to FB = 20 i − 15 j + 30 k (kN), and you want to represent system 1 by an equivalent system consisting of a force F acting at the point P with coordinates (4, 3, − 2) m and a couple M (system 2). Determine F and M. Solution:

From Practice Example 4.12 we know that

F A = (−10 i + 10 j − 15k) kN M C = (−90 i + 150 j + 60 k) kN-m The force F must equal the sum of the forces in system 1: ( ∑ F) 2 = ( ∑ F) 1: F = F A + FB = (10 i − 5 j + 15k) kN In system 2, the sum of the moments about P is M. Therefore equivalence requires that M be equal to the sum of the moments about point P due to the forces and moments in system 1: (∑ M P ) 2 = (∑ M P ) 1:  i j k i j k  M =  −4 −3 2 + 2 −3 2  −10 10 −15 20 −15 30  M = (−125i + 50 j + 20 k) kN-m

  + (−90 i + 150 j + 60 k)  kN-m  

Thus F = (10 i − 5 j + 15k) kN, M = (−125i + 50 j + 20 k) kN-m

Problem 4.152 The wall bracket is subjected to the force shown. (a) Determine the moment exerted by the force about the z-axis. (b) Determine the moment exerted by the force about the y-axis. (c) If you represent the force by a force F acting at O and a couple M, what are F and M? Solution:

y O z

10i 2 30j 1 3k (lb) 12 in

(a) The moment about the z-axis is negative, M Z = −1(30) = −30 ft lb, (b) The moment about the y-axis is negative, M Y = −1(3) = −3 ft lb (c) The equivalent force at O must be equal to the force at x = 12 in, thus FEQ = 10 i − 30 j + 3k (lb) The couple moment must equal the moment exerted by the force at x = 12 in, This moment is the product of the moment arm and the y- and z-components of the force: M = −1(30)k − 1(3) j = −3 j −30 k (ft lb).

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Problem 4.153 The forces exerted on the runway by the airplane’s five wheels are F A = 8.22k (kN), FB = 7.94 k (kN), and FC = 1.12k (kN). If you represent these forces by an equivalent system consisting of a force F, what is F, and where does its line of action intersect the x– y plane? Solution:

The force F must equal the sum of the forces on the

wheels: F = F A + FB + FC = 8.22k + 7.94 k + 1.12k (kN) = 17.28k (kN). Assume that the line of action of F intersects the x -y plane at a point ( x, y ). Moment of F about the origin must equal the sum of the moments of the three forces on the runway about the origin: i j k x y 0 0 0 17.28 kN

i j k 0 −1.5 m 0 0 0 7.94 kN

1.5 m

i j k 0 1.5 m 0 0 0 8.22 kN +

1.5 m

i j k 1.86 m 0 0 . 0 0 1.12 kN 1.86 m

The i and j components of this equations give the equations (17.28 kN) y = (1.5 m)(8.22 kN) + (−1.5 m)(7.94 kN), (17.28 kN) x = (1.86 m)(1.12 kN).

We obtain x = 0.121 m, y = 0.0243 m. F = 17.3k (kN) at x = 121 mm, y = 24.3 mm.

Problem 4.154 In Example 4.14, suppose that the 30-lb upward force in system 1 is changed to a 25-lb upward force. If you want to represent system 1 by a single force F (system 2), where does the line of action of F intersect the x– z plane? Solution:

The sum of the forces in system 2 must equal the sum of the forces in system 1: (∑ F) 2 = (∑ F) 1 F = (20 + 25 − 10) j lb F = 35 j 1b The sum of the moments about a point in system 2 must equal the sum of the moments about the same point is system 1. We sum moments about the origin. ( ∑ M) 2 = ( ∑ M) 1 i

35 0

i =

6 0 2 0 25 0

i +

2 0 4 0 −10 0

i +

−3 0 −2 0 20 0

Expanding the determinants results in the equations −35z = −50 + 40 + 40 35 x = 150 − 20 − 60 Solving yields x = 2.00 ft, z = −0.857 ft

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Problem 4.155 The normal forces exerted on the car’s tires by the road are

A 0.8 m

N A = 5104 j (N),

N B = 5027 j (N),

0.8 m

N C = 3613 j (N),

N D = 3559 j (N).

1.4 m

If you represent these forces by a single equivalent force N, what is N and where does its line of action intersect the x−z plane? Solution:

1.4 m

We must satisfy the following three equations

∑ Fy : 5104 N + 5027 N + 3613 N + 3559 N = R y ∑ M x : (5104 N + 3613 N)(0.8 m)

− (5027 N + 3559 N)(0.8 m) = −R y z ∑ M z : (5104 N + 5027 N)(1.4 m) − (3613 N + 3559 N)(1.4 m) = R y x Solving we find R y = 17303 N, x = 0.239 m, z = −0.00606 m

Problem 4.156 Two forces act on the beam. If you represent them by a force F acting at C and a couple M, what are F and M?

100 N 80 N

Solution:

The equivalent force must equal the sum of forces: F = 100 j + 80 k. The equivalent couple is equal to the moment about C:

∑ M = (3)(80) j − (3)(100)k = 240 j − 300 k

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Problem 4.157 An axial force of magnitude P acts on the beam. If you represent it by a force F acting at the origin O and a couple M, what are F and M?

Pi z h O

Solution:

The equivalent force at the origin is equal to the applied force F = Pi. The position vector of the applied force is r = −h j + b k. The moment is

M = (r × P ) =

i j k 0 −h +b P 0 0

= bPj + hPk.

This is the couple at the origin. (Note that in the sketch the axis system has been rotated 180 about the x-axis; so that up is negative and right is positive for y and z.)

Problem 4.158 The brace is being used to remove a screw. (a) If you represent the forces acting on the brace by a force F acting at the origin O and a couple M, what are F and M? (b) If you represent the forces acting on the brace by a force F′ acting at a point P with coordinates ( x P , y P , z P ) and a couple M′, what are F′ and M′ ?

(a) Equivalent force at the origin O has the same value as the sum of forces, ∑ FX = ( B − B) i = 0, ∑ FY = (− A + 12 A + 12 A) j = 0,

thus F = 0. The equivalent couple moment has the same value as the moment exerted on the brace by the forces, ∑ M O = (rA) i.

Thus the couple at O has the moment M = rAi. (b) The equivalent force at ( x P , y P , z P ) has the same value as the sum of forces on the brace, and the equivalent couple at ( x P , y P , z P ) has the same moment as the moment exerted on the brace by the forces: F = 0, M = rAi.

h r

B z

Solution:

O 1 A 2

1 A 2

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Problem 4.159 The wheel is subjected to four forces. Assume that the force vectors lie in the x– y plane. Represent them by an equivalent system consisting of a force F acting at the center of the wheel and a couple M . What are F and M ?

Solution:

We want to represent the system y 8 kN 0.4 m 4 kN

0.75 m 6 kN

6 kN

8 kN

0.75 m

0.4 m 4 kN

by the system

y 6 kN

6 kN

The sums of the forces must be equal, −8i + 6 i − 4 j + 6 j (kN) = F: −2 i + 2 j (kN) = F, and the sums of the moments about the origin must be equal, (0.4 m)(8 kN) + (0.75 m)(6 kN) = M : 7.7 kN-m = M . F = −2 i + 2 j (kN), M = 7.7 kN-m counterclockwise.

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Problem 4.160 The wheel is subjected to four forces. Assume that the force vectors lie in the x– y plane. If you represent them by an equivalent system consisting of a force F, what is F, and where does its line of action intersect the x-axis?

by the system y F x x

The sums of the forces must be equal, 8 kN

0.75 m

0.4 m

4 kN

and the sums of the moments about the origin must be equal, (0.4 m)(8 kN) + (0.75 m)(6 kN) counterclockwise = −x (2 kN): 3.85 m = x.

6 kN 6 kN

Solution:

−8i + 6 i − 4 j + 6 j (kN) = F: −2 i + 2 j (kN) = F,

F = −2 i + 2 j (kN), x = 3.85 m.

We want to represent the system y 8 kN 0.4 m 4 kN

0.75 m 6 kN

6 kN

Problem 4.161 The two systems of forces and moments acting on the bar are equivalent. If

F A = 30 i + 30 j − 20 k (kN), FB = 40 i − 20 j + 25k (kN), M B = 10 i + 40 j − 10 k (kN-m),

what are F and M?

FB System 1

Solution: y F = F A + FB = (70 i + 10 j + 5k) kN M = (2 mi) × F A + (4 mi) × FB + M B

= (10 i − 20 j − 30 k) kNm z

M System 2

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Problem 4.162 The forces are

Point G is at the center of the block.

Solution:

The equivalent force is the sum of the forces:

∑ F = (−20) i + (10 + 10) j + (20 − 10)k = −20 i + 20 j + 10 k (lb).

F A = −20 i + 10 j + 20 k (lb), FB = 10 j − 10 k (lb). If you represent the two forces by a force F acting at G and a couple M , what are F and M ?

The equivalent couple is the sum of the moments about G. The position vectors are: r A = −15i + 5 j + 10 k (in), rB = 15i + 5 j − 10 k. The sum of the moments:

y FB

∑ M G = (r A × F A ) + (rB × FB )

FA =

10 in x

i j k −15 5 10 −20 10 20

j k i 15 5 −10 0 10 −10

= 50 i + 250 j + 100 k (in 1b) 20 in

30 in

Problem 4.163 The engine above the airplane’s fuselage exerts a thrust T0 = 16 kip, and each of the engines under the wings exerts a thrust TU = 12 kip. The dimensions are h = 8 ft, c = 12 ft, and b = 16 ft. If you represent the three thrust forces by a force F acting at the origin O and a couple M, what are F and M? Solution:

T0 c

2 TU y

The equivalent thrust at the point G is equal to the sum

of the thrusts: ∑ T = 16 + 12 + 12 = 40 kip The sum of the moments about the point G is ∑ M = (r1U × TU ) + (r2U × TU ) + (rO × TO ) = (r1U + r2U ) × TU + (rO × TO ).

The position vectors are r1U = +b i − h j, r2U = −b i − h j, and rO = +cj. For h = 8 ft, c = 12 ft, and b = 16 ft, the sum of the moments is

∑M =

i j k 0 −16 0 0 0 12

i j k 0 12 0 0 0 16

= (−192 + 192) i = 0.

Thus the equivalent couple is M = 0

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Problem 4.164 The cables AB, AC , and AD exert forces F AB , F AC , and F AD that support the 40-kg object. If you represent F AB , F AC , and F AD by an equivalent system consisting of a force F, what is F, and where does its line of action intersect the x–z plane?

Solution:

We know that

F AB + F AC + F AD = mg j, so F = mg j

= (40 kg)(9.81 m/s 2 ) j = 392 j (N). We know that the moment about point A due to the forces F AB , F AC and F AD is zero. Therefore if we place the force F at point A, it is equivalent. Its line of action intersects the x−z plane at x = 0, z = 0.

C 0.3 m 0.8 m 0.3 m

F = 392 j (N) at x = 0, z = 0.

0.6 m D B

0.4 m

0.3 m

z 0.6 m

FAC

FAD

FAB A

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Problem 4.165 The cables AB, AC , and AD exert forces F AB , F AC , and F AD that support the 40-kg object. If you represent F AB , F AC , and F AD by an equivalent system consisting of a force F acting at the point x = 1 m , y = 0 , and z = 2 m and a couple M, what are F and M?

Solution: y

y FAC

0.3 m 0.3 m

Let us refer to the three cable forces as System 1. We want to find a System 2 consisting of a force F acting at the point x = 1 m, y = 0, z = 2 m and a couple M that is equivalent to System 1:

0.6 m D B

0.4 m

y 0.3 m

z 0.6 m

A (0, 20.6, 0) m

System 1

0.8 m

FAB

FAD

M System 2

FAC

FAD

(1, 0, 2) m

FAB

(0, 20.6, 0) m

To be equivalent, the force F must equal the sum of the three cable forces. We know from equilibrium that F AB + F AC + F AD = mg j, so the force F must be F = mg j = (40 kg)(9.81 m/s 2 ) j = 392 j (N). We know that the moment about point A due to System 1 is zero. To be equivalent, the moment about point A due to System 2 must also be zero: i j k 1 m 0.6 m 2 m 0 392 N 0

+ M = 0.

Solving, we obtain M = 785 i − 392 k (N-m). F = 392 j (N), M = 785 i − 392 k (N-m).

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Problem 4.166 The distance s = 4 m. If you represent the force and the 200-N-m couple by a force F acting at the origin O and a couple M, what are F and M?

y (2, 6, 0) m

s 100i 1 20j 2 20k (N)

Solution:

The equivalent force at the origin is

F = 100 i + 20 j − 20 k. The strategy is to establish the position vector of the action point of the force relative to the origin O for the purpose of determining the moment exerted by the force about the origin. The position of the top of the bar is rT = 2 i + 6 j + 0 k. The vector parallel to the bar, pointing toward the base, is rTB = 2 i − 6 j + 3k, with a magnitude of rTB = 7. The unit vector parallel to the bar is

200 N-m

(4, 0, 3) m z

e TB = 0.2857 i − 0.8571 j + 0.4286k. The vector from the top of the bar to the action point of the force is rTF = se TB = 4 e TB = 1.1429 i − 3.4286 j + 1.7143k. The position vector of the action point from the origin is rF = rT + rTF = 3.1429 i + 2.5714 j + 1.7143k. The moment of the force about the origin is

MF = r × F =

i j k 3.1429 2.5714 1.7143 −20 100 20

= −85.71i + 234.20 j − 194.3k. The couple is obtained from the unit vector and the magnitude. The sense of the moment is directed positively toward the top of the bar. M C = −200 e TB = −57.14 i + 171.42 j − 85.72k. The sum of the moments is M = M F + M C = −142.86 i + 405.72 j − 280 k. This is the moment of the equivalent couple at the origin.

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Problem 4.167 1 are

The force F and couple M in system

System 1

System 2 y

F = 12 i + 4 j − 3k (lb), M = 4 i + 7 j + 4 k (ft-lb). Suppose you want to represent system 1 by a wrench (system 2). Determine the couple M p and the coordinates x and z where the line of action of the force intersects the x−z plane.

M F

O z

x z

F x

(x, 0, z)

Solution:

The component of M that is parallel to F is found as follows: The unit vector parallel to F is eF =

F = 0.9231i + 0.3077 j − 0.2308k. F

The component of M parallel to F is M P = (e F ⋅ M)e F = 4.5444 i + 1.5148 j − 1.1361k (ft-lb). The component of M normal to F is M N = M − M P = −0.5444 i + 5.4858 j + 5.1361k (ft-lb). The moment of F must produce a moment equal to the normal component of M. The moment is

MF = r × F =

i j k x 0 z 12 4 −3

= −(4 z ) i + (3 x + 12 z ) j + (4 x )k,

from which −0.5444 = 0.1361 ft −4 5.1362 x = = 1.2840 ft 4

z =

Problem 4.168 A system consists of a force F acting at the origin O and a couple M, where F = 10 i (lb),

M = 20 j (ft-lb).

If you represent the system by a wrench consisting of the force F and a parallel couple M p , what is M p , and where does the line of action of F intersect the y−z plane? Solution:

The component of M parallel to F is zero, since M P = (e F ⋅ M)e F = 0. The normal component is equal to M. The equivalent force must produce the same moment as the normal component

M = r×F =

i j k 0 y z 10 0 0

from which z =

20 = 2 ft and y = 0 10

= (10 z ) j − (10 y )k = 20 j,

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Problem 4.169 Represent the two forces acting on the bar by a wrench consisting of a force F and parallel couple M p . What are F and M p , and where does the line of action of F intersect the x–z plane?

We need to determine the components of M that are parallel and normal (perpendicular) to F. We divide F by its magnitude to obtain a unit vector e that is parallel to F, e =

F = 0.6 i + 0.8 j, F

and use it to determine the component of M parallel to F:

M p = ( e ⋅ M) e = [(0.6)(0) + (0.8)(3 kN-m) + (0)(8 kN-m)](0.6 i + 0.8 j) = 1.44 i + 1.92 j (kN-m). 1m

Then the component of M that is normal to F is 4 j (kN)

Mn = M − Mp

= (3 j + 8k) − (1.44 i + 1.92 j) kN-m = −1.44 i + 1.08 j + 8k (kN-m). 2m x

Now we want to find a new equivalent system consisting of the force F acting at some point in the x -z plane and the couple M p :

3 i (kN)

Solution:

The force F must equal the sum of the forces acting

on the bar:

F = 3i + 4 j (kN). Let’s begin by representing the forces on the bar by a system consisting of the force F acting at the origin and a couple M: y

x To do so, we must choose the point so that the moment due to F about the origin is equal to M n . We set

This yields the equations

z x

For equivalence, M must equal the moment about the origin due to the two forces on the bar:

240

(x, 0, z)

 i j k   M n = −1.44 i +1.08 j + 8k (kN-m) =  x 0 z  .  3 kN 4 kN 0   

M = (1 m)(3 kN) j + (2 m)(4 kN)k = 3 j + 8k (kN-m).

−1.44 kN-m = −(4 kN) z, 1.08 kN-m = (3 kN) z, 8 kN-m = (4 kN) x. We obtain x = 2 m, z = 0.36 m. F = 3i + 4 j (kN), M p = 1.44 i + 1.92 j (kN-m), x = 2 m, z = 0.36 m.

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Problem 4.170 Consider the force F acting at the origin O and the couple M given in Example 4.15. If you represent this system by a wrench, where does the line of action of the force intersect the x−y plane?

Solution:

From Example 4.15 the force and moment are F = 3i + 6 j + 2k (N), and M = 12 i + 4 j + 6k (N-m).

The normal component of the moment is M N = 7.592 i − 4.816 j + 3.061k (N-m). The moment produced by the force must equal the normal component:

MN = r × F =

i j k x y 0 3 6 2

= (2 y ) i − (2 x ) j + (6 x − 3 y )k = 7.592 i − 4.816 j + 3.061k, from which x =

Problem 4.171 Consider the force F acting at the origin O and the couple M given in Example 4.15. If you represent this system by a wrench, where does the line of action of the force intersect the plane y = 3 m?

7.592 4.816 = 2.408 m and y = = 3.796 m 2 2

Solution:

From Example 4.15 (see also Problem 4.170) the force is F = 3i + 6 j + 2k, and the normal component of the moment is M N = 7.592 i − 4.816 j + 3.061k.

The moment produced by the force must be equal to the normal component:

MN = r × F =

i j k x 3 z 3 6 2

= (6 − 6 z ) i − (2 x − 3z ) j + (6 x − 9)k

= 7.592 i − 4.816 j + 3.061k, from which x =

Problem 4.172 A wrench consists of a force of magnitude 100 N acting at the origin O and a couple of magnitude 60 N-m. The force and couple point in the direction from O to the point (1, 1, 2) m. If you represent the wrench by a force F acting at the point (5, 3, 1) m and a couple M, what are F and M?

9 + 3.061 6 − 7.592 = 2.01 m and z = = −0.2653 m 6 6

Solution: The vector parallel to the force is rF = i + j + 2k, from which the unit vector parallel to the force is e F = 0.4082 i + 0.4082 j + 0.8165k. The force and moment at the origin are F = F e OF = 40.82 i + 40.82 j + 81.65k (N), and M = 24.492 i + 24.492 j + 48.99k (N-m). The force and moment are parallel. At the point (5, 3, 1) m the equivalent force is equal to the force at the origin, given above. The moment of this force about the origin is

MF = r × F =

i j k 5 3 1 40.82 40.82 81.65

= 204.13i − 367.43 j + 81.64 k. For the moments to be equal in the two systems, the added equivalent couple must be M C = M − M F = −176.94 i + 391.92 j − 32.65k (N-m)

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Problem 4.173 System 1 consists of two forces and a couple. Suppose that you want to represent it by a wrench (system 2). Determine the force F, the couple M p , and the coordinates x and z where the line of action of F intersects the x−z plane.

Solution:

The sum of the forces in System 1 is F = 300 j + 600 k (N). The equivalent force in System 2 must have this value. The unit vector parallel to the force is e F = 0.4472 j + 0.8944 k. The sum of the moments in System 1 is M = 600(3) i + 300(4)k + 1000 i + 600 j = 2800 i + 600 j + 1200 k (kN m).

System 1 y 1000i

600j (kN-m)

600k (kN)

System 2 y

M P = 599.963 j + 1199.93k (kN-m) = 600 j + 1200 k (kN-m).

300j (kN)

The normal component is M N = M − M P = 2800 i. The moment of the force

F x

The component parallel to the force is

MN =

(x, 0, z)

i j k x 0 z 0 300 600

= −300 zi − 600 xj + 300 xk = 2800 i,

from which x = 0, z =

Problem 4.174 A plumber exerts the two forces shown to loosen a pipe. (a) What total moment does the plumber exert about the axis of the pipe? (b) If you represent the two forces by a force F acting at O and a couple M, what are F and M? (c) If you represent the two forces by a wrench consisting of the force F and a parallel couple M p , what is M p , and where does the line of action of F intersect the x–y plane?

2800 = −9.333 m −300

Solution:

The sum of the forces is

∑ F = 50 k − 70 k = −20 k (lb). (a) The total moment exerted on the pipe is M = 16(20) i = 320 i (ft lb). (b) The equivalent force at O is F = −20 k. The sum of the moments about O is ∑ M O = (r1 × F1 ) + (r2 × F2 )

y 12 in O

= 6 in

i j k 12 −16 0 0 0 50

i j k 18 −16 0 0 0 −70

= 320 i + 660 j.

z x 16 in 16 in

(c) The unit vector parallel to the force is e F = k, hence the moment parallel to the force is M P = (e F ⋅ M)e F = 0, and the moment normal to the force is M N = M − M P = 320 i + 660 j. The force at the location of the wrench must produce this moment for the wrench to be equivalent. MN =

50k (lb) 270k (lb)

i j k x y 0 0 0 −20

from which x =

242

= −20 yi + 20 xj = 320 i + 660 j,

660 320 = 33 in, y = = −16 in 20 −20

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Problem 4.175 The weights of the masses m1 and m 2 exert forces on the bar AB. The magnitude of the total moment due to the two forces about A is 275 N-m . The magnitude of the total moment due to the two forces about B is 343 N-m. Determine the masses m1 and m 2 .

Solution: 0.35 m

0.35 m A

m1 g 0.35 m

0.35 m

m2 g

0.35 m

A B

The weights of the masses are m1g = m1 (9.81 m/s 2 ) and m 2 g = m 2 (9.81 m/s 2 ). The magnitude of the moment exerted about point A is (0.35 m)m1 (9.81 m/s 2 ) + (0.70 m)m 2 (9.81 m/s 2 ) = 275 N-m.

The magnitude of the moment exerted about point B is (0.70 m)m1 (9.81 m/s 2 ) + (0.35 m)m 2 (9.81 m/s 2 ) = 343 N-m. Solving these two equations, we obtain m1 = 39.9 kg, m 2 = 20.1 kg. m1 = 39.9 kg, m 2 = 20.1 kg.

Problem 4.176 The cable AB exerts a 300-N force on the support A that points from A toward B. Determine the magnitude of the moment the force exerts about point P .

Solution: 0.7 i + 0.3 j F = (300 N) , 0.58 rPA = (−0.9 i + 0.5 j) m

(0.3, 0.6) m

A ( 0.4, 0.3) m x

M P = rPA × F = −(244 Nm)k ⇒ M P = 244 Nm P (0.5,

Problem 4.177 Three forces act on the structure. The sum of the moments due to the forces about A is zero. Determine the magnitude of the force F .

0.2) m

2 kN

4 kN

Solution: ∑ M A = −(4 kN)( 2b) − (2 kN cos30 °)3b

+ (2 kN sin 30 °)b + F (4 b) = 0 Solving we find

F b

F = 2.463 kN

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Problem 4.178 The cables A, B , and C exert forces on the vertical post at D . The force exerted by cable A is FA = 8 kN. The total downward force (the vertical component of the total force) exerted at D by the three cables is 12 kN. The magnitude of the moment about E due to the forces exerted at D by the three cables is 76 kN-m . Determine the forces FB and FC .

Solution: D a b

FC FB

6m FA E

D FC

D FA

The angles α = arctan(6/4) = 56.3 °, β = arctan(6/8) = 36.9 °,

γ = arctan(6/12) = 26.6 °.

The total downward force is

E 4m

FA sin α + FB sin β + FC sin γ = 12 kN. 4m

The magnitude of the total moment about E is (6 m)( FA cos α + FB cos β + FC cos γ ) = 76 kN-m. Solving these two equations with FA = 8 kN, we obtain FB = 6.15 kN, FC = 3.70 kN. FB = 6.15 kN, FC = 3.70 kN.

Problem 4.179 Determine the sum of the moments exerted about A by the three forces and the couple.

Solution:

Establish coordinates with origin at A, x horizontal, and y vertical with respect to the page. The moment exerted by the couple is the same about any point. The moment of the 300 lb force about A is M 300 = (−6 i − 5 j) × (300 j) = −1800 k ft-1b.

5 ft

300 lb

800 ft-lb 200 lb

The moment of the downward 200 lb force about A is zero since the line of action of the force passes through A. The moment of the 200 lb force which pulls to the right is

200 lb 6 ft

M 200 = (3i − 5 j) × (200 i) = 1000 k (ft-lb).

3 ft

The moment of the couple is M C = −800 k (ft-lb). Summing the four moments, we get M A = (−1800 + 0 + 1000 − 800)k = −1600 k (ft-lb)

Problem 4.180 If you represent the three forces and the couple by an equivalent system consisting of a force F acting at A and a couple M, what are the magnitudes of F and M? Solution:

The equivalent force will be equal to the sum of the forces and the equivalent couple will be equal to the sum of the moments about A. From the solution to Problem 4.189, the equivalent couple will be C = M A = −1600 k (ft-lb). The equivalent force will be FEQUIV. = 200 i − 200 j + 300 j = 200 i + 100 j (lb).

A 5 ft

300 lb

800 ft-lb 200 lb 200 lb 6 ft

244

3 ft

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Problem 4.181 The vector sum of the forces acting on the beam is zero, and the sum of the moments about A is zero. (a) What are the forces A x , A y , and B? (b) What is the sum of the moments about B? Solution:

30 220 mm

The vertical and horizontal components of the 400 N

force are:

400 N

260 mm

FX = 400 cos30 ° = 346.41 N, FY = 400 sin 30° = 200 N.

500 mm

The sum of the forces is

∑ FX = A X + 346.41 = 0, from which A X = −346.41 N ∑ FY = AY + B − 200 = 0. The sum of the moments about A is ∑ M A = 0.5B − 0.22(346.41) = 0, from which B = 152.42 N. Substitute into the force equation to get AY = 200 − B = 47.58 N (b) The moments about B are M B = −0.5 AY − 0.48(346.41) − 0.26 A X + 0.5(200) = 0

Problem 4.182 The hydraulic piston BC exerts a 970-lb force on the boom at C in the direction parallel to the piston. The angle α = 40°. The sum of the moments about A due to the force exerted on the boom by the piston and the weight of the suspended load is zero. What is the weight of the suspended load?

Solution: The horizontal (x) and vertical (y) coordinates of point C relative to point B are x = (9 ft) cos α − (6 ft) = 0.894 ft y = (9 ft)sin α = 5.79 ft The angle between the piston BC and the horizontal is β = tan −1 ( y /x ) = 81.2 °

The horizontal and vertical components of the force exerted by the piston at C are C x = (970 lb) cos β = 148 lb C y = (970 lb)sin β = 959 lb

The sum of the moments about A due to the pistion force and the suspended weight W is ∑ M A = −W (15 ft) cos α + C y (9 ft) cos a − C x (9 ft)sin α = 0

Solving, yields W = 501 lb

6 ft

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Problem 4.183 The tension in cable BC is 200 N. Determine the magnitude of the moment about A due to the force cable BC exerts on bar AB at B.

Solution: y (0, 7, 3) m

(0, 7, 3) m

rBC

(0, 6, 25) m y

rAB

B (4, 4, 1) m

C B (4, 4, 1) m

A z

The vector from A to B is

r AB = 4 i + 4 j + k (m). x z

The vector from B to C is rBC = (0 − 4) i + (7 − 4) j + (3 − 1)k (m) = −4 i + 3 j + 2k (m). We can express the force exerted at B by cable BC as TBC = (200 N)

rBC rBC

= −149 i + 111 j + 74.3k (N). The moment due to this force about A is M A = r AB × TBC   i j k    4m 4m 1 m  =   −149 N 111 N 74.3 N    = 186 i − 446 j + 1040 k (N-m). Its magnitude is M A = 1150 N-m. 1.15 kN-m.

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Problem 4.184 The tension in cable BD is 200 N. Determine the magnitude of the moment about A due to the force cable BD exerts on bar AB at B.

Solution: y

D (0, 6, 25) m rBD

(0, 6, 25) m y

(0, 7, 3) m

rAB

B (4, 4, 1) m

C B (4, 4, 1) m

A z

The vector from A to B is r AB = 4 i + 4 j + k (m). x m

The vector from B to D is rBD = (0 − 4) i + (6 − 4) j + (−5 − 1)k (m) = −4 i + 2 j − 6k (m). We can express the force exerted at B by cable BD as TBD = (200 N)

rBD rBD

= −107 i + 53.5 j − 160 k (N). The moment due to this force about A is M A = r AB × TBD   i j k    4m 4m 1m =    −107 N 53.5 N −160 N    = −695i + 535 j + 641k (N-m). Its magnitude is M A = 1090 N-m. 1.09 kN-m.

Problem 4.185 What is the total moment due to the two couples? (a) Express the answer by giving the magnitude and stating whether the moment is clockwise or counterclockwise. (b) Express the answer as a vector. Solution: (a) The couple in which the forces are 4 m apart exerts a counterclockwise moment of magnitude (100 N)(4 m) = 400 N-m. The couple in which the forces are 8 m apart exerts a clockwise moment of magnitude (100 N)(8 m) = 800 N-m. The sum of their moments is a clockwise moment of 400 N-m. (b) The vector representation of the clockwise moment of 400 N-m magnitude is −400 k (N-m). This expression can also be obtained by calculating the sum of the moments of the four forces about any point. The sum of the moments about the origin is

y 100 N

100 N 2m x 2m

100 N

M = (2 m) i × (100 N) j + (−2 m) i × (−100 N) j + (4 m) j × (100 N) i + (−4 m) j × (−100 N) i = (−400 N-m)k (a) 400 N-m clockwise (b) − 400 k N-m

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Problem 4.186 The bar AB supporting the lid of the grand piano exerts a force F = −6 i + 35 j − 12k (lb) at B . The coordinates of B are (3, 4, 3) ft . What is the moment of the force about the hinge line of the lid (the x-axis )?

y B

x A

Solution:

The position vector of point B is rOB = 3i + 4 j + 3k. The moment about the x-axis due to the force is

M X = e X ⋅ (rOB × F) = i ⋅ (rOB × F) MX =

1 0 0 3 4 3 −6 35 −12

= −153 ft lb

Problem 4.187 Determine the moment of the vertical 800-lb force about point C .

y 800 lb

Solution:

The force vector acting at A is F = −800 j (lb) and the position vector from C to A is

A (4, 3, 4) ft

rCA = ( x A − x C ) i + ( y A − y C ) j + ( z A − z C )k

D (6, 0, 0) ft

= (4 − 5) i + (3 − 0) j + (4 − 6)k = −1i + 3 j − 2k (ft).

The moment about C is

MC =

i j k −1 3 −2 0 −800 0

C (5, 0, 6) ft

= −1600 i + 0 j + 800 k (ft-1b)

Problem 4.188 Determine the moment of the vertical 800-lb force about the straight line through points C and D .

Solution:

In Problem 4.197, we found the moment of the 800 lb force about point C to be given by M C = −1600 i + 0 j + 800 j (ft-1b). The vector from C to D is given by

rCD = ( x D − x C ) i + ( y D − y C ) j + ( z D − z C )k

800 lb

= (6 − 5) i + (0 − 0) j + (0 − 6)k = 1i + 0 j − 6 j (ft),

A (4, 3, 4) ft

and its magnitude is

D (6, 0, 0) ft x

rCD =

12 + 6 2 =

37 (ft).

The unit vector from C to D is given by

C (5, 0, 6) ft

e CD =

1 i− 37

6 k. 37

The moment of the 800 lb vertical force about line CD is given by

( 137 i − 637 k ) ⋅ (−1600i + 0 j + 800 j (ft-lb)) −1600 − 4800 = ( )(ft-lb). 37

M CD =

Carrying out the calculations, we get M CD = −1052 (ft-lb)

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Problem 4.189 The system of cables and pulleys supports the 300-lb weight of the work platform. If you represent the upward force exerted at E by cable EF and the upward force exerted at G by cable GH by a single equivalent force F, what is F, and where does its line of action intersect the x-axis?

H F

Solution: The cable-pulley combination does not produce a moment. Hence the equivalent force does not. The equivalent force is 600 equal to the total supported weight, or F = + j = 300 j (lb). The 2 8 force occurs at midpoint of the platform width, x = = 4 ft. 2

608

C x

8 ft

Problem 4.190 The system of cables and pulleys supports the 300-lb weight of the work platform. (a) What are the tensions in cables AB and CD? (b) If you represent the forces exerted on the work platform by the cables at A and C by a single equivalent force F, what is F and where does its line of action intersect the x-axis?

H F

Solution:

The vertical component of the tension is each cable must equal half the weight supported. 150 = 173.2 lb. By T AB sin 60 ° = 150 lb, from which T AB = sin 60 ° symmetry, the tension TCD = 173.2 lb.

608

The single force must equal the sum of the vertical components; since there is no resultant moment produced by the cables, the force is F = 300 j 1b and it acts at the platform width midpoint x = 4 ft.

C x

8 ft

Problem 4.191 The two systems are equivalent. Determine the forces A x and A y , and the couple M A . Solution:

608

System 1 y 20 N

The sum of the forces for System 1 is

∑ FX = ( A X + 20) i, 400 mm

∑ FY = ( AY + 30) j. The sum of forces for System 2 is

∑ FX = (−20) i and ∑ FY = (80 − 10) j.

30 N 600 mm

Equating the two systems: A X + 20 = −20 from which A X = −40 N

400 mm

System 2

AY + 30 = 80 − 10 from which AY = 40 N

The sum of the moments about the left end for System 1 is

8 N-m

∑ M 1 = −(0.4)(20) + 30(1) = 22 N-m. 400 mm

The sum of moments about the left end for System 2 is ∑ M 2 = M A − 10(1) − 8 = M A − 18.

20 N

10 N x

Equating the moments for the two systems: M A = 18 + 22 = 40 N-m

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80 N

600 mm

400 mm

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Problem 4.192 If you represent the equivalent systems in Problem 4.191 by a force F acting at the origin and a couple M , what are F and M ?

System 1 y 20 N

Solution:

Summing the forces in System 1, F = ( A X + 20) i + ( AY + 30) j. Substituting from the solution in Problem 4.201, F = −20 i + 70 j. The moment is M = −20(0.4)k + 30 k = 22k (N-m).

400 mm Ax

x Ay

30 N 600 mm

400 mm

System 2 y 8 N-m

400 mm MA

20 N

x 80 N

Problem 4.193 If you represent the equivalent systems in Problem 4.191 by a force F, what is F, and where does its line of action intersect the x-axis? Solution:

10 N

600 mm

400 mm

System 1 y 20 N

The force is F = −20 i + 70 j. The moment to be rep-

resented is M = (r × F) = 22k =

from which x =

i j k x 0 0 −20 70 0

400 mm = 70 xk,

x Ay

22 = 0.3143 m 70

30 N 600 mm

400 mm

System 2 y 8 N-m

400 mm 20 N

x 80 N

250

10 N

600 mm

400 mm

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Problem 4.194

Solution:

The two systems are equivalent. If

The sum of forces in the two systems must be equal, thus F′ = F = −100 i + 40 j + 30 k (lb).

F = −100 i + 40 j + 30 k (lb),

The moment for the unprimed system is M T = r × F + M.

M ′ = −80 i + 120 j + 40 k (in-lb),

The moment for the primed system is M′T = r ′ × F + M′.

determine F′ and M.

M F

System 1 y

System 2 y

4 in

The position vectors are r = 0 i + 6 j + 6k, and r ′ = 4 i + 6 j + 6k. Equating the moments and solving for the unknown moment M = M′ + (r ′ − r ) × F = −80 i + 120 j + 40 k + F9

6 in

= −80 i + 120 j + 40 k − 120 j + 160 k

6 in

i j k 4 0 0 −100 40 30

= −80 i + 200 k (in-1b) x

6 in

6 in z

Problem 4.195 The tugboats A and B exert forces FA = 1 kN and FB = 1.2 kN on the ship. The angle θ = 30°. If you represent the two forces by a force F acting at the origin O and a couple M , what are F and M ?

Solution:

The sums of the forces are:

∑ FX = (1 + 1.2 cos30 °) i = 2.0392 i (kN) ∑ FY = (1.2sin 30 °) j = 0.6 j (kN). The equivalent force at the origin is FEQ = 2.04 i + 0.6 j

The moment about O is M O = r A × F A + rB × FB . The vector positions are r A = −25i + 60 j (m), and rB = −25i − 60 j (m).

A FA

The moment:

60 m O 60 m

MO =

i j k −25 60 0 1 0 0

i j k −25 −60 0 1.0392 0.6 0

= −12.648k = −12.6k (kN-m) Check: Use a two dimensional description: The moment is

M O = −(25) FB sin 30 ° + (60)( FB cos30°) − (60)( FA ) = 39.46 FB − 60 FA = −12.6 kN-m 25 m

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Problem 4.196 The tugboats A and B exert forces FA = 600 N and FB = 800 N on the ship. The angle θ = 45°. If you represent the two forces by a force F, what is F, and where does its line of action intersect the y-axis? Solution:

The equivalent force is

A FA

F = (0.6 + 0.8cos 45 °) i + 0.8sin 45 ° j = 1.1656 i + 0.5656 j (kN). 60 m

The moment produced by the two forces is

M O = r A × F A + rB × FB . The vector positions are

60 m

r A = −25i + 60 j (m), and rB = −25i − 60 j (m). The moment: MO =

i j k −25 60 0 0.6 0 0

u +

i j k −25 −60 0 0.5656 0.5656 0

= −16.20 k (kN-m)

25 m

Check: Use a two dimensional description: M O = −(25) FB sin 45 ° + (60) FB cos 45 ° − 60 FA = 24.75FB − 60 FA = −16.20 kN-m. The single force must produce this moment.

MO =

i j k 0 y 0 1.1656 0.5656 0

= −1.1656 yk = −16.20 k,

from which y =

252

16.20 = 13.90 m 1.1656

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Problem 4.197 The tugboats A and B want to exert two forces on the ship that are equivalent to a force F acting at the origin O of 2-kN magnitude. If FA = 800 N, determine the necessary values of FB and θ.

Solution:

The equivalent force at the origin is ( FA + FB cos θ) 2 + ( FB sin θ) 2 = 2000 2. The moment about the origin due to FA and FB must be zero: ∑ M O = −60 FA + 60 FB cos θ − 25FB sin θ = 0. These are two equations in two unknowns FB sin θ and FB cos θ For brevity write x = FB cos θ, y = FB sin θ, so that the two equations become x 2 + 2 FA x + FA 2 + y 2 = 2000 2 and 60 x − 25 y − 60 FA = 0. Eliminate y by solving each equation for y 2 and equating the results:

60 F + x) . ( 60 25 25

y 2 = 2000 2 − x 2 − 2 FA x − FA 2 = − Reduce to obtain the quadratic in x:

   60 2  2 60 2  60 2  2 1 +  x + 2 FA  1 −  x + 1 +  F − 2000 2 = 0.    25  25  25  A

( )

FA 60 m O 60 m

( )

Substitute FA = 800 N to obtain 6.76 x 2 − 7616 x + 326400 = 0. In canonical form: x 2 + 2bx + c = 0, where b = −563.31, and c = 48284.0, with the solutions x = −b ± b 2 − c = 1082.0, = 44.62. From the second equation, y = −1812.9, = 676.81. The force FB has two solutions: Solve for FB and θ: (1) FB =

( )

44.6 2 + 1812.9 2 = 1813.4 N

at the angle 25 m

θ = tan −1 FB =

( −1812.9 ) = −88.6°, and (2) 44.6

676.8 2 + 1082.0 2 = 1276.2 N,

at the angle θ = tan −1

Problem 4.198 If you represent the forces exerted by the floor on the table legs by a force F acting at the origin O and a couple M, what are F and M?

676.8 ( 1082.0 ) = 32.0°

y 1m

Solution:

The sum of the forces is the equivalent force at the origin. F = (50 + 48 + 50 + 42) j = 190 j (N). The position vectors of the legs are, numbering the legs counterclockwise from the lower left in the sketch: r1 = +1k,

50 N

r2 = 2 i + 1k, r3 = 2 i, r4 = 0.

i j k 0 0 1 0 48 0

i j k 2 0 1 0 50 0

50 N

The sum of the moments about the origin is MO =

42 N

48 N

i j k 2 0 0 0 42 0

= −98i + 184 k (N-m). This is the couple that acts at the origin.

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Problem 4.199 If you represent the forces exerted by the floor on the table legs by a force F, what is F, and where does its line of action intersect the x–z plane?

y 1m

Solution:

From the solution to Problem 4.198 the equivalent force is F = 190 j. This force must produce the moment M = −98i + 184 k obtained in Problem 4.198. i j k x 0 z 0 190 0

M =

= −190 zi + 190 xk = −98i + 184 k,

50 N

from which

42 N

48 N

50 N

184 = 0.9684 m and 190 98 z = = 0.5158 m 190 x =

Problem 4.200 Two forces are exerted on the crankshaft by the connecting rods. The direction cosines of F A are cos θ x = −0.182, cos θ y = 0.818, and cos θ z = 0.545, and its magnitude is 4 kN. The direction cosines of FB are cos θ x = 0.182, cos θ y = 0.818, and cos θ z = −0.545, and its magnitude is 2 kN. If you represent the two forces by a force F acting at the origin O and a couple M, what are F and M? y

Solution:

F A = 4(−0.182 i + 0.818 j + 0.545k) = −0.728i + 3.272 j + 2.18k (kN) FB = 2(0.182 i + 0.818 j − 0.545k) = 0.364 i + 1.636 j − 1.09k (kN). The sum: F A + FB = −0.364 i + 4.908 j + 1.09k (kN) The equivalent couple is the sum of the moments. M = r A × F A + rB × FB . The position vectors are: r A = 0.16 i + 0.08k,

The sum of the moments:

360 mm

O M =

254

The equivalent force is the sum of the forces:

rB = 0.36 i − 0.08k.

160 mm

80 mm 80 mm x

i j k 0.16 0 0.08 −0.728 3.272 2.180

j k i 0 −0.08 0.36 0.364 1.636 −1.090

M = −0.1309 i − 0.0438 j + 1.1125k (kN-m)

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Problem 4.201 If you represent the two forces exerted on the crankshaft in Problem 4.200 by a wrench consisting of a force F and a parallel couple M p , what are F and M p , and where does the line of action of F intersect the x–z plane? Solution:

FA 360 mm

From the solution to Problem 4.200,

F = −0.364 i + 4.908 j + 1.09k (kN) and M = −0.1309 i − 0.0438 j + 1.1125k (kN-m). The unit vector parallel to F is eF =

160 mm

80 mm 80 mm x

F = −0.0722 i + 0.9737 j + 0.2162k. F

The moment parallel to the force is M P = ( e F ⋅ M) e F . Carrying out the operations: M P = 0.2073e F = −0.01497 i + 0.2019 j + 0.0448k (kN-m). This is the equivalent couple parallel to F. The component of the moment perpendicular to F is M N = M − M P = −0.1159 i − 0.2457 j + 1.0688k. The force exerts this moment about the origin.

MN =

i j k x 0 z −0.364 4.908 1.09

= (−4.908z ) i − (1.09 x + 0.364 z ) j + (4.908 x )k = −0.1159 i − 0.2457 j + 1.06884 k. From which 1.0688 = 0.2178 m, 4.908 0.1159 z = = 0.0236 m 4.908 x =

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Chapter 5 Problem 5.1 In Practice Example 5.1, suppose that the beam is subjected to a 6 kN-m counterclockwise couple at the right end in addition to the 4-kN downward force. Draw a sketch of the beam showing its new loading. Draw the free-body diagram of the beam and apply the equilibrium equations to determine the reactions at A.

Solution:

The equilibrium equations are

ΣFx : A x = 0 ΣFy : A y − 4 kN = 0 ΣM A : M A − (4 kN)(2 m) + (6 kN-m) = 0 Solving yields Ax = 0 A y = 4 kN M A = 2 kN-m

Problem 5.2 The beam has a fixed support at A and is loaded by two forces and a couple. Draw the free-body diagram of the beam and apply equilibrium to determine the reactions at A. 4 kN A

6 kN-m

256

1.5 m

Solution:

The free-body diagram is drawn.

The equilibrium equations are ΣFx : A x + (2 kN) cos60 ° = 0 ΣFy : A y + (4 kN) + (2 kN)sin 60 ° = 0

2 kN 608

ΣM A : M A + (6 kN-m) + (4 kN)(2.5 m) + (2 kN)sin 60 °(4 m = 0) We obtain: A x = −1 kN, A y = −5.73 kN, M A = −22.9 kN-m

1.5 m

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Problem 5.3 The structure is loaded by three forces. Determine the reactions at its supports A and B.

Solution:

We draw the free-body diagram of the structure:

308 A

10 kN

Ax 0.5 m

308

10 kN

0.5 m

D 2 kN

2 kN

0.6 m 0.6 m

0.6 m

2 kN

x 2 kN

0.6 m

Writing the equilibrium equations, ΣFx = A x + B − (10 kN)sin 30 ° = 0, ΣFy = A y + 2 kN + 2 kN − (10 kN) cos30 ° = 0, ΣM point B = −(0.5 m) A x + (0.6 m)(2 kN) + (1.2 m)(2 kN) + (0.5 m)(10 kN)sin 30° − (0.6 m)(10 kN) cos30° = 0, and solving, we obtain A x = 1.81 kN, A y = 4.66 kN, B = 3.19 kN. A x = 1.81 kN, A y = 4.66 kN, B = 3.19 kN.

Problem 5.4 (a) Draw the free-body diagram of the beam. (b) Determine the tension in the rope and the reactions at B.

308

308 600 lb B

5 ft

Solution:

9 ft

Let T be the tension in the rope.

The equilibrium equations are: ΣFx : −T sin 30 ° − (600 lb)sin 30 ° + B x = 0 ΣFy : T cos30 ° − (600 lb) cos30 ° + B y = 0 ΣM B : (600 lb) cos30 °(9 ft) − T cos30 °(14 ft) = 0 Solving yields T = 368 lb, B x = 493 lb, B y = 186 lb

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Problem 5.5 The forces acting on the free-body diagram of a person’s forearm as they hold a suspended mass m are shown. The mass m = 10 kg, and the mass of his forearm is m F = 2.2 kg. Determine the force FB exerted by the biceps muscle and the reactive force FE on the elbow joint.

Solution:

The sum of the forces on the free-body diagram of the arm equals zero, FB − FE − m F g − mg = 0,

mF g

and the sum of the moments about the elbow joint equals zero,

(30 mm) FB − (130 mm)m F g − (350 mm)mg = 0. Solving with m = 10 kg, m F = 2.2 kg, and g = 9.81 m/s 2 , we obtain FB = 1.24 kN, FE = 1.12 kN.

30 mm

FB = 1.24 kN, FE = 1.12 kN.

Problem 5.6 The forces acting on the free-body diagram of a person’s forearm as they hold a suspended mass m are shown. (This is a more complete model of the situation in Problem 5.5.) The dimensions and angles are not drawn to scale so that the forces can be distinguished more clearly. The mass m = 10 kg, and the mass of his forearm is m F = 2.2 kg. In addition to the force FB exerted by the biceps muscle and the reactive force FE on the elbow joint, the forearm is subjected to the force FC exerted by the brachialis muscle and the force FD exerted by the brachioradialis muscle. Assume that FC = 0.25FB and FD = 0.16 FB . Determine the magnitudes of the forces FB and FE and the angle α.

Solution:

100 mm

220 mm

The sum of the moments about the elbow joint equals zero,

(30 mm) FB + (45 mm)FC sin 75 ° + (220 mm)FD sin12 ° − (130 mm)m F g − (350 mm)mg = 0. Setting FC = 0.25FB , FD = 0.16 FB , m = 10 kg, m F = 2.2 kg, and g = 9.81 m/s 2 and solving, we obtain FB = 771 N. Now we set the sums of the forces perpendicular and parallel to the arm equal to zero: FB + FC sin 75 ° + FD sin12 ° − FE sin α − m F g − mg = 0, −FC cos 75 ° − FD cos12 ° + FE cos α = 0. With FB , FC , and FD known, we can solve these two equations for FE sin α and FE cos α, obtaining FE sin α = 863 N and FE cos α = 171 N. Using these values, we determine that FE = 880 N, α = 78.8 °. FB = 771 N, FE = 880 N, α = 78.8 °.

758

128 FC

m mF g 90 30 15 85 mm mm mm mm

258

mg 130 mm

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Problem 5.7 A person holds a mass m with their arm extended horizontally. The mass m = 5 kg, and the mass of his arm is m A = 3.6 kg. The angle α = 15 °. Determine the force FD exerted by the deltoid muscle and the components S x and S y of the reaction at the shoulder joint. Solution:

The sum of the moments about the shoulder joint

equals zero, (120 mm)FD sin α − (300 mm)m A g − (620 mm)mg = 0.

Setting α = 15 °, m = 5 kg, m A = 3.6 kg, and g = 9.81 m/s 2 and solving, we obtain FD = 1290 N. Now we set the sums of the forces perpendicular and parallel to the arm equal to zero:

S x − FD cos α = 0, −S y + FD sin α − m A g − mg = 0.

mAg

Solving yields S x = 1240 N, S y = 249 N.

120 mm

FD = 1320 N, S x = 1280 N, S y = 257 N.

180 mm

mg 320 mm x

Problem 5.8 The distance x = 9 m. (a) Draw the free-body diagram of the beam. (b) Determine the reactions at the supports.

Solution: 10 kN

10 kN

x59m

Ax A

B 6m x

6m Ay

(a) The FBD (b) The equilibrium equations ΣFx : A x = 0 ΣFy : A y + B y − 10 kN = 0 ΣM A : B y (6 m) − (10 kN)(9 m) = 0 Solving we find A x = 0, A y = −5 kN, B y = 15 kN

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Problem 5.9 The mass of bar AB is 14 kg. Assume that the weight of the bar acts at its midpoint. Determine the reactions on the bar at A, and the tension in the cable BC.

Solution: The weight of the bar is W = (14 kg) (9.81 m/s 2 ) = 137 N. Denote the tension in the cable by T. The free-body diagram of the bar is y T 258

258

0.8 m B

W Ax 0.8 m

x Ay 0.6 m

0.6 m

The equilibrium equations are ΣFx = A x − T cos 25 ° = 0,

1.2 m

ΣFy = A y + T sin 25 ° − W = 0, ΣM point A = (1.2 m)T sin 25 ° + (0.8 m)T cos 25 ° − (0.6 m)W = 0. Solving, we obtain A x = 60.6 N, A y = 109 N, T = 66.9 N. A x = 60.6 N, A y = 109 N, tension = 66.9 N.

Problem 5.10 (a) Draw the free-body diagram of the beam. (b) Determine the reactions at the supports. 400 lb

100 lb

Solution: (a) Both supports are roller supports. The free body diagram is shown. (b) The sum of the forces: ΣFX = 0,

900 ft-lb B A 3 ft

and ΣFY = FA + FB + 100 − 400 = 0. The sum of the moments about A is

4 ft

3 ft

4 ft

ΣM A = −3(100) + 900 − 7(400) + 11FB = 0. From which FB =

2200 = 200 lb 11

Substitute into the force balance equation to obtain FA = 300 − FB = 100 lb 100 lb 3 ft

400 lb 4 ft

A 100 lb 3 ft

4 ft

900 ft lb

4 ft

260

3 ft

400 lb 4 ft

900 ft lb

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Problem 5.11 The mass of the pulley is 20 kg, and the mass of the suspended object is 10 kg. Determine the reactions exerted on the pulley at C by the horizontal bar.

Solution:

We draw the free-body diagram of the pulley, y

0.4 m Cy

0.4 m T

mP g T

0.8 m

where m P = 20 kg is the mass of the pulley. From the freebody diagram of the suspended mass (not shown), we know that the tension in the cable is equal to the weight of the mass, T = (10 kg) (9.81 m/s 2 ) = 98.1 N. By summing moments about the center of the pulley, we also know that the tension T in the cable is the same on both sides of the pulley. The angle α = arc sin(0.4 m/0.8 m) = 30 °. From the equilibrium equations ΣFx = C x − T cos α = 0, ΣFy = C y − T − T sin α − m P g = 0, we obtain C x = 85.0 N, C y = 343 N. C x = 85.0 N, C y = 343 N.

Problem 5.12 The mass of the pulley is 20 kg, and the mass of the suspended object is 10 kg. The mass of the horizontal bar is 6 kg. Assume that the bar’s weight acts at B. What are the reactions at the built-in support A?

Solution: Denote the masses by m B (bar), m P (pulley), and m S (suspended mass). The free-body diagram of the system is y MA

0.4 m A

0.4 m

x Ay 0.8 m

0.8 m

mP g

mB g 0.8 m

0.8 m mS g

The equilibrium equations are ΣFx = A x = 0, ΣFy = A y − m B g − m P g − m S g = 0, ΣM point A = M A − (0.8 m)m B g − (1.6 m)m P g − (2.0 m)m S g = 0. Solving, we obtain A x = 0, A y = 353 N, M A = 557 N-m. A x = 0, A y = 353 N, M A = 557 N-m.

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Problem 5.13 Determine the tension T in the rope AB and the reactions on the bar at C.

Solution:

We draw the free-body diagram of the bar: y

0.3 m

30 N-m

0.5 m

30 N-m

50 N

0.3 m

1.0 m

0.4 m

50 N 0.4 m

The angle α = arctan(7/13) = 28.3 °. The equilibrium equations are ΣFx = C x − T cos α = 0,

0.5 m

ΣFy = C y + T sin α − 50 N = 0, ΣM point C = −(0.4 m) (50 N) + (1.4 m)T sin α + (0.3 m)T cos α − 30 N-m = 0. Solving, we obtain T = 53.9 N, C x = 47.4 N, C y = 24.5 N. T = 53.9 N, C x = 47.4 N, C y = 24.5 N.

Problem 5.14 bar at B.

Determine the reactions on the L-shaped

Solution:

We draw the free-body diagram of the bar: y 0.7 m

608

4 kN-m

20 kN

308

4 kN-m

0.4 m

Bx By

B The equilibrium equations are 0.7 m

ΣFx = A cos30 ° + B x − 20 kN = 0, ΣFy = A sin 30 ° + B y = 0, ΣM point A = (0.4 m) B x + (0.7 m) B y − 4 kN-m = 0. Solving, we obtain A = 5.74 kN, B x = 15.0 kN, B y = −2.87 kN. B x = 15.0 kN, B y = −2.87 kN.

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Problem 5.15* The mass of the rectangular plate is 20 kg. Assume that its weight acts at its midpoint. By solving the equilibrium equations for the plate, determine the angle α and the tensions in the cables AB and CD.

The equilibrium equations are (notice that we obtain a simpler moment equation by summing moments about point A) ΣFx = −T AB cos α + TCD cos 45 ° = 0, ΣFy = T AB sin α + TCD sin 45 ° − W = 0,

458

0.4 m

ΣM point A = (1.0 m) TCD sin 45 ° − (0.3 m)W = 0. Solving the third equation, we obtain TCD = 83.2 N. The first two equations can be written T AB sin α = W − TCD sin 45 °,

(1)

T AB cos α = TCD cos 45 °.

(2)

Dividing Eq. (1) by Eq. (2) yields

0.4 m

tan α =

1.0 m

Solution: The weight of the plate is W = (20 kg) (9.82 m/s 2 ) = 196 N. Denote the tensions in the cables by T AB and TCD . The freebody diagram of the plate is

W − TCD sin 45 ° . TCD cos 45 °

From this equation we determine that α = 66.8 °. Then from Eq. (2) we obtain T AB = 149 N. α = 66.8 °, AB : 149 N, CD : 83.2 N.

y TAB a

TCD

0.2 m

458

W 0.3 m

0.7 m

Problem 5.16 A person doing push-ups pauses in the position shown. His 180-lb weight W acts at the point shown. The dimensions a = 15 in, b = 42 in, and c = 16 in. Determine the normal force exerted by the floor on each of his hands and on each of his feet.

Solution:

The free-body diagram is shown. The equilibrium equa-

tions are ΣFy : 2 H + 2 F − 180 lb = 0 ΣM H : −Wa + 2 F (a + b) 0 We find that H = 66.3 lb, F = 23.7 lb Thus

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66.3 lb on each hand 23.7 lb on each foot

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Solution:

Problem 5.17 The hydraulic piston AC exerts a 6-kN force on the horizontal beam at C that points in the direction from A toward C. Determine the reactions on the beam at B and D.

We draw the free-body diagram of the bar:

y 6 kN a

D 0.8 m

0.5 m C

0.4 m

The angle α = arctan(0.5 m/0.8 m) = 32.0 °. From the equilibrium equations

ΣFx = B x + (6 kN) cos α = 0, ΣFy = B y + D − (6 kN)sin α = 0, ΣM point B = (1.2 m) D − (0.8 m)(6 kN)sin α = 0,

0.8 m

we obtain B x = −5.09 kN, B y = 1.06 kN, D = 2.12 kN.

0.4 m

B x = −5.09 kN, B y = 1.06 kN, D y = 2.12 kN.

Problem 5.18 A and B.

Solution:

Determine the reactions on the beam at

We draw the free-body diagram of the bar:

3 ft

4 ft

140 ft-lb x

140 ft-lb

6 ft

50 lb

20 lb

50 lb

20 lb

3 ft

B 6 ft

4 ft

The equilibrium equations are ΣFx = A x = 0, ΣFy = A y + B + 20 lb − 50 lb = 0, ΣM point A = (10 ft) B − (3 ft)(20 lb) − (6 ft)(50 lb) + 140 ft-lb = 0. Solving, we obtain A x = 0, A y = 8 lb, B = 22 lb. A x = 0, A y = 8 lb, B = 22 lb.

264

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Problem 5.19 Each pulley weighs 8 lb and has a 5-in radius. The horizontal bar weighs 20 lb. Assume that the bar’s weight acts at its midpoint. Determine the tension in the cord and the reactions on the bar at A.

Solution:

By summing moments about the center of the free-body diagram of one of the pulleys, we can see that the tension in the cord is the same on both sides. Let us obtain a free-body diagram by “cutting” the cord at three places:

8 lb Ax

x Ay

20 lb

10 in

5 in 5 in

21 in 42 in A

From the equilibrium equations ΣFx = A x = 0, ΣFy = A y + 3T − 20 lb − 8 lb = 0,

42 in

ΣM point A = (27 in)T + (37 in)T + (47 in)T − (21 in)(20 lb) − (42 in)(8 lb) = 0, we obtain T = 6.81 lb, A x = 0, A y = 7.57 lb. T = 6.81 lb, A x = 0, A y = 7.57 lb.

Problem 5.20 The unstretched length of the spring CD is 350 mm. Suppose that you want the lever ABC to exert a 120-N normal force on the smooth surface at A. Determine the necessary value of the spring constant k and the resulting reactions at B. C k D

230 mm 450 mm

Solution:

We have

F = k ( (0.23 m) 2 + (0.3 m 2 ) − 0.35 m) A = 120 N 30 F (0.45 m) + A cos 20 °(0.18 m) 1429 + A sin 20 °(0.33 m) = 0

ΣM B : −

ΣFx : A cos 20 ° + B x +

30 F = 0 1429

ΣFy : − A sin 20 ° + B y −

23 F = 0 1429

Solving we find: k = 3380 N/m, B x = −188 N, B y = 98.7 N

208 180 mm

F 30

A 330 mm

300 mm Bx A By 20°

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Problem 5.21 The mobile is in equilibrium. The fish B weighs 27 oz. Determine the weights of the fish A, C, and D. (The weights of the crossbars are negligible.)

For the crossbar supporting B, the sum of the moments is ΣM BCD = 6 FCD − 2 B = 0, from which, substituting from above

12 in

FCD =

3 in

6 in

or C = 7 B /27 = 7 oz,

2 in B

7 in

2 in

and D = 2C /7 = 2 oz. The sum of the moments about the crossbar supporting A is

2B 2C 9C = C + D = C + = , 6 7 7

ΣM AB = 12 A − 3FBCD = 0, from which, substituting from above, A =

Solution:

Denote the reactions at the supports by FAB , FCD , and FBCD as shown. Start with the crossbar supporting the weights C and D. The sum of the forces is

3( B + C + D) 27 + 7 + 2 = = 9 oz 12 4

FCD

ΣFY = −C − D + FCD = 0,

from which FCD = C + D.

7 in

2 in FCD

For the cross bar supporting the weight B, the sum of the forces is

FCD

ΣFY = −B + FBCD − FCD = 0,

6 in

from which, substituting, FBCD = B + C + D.

2 in FAB

For the crossbar supporting C and D, the sum of the moments about the support is

FBCD 12 in

3 in

ΣM CD = 7 D + 2C = 0, from which D =

Problem 5.22 and B.

2C . 7

Determine the reactions on the bar at A

Solution:

We draw the free-body diagram of the bar:

y 400 in-lb A

400 in-lb

11 in

308 18 in

11 in

50 lb

308

50 lb

18 in The equilibrium equations are (notice that it is advantageous to sum moments about point B) ΣFx = A x − (50 lb)sin 30 ° = 0, ΣFy = A y + B − (50 lb) cos30 ° = 0, ΣM point B = −(18 in) A y − (11 in)(50 lb) − 400 in-lb = 0. Solving, we obtain A x = 25 lb, A y = −52.8 lb, B = 96.1 lb. A x = 25 lb, A y = −52.8 lb, B = 96.1 lb.

266

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Problem 5.23 The structural element has roller supports at A, B, and C and is subjected to a vertical 800-lb load. The dimension h = 2 ft. Determine the magnitudes of the reactions at A, B, and C.

Solution:

The free-body diagram is

y 2 ft

308 308

158

800 lb

2 ft x

2 ft

158

800 lb

From the equilibrium equations

ΣFx = A sin 30 ° − B − C sin15 ° = 0, ΣFy = A cos30° + C cos15 ° − 800 lb = 0, ΣM point A = −hB + (6 ft)C cos15 ° − (2 ft)(800 lb) = 0, we obtain A = 545 lb, B = 184 lb, C = 340 lb. A: 545 lb; B: 184 lb; C: 340 lb.

Solution:

Problem 5.24 The 14.5-lb chain saw is subjected to the loads at A by the log it cuts. Determine the reactions R, B x , and B y that must be applied by the person using the saw to hold it in equilibrium.

The sum of the forces are

ΣFX = −5 + B X − R cos60° = 0. ΣFY = 10 − 14.5 + BY − R sin 60 ° = 0. The sum of the moments about the origin is

ΣM O = 7 R cos60 ° + 8BY − 2(14.5) − 13(10) − 5(1.5) = 0.

From which 7 R cos60 ° + 8BY − 166.5 = 0. Collecting equations and reducing to 3 equations in 3 unknowns:

608 By

1.5 in A 5 lb

Bx 14.5 lb

10 lb 13 in

B X + 0 BY − 0.5 R = 5

7 in

6 in 2 in

0 B X + BY − 0.866 R = 4.5 0 B X + 8BY + 3.5 R = 166.5. Solving: B X = 11.257 lb, BY = 15.337 lb, and R = 12.514 lb

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Problem 5.25 The mass of the trailer is 3600 kg. The dimensions a = 2.0 m and b = 5.3 m. The truck is stationary, and the trailer’s wheels can turn freely, which means the road exerts no horizontal force on them. The hitch at B can be modeled as a pin support. Determine the normal force on the wheels at A and the reactions at B.

Solution:

The equilibrium equations are

ΣFx = B x = 0, ΣFy = A + B y − mg = 0, ΣM A = (a + b) B y − amg = 0. Solving, we obtain A = 25,600 N, B x = 0, B y = 9680 N. A = 25,600 N, B x = 0, B y = 9680 N.

A a

b a

Problem 5.26 The total weight of the wheelbarrow and its load is W = 100 lb. (a) What is the magnitude of the upward force F necessary to lift the support at A off the ground? (b) What is the magnitude of the downward force necessary to raise the wheel off the ground?

Solution:

The free-body diagram is shown. The equilibrium

equations are ΣFy : A + B + F − W = 0 ΣM A : B(26 in) − W (12 in) − F (40 in) = 0 (a)

Set A = 0 and solve. We find that F = 21.2 lb

(b) Set B = 0 and solve. We find that F = −30 lb

So we have (a) 21.2 lb, (b) 30 lb.

A 40 in

268

12 in 14 in

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Problem 5.27 The weight of the airplane and its occupants is W = 1700 lb. While applying the brakes so that the rear wheels are locked, the pilot advances the throttle. The front (nose) wheel can turn freely, so the ground does not exert a horizontal force on it. The vertical force exerted on the nose wheel by the runway is 680 lb. What is the airplane’s thrust T?

Solution:

The equilibrium equations are

ΣFx = B x − T = 0, ΣFy = A + B y − W = 0, ΣM A = (4.5 ft)T − (6 ft)W + (8 ft) B y = 0. Setting A = 680 lb and solving, we obtain T = B x = 453 lb, B y = 1020 lb. T = 453 lb.

4 ft 6 in

W A

6 ft

4.5 ft

2 ft

By 6 ft

Problem 5.28 A safety engineer establishing limits on the loads that can be carried by a forklift analyzes the situation shown. The dimensions are a = 32 in, b = 30 in, and c = 26 in. The combined weight of the forklift and operator is W F = 1200 lb. As the weight W L supported by the forklift increases, the normal force exerted on the floor by the rear wheels at B decreases. The forklift is on the verge of tipping forward when the normal force at B is zero. Determine the value of W L that will cause this condition.

2 ft

Solution:

The equilibrium equations and the special condition for this problem are ΣFy : A + B − W L − (1200 lb) = 0 ΣM A : W L a − W F b + Bc = 0 B = 0 We obtain W L = 1125 lb

WF A a

B b

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Problem 5.29* Paleontologists speculate that the stegosaurus could stand on its hind limbs for short periods to feed. Assuming a mass of 5000 kg, determine the magnitudes of the forces B and C exerted by the ligament-muscle brace and vertebral column, and determine the angle α.

580 mm

160 mm B 248

C a 415 mm

790 mm

Solution:

In terms of the coordinate system shown, the forces in terms of their components are

These two equations can be solved by treating C sin α and C cosα as two unknowns, obtaining

FB = −B sin 24 °i − B cos 24 ° j,

C sin α = 3.55E4 N,

FC = C sin αi + C cos α j.

Dividing these results to determine tan α, we obtain α = 15.4 °, and then either equation can be used to obtain C = 1.34E5 N. Finally, Eq. (1) gives B = 8.72E4 N.

The position vectors of the points of application of the forces are rB /G = −1.205i − 0.420 j (m), rC /G = −0.790 i − 0.580 j (m).

C cos α = 1.29E5 N.

B = 87.2 kN, C = 134 kN, α = 15.4 °.

The equilibrium equations are

ΣFx = −B sin 24 ° + C sin α = 0,

(1)

ΣFy = −B cos 24 ° + C cos α − mg = 0,

(2) G

ΣM G = rB /G × FB + rC /G × FC =

i j k −1.205 m −0.420 m 0 −B sin 24 ° −B cos 24 ° 0

i j k −0.790 m −0.580 m 0 C sin α C cos α 0

= 0.

mg B

Expanding the moment equation, it can be written 1.205B cos 24 ° − 0.420 B sin 24 ° − 0.790C cos α + 0.580C sin α = 0.

C 248

(3)

By solving Eq. (1) for B and substituting the result into Eqs. (2) and (3), they become −

cos 24 ° C sin α + C cos α = mg, sin 24 °

1.205

270

cos 24 ° C sin α − 0.420C sin α − 0.790C cos α + 0.580C sin α = 0. sin 24 °

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Problem 5.30 The weight of the fan is W = 20 lb. Its base has four equally spaced legs of length b = 12 in. Each leg has a pad near the end that contacts the floor and supports the fan. The height h = 32 in. If the fan’s blade exerts a thrust T = 2 lb, what total normal force is exerted on the two legs at A?

Solution:

T b

W h

T A

The free-body diagram is shown.

The equilibrium equations are Side View

Top View

ΣFy : A + B − W = 0 ΣM B : W

b 2b − A − Th = 0 2 2

We obtain A = 6.23 lb

Solution:

Problem 5.31 The person and the upper arm ABG of the “cherry picker” lift have a combined mass of 280 kg. Their weight acts at G. Determine the reactions on the arm at the pin support A and the magnitude of the force the hydraulic cylinder BC exerts on the arm at B. 2.5 m

The weight of the person and the upper arm is W = (280 kg) (9.81 m/s 2 ) = 2750 N. Let C denote the magnitude of the compressive force exerted at B by the hydraulic cylinder. The free-body diagram of the upper arm is

1.5 m

G B A 1.2 m

The angle α = arctan(0.7 m/1.2 m) = 30.3 °. From the equilibrium equations

0.6 m

y C

ΣFx = A x − C sin α = 0, ΣFy = A y + C cos α − W = 0, 0.8 m

ΣM point A = (0.6 m)C sin α − (1.5 m)C cos α + (4.0 m)W = 0, we obtain A x = 5.57 kN, A y = −6.81 kN, C = 11.1 kN. A x = 5.57 kN, A y = −6.81 kN, magnitude = 11.1 kN. x

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Problem 5.32 The lifting mechanism of the upper portion of a bed designed for hospital patients is shown. The mass of the rotating portion of the bed is 18 kg. Assume that its weight acts at its midpoint. The force F = 130 N, which is normal to the rotating portion of the bed, is exerted by the patient and bedding. The hydraulic piston AC exerts a force at A that is directed from C toward A. Determine the reactions on the rotating portion of the bed at B.

Solution

The weight of the rotating part of the bed is W = (18 kg) (9.81 m/s 2 ) = 177 N. The free-body diagram of the rotating part is

P a

300 mm

520 mm

458

300 mm A

290 mm

180 mm

The angle α = arctan(620 mm/320 mm) = 62.7 °. The magnitude of the compressive force exerted on the bed by the hydraulic cylinder is denoted by A.

620 mm

The equilibrium equations are 290 mm

ΣFx = B x − F cos 45 ° − A cos α = 0, ΣFy = B y − F sin 45 ° − W + A sin α = 0, ΣM point B = (300 mm)F + [(410 mm) cos 45 °]W − (500 mm) A sin α

320 mm

+ (330 mm) A cos α = 0. Solving, we obtain A = 308 N, B x = 233 N, B y = −5.08 N.

180 mm

B x = 233 N, B y = −5.08 N.

Problem 5.33 A force F = 400 N acts on the bracket. What are the reactions at A and B?

Solution

The joint A is a pinned joint; B is a roller joint. The pinned joint has two reaction forces A X , AY . The roller joint has one reaction force B X . The sum of the forces is ΣFX = A X + B X = 0,

F A

ΣFY = AY − F = 0, from which AY = F = 400 N.

80 mm

The sum of the moments about A is B

320 mm

ΣM A = 0.08B X − 0.320 F = 0, from which BX =

0.320(400) = 1600 N. 0.08

Substitute into the sum of forces equation to obtain: A X = −B X = −1600 N

AY AX 80 mm BX

272

320 mm

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Problem 5.34 The boom BC is 20 ft in length and weighs 200 lb. Assume that the boom’s weight acts at its midpoint. The suspended object weighs 500 lb. The angle α = 50 °. Determine the tension in cable AC and the reactions on the boom at B.

Solution:

Let T denote the tension in the cable. Consider the freebody diagram of the boom:

y C b

500 lb

C a

200 lb x

The horizontal distance from A to C is 12 ft + (20 ft) cos α, so the angle β is

a A

12 ft

  (20 ft) sin α β = arctan  .  12 ft + (20 ft) cos α  From the equilibrium equation ΣM point B = (20 ft)  sin α(T cos β ) − cos α(T sin β ) − cos α(500 lb)  1 − cos α(200 lb)  = 0,  2 we obtain T = 1225 lb. Then we can see from the free-body diagram that B x = T cos β = 1043 lb, B y = T sin β + 200 lb + 500 lb = 1343 lb. Tension = 1225 lb, B x = 1043 lb, B y = 1343 lb.

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Problem 5.35* The boom BC is 20 ft in length and weighs 200 lb. Assume that the boom’s weight acts at its midpoint. The suspended object weighs 500 lb. Suppose that you don’t want to subject cable AC to a tension greater than 4000 lb. What is the minimum allowable value of the angle α ?

Solution:

Let T denote the tension in the cable. Consider the freebody diagram of the boom: y

C b

500 lb

C a

200 lb x

The horizontal distance from A to C is 12 ft + (20 ft) cos α, so the angle β is   (20 ft)sin α β = arctan  (1) .  12 ft + (20 ft) cos α  a A

12 ft

From the equilibrium equation ΣM point B = (20 ft)  sin α(T cos β ) − cos α(T sin β ) − cos α(500 lb)  1 − cos α(200 lb)  = 0,  2 we obtain T =

600 lb . tan α cos β − sin β

(2)

Using Eqs. (1) and (2), we can draw a graph of the tension T as a function of the angle α and use it to estimate the angle at which the tension is equal to 4000 lb: 6000

tension, pounds

5500 5000 4500 4000 3500 3000 15

alpha, degrees From the graph we estimate that α = 21.5 °. Alternatively, we can apply software designed to solve nonlinear algebraic equations to Eqs. (1) and (2). By doing so we obtain α = 21.4765 °. Answer: α = 21.5 °.

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Problem 5.36 This structure, called a truss, has a pin support at A and a roller support at B, and is loaded by two forces. (Assume that the weight of the structure is negligible.) The dimensions b = 0.25 m, h = 0.35 m. Determine the reactions at the supports. Strategy: Draw a free-body diagram treating the entire truss as a single object. 308

458

4 kN

Solution:

2 kN

4 kN

2 kN

308

458

The equilibrium equations are h B

A b

ΣFy = A y + B − (4 kN) cos 45 ° − (2 kN) cos30 ° = 0, ΣM A = 4 bB − b (4 kN) cos 45 ° − h (4 kN) cos 45 °

ΣFx = A x + (4 kN) cos 45 ° − (2 kN)sin 30 ° = 0,

− 3b (2 kN) cos30 ° + h (2 kN)sin 30 ° = 0. Solving, we obtain A x = −1.83 kN, A y = 1.91 kN, B = 2.65 kN. A x = −1.83 kN, A y = 1.91 kN, B = 2.65 kN.

Problem 5.37 An Olympic gymnast is stationary in the “iron cross” position. The weight of their left arm and the weight of their body not including their arms are shown. The distances are a = b = 9 in and c = 13 in. Treat their shoulder S as a fixed support, and determine the magnitudes of the reactions at their shoulder. That is, determine the force and couple their shoulder must support.

Solution: FH

8 lb

8 lb 144 lb SX M

FH 8 lb

SY b

The shoulder as a built-in joint has two-force and couple reactions. The left hand must support the weight of the left arm and half the weight of the body:

8 lb

144 lb

FH =

144 + 8 = 80 lb. 2

The sum of the forces on the left arm is the weight of his left arm and the vertical reaction at the shoulder and hand: ΣFX = S X = 0. ΣFY = FH − S Y − 8 = 0,

from which S Y = FH − 8 = 72 lb. The sum of the moments about the shoulder is ΣM S = M + (b + c) FH − b8 = 0, where M is the couple reaction at the shoulder. Thus M = b8 − (b + c) FH = −1688 in lb = 1688 (in lb)

( 121 ftin )

= 140.67 ft lb

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Problem 5.38 The 40-lb weight of the ladder acts at the midpoint of its 136-in length. The person’s 150-lb weight effectively acts at the point shown. Assume that the vertical wall is smooth and that the system is in equilibrium. If the angle α = 30 °, what are the magnitudes of the normal and friction forces exerted on the ladder by the floor?

Solution:

We draw the free-body diagram of the ladder:

y P 150 lb a

40 lb 150 lb

N 44 in Here N is the normal force and f is the friction force exerted by the floor, and P is the normal force exerted by the wall. From the equilibrium equations ΣFx = f − P = 0,

40 lb

ΣFy = N − 40 lb − 150 lb = 0, ΣM origin = [(136 in) cos α]P − (44 in) (150 lb) − [(68 in)sin α](40 lb) = 0, we obtain N = 190 lb, f = P = 67.6 lb. 44 in

Normal force = 190 lb, friction force = 67.6 lb.

Problem 5.39* The 40-lb weight of the ladder acts at the midpoint of its 136-in length. The person’s 150-lb weight effectively acts at the point shown. Assume that the vertical wall is smooth. If you know that the ratio of the friction force exerted on the ladder by the floor to the normal force exerted on the ladder by the floor cannot exceed 0.4, what is the largest value of the angle α for which the system can remain in equilibrium?

Solution:

We draw the free-body diagram of the ladder:

y P 150 lb a

40 lb f

x N

150 lb a

44 in Here N is the normal force and f is the friction force exerted by the floor, and P is the normal force exerted by the wall. The equilibrium equations are ΣFx = f − P = 0,

(1)

ΣFy = N − 40 lb − 150 lb = 0,

(2)

ΣM origin = [(136 in) cos α]P − (44 in)(150 lb) 40 lb

− [(68 in)sin α] (40 lb) = 0.

(3)

It is given that f /N cannot exceed 0.4. From Eq. (2), N = 190 lb. Let us determine the value of α for which f = 0.4 N = 76 lb. From Eq. (1), P = f = 76 lb, so Eq. (3) becomes 2584 cos α − 1650 − 680 sin α = 0. 44 in

276

Solving this equation, we obtain α = 37.1 °. α = 37.1 °.

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Problem 5.40 The length of the bar is L = 4 ft. Its weight W = 6 lb acts at the midpoint of the bar. The floor and wall are smooth. The spring is unstretched when the angle α = 0. If the bar is in equilibrium when α = 40 °, what is the spring constant k ?

Solution:

The free-body diagram is shown. The stretch in the spring is L − L cos α, so the upward force exerted on the bar by the spring is F = kL(1 − cos α). Let N and R be the normal forces exerted by the floor and the wall, respectively. The equilibrium equations for the bar are ΣFx : R = 0 ΣFy : F + N − W = 0 L sin α − RL cos α 2 − FL sin α = 0

ΣM bottom : W

Because R = 0, the moment equation can be solved for the force exerted by the spring. F = 0.5W = 3 lb = kL (1 − cos α) Solving yields k = 3.21 lb/ft

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Problem 5.41 The weight W of the bar acts at its midpoint. The floor and wall are smooth. The spring is unstretched when the angle α = 0. Determine the angle α at which the bar is in equilibrium in terms of W, k, and L.

Solution:

The free-body diagram is shown.

The stretch in the spring is L − L cos α, so the upward force exerted on the bar by the spring is F = kL(1 − cos α). Let N and R be the normal forces exerted by the floor and the wall, respectively. The equilibrium equations for the bar are ΣFx : R = 0 ΣFy : F + N − W = 0 L sin α − RL cos α 2 − FL sin α = 0

ΣM bottom : W

Because R = 0, the moment equation can be solved for the force exerted by the spring. F = L

W = kL (1 − cos α) 2

(

Solving yields α = cos −1 1 −

Problem 5.42 The plate is supported by a pin in a smooth slot at B. The mass of the plate is 140 kg, and its weight acts at the center. What are the reactions at the pin support A? 2 kN-m

Solution:

W 2L

)

The free-body diagram is y 2 kN-m

6 kN-m

6 kN-m mg mg

Ax A

B 2m

608

The equilibrium equations are ΣFx = A x − B sin 60 ° = 0, ΣFy = A y + B cos60 ° − mg = 0, ΣM A = (2 m) B cos60 ° − (1 m)mg + 2000 N-m − 6000 N-m = 0. Solving, we obtain A x = 4650 N, A y = −1310 N, B = 5370 N. A x = 4.65 kN, A y = −1.31 kN.

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Problem 5.43 port A.

Solution:

Determine the reactions at the fixed sup-

The free-body diagram is y

2 kN

6 kN-m

458

6 kN-m

458

4 kN

2 kN

4 kN x

The equilibrium equations are

ΣFx = A x + (2 kN) cos 45° = 0, ΣFy = A y + (2 kN) cos 45° + 4 kN = 0, ΣM point A = M A − 6 kN-m + (1 m)(2 kN) cos 45 ° + (4 m)(4 kN) = 0. Solving, we obtain A x = −1.41 kN, A y = −5.41 kN, M A = −11.4 kN-m. A x = −1.41 kN, A y = −5.41 kN, M A = 11.4 kN-m clockwise.

Problem 5.44 Suppose that you want to represent the two forces and couple acting on the beam in Problem 5.43 by an equivalent force F as shown. Determine F and the distance D at which its line of action crosses the x-axis.

Solution: For the force F = Fx i + Fy j to be equivalent to the system of forces and couple in Problem 5.43, the total force in the two systems must be the same, Fx = (2 kN) cos 45 ° = 1.41 kN, Fy = (2 kN) cos 45 ° + 4 kN = 5.41 kN,

and the sum of the moments about a point must be the same. The counterclockwise moments about A are

DFy = −6 kN-m + (1 m)(2 kN) cos 45 ° + (4 m)(4 kN) = 11.4 kN-m.

x D

This equation yields D = 2.11 m. F = 1.41i + 5.41 j (kN), D = 2.11 m. y

2 kN

458 Ax

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4 kN x

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Problem 5.45 The bicycle brake on the right is pinned to the bicycle’s frame at A. A tension T in the cable at the top exerts a horizontal force on the wheel at B, which results in a friction force that slows the wheel. Determine the value of T necessary to cause a 20-N horizontal force on the brake at B.

Solution:

Consider the free-body diagram of the joint where the three cables are joined: T

358

We have the equilibrium equation in the vertical direction: T − 2C sin 35 ° = 0.

(1)

Now consider the free-body diagram of the brake: y C

358

358 40 mm

45 mm

Brake pad

Ay x

Wheel rim

The equilibrium equations are 40 mm

ΣFx = A x + B − C cos35° = 0,

(2)

ΣFy = A y + C sin 35 ° = 0,

(3)

ΣM A = −(45 mm) B + (85 mm)C cos35 ° + (40 mm)C sin 35 ° = 0.

(4)

Setting B = 20 N, from Eqs. (1) and (4) we obtain C = 9.72 N, T = 11.2 N. T = 11.2 N.

Problem 5.46 The mass of each of the suspended weights is 80 kg. Determine the reactions at the supports at A and E.

Solution:

From the free body diagram, the equations of equilibrium for the rigid body are ΣFx = A X + E X = 0, ΣFy = AY − 2(80)(9.81) = 0,

and ΣM A = 0.3E X − 0.2(80)(9.81) − 0.4(80)(9.81) = 0. We have three equations in the three components of the support reactions. Solving for the unknowns, we get the values A X = −1570 N,

300 mm

AY = 1570 N,

and E X = 1570 N. E

y AY

200 mm

0.2 m

AX A

x mg

0.3 m E

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Problem 5.47 The suspended weights are each of mass m. The supports at A and E will each safely support a force of 6 kN magnitude. Based on this criterion, what is the largest safe value of m?

Solution:

Written with the mass value of 80 kg replaced by the symbol m, the equations of equilibrium from Problem 5.46 are ΣFx = A X + E X = 0, ΣFy = AY − 2 m(9.81) = 0, and ΣM A = 0.3E X − 0.2 m(9.81) − 0.4 m(9.81) = 0.

We also need the relation A =

A X2 + AY2 = 6000 N.

We have four equations in the three components of the support reactions plus the magnitude of A. This is four equations in four unknowns. Solving for the unknowns, we get the values

300 mm D

A X = −4243 N, AY = 4243 N, E X = 4243 N,

and m = 216.5 kg. 200 mm

Note: We could have gotten this result by a linear scaling of all of the numbers in Problem 5.46.

200 mm

Problem 5.48 The masses of the pulleys and the horizontal bar are negligible. Determine the reactions on the bar at A.

Solution:

The key to the solution is recognizing that the tension in a rope that goes around a pulley is the same on both sides; if it were not, the pulley would rotate. (You can draw the free-body diagram of a pulley and sum moments about the center to get this result.) Let’s begin with this information and draw the free-body diagrams of the two lower pulleys: TT

TT TB TB

TB TC

(a)

B 0.4 m

0.4 m 180 kg

(b)

In the case of the left pulley (Fig. a), TB is the tension in the rope attached at point B and TT is the tension in the rope that goes over the top pulley. Since we are neglecting the masses of the pulleys, we see that TT = 2TB . For the right pulley (Fig. b), TC is the tension in the rope attached at point C. From this free-body diagram we see that TC = 2TB + TT = 4TB . It isn’t obvious that this is enough information to solve the problem, but if we proceed to the free-body diagram of the bar we will find that it is. Using our information from the analysis of the pulleys, the freebody diagram of the bar is y

0.4 m

4TB

Ax Ay

W 0.4 m

0.4 m

The weight of the suspended mass is W = (180 kg)(9.81 m/s 2 ) = 1770 N. From the equilibrium equations ΣFx = A x = 0, ΣFy = A y + TB + 4TB − W = 0, ΣM point A = (0.4 m)TB + (0.8 m)(4TB ) − (1.2 m)W = 0, we obtain A x = 0, A y = −1180 N, TB = 589 N. A x = 0, A y = −1.18 kN.

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Problem 5.49 The tension in cable AB is 2 kN. What are the reactions at C in the two cases?

Solution: T

608 MC

2m 1m Case (a) A

608

B C

608

B C CY

(a)

Case (b)

(b)

First Case: The sum of the forces: ΣFX = C X − T cos60 ° = 0, from which C X = 2(0.5) = 1 kN ΣFY = C Y + T sin 60 ° + T = 0, from which C Y = −1.866(2) = −3.732 kN. The sum of the moments about C is ΣM = M C − T sin 60 ° − 3T = 0, from which M C = 3.866(2) = 7.732 kN Second Case: The weight of the beam is ignored, hence there are no external forces on the beam, and the reactions at C are zero.

Problem 5.50

Determine the reactions at the supports. 6 in

The reaction at A is a two-force reaction. The reaction at B is one-force, normal to the surface.

5 in 50 lb

A 3 in

Solution:

The sum of the forces: ΣFX = A X − B cos60 ° − 50 = 0.

100 in-lb

ΣFY = AY + B sin 60 ° = 0. The sum of the moments about A is

3 in B

ΣM A = −100 + 11B sin 60° − 6 B cos60 ° = 0, 308

from which B =

100 = 15.3 lb. (11sin 60 ° − 6 cos60 °)

Substitute into the force equations to obtain AY = −B sin 60 ° = −13.3 lb and A X = B cos60 ° + 50 = 57.7 lb AX 50 lb

6 in. 100 11 in.

B 608

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Problem 5.51 The weight W = 2 kN. Determine the tension in the cable and the reactions at A.

Solution: AY T AX

308

308 0.6 m

0.6 m W 5 2 kN 5 2000 N

Equilibrium Eqns:

0.6 m

ΣFX = 0 : A X + T cos30 ° = 0 ΣFY = 0 : AY + T + T sin 30° − W = 0

a + ΣM A = 0 : (−0, 6)(W ) + (0.6)(T sin 30°) + (1, 2)(T ) = 0 Solving, we get A X = −693 N, AY = 800 N, T = 800 N

Problem 5.52 The cable will safely support a tension of 6 kN. Based on this criterion, what is the largest safe value of the weight W ?

Solution:

The equilibrium equations in the solution of problem

are ΣFX = 0 : A X + T cos30 ° = 0 ΣFY = 0 : AY + T + T sin 30° − W = 0

a + ΣM A = 0 : (−0, 6)(W ) + (0, 6)(T sin 30 °) + (1, 2)(T ) = 0 We previously had 3 equations in the 3 unknowns A X , AY and T (we knew W ). In the current problem, we know T but don’t know W. We again have three equations in three unknowns ( A X , AY , and W ). Setting T = 6 kN, we solve to get

308

A X = −5.2 kN AY = 6.0 kN

W 0.6 m

W = 15.0 kN

0.6 m

Problem 5.53 The threaded shaft BE exerts a 130-N force on the pin at E that points from B toward E. Assume that the pin at C behaves like a pin support. By drawing a free-body diagram of the arm DCE of the clamp, determine the vertical force exerted on the blocks at D and the magnitude of the reaction at C. 125 mm

125 mm B

50 mm

Solution:

The free-body diagram is E

Cy y

D Cx

125 mm E x

50 mm

The equilibrium equations are C

50 mm D

ΣFx = C x + E = 0, ΣFy = −D + C y = 0, ΣM C = (250 mm) D − (100 mm)E = 0. Setting E = 130 N and solving, we obtain C x = −130 N, C y = 52 N, D = 52 N. The magnitude of the reaction at C is C x2 + C y2 = 140 N. Force on blocks is 52 N, reaction at C is 140 N.

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Problem 5.54 The threaded shaft BE exerts a 130-N force on the pin at B that points from E toward B. Assume that the pin at C behaves like a pin support. By drawing a free-body diagram of the arm ABC of the clamp, determine the vertical force exerted on the blocks at A and the magnitude of the reaction at C.

Solution:

The free-body diagram is y

B Cy A

Cx x

125 mm

125 mm B

The equilibrium equations are

125 mm E

ΣFx = −B + C x = 0, ΣFy = A + C y = 0,

50 mm

ΣM C = −(250 mm) A + (100 mm)B = 0.

50 mm

Setting B = 130 N and solving, we obtain C x = 130 N, C y = −52 N,

50 mm

A = 52 N. The magnitude of the reaction at C is C x2 + C y2 = 140 N. D

Force on blocks is 52 N, reaction at C is 140 N.

Problem 5.55 Suppose that you want to design the safety valve to open when the difference between the pressure p in the circular pipe (diameter = 150 mm) and atmospheric pressure is 10 MPa (megapascals; a pascal is 1 N/m 2). The spring is compressed 20 mm when the valve is closed. What should the value of the spring constant be?

Solution:

The area of the valve is

0.15 2 a = π = 17.671 × 10 −3 m 2 . 2

(

)

The force at opening is F = 10 a × 10 6 = 1.7671 × 10 5 N. The force on the spring is found from the sum of the moments about A, ΣM A = 0.15F − (0.4)k∆L = 0. Solving, 0.15(1.7671 × 10 5 ) 0.15F = (0.4)∆L (0.4)(0.02) N = 3.313 × 10 6 m

k = 250 mm

150 mm

k 150 mm

250 mm

A k A p 150 mm 150 mm kDL

A F 0.15 m

284

0.25 m

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Problem 5.56 The 10-lb weight of the bar AB acts at the midpoint of the bar. The length of the bar is 3 ft. Determine the tension in the string BC and the reactions at A.

Solution:

Geometry:

3 ft − 3 ft sin 30 ° = 0.4169 ⇒ θ = 22.63 ° 1 ft + 3 ft cos30 °

tan θ =

The equilibrium equations ΣM A : TBC cos θ(3 ft sin 30 °) + TBC sin θ(3 ft cos30°) − (10 lb)(1.5 ft cos30°) = 0

ΣFx : −TBC cos θ + A x = 0 ΣFy : TBC sin θ − 10 lb + A y = 0 Solving: A x = 5.03 lb, A y = 7.90 lb, T = 5.45 lb B

3 ft

TBC

308

1 ft

10 lb

Ax Ay

Problem 5.57 The crane’s arm has a pin support at A. The hydraulic cylinder BC exerts a force on the arm at C in the direction parallel to BC. The crane’s arm has a mass of 200 kg, and its weight can be assumed to act at a point 2 m to the right of A. If the mass of the suspended box is 800 kg and the system is in equilibrium, what is the magnitude of the force exerted by the hydraulic cylinder?

Solution:

The free-body diagram of the arm is shown.

The angle θ = tan −1

2.4 ( 1.2 ) = 63.4°

The equilibrium equations are ΣFx : A x + FH cos θ = 0 ΣFy : A y + FH sin θ − (200 kg)(9.81 m/s 2 ) − (800 kg)(9.81 m/s 2 ) = 0 ΣM A : FH sin θ(3 m) − FH cos θ(1.4 m) − (200 kg)(9.81 m/s 2 )(2 m) − (800 kg)(9.81 m/s 2 )(7 m) = 0 We obtain A x = −12.8 kN, A y = −15.8 kN, FH = 28.6 kN

Thus FH = 28.6 kN

2.4 m 1m

1.8 m

1.2 m 7m

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Problem 5.58 In Problem 5.57, what is the magnitude of the force exerted on the crane’s arm by the pin support at A?

Solution:

See the solution to Problem 5.57.

A x = −12.8 kN, A y = −15.8 kN, FH = 28.6 kN A =

(−12.8 kN) 2 + (−15.8 kN) 2 = 20.3 kN

Thus A = 20.3 kN

C A

2.4 m 1m

1.8 m

1.2 m 7m

Problem 5.59 A speaker system is suspended by the cables attached at D and E. The mass of the speaker system is 130 kg, and its weight acts at G. Determine the tensions in the cables and the reactions at A and C.

Solution: The weight of the speaker is W = mg = 1275 N. The equations of equilibrium for the entire assembly are ΣFx = C X = 0, ΣFy = AY + C Y − mg = 0 (where the mass m = 130 kg ), and

0.5 m 0.5 m 0.5 m

0.5 m

ΣM C = −(1) AY − (1.5)mg = 0. Solving these equations, we get C X = 0, E

C A

C Y = 3188 N, and AY = −1913 N. From the free body diagram of the speaker alone, we get

ΣFy = T1 + T2 − mg = 0, and ΣM left support = −(1)mg + (1.5)T2 = 0. B

Solving these equations, we get T1 = 425. N and T2 = 850 N G 1m AY A

1.5 m CY E

C CX B

1.5 m 1m T2 T1

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Problem 5.60 The weight W1 = 1000 lb. Neglect the weight of the bar AB. The cable goes over a pulley at C. Determine the weight W2 and the reactions at the pin support A.

358 T2 rB

T1 508

AY AX The tension exerted by W1 is T1 = −1000 j. The sum of the moments about A is: 508

358

ΣM A = (rB × T1 ) + (rB × T2 ) = rB × (T1 + T2 ) i −0.6428 0.8191W2

C = L

j 0.7660 −0.5736W2 − 1000

ΣM A = (−0.2587W2 + 642.8)k = 0, from which W2 = 2483.5 lb The sum of the forces:

Solution:

The strategy is to resolve the tensions at the end of bar AB into x- and y-components, and then set the moment about A to zero. The angle between the cable and the positive x-axis is −35 °. The tension vector in the cable is T2 = W2 ( i cos(−35 °) + j sin(−35 °)).

ΣFX = ( A X + W2 (0.8192)) i = 0, from which A X = −2034.4 lb ΣFY = ( AY − W2 (0.5736) − 1000) j = 0, from which AY = 2424.5 lb

= W2 (0.8192 i − 0.5736 j)(lb). Assume a unit length for the bar. The angle between the bar and the positive x-axis is 180 ° − 50 ° = 130 °. The position vector of the tip of the bar relative to A is rB = i cos(130 °) + j sin(130 °) = −0.6428i + 0.7660 j.

Problem 5.61 The unstretched length of the spring is 0.2 m, and the spring constant is k = 8 kN/m. Suppose that the vertical force F holds the system in equilibrium when the angle α = 0. The bar’s weight is negligible. Determine F and the reactions on the bar at A.

Solution:

Let T denote the tension in the spring. When α = 0

it is T = k  (0.24 m) 2 + (0.18 m) 2 − 0.2 m  = 800 N. We draw the free-body diagram of the bar: y

T 0.18 m

a 0.24 m

F 0.24 m

0.24 m

B 0.24 m

The angle β = arctan(0.18 m/0.24 m) = 36.9 °. From the equilibrium equations ΣFx = A x − T cos β = 0, ΣFy = A y + T sin β − F = 0, ΣM point A = (0.24 m)T sin β − (0.48 m) F = 0, we obtain F = 240 N, A x = 640 N, A y = −240 N. F = 240 N, A x = 640 N, A y = −240 N.

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Problem 5.62* The unstretched length of the spring is 0.2 m, and the spring constant is k = 8 kN/m. The vertical force F = 300 N, and the bar’s weight is negligible. What is the angle α ? D

Consider the equilibrium equation ΣM point A = [(0.24 m) cos α]T sin β − [(0.24 m)sin α]T cos β − [(0.48 m) cos α]F = 0.

(2)

If we choose a value of the angle α, we can use Eqs. (1) and (2) to determine F. By drawing a graph of F as a function of α we can estimate the value of the angle at which F = 300 N: 320

0.18 m

k 315

310

0.24 m

force, newtons

B C

0.24 m

305

300

Solution:

Let T denote the tension in the spring. We need to determine the magnitude and direction of the spring force in terms of the angle α. Consider the geometry:

295

290

21 19 20 alpha, degrees

From the graph we estimate that α = 18.6 °. Alternatively, we can apply software designed to obtain roots of nonlinear algebraic equations to Eqs. (1) and (2). By doing so we obtain α = 18.5951 °.

α = 18.6 °.

The distance BE is (0.24 m) cos α. The distance DE is (0.24 m)sin α + 0.18 m. The length BD = BE 2 + DE 2 and the angle β = arctan ( DE /BE ). The tension in the spring is T = k ( BD − 0.2 m).

(1)

Now we draw the free-body diagram of the bar: y T Ax

288

b Ay

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Problem 5.63 The length of the bar is 14 in. The unstretched length of the spring is 4 in, and the spring constant is k = 2 lb/in. The system is in equilibrium when the angle between the bar’s axis and the horizontal is α = 30 °. Determine the weight of the bar and the reactions at A. (Assume that the bar’s weight acts at its midpoint.)

Solution: We need to determine the magnitude and direction of the spring force. Consider the geometry: 14 in

4 in A

14 in C D

4 in

The distance BD is 14 in − (14 in) cos α. The distance CD is 4 in + (14 in)sin α. The angle β = arctan( BD /CD). The tension T in the spring is T = k  BD 2 + CD 2 − 4 in  . We draw the free-body diagram of the bar: y

B x

a Ay T W

From the equilibrium equations ΣFx = A x + T sin β = 0, ΣFy = A y + T cos β − W = 0, ΣM point A = [(14 in) cos α]T cos β + [(14 in)sin α]T sin β − [(7 in) cos α]W = 0, we obtain W = 31.0 lb, A x = −2.41 lb, A y = 16.9 lb. W = 31.0 lb, A x = −2.41 lb, A y = 16.9 lb.

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Problem 5.64* The length of the bar is 14 in, and it weighs 60 lb. The unstretched length of the spring is 4 in, and the spring constant is k = 2 lb/in. The bar’s weight acts at its midpoint. Determine the angle α at which the system is in equilibrium. 14 in

Using the equilibrium equation ΣM point A = [(14 in) cos α]T cos β + [(14 in)sin α]T sin β − [(7 in) cos α]W = 0, we can choose a value of the angle α and determine W. By drawing a graph of W as a function of α we can estimate the value of the angle at which W = 60 lb: 90

4 in

80 75

weight, pounds

70 65 60 55

Solution:

We need to determine the magnitude and direction of the spring force. Consider the geometry:

14 in

4 in A

48 50 52 alpha, degrees

From the graph we estimate that α = 49.3 °. Alternatively, we can use software designed to obtain roots of nonlinear algebraic equations. By doing so we obtain α = 49.2773 °.

α = 49.3 °.

a Ay T W

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Problem 5.65 In Example 5.4, suppose that α = 40 °, d = 1 m, a = 200 mm, b = 500 mm, R = 75 mm, and the mass of the luggage is 40 kg. Determine F and N.

Solution:

(See Example 5.4.)

The sum of the moments about the center of the wheel: ΣM C = dF cos α + aW sin α − bW cos α = 0, (b − a tan α)W = 130.35 N. d The sum of the forces:

from which F =

ΣFY = N − W + F = 0, from which N = 262.1 N

d F b

a W

R C

Problem 5.66 In Example 5.4, suppose that α = 35 °, d = 46 in, a = 10 in, b = 14 in, R = 3 in, and you don’t want the user to have to exert a force F larger than 20 lb. What is the largest luggage weight that can be placed on the carrier?

Solution:

(See Example 5.4.) From the solution to Problem 5.65,

the force is F =

(b − a tan α)W . d

Solve for W: W =

Fd . (b − a tan α)

For F = 20 lb, W = 131.47 = 131.5 lb

Problem 5.67 One of the difficulties in making design decisions is that you don’t know how the user will place the luggage on the carrier in Example 5.4. Suppose you assume that the point where the weight acts may be anywhere within the “envelope” R ≤ a ≤ 0.75c and 0 ≤ b ≤ 0.75d . If α = 30 °, c = 14 in, d = 48 in, R = 3 in, and W = 80 lb, what is the largest force F the user will have to exert for any luggage placement?

Solution: F =

(b − a tan α)W . d

The force is maximized as b → 0.75d , and a → R. Thus FMAX =

BandF_6e_ISM_CH05.indd 291

(See Example 5.4.) From the solution to Problem 5.65,

the force is

(0.75d − R tan α)W = 57.11 lb d

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Problem 5.68 In Example 5.4, assume a user that would hold the carrier’s handle at h = 36 in above the floor. Assume that R = 3 in, a = 6 in, b = 12 in, and d = 4 ft. The resulting ratio of the force the user must exert to the weight of the luggage is F /W = 0.132. Suppose that people with a range of heights use this carrier. Obtain a graph of F /W as a function of h for 24 ≤ h ≤ 36 in.

Solution:

(See Example 5.4.) From the solution to Problem 5.67, the force that must be exerted is F =

(b − a tan α)W , d

from which

(b − a tan α) F = . W d

The angle a is given by α = sin −1

( h −d R ).

The commercial package TK Solver Plus was used to plot a graph of F as a function of h. W

F .2 / W .19 , d .18 i m .17 e n .16 s i o .15 n l .14 e x .13 e 24

Problem 5.69 (a) Draw the free-body diagram of the beam and show that it is statically indeterminate. (See Practice Example 5.5.) (b) Determine as many of the reactions at the supports. 20 N-m

F/W versus height

28 30 32 height h, in

Solution: (a) The free body diagram shows that there are four unknowns, whereas only three equilibrium equations can be written. (b) The sum of moments about A is

ΣM A = M + 1.1BY = 0, from which BY = −

20 = −18.18 N. 1.1

The sum of forces in the vertical direction is

800 mm

300 mm

ΣFY = AY + BY = 0, from which AY = −BY = 18.18 N. The sum of forces in the horizontal direction is ΣFX = A X + B X = 0, from which the values of A X and B X are indeterminate. 20 N-m

800 mm

300 mm BX

AX AY

292

800 mm

300 mm

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Problem 5.70 Choose supports at A and B so that the beam is not statically indeterminate. Determine the reactions at the supports. 20 N-m

Solution:

One possibility is shown: the pinned support at B is replaced by a roller support. The equilibrium conditions are:

ΣFX = A X = 0. The sum of moments about A is ΣM A = M + 1.1BY = 0,

800 mm

300 mm

from which BY = −

20 = −18.18 N. 1.1

The sum of forces in the vertical direction is ΣFY = AY + BY = 0, from which AY = −BY = 18.18 N. 20 N-m

B 800 mm AX AY

Problem 5.71 (a) Draw the free-body diagram of the beam and show that it is statically indeterminate. (The external couple M 0 is known.) (b) By an analysis of the beam’s deflection, it is determined that the vertical reaction B exerted by the roller support is related to the couple M 0 by B = 2 M 0 /L. What are the reactions at A? M0 A

300 mm 20 N-m 300 mm BY

800 mm

Solution: (a)

Σ FX : A X = 0

(1)

ΣFY : AY + B = 0

(2)

a + MFA : M A − M O + BL = 0

(3)

Unknowns: M A , A X , AY , B. 3 Eqns in 4 unknowns ∴ Statistically indeterminate (b) Given B = 2 M O /L

(4)

We now have 4 eqns in 4 unknowns and can solve.

Eqn (1) yields A X = 0

Eqn (2) and Eqn (4) yield AY = −2 M O /L Eqn (3) and Eqn (4) yield M A = M O − 2M O M A = −M O M A was assumed counterclockwise M A = M O clockwise AX = 0 AY = −2 M O /L

MA AX

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Problem 5.72 Choose supports at A and B so that the beam is not statically indeterminate. Determine the reactions at the supports.

MO L

M0 A

B L

Solution:

This result is not unique. There are several possible

answers Σ FX : A X = 0 ΣFY : AY + BY = 0 ΣFA : −M o + BL = 0 AX = 0 B = M O /L AY = −M O /L

Problem 5.73 Draw the free-body diagram of the L-shaped pipe assembly and show that it is statically indeterminate. Determine as many of the reactions as possible. Strategy: Place the coordinate system so that the x-axis passes through points A and B.

The length from A to B is L =

0.3 2 + 0.7 2 = 0.76157 m.

The angle between the new axis and the horizontal is θ = tan −1

0.3 ( 0.7 ) = 23.2°.

The moment about the point A is M A = LB N − 0.3F + M = 0, from which B N =

B 80 N A

100 N-m

300 mm

−M + 0.3F −76 = = −99.79 N, L 0.76157

from which The sum of the forces normal to the new axis is ΣFN = A N + B N + F cos θ = 0, from which

300 mm

700 mm

A N = −B N − F cos θ = 26.26 lb The reactions parallel to the new axis are indeterminate.

Solution:

The free body diagram shows that there are four reactions, hence the system is statically indeterminate. The sum of the forces: ΣFX = ( A X + B X ) = 0,

and ΣFY = AY + BY + F = 0. A strategy for solving some statically indeterminate problems is to select a coordinate system such that the indeterminate reactions vanish from the sum of the moment equations. The choice here is to locate the x-axis on a line passing through both A and B, with the origin at A. Denote the reactions at A and B by A N , AP , B N , and B P , where the subscripts indicate the reactions are normal to and parallel to the new x-axis. Denote

BN 80 N

300 mm

AP 100 N-m 700 mm

F = 80 N, M = 100 N-m.

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Problem 5.74 Choose supports at A and B so that the pipe assembly is not statically indeterminate. Determine the reactions at the supports.

Solution:

This problem has no unique solution.

B 80 N 300 mm

100 N-m

300 mm

700 mm

Problem 5.75 State whether each of the L-shaped bars shown is properly or improperly supported. If a bar is properly supported, determine the reactions at its supports. (See Practice Example 5.6.)

Solution: (1)

is properly constrained. The sum of the forces ΣFX = −F + B X = 0, from which B X = F.

(1)

(2)

ΣM B = −LAY + LF = 0, from which AY = F, and B y = −F (2) is improperly constrained. The reactions intersect at B, while the force produces a moment about B. (3) is properly constrained. The forces are neither concurrent nor parallel. The sum of the forces: ΣFY = C sin 45 ° − A sin 45 ° = 0

(3)

BandF_6e_ISM_CH05.indd 295

from which B y = − A y . The sum of the moments about B:

ΣFX = −C cos 45 ° − B − A cos 45 ° + F = 0. 1 L 2 1 L 2

458

1 L 2 1 L 2

L A

ΣFY = BY + A y = 0,

from which A = C. The sum of the moments about A: ΣM A = − 12 LF + LC cos 45 ° + LC sin 45 ° = 0, from which C = B =

F 2

F F . Substituting and combining: A = , 2 2 2 2

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Problem 5.76 State whether each of the L-shaped bars shown is properly or improperly supported. If a bar is properly supported, determine the reactions at its supports. (See Practice Example 5.6.)

1 L 2 1 L 2

A L

458

(1)

is improperly constrained. The reactions intersect at a point P, and the force exerts a moment about that point. (2) is improperly constrained. The reactions intersect at a point P and the force exerts a moment about that point. (3) is properly constrained. The sum of the forces:

Solution:

1 L 2 1 L 2

(2) C 1 L 2 1 L 2

ΣFX = C − F = 0, A

from which C = F.

ΣFY = − A + B = 0,

from which A = B. The sum of the moments about B : LA + L 1 1 F − LC = 0, from which A = F , and B = F 2 2 2

(3)

Problem 5.77 The bar AB has a fixed support at A and is loaded by the forces

Solution:

FB = 2 i + 6 j + 3k (kN),

(b)

M A = M A X i + M AY j + M AZ k

FC = i − 2 j + 2k (kN).

Equilibrium Eqns (Forces) ΣFX : A X + FB X + FC X = 0 ΣFY : AY + FBY + FCY = 0

(a) Draw the free-body diagram of the bar. (b) Determine the reactions at A. Strategy: (a) Draw a diagram of the bar isolated from its supports. Complete the free-body diagram of the bar by adding the two external forces and the reactions due to the fixed support (see Table 5.2). (b) Use the scalar equilibrium equations (5.9–5.14) to determine the reactions.

ΣFZ : AZ + FBZ + FC Z = 0 Equilibrium Equations (Moments) Sum moments about A r AB × FB = 1i × (2 i + 6 j + 3k) kN-m r AB × FB = −3 j + 6k (kN-m) r AC × FC = 2 i × (1i − 2 j + 2k) kN-m r AC × FC = −4 j − 4 k (kN-m) x : ΣM A : M A X = 0

y : ΣM A : M AY − 3 − 4 = 0 z : ΣM A : M Az + 6 − 4 = 0 A

FB AX

1m 1m

MA5 MAX i 1 MAY j 1 MAZ K

1m AZ

C 1m

x FC

Solving, we get A X = −3 kN, AY = −4 kN, AZ = −5 kN M Ax = 0, M Ay = 7 kN-m, M Az = −2 kN-m

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Problem 5.78 The bar AB has a fixed support at A. The tension in cable BC is 8 kN. Determine the reactions at A.

A z

C B

(3, 0.5, 20.5) m x

Solution: AX

M A = M Ax i + M Ay j + M Az k

MA5 MAX i1 MAY j 1 MAZ K

We need the unit vector e BC e BC =

( x C − x B ) i + ( y C − y B ) j + ( z C − z B )k ( x C − x B ) 2 + ( y C − y B ) 2 + (z C − z B ) 2

e BC = 0.816 i + 0.408 j − 0.408k

TBC B (2, 0, 0)

C (3, 0.5,20.5) x

TBC = (8 kN)e BC TBC = 6.53i + 3.27 j − 3.27k (kN) The moment of TBC about A is

M BC = r AB × TBC =

6.53 3.27 − 3.27 M BC = r AB × TBC = 0 i + 6.53 j + 6.53k (kN-m) Equilibrium Eqns. ΣFX : A X + TBC X = 0 ΣFY : AY + TBCY = 0 ΣFZ : AZ + TBC Z = 0 ΣM X : M AX + M BC X = 0 ΣM Y : M AY + M BCY = 0 ΣM Z : M AZ + M BC Z = 0 Solving, we get A X = −6.53 (kN), AY = −3.27 (kN), AZ = 3.27 (kN) M Ax = 0, M Ay = −6.53 (kN-m), M Az = −6.53 (kN-m)

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Problem 5.79 The bar AB has a fixed support at A. The collar at B is fixed to the bar. The tension in the rope BC is 300 lb. (a) Draw the free-body diagram of the bar. (b) Determine the reactions at A.

B (6, 6, 2) ft

A x C (8, 0, 3) ft z

Solution: (a) The free-body diagram is shown. (b) We need to express the force exerted by the rope in terms of its components. The vector from B to C is rBC = [(8 − 6) i + (0 − 6) j + (3 − 2)k] ft = (2 i − 6 j + k) ft The force in the rope can now be written T = TBC

rBC = TBC (0.312 i − 0.937 j + 0.156k) rBC

The equilibrium equations for the bar are ΣFx : A x + 0.312TBC = 0 ΣFy : A y − 0.937TBC = 0 ΣFz : Az + 0.156TBC = 0

ΣM A : M Ax i + M Ay j + M Az k +

i j k 6 ft 6 ft 2 ft = 0 0.312TBC − 0.937TBC 0.156TBC

These last equations can be written as M Ax = −(0.218 ft)TBC , M Ay = (0.312 ft)TBC , M Az = (7.50 ft)TBC Setting TBC = 300 lb, expanding and solving we have A x = −93.7 lb, A y = 281 lb, Az = −46.9 lb M Ax = −843 ft-lb, M Ay = 93.7 ft-lb, M Az = 2250 ft-lb

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Problem 5.80 The bar AB has a fixed support at A. The collar at B is fixed to the bar. Suppose that you don’t want the support at A to be subjected to a couple of magnitude greater than 3000 ft-lb. What is the largest allowable tension in the rope BC ? Solution:

See the solution to Problem 5.79. The magnitude of the couple at A can be expressed in terms of the tension in the rope as MA = =

B (6, 6, 2) ft

A x

2 + M2 + M2 M Ax Ay Az

C (8, 0, 3) ft

(−2.81 ft) 2 + (0.312 ft) 2 + (7.50 ft) 2 TBC

Setting M A = 3000 ft-1b and solving for TBC yields TBC = 374 lb

Problem 5.81 The total force exerted on the highway sign by its weight and the most severe anticipated winds is F = 2.8i − 1.8 j (kN). Determine the reactions at the fixed support.

Solution:

The applied load is F = (2.8i − 1.8 j) kN applied at r = (8 j + 8k) m

The force reaction at the base is R = Ox i + Oy j + Ozk The moment reaction at the base is

M O = M Ox i + M Oy j + M Oz k

For equilibrium we need  ΣF : 2.8 kN + O = 0  x x ΣF = F + R = 0 ⇒  ΣFy : − 1.8 kN + O y = 0   ΣFz : 0 + O z = 0 

O x = −2.8 kN

O x

⇒ O y = 1.8 kN Oz = 0

 ΣM : 14.4 kN-m + M = 0  x Ox  ΣM = r × F + M O = 0 ⇒  ΣM y : 22.4 kN-m + M Oy = 0   ΣM z : − 22.4 kN-m + M Oz = 0 M Ox = −14.4 kN-m ⇒ M Oy = −22.4 kN-m M Oz = 22.4 kN-m

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Problem 5.82 The segment BC of the bar is parallel to the z-axis. The tension in cable CD is 400 N. Determine the reactions on the bar at the fixed support A. y

We divide this vector by its magnitude and multiply by the tension T to express the cable force in terms of components:  r  T = T  CD  = T (0.557 i − 0.743 j + 0.371k)  rCD  = T (c x i + c y j + c z k).

(In the last line we introduce notation to clarify the subsequent steps.) The force equilibrium equations are

ΣFx = A x + Tc x = 0, ΣFy = A y + Tc y = 0, ΣFz = Az + Tc z = 0. 2m

The sum of the moments about point A is i j k ΣM point A = M Ax i + M Ay j + M Az k + 1.5 m 0 0 = 0. Tc x Tc y Tc z

A z

1.5 m

Solution:

We draw the free-body diagram of the bar, showing the tension T in the cable and the six components of the force and couple exerted at A:

(Because the moment due to a force is unchanged if it acts at any point along its line of action, we obtain a slightly simpler equation by using D as the point of application of the tension force instead of C.) This yields the three equations M Ax = 0, M Ay − (1.5 m)Tc z = 0,

M Az + (1.5 m)Tc y = 0.

Because T is given, from the six equilibrium equations we obtain A x = −223 N, A y = 297 N, Az = −149 N, M Ax = 0, M Ay = 223 N-m, M Az = 446 N-m.

A x = −223 N, A y = 297 N, Az = −149 N, M Ax = 0, M Ay = 223 N-m, M Az = 446 N-m.

MAz z

MAx Ay

Ax MAy

D x

We need to express the force exerted on the bar by the cable in terms of its components. The position vector from point C to point D is rCD = (1.5 m − 0) i + (0 − 2 m) j + (0 − (−1) m)k = 1.5i − 2 j + k (m).

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Problem 5.83 The tension in cable AB is 24 kN. Determine the reactions at the fixed support D.

Solution:

The force acting on the device is

F = FX i + FY j + FZ k = (24 kN)e AB , and the unit vector from A toward B is given by z

1i − 2 j + 1k e AB = . 6

The force, then, is given by F = 9.80 i − 19.60 j + 9.80 k kN. The position from D to A is r = 2 i + 2 j + 0 k m. The force equations of equilibrium are D X + FX = 0, DY + FY = 0, and D Z + FZ = 0. The moment equation, in vector form, is ΣM = M D + r × F. Expanded, we get ΣM = M DX i + M DY j + M DZ k +

= 0.

9.80 −19.60 9.80 The corresponding scalar equations are M DX + (2)(9.80) = 0, M DY − (2)(9.80) = 0, and M DZ + (2)(−19.60) − (2)(9.80) = 0. Solving for the support reactions, we get D X = −9.80 kN, OY = 19.60 kN, O Z = −9.80 kN. M DX = −19.6 kN-m, M DY = 19.6 kN-m, and M DZ = 58.8 kN-m.

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Problem 5.84 The robotic manipulator is stationary and the y -axis is vertical. The weights of the arms AB and BC act at their midpoints. The direction cosines of the centerline of arm AB are cos θ x = 0.174, cos θ y = 0.985, cos θ z = 0, and the direction cosines of the centerline of arm BC are cos θ x = 0.743, cos θ y = 0.557, cos θ z = −0.371. The support at A behaves like a fixed support. (a) What is the sum of the moments about A due to the weights of the two arms? (b) What are the reactions at A?

and r AD2 = r AB + rBD2 r AD2 = 0.3273i + 0.7581 j − 0.1113k m (a)

We now have the geometry determined and are ready to determine the moments of the weights about A. ΣM W = r AD1 × W1 + r AD 2 × W2 where i

r AD1 × W1 = 0.0522 0.2955 0 − 200

r AD1 × W1 = −10.44 k N-m and

m 0m

r AD 2 × W2 = 0.3273 0.7581 − 0.1113 −160

0 B

r AD 2 × W2 = −17.81i − 52.37k

160 N

600 mm

Thus, ΣM W = −17.81i − 62.81k (N-m) (b) Equilibrium Eqns Σ FX : A X = 0

200 N

ΣFY : AY − W1 − W2 = 0

A z

ΣFZ : AZ = 0 Sum Moments about x

A : M A + ΣM W = 0 ΣM X : M Ax − 17.81 = 0 (N-m) ΣM Y : M Ay + 0 = 0

Solution:

Denote the center of mass of arm AB as D1 and that of BC as D 2 . We need r AD , r AB ,

and rB D 2 .

ΣM Z : M Z − 62.81 = 0 (N-m) Thus: A X = 0, AY = 360 (N), AZ = 0, M Ax = 17.81 (N-m), M Ay = 0, M Az = 62.81 (N-m)

To get these, use the direction cosines to get the unit vectors e AB and e BC . Use the relation e = cos θ X i + cos θY j + cos θ Z k

C W2

e AB = 0.174 i + 0.985 j + 0 k e BC = 0.743i + 0.557 j − 0.371k r AD1 = 0.3e AB m W1

r AB = 0.6e AB m

rBC = 0.6e BC m rBD2 = 0.3e BC m W AB = −200 j N WBC = −160 j N

MA5 MAX i 1 MAY j 1 MAZ k W1 5 200 N W2 5 160 N

Thus r AD1 = 0.0522 i + 0.2955 j m r AB = 0.1044 i + 0.5910 j m rBD2 = 0.2229 i + 0.1671 j − 0.1113k m rBC = 0.4458i + 0.3342 j − 0.2226k m

302

AZ AY

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Problem 5.85 The force exerted on the grip of the exercise machine is F = 260 i − 130 j (N). What are the reactions at the fixed support at O?

Solution: M O = M Ox i + M Oy j + M Oz k rOP = 0.25i + 0.2 j − 0.15k Equilibrium (Forces)

150 mm

ΣFX : O X + FX = O X + 260 = 0 (N) ΣFY : OY + FY = OY − 130 = 0 (N) ΣFZ : O Z + FZ = O Z = 0 (N) Thus, O X = −260 N, OY = 130 N, O Z = 0 F

Σ M X : M O X + M FX = 0

200 mm

Summing Moments about O

ΣM Y : M OY + M FY = 0

250 mm

ΣM Z : M O Z + M FZ = 0 where x

i M F = rOP × F = 0.25

0.2

− 0.15

260 −130

M F = −19.5i − 39 j − 84.5k (N-m) and from the moment equilibrium eqns, M O X = 19.5 (N-m) M OY = 39.0 (N-m) M O z = 84.5 (N-m)

0.15 P m

y MO OX z

0.2

F 5 260i 2 130 j (N)

0.2 m x

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Problem 5.86 In Practice Example 5.7, suppose that cable BD is lengthened and the attachment point D moved from (0, 600, 400) mm to (0, 600, 600) mm. (The end B of bar AB remains where it is.) Draw a sketch of the bar and its supports showing cable BD in its new position. Draw the free-body diagram of the bar and apply equilibrium to determine the tensions in the cables and the reactions at A.

Solution: The sketch and free-body diagram are shown. We must express the force exerted on the bar by cable BD in terms of its components. The vector from B to D is rBD = [(0 − 1000) i + (600 − 600) j + (600 − 400)k] mm = (−1000 i + 200 j) mm The force exerted by cable BD can be expressed as TBD

rBD = TBD (−0.981i + 0.196k) rBD

The equilibrium equations are ΣFx : A x − 0.981TBD = 0 ΣFy : A y − 200 N = 0 ΣFz : Az + 0.196TBD − TBC = 0

ΣM A :

0.6

0.4

+ 0.5 0.3 0.2

−0.981TBD 0 0.196TBD − TBC

= 0

0 −200 0

Expanding and solving these equations, we find A x = 166.7 N, A y = 200 N, Az = 66.7 N, TBC = 100 N, TBD = 170 N

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Problem 5.87 The force F acting on the boom ABC at C points in the direction of the unit vector 0.512 i − 0.384 j + 0.768k and its magnitude is 8 kN. The boom is supported by a ball-and-socket at A and the cables BD and BE. The collar at B is fixed to the boom. (a) Draw the free-body diagram of the boom. (b) Determine the tensions in the cables and the reactions at A.

1.5 m

D E 1m 2m

A B

C 2m

x F

Solution:

(a) The free-body diagram (b) We identify the following forces, position vectors, and reactions

TBE

r AC = 4 mi, F = 8 kN(0.512 i − 0.384 j + 0.768k)  −2 i + 2 j + 1.5k  TBD = TBD 10.25  r AB = 2 mi,   − 2 i + j − 2k  TBE = TBE 3 

(

)

B Ay

TBD

R = A x i + A y j + Az k Force equilibrium requires: ΣF = R + TBD + TBE + F = 0.

In component form we have ΣFx : A x + 8 kN(0.512) −

2 2 TBD − TBE = 0 3 10.25

ΣFy : A y − 8 kN(0.384) +

2 1 TBD + TBE = 0 3 10.25

ΣFz : Az + 8 kN(0.768) +

1.5 2 TBD − TBE = 0 3 10.25

Moment equilibrium requires: ΣM A = r AB × (TBD + TBE ) + r AC × F = 0. In components: ΣM x : 0 = 0 ΣM y : − 8 kN(0.768)(4 m) −

1.5 2 TBD (2 m) + TBE (2 m) = 0 3 10.25

ΣM z : − 8 kN(0.384)(4 m) +

2 1 TBD (2 m) + TBE (2 m) = 0 3 10.25

Solving five equations for the five unknowns we find A x = 8.19 kN, A y = −3.07 kN, Az = 6.14 kN, TBD = 0, TBE = 18.43 kN

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Problem 5.88 The cables BD and BE in Problem 5.87 will each safely support a tension of 25 kN. Based on this criterion, what is the largest acceptable magnitude of the force F?

1.5 m

D E 1m

Solution:

2m We have the force and distances:

r AC = 4 mi, F = F (0.512 i − 0.384 j + 0.768k)  −2 i + 2 j + 1.5k  T = T BD  BD 10.25 r AB = 2 mi,   − 2 i + j − 2k  TBE = TBE 3 

(

)

A B

C 2m

x F

The moment equations are ΣM y : −F (0.768)(4 m) − ΣM z : −F (0.384)(4 m) +

1.5 2 TBD (2 m) + TBE (2 m) = 0 3 10.25 2 1 TBD (2 m) + TBE (2 m) = 0 3 10.25

Solving we find TBE = 2.304 F , TBD = 0 Thus: 25 kN = 2.304 F ⇒ F = 10.85 kN

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Problem 5.89 The suspended load exerts a force F = 600 lb at A, and the weight of the bar OA is negligible. Determine the tensions in the cables and the reactions at the ball-and-socket support O. y

M = r×H =

i 8 HX

j 6 HY

k 0 HZ

where H can be any of the three forces acting at point A. The vector moment equation provides another three equations of equilibrium. Once we have evaluated and applied the unit vectors, we have six vector equations of equilibrium in the five unknowns T AB , T AC , S X , S Y , and S Z (there is one redundant equation since all forces pass through the line OA). Solving these equations yields the required values for the support reactions at the origin.

C (0, 6, 210) ft A (8, 6, 0) ft

The cross products are evaluated using the form

(0, 10, 4) ft

If we carry through these operations in the sequence described, we get the following vectors: 2F j

e AB = −0.816 i + 0.408 j + 0.408k, x

e AC = −0.625i + 0 j − 0.781k, T AB = −387.1i + 193.5 j + 193.5k lb, T AB = 474.1 lb,

T AC = −154.8i + 0 j − 193.5k lb, T AC = 247.9 lb, M AB = rOA × T AB = 1161i − 1548 j + 3871k ft -lb, M AC = rOA × T AC = −1161i + 1548 j + 929k ft -lb,

Solution: From the diagram, the important points in this problem are A (8, 6, 0), B (0, 10, 4), C (0, 6, −10), and the origin O (0, 0, 0) with all dimensions in ft. We need unit vectors in the directions A to B and A to C. Both vectors are of the form

and S = 541.9 i + 406.5 j + 0 k lb

e AP = ( x P − x A ) i + ( y P − y A ) j + ( z P − z A )k, where P can be either A or B. The forces in cables AB and AC are T AB = T AB e AB = T ABX i + T ABY j + T ABZ k, and T AC = T AC e AB = T ACX i + T ACY j + T ACZ k. The weight force is F = 0 i − 600 j + 0 k, and the support force at the ball joint is S = S X i + S Y j + S Z k. The vector form of the force equilibrium equation (which gives three scalar equations) for the bar is T AB + T AC + F + S = 0. Let us take moments about the origin. The moment equation, in vector form, is given by ΣM O = rOA × T AB + rOA × T AC + rOA × F = 0, where rOA = 8i + 6 j + 0 k.

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Problem 5.90 The suspended load exerts a force F = 600 lb at A and bar OA weighs 200 lb. Assume that the bar’s weight acts at its midpoint. Determine the tensions in the cables and the reactions at the ball-andsocket support O. y

Solution:

Point G is located at (4, 3, 0) and the position vector of G with respect to the origin is rOG = 4 i + 3 j + 0 k ft.

The weight of the bar is WB = 0 i − 200 j + 0 k lb,

and its moment around the origin is

(0, 6, 210) ft

M WB = 0 i + 0 j − 800 k ft-1b. A (8, 6, 0) ft

The mathematical representation for all other forces and moments from Problem 5.89 remain the same (the numbers change!). Each equation of equilibrium has a new term reflecting the addition of the weight of the bar. The new force equilibrium equation is

(0, 10, 4) ft

2F j

T AB + T AC + F + S + WB = 0. x

The new moment equilibrium equation is ΣM O = rOA × T AB + rOA × T AC + rOA × F + rOG × WB = 0. As in Problem 5.89, the vector equilibrium conditions can be reduced to six scalar equations of equilibrium. Once we have evaluated and applied the unit vectors, we have six vector equations of equilibrium in the five unknowns T AB , T AC , S X , S Y , and S Z (As before, there is one redundant equation since all forces pass through the line OA). Solving these equations yields the required values for the support reactions at the origin.

If we carry through these operations in the sequence described, we get the following vectors: e AB = −0.816 i + 0.408 j + 0.408k, e AC = −0.625i + 0 j − 0.781k, T AB = −451.6 i + 225.8 j + 225.8k lb, T AB = 553.1 lb, T AC = −180.6 i + 0 j − 225.8k lb, T AC = 289.2 lb, M AB = rOA × T AB = 1355i − 1806 j + 4516k ft-lb, M AC = rOA × T AC = −1354 i + 1806 j + 1084 k ft-lb, and S = 632.3i + 574.2 j + 0 k lb

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Problem 5.91 The 158,000-kg airplane is at rest on the ground (z = 0 is ground level). The landing gear carriages are at A, B, and C. The coordinates of the point G at which the weight of the plane acts are (3, 0.5, 5) m. What are the magnitudes of the normal reactions exerted on the landing gear by the ground?

Solution: ΣFY = ( N L + N R ) + N F − W = 0 ΣM R = −3 mg + 21N F = 0 Solving, N F = 221.4 kN

(1)

( N L + N R ) = 1328.6 kN

(2)

ΣFY = N R + N L + N F − W = 0

21 m

(same equation as before) a + ΣM O = 0.5 W − 6( N R ) + 6( N L ) = 0

6m B G

(3)

Solving (1), (2), and (3), we get N F = 221.4 kN N R = 728.9 kN

N L = 599.7 kN

mg 3m

Side View

21 m x R

(NL1 NR)

Z 0.5 m

Front View

6 NF

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Problem 5.92 The horizontal triangular plate is suspended by the three vertical cables A, B, and C. The tension in each cable is 80 N. Determine the x and z coordinates of the point where the plate’s weight effectively acts.

Solution: ΣM x : (240 N) z − (80 N)(0.4 m) = 0 ΣM z : (80 N)(0.3 m) − (240 N) x = 0 Solving

x = 0.1 m, z = 0.1333 m

A 80 N

B C 0.4 m

(x, 0, z)

80 N

0.3 m x

80 N

240 N Z

Problem 5.93 The 800-kg horizontal wall section is supported by the three vertical cables A, B, and C. What are the tensions in the cables?

Solution:

All dimensions are in m and all forces are in N. Forces A, B, C , and W act on the wall at (0, 0, 0), (5, 14, 0), (12, 7, 0), and (4, 6, 0), respectively. All forces are in the z direction. The force equilibrium equation in the z direction is A + B + C − W = 0. The moments are calculated from M B = rOB × Bk, M C = rOC × Ck,

and M G = rOG × (−W )k.

7m C

6m 4m

The moment equilibrium equation is ΣM O = M B + M C + M G = 0. Carrying out these operations, we get A = 3717 N,

B = 2596 N, C = 1534 N, and W = 7848 N.

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Problem 5.94 The 260-lb horizontal rectangular plate rests on three posts. Assume that the weight of the plate acts at its center. What are the magnitudes of the vertical forces exerted on the plate by the posts at A, B, and C ?

Solution:

The sum of the forces equals zero:

ΣFz = FA + FB + FC − 260 lb = 0. The geometry of the problem makes it simple to determine the moments about the coordinate axes. The moment about the x-axis is (20 in) FA + (30 in) FB − (15 in)(260 lb) = 0,

50 in

6 in

and the moment about the y-axis is

10 in

(6 in) FA + (50 in) FB + (40 in) FC − (25 in)(260 lb) = 0.

30 in B

Solving these three equations, we obtain FA = 128 lb, FB = 44.8 lb, FC = 87.4 lb. FA = 128 lb, FB = 44.8 lb, FC = 87.4 lb.

40 in 50 in

30 in 20 in

260 lb 6 in

Problem 5.95 The L-shaped bar is supported by a bearing at A and rests on a smooth horizontal surface at B. The vertical force F = 4 kN and the distance b = 0.15 m. Determine the reactions at A and B.

Solution:

40 in

Equilibrium Eqns:

Σ FX : O = 0 ΣFY :

AY + B − F = 0

ΣFZ :

AZ = 0

Sum moments around A

x : Fb − 0.3B = (4)(0.15) − 0.3B = 0 y : M AY = 0

z : M AZ + 0.2 F − 0.2 B = 0

b A x

B z

Solving, A X = 0, AY = 2 (kN),

0.2 m 0.3 m

AZ = 0 M A X = 0, M AY = 0, M AZ = −0.4 (kN-m) y F

B z

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0.2 m

MZ A AY

(MAX ; 0) AX ; 0 x b 5 0.15 m F 5 4 kN

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Problem 5.96 The vertical force F = 4 kN and the distance b = 0.15 m. If you represent the reactions at A and B by an equivalent system consisting of a single force, what is the force and where does its line of action intersect the x−z plane?

Solution: We want to represent the forces at A & B by a single force. From Prob. 5.95 A = +2 j (kN), B = +2 j (kN) M A = −0.4 k (kN-m) We want a single equivalent force, R that has the same resultant force and moment about A as does the set A, B, and M A .

R = A + B = 4 j (kN) F

Let R pierce the x−z plane at ( x R , z R )

b A x

0.2 m

− z R R = −0.3B

ΣM Z :

− x R R = 0.2 AY

z R (4) = (+0.3)(2)

0.3 m

ΣM X :

z R = +0.15 m x R (4) = 0.2(2) x R = 0.1 m y R F

b A x

B z

Problem 5.97 The vertical force F = 4 kN. The bearing at A will safely support a force of 2.5-kN magnitude and a couple of 0.5 kN-m magnitude. Based on these criteria, what is the allowable range of the distance b?

b x

0.2 m 0.3 m

AY + B − F = 0 ( F = 4 kN ) Fb − 0.3B = 0 M AZ + 0.2 F − 0.2 B = 0

Set the force at A to its limit of 2.5 kN and solve for b. In this case, M AZ = −0.5 (kN-m) which is at the moment limit. The value for b is b = 0.1125 m A

The solution to Prob. 5.95 produced the relations

A X = AZ = M A X = M AY = 0

Solution:

0.2 m 0.3 m

We make AY unknown, b unknown, and B unknown ( F = 4 kN, M AY = +0.5 (kN-m), and solve we get AY = −2.5 at b = 0.4875 m However, 0.3 is the physical limit of the device. Thus, 0.1125 m ≤ b ≤ 0.3 m y

A x

B z

312

0.2 m 0.3 m

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Problem 5.98 The 1.1-m bar is supported by a balland-socket support at A and the two smooth walls. The tension in the vertical cable CD is 1 kN. (a) Draw the free-body diagram of the bar. (b) Determine the reactions at A and B.

The reaction at B is horizontal, with unknown x-component and z-components. The sum of the moments about A is

ΣM A = r AB × B + r AD × D = 0 =

i j k + −0.4455 0.3819 −0.3819 0 0 −1

Expand and collect like terms:

ΣM A = (0.6 B Z − 0.3819) i − (0.6 B X − 0.7 B Z ) j + (−0.6 B X + 0.4455)k = 0.

x A

600 mm

700 mm

Solution:

From which, BZ =

0.3819 = 0.6365 kN, 0.6

BX =

0.4455 = 0.7425 kN. 0.6

The reactions at A are determined from the sums of the forces:

(a) The ball and socket cannot support a couple reaction, but can support a three force reaction. The smooth surface supports oneforce normal to the surface. The cable supports one force parallel to the cable. (b) The strategy is to determine the moments about A, which will contain only the unknown reaction at B. This will require the position vectors of B and D relative to A, which in turn will require the unit vector parallel to the rod. The angle formed by the bar with the horizontal is required to determine the coordinates of B:  0.6 2 + 0.7 2  α = cos −1   = 33.1 °.   1.1 The coordinates of the points are: A (0.7, 0, 0.6), B (0, 1.1 (sin 33.1 °), 0) = (0, 0.6, 0), from which the vector parallel to the bar is r AB = rB − r A = −0.7 i + 0.6 j − 0.6k (m). The unit vector parallel to the bar is e AB =

= 0

400 mm

C z

i j k −0.7 0.6 −0.6 0 BX BZ

ΣFX = ( B X + A X ) i = 0, from which A X = −0.7425 kN. ΣFY = ( AY − 1) j = 0, from which AY = 1 kN. ΣFZ = ( B Z + AZ )k = 0, from which AZ = −0.6365 kN

r AB = −0.6364 i + 0.5455 j − 0.5455k. 1.1

The vector location of the point D relative to A is r AD = (1.1 − 0.4)e AB = 0.7e AB = −0.4455i + 0.3819 j − 0.3819k.

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Problem 5.99 The 8-ft bar is supported by a ball-andsocket support at A, the cable BD, and a roller support at C. The collar at B is fixed to the bar at its midpoint. The force F = −50 k (lb). Determine the tension in cable BD and the reactions at A and C.

The tension in the cable is T = T e BD . The reaction at the roller support C is normal to the x−z plane. The sum of the moments about A ΣM A = r AB × F + r AB × T + r AC × C = 0 i =

−50

+ T

3.7081

−1.5

y A

3.7081 −1.5

0.1160 −0.5960 0.7946 3 ft

i B

7.4162 −3 0 CY

4 ft

k = 0

= 75i + 185.4 j + T (−1.192 i − 2.9466 j − 2.036k)

D 2 ft

+ 7.4162C Y k = 0, 75 = 62.92 lb 1.192

C x

from which T = CY =

2.036 T = 17.27 lb. 7.4162

The reaction at A is determined from the sums of forces:

Solution:

The strategy is to determine the sum of the moments about IEQ. which will involve the unknown reactions at B and C. This will require the unit vectors parallel to the rod and parallel to the cable. The angle formed by the rod is 3 α = sin −1 = 22 °. 8 The vector positions are:

( )

r A = 3 j,

ΣFX = ( A X + 0.1160 T ) i = 0, from which A X = −7.29 lb, ΣFY = ( AY − 0.5960 T + C Y ) j = 0, from which AY = 20.23 lb ΣFZ = ( AZ + 0.7946 T − 50)k = 0, from which AZ = 0 lb

r D = 4 i + 2k y

and rC = (8cos 22 °) i = 7.4162 i.

The vector parallel to the rod is r AC = rC − r A = 7.4162 i − 3 j. The unit vector parallel to the rod is

3 ft

e AC = 0.9270 i − 0.375 j. The location of B is

4 ft

r AB = 4 e AC = 3.7081i − 1.5 j.

2 ft

The vector parallel to the cable is rBD = rD − (r A + r AB ) = 0.2919 i − 1.5 j + 2k.

The unit vector parallel to the cable is AY

e BD = 0.1160 i − 0.5960 j + 0.7946k.

AX T

z D CY

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Problem 5.100 The bar is 8 ft in length. The force F = Fy j − 50 k (lb). What is the largest value of Fy for which the roller support at C will remain on the floor?

Solution:

From the solution to Problem 5.99, the sum of the moments

about A is i ΣM A =

3.7081 −1.5

−50

0 y + T

3.7081

−1.5

0.1160 −0.5960 0.7946 i

3 ft B

4 ft

7.4162 −3 0 0

= 0

= 75i + 185.4 j + 3.7081FY k + T (−1.192 i − 2.9466 j − 2.036k)

D 2 ft

+ 7.4162C Y k = 0,

C x

from which, T =

75 = 62.92 lb. 1.192

Collecting terms in k, 3.7081FY + 2.384 T − 7.4162C Y = 0. For C Y = 0, FY =

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128.11 = 34.54 lb 3.708

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Problem 5.101 The tower is 70 m tall. The tension in each cable is 2 kN. Treat the base of the tower A as a fixed support. What are the reactions at A?

The sum of the moments about A is ΣM A = M A + r AB × TBE + r AB × TBD + r AB × TBC + 0 = M A + r AB × (TBE + TBC + TBD )

ΣM A =

= 0

0.1793 −4.7682 0.2434 B

= ( M X A + 17.038) i + ( M Y A + 0) j + ( M Z A − 12.551)k = 0 from which

M X A = −17.038 kN-m, 40 m

M Y A = 0, M Z A = 12.551 kN-m.

50 m

40 m

D z

20 m

50 m

The force reactions at A are determined from the sums of forces. (Note that the sums of the cable forces have already been calculated and used above.) ΣFX = ( A X + 0.17932) i = 0, from which A X = −0.179 kN, ΣFY = ( AY − 4.7682) j = 0,

Solution:

The strategy is to determine moments about A due to the cables. This requires the unit vectors parallel to the cables. The coordinates of the points are:

from which AY = 4.768 kN, ΣFZ = ( AZ + 0.2434)k = 0, from which AZ = −0.2434 kN

A(0, 0, 0), B(0, 70, 0), C (−50, 0, 0), D(20, 0, 50), E (40, 0, −40). The unit vectors parallel to the cables, directed from B to the points E, D, and C

y B

rBE = 40 i − 70 j − 40 k,

TBC

rBD = 20 i − 70 j + 50 k,

TBD

rBC = −50 i − 70 j. C

The unit vectors parallel to the cables, pointing from B, are: e BE = 0.4444 i − 0.7778 j − 0.4444 k, e BD = 0.2265i − 0.7926 j + 0.5661k,

AZ ,MA

e BC = −0.5812 i − 0.8137 j + 0 k.

The tensions in the cables are:

TBE AY , A

EA AX , MA X

TBD = 2e BD = 0.4529 i − 1.5852 j + 1.1323k (kN), TBE = 2e BE = 0.8889 i − 1.5556 j − 0.8889k (kN), TBC = 2e BC = −1.1625i − 1.6275 j − 0 k.

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Problem 5.102 The tower is 70 m tall. If the tension in cable BC is 2 kN, what must the tensions in cables BD and BE be if you want the couple exerted on the tower by the fixed support at A to be zero? What are the resulting reactions at A?

Solution:

From the solution to Problem 5.101, the sum of the moments about A is given by ΣM A = M A + r AB × (TBE + TBC + TBD ) = 0. If the couple M A = 0, then the cross product is zero, which is possible only if the vector sum of the cable tensions is zero in the x and z directions. Thus, from Problem 5.101, e x ⋅ (TBC + TBE e BE + TBD e BD ) = 0,

and e z ⋅ (TBC + TBE e BE + TBD e BD ) = 0. B

Two simultaneous equations in two unknowns result; 0.4444 TBE + 0.2265 TBD = 1.1625 −0.4444 TBE + 0.5661 TBD = 0.

Solve: 40 m

TBE = 1.868 kN, TBD = 1.467 kN.

50 m

40 m

D z

20 m

50 m

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The reactions at A oppose the sum of the cable tensions in the x-, y-, and z-directions. A X = 0, AY = 4.243 kN, AZ = 0. (These results are to be expected if there is no moment about A.)

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Problem 5.103 The space truss has roller supports at B, C, and D and is subjected to a vertical force F = 20 kN at A. What are the reactions at the roller supports?

y F A (4, 3, 4) m

B D (6, 0, 0) m x

Solution:

The key to this solution is expressing the forces in terms of unit vectors and magnitudes-then using the method of joints in three dimensions. The points A, B, C , and D are located at A(4, 3, 4) m, B(0, 0, 0) m, C (5, 0, 6) m,

C (5, 0, 6) m

D(6, 0, 0) m

we need e AB , e AC , e AD , e BC , e BD , and e CD . Use the form e PQ =

( x Q − x P ) i + ( y Q − y P ) j + ( z Q − z P )k [( x Q − x P ) 2 + ( y Q − y P ) 2 + ( z Q − z P ) 2 ]1/ 2

e AB = −0.625i − 0.469 j − 0.625k

e BC = 0.640 i + 0 j + 0.768k

TAD

TAB

e AC = 0.267 i − 0.802 j + 0.535k e AD = 0.371i − 0.557 j − 0.743k

Joint A :

TAC

e BD = 1i + 0 j + 0 k

e CD = 0.164 i + 0 j − 0.986k We will write each force as a magnitude times the appropriate unit vector.

Joint B : 2TAB TBD

T AB = T AB e AB , T AC = T AC e AC T AD = T AD e AD , TBC = TBC e BC

NBJ

TBD = TBD e BD , TCD = TCD e CD Each force will be written in component form, i.e.  T AB X = T AB e AB X   T ABY = T AB e ABY  etc.  T ABZ = T AB e ABZ  

Joint C : 2TAC

T ABY + T ACY + T ADY − 20 = 0 T ABZ + T AC Z + T AD Z = 0 Joint B: −T AB + TBC + TBD + N B j = 0

TCD

2TBC NCJ

Joint A: T AB + T AC + T AD + F = 0 T AB X + T AC X + T AD X = 0

TBC

Joint D :

2TAD 2TBD 2TCD

NDJ

Joint C : −T AC − TBC + TCD + N C j = 0 Joint D: −T AD − TBD − TCD + N D j = 0 Solving for all the unknowns, we get N B = 4.44 kN N C = 2.22 kN N D = 13.33 kN Also, T AB = −9.49 kN, T AC = −16.63 kN T AD = −3.99 kN, TBC = 7.71 kN TBD = 0.99 kN, TCD = 3.00 kN

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Problem 5.104 In Example 5.8, suppose that the cable BD is lengthened and the attachment point B is moved to the end of the bar at C. The positions of the attachment point D and the bar are unchanged. Draw a sketch of the bar showing cable BD in its new position. Draw the free-body diagram of the bar and apply equilibrium to determine the tension in the cable and the reactions at A.

Solution: The sketch and free-body diagram are shown. We must express the force exerted on the bar by cable BD in terms of its components. The bar AC is 4 ft long. The vector from C to D is rCD = [(2 − 4 cos30 °) i + (2 − {−4 sin 30 °}) j + (−1 − 0) j] ft rCD = (−1.46 i + 4 j − k) ft The force exerted by the cable CD can be expressed T

rCD = T (−0.335i + 0.914 j − 0.229k) rCD

The equilibrium equations are ΣFx : A x − 0.335T = 0 ΣFy : A y + 0.914T − 100 1b = 0 ΣFz : Az − 0.229T = 0 ΣM A : M Ax i + M Ay j +

3.464

−2

k 0

−0.335T

0.914T − 100

−0.229T

= 0

Expanding the determinant and solving the six equations, we obtain T = 139 lb, A x = 46.4 lb, A y = −26.8 lb, Az = 31.7 lb M Ax = −63.4 ft-1b, M Ay = −110 ft-1b

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Problem 5.105 The 40-lb door is supported by hinges at A and B. The y-axis is vertical. The hinges do not exert couples on the door, and the hinge at B does not exert a force parallel to the hinge axis. The weight of the door acts at its midpoint. What are the reactions at A and B?

1 ft

4 ft B

5 ft

Solution:

The position vector of the midpoint of the door:

rCM = (2cos50 °) i + 3.5 j + (2cos 40 °)k 1 ft

= 1.2856 i + 3.5 j + 1.532k.

The position vectors of the hinges: r A = j,

408

rB = 6 j. z

The forces are: W = −40 j, A = A X i + AY j + AZ k, B = B X i + B Z k.

The position vectors relative to A are BX

r ACM = rCM − r A = 1.2856 i + 2.5 j + 1.532k, BZ

r AB = rB − r A = 5 j. The sum of the moments about A

ΣM A = r ACM × W + r AB × B i = 1.2856 0

2.5 1.532 + −40

i j

0 5

AZ = 0

W AX

B X 0 BZ

ΣM A = (5B Z + 40(1.532)) i + (−5B X − 40(1.285))k = 0, from which B Z =

−40(1.532) = −12.256 lb 5

BX =

−40(1.285) = −10.28 lb. 5

and

The reactions at A are determined from the sums of forces: ΣFX = ( A X + B X ) i = 0, from which A X = 10.28 lb, ΣFY = ( AY − 40) j = 0, from which AY = 40 lb, ΣFZ = ( AZ + B Z )k = 0, from which AZ = 12.256 lb

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Problem 5.106 The vertical cable is attached at A. Determine the tension in the cable and the reactions at the bearing B due to the force F = 10 i − 30 j − 10 k (N).

y 200 mm 100 mm 100 mm B

Solution:

200 mm

The position vector of the point of application of the

force is

rF = 0.2 i − 0.2k.

x A

The position vector of the bearing is rB = 0.1i. The position vector of the cable attachment to the wheel is rC = 0.1k. he position vectors relative to B are: rBC = rC − rB = −0.1i + 0.1k, rBF = rF − rB = 0.1i − 0.2k. The sum of the moments about the bearing B is ΣM B = M B + rBF × F + rBC × C = 0,

or ΣM B = M B +

0.1

0 −0.2 + −0.1

10 −30 −10

i 0

0 0.1 −T

= (−6 + 0.1T ) i + ( M BY − 1) j + ( M BZ − 3 + 0.1T )k = 0, from which T =

6 = 60 N, 0.1

M BY = +1 N-m, M BZ = −0.1T + 3 = −3 N-m. The force reactions at the bearing are determined from the sums of forces: ΣFX = ( B X + 10) i = 0, from which B X = −10 N. ΣFY = ( BY − 30 − 60) j = 0, from which BY = 90 N. ΣFZ = ( B Z − 10) j = 0, from which B Z = 10 N.

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Problem 5.107 Suppose that the z component of the force F is zero, but otherwise F is unknown. If the couple exerted on the shaft by the bearing at B is M B = 6 j − 6k N-m, what are the force F and the tension in the cable?

y 200 mm 100 mm 100 mm B

200 mm F

z x A

Solution:

From the diagram of Problem 5.106, the force equilibrium equation components are ΣFx = B X + FX = 0, ΣFy = BY + FY = 0, and ΣFz = B Z + FZ = 0,

where FZ = 0 is given in the problem statement. The moment equations can be developed by inspection of the figure also. They are ΣM x = M BX + M AX + M FX = 0, ΣM Y = M BY + M AY + M FY = 0, and ΣM Z = M BZ + M AZ + M FZ = 0, where M B = 6 j − 6k N-m. Note that M BX = 0 can be inferred. The moments which need to be substituted into the moment equations are M A = (0.1) Ai + 0 j + (0.1) Ak N-m, and M F = (0.2) FY i − (0.2) FX j + (0.1) FY k N-m. Substituting these values into the equilibrium equations, we get F = 30 i − 60 j + 0 k N, and A = 120 N.

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Problem 5.108 The device in Problem 5.106 is badly designed because of the couples that must be supported by the bearing at B, which would cause the bearing to “bind.” (Imagine trying to open a door supported by only one hinge.) In this improved design, the bearings at B and C support no couples, and the bearing at C does not exert a force in the x direction. If the force F = 10 i − 30 j − 10 k (N), what are the tension in the vertical cable and the reactions at the bearings B and C ?

y 200 mm

50 mm 100 mm 50 mm

200 mm B

C F

z x

Solution:

The position vectors relative to the bearing B are: the position vector of the cable attachment to the wheel is

rBT = −0.05i + 0.1k. BY

The position vector of the bearing C is: rBC = 0.1i.

BX CY

The position vector of the point of application of the force is:

z T

rBF = 0.15i − 0.2k.

F CZ

The sum of the moments about B is ΣM B = rBT × T + rBC × C + rBF × F = 0

ΣM B =

−0.05

0.1

−T

0.15

−0.2

−30

−10

0.1

k 0

= 0

ΣM B = (0.1T − 6) i + (−0.1C Z + 1.5 − 2) j + (0.05T + 0.1C Y − 4.5)k = 0. From which: T = 60 N, CZ =

−0.5 = −5 N, 0.1

CY =

4.5 − 0.05T = 15 N. 0.1

The reactions at B are found from the sums of forces: ΣFX = ( B X + 10) i = 0, from which B X = −10 N. ΣFY = ( BY + C Y − T − 30) j = 0, from which BY = 75 N. ΣFZ = ( B Z + C Z − 10)k = 0, from which B Z = 15 N

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Problem 5.109 Assume that the 100-lb weight of the hinged rectangular door acts at the center of the door. The y-axis is vertical. The hinges at C and D do not exert couples on the door. Determine the tension in the rope AB. Strategy: Evaluate the moment about the x-axis.

y 15 in A

30 in

12 in

12 in C

Solution:

The hinges C and D exert only forces on the door, so they exert no moment about the x-axis. We can determine the tension T in the rope by evaluating the moment about the x-axis due to the tension and the door’s weight:

358

x B

30 in

60 in

First we will determine the components of the force exerted on the door by the tension. The position vector of point B is y

rB = (60 in) i + [−(30 in)sin35 °] j + [(30 in)cos35 °]k = 60 i − 17.2 j + 24.6k in,

so the position vector from point B to point A is rBA = r A − rB = [15 in − 60 in]i + [30 in + 17.2 in] j + [0 − 24.6 in]k

= −45i + 47.2 j − 24.6k (in).

The force exerted on the door by the rope is z

 r  T = T  BA  = T (−0.646 i + 0.677 j − 0.353k).  rBA 

B 100 lb

The position vector of point G is rG = (30 in) i + [−(15 in)sin35 °] j + [(15 in)cos35 °]k = 30 i − 8.60 j + 12.3k in. The sum of the moments due to the two forces about the origin is i rG × (−100 lb) j + rB × T =

30 in −8.60 in 12.3 in −100 lb

k 0

60 in

−17.2 in

24.6 in

−0.646T

0.677T

−0.353T

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Problem 5.110 Assume that the 100-lb weight of the hinged rectangular door acts at the center of the door. The y-axis is vertical. The hinges at C and D do not exert couples on the door, and the hinge at D does not exert a force parallel to the hinge axis. Determine the reactions at the hinges.

The i component of this expression is the sum of the moments about the x-axis, which must equal zero: −(12.3 in)(−100 lb) + (−17.2 in)(−0.353T ) − (24.6 in)(0.677T ) = 0. Solving yields T = 116 lb. Now we want to write the equilibrium equations for the door. The sums of the forces in the three coordinate directions equal zero, ΣFx = C x − 0.646T = 0,

ΣFy = C y + D y + 0.677T − 100 lb = 0,

15 in

ΣFz = C z + D z − 0.353T = 0,

and the sum of the moments about the origin equals zero: 30 in

(12 i in) × C + (48i in) × D + rG × (−100 lb) j + rB × T 12 in

12 in C

358

x B

30 in

12 in

Cy i

60 in

Solution:

Consider the free-body diagram of the door:

First we will determine the components of the force exerted on the door by the tension. The position vector of point B is

j 0

−100 lb

60 in

−17.2 in

24.6 in

−0.646T

0.677T

−0.353T

k = 0.

We’ve already used the i component of the moment equation. The other two components give the two equations −(12 in)C z − (48 in) D z + (24.6 in)(−0.646T )

rB = (60 in) i + [−(30 in)sin35 °] j + [(30 in)cos35 °]k = 60 i − 17.2 j + 24.6k in,

30 in −8.60 in 12.3 in 0

i 48 in

− (60 in)(−0.353T ) = 0, (12 in)C y + (48 in) D y + (30 in)(−100 lb)

so the position vector from point B to point A is rBA = r A − rB = [15 in − 60 in]i + [30 in + 17.2 in] j + [0 − 24.6 in]k = −45i + 47.2 j − 24.6k (in).

+ (60 in)(0.677T ) − (−17.2 in)(−0.646T ) = 0. Solving these equation together with the three force equations, we obtain C x = 75.0 lb, C y = 40.4 lb, C z = 37.5 lb, D y = −19.1 lb, and D z = 3.41 lb.

The force exerted on the door by the rope is

C x = 75.0 lb, C y = 40.4 lb, C z = 37.5 lb,

 r  T = T  BA  = T (−0.646 i + 0.677 j − 0.353k).  rBA 

D y = −19.1 lb, D z = 3.41 lb. y

The position vector of point G is rG = (30 in) i + [−(15 in)sin35 °] j + [(15 in)cos35 °]k

= 30 i − 8.60 j + 12.3k in. The sum of the moments due to the two forces about the origin is

rG × (−100 lb) j + rB × T =

i j k 30 in −8.60 in 12.3 in 0 −100 lb 0 i j k 60 in −17.2 in 24.6 in . −0.646T 0.677T −0.353T

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Cy Cz

T Dz

B 100 lb

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Problem 5.111 The crane’s cable CD is attached to a stationary object at D. The crane is supported by the bearings E and F and the horizontal cable AB. The tension in cable AB is 8 kN. Determine the tension in the cable CD. Strategy: Since the reactions exerted on the crane by the bearings do not exert moments about the z-axis, the sum of the moments about the z-axis due to the forces exerted on the crane by the cables AB and CD equals zero.

Solution:

rCD = 3i − 6 j − 3k (m), so we can write the force exerted at C by cable CD as rCD = TCD (0.408i − 0.816 j − 0.408k). rCD 4 The coordinates of pt. B are x = (3) = 2 m, y = 4 m. 6 The moment about the origin due to the forces exerted by the two cables is TCD = TCD

y C

MO =

0.408TCD −0.816TCD −0.408TCD

The moment about the z-axis is k ⋅ M O = 32 − 4.896TCD = 0,

so TCD = 6.54 kN.

3m y

j k 4 0

= 32k − 2.448TCD i + 1.224TCD j − 4.896TCD k. F

E 2m

i 2

−8 0 0

The position vector from C to D is

C B

6m 4m D 3m

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Problem 5.112 In Example 5.9, suppose that the cable CE is shortened and its attachment point E is moved to the point (0, 80, 0) mm. The plate remains in the same position. Draw a sketch of the plate and its supports showing the new position of cable CE. Draw the free-body diagram of the plate and apply equilibrium to determine the reactions at the hinges and the tension in the cable.

Solution:

The sketch and free-body diagram are shown. The vector from C to E is rCE = [(0 − 200) i + (80 − 0) j + (0 − 0)k] mm = (−200 i + 80 j) mm

The force exerted by cable CE can be expresses as T

rCE = T (−0.928i + 0.371 j) rCE

The equilibrium equations for the plate are ΣFx : A x + B x − 0.928T = 0 ΣFy : A y + B y + 0.371T − 400 N = 0 ΣFz : Az + B z = 0 i

0.2

−0.928T

0.371T

ΣM B :

i +

0 0

0.2 + 0.2

0.2 = 0

Ax Ay 0

0 −400

Expanding and solving we find A x = 0, A y = 400 N, T = 1080 N B x = 1000 N, B y = −400 N, B z = 0,

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Problem 5.113 Assume that the weight of the 14-kg horizontal door acts at the center of the door. The door is held in place by hinges at A and B and by the cord DCE. The hinges do not exert couples on the door. Neglect the size of the pulley at C. Determine the tension in cord DCE.

0.4 m 0.15 m

A 0.3 m z

Solution:

The hinges at A and B exert only forces on the door, so they exert no moment about the x-axis. We can determine the tension T in the cord by evaluating the moment about the x-axis due to the two tension forces and the door’s weight:

B x 0.5 m

The force exerted on the door by the cord is  r  T = T  EC  = T (−0.286 i + 0.429 j − 0.857k)  rEC  = T (e x i + e y j + e z k).

The position vector of point G is rG = 0.25i + 0.15k (m). The door’s weight is W = (14 kg)(9.81 m/s 2 ) = 137 N. The sum of the moments about the origin due to the weight and the two tension forces is

C A T z

B G W

ΣM point A =

i j 0.25 m 0 0 −W

k 0.15 m 0

i j 0.5 m 0 Te x Te y

k 0.3 m Te z

First we need to determine the forces exerted on the door by the tension in terms of their components. The position vector from point D to point C is rDC = rC − rD

i 0 Td x

j 0 Td y

k 0.3 m Td z

= 0.

The i component of this expression is the sum of the moments about the x-axis: −(0.15 m)(−W ) − (0.3 m)(Td y ) − (0.3 m)(Te y ) = 0.

= [0.4 m − 0]i + [0.15 m − 0] j + [0 − 0.3 m]k

Solving yields T = 95.9 N.

= 0.4 i + 0.15 j − 0.3k (m).

95.9 N.

The force exerted on the door by the cord is   r T = T  DC  = T (0.766 i + 0.287 j − 0.575k)  rDC  = T (d x i + d y j + d z k). (In the last line we introduce some notation to make it easier to follow subsequent steps.) The position vector from point E to point C is rEC = rC − rE = [0.4 m − 0.5 m]i + [0.15 m − 0] j + [0 − 0.3 m]k = −0.1i + 0.15 j − 0.3k (m).

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Problem 5.114 Assume that the weight of the 14-kg horizontal door acts at the center of the door. The door is held in place by hinges at A and B and by the cord DCE. The hinges do not exert couples on the door, and the hinge at B does not exert a force on the door in the direction of the hinge axis. Neglect the size of the pulley at C. Determine the reactions at the hinges.

0.4 m 0.15 m 0.3 m z

Solution:

We draw the free-body diagram of the door:

B x

0.5 m

The position vector of point G is rG = 0.25i + 0.15k (m).

The door’s weight is W = (14 kg)(9.81 m/s 2 ) = 137 N. The force equilibrium equations are Ay Ax

By Bz

T z

ΣFx = A x + Td x + Te x = 0,

(1)

ΣFy = A y + B y + Td y + Te y − W = 0,

(2)

ΣFz = Az + B z + Td z + Te z = 0.

(3)

The sum of the moments about the origin is W

We need to determine the forces exerted on the door by the tension in the cord in terms of their components.

i ΣM point A =

The position vector from point D to point C is rDC = rC − rD = [0.4 m − 0]i + [0.15 m − 0] j + [0 − 0.3 m]k = 0.4 i + 0.15 j − 0.3k (m). The force exerted on the door by the cord is   r T = T  DC  = T (0.766 i + 0.287 j − 0.575k)  rDC  = T (d x i + d y j + d z k).

0.5 m

0.3 m

Td x

Td y

Td z

i +

0.25 m

0.15 m

−W

0.5 m

0.3 m

Te x

Te y

Te z

= 0.

−(0.15 m)(−W ) − (0.3 m)Td y − (0.3 m)Te y = 0, −(0.5 m) B z + (0.3 m)Td x − (0.5 m)Te z + (0.3 m)Te x = 0, (0.5 m) B y + (0.25 m)(−W ) + (0.5 m)Te y = 0. This yields the three equations −(0.15 m)(−W ) − (0.3 m)Td y − (0.3 m)Te y = 0,

(4)

(In the last line we introduce some notation to make it easier to follow subsequent steps.)

−(0.5 m) B z + (0.3 m)Td x − (0.5 m)Te z + (0.3 m)Te x = 0,

(5)

(0.5 m) B y + (0.25 m)(−W ) + (0.5 m)Te y = 0.

(6)

The position vector from point E to point C is

Eq. (4) is the sum of the moments about the x-axis. Solving it yields T = 95.9 N. Solving the other equations, which is simplified because we now know T, we obtain A x = −46.1 N, A y = 41.1 N, Az = 27.5, B y = 27.6 N, B z = 110 N.

rEC = rC − rE = [0.4 m − 0.5 m]i + [0.15 m − 0] j + [0 − 0.3 m]k = −0.1i + 0.15 j − 0.3k (m). The force exerted on the door by the cord is

A x = −46.1 N, A y = 41.1 N, Az = 27.5, B y = 27.6 N, B z = 110 N.

 r  T = T  EC  = T (−0.286 i + 0.429 j − 0.857k)  rEC  = T (e x i + e y j + e z k).

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Problem 5.115 The force F = 2 kN applied to the bar at B is parallel to the y-axis. The mass of the bar is negligible. It has ball-and-socket supports at A and D. What is the tension in cable CE ? Strategy: Use the fact that the sum of the moments about the line that passes through points A and D equals zero.

0.32 m 0.4 m

0.52 m

1.0 m

0.28 m F

Solution:

Consider the free-body diagram of the bar (the reactions at A and D are not shown): y

We need a unit vector parallel to the line through points A and D. The position vector from point A to point D is r AD = (1.0 m − 0) i + (0 − 0) j + (0 − 0.8 m)k = 1.0 i − 0.8k (m). Dividing this vector by its magnitude gives a unit vector that points from A toward D :

e AD = 0.781i − 0.625k. The moment about point A due to the cable tension T and the force F is

x C

MA =

1.0 m

−F

1.0 m

−0.28 m

−0.720T

0.423T −0.550T

= −Fk + [0 − (−0.28 m)(0.423T )]i − [(1.0 m)(−0.550T ) − (−0.28 m)(−0.720T )] j

Because the supports at A and D exert only forces on the bar, they exert no moment about the straight line that passes through A and D. By evaluating the moment about that line, we can determine the tension T in the cable in terms of the force F. We need to express the force exerted on the bar by the cable in terms of its components. The position vector from point C to point E is rCE = (0.32 m − 1.0 m) i + (0.4 m − 0) j + (0 − 0.52 m)k = −0.68i + 0.4 j − 0.52k (m). We divide this vector by its magnitude and multiply by the tension T to express the cable force in terms of components:

+ [(1.0 m)(0.423T ) − 0]k = 0.119Ti − 0.349Tj + (0.423T − F )k. The moment of the two forces about the line through points A and D must equal zero, so e AD ⋅ M A = (0.781 )(0.119T ) + (−0.625)(0.423T − F ) = 0. Solving this equation, we obtain T = 3.63F = 7.27 kN. 7.27 kN.

 r  T = T  CE  = T (−0.720 i + 0.423 j − 0.550 k).  rCE 

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Problem 5.116 The force F = 2 kN applied to the bar at B is parallel to the y-axis. The mass of the bar is negligible. It has ball-and-socket supports at A and D. The support at D is designed so that it does not exert a reaction parallel to the x-axis. Determine the reactions on the bar at A and D.

The moment about point A due to the cable tension T and the force F is 0.32 m MA =

0.4 m

1.0 m

−F

1.0 m

−0.28 m

−0.720T

0.423T −0.550T

= −Fk + [0 − (−0.28 m)(0.423T )]i − [(1.0 m)(−0.550T ) − (−0.28 m)(−0.720T )] j + [(1.0 m)(0.423T ) − 0]k = 0.119T i − 0.349Tj + (0.423T − F )k.

0.52 m

e AD ⋅ M A = (0.781 )(0.119T ) + (−0.625)(0.423T − F ) = 0.

1.0 m

Solving this equation, we obtain T = 3.63F = 7.27 kN.

0.28 m F

Solution:

The moment of the two forces about the line through points A and D must equal zero, so

Now we write the equilibrium equations for the bar. The force equations are ΣFx = A x − 0.720T = 0,

Consider the free-body diagram of the bar:

ΣFy = A y + D y + 0.423T − F = 0,

ΣFz = Az + D z − 0.550T = 0. The sum of the moments about the origin is E

Dz x

0.8 m

1.0 m

0.8 m

−F

ΣM origin =

C B F

i +

1.0 m

i +

1.0 m −0.720T

0.52 m

= 0.

0.423T −0.550T

This yields the three equations −(0.8 m) A y − (0.8 m)(−F ) − (0.52 m)(0.423T ) = 0,

We will first determine the tension T in the cable. Because the supports at A and D exert only forces on the bar, they exert no moment about the straight line that passes through A and D. By evaluating the moment about that line, we can determine the tension T in terms of the force F. We need to express the force exerted on the bar by the cable in terms of its components. The position vector from point C to point E is rCE = (0.32 m − 1.0 m) i + (0.4 m − 0) j + (0 − 0.52 m)k = −0.68i + 0.4 j − 0.52k (m). We divide this vector by its magnitude and multiply by the tension T to express the cable force in terms of components:

(0.8 m) A x − (1.0 m) D z + (0.52 m)(−0.720T ) − (1.0 m)(−0.550T ) = 0, (1.0 m) D y + (1.0 m)(−F ) + (1.0 m)(0.423T ) = 0. Solving the six equilibrium equations, which is simplified because we already know T, we obtain A x = 5.23 kN, A y = 0, Az = −1.46 kN, D y = −1.08 kN, D z = 5.46 kN. A x = 5.23 kN, A y = 0, Az = −1.46 kN, D y = −1.08 kN, D z = 5.46 kN.

 r  T = T  CE  = T (−0.720 i + 0.423 j − 0.550 k).  rCE 

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Problem 5.117 The bearings at A, B, and C do not exert couples on the bar and do not exert forces in the direction of the axis of the bar. Determine the reactions at the bearings due to the two forces on the bar.

y 200i (N) 300 mm

180 mm

B z A

150 mm

100k (N) 150 mm

Solution: The strategy is to take the moments about A and solve the resulting simultaneous equations. The position vectors of the bearings relative to A are:

(The HP-28S hand held calculator was used to solve these equations.) The Solution:

r AB = −0.15i + 0.15 j,

B X = 750 N, B Z = 1800 N,

r AC = −0.15i + 0.33 j + 0.3k.

C X = −950 N,

Denote the lower force by subscript 1, and the upper by subscript 2:

C Y = 900 N.

r A1 = −0.15i,

The reactions at A are determined by the sums of forces: ΣFY = ( AY + C Y ) j = 0, from which AY = −C Y = −900 N

r A 2 = −0.15i + 0.33 j.

ΣFZ = ( AZ + B Z + 100)k = 0, from which AZ = −1900 N

The sum of the moments about A is:

ΣM A = r A1 × F1 + r AB × B + r A 2 × F2 + r AC × C = 0 i

ΣM A = −0.15 0 0

z k

+ −0.15 0.33 0 + −0.15 0.33 0.3 = 0 200

200 N

0 + −0.15 0.15

0 100 i

BZ 100 N

AY AZ

ΣM A = (0.15B Z − 0.3C Y ) i + (15 + 0.15B Z + 0.3C X ) j + (−0.15B X − 66 − 0.15C Y − 0.33C X )k = 0. This results in three equations in four unknowns; an additional equation is provided by the sum of the forces in the x-direction (which cannot have a reaction term due to A) ΣFX = ( B X + C X + 200) i = 0. The four equations in four unknowns: 0 B X + 0.15B Z + 0C X − 0.3C Z = 0 0 B X + 0.15B Z + 0.3C X + 0C Y = −15 −0.15B X + 0 B Z − 0.33C X − 0.15C Y = 66 B X + 0 B Z + C X + 0C Z = −200.

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Problem 5.118 The support that attaches the sailboat’s mast to the deck behaves like a ball-and-socket support. The line that attaches the spinnaker (the sail) to the top of the mast exerts a 200-lb force on the mast. The force is in the horizontal plane at 15° from the centerline of the boat. (See the top view.) The spinnaker pole exerts a 50-lb force on the mast at P. The force is in the horizontal plane at 45° from the centerline. (See the top view.) The mast is supported by two cables, the backstay AB and the port shroud ACD. (The forestay AE and the starboard shroud AFG are slack, and their tensions can be neglected.) Determine the tensions in the cables AB and CD and the reactions at the bottom of the mast.

y A

A Spinnaker 50 ft

P 6 ft

Side View

15 ft

21 ft

Aft View

Top View 200 lb 158 E 458

Solution:

Although the dimensions are not given in the sketch, assume that the point C is at the midpoint of the mast (25 ft above the deck). The position vectors for the points A, B, C , D, and P are: r A = 50 j, rB = −21i, rP = 6 j, rC = 25 j − 7.5k.

z (Spinnaker not shown) F G A B P 50 lb C D

The sum of the moments about the base of the mast is M Q = r A × F A + r A × T AB + r A × T AC + rC × TCE + rP × FP = 0 ΣM Q = r A × (F A + T AB + T AC ) + rC × TCE + rP × FP = 0. From above, F A + T AB + T AC = FTX i + FTY j + FTZ k = (193.2 − (0.3872) T AB ) i + (−0.922 T AB

The vector parallel to the backstay AB is

− 0.9578 T AC ) j + (51.76 − 0.2873 T AC )k

r AB = rB − r A = −21i − 50 j. The unit vector parallel to backstay AB is

i j k 0 50 50 FTX FTY FTZ

ΣM Q =

e AB = −0.3872 i − 0.9220 j. The vector parallel to AC is +

r AC = rC − r A = −25 j − 7.5k. The forces acting on the mast are: (1) The force due to the spinnaker at the top of the mast:

i j k −7.5 0 25 0 0 0.2873 T AC

i j k 0 6 6 −35.35 0 35.35

= 0

= (50 FTZ + (25)(0.2873) T AC + 212.1) i + (−50 FTX + 212.1)k = 0.

F A = 200( i cos15 ° + k cos 75 °) = 193.19 i + 51.76k. (2) The reaction due to the backstay:

Substituting and collecting terms:

T AB = T AB e AB

(2800 − 7.1829 T AC ) i + (−9447.9 + 19.36 T AB )k = 0,

(3) The reaction due to the shroud:

from which

T AC = T AC e AC (4) The force acting on the cross spar CE: TCE = −(k ⋅ T AC )k = 0.2873 T AC k.

T AC =

2800 = 389.81 lb, 7.1829

T AB = 488.0 lb.

(5) The force due to the spinnaker pole:

The tension in cable CD is the vertical component of the tension in AC,

FP = 50(−0.707 i + 0.707k) = −35.35i + 35.35k.

TCD = T AC ( j ⋅ e AC ) = T AC (0.9578) = 373.37 1b.

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Problem 5.118

(Continued)

The reaction at the base is found from the sums of the forces:

y FA

ΣFX = (Q X + 193.19 − 35.35 − T AB (0.3872)) = 0,

TAB

from which Q X = 31.11 lb ΣFY = (QY − 0.922 T AB − 0.9578 T AC ) j = 0, z

from which QY = 823.24 lb

FP QX

ΣFZ = (Q Z + 51.76 + (0.2873 T AC

TAC FA TCD TCE

TCD FP QY

SIDE VIEWz AFT VIEW FA Q

− 0.2873 T AC + 35.35))k = 0,

from which Q Z = −87.11 lb Collecting the terms, the reaction is

FP TAB TOP VIEW

Q = 31.14 i + 823.26 j − 87.12k (1b)

Problem 5.119* The bar AC is supported by the cable BD and a bearing at A that can rotate about the axis AE. The person exerts a force F = 50 j (N) at C. Determine the tension in the cable. Strategy: Use the fact that the sum of the moments about the axis AE due to the forces acting on the free-body diagram of the bar must equal zero.

Solution: We will take moments about the line AE in order to eliminate all of the reactions at the bearing A. We have: e AE =

0.1 j − 0.3k = 0.316 j − 0.949k 0.1

r AB = (0.16 i + 0.06 j + 0.03k)m, TBD = TBD

j + 0.17k ( 0.24i − 0.46 ) 0.2981

r AC = (0.52 i + 0.2 j + 0.1k)m, F = (50 j) N Then the equilibrium equation is

ΣM AE = e AE ⋅ (r AB × TBD + r AC × F) = 0 (0.3, 0.5, 0) m

This reduces to the single scalar equation E

(0.3, 0.4, 0.3) m A

TBD = 174.5 N

(0.82, 0.60, 0.40) m B (0.46, 0.46, 0.33) m x

z D

334

(0.7, 0, 0.5) m

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Problem 5.120* In Problem 5.119, determine the reactions at the bearing A. Strategy: Write the couple exerted on the free-body diagram of the bar by the bearing as M A = M Ax i + M Ay j + M Az k. Then, in addition to the equilibrium equations, obtain an equation by requiring the component of M A parallel to the axis AE to equal zero.

(0.3, 0.5, 0) m

(0.3, 0.4, 0.3) m A

See the previous problem for setup. We add the reactions (force, moment) A = A x i + A y j + Az k, M A = M Ax i + M Ay j + M Az k This gives us too many reaction moments. We will add the constrain that M A ⋅ e AE = 0 We have the following 6 equilibrium equations:  ΣF : A + 0.440T = 0 x BD  x  F A T F 0 F : A 0.843 T Σ = + BD + = ⇒  Σ y y − BD + 50 N = 0   ΣF : A + 0.311T = 0 z BD  x

Solution:

ΣM A = M A + r AB × TBD + r AC × F = 0

(0.82, 0.60, 0.40) m B (0.46, 0.46, 0.33) m x

 ΣM : M − 5 N-m + (0.0440 m)T = 0 Ax Ax BD   ⇒  ΣM Ay : M Ay − (0.0366 m)TBD = 0   ΣM : M + 26 N-m − (0.161 m)T = 0 Az Az BD  e AE ⋅ M A = 0 ⇒ 0.316 M Ay − 0.949 M Az = 0

z D

Solving these 7 equations we find (0.7, 0, 0.5) m

A x = −76.7 N, A y = 97.0 N, Az = −54.3 N M Ax = −2.67 N-m, M Ay = 6.39 N-m, M Az = 2.13 N-m

Problem 5.121 In Practice Example 5.10, suppose that the support at A is moved so that the angle between the bar AB and the vertical decreases from 45 ° to 30 °. The position of the rectangular plate does not change. Draw the free-body diagram of the plate showing the point P where the lines of action of the three forces acting on the plate intersect. Determine the magnitudes of the reactions on the plate at B and C.

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Solution:

The equilibrium equations are

ΣFx : B sin 30 ° − C sin 30 ° = 0 ΣFy : B cos30 ° + C cos30 ° − 100 lb = 0 Solving yields B = C = 57.7 lb

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Problem 5.122 The magnitude of the reaction exerted on the L-shaped bar at B is 60 lb. (See Example 5.11.) (a) What is the magnitude of the reaction exerted on the bar by the support at A? (b) What are the x and y components of the reaction exerted on the bar by the support at A?

y 14 in B 17 in A

Solution:

The angle between the line AB and the x-axis is

θ = tan −1 (17/14) = 50.5 ° (a) The bar is a two-force member, so the magnitude of the reaction at A is A = 60 lb (b) The reaction at A must be parallel to the line from A to B, but it may point either from A toward B or from B toward A. In the first case, the components are A x = (60 lb) cos θ, A y = (60 lb)sin θ In the second case the components are A x = −(60 lb) cos θ, A y = −(60 lb) sin θ Thus A x = 38.1 lb, A y = 46.3 lb or A x = −38.1 lb, A y = −46.3 lb

Problem 5.123 The suspended load weighs 1000 lb. The structure is a three-force member if its weight is neglected. Use this fact to determine the magnitudes of the reactions at A and B.

Solution: The pin support at A is a two-force reaction, and the roller support at B is a one force reaction. The moment about A is M A = 5B − 10(1000) = 0, from which the magnitude at B is B = 2000 lb. The sums of the forces: ΣFX = A X + B = A X + 2000 = 0, from which A X = −2000 lb. ΣFY = AY − 1000 = 0, from which AY = 1000 lb.

The magnitude at A is A =

2000 2 + 1000 2 = 2236 lb A

5 ft 5 ft

B 10 ft

10 ft 1000 lb

AX AY

5 ft

1000 lb 10 ft

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Problem 5.124 The weight W = 50 lb acts at the center of the disk. Use the fact that the disk is a threeforce member to determine the tension in the cable and the magnitude of the reaction at the pin support.

608

Solution: Denote the magnitude of the reaction at the pinned joint by B. The sums of the forces are:

608

ΣFX = B X − T sin 60 ° = 0, and ΣFY = BY + T cos60 ° − W = 0.

The perpendicular distance to the action line of the tension from the center of the disk is the radius R. The sum of the moments about the center of the disk is M C = −RBY + RT = 0, from which BY = T. Substitute into the sum of the forces to obtain: T + T (0.5) − W = 0, from which T =

608

T R

2 W = 33.33 lb. 3

Substitute into the sum of forces to obtain

W BY

B X = T sin 60 ° = 28.86 lb. The magnitude of the reaction at the pinned joint is B =

33.33 2 + 28.86 2 = 44.1 lb

Problem 5.125 The weight W = 40 N acts at the center of the disk. The surfaces are rough. What force F is necessary to lift the disk off the floor?

150 mm

W 50 mm

Solution: The reaction at the obstacle acts through the center of the disk (see sketch) Denote the contact point by B. When the moment is zero about the point B, the disk is at the verge of leaving the floor, hence the force at this condition is the force required to lift the disk. The perpendicular distance from B to the action line of the weight is d = R cos α, where α is given by (see sketch) α = sin −1

150 mm

50 mm W

( R −R h ) = sin ( 150150− 50 ) = 41.81°. −1

The perpendicular distance to the action line of the force is D = 2 R − h = 300 − 50 = 250 mm. F

The sum of the moments about the contact point is M B = −( R cos α)W + (2 R − h) F = 0, from which F =

(150 cos 41.81 °)W = 0.4472W = 17.88 N 250

W h b

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Problem 5.126 Use the fact that the horizontal bar is a three-force member to determine the angle α and the magnitudes of the reactions at A and B. Assume that 0 ≤ α ≤ 90 ° .

2m a

3 kN

Solution: The forces at A and B are parallel to the respective bars since these bars are 2-force members. Since the horizontal bar is a 3-force member, all of the forces must intersect at a point. Thus we have the following picture:

608 B 308

From geometry we see that

d = 1 m cos30 ° d sin 30 ° = e sin α e

d cos30 ° + e cos α = 3 m Solving we find α = 10.89 °

608

308

d 308

3 kN

To find the other forces we look at the force triangle

3 kN

FB = 3 kN cos 40.89 ° = 2.27 kN FA = 3 kN sin 40.89 ° = 1.964 kN

40.898 FB

338

FA 908

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Problem 5.127 The suspended load weighs 600 lb. Use the fact that ABC is a three-force member to determine the magnitudes of the reactions at A and B.

3 ft B

4.5 ft 308

C 458 A

Solution:

All of the forces must intersect at a point.

From geometry tan θ =

3 ft = 0.435 (3 + 4.5cos30 °) ft

⇒ θ = 23.5 ° Now using the force triangle we find FB = 600 lb cot θ = 1379 lb FA = 600 lb csc θ = 1504 lb

FB u

600 lb FB u

600 lb

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Problem 5.128 (a) Is the L-shaped bar a three-force member? (b) Determine the magnitudes of the reactions at A and B. (c) Are the three forces acting on the L-shaped bar concurrent?

2 kN

3 kN-m B

300 mm

150 mm

700 mm

A 250 mm

500 mm

Solution: (a) No. The reaction at B is one-force, and the reaction at A is twoforce. The couple keeps the L-shaped bar from being a three force member. (b) The angle of the member at B with the horizontal is α = tan −1

( 150 ) = 30.96°. 250

The sum of the moments about A is

0.5 m

2 kN

3 kN-m

0.3 m

0.7 m Ay

ΣM A = −3 − 0.5(2) + 0.7 B cos α = 0, from which B = 6.6637 kN. The sum of forces:

ΣFX = A X + B cos α = 0, from which A X = −5.7143 kN. ΣFY = AY − B sin α − 2 = 0, from which AY = 5.4281 kN. The magnitude at A: A =

340

5.71 2 + 5.43 2 = 7.88 kN (c) No, by inspection.

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Problem 5.129 The hydraulic piston exerts a horizontal force at B to support the weight W = 1500 lb of the bucket of the excavator. Determine the magnitude of the force the hydraulic piston must exert. (The vector sum of the forces exerted at B by the hydraulic piston, the twoforce member AB, and the two-force member BD must equal zero.)

Solution: TAB a A

C B a W

Hydraulic piston 14 in

16 in

TAB

(a)

b TBD (b)

12 in

Consider the free-body diagram of the bucket (Fig. a). The force T AB is the axial force exerted by the two-force member AB. The angle α = arctan (12 in/14 in) = 40.6 °. The sum of the moments about C is

ΣM point C = (8 in)W − (16 in)T AB sin α − (4 in)T AB cos α = 0.

4 in

Bucket W

8 in

Solving yields T AB = 892 lb. Now consider the forces acting at B (Fig. b). The force F is the force exerted by the piston, and the force TBD is the force exerted by the two-force member BD. The angle β = arctan (16 in/12 in) = 53.1 °. From the equilibrium equations ΣFx = F − T AB cos α + TBD cos β = 0, ΣFy = −T AB sin α − TBD sin β = 0, we obtain TBD = −726 lb, F = 1110 lb. 1110 lb.

Problem 5.130 The member ACG of the front-end loader is subjected to a load W = 2 kN and is supported by a pin support at A and the hydraulic cylinder BC. Treat the hydraulic cylinder as a two-force member. (a) Draw the free-body diagrams of the hydraulic cylinder and the member ACG. (b) Determine the reactions on the member ACG.

A 0.75 m B

1m G

0.5 m

W 1.5 m

Solution:

This is a very simple Problem. The free body diagrams are shown at the right. From the free body diagram of the hydraulic cylinder, we get the equation B X + C X = 0. This will enable us to find B X once the loads on member ACG are known. From the diagram of ACG, the equilibrium equations are Σ Fx = A X + C X = 0, ΣFy = AY − W = 0, and ΣM A = (0.75)C X − (3)W = 0. Using the given value for W and solving these equations, we get A X = −8 kN,

1.5 m

BX B AX

0.75 m 1m

0.5 m 1.5 m

1.5 m

AY = 2 kN, C X = 8 kN, and B X = −8 kN.

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Problem 5.131 In Problem 5.130, determine the reactions on the member ACG by using the fact that it is a three-force member.

A 0.75 m B

1m G

0.5 m

W 1.5 m

Solution:

The easiest way to do this is take advantage of the fact that for a three force member, the three forces must be concurrent. The fact that the force at C is horizontal and the weight is vertical make it very easy to find the point of concurrency. We then use this point to determine the direction of the force through A. We can even know which direction this force must take along its line—it must have an upward component to support the weight—which is down. From the geometry, we can determine the angle between the force A and the horizontal.

1.5 m

A A

0.75 m

u CX C

1.5 m

x 1.5 m G

W5 2 kN

tan θ = 0.75/3, or θ = 14.04 °. Using this, we can write force equilibrium equations in the form ΣFx = − A cos θ + C X = 0, and ΣFy = A sin θ − W = 0. Solving these equations, we get A = 8.246 kN, and C X = 8 kN. The components of A are as calculated in Problem 5.130.

342

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Problem 5.132 A rectangular plate is subjected to two forces A and B (Fig. a). In Fig. b, the two forces are resolved into components. By writing equilibrium equations in terms of the components A x , A y , B x , and B y , show that the two forces A and B are equal in magnitude, opposite in direction, and directed along the line between their points of application. B B A

Solution:

The sum of forces:

ΣFX = A X + B X = 0, from which A X = −B X ΣFY = AY + BY = 0, from which AY = −B y . These last two equations show that A and B are equal and opposite in direction, (if the components are equal and opposite, the vectors are equal and opposite). To show that the two vectors act along the line connecting the two points, determine the angle of the vectors relative to the positive x-axis. The sum of the moments about A is M A = B x (h) − bB y = 0,

from which the angle of direction of B is A

B  h tan −1  Y  = tan −1 = αB .  BX  b

( )

b (a)

or (180 + α B ). Similarly, by substituting A:

( )

or (180 + α A ). But

Ay A

A  h tan −1  Y  = tan −1 = α A,  AX  b

α = tan −1 x

b (b)

( hb )

describes direction of the line from A to B. The two vectors are opposite in direction, therefore the angles of direction of the vectors is one of two possibilities: B is directed along the line from A to B, and A is directed along the same line, oppositely to B. b B h A

y Ay

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Fig a

Fig b x

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Problem 5.133 An object in equilibrium is subjected to three forces whose points of application lie on a straight line. Prove that the forces are coplanar.

F3 F1

Solution:

The strategy is to show that for a system in equilibrium under the action of forces alone, any two of the forces must lie in the same plane, hence all three must be in the same plane, since the choice of the two was arbitrary. Let P be a point in a plane containing the straight line and one of the forces, say F2 . Let L also be a line, not parallel to the straight line, lying in the same plane as F2 , passing through P. Let e be a vector parallel to this line L. First we show that the sum of the moments about any point in the plane is equal to the sum of the moments about one of the points of application of the forces. The sum of the moments about the point P: M = r1 × F1 + r2 × F2 + r3 × F3 = 0, where the vectors are the position vectors of the points of the application of the forces relative to the point P. (The position vectors lie in the plane.) Define d 12 = r2 − r1 , and d 13 = r3 − r1 . Then the sum of the moments can be rewritten, M = r1 × (F1 + F2 + F3 )

which is the moment about the point of application of F1 . (The vectors d 12 , d 13 are parallel to the line L.) The component of the moment parallel to the line L is e ⋅ (d 12 × F2 )e + e ⋅ (d 13 × F3 )e = 0, or F2 ⋅ (d 12 × e)e + F3 ⋅ (d 13 × e)e = 0. But by definition, F2 lies in the same plane as the line L, hence it is normal to the cross product d 12 × e ≠ 0, and the term F2 ⋅ (d 12 × e) = 0. But this means that F3 ⋅ (d 13 × e)e = 0, which implies that F3 also lies in the same plane as F2 , since d 13 × e ≠ 0. Thus the two forces lie in the same plane. Since the choice of the point about which to sum the moments was arbitrary, this process can be repeated to show that F1 lies in the same plane as F2 . Thus all forces lie in the same plane.

+ d 12 × F2 + d 13 × F3 = 0.

Since the system is in equilibrium, F3

F1 + F2 + F3 = 0, and the sum of moments reduces to M = d 12 × F2 + d 13 × F3 = 0,

344

F1 P

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Problem 5.134 The suspended cable weighs 12 lb. (a) Draw the free-body diagram of the cable. (The tensions in the cable at A and B are not equal.) (b) Determine the tensions in the cable at A and B. (c) What is the tension in the cable at its lowest point?

B 508 A 328

Solution:

(a) The FBD (b) The equilibrium equations ΣFx : −T A cos32 ° + TB cos 50 ° = 0 ΣFy : T A sin 32 ° + TB sin 50 − 12 lb = 0

508

TA 328

Solving we find T A = 7.79 lb, TB = 10.28 lb (c) Consider the FBD where W represents only a portion of the total weight. We have

12 lb

ΣFx : − T A cos32 ° + T = 0 Solving

TA 328

T = 6.61 lb

Problem 5.135 support.

Determine the reactions at the fixed

4 kN 3m A 20 kN-m 2 kN

Solution:

3 kN

The equilibrium equations

4 kN

ΣFx : A x + 4 kN = 0 ΣFy : A y − 2 kN − 3 kN = 0 ΣM A : M A − (2 kN)(5 m) − (4 kN)(3 m)

20 kN-m

− (3 kN)(8 m) + 20 kN-m = 0 Solving A x = −4 kN, A y = 5 kN, M A = 26 kN-m

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2 kN

3 kN

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Problem 5.136 (a) Draw the free-body diagram of the 50-lb plate, and explain why it is statically indeterminate. (b) Determine as many of the reactions at A and B as possible.

A 12 in 8 in

x 20 in

Solution:

50 lb

(a) The pin supports at A and B are two-force supports, thus there are four unknown reactions A X , AY , B X , and BY , but only three equilibrium equations can be written, two for the forces, and one for the moment. Thus there are four unknowns and only three equations, so the system is indeterminate. (b) Sums the forces:

12 in 8 in

B 20 in

ΣFX = A X + B X = 0, or A X = −B X , and

ΣFY = AY + BY − 50 = 0.

12 in.

The sum of the moments about B

BY 8 in. BX

M B = −20 A X − 50(20) = 0,

50 lb

20 in.

from which A X = −50 lb,

50 lb

and from the sum of forces B X = 50 lb.

Problem 5.137 The mass of the truck is 4000 kg . Its wheels are locked, and the tension in its cable is T = 10 kN. (a) Draw the free-body diagram of the truck. (b) Determine the normal forces exerted on the truck’s wheels at A and B by the road.

308

2.2 m

2.5 m

2m mg

Solution: The weight is 4000(9.81) = 39.24 kN. The sum of the moments about B ΣM B = −3T sin 30 ° − 2.2T cos30 ° + 2.5W − 4.5 A N = 0

308

from which AN = =

2.5W − T (3sin 30 ° + 2.2 cos30 °) 4.5 64.047 = 14.23 N 4.5

AX A AN

B BN

The sum of the forces: ΣFY = A N − W + B N − T cos30 ° = 0, from which B N = T cos30 ° − A N + W = 33.67 N

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Problem 5.138 Assume that the force exerted on the head of the nail by the hammer is vertical, and neglect the hammer’s weight. (a) Draw the free-body diagram of the hammer. (b) If F = 10 lb, what are the magnitudes of the force exerted on the nail by the hammer and the normal and friction forces exerted on the floor by the hammer?

Solution: Denote the point of contact with the floor by B. The perpendicular distance from B to the line of action of the force is 11 in. The sum of the moments about B is M B = 11F − 2 FN = 0, from 11F which the force exerted by the nail head is FN = = 5.5F . The 2 sum of the forces: ΣFX = −F cos 25 + H x = 0, from which the friction force exerted on the hammer is H X = 0.9063F . ΣFY = N H − FN + F sin 25 ° = 0, from which the normal force exerted by the floor on the hammer is N H = 5.077 F

F 11 in

If the force on the handle is F = 10 lb, then FN = 55 lb,

658

H X = 9.063 lb, and N H = 50.77 lb

2 in F

11 in.

658 2 in.

F 11 in. 658 HX B NH FN

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Problem 5.139 The spring constant is k = 9600 N/m and the unstretched length of the spring is 30 mm. Treat the bolt at A as a pin support and assume that the surface at C is smooth. Determine the reactions at A and the normal force at C.

Solution: AY AX 24 mm B kd

15 mm

308

30 mm

308

458

30 mm

15 mm

30 mm

24 mm

608 50 mm

The length of the spring is l =

30 2 + 30 2 mm =

1800 mm

l = 42.4 mm = 0.0424 m

30 mm

50 mm

The spring force is kδ where δ = l − l 0 . l 0 is give as 30 mm. (We must be careful because the units for k are given as N/m ) We need to use length units as all mm or all meters). k is given as 9600 N/m. Let us use l 0 = 0.0300 m and l = 0.0424 m Equilibrium equations: ΣFX = 0:

A X − k (l − l 0 )sin 45 ° − N C cos60 ° = 0

ΣFY = 0:

AY − k (l − l 0 ) cos 45 ° + N C sin 60 ° = 0

ΣM B = 0: (−0.024) A X + (0.050)( N C sin 60 °) − (0.015)( N C cos60 °) = 0 Solving, we get A X = 126.7 N AY = 10.5 N N C = 85.1 N

Problem 5.140 The engineer designing the release mechanism wants the normal force exerted at C to be 120 N. If the unstretched length of the spring is 30 mm, what is the necessary value of the spring constant k ?

Solution:

Refer to the solution of Problem 5.139. The equilibrium equations derived were ΣFX = 0: A X − k (l − l 0 )sin 45 − N C cos60 ° = 0 ΣFY = 0: AY − k (l − l 0 ) cos 45 + N C sin 60 ° = 0 ΣM B = 0: −0.024 A X + 0.050 N C sin 60 ° − 0.015 N C cos60 ° = 0

where l = 0.0424 m, l 0 = 0.030 m, N C = 120 N, and A X , AY , and k are unknowns.

Solving, we get

24 mm

30 mm

308

A X = 179.0 N, AY = 15.1 N, k = 13500 N/m

30 mm

348

15 mm

50 mm

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Problem 5.141 The truss supports a 90-kg suspended object. What are the reactions at the supports A and B? 400 mm

700 mm

Solution: Treat the truss as a single element. The pin support at A is a two force reaction support; the roller support at B is a single force reaction. The sum of the moments about A is M A = B(400) − W (1100) = 0, from which B =

300 mm

1100W = 2.75W 400

B = 2.75(90)(9.81) = 2427.975 = 2.43 kN. The sum of the forces: Σ FX = A X = 0 ΣFY = AY + B − W = 0, from which AY = W − B = 882.9 − 2427.975 = −1.545 kN

400 mm

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700 mm

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Problem 5.142 The trailer is parked on a 15° slope. Its wheels are free to turn. The hitch H behaves like a pin support. Determine the reactions at A and H.

1.4 ft 870 lb

2.8 ft

H 1.6 ft 8 ft

158

Solution: The coordinate system has the x-axis parallel to the road. The wheels are a one force reaction normal to the road, the pin H is a two force reaction. The position vectors of the points of the center of mass and H are:

from which AY =

5816.55 = 727.1 lb. 8

The sum of the forces is

rW = 1.4 i + 2.8 j ft and

ΣFX = ( H X − 225.173) i = 0, from which H X = 225.2 lb,

rH = 8i + 1.6 j.

ΣFY = ( AY + H Y − 840.355) j = 0, from which H Y = 113.3 lb

The angle of the weight vector realtive to the positive x-axis is α = 270 ° − 15 ° = 255 °. 1.4 ft

The weight has the components

1.2 ft

W = W ( i cos 255 ° + j sin 255 °) = 870(−0.2588i − 0.9659 j) W

= −225.173i − 840.355 j (lb).

158

The sum of the moments about H is M H = (rW − rH ) × W + (r A − rH ) × A, i −6.6 −225.355

MH =

j 1.2 −840.355

k 0 0

AY i −8 0

j −1.6 AY

k 0 0

HX HY

8 ft

= 0

= 5816.55 − 8 AY = 0,

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Problem 5.143 To determine the location of the point where the weight of a car acts (the center of mass), an engineer places the car on scales and measures the normal reactions at the wheels for two values of α, obtaining the following results.

Solution:

The position vectors of the cm and the point B are

rCM = (2.7 − b) i + h j, rB = 2.7 i. The angle between the weight and the positive x-axis is β = 270 − α. The weight vector at each of the two angles is

A y (kN)

B (kN)

W10 = W ( i cos 260 ° + j sin 260 °)

10°

10.134

4.357

W10 = W (−0.1736 i − 0.9848 j)

20°

10.150

3.677

W20 = W ( i cos 250 ° + j sin 250 °) or W20 = W (−0.3420 i − 0.9397 j) The weight W is found from the sum of forces:

What are the distances b and h?

ΣFY = AY + BY + W sin β = 0, y

from which Wβ =

AY + BY . sin β

Taking the values from the table of measurements: B

Ax a

W10 = −

10.134 + 4.357 = 14.714 kN, sin 260 °

[check : W20 = −

2.7 m

10.150 + 3.677 = 14.714 kN check ] sin 250 °

The moments about A are M A = rCM × W + rB × B = 0. Taking the values at the two angles: M A 10 =

i 2.7 − b −2.5551

j h −14.4910

k 0 0

i 2.7 0

j 0 4.357

k 0 0

i 2.7 0

j 0 3.677

k 0 0

= 0

= 14.4903b + 2.5551h − 27.3618 = 0

M A 20 =

i 2.7 − b −5.0327

j h −13.8272

k 0 0

= 013.8272b + 5.0327h − 27.4054 = 0 These two simultaneous equations in two unknowns were solved using the HP-28S hand held calculator. b = 1.80 m, h = 0.50 m

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Problem 5.144 The bar is attached by pin supports to collars that slide on the two fixed bars. Its mass is 10 kg, it is 1 m in length, and its weight acts at its midpoint. Neglect friction and the masses of the collars. The spring is unstretched when the bar is vertical (α = 0), and the spring constant is k = 100 N/m. Determine the values of α in the range 0 ≤ α ≤ 60 ° at which the bar is in equilibrium.

Solution:

The force exerted by the spring is given by FS = k ( L − L cos α). The equations of equilibrium, from the free body diagram, are

Spring Constant (K) in N/m vs Alpha (deg)

90000 80000

ΣFx = N B = 0,

70000

ΣFy = FS + N A − mg = 0, and ΣM B = (−L sin α) N A +

60000 K

( L2 sin α )mg = 0.

These equations can be solved directly with most numerical solvers and the required plot can be developed. The plot over the given range is shown at the left and a zoom-in is given at the right. The solution and the plot were developed with the TK Solver Plus commercial software package. From the plot, the required equilibrium value is α ≅ 59.4 °.

550000 N

240000 m 30000 20000 10000 0 0

Alpha (deg)

FS B y

114

NB a mg

Spring Constant (K) in N/m vs Alpha (deg)

116

112 110 K 5108 N 2106 m 104

102 100 98 55

55.5

56.5

57.5

58.5

59.5

Alpha (deg)

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Problem 5.145 With each of the devices shown you can support a load R by applying a force F. They are called levers of the first, second, and third class. (a) The ratio R /F is called the mechanical advantage. Determine the mechanical advantage of each lever. (b) Determine the magnitude of the reaction at A for each lever. (Express your answers in terms of F.) R

(a)

First Class ΣM A : LF − LR = 0 ⇒ F = R R /F = 1 Second Class ΣM A : 2 LF − LR = 0 ⇒ F = R /2 R /F = 2 Third Class ΣM A : LF − 2 LR = 0 ⇒ F = 2 R R /F = 1/2

(b) First Class ΣFy : −F + A − R = 0 ⇒ A = 2 F Second Class ΣFy : A − R + F = 0 ⇒ A = F Third Class ΣFy : A + F − R = 0 ⇒ A = −F /2

Solution:

First-class lever

Second-class lever F

Third-class lever

Problem 5.146 The force exerted by the weight of the horizontal rectangular plate is 800 N. The weight of the rectangular plate acts at its midpoint. If you represent the reactions exerted on the plate by the three cables by a single equivalent force, what is the force, and where does its line of action intersect the plate?

Solution: y TA TC

z 2m

TB 0.5 m

800 N

2m 0.5 m 1m

The equivalent force must equal the sum of the reactions: FEQ = T A + TB + TC . FEQ = 300 + 100 + 400 = 800 N. The moment due to the action of the equivalent force must equal the moment due to the reactions: The moment about A is

MA =

i 2 0

j 0 100

k 0 0

i 1.5 0

j 0 400

k −1 0

i x 0

j 0 800

k z 0

M A = 400 i + 800 k = −(800 z ) i + (800 x )k, from which z = −0.5 m, and x = 1 m, which corresponds to the midpoint of the plate. Thus the equivalent force acts upward at the midpoint of the plate.

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Problem 5.147 The 20-kg mass is suspended by cables attached to three vertical 2-m posts. Point A is at (0, 1.2, 0) m. Determine the reactions at the fixed support at E. y C

Now that we know the tensions in the cables, we are ready to tackle the reactions at E (also G and H ). The first step is to draw the free body diagram of the post EB and to write the equations of equilibrium for the post. A key point is to note that the force on the post from cable AB is opposite in direction to the force found in the first part of the problem. The equations of equilibrium for post EB are ΣFx = E X − T ABX = 0,

ΣFy = E Y − T ABY = 0,

ΣFz = E Z − T ABZ = 0,

and, summing moments around the base point E, ΣM = M E + (2 j) × (−T AB ) = 0.

0.3 m z

The couple M E is the couple exerted on the post by the built in support. Solving these equations, we get

E 2m

E = −34.2 i + 91.3 j + 114.1k N and M E = 228.1i + 0 j + 68.44 k N-m. Also, M E = 238.2 N-m.

Solution:

All distances will be in meters, all forces in Newtons, and all moments in Newton-meters. To solve the three dimensional point equilibrium problem at A, we will need unit vectors and e AB , e AC , and e AD . To determine these, we need the coordinates of points A, B, C , and D. The rest of the problem will require knowing where points E , G, (under C), and H (under D) are located. From the diagram, the required point locations are A (0, 1.2, 0), B (−0.3, 2, 1), C (0, 2, −1), D (2, 2, 0), E (−0.3, 0, 1), G(0, 0, −1), and H(2, 0, 0). The required unit vectors are calculated from the coordinates of the points of the ends of the lines defining the vector. These are e AB = −0.228i + 0.608 j + 0.760 k, e AC = 0 i + 0.625 j − 0.781k,

Using a procedure identical to that followed for post EB above, we can find the built-in support forces and moments for posts CG and DH. The results for CG are: G = 0 i + 91.3 j − 114.1k N and M G = −228.1i + 0 j + 0 k N-m. Also, M G = 228.1 N-m. The results for DH are: H = 34.2 i + 13.7 j + 0 kN and M H = 0 i + 0 j + 68.4 k N-m. Also, M G = 68.4 N-m

and e AD = 0.928i + 0.371 j + 0 k. The force T AB in cable AB can be written as

C 2TAB

T AB = T ABX i + T ABY j + T ABZ k, where T ABX = T AB e ABX , etc. Similar equations can be written for the forces in AC and AD. The free body diagram of point A yields the following three equations of equilibrium.

and ΣFz = T ABZ + T ACZ + T ADZ = 0,

EY EX EZ

MG ME

ΣFx = T ABX + T ACX + T ADX = 0, ΣFy = T ABY + T ACY + T ADY − W = 0,

2TAD

2TAC

H GZ

G GY GX

HZ HY

where W = mg = (20)(9.81) = 196.2 N. Solving the equations above after making the substitutions related to the force components yields the tensions in the cables. They are T AB = 150 N, T AC = 146 N, and T AD = 36.9 N.

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Problem 5.148 In Problem 5.147, the fixed support of each vertical post will safely support a couple of 800 N-m magnitude. Based on this criterion, what is the maximum safe value of the suspended mass? y C B

Solution: We have all of the information necessary to solve this problem in the solution to Problem 5.147 above. All of the force and moment equations are linear and we know from the solution that a 20 kg mass produces a couple of magnitude 238.2 N-m at support E and that the magnitudes of the couples at the other two supports are smaller than this. All we need to do is scale the Problem. The scale factor is f = 800/238.2 = 3.358 and the maximum value for the suspended mass is m max = 20 f = 67.16 kg .

D A

1m 1m

E 2m

0.3 m z

Problem 5.149 The 80-lb bar is supported by a balland-socket support at A, the smooth wall it leans against, and the cable BC. The weight of the bar acts at its midpoint. (a) Draw the free-body diagram of the bar. (b) Determine the tension in cable BC and the reactions at A. y

r AW = 2.6936e AB r AW = 1.000 i + 2.000 j − 1.500k . The reaction at B is B = B k , since it is normal to a wall in the y − z plane. The sum of the moments about A is M A = r AW × W + r AB × B + r AB × T = 0

5 ft

3 ft

The cable tension is T = T e BC . The point of application of the weight relative to A is

i 1 0

MA = B C +

4 ft 3 ft

3 ft

k −1.5 0

i 2 −0.8575

j 4 0

k −3 0.5145

i 2 0

j 4 0

k −3 B

T = 0

M A = (−120 + 4 B + 2.058 T ) i − (2 B − 1.544 T ) j + (3.43 T − 80)k = 0.

A Solve:

Solution: (a)

The ball and socket is a three reaction force support; the cable and the smooth wall are each one force reaction supports. (b) The coordinates of the points A, B and C are A (3, 0, 3), B (5, 4, 0), and C (0, 4, 3). The vector parallel to the bar is r AB = rB − r A = 2i + 4 j − 3k . The length of the bar is r AB =

j 2 −80

80 = 23.32 lb 3.43 120 − 2.058 T B = = 18.00 lb. 4 T =

The reactions at A are found from the sums of forces: ΣFX = A X − T (0.8575) = 0 from which A X = 20 lb ΣFY = AY − 80 = 0, from which AY = 80 lb ΣFZ = AZ + T (0.5145) + B = 0, from which AZ = −30 lb

2 2 + 4 2 + 3 2 = 5.3852.

The unit vector parallel to the bar is

e AB = 0.3714 i + 0.7428 j − 0.5571k . The vector parallel to the cable is rBC = rC − rB = −5i + 3k .

The unit vector parallel to the cable is e BC = −0.8575i + 0.5145k .

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Problem 5.150 The horizontal bar of weight W is supported by a roller support at A and the cable BC. Use the fact that the bar is a three-force member to determine the angle α, the tension in the cable, and the magnitude of the reaction at A.

Solution: T a AY

C A

B W L/ 2

L/ 2

L /2

The sum of the moments about B is L M B = −LAY + W = 0, 2 W from which AY = . The sum of the forces: 2

( )

ΣFX = T cos α = 0, from which T = 0 or cos α = 0. The choice is made from the sum of forces in the y-direction: ΣFY = AY − W + T sin α = 0, W . This equation cannot be satis2 W fied if T = 0, hence cos α = 0, or α = 90 °, and T = 2 from which T sin α = W − AY =

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Chapter 6 Problem 6.1 The free-body diagrams of three truss joints are shown. They are in equilibrium. Determine the unknown axial forces T AB and T AC in each case. 200 lb T

458

608

308

400 lb

400 lb 400 lb

(b)

(c)

(a)

Solution:

Case (a) For this free-body diagram we can use inspection. The only way the sum of the forces on the free-body diagram in the direction perpendicular to the two parallel members can be zero is if T AC = 0. Once we know that, it’s clear that T AB = 400 lb. To obtain these results formally, we introduce the coordinate system

Case (c) Introduce the coordinate system y 200 lb TAB

TAC

608 TAB

TAC 308

458 400 lb

From the equilibrium equations ΣFx = −T AB cos60 ° + T AC cos 45 ° + 400 lb = 0, ΣFy = T AB sin 60 ° + T AC sin 45 ° + 200 lb = 0,

400 lb From the equilibrium equations

we obtain T AB = 146 lb, T AC = −462 lb.

ΣFx = T AC cos30 ° = 0,

(a) T AB = 400 lb, T AC = 0. (b) T AB = −566 lb, T AC = 400 lb.

ΣFy = T AB − 400 lb = 0,

we obtain T AB = 400 lb, T AC = 0. Case (b) Introduce the coordinate system y TAB 458 400 lb

TAC

From the equilibrium equations ΣFx = T AB cos 45 ° + T AC = 0, ΣFy = T AB sin 45 ° + 400 lb = 0, we obtain T AB = −566 lb, T AC = 400 lb.

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Problem 6.2 The dimension h = 2 ft. Determine the axial forces in the members of the truss and indicate whether they are in tension (T) or compression (C).

Solution:

We draw the free-body diagram of joint B: y

200 lb

a TAB

h C

TBC

The angles α = arctan(2 ft/4 ft) = 26.6 ° and β = arctan(2 ft/2 ft) = 45°. The equilibrium equations for the joint are

2 ft

4 ft

ΣFx = −T AB cos α + TBC cos β = 0, ΣFy = −T AB sin α − TBC sin β + 200 lb = 0. Solving, we obtain T AB = 149 lb, TBC = 189 lb. AB: 149 lb (T); BC: 189 lb (T).

Problem 6.3 Member AB of the truss is subjected to a 1000-lb tensile force. Determine the weight W and the axial force in member AC.

Solution: A

2 1

1000 lb 1 FAC

60 in W

C 60 in

W 60 in

Using joint A 2 1 (1000 lb) − FAC = 0 2 5 1 1 (1000 lb) − ΣFy : − FAC − W = 0 2 5 ΣFx : −

Solving we have FAC = −1265 lb, W = 447 lb In summary we have W = 447 lb, FAC = 1265 lb(C )

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Problem 6.4 Determine the axial forces in the members of the truss. A

ΣFx = −T AB − TBC cos 45 ° = 0, ΣFy = −TBC sin 45 ° − 150 lb = 0,

2 ft

From the equilibrium equations

the axial forces T AB = 150 lb, TBC = −212 lb. Consider the free-body diagram of joint D: y

2 ft 150 lb

TAD

TCD

458 2 ft

200 lb

100 lb

x D

Solution:

From the equilibrium equations ΣFx = TCD sin 45 ° + 200 lb = 0,

Consider the free-body diagram of the entire truss: y

ΣFy = T AD + TCD cos 45 ° = 0, the axial forces T AD = 200 lb, TCD = −283 lb. Finally, consider the free-body diagram of joint C:

Ay Ax

TAC

150 lb

212 lb

4 ft

458

100 lb D

C 100 lb

x 2 ft

2 ft

283 lb

The equilibrium equations are

From this diagram we can see that T AC = (100 lb) cos 45 ° = 70.7 lb.

ΣFx = A x + D = 0,

AB: 150 lb (T); AC: 70.7 lb (T); AD: 200 lb (T); BC: 212 lb (C);

ΣFy = A y − 100 lb − 150 lb = 0,

CD: 283 lb (C).

ΣM point A = (4 ft) D − (2 ft)(100 lb) − (4 ft)(150 lb) = 0. Solving, we obtain A x = −200 lb, A y = 250 lb, D = 200 lb. Consider the free-body diagram of joint B: y B TAB

458 150 lb TBC

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Problem 6.5 Each suspended weight has mass m = 20 kg. Determine the axial forces in the members of the truss and indicate whether they are in tension (T) or compression (C). A

Now work with joint C 5 T AC − 196.2 N = 0 29 2 T AC − TBC + TCD = 0 ΣFx : − 29 ΣFy :

Solving:

T AC = 211 N, TBC = −313 N TAC

0.4 m C

D 5 m

0.16 m 0.16 m

Solution:

TBC

0.32 m

TCD

196.2 N

Assume all bars are in tension. Start with joint D

5 T AD − 196.2 N = 0 61 6 T AD − TCD = 0 ΣFx : − 61 ΣFx :

Finally work with joint A ΣFy : −

Solving: T AD = 306 N, TCD = −235 N

5 (T AB + T AC ) − 29

5 T AD = 0 61

⇒ T AB = −423 N

TAD

2 D 5

TCD

5 2

TAD TAB

TAC

196.2 N In summary: T AB = 423 N(C ) T AC = 211 N(T ) T AD = 306 N(T ) TBC = 314 N(C ) TCD = 235 N(C )

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Problem 6.6 Determine the axial forces in members AB, BC, and CD of the truss.

Now consider the free-body diagram of joint A: y

4.66 KN

308

1.81 KN

10 kN

TAC

A 0.5 m

TAB D

E 2 kN

0.6 m

Solution:

We see that T AB = 4.66 kN and T AC = −1.81 kN. Next we draw the free-body diagram of joint B:

2 kN

0.6 m 4.66 KN

We begin by drawing the free-body diagram of the

TBC

entire truss: a

TBD

308

From the equilibrium equations

10 kN

ΣFx = TBD + TBC cos α = 0, ΣFy = TBC sin α + 4.66 kN = 0,

0.5 m a

0.6 m

2 kN

x 2 kN

we obtain TBC = −7.28 kN, TBD = 5.59 kN. By inspection of joint D we can see that TCD = −2 kN. AB: 4.66 kN (T), BC: 7.28 kN (C), CD: 2 kN (C).

0.6 m

The angle α = arctan(0.5 m/0.6 m) = 39.8 °. Writing the equilibrium equations, ΣFx = A x + B − (10 kN)sin 30 ° = 0, ΣFy = A y + 2 kN + 2 kN − (10 kN) cos30 ° = 0, ΣM point B = −(0.5 m) A x + (0.6 m)(2 kN) + (1.2 m)(2 kN) + (0.5 m)(10 kN)sin 30 ° − (0.6 m)(10 kN) cos30 ° = 0, and solving, we obtain the reactions A x = 1.81 kN, A y = 4.66 kN, B = 3.19 kN.

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Problem 6.7 This steel truss bridge is in the Gallatin National Forest south of Bozeman, Montana. Suppose that one of the tandem trusses supporting the bridge is loaded as shown. Determine the axial forces in members AB, AC , and BC.

Solution:

Because of the symmetry of the truss and loading, we can see that the vertical forces exerted on the truss at the supports A and H each equal (30 kip)/2 = 15 kip. We begin with the free-body diagram of joint A: TAB y

A x

TAC

15 kip

The angle α = arctan(9 ft/20 ft) = 24.2 °. From the equilibrium equations ΣFx = T AB cos α + T AC = 0, ΣFy = T AB sin α + 15 kip = 0, B

H C

10 kip 20 ft

20 ft

9 ft

10 kip 20 ft

we obtain T AB = −36.6 kip, T AC = 33.3 kip. We can see by inspection of joint C that member BC is subjected to a tensile load of 10 kip. (Draw a free-body diagram of joint C to confirm this if it doesn’t seem obvious.) AB: 36.6 kip (C), AC: 33.3 kip (T), BC: 10 kip (T).

Problem 6.8 Determine the largest tensile and compressive forces in the members of the bridge truss, and indicate the members in which they occur.

We can see by inspection of joint C that member BC is subjected to a tensile load of 10 kip. (Draw a free-body diagram of joint C to confirm this if it doesn’t seem obvious.) We can also see from joint C that TCE = T AC = 33.3 kip. We next draw the free-body diagram of joint B: B y

TBD TBE

x TAB TBC Its equilibrium equations are ΣFx = −T AB cos α + TBD + TBE cos α = 0, B

ΣFy = −T AB sin α − TBC − TBE sin α = 0.

H C 20 ft

E 10 kip 20 ft

9 ft

G 10 kip 20 ft

10 kip 20 ft

Solution: Because of the symmetry of the truss and loading, we can see that the vertical forces exerted on the truss at the supports A and H each equal (30 kip)/2 = 15 kip. We begin with the free-body diagram of joint A:

We know that T AB = −36.6 kip and TBC = 10 kip, so we can solve these equilibrium equations to obtain TBD = −44.4 kip, TBE = 12.2 kip. Finally, we can see from joint D that T DE = 0. Because of the symmetry of the truss and loading, we can now see that the largest tensile load is 33.3 kip in members AC , CE, EG, and GH, and the largest compressive load is 44.4 kip in members BD and DF. AC , CE , EG, GH : 33.3 kip (T), BD, DF: 44.4 kip (C).

TAB y

A x

TAC

15 kip

The angle α = arctan(9 ft/20 ft) = 24.2 °. From the equilibrium equations ΣFx = T AB cos α + T AC = 0, ΣFy = T AB sin α + 15 kip = 0, we obtain T AB = −36.6 kip, T AC = 33.3 kip.

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Problem 6.9 The trusses supporting the bridge in Problems 6.7 and 6.8 are called Pratt trusses. Suppose that the bridge designers had decided to use the truss shown instead, which is called a Howe truss. Determine the largest tensile and compressive forces that occur in the members, and indicate the members in which they occur. Compare your answers to the answers to Problem 6.8. B

ΣFy = −T AB sin α − TBC = 0. We know that T AB = −36.6 kip, so we can solve these equilibrium equations to obtain TBC = 15 kip, TBD = −33.3 kip. Now we consider joint C: TBC

20 ft

ΣFx = −T AB cos α + TBD = 0,

A C

Its equilibrium equations are

9 ft

10 kip 20 ft

TCD x

10 kip 20 ft

TAC

a C

TCE

10 kip

Solution:

A x

Its equilibrium equations are ΣFx = −T AC + TCD cos α + TCE = 0, ΣFy = TBC + TCD sin α − 10 kip = 0. We know that TBC = 15 kip, T AC = 33.3 kip, so we can solve these equilibrium equations to obtain TCD = −12.2 kip, TCE = 44.4 kip.

TAC

Finally, we can see from joint E that T DE = 10 kip.

15 kip

The angle α = arctan(9 ft/20 ft) = 24.2 °. From the equilibrium equations

Because of the symmetry of the truss and loading, we can now see that the largest tensile load is 44.4 kip in members CE and EG, and the largest compressive load is 36.6 kip in members AB and FH. CE, EG: 44.4 kip (T), AB, FH : 44.4 kip (C).

ΣFx = T AB cos α + T AC = 0, ΣFy = T AB sin α + 15 kip = 0, we obtain T AB = −36.6 kip, T AC = 33.3 kip. We next draw the free-body diagram of joint B: B

y a x

TBD

TAB TBC

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Problem 6.10 The system is in equilibrium. Gas pressure exerts a force FL = 20 kN on the left cylinder. Determine the force FR on the right cylinder.

The angles β = arctan(0.6 m/0.3 m) = 63.4 ° and γ = arctan (0.6 m/0.6 m) = 45 °. From the equilibrium equations ΣFx = −T AC cos β + T AD cos γ + (24.0 kN) cos α = 0, ΣFy = −T AC sin β − T AD sin γ + (24.0 kN)sin α = 0,

we obtain T AC = 24.8 kN and T AD = −12.6 kN. Finally, consider the free-body diagram of the right end of member AD: 0.6 m

D 0.6 m

12.6 kN FR

0.6 m

0.3 m

Solution:

Consider the free-body diagram of the left end of mem-

ber AB:

The equilibrium equation ΣFx = −FR + (12.6 kN) cos γ = 0 yields FR = 8.89 kN.

y TAB B

20 kN

FR = 8.89 kN.

The angle α = arctan(0.6 m/0.9 m) = 33.7 °. From the equilibrium equation ΣFx = T AB cos α + 20 kN = 0, we obtain T AB = −24.0 kN. Now consider the free-body diagram of jont A: y

a 24.0 kN

364

g TAD TAC

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Problem 6.11 The loads F1 = F2 = 8 kN. Determine the axial forces in members BD, BE, and BG. D

Solving, BD = 10 kN (T ) DE = −6 kN (C ) Joint E: y

B E

F2 5 8 kN

A C

Solution:

First find the external support loads and then use the method of joints to solve for the required unknown forces. (Assume all unknown forces in members are tensions). External loads: D

ΣFy = −BE + F2 = 0 Solving: EG = −6 kN (C ) BE = 8 kN (T )

F15 8 kN Joint G:

E G

A AX

DE = −6 kN ΣFx = DE − EG = 0

F2 5 8 kN

x CG

ΣFx : A x + F1 + F2 = 0 (kN)

ΣFy : A y + G y = 0

a +∑ M A : 8G y − 3F2 − 6 F1 = 0

Solving for the external loads, we get

(EG = −6 kN (C )) G y = 9 kN

A x = −16 kN (to the left) A y = −9 kN (downward)

ΣFx : −CG − BG cos θ = 0

G y = 9 kN (upward)

ΣFy : BG sin θ + EG + G y = 0

Now use the method of joints to determine BD, BE , and BG. Start with joint D.

Solving, we get BG = −5 kN (C ) CG = 4 kN (T )

Joint D: y

Thus, we have D

F1 5 8 kN x DE

BD = 10 kN (T ) BE = 8 kN (T ) BG = −5 kN (C )

u BD cos θ = 0.8 sin θ = 0.6 θ = 36.87 ° ΣFx : F1 − BD cos θ = 0 ΣFy : −BD sin θ − DE = 0

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Problem 6.12 The loads F1 = F2 = 800 lb. Determine the axial forces in members BD, CD, and CE.

6 ft

6 ft A

4 ft

F1 C

4 ft

F2 E

Solution:

We start with the free-body diagram of the entire truss:

Now we move to the free-body diagram of joint D,

1600 lb

800 lb

1800 lb

8 ft x 6 ft

TBD

TCD

800 lb E

6 ft

1200 lb and from the equilibrium equations

From the equilibrium equations

ΣFx = TBD + TCD cos α − 1800 lb = 0,

ΣFx = D x + E = 0,

ΣFy = −TCD sin α + 1600 lb − 1200 lb = 0,

ΣFy = D y − 800 lb − 800 lb = 0, ΣM point E = −(8 ft)D x − (6 ft)(800 lb) − (12 ft)(800 lb) = 0,

we obtain TBD = 1200 lb, TCD = 721 lb.

the reactions are D x = −1800 lb, D y = 1600 lb, E = 1800 lb. Now we consider the free-body diagram of joint E:

BD: 1200 lb (T), CD: 721 lb (T), CE: 2160 lb (C).

TDE TCE a 1800 lb

The angle α = arctan(4 ft/6 ft) = 33.7 °. From the equilibrium equations ΣFx = TCE cos α + 1800 lb = 0, ΣFy = TCE sin α + T DE = 0, the axial forces TCE = −2160 lb, T DE = 1200 lb.

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Problem 6.13 The truss supports loads at C and E. If F = 3 kN, what are the axial forces in members BC and BE ?

from which 5 BD = − F = −5 kN (C ). 3

(3) Joint E: 1m

BE − 2 F + DE = 0, 2

ΣFy =

from which BE = 2 2F − 1m G C

ΣFx = −CE −

CE = EG − 2F

The moment about A is

ΣM A = −1F − 4 F + 3G = 0, 5 F = 5 kN. The sums of forces: 3

ΣFY = AY − 3F + G = 0, from which AY =

4 F = 4 kN. 3

ΣFX = A X = 0, from which A X = 0. The interior angles GDE , EBC are 45 °, 1 . from which sin α = cos α = 2 Denote the axial force in a member joining I , K by IK.

BE + EG = 0, 2

BE 4 = F = 4 kN (T ). 3 2

AC = 0, 2

from which AC =

4 2 F = 4 2 kN (T ). 3

AC = 0, 2 4 from which AB = − F = −4 kN (C ). 3 ΣFx = AB +

(5) Joint C: ΣFy = BC +

AC − F = 0, 2

from which BC = F −

(1) Joint G: ΣFY =

2 kN (T ).

(4) Joint A: ΣFy = A y −

from which G =

2 F = 3

from which

E F

Solution:

2 DE =

1 AC = − F = −1 kN (C ). 3 2

AY DG + G = 0, 2

AX 1m

from which

5 2 DG = − 2G = − F = −5 2 kN (C ). 3 DG ΣFx = − − EG = 0, 2

DG 5 = F = 5kN (T ). 3 2

(2) Joint D: ΣFy = −DE −

AY AC

DG = 0, 2

458

G Joint G

from which EG = −

AB 458

Joint A

DG DE 458 Joint D

2F 1m

BE 458 CE

EG Joint E

458 BC AC

CE F Joint C

from which 5 F = 5 kN (T ). 3 DG ΣFx = −BD + = 0, 2 DE =

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Problem 6.14 The dimensions b = h = 2 m. Determine the axial forces in the members of the truss and indicate whether they are in tension (T) or compression (C). B

Now we analyze free-body diagrams of the joints, beginning with the ones that appear simplest.

TAB

C 6 kN

TAD

6 kN Joint A D

A b

Solution:

Consider the free-body diagram of the entire truss: y C

6 kN

TBC

TCD Joint C

6 kN

TBD

Joint B

Only two members are joined at joint A and the external reaction is now known. From its free-body diagram we see that the axial forces T AB = −6 kN and T AD = 0. Then from the free-body diagram of joint C we see that TBC = −6 kN and TCD = 0. Finally, consider the freebody diagram of joint B. Three members are joined there, but we already know the axial forces in members AB and BC. From the equilibrium equation ΣFx = −6 kN + TBD cos 45 ° = 0,

6 kN

we obtain TBD = 8.49 kN.

2m D A

Dx Dy

AB: 6 kN (C); AD: zero; BC: 6 kN (C); BD: 8.49 kN (T); CD: zero. x

2m Applying the equilibrium equations, ΣFx = D x − 6 kN = 0, ΣFy = A + D y = 0, ΣM point A = (2 m) D y + (2 m)(6 kN) = 0, The reactions are A = 6 kN, D x = 6 kN, D y = −6 kN.

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Problem 6.15 The truss is a preliminary design for a structure to attach one end of a stretcher to a rescue helicopter. Based on dynamic simulations, the design engineer estimates that the downward forces the stretcher will exert will be no greater than 1.6 kN at A and at B. What are the resulting axial forces in members CF , DF , and FG ?

300 mm

290 mm

48 FCF − 1.6 kN = 0 ⇒ FCF = 2.06 kN 3825 FCF 39 48 C

FCD 150 mm

480 mm

D B

ΣFy :

Start with joint C

390 mm F

Solution:

1.6 kN Now use joint F

200 mm ΣFx : − ΣFy :

59 29 39 FFG − FDF + FCF = 0 3706 3145 3825 15 48 48 FFG − FDF + FCF = 0 3706 3145 3825

Solving we find FDF = −1.286 kN, FCF = 2.03 kN 59 15

FFG

F 39

29 FDF

FCF

In Summary FCF = 2.06 kN(T ), FDF = 1.29 kN(C ), FCF = 2.03 kN(T )

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Problem 6.16 Member AB of the bridge truss is subjected to a compressive axial load of 180 kN. Member AD is subjected to a compressive axial load of 6 kN. Determine the axial loads in members AC and AE and state whether they are in tension (T) or compression (C).

Solution:

The free-body diagram of joint A is y

TAC TAD 528 TAE TAB

108

The equilibrium equations for the joint are ΣFx = −T AB cos10 ° + T AD cos 52 ° + T AE cos 4 ° = 0, ΣFy = −T AB sin10 ° + T AC + T AD sin 52 ° + T AE sin 4 ° = 0. We know that T AB = −180 kN and T AD = −6 kN. Solving the equations, we obtain T AC = −14.4 kN and T AE = −174 kN. AC: 14.4 kN (C); AE: 174 kN (C). D

388 1008

488 E A

Source: Courtesy of Vince Streano/ Corbis Documentary/ Getty Images.

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Problem 6.17 Determine the axial forces in the members in terms of the weight W. E

Solution:

Denote the axial force in a member joining two points I , K by IK. The angle between member DE and the positive x-axis is α = tan −1 0.8 = 38.66 °. The angle formed by member DB with the positive x-axis is 90 ° + α. The angle formed by member AB with the positive x-axis axis is α.

Joint E: ΣFy = −DE cos α − W = 0,

from which DE = −1.28W (C ) . A

ΣFy = −BE − DE sin α = 0, from which BE = 0.8W (T ) Joint D:

ΣFx = DE cos α + BD cos α − CD cos α = 0, 0.8 m

0.8 m

from which BD − CD = −DE. ΣFy = −BD sin α + DE sin α − CD sin α = 0, from which BD + CD = DE. Solving these two equations in two unknowns: CD = DE = −1.28W (C ) , BD = 0 Joint B: ΣFx = BE − AB sin α − BD sin α = 0, from which AB =

BE = 1.28W (T ) sin α

ΣFy = − AB cos α − BC = 0, from which BC = − AB cos α = −W (C )

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Problem 6.18 The lengths of the members of the truss are shown. The mass of the suspended crate is 900 kg. Determine the axial forces in the members.

Next work with joint C ΣFx : −FCD cos 40 ° − FBC cos 50 ° + FAC sin 27.4 ° = 0 ΣFy : −FCD sin 40 ° + FBC sin 50 ° + FAC cos 27.4 ° = 0

A FAC

FBC

27.48

12 m 508 B

13 m

408

FCD

13 m

Finally work with joint B

12 m

ΣFy : FAB cos 50 ° − FBC sin 50 ° − FBD cos 27.4 ° = 0 D

408 FAB

508

Solution:

Start with joint A

B 508

ΣFx : −FAB cos 40 ° − FAC sin 27.4 ° = 0 ΣFy : −FAB sin 40 ° − FAC cos 27.4 ° − (900 kg)(9.81 m/s 2 ) = 0 27.48 A 408

FBD

FBC

Solving we find FAB = 10.56 kN = 10.56 kN(T )

27.48

FAB

FAC = −17.58 kN = 17.58 kN(C ) FCD = −16.23 kN = 16.23 kN(C ) FAC

372

FBC = 6.76 kN = 6.76 kN(T ) 8829 N

FBD = 1.807 kN = 1.807 kN(T )

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Problem 6.19 Determine the axial forces in members AC , BC , CD, and CE.

The angle α = arctan(0.4 m/0.6 m) = 33.7 °. From the equilibrium equations ΣFx = T AB + T AC cos α + A = 0, ΣFy = −T AC sin α = 0,

6 kN B

we obtain T AB = −10 kN, T AC = 0. The free-body diagram of joint B is

0.4 m

8 kN D

TAB

C 0.4 m

6 kN

TBD

x TBC

0.4 m G

Solution:

0.6 m

6 kN

The free-body diagram of the entire structure is

From the equilibrium equations ΣFx = −T AB + TBD cos α = 0, ΣFy = −TBC − TBD sin α − 6 kN = 0, we obtain TBC = 0.667 kN, TBD = −12.0 kN. The free-body diagram of joint C is TBC

6 kN A

8 kN

6 kN x

TCD TCE

From the equilibrium equations Gx

ΣFx = TCD + TCE cos α = 0, ΣFy = TBC − TCE sin α = 0,

From the equilibrium equations ΣFx = A + G x = 0,

we obtain TCD = −1 kN, TCE = 1.20 kN.

ΣFy = G y − 6 kN − 8 kN − 6 kN = 0, ΣM point G = −(1.2 m) A + (1.2 m)(6 kN) + (0.6 m)(8 kN) = 0,

AC: zero; BC: 0.667 kN (T); CD: 1 kN (C); CE: 1.20 kN (T).

The reactions A = 10 kN, G x = −10 kN, G y = 20 kN. The free-body diagram of joint A is

TAB

a x

TAC

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Problem 6.20 Determine the axial forces in members CE , DE , EF , and EG.

we obtain TEG = −12.0 kN, TFG = −13.3 kN. The free-body diagram of joint F is TDF 6 kN

6 kN B

TEF

0.4 m

8 kN

TFG

C 0.4 m E

6 kN

From the equilibrium equations

ΣFx = −TEF − T DF cos α = 0, ΣFy = −TFG + T DF sin α − 6 kN = 0,

0.4 m G 0.6 m

we obtain T DF = −13.2 kN, TEF = 11 kN. The free-body diagram of joint E is TDE

0.6 m

TCE

Solution:

The free-body diagram of the entire structure is

y E

6 kN

8 kN

TEF TEG

From the equilibrium equations

6 kN

ΣFx = −TCE cos α + TEF + TEG cos α = 0, ΣFy = TCE sin α + T DE − TEG sin α = 0,

we obtain TCE = 1.20 kN, T DE = −7.33 kN. Gx

CE: 1.20 kN (T); DE: 7.33 kN (C); EF: 11 kN (T); EG: 12.0 kN (C).

From the equilibrium equations ΣFx = A + G x = 0, ΣFy = G y − 6 kN − 8 kN − 6 kN = 0, ΣM point G = −(1.2 m) A + (1.2 m)(6 kN) + (0.6 m)(8 kN) = 0, The reactions A = 10 kN, G x = −10 kN, G y = 20 kN. The free-body diagram of joint G is TFG TEG y

a x

The angle α = arctan(0.4 m/0.6 m) = 33.7 °. From the equilibrium equations ΣFx = −TEG cos α + G x = 0, ΣFy = TEG sin α + TFG + G y = 0,

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Problem 6.21 Determine the axial forces in members BC, CD, and CE of the truss.

E 4 ft G

12 kip

4 ft H

A 4 ft

4 ft

Solution:

The free-body diagrams for the entire truss as well as for joints A, B and C are shown. From the entire truss:

ΣFx : A x = 0 ΣFH : (12 kip)(8 ft) − A y (12 ft) = 0 Solving, yields A x = 0, A y = 8 kip From joint A: ΣFx : A x + T AD cos 45 ° = 0 ΣFy : A y + T AB + T AD sin 45 ° = 0 Solving yields T AB = −8 kip, T AD = 0 From joint B: ΣFx : TBD + TBC cos 45 ° = 0 ΣFy : TBC + sin 45 ° − T AB = 0 Solving yields TBD = 8 kip, TBC = −11.3 kip From joint C: ΣFx : TCE − TBC cos 45 ° = 0 ΣFy : −TBC sin 45 ° − TCD = 0 Solving yields TCD = 8 kip, TCE = −8 kip Thus we have BC: 11.3 kip (C), CD: 8 kip (T), CE: 8 kip (C)

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Problem 6.22 The Warren truss supporting the walkway is designed to support vertical 50-kN loads at B, D, F , and H. If the truss is subjected to these loads, what are the resulting axial forces in members BC , CD, and CE ?

Joint B: y 50 kN B

u BC

AB = −180.3 kN B

θ = 33.69 °

H 2m

ΣFx : BC cos θ + BD − AB cos θ = 0 ΣFy : −50 − AB sin θ − BC sin θ = 0 Solving, BC = 90.1 kN (T ) BD = −225 kN (C )

Solution: Assume vertical loads at A and I Find the external loads at A and I, then use the method of joints to work through the structure to the members needed. 50 kN

50 kN

Joint C: y

50 kN

x IY

ΣM A : −3(50) − 9(50) − 15(50) − 21(50) + 24 I y = 0

AC = 150 kN (T )

θ = 33.69 °

ΣFy : A y + I y − 4(50) = 0 (kN)

Solving A y = 100 kN

BC = 90.1 kN (T ) ΣFx : CE − AC + CD cos θ − BC cos θ = 0 ΣFy : CD sin θ + BC sin θ = 0

I y = 100 kN

Solving,

Joint A:

CE = 300 kN (T ) CD = −90.1 kN (C )

y AB

Hence BC = 90.1 kN (T ) CD = −90.1 kN (C )

u A AC

CE = 300 kN (T )

tan θ = 23 θ = 33.69 ° ΣFx : AB cos θ + AC = 0 ΣFy : AB sin θ + A y = 0 Solving, AB = −180.3 kN (C ) AC = 150 kN (T )

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Problem 6.23 Determine the axial forces in members AB, BC , BD, and BE of the truss.

Solution:

Consider the free-body diagram of the entire truss:

3 ft

Ax 400 lb

600 lb

A C

The equilibrium equations are

E 400 lb

600 lb

ΣFx = A x = 0, ΣFy = A y + F − 400 lb − 600 lb = 0,

3 ft

ΣM point A = (9 ft) F − (3 ft)(400 lb) − (6 ft)(600 lb) = 0. Solving, the reactions are A x = 0, A y = 467 lb, F = 533 lb. Only two members are joined at A: TAB

TAC

From the equilibrium equations ΣFx = T AB cos 45 ° + T AC = 0, ΣFy = T AB sin 45 ° + A y = 0, the axial forces T AB = −660 lb, T AC = 467 lb. By examination of joint C, we can see that the tension in member BC is TBC = 400 lb. As a result, there now are only two unknown axial forces at joint B: y B

TAB

TBC

TBD

TBE

From the equilibrium equations ΣFx = −T AB cos 45 ° + TBD + TBE cos 45 ° = 0, ΣFy = −T AB sin 45 ° − TBC − TBE sin 45 ° = 0, the axial forces TBD = −533 lb, TBE = 94.3 lb. AB: 660 lb (C); BC: 400 lb (T); BD: 533 lb (C); BE: 94.3 lb (T).

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Problem 6.24 The Pratt bridge truss supports five forces ( F = 300 kN). The dimension L = 8 m. Determine the axial forces in members BC, BI, and BJ. L

Solution:

Find support reactions at A and H. From the free body

diagram,

ΣFx = A X = 0,

u I

ΣFy = AY + H Y − 5(300) = 0,

L 8

J L 8

F F L5 8 m

and ΣM A = 6(8)H Y − 300(8 + 16 + 24 + 32 + 40) = 0.

L 8

TIJ

L 8 F

F F F 5 300 kN

From these equations, AY = H Y = 750 kN. Joint A

From the geometry, the angle θ = 45 °

TAB A

ΣFy = T AB sin θ + AY = 0. From these equations,

Joint A: From the free body diagram, ΣFx = A X + T AB cos θ + T AI = 0,

Joint I

TBI

u TAI

I x

TAI F

T AB = −1061 kN

Joint B y

and T AI = 750 kN. Joint I: From the free body diagram, ΣFx = TIJ − T AI = 0, ΣFy = TBI − 300 = 0.

u TAB

TBC x TBJ

TBI

From these equations, TBI = 300 kN and TIJ = 750 kN. Joint B: From the free body diagram, ΣFx = TBC + TBJ cos θ − T AB cos θ = 0, ΣFy = −TBI − TBJ sin θ − T AB sin θ = 0. From these equations, TBC = −1200 kN and TBJ = 636 kN.

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Problem 6.25 For the roof truss shown, determine the axial forces in members AD, BD, DE, and DG. Model the supports at A and I as roller supports.

Solving: FBC = −43.1 kN, FBD = −8.08 kN 6 kN FBC

6 kN C

8 kN E

10 kN 8 kN 6 kN H

3.6 m I

A D 3m

FBD

FAB

Next go to joint C ΣFy : −(8 kN) − FCD + ( FCE − FBC )sin 21.8 ° = 0

Solution:

Use the whole structure to find the reaction at A.

ΣM I : (6 kN)(3 m) + (8 kN)(6 m) + (10 kN)(9 m)

ΣFx : ( FCE − FBC ) cos 21.8 ° = 0 Solving: FCD = −8 kN, FCE = −43.1 kN

+ (8 kN)(12 m) + (6 kN)(15 m)

8 kN

+ A(18 m) = 0 ⇒ A = 19 kN

FCD

10 kN 8 kN

6 kN 8 kN

6 kN

FBC

FCD

Finally examine joint D A

Now work with joint A ΣFy : FAB sin 21.8 ° + A = 0 ⇒ FAB = −51.2 kN

ΣFx : −FAD + FDG − FBD cos 21.8 ° + FDE cos 50.19 ° = 0 ΣFy : FBD sin 21.8 ° + FCD + FDE sin 50.19 ° = 0 Solving: FDE = 14.3 kN, FDG = 30.8 kN FCD

ΣFx : FAD + FAB cos 21.8° = 0 ⇒ FAD = 47.5 kN FBD

FAB 21.88

50.198

FAD FAD

A Next use joint B

FDE

FDG

In Summary FAD = 47.5 kN(T ), FBD = 8.08 kN(C ), FDE = 14.32 kN(T ), FDG = 30.8 kN(T )

ΣFx : (−FAB + FBC + FBD ) cos 21.8 ° = 0 ΣFy : (−FAB + FBC − FBD )sin 21.8 ° − (6 kN) = 0

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Problem 6.26 The truss supporting the bridge in Problem 6.8 is called a Pratt truss. Suppose that the bridge designers had decided to use the truss shown here instead, which is called a Howe truss. Determine the axial forces in members AB, BD, and CD. B

20 ft

A x H

10 kip 20 ft

A C

Solution:

TAC

15 kip

9 ft

G 10 kip 20 ft

10 kip 20 ft

The angle α = arctan(9 ft/20 ft) = 24.2 °. From the equilibrium equations ΣFx = T AB cos α + T AC = 0, ΣFy = T AB sin α + 15 kip = 0, we obtain T AB = −36.6 kip, T AC = 33.3 kip. Four members are pinned at joint C, so we next consider joint B: B

TBD

a x TAB TBC From the equilibrium equations ΣFx = −T AB cos α + TBD = 0, ΣFy = −T AB sin α − TBC = 0,

we obtain TBC = 15 kip, TBD = −33.3 kip. Now we move to joint C: TBC y

TCD x

TAC

TCE

From the equilibrium equation ΣFy = TBC + TCD sin α = 0, we obtain TCD = −36.6 kip. Comparing these results to those of the Pratt truss in Problem 6.8, notice that the diagonal member BE of the Pratt truss is subjected to a tensile load, whereas the diagonal member CD of the Howe truss is subjected to a substantial compressive load. AB: 36.6 kip (C), BD: 33.3 kip (C), CD: 36.6 kip (C).

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Problem 6.27 The plane truss forms part of the supports of a crane on an offshore oil platform. The crane exerts vertical 75-kN forces on the truss at B, C, and D. You can model the support at A as a pin support and model the support at E as a roller support that can exert a force normal to the dashed line but cannot exert a force parallel to it. The angle α = 45 °. Determine the axial forces in the members of the truss.

75 kN 75 kN 75 kN

AY 3.4 m

AB AX

B 1.8 m 2.2 m

CD u DG

3.4 m

g AY b Joint A 75 kN BC g u BF BG AB Joint B

EY Joint E 75 kN

g DE DH Joint D

3.4 m

b AF

Joint F 75 kN BC CD

b EH

Joint H

CG Joint C

a from which two simultaneous equations are obtained.

3.4 m

Solve: EH = −44.67 kN (C ) , and DE = −115.8 kN (C )

Solution:

The included angles

( ) ( ) ( )

4 γ = = 49.64 °, 3.4 2.2 = 32.91 °, β = tan −1 3.4 1.8 = 27.9 °. θ = tan −1 3.4 tan −1

Joint F: ΣFx = − AF cos β + FG = 0, from which FG = −37.5 kN (C ) ΣFy = − AF sin β + BF = 0, from which BF = −24.26 kN (C )

The complete structure as a free body: The sum of the moments about A is

Joint H:

M A = −(75)(3.4)(1 + 2 + 3) + (4)(3.4) E y = 0.

ΣFx = EH cos β − GH = 0,

with this relation and the fact that E x cos 45 ° + E y cos 45 ° = 0, we obtain E x = −112.5 kN and E y = 112.5 kN. From

from which GH = −37.5 kN (C ) ΣFy = −EH sin β + DH = 0,

ΣFxA

= A x + E x = 0, A X = −E X = 112.5. kN.

from which DH = −24.26 kN (C )

ΣFxA

= A y − 3(75) + E y = 0,

Joint B:

from which A y = 112.5 kN. Thus the reactions at A and E are symmetrical about the truss center, which suggests that symmetrical truss members have equal axial forces.

ΣFy = − AB sin γ − BF + BG sin θ − 75 = 0,

The method of joints: Denote the axial force in a member joining two points I , K by IK.

ΣFx = − AB cos γ + BC + BG cos θ = 0,

Joint A: ΣFx = AB cos γ + A x + AF cos β = 0, ΣFy = AB sin γ + A y + AF sin β = 0, from which two simultaneous equations are obtained. Solve: AF = −44.67 kN (C ) , and AB = −115.8 kN (C )

from which BG = 80.1 kN (T )

from which BC = −145.8 kN (C ) Joint D: ΣFy = −DE sin γ − DH − DG sin θ − 75 = 0, from which DG = 80.1 kN (T ) ΣFx = DE cos γ − CD − DG cos θ = 0, from which CD = −145.8 kN (C ) Joint C:

Joint E:

ΣFx = CD − BC = 0,

ΣFy = −DE cos γ + E x − EH cos β = 0.

from which CD = BC Check.

ΣFy = DE sin γ + E y + EH sin β = 0,

ΣFy = −CG − 75 = 0, from which CG = −75 kN (C )

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Problem 6.28 The loads F1 = 2 kip, F2 = 4 kip, and F3 = 6 kip. Determine the axial forces in members CD, DE, and DG.

8 in

F1 G

F2 F3

8 in A

18 in

Solution:

The free-body diagram of the entire truss is F1

ΣFx = −T AB − TBC cos β + B x = 0, ΣFy = TBC sin β + TBD + B y = 0,

we obtain TBC = 6.41 kip, TBD = 5.33 kip.

The angle β = arctan(8 in/12 in) = 33.7 °. From the equilibrium equations

Now we consider joint C: TCE

Bx y

From the equilibrium equations x

ΣFx = B x − F1 − F2 − F3 = 0, ΣFy = A + B y = 0,

b TBC

TAC

ΣM point B = −(18 in) A + (24 in) F1 + (16 in) F2 + (8 in) F3 = 0, the reactions are A = 8.89 kip, B x = 12 kip, B y = −8.89 kip.

From the equilibrium equations ΣFx = −T AC cos α + TCD + TBC cos β + TCE cos α = 0,

Only two members are joined at joint A:

ΣFy = −T AC sin α − TBC sin β + TCE sin α = 0,

TAC

we obtain TCD = −8 kip, TCE = −6.67 kip.

A x

TAB

There are now only two unknown axial forces at joint D: TDE

The angle α = arctan(8 in/6 in) = 53.1 °. From the equilibrium equations

y TCD

ΣFx = T AB + T AC cos α = 0,

TDG

a D

ΣFy = T AC sin α + A = 0, the axial forces T AB = 6.67 kip, T AC = −11.1 kip. Now we move to the free-body diagram of joint B: TBD TBC

From the equilibrium equations ΣFx = −TCD − T DE cos α − F3 = 0, ΣFy = −TBD + T DE sin α + T DG = 0, we obtain T DE = 3.33 kip, T DG = 2.67 kip.

TAB

382

TBD

B Bx

CD: 8 kip (C); DE: 3.33 kip (T); DG: 2.67 kip (T).

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Problem 6.29 (a) Design a truss attached to the supports A and B that goes over the obstacle and supports the load applied at C. (b) Determine the axial forces in the members of the truss you designed in (a).

Obstacle

Solution:

10 kN

This is a design problem with many possible solutions. 6m

Problem 6.30 Suppose that you want to design a truss supported at A and B (Fig. a) to support a 3-kN downward load at C. The simplest design (Fig. b) subjects member AC to a 5-kN tensile force. Redesign the truss so that the largest tensile force is less than 3 kN. A

3.5 m

4.5 m

Reference Solution (Fig. (b)) Joint C: AC

BC 3 kN

1.2 m C

C B

3 kN

1.6 m

3 kN

(a)

(b)

θ = 36.87 ° ΣFx : −BC − AC cos θ = 0 ΣFy : AC sin θ − 3 kN = 0 Solving: BC = −4 kN (C ) AC = 5 kN (C ) Thus, AC is beyond the limit, but BC (in compression) is not,

Solution:

There are many possible designs. To better understand the problem, let us calculate the support forces in A and B and the forces in the members in Fig. (b).

Joint B:

Ay Ax

1.2 m

ΣFx : B x + BC = 0

C Bx

1.6 m 3 kN

1.2 tan θ = 1.6 θ = 36.87 °

ΣFy : AB = 0 Solving, BC and B x are both already known. We get AB = 0 Thus, we need to reduce the load in AC. Consider designs like that shown below where D is inside triangle ABC. Move D around to adjust the load. A

sin θ = 0.6 cos θ = 0.8 ΣFx : A x + B x = 0

ΣFy : A y − 3 kN = 0

a +ΣM A : 1.2 B x − 1.6(3) = 0

Solving, we get A x = −4 kN

B x = 4 kN A y = 3 kN Note: These will be the external reactions for every design that we produce (the supports and load do not change).

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However, the simplest solution is to place a second member parallel to AC, reducing the load by half.

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Problem 6.31 The bridge structure shown in Example 6.2 can be given a higher arch by increasing the 15 ° angles to 20 °. If this is done, what are the axial forces in members AB, BC , CD, and DE ? Solution:

Follow the solution method in Example 6.3. F is known

For joint C, ΣFx : −TBC cos 20 ° + TCD cos 20 ° = 0 ΣFy : −F − TBC sin 20° − TCD sin 20 ° = 0 TBC = TCD = −1.46 F (C ) For joint B.

Joint B:

ΣFx : TBC cos 20 − T AB cos α = 0

y F

TBC

ΣFy : TBC sin 20 ° − F − T AB sin α = 0 Solving, we get α = 47.5 ° and T AB = −2.03F (C ) For the new truss (using symmetry)

208 x a TAB

Members

Forces

AG, BH , CI , DJ , EK

AB, DE

2.03F (C )

BC , CD

1.46 F (C )

Joint C: F C 208 TBC

208 TCD

Problem 6.32 In Practice Example 6.3, use the method of sections to determine the axial forces in members BC, BI, and HI.

Solution:

The horizontal members of the truss are each 1 m in length. We cut through the relevant members and draw a free-body diagram of the section to the right of the cut. We will use equilibrium equations for this section that are designed to allow us to easily solve for the unknowns. The equilibrium equations ΣM I : TBC (1 m) − (100 kN)(4 m) = 0 ⇒ TBC = 400 kN ΣM B : −T HI (1 m) − (100 kN)(5 m) = 0 ⇒ T HI = −500 kN ΣFy : TBI sin 45 ° − 100 kN = 0 ⇒ TBI = 141 kN

In summary we have BC: 400 kN (T), BI : 141 kN (T), HI : 500 kN (C)

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Problem 6.33 Use the method of sections to determine the axial forces in members AC , BC , and BD.

Solution:

We obtain a section by passing a plane through members AC , BC , and BD: y

308 A

308

10 kN

TAC 0.5 m

0.5 m

a TBC TBD

D 2 kN

2 kN

x D

2 kN

0.6 m

The angle α = arctan(0.5 m/0.6 m) = 39.8 °. Writing the equilibrium equations,

0.6 m

10 kN

ΣFx = −T AC − TBC cos α − TBD − (10 kN)sin 30 ° = 0, ΣFy = −TBC sin α + 2 kN + 2 kN − (10 kN) cos30° = 0, ΣM point C = −(0.5 m)TBD + (0.6 m)(2 kN) = 0, and solving, we obtain T AC = −1.81 kN, TBC = −7.28 kN, TBD = 2.40 kN. AC: 1.81 kN (C), BC: 7.28 kN (C), BD: 2.40 kN (T).

Problem 6.34 The truss supports a 100-kN load at J. The horizontal members are each 1 m in length. (a) Use the method of joints to determine the axial force in member DG. (b) Use the method of sections to determine the axial force in member DG. A

1m J 100 kN

Solution: (a) We draw free-body diagrams of joints J , H , and D. From joint J we have ΣFy : T DJ sin 45 ° − (100 kN) = 0 ⇒ T DJ = 141 kN From joint H we have ΣFy : T DH = 0 From joint D we have ΣFy : −T DG sin 45 ° − T DH − T DJ sin 45 ° = 0 Solving yields T DG = −141 kN (b) We cut through CD, DG and GH. The free-body diagram of the section to the right of the cut is shown. From this diagram we have ΣFy : −T DG sin 45 ° − (100 kN) = 0 ⇒ T DG = −141 kN In summary (a), (b) DG: 141 kN (C)

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Problem 6.35 The horizontal members are each 1 m in length. Use the method of sections to determine the axial forces in members BC , CF , and FG.

Solution: BC

458 A

1m E

1m 100 kN

100 kN ΣFx : −BC − CF cos 45 − FG = 0 ΣFy : −CF sin 45 ° − 100 = 0 ΣM C : −(1) FG − 2(100) = 0 Solving BC = 300 kN (T ) CF = −141.4 kN (C ) FG = −200 kN (C )

Problem 6.36 The loads F1 = F2 = 800 lb. Use the method of sections to determine the axial forces in members BD, CD, and CE. 6 ft

Solution:

We obtain a section of the truss by passing a plane through members BD, CD, and CE: y TBD

6 ft

4 ft TCD

TCE

4 ft

A 800 lb

a a

C 800 lb

6 ft

4 ft

The angle α = arctan(4 ft/6 ft) = 33.7 °. From the equilibrium equation

ΣM point C = (4 ft)TBD − (6 ft)(800 lb) = 0, we obtain TBD = 1200 lb. Then from the equations ΣFx = −TCD cos α − TCE cos α − 1200 lb = 0,

ΣFy = TCD sin α − TCE sin α − 1600 lb = 0, the axial forces TCD = 721 lb, TCE = −2160 lb. BD: 1200 lb (T); CD: 721 lb (T); CE: 2160 lb (C).

Problem 6.37 The dimensions b = h = 2 m. Obtain a section of the truss by passing a plane through members AB, BD, and CD, and use it to determine the axial force in member BD. B

Solution:

Passing a plane through members AB, BD, and CD, we obtain the section

y C

C 6 kN

TAB D

A b

6 kN

458 TBD T

From the equilibrium equation ΣFx = TBD cos 45 ° − 6 kN = 0, we obtain TBD = 8.49 kN. 8.49 kN (T).

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Problem 6.38 Use the method of sections to determine the axial forces in members BD, CD, and CE.

ΣFx = A + G x = 0, ΣFy = G y − 6 kN − 8 kN − 6 kN = 0,

6 kN A

From the equilibrium equations

ΣM point G = −(1.2 m) A + (1.2 m)(6 kN) + (0.6 m)(8 kN) = 0, The reactions A = 10 kN, G x = −10 kN, G y = 20 kN.

Let us obtain a section by passing a plane through members BD, CD, and CE:

8 kN

0.4 m

6 kN

0.4 m

TBD

0.4 m G 0.6 m

0.6 m

Solution:

TCD TCE

The angle α = arctan(0.4 m/0.6 m) = 33.7 °. From the equilibrium equations

0.6 m

ΣFx = TBD cos α + TCD + TCE cos α + A = 0,

The free-body diagram of the entire structure is

ΣFy = −TBD sin α − TCE sin α − 6 kN = 0, ΣM point C = −(0.4 m)TBD cos α − (0.4 m) A = 0,

6 kN

we obtain TBD = −12.0 kN, TCD = −1 kN, TCE = 1.20 kN.

8 kN

BD: 12.0 kN (C); CD: 1 kN (C); CE: 1.20 kN (T).

6 kN x Gx

Problem 6.39 The Howe bridge truss is loaded as shown. Use the method of sections to determine the axial forces in members BD, CD, and CE. B

17 ft

E 10 kip 17 ft

8 ft

G 30 kip 17 ft

Now cut through BD, CD, and CE and use the left section. ΣM C : − A(17 ft) − FBD (8 ft) = 0 ⇒ FBD = −58.4 kip

A C

Solution: Use the whole structure to find the reaction at A (same as 6.38) A = 27.5 kip

20 kip 17 ft

ΣM D : − A(34 ft) + (10 kip)(17 ft) + FCE (8 ft) = 0 ⇒ FCE = 95.6 kip ΣFy : A − 10 kip +

8 FCD = 0 ⇒ FCD = −41.1 kip 353

In Summary FBD = 58.4 kip(C ), FCE = 95.6 kip(T ), FCD = 41.1 kip(C ) B

FBD 17

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FCD FCE

10 kip

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Problem 6.40 For the Howe bridge truss, use the method of sections to determine the axial forces in members DF, DG, and EG.

Solution:

Same truss as 6.39. D

FDF

8 B

H C

E 10 kip 17 ft

17 ft

G 30 kip 17 ft

FDG

8 ft FEG

20 kip 17 ft 30 kip

10 kip

Cut through DF , DG, and EG and use left section ΣM D : − A(34 ft) + (10 kip)(17 ft) + FEG (8 ft) = 0 ⇒ FEG = 95.6 kip ΣM G : − A(51 ft) + (10 kip)(34 ft) + (30 kip)(17 ft) − FDF (8 ft) = 0 ⇒ FDF = −69.1 kip 8 FDG = 0 ⇒ FDG = −29.4 kip 353

ΣFy : A − 10 kip − 30 kip − In summary

FEG = 95.6 kip(T ), FDF = 69.1 kip(C ), FDG = 29.4 kip(C )

Problem 6.41 The Pratt bridge truss supports five forces F = 340 kN. The dimension L = 8 m. Use the method of sections to determine the axial force in member JK.

Note the symmetry: Method of sections to find axial force in member JK. C

u L

L B

A L

CK J

AY F

A I

θ = 45 ° L = 8M

F = 340 kN A y = 850 kN

Solution:

First determine the external support forces.

ΣFx : CD + JK + CK cos θ = 0 ΣFy : A y − 2 F − CK sin θ = 0

a +ΣM C : L ( JK ) + L ( F ) − 2 L ( A y ) = 0

L AY

L F

Solving, JK = 1360 kN (T )

F 5 340 kN, L5 8 M

Also, CK = 240.4 kN (T ) CD = −1530 kN (C )

ΣFx : A x = 0 ΣFy : A y − 5F + H y = 0

a +ΣM A : 6 LH y − LF − 2 LF − 3 LF − 4 LF − 5 LF = 0

Solving: A x = 0, A y = 850 kN H y = 850 kN

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Problem 6.42 For the Pratt bridge truss in Problem 6.41, use the method of sections to determine the axial force in member EK.

Solution:

From the solution to Problem 6.41, the support forces are A x = 0, A y = H y = 850 kN.

Method of Sections to find axial force in EK. E

u L

L B

L G

A I F

L F

ΣFx : −DE − EK cos θ − KL = 0

ΣFy : H y − 2 F − EK sin θ = 0 ΣM E : −( L )( KL ) − ( L )( F ) + (2 L ) H y = 0 Solution: EK = 240.4 kN (T ) Also, KL = 1360 kN (T ) DE = −1530 kN (C )

Problem 6.43 The walkway exerts vertical 50-kN loads on the Warren truss at B, D, F , and H. Use the method of sections to determine the axial force in member CE.

Solution:

First, find the external support forces. By symmetry, A y = I y = 100 kN (we solved this problem earlier by the method of joints). y

D BD

6m 2m

50 kN 2m

A B

CD u CE

AY 2 3 θ = 33.69 °

tan θ =

ΣFx : BD + CD cos θ + CE = 0 ΣFy : A y − 50 + CD sin θ = 0 ΣM C : −6 A y + 3(50) − 2 BD = 0 Solving: CE = 300 kN (T ) Also, BD = −225 kN (C ) CD = −90.1 kN (C )

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Problem 6.44 Use the method of sections to determine the axial forces in members BD, BE, and CE. B

From the equilibrium equations ΣFx = A x = 0, ΣFy = A y + F − 400 lb − 600 lb = 0,

ΣM point A = (9 ft) F − (3 ft)(400 lb) − (6 ft)(600 lb) = 0, 3 ft

the reactions are A x = 0, A y = 467 lb, F = 533 lb. We obtain a section by passing a plane through members BD, BE , and CE: y

A C

400 lb 3 ft

600 lb

TBE 3 ft

3 ft

Solution:

TBD

458

The free-body diagram of the entire truss is

TCE 400 lb

From the equilibrium equations

ΣFx = TBD + TBE cos 45 ° + TCE = 0, ΣFy = −TBE sin 45 ° + A y − 400 lb = 0, ΣM point B = (3 ft)TCE − (3 ft) A y = 0,

400 lb

x 600 lb

we obtain TBD = −533 lb, TBE = 94.3 lb, TCE = 467 lb. BD: 533 lb (C); BE: 94.3 lb (T); CE: 467 lb (T).

Problem 6.45 Use the method of sections to determine the axial forces in members FH, GH, and GI.

Solution:

The free-body diagram of the entire truss is

x 300 mm

G H

Ax Ay

4 kN

6 kN

300 mm From the equilibrium equations

A B 400 mm

D 400 mm

6 kN 400 mm

4 kN 400 mm

ΣFx = A x − N = 0, ΣFy = A y − 6 kN − 4 kN = 0, ΣM point A = (0.6 m) N − (0.8 m)(6 kN) − (1.2 m)(4 kN) = 0, the reactions are A x = 16 kN, A y = 10 kN, N = 16 kN. We obtain a section by cutting members FH , GH , and GI:

TGI

y x

TGH

TFH The angle α = arctan(3/4) = 37.0 °. From the equilibrium equations ΣFx = −N − TFH cos α − TGH − TGI cos α = 0, ΣFy = −TFH sin α − TGI sin α = 0, ΣM point H = (0.3 m) N + (0.3 m)TGI cos α = 0, we obtain TFH = 20 kN, TGH = −16 kN, TGI = −20 kN. FH : 20 kN (T); GH : 16 kN (C); GI : 20 kN (C).

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Problem 6.46 Use the method of sections to determine the axial forces in members BD, CD, and CE.

From the equilibrium equations ΣFx = A x − N = 0, ΣFy = A y − 6 kN − 4 kN = 0,

ΣM point A = (0.6 m) N − (0.8 m)(6 kN) − (1.2 m)(4 kN) = 0, E

300 mm

the reactions are A x = 16 kN, A y = 10 kN, N = 16 kN. We obtain a section by cutting members BD, CD, and CE:

300 mm

TCE

a B

400 mm

Solution:

400 mm

6 kN

4 kN

400 mm

TBD

The angle α = arctan(3/4) = 37.0 °. From the equilibrium equations

The free-body diagram of the entire truss is

TCD

ΣFx = A x + TBD + TCD cos α + TCE = 0, ΣFy = A y − TCD sin α = 0, ΣM point C = (0.3 m) A x + (0.3 m)TBD − (0.4 m) A y = 0,

we obtain TBD = −2.67 kN, TCD = 16.7 kN, TCE = −26.7 kN.

Ax Ay

BD: 2.67 kN (C); CD: 16.7 kN (T); CE: 26.7 kN (C).

4 kN

6 kN

Problem 6.47 The loads F1 = 2 kip, F2 = 4 kip, and F3 = 6 kip. Use the method of sections to determine the axial forces in members AC , BC , and BD.

From the equilibrium equations ΣFx = B x − F1 − F2 − F3 = 0, ΣFy = A + B y = 0, ΣM point B = −(18 in) A + (24 in) F1 + (16 in) F2 + (8 in) F3 = 0,

the reactions are A = 8.89 kip, B x = 12 kip, B y = −8.89 kip. We obtain a section by passing a plane through members AC , BC , and BD:

8 in G

TAC

F2 y

8 in

a D

TBC b

x A

8 in A

TBD

The angles α = arctan(8 in/6 in) = 53.1 ° and β = arctan(8 in/12 in) = 33.7 °. From the equilibrium equations ΣFx = T AC cos α − TBC cos β + B x = 0,

18 in

Solution:

The free-body diagram of the entire truss is F1

y F2

ΣFy = T AC sin α + TBC sin β + TBD + A + B y = 0, ΣM point B = −(18 in)T AC sin α − (18 in) A = 0, we obtain T AC = −11.1 kip, TBC = 6.41 kip, TBD = 5.33 kip. AC: 11.1 kip (C); BC: 6.41 kip (T); BD: 5.33 kip (T).

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Problem 6.48 The loads F1 = 2 kip, F2 = 4 kip, and F3 = 6 kip. Use the method of sections to determine the axial forces in members CE, DE , and DG. H

Solution:

The free-body diagram of the entire truss is F1

8 in E

8 in D

From the equilibrium equations F3

ΣFx = B x − F1 − F2 − F3 = 0, ΣFy = A + B y = 0,

8 in

ΣM point B = −(18 in) A + (24 in) F1 + (16 in) F2 + (8 in) F3 = 0,

the reactions are A = 8.89 kip, B x = 12 kip, B y = −8.89 kip. We obtain a section by passing a plane through members CE , DE , and DG:

18 in

H y

E x TCE

a TDE

F2 TDG

The angle α = arctan(8 in/6 in) = 53.1 °. From the equilibrium equations ΣFx = −TCE cos α + T DE cos α − F1 − F2 = 0, ΣFy = −TCE sin α − T DE sin α − T DG = 0, ΣM point E = −(6 in)T DG + (8 in) F1 = 0, we obtain TCE = −6.67 kip, T DE = 3.33 kip, T DG = 2.67 kip. CE: 6.67 kip (C); DE: 3.33 kip (T); DG: 2.67 kip lb (T).

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Problem 6.49 Use the method of sections to determine the axial forces in members CE, DE, and DF.

E 4 ft G

D 12 kip

F 4 ft H

A 4 ft

4 ft

Solution:

The free-body diagrams for the entire structure and the section to the right of the cut are shown. From the entire structure: ΣM A : −(12 kip) (4 ft) H (12 ft) = 0 ⇒ H = 4 kip

Using the section to the right of the cut we have ΣM E : H (4 ft) − T DF (4 ft) = 0 ΣM D : H (8 ft) + TCE (4 ft) = 0 ΣFy : H − T DE sin 45 ° = 0 Solving yields T DF = 4 kip, TCE = −8 kip, T DE = 5.66 kip Thus we have DF : 4 kip (T) CE : 8 kip (C) DE : 5.66 kip (T)

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Problem 6.50 For the bridge truss shown, use the method of sections to determine the axial forces in members CE, CF , and DF. 200 kN B

200 kN

Solution:

Because of the symmetry of the truss and loading, we can see that the vertical reactions at A and I are each equal to 1 5(200 kN) = 500 kN. 2 Also, because of the roller support there is no horizontal reaction at I, and because there are no horizontal loads the horizontal reaction at A is also zero. We obtain a section by cutting members CE , CF , and DF:

200 kN

TDF

2.5 m

TCF

b 5m

a TCE

C 500 kN

The angles α = arctan(1.5 m/5 m) = 16.7 °, β = arctan(4.5 m/5 m) = 42.0 ° From the equilibrium equations ΣFx = TCE cos α + TCF cos β + T DF = 0, ΣFy = TCE sin α + TCF sin β + 100 kN = 0, ΣM point C = (5 m)(200 kN) − (5 m)(500 kN) − (4.5 m)T DF = 0, we obtain TCE = 696 kN, TCF = −448 kN, T DF = −333 kN. CE: 696 kN (T); CF : 448 kN (C); DF: 333 kN (C).

Problem 6.51 The load F = 20 kN and the dimension L = 2 m. Use the method of sections to determine the axial force in member HK. Strategy: Obtain a section by cutting members HK , HI , IJ , and JM. You can determine the axial forces in members HK and JM even though the resulting freebody diagram is statically indeterminate.

from which K X = −2 F = −40 kN. The section as a free body: Denote the axial force in a member joining I , K by IK. The sum of the forces: ∑ Fx = K x − HI + IJ = 0, from which HI − IJ = K x . Sum moments about K to get M K = M ( L )(2) + JM ( L )(2) − IJ ( L ) + HI ( L ) = 0. Substitute HI − IJ = K x , to obtain JM = −M −

L F

L B

Kx = −30 kN (C ). 2

∑ Fy = K y + M + JM + HK = 0,

from which HK = −JM = 30 kN(T ) D

L I H

J 2L

L K

M KX

Solution: The complete structure as a free body: The sum of the moments about K is M K = −FL (2 + 3) + ML (2) = 0, from which 5F M = = 50 kN. The sum of forces: 2 ΣFY = K Y + M = 0, from which K Y = −M = −50 kN.

KY HI

HK KX

IJ JM

KY 2L

ΣFX = K X + 2 F = 0,

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Problem 6.52 The weight of the bucket is W = 1000 lb. The cable passes over pulleys at A and D. (a) Determine the axial forces in members FG and HI. (b) By drawing free-body diagrams of sections, explain why the axial forces in members FG and HI are equal.

W 3.25 ft

3 ft

3.5 ft

C W F H 3 ft L

3 ft 3 in

3 ft 6 in HI

G I 358

3 ft

from which the two simultaneous equations: −0.9848 HI − 0.5736 JH = 257.22,

Solution:

The truss is at angle α = 35 ° relative to the horizontal. The angles of the members FG and HI relative to the horizontal are β = 45 ° + 35 ° = 80 °. (a) Make the cut through FH , FG, and EG, and consider the upper section. Denote the axial force in a member joining, α, β by αβ . The section as a free body: The perpendicular distance from point F is L FW = 3 2 sin β + 3.5 = 7.678 ft. The sum of the moments about F is M F = −WL FW + W (3.25) − EG (3) = 0, from which EG = −1476.1 lb (C ). The sum of the forces: ΣFY = −FG sin β − FH sin α − EG sin α − W sin α − W = 0, ΣFX = −FG cos β − FH cos α − EG cos α − W cos α = 0,

and −0.1736 HI − 0.8192 JH = −1060.8. Slove: HI = −1158.5 lb(C ) , and JH = 1540.6 lb(T ) . (b) Choose a coordinate system with the y-axis parallel to JH. Isolate a section by making cuts through FH , FG, and EG, and through HJ , HI , and GI. The free section of the truss is shown. The sum of the forces in the x- and y-direction are each zero; since the only external x-components of axial force are those contributed by FG and HI, the two axial forces must be equal: ΣFx = HI cos 45 ° − FG cos 45 ° = 0, from which HI = FG

from which the two simultaneous equations: −0.9848FG − 0.5736 FH = 726.9, and −0.1736 FG − 0.8192 FH = −389.97. Solve: FG = −1158.5 lb (C ) , and FH = 721.64 lb (T ). Make the cut through JH , HI , and GI, and consider the upper section. The section as a free body: The perpendicular distance from point H to the line of action of the weight is L HW = 3cos α + 3 2 sin β + 3.5 = 10.135 ft. The sum of the moments about H is M H = −W ( L ) − GI (3) + W (3.25) = 0, from which GI = −2295 lb (C ). ∑ FY = −HI sin β − JH sin α − GI sin α − W sin α − W = 0, ∑ FX = −HI cos β − JH cos α − GI cos α − W cos α = 0,

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Problem 6.53 The weight of the bucket is W = 1000 lb. The cable passes over pulleys at A and D. Determine the axial forces in members IK and JL. D

Solution: W

3 ft

3 ft 3 in

JK IK

Make a cut through JL, JK , and IK, and consider the upper section. Denote the axial force in a member joining, α, β by αβ . The section as a free body: The perpendicular distance from point J to the line of action of the weight is L = 6 cos α + 3 2 sin β + 3.5 = 12.593 ft. The sum of the moments about J is M J = −W ( L ) + W (3.25) − IK (3) = 0, from which IK = −3114.4 lb(C ).

3 ft 6 in

E G I 358

3 ft

3.5 ft

3.25 ft

3 ft L

The sum of the forces: ΣFx = JL cos α − IK cos α − W cos α − JK cos β = 0, and ΣFy = −JL sin α − IK sin α − W sin α − W − JK sin β = 0, from which two simultaneous equations: 0.8192 JL + 0.1736 JK = −1732 and 0.5736 JL + 0.9848JK = 212.75. Solve: JL = 2360 lb(T ) , and JK = −1158.5 lb(C ) .

Problem 6.54 The truss supports loads at N , P, and R. Determine the axial forces in members LN , LO, and MO. 2m

2m K

2m A

M L

2m O N

TLO

TLN

Q P

We obtain a section by cutting members LN , LO, and MO: TMO

x R

1 kN 2 kN 1 kN

a R N

1 kN

2 kN

1 kN

The angle α = arctan(1 m/2 m) = 26.6 °. From the equilibrium equations ΣFx = −T LN − T LO cos α − T MO = 0,

Solution:

ΣFy = −T LO sin α − 4 kN = 0,

ΣM point O = −(1 m)T LN − (2 m)(2 kN) − (4 m)(1 kN) = 0,

we obtain T LN = −8 kN, T LO = −8.94 kN, T MO = 16 kN. LN : 8 kN (C); LO: 8.94 kN (C); MO: 16 kN (T).

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Problem 6.55 Determine the axial forces in members HJ and GI.

Next, we obtain a section by cutting members GI , HI , and HJ: THJ

2m A

THI

y x

1 kN 2 kN 1 kN

B From this diagram we can see that summing moments about B implies that T HI = 0.

Solution:

TGI

The angle α = arctan(2 m/8 m) = 14.0 °. From the equilibrium equations

First we consider the free-body diagram of the entire

truss:

ΣFx = B x − T HJ sin α = 0, ΣFy = B y + T HJ cos α + TGI = 0, we obtain TGI = −16 kN, T HJ = 8.25 kN.

1 kN

TAJ b

1 kN

GI : 16 kN (C); HJ : 8.25 kN (T).

2 kN

y x

Bx By

The angle β = arctan(4 m/8 m) = 26.6 °. From the equilibrium equations ΣFx = −T AJ sin β + B x = 0, ΣFy = −T AJ cos β + B y − 4 kN = 0, ΣM point J = (8 m) B x + (2 m) B y − (6 m)(1 kN) − (8 m)(2 kN) − (10 m)(1 kN) = 0, we obtain T AJ = 4.47 kN, B x = 2 kN, B y = 8 kN.

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Problem 6.56 By drawing the free-body diagram of a section, explain why the axial force in member EF is zero.

Solution:

We obtain a section by cutting members EF , DF , and EG: TEG TDF

2m A

TEF

1 kN 2 kN 1 kN By

Observe that only the axial force TEF exerts a moment about point B. Therefore TEF = 0.

Cut members EF , DF , and EG.

Problem 6.57 In Example 6.5, draw the free-body diagram of joint A and use it to determine the axial forces in members AB, AC , and AD.

Solution:

The position vectors from point A to points B, C , and D are

r AB = −5i − 3 j − 2k ft, r AC = i − 3 j + 4 k ft, r AD = 5i − 3 j − 2k ft. Dividing each vector by its magnitude, we obtain unit vectors that point from point A toward points B, C , and D : e AB = −0.811i − 0.487 j − 0.324 k, e AC = 0.196 i − 0.588 j + 0.784 k, e AD = 0.811i − 0.487 j − 0.324 k. Let T AB , T AC , and T AD denote the tensile axial loads in the three bars: y 1200 lb TAB

TAD x TAC

z The equilibrium equation for the joint A is T AB e AB + T AC e AC + T AD e AD − (1200 lb) j = 0: T AB (−0.811i − 0.487 j − 0.324 k) + T AC (0.196 i − 0.588 j + 0.784 k) + T AD (0.811i − 0.487 j − 0.324 k) − (1200 lb) j = 0 . This yields the three equations −0.811T AB + 0.196T AC + 0.811T AD = 0, −0.487T AB − 0.588T AC − 0.487T AD = 1200 lb, −0.324T AB + 0.784T AC − 0.324T AD = 0. Solving, we obtain T AB = −904 lb, T AC = −680 lb, T AD = −740 lb. AB: 904 lb (C); AC: 680 lb (C); AD: 740 lb (C).

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Problem 6.58 The space truss supports a vertical 10-kN load at D. The reactions at the supports at joints A, B, and C are shown. What are the axial forces in members AD, BD, and CD? 10 kN

The unit vectors parallel to the members from D are:

D (4, 3, 1) m

r DA = −0.7845i − 0.5883 j − 0.1961k r DA r e DB = DB = 0.2673i − 0.8018 j + 0.5345k r DB r e DC = DC = 0.5345i − 0.8018 j − 0.2673k r DC e DA =

Ay Cy C (6, 0, 0) m

rDA = −4 i − 3 j − k, rDB = i − 3 j + 2k, rDC = 2 i − 3 j − k.

Ax A

Solution: Consider the joint D only. The position vectors parallel to the members from D are

Cz B (5, 0, 3) m

The equilibrium conditions for the joint D are ΣF = T DA e DA + T DB e DB + T DC e DC − FD = 0, from which ΣFx = −0.7845T DA + 0.2673T DB + 0.5345T DC = 0 ΣFy = −0.5883T DA − 0.8018T DB − 0.8108T DC − 10 = 0 ΣFz = −0.1961T DA + 0.5345T DB − 0.2673T DC = 0. Solve: T DA = −4.721 kN (C ) , T DB = −4.157 kN (C ) T DC = −4.850 kN (C )

10 kN

TDA

TDC TDB

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Problem 6.59 The reactions at the supports at joints A, B, and C are shown. What are the axial forces in members AB, AC , and AD?

TAD

Ax TAC

10 kN

TAB

D (4, 3, 1) m

Ax A

These equations for the forces and moments are to be solved for the unknown reactions. The Solution:

Cy C (6, 0, 0) m

Az By

Cz B (5, 0, 3) m

A x = C z = 0, A y = 2.778 kN, B y = 3.333 kN, and C y = 3.889 kN

Solution: The reactions at A are required for a determination of the equilibrium conditions at A. The complete structure as a free body: The position vectors are r AB = 5i + 3k, r AC = 6 i, r AD = 4 i + 3 j + k. The sum of the forces: ∑ Fx = A x = 0,

The method of joints: Joint A: The position vectors are given above. The unit vectors are: e AB = 0.8575i + 0.5145k, e AC = i, e AD = 0.7845i + 0.5883 j + 0.1961k. The equilibrium conditions are: ΣF = T AB e AB + T AC + e AC + T AD e AD + A = 0,

∑ Fy = A y + C y + B y − 10 = 0,

from which

and ∑ Fz = Az + C z = 0.

ΣFx = 0.8575T AB + T AC + 0.7845T AD = 0

The moments due to the reactions:

ΣFy = 0T AB + 0T AC + 0.5883T AD + 2.778 = 0

M = r AB × FB + r AC × FC + r AD × FD = 0

ΣFz = 0.5145 T AB + 0 T AC + 0.1961 T AD = 0.

M =

i j 5 0 0 By

k 3 0

i j 6 0 0 Cy

k 0 Cz

i j k 4 3 1 0 −10 0

= 0

Solve: T AB = 1.8 kN (T ) , T AC = 2.16 kN (T ) T AD = −4.72 kN (C )

= (−3B y + 10) i − (6C z ) j + (5B y + 6C y − 40)k = 0.

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Problem 6.60 The ( x, y , z ) coordinates (in meters) of joints B, C , and D of the space truss are (0.6, 0, 0.5), (0.8, 0, 0), and (0.5, 0.4, 0.3), respectively. The truss is supported at joints A, B, and C. The reactions at the supports are shown. The truss is loaded at joint D by the force F = −2 i − 12 j − 3k (kN). Determine the axial forces in members AB, AC , and AD .

Solving the six scalar equilibrium equations, we obtain A x = 2.00 kN, A y = 4.30 kN, Az = 1.88 kN, B y = 4.80 kN, C y = 2.90 kN, C z = 1.13 kN. Dividing the position vectors by their magnitudes, we obtain unit vectors that point from point A toward points B, C , and D : e AB = 0.768i + 0.640 k, e AC = i, e AD = 0.707 i + 0.566 j + 0.707k. Let T AB , T AC , and T AD denote the tensile axial loads in the three bars:

F D

Ax A

TAD

Ay Cy x

Cz By

TAC

Az B

x Az

TAB

The equilibrium equation for the joint A is T AB e AB + T AC e AC + T AD e AD + A x i + A y j + Az k = 0.

Solution:

The position vectors from point A to points B, C , and D

Solving the three scalar equations yielded by this equation, we obtain T AB = 2.11 kN, T AC = 1.76 kN, T AD = −7.60 kN.

are r AB = 0.6 i + 0.5k (m),

AB: 2.11 kN (T); AC: 1.76 kN (T); AD: 7.60 kN (C).

r AC = 0.8i (m), r AD = 0.5i + 0.4 j + 0.3k (m). The equilibrium equations for the entire truss are ΣFx = A x − 2 kN = 0, ΣFy = A y + B y + C y − 12 kN = 0, ΣFz = Az + C z − 3 kN = 0, ΣM point A = r AB × B y j + r AC × (C y j + C z k) + r AD × F i = 0.6 m 0

j 0 By

0.5 m + 0.8 m 0

0 +

Cy Cz

0.5 m

0.4 m

0.3 m

= 0.

−2 kN −12 kN − 3 kN

The moment equation yields the three scalar equations −(0.5 m) B y + (0.4 m)(−3 kN) − (0.3 m)(−12 kN) = 0, −(0.8 m)C z + (0.3 m)(−2 kN) − (0.5 m)(−3 kN) = 0, (0.6 m) B y + (0.8 m)C y + (0.5 m)(−12 kN) − (0.4 m)(−2 kN) = 0.

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Problem 6.61 The (x, y, z) coordinates (in feet) of joints B, C , and D of the space truss are (5, 0, 4), (6, 0, 0), and (4, 4, 2), respectively. The truss is supported at joints A, B, and C. The reactions at the supports are shown. The truss is loaded at joint D by the force F = −50 i − 320 j − 40 k (lb). Determine the axial forces in members AC , BC, and CD.

The moment equation yields the three scalar equations −(4 ft) B y + (4 ft)(−40 lb) − (2 ft)(−320 lb) = 0, −(−6 ft) Az + (2 ft)(−50 lb) − (−2 ft)(−40 lb) = 0, (−6 ft) A y + (−1 ft) B y + (−2 ft)(−320 lb) − (4 ft)(−50 lb) = 0. Solving the six scalar equilibrium equations, we obtain A x = 50 lb, A y = 120 lb, Az = 30 lb, B y = 120 lb, C y = 80 lb, C z = 10 lb. Dividing the position vectors by their magnitudes, we obtain unit vectors that point from point C toward points A, B, and D:

e CA = −i,

e CB = −0.243i + 0.970 k,

Ax A

e CD = −0.408i + 0.817 j + 0.408k. Let T AC , TBC , and TCD denote the tensile axial loads in the three bars:

Ay Cy

Cz By

TDC TAC

Solution:

The position vectors from point C to points A, B, and

D are

Cy Cz

rCA = −6 i (ft),

The equilibrium equation for the joint C is

rCB = − i + 4 k (ft),

T AC e CA + TBC e CB + TCD e CD + C y j + C z k = 0.

rCD = −2 i + 4 j + 2k (ft).

TBC

Solving the three scalar equations yielded by this equation, we obtain T AC = 32.5 lb, TBC = 30.9 lb, TCD = −98.0 lb.

The equilibrium equations for the entire truss are

AC: 32.5 lb (T); BC: 30.9 lb (T); CD: 98.0 lb (C).

ΣFx = A x − 50 lb = 0, ΣFy = A y + B y + C y − 320 lb = 0, ΣFz = Az + C z − 40 lb = 0, ΣM point C = rCA × ( A x i + A y j + Az k) + rCB × B y j + rCD × F i

402

= −6 ft

0 + −1 ft

4 + −2 ft

By 0

4 ft

2 ft

= 0.

−50 lb − 320 lb −40 lb

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Problem 6.62 The space truss has roller supports at B, C , and D and supports a vertical 800-lb load at A. What are the axial forces in members AB, AC, and AD? y 800 lb A (4, 3, 4) ft B

D (6, 0, 0) ft x

Joint A: The unit vectors parallel to members AB, AC , and AD are e AB =

r AB = −0.6247 i − 0.4685 j − 0.6247k, r AB

e AC =

r AC = 0.2673i − 0.8018 j − 0.5345k, r AC

and e AD =

r AD = 0.3714 i − 0.5570 j − 0.7428k. r AD

The equilibrium conditions at point A: ΣFx = −0.6247T AB + 0.2673T AC + 0.3714T AD = 0

C (5, 0, 6) ft

ΣFy = −0.4685T AB − 0.8018T AB − 0.5570T AD − 800 = 0 ΣFz = −0.6247T AB + 0.5345T AC − 0.7428T AD = 0.

Solution:

800 lb

The position vectors of the points A, B, C , and D are

r A = 4 i + 3 j + 4k , rC = 5i + 6k , rD = 6i.

TAB

TAD

The position vectors from joint A to the vertices are: TAC

r AB = rB − r A = −4 i − 3 j − 4k , r AC = rC − r A = 1i − 3 j + 2k , r AD = rD − r A = 2i − 3 j − 4k

Solve: and

T AB = −379.4 lb (C ) , T AC = −665.2 lb (C ) , T AD = −159.6 lb (C )

Problem 6.63 The space truss shown models an airplane’s landing gear. It has ball-and-socket supports at C , D , and E. If the force exerted at A by the wheel is F = 40 j (kN), what are the axial forces in members AB, AC, and AD?

y E (0, 0.8, 0) m

D 0.4 m

Solution:

The important points in this problem are A(1.1, −0.4, 0), B(1, 0, 0), C (0, 0, 6), and D(0, 0, −0.4). We do not need point E as all of the needed unknowns converge at A and none involve the location of point E. The unit vectors along AB, AC , and AD are u AB = −0.243i + 0.970 j + 0 k,

B x (1, 0, 0) m

0.6 m

A (1.1, 20.4, 0) m

C z

u AC = −0.836 i + 0.304 j + 0.456k, and u AD = −0.889 i + 0.323 j − 0.323k. The forces can be written as

y E (0, 0.8, 0) m

TRS = TRS u RS = TRSX i + TRSY j + TRSZ k, where RS takes on the values AB, AC , and AD. We now have three forces written in terms of unknown magnitudes and known directions. The equations of equilibrium for point A are ΣFx = T ABu ABX + T AC u ACX + T AD u ADX + FX = 0, ΣFy = T ABu ABY + T AC u ACY + T AD u ADY + FY = 0, and ΣFz = T ABu ABZ + T AC u ACZ + T AD u ADZ + FZ = 0, where F = FX i + FY j + FZ k = 40 j kN. Solving these equations for the three unknowns, we obtain T AB = −45.4 kN (compression), T AC = 5.26 kN (tension), and T AD = 7.42 kN (tension).

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0.4 m 0.6 m

TAD C

TAC

B x (1, 0, 0) m TAB F

A (1.1,20.4, 0) m

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Problem 6.64 The truss is loaded at joint E by the force F = 8i − 4 j + 6k kN. The reactions due to the supports at joints B, C , and D are shown. Determine the axial forces in members AE , BE, CE , and DE. Strategy: Begin by considering what the free-body diagram of joint D tells you about the axial force in member DE.

Solution: On the free-body diagram of joint D, the only force in the x direction is the axial force in member DE. Therefore it must be zero. In analyzing joint E, we can ignore member DE. The position vectors from point E to points A, B, and C are rEA = −i + 0.8 j − 1.2k (m), rEB = −i − 1.2k (m), rEC = −1.2k (m). Dividing these position vectors by their magnitudes, we obtain unit vectors that point from point A toward points B, C , and D:

e EA = −0.570 i + 0.456 j − 0.684 k,

e EB = −0.640 i − 0.768k, e EC = −k. Let T AE , TBE , and TCE denote the tensile axial loads in the three bars:

0.8 m Bz

Bx Cy

1.2 m

TAE

D z

Dy 1.0 m

TCE

TBE

F z

The equilibrium equation for the joint E is T AE e EA + TBE e EB + TCE e EC + Fx i + Fy j + Fz k = 0. Solving the three scalar equations yielded by this equation, we obtain T AE = 8.78 kN, TBE = −4.69 kN, TCE = 3.60 kN. AE: 8.78 kN (T); BE: 4.69 kN (C); CE: 3.60 kN (T) DE: Zero.

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Problem 6.65 The truss is loaded at joint E by the force F = 8i − 4 j + 6k kN. The reactions due to the supports at joints B, C , and D are shown. Determine |the axial forces in members AB, AC , AD , and AE. Strategy: Begin by determining the reaction D y and using it to obtain the axial force in member AD.

Solution: Considering the free-body diagram of the entire truss, the sum of the moments about the x-axis is ΣM x -axis = −(1.2 m) D y − (1.2 m) Fy = 0, so D y = −Fy = 4 kN. The angle between members AD and BD is α = arctan(0.8 m/1.2 m) = 33.7 °. Let T AD denote the tensile axial force in member AD. For joint D we have ΣFy = D y + T AD sin α,

which yields T AD = −7.21 kN.

Now we can analyze joint A. The position vectors from point A to points B, C , D and E are r AB = −0.8 j (m), r AC = i − 0.8 j (m),

0.8 m Bz

Bx Cy

1.2 m

r AE = i − 0.8 j + 1.2k (m). Dividing these position vectors by their magnitudes, we obtain unit vectors that point from point A toward points B, C , D and E: e AB = − j, e AC = 0.781i − 0.625 j,

D z

r AD = −0.8 j + 1.2k (m),

e AD = −0.8 j + 1.2k, F

1.0 m

e AE = 0.570 i − 0.456 j + 0.684 k. Let T AB , T AC , T AD , and T AE denote the tensile axial loads in the four bars:

y A TAC TAD TAB TAE

z The equilibrium equation for the joint A is T AB e AB + T AC e AC + T AD e AD + T AE e AE = 0. (Recall that T AD = −7.21 kN. ) Solving the three scalar equations yielded by this equation, we obtain T AB = 4 kN, T AC = −6.40 kN, T AD = −7.21 kN , T AE = 8.78 kN. AB: 4 kN (T); AC: 6.40 kN (C); AD: 7.21 kN (C) AE: 8.78 kN (T).

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Problem 6.66 The truss is loaded at joint E by the force F = 12 i + 5 j − 5k kip. The reactions due to the supports at joints A, B, and C are shown. Determine the axial forces in members BG, CG, DG, and EG. (It isn’t necessary to determine the support reactions.)

We see that TBE = 5 kip, T DE = 7 kip, TEG = 7.07 kip. Now there will be only three unknown axial forces when we proceed to joint G: y TDG

TEG

TCG

TBG

F x z

8 ft Ax

From Eq. (1), the force exerted on joint G by the tensile load in member EG is TEG (0.707 i − 0.707k). The position vector from G to B is

A Az

6 ft

rGB = 6 i − 8 j − 6k (ft).

Ay By

The force exerted on joint G by the tensile load in member BG is x

TBG

rGB = TBG (0.514 i − 0.686 j − 0.514 k). rGB

The equilibrium equations for joint G are ΣFx = 0.514TBG + 0.707TEG = 0,

Solution:

ΣFy = −0.686TBG − TCG = 0,

We will begin by analyzing joint E, which has only three joined members and a known external force. Then joint G will have only three unknown axial forces.

We see that TBG = −9.72 kip, TCG = 6.67 kip, T DG = 0.

Let TBE , T DE , and TEG denote the tensile axial loads in the joined bars:

BG: 9.72 kip (C), CG: 6.67 kip (T), DG: zero, EG: 7.07 kip (T).

ΣFz = −0.514TBG − T DG − 0.707TEG = 0.

y TDE

F E

TEG

TBE

The position vector from E to G is rEG = −6 i + 6k (ft). The force exerted on joint E by member EG is TEG

rEG = TEG (−0.707 i + 0.707k). (1) rEG

The equilibrium equations for joint E are ΣFx = −T DE − 0.707TEG + 12 kip = 0, ΣFy = −TBE + 5 kip = 0, ΣFz = 0.707TEG − 5 kip = 0.

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Problem 6.67 The truss is loaded at joint E by the force F = 12 i + 5 j − 5k kip. The reactions due to the supports at joints A, B, and C are shown. (a) Determine the reactions A x , A y , Az , B y , B z , and C y . (b) Determine the axial forces in members BC , BD , and BE.

so the axial force in member AB is T AB = − A x = 12 kip. For joint E, we write the equilibrium equation ΣFy = −TBE + 5 kip, and we see that TBE = 5 kip. There are now only three unknown axial forces at joint B. The position vectors from point B to points A, C , D, E, and G are

rBA = −6 i (ft), rBC = −6 i + 6k (ft), rBD = −6 i + 8 j (ft),

rBE = 8 j (ft), rBG = −6 i + 8 j + 6k (ft). E

Dividing each vector by its magnitude, we obtain unit vectors that point from point B toward points A, C , D, E, and G:

e BA = −i (ft), e BC = −0.707 i + 0.707k (ft), e BD = −0.6 i + 0.8 j (ft), e BE = j (ft), e BG = −0.514 i + 0.686 j + 0.514 k (ft). Using these expressions, the equilibrium equations for joint B are 8 ft Ax

−T AB − 0.707TBC − 0.6TBD − 0.514TBG = 0,

0.8TBD + TBE + 0.686TBG + B y = 0,

Solving these three equations, we obtain TBC = 0, TBD = −11.7 kip,

B Cy

0.707TBC + 0.514TBG + B z = 0.

Ay By

6 ft

TBG = −9.72 kip. (a) A x = −12 kip, A y = −9.33 kip, Az = 0, B y = 11 kip,

Solution:

B z = 5 kip, C y = 1.13 kip. (b) BC : zero; BD : 11.7 kip (C); The force equilibrium equations for the entire structure

BE : 5 kip (T).

are ΣFx = A x + 12 kip = 0, ΣFy = A y + B y + C y + 5 kip = 0, ΣFz = Az + B z − 5 kip = 0. The moment equation is ΣM point A =

i j 6 ft 0 0 By

k 0 Bz

i j 0 0 0 Cy

k 6 ft 0

i j k 6 ft 8 ft 0 12 kip 5 kip −5 kip

= 0.

This yields the three equations −(6 ft)C y + (8 ft)(−5 kip) = 0, −(6 ft) B z − (6 ft)(−5 kip) = 0, (6 ft) B y + (6 ft)(5 kip) − (8 ft)(12 kip) = 0. Solving the six equilibrium equations yields A x = −12 kip, A y = −9.33 kip, Az = 0, B y = 11 kip, B z = 5 kip, C z = −6.67 kip. Five members are joined at B, but we can obtain some preliminary information before we begin to analyze it. For joint A, we write the equilibrium equation ΣFx = T AB + A x = 0,

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Problem 6.68* The mirror housing of the telescope is supported by a six-bar space truss. The mass of the housing is 360 kg and its weight acts at G. The distance from the z-axis to points A, B, and C is 0.85 m, and the distance from the z-axis to points D, E , and F is 2.4 m. If the telescope is rotated so that the z-axis is vertical (α = 90 °), what are the axial forces in the members of the truss?

Solution: Let r1 = 0.85 m be the radial distance from the axis to point A, and let r2 = 2.4 m be the radial distance to point D. The x and y coordinates of point A are x A = −r1 cos60 °, y A = r1 sin 60 °. The x and y coordinates of point D are x D = −r2 , y D = 0. The position vector from point A to point D is r AD = ( x D − x A ) i + ( y D − y A ) j + ( z D − z A )k

END VIEW 608

Mirror housing

e AD =

F G

B 608

608

C E

608

When α = 90 °, the telescope points vertically and symmetry comes to the rescue. The axial forces in the six members are equal. Let T AD be the tensile axial load in member AD. The z component of the force exerted on the mirror housing by member AD is T AD e ADz . If m is the mass of the mirror housing, the z component of the force equilibrium equation is

T AD =

D a 4m

408

= e ADx i + e ADy j + e ADz k.

The axial load in each member is

A C

r AD = −0.437 i − 0.163 j − 0.885k r AD

6T AD e ADz − mg = 0.

Dividing this vector by its magnitude yields a unit vector that points from A toward D:

608

D 608

= (−r2 + r1 cos60 °) i + (0 − r1 sin 60 °) j + (−4 m − 0)k.

mg 6e ADz

(360 kg)(9.81 m/s 2 ) 6(−0.885) = −665 N. =

665 N (C) in each member.

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Problem 6.69* If the telescope described in Problem 6.68 is rotated so that the angle α = 20 °, what are the axial forces in the members of the truss?

By dividing each of these vectors by its magnitude, we obtain unit vectors parallel to the members. We will denote the unit vectors by the notation, for example for member AD, e AD = e ADx i + e ADy j + e ADz k.

END VIEW 608 D

608

W = −mg cos α j − mg sin αk.

Let T AD , T AF , etc. denote the tensile axial forces in the members. The force equilibrium equations are

608

B 608

608

C E

Let m denote the mass of the mirror housing. The force exerted on the housing by its weight is

608

T AD e ADx + T AF e AFx + TBD e BDx + TBE e BEx + TCE e CEx + TCF e CFx = 0, T AD e ADy + T AF e AFy + TBD e BDy + TBE e BEy + TCE eCEy + TCF eCFy = mg cos α,

Mirror housing

T AD e ADz + T AF e AFz + TBD e BDz + TBE e BEz + TCE eCEz

+ TCF eCFz = mg sin α.

The position vectors of points A, B, C , and G relative to the origin are

A F

r A = −r1 cos60 °i + r1 sin 60 ° j = x A i + y A j, rB = −r1 cos60°i − r1 sin 60° j = x B i + y B j,

a 4m

rC = r1i = x C i, rG = (1 m)k. The moment equation is

E ΣM origin = T AD

0 + T AF

e AFx

e AFy

e AFz

e ADx e ADy e ADz

+ TBD

Solution: Let r1 = 0.85 m be the radial distance from the z-axis to points A, B, and C, and let r2 = 2.4 m be the radial distance from the z-axis to points D, E, and F. The x and y coordinates of points A, B, and C are x A = −r1 cos60 °,

+ TBE x B

e BEx e BEy e BEz

e BDx e BDy e BDz i

+ TCE x C

+ TCF x C

e CFx e CFy e CFz

e CEx e CEy e CEz

y A = r1 sin 60 °,

x B = −r1 cos60°, y B = −r1 sin 60 °, x C = r1 , y C = 0,

+ 0

and the x and y coordinates of points D, E, and F are x D = −r2 ,

y D = 0,

x E = r2 cos60°,

y E = −r2 sin 60 °,

x F = r2 cos60°,

y F = r2 sin 60 °.

We need position vectors along each of the members. That is, we need position vectors from point A to points D and F, from point B to points D and E, and from point C to points E and F:

= 0.

0 −mg cos α − mg sin α Solving the six equilibrium equations, we obtain T AD = −32.9 N, T AF = 2130 N, TBD = −422 N, TBE = −2580 N, TCE = −2390 N, TCF = 1930 N. AD: 32.9 N (C), AF : 2130 N (T), BD: 422 N (C), BE: 2580 N (C), CE: 2390 N (C), CF : 1930 N (T).

r AD = (−r2 + r1 cos60 °) i + (0 − r1 sin 60 °) j + (−4 m − 0)k, r AF = (r2 cos60 ° + r1 cos60°) i + (r2 sin 60 ° − r1 sin 60 °) j + (−4 m − 0)k, rBD = (−r2 + r1 cos60°) i + (0 + r1 sin 60°) j + (−4 m − 0)k, rBE = (r2 cos60° + r1 cos60°) i + (−r2 sin 60° + r1 sin 60°) j + (−4 m − 0)k, rCE = (r2 cos60° − r1 ) i + (−r2 sin 60° − 0) j + (−4 m − 0)k, rCF = (r2 cos60° − r1 )i + (r2 sin 60° − 0) j + (−4 m − 0)k.

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Problem 6.70 Determine the reactions on member AB at A and at B.

Solution:

We draw the free-body diagrams of the members:

By Bx

Bx 2 kN

6 kN

0.4 m 2 kN

6 kN

Ax 0.4 m

The equilibrium equations for member AB are ΣFx = A x + B x + 6 kN = 0,

ΣFy = A y + B y = 0, ΣM point A = −(0.8 m) B x − (0.4 m)(6 kN) = 0,

0.4 m

and the equilibrium equations for member BC are ΣFx = −B x + C x = 0, ΣFy = −B y + C y − 2 kN = 0, ΣM point C = (0.8 m) B x + (0.8 m) B y + (0.4 m)(2 kN) = 0. Solving,

obtain

A x = −3 kN, A y = −2 kN, B x = −3 kN,

B y = 2 kN. A x = −3 kN, A y = −2 kN, B x = −3 kN, B y = 2 kN.

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Problem 6.71 The object suspended at E weighs 200 lb. Determine the reactions on member ACD at A and C.

ΣFx : A x = 0

ΣFy : A y − 200 lb = 0

3 ft E

B 5 ft

A 4 ft

Solution: We start with the free-body diagram of the entire frame. We have the equilibrium equations:

ΣM A : M A − (200 lb) (6 ft) = 0 Next we use the free-body diagram of the post ACD. Notice that BD is a two-force body and the angle α is α = tan −1 (3/4) = 36.9 ° The equilibrium equations are ΣM C : M A + A x (5 ft) + TBD cos α(3 ft) = 0

6 ft

ΣFx : A x + C x − TBD cos α = 0 ΣFy : A y + C y − TBD sin α = 0 Solving these six equations we find TBD = −500 lb and A x = 0, A y = 200 lb C x = −400 lb, C y = −500 lb M A = 1200 ft-lb

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Problem 6.72 The hydraulic piston BC adjusts the height of the platform on which the 180-kg mass rests. When the system is in the position shown, what are the axial forces in members AB and CD?

Solution:

The weight of the supported mass is W = (180 kg) (9.81 m/s 2 ) = 1.77 kN. Members AB, BC , and CD are two-force members. The free-body diagram of the member that supports the mass is B

a TAB

B 0.2 m

458

TBC

a D

TCD

180 kg

0.4 m

The angle α = arctan(0.2 m/0.8 m) = 14.0 °. From the equilibrium equations

0.2 m

ΣFx = −T AB cos α − TBC sin 45 ° − TCD cos α = 0,

ΣFy = −T AB sin α − TBC cos 45 ° − TCD sin α − W = 0, ΣM point B = −(0.6 m)TCD cos α − (0.45 m)W = 0,

we obtain T AB = 3.79 kN, TBC = −3.33 kN, TCD = −1.37 kN.

0.45 m

0.8 m

AB: 3.79 kN (T); CD: 1.37 kN (C).

Problem 6.73 The frame is loaded by a horizontal 6-kN force. Determine the reactions on member ABC at A, B , and C.

we obtain A x = −6 kN, A y = −7.5 kN, E = 7.5 kN. Now we consider the free-body diagrams of members ABC (Fig. b) and CD (Fig. c):

0.2 m

6 kN

6 kN C

By 0.4 m B

Dx Dy

y Ax

0.4 m A

(b) 0.4 m

(c)

From Fig. (b) we obtain the equilibrium equations

0.4 m

ΣFx = A x + B x + C x = 0,

Solution:

ΣFy = A y + B y + C y = 0,

Consider the free-body diagram of the entire structure

ΣM point B = (0.4 m) A x − (0.4 m)C x = 0.

(Fig. a): y

From Fig. (c) we use the equation

6 kN

ΣM point D = (0.4 m)C x + (0.4 m)C y − (0.6 m)(6 kN) = 0.

Solving these four equations, we obtain B x = 12 kN, B y = −7.5 kN,

C x = −6 kN, C y = 15 kN.

A x = −6 kN, A y = −7.5 kN, B x = 12 kN, B y = −7.5 kN, C x = −6 kN, C y = 15 kN.

E (a)

From the equilibrium equations ΣFx = A x + 6 kN = 0, ΣFy = A y + E = 0, ΣM point A = (0.8 m) E − (1.0 m)(6 kN) = 0,

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Problem 6.74 Determine the reactions on the frame at A and C. 600 in-lb

Solution:

600 in-lb

We draw the free-body diagrams of the members: By Bx

600 in-lb

Bx By

8 in

Cx Cy

The equilibrium equations for member AB are

12 in

ΣFx = A x + B x = 0, ΣFy = A y + B y = 0, ΣM point A = −(8 in) B x + (12 in) B y + 600 in-lb = 0, and the equilibrium equations for member BC are ΣFx = −B x + C x = 0, ΣFy = −B y + C y = 0, ΣM point C = (8 in) B x + (12 in) B y + 600 in-lb = 0. Solving the two moment equations yields B x = 0, B y = −50 lb. Then from the force equations we obtain A x = 0, A y = 50 lb, C x = 0, C y = −50 lb. A x = 0, A y = 50 lb, C x = 0, C y = −50 lb.

Problem 6.75 The tension in cable BD is 500 lb. Determine the reactions at A for cases (1) and (2). G

Case (a) The complete structure as a free body: The sum of the moments about G: ΣM G = −16(300) + 12 A x = 0, from which A x = 400 lb . The sum of the forces:

ΣFx = A x + G x = 0, 6 in

6 in

D A

6 in

C 8 in

300 lb

8 in

(1)

8 in

300 lb

(2)

(b)

Gy Ay

Gx 16 in

300 lb

Ax 8 in

from which A y − 300 − G y . Element GE: The sum of the moments about E: ΣM E = −16G y = 0, from which G y = 0, and from above A y = 300 lb.

8 in

ΣM G = −16(300) + 12 A x = 0, from which A x = 400 lb . Element ABC: The tension at the lower end of the cable is up and to the right, so that the moment exerted by the cable tension about point C is negative. The sum of the moments about C:

Cy Cx 300 lb

BandF_6e_ISM_CH06.indd 413

ΣFy = A y − 300 + G y = 0,

Case (b) The complete structure as a free body: The free body diagram, except for the position of the internal pin, is the same as for case (a). The sum of the moments about G is

Solution: Gx (a) 12 in

from which G x = −400 lb.

ΣM C = −8B sin α − 16 A y = 0, noting that B = 500 lb and α = tan −1 then A y = −150 lb.

( 86 ) = 36.87°,

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Problem 6.76 The mass of the pulley is 5 kg, the mass of the suspended object is 20 kg, and the mass of the horizontal bar is 8 kg. Determine the reactions on the bar at A and B.

0.4 m

Solution:

The weight of the pulley is W P = (5 kg)(9.81 m/s 2 ) = 49.1 N, the weight of the suspended object is WS = (20 kg) (9.81 m/s 2 ) = 196 N, and the weight of the bar is W B = (8 kg) (9.81 m/s 2 ) = 78.5 N. Let’s begin with the free-body diagram of the pulley. The tension in the rope is equal to the weight of the suspended object:

0.4 m

WP WS

1.6 m

Here B x and B y are the forces exerted on the pulley by the bar. From equilibrium we see that B x = −WS = −196 N, B y = −W P − WS = −245 N. The free-body diagram of the bar is y

MA Ax

Bx Ay

From the equilibrium equations ΣFx = A x + B x = 0, ΣFy = A y + B y − W B = 0, ΣM point A = (1.6 m) B y − (0.8 m)W B + M A = 0, we obtain A x = 196 N, A y = 324 N, M A = 455 N-m. A x = 196 N, A y = 324 N, M A = 455 N-m, B x = −196 N, B y = −245 N.

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Problem 6.77 Determine the forces exerted on member ABC at A and C.

Solution:

D 400 lb

2 ft

1 ft

C 100 lb

1 ft

2 ft

We start with the free-body diagram of the entire frame. Two of the equilibrium equations for the whole frame are ΣFx : A x + 100 lb = 0 ΣM E : − A x (2 ft) − A y (4 ft) − (100 lb)(1 ft) − (400 lb)(2 ft) = 0 Next we examine the free-body diagram of bar ABC. Note that BD is a two-force body and that the angle α = 45 °. The equilibrium equations are ΣM C : − A y (4 ft) − TBD sin α(2 ft) − (400 lb)(2 ft) = 0 ΣFx : A x + TBD cos α + C x = 0 ΣFy : A y + TBD sin α + C y − 400 lb = 0 Solving, we find that TBD = −70.7 lb and A x = −100 lb, A y = −175 lb C x = 150 lb, C y = 625 lb

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Problem 6.78 An athlete works out with a squat thrust machine. To rotate the bar ABD, the athlete must exert a vertical force at A that causes the magnitude of the axial force in the two-force member BC to be 1800 N. When the bar ABD is on the verge of rotating, what are the reactions on the vertical bar CDE at D and E? 0.6 m

Solution: Member BC is a two force member. The force in BC is along the line from B to C. C y

FBC

0.6 m tan Θ =

0.6 m

Dy 0.42 m x (FBC 51800 N)

D Dx

0.42 0.6

Θ = 34.990

C A

( FBC = 1800 N) 0.42 θ = 34.99 °. tan θ = 0.6

0.42 m

B D

ΣFx : D x − FBC cos θ = 0 ΣFy : A y − FBC sin θ + D y = 0

a +ΣM D : −1.2 A y + 0.6 FBC sin θ = 0 Solving, we get D x = 1475 N

1.65 m

D y = 516 N A y = 516 N

Problem 6.79 The frame supports a 6-kN vertical load at C. The bars ABC and DEF are horizontal. Determine the reactions on the frame at A and D.

Member DEF

Dx 0.4 m

6 kN

1.0 m

0.4 m

Member ABC: ΣFx : A x + FBE cos θ − FCF cos φ = 0

E F 0.8 m

ΣFy : A y − FBE sin θ − FCF sin φ − 6 = 0

a +ΣM A : −(0.4) FBE sin θ − (1.4) FCF sin φ − 1.4(6) = 0

0.4 m

Member DEF:

Solution:

Note that members BE and CF are two force members. Consider the 6 kN load as being applied to member ABC. Ay

1.0 m

B 0.4 m

FBE u

0.5 θ = 51.34 ° 0.4 0.5 tan φ = φ = 68.20 ° 0.2

6 kN C

ΣFx : D x − FBE cos θ + FCF cos φ = 0 ΣFy : D y + FBE sin θ + FCF sin φ = 0

a +ΣM D : (0.8)( FBE sin θ) + 1.2 FCF sin φ = 0 Unknowns A x , A y , D x , D y , FBE , FCF we have 6 eqns in 6 unknowns.

FCF f

A x = −16.8 kN Solving, we get A y = 11.25 kN D x = 16.3 kN D y = −5.25 kN

tan θ =

416

Equations of equilibrium:

0.5 m

FCF

0.8 m

FBE

Also, FBE = 20.2 kN (T ) FCF = −11.3 kN (C )

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Problem 6.80 The frame is loaded by a vertical 600-lb force. Determine the reactions on the frame at A and D. 8 in

8 in A

Solution: The key to this solution is recognizing that BE and CF are two-force members. We can draw the free-body diagrams of the two horizontal bars like this: y

8 in C

Ax 8 in

F 600 lb

T x

Dx Dy

600 lb

From the two equilibrium equations ΣM point A = −(8 in) P sin 45 ° − (16 in)T sin 45 ° = 0, ΣM point D = (16 in) P sin 45 ° + (24 in)T sin 45 ° − (24 in)(600 lb) = 0, P = −2T , 1800 = −2550 lb, sin 45 ° P = 5090 lb.

T = −

we obtain P = 5090 lb, T = −2550 lb. Then by summing the forces on the two free-body diagrams we see that A x = −P sin 45 ° − T sin 45 ° = −1800 lb, A y = P sin 45 ° + T sin 45 ° = 1800 lb, D x = P sin 45 ° + T sin 45 ° = 1800 lb, D y = −P sin 45 ° − T sin 45 ° + 600 lb = −1200 lb. A x = −1800 lb, A y = 1800 lb, D x = 1800 lb, D y = −1200 lb.

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Problem 6.81 Determine the reactions on member BCD.

30 lb F

G 8 in

E 40 lb

8 in

C 8 in

A 18 in

12 in

8 in

Solution:

We will use frame ADG, bar DFG and bar BCD. The free-body diagrams ares shown. The angle α = tan −1 (18/24) = 36.9 °

From ADG we have ΣM D : B x (24 in) + (40 lb)(8 in) − (30 lb) (20 in) = 0 ΣΜ A : B y (18 in) − (40 lb)(16 in) − (30 lb)(38 in) = 0 ΣFy : By − (30 lb) − T AD cos α = 0 From DFG we have ΣΜ F : ( D y + T AD cos α) (12 in) − (30 lb)(8 in) = 0 And finally from BCD we have ΣFy : B y + C y + D y = 0 ΣM D : B x (24 in) + C x (16 in) = 0 ΣFx : D x + C x + B x = 0 Solving these seven equations, we find T AD = 86.1 lb and B x = 11.7 lb, B y = 98.9 lb C x = −17.5 lb, C y = −50 lb D x = 5.83 lb, D y = −48.9 lb

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Problem 6.82 The weight of the suspended object is W = 50 lb. Determine the tension in the spring and the reactions at F. (The slotted member DE is vertical.) A

1 1 FB

4 in

6 in

W C

10 in F

Fy Finally examine DCE

8 in

10 in

ΣM D : −T (16 in) + FC (10 in) = 0 ⇒ T = 62.5 lb

Solution:

Start with member AB 1 ΣM A : −(50 lb)(8 in) + FB (16 in) = 0 ⇒ FB = 35.4 lb 2 FC

FB 1 1 Ax Ay

50 lb Now examine BCF ΣM F : FB (20 2) in − FC (10 in) = 0 ⇒ FC = 100 lb 1 ΣFx : − FB + FC + Fx = 0 ⇒ Fx = −75 lb 2 1 FB + Fy = 0 ⇒ Fy = 25 lb ΣFy : − 2

BandF_6e_ISM_CH06.indd 419

Dy Summary Tension in Spring = 62.5 lb Fx = 25 lb, Fy = −75 lb

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Problem 6.83 The frame is loaded by a 2-kN force. The dimension h = 0.6 m. The pin at E slides in the smooth vertical slot. Determine the axial load in the twoforce member CD and state whether it is in tension (T) or compression (C).

Solution:

The free-body diagrams of members AC and BDE are

2 kN E

a 2 kN

0.3 m E

Ay C

0.6 m

(a)

0.3 m A

(b)

The angle α = arctan[(h − 0.3 m)/0.6 m] = 26.6 °. From Fig. (a) we write the equilibrium equation

ΣM point A = −hT cos α − (0.9 m) E = 0,

and from Fig. (b) we write the equation ΣM point B = (0.3 m)T cos α + (0.9 m)E − (1.2 m)(2 kN) = 0. Solving these equations, we obtain T = −8.94 kN, E = 5.33 kN.

0.6 m

8.94 kN (C).

Problem 6.84 The frame is loaded by a 400-lb force. Determine the reactions on member ABC at A and B.

Solution:

The free-body diagram of the entire frame is

B Ax

8 in

B 400 lb

8 in

400 lb

E From the equilibrium equations ΣFx = A x + E = 0, ΣFy = A y − 400 lb = 0, ΣM point A = (8 in)E − (20 in)(400 lb) = 0,

we obtain A x = −1000 lb, A y = 400 lb, E = 1000 lb. 12 in

8 in

Member CD is a two-force member. The free-body diagrams of the other two members are D T

458 1000 lb 400 lb

Bx By

458

400 lb 1000 lb

For member ABC we write the equilibrium equations ΣFx = B x − T cos 45 ° − 1000 lb = 0, ΣFy = B y + T sin 45 ° + 400 lb − 400 lb = 0, ΣM point B = (16 in)T sin 45 ° − (12 in)(400 lb) − (8 in)(400 lb) = 0. Solving these equations, we obtain T = 707 lb, B x = 1500 lb, B y = −500 lb. A x = −1000 lb, A y = 400 lb, B x = 1500 lb, B y = −500 lb.

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Problem 6.85 Determine the forces on member ABC.

Solution:

E 6 kN

EY E DY DX D DX D DY BX B

AX A

B 2m

A AY

BY B

6 kN

C CY BX

1m The frame as a whole: The equations of equilibrium are ΣFX = A X + E X = 0, ΣFY = AY + E Y − 6000 N = 0, and, with moments about E, ΣM E = 2 A X − (5)6000 = 0. Solving for the support reactions, we get A X = 15,000 N and E X = −15,000 N. We cannot yet solve for the forces in the y direction at A and E. Member ABC: The equations of equilibrium are ΣFX = A X − B X = 0, ΣFY = AY − BY − C Y = 0, and summing moments about A, ΣM A = −2 BY − 4C Y = 0. Member BDE: The equations of equilibrium are ΣFX = E X − D X + B X = 0, ΣFY = E Y − DY − BY = 0, and, summing moments about E, ΣM E = (1)DY + (1)D X + (2)BY + (2)B X = 0. Member CD: The equations of equilibrium are ΣFX = −D X = 0, ΣFY = −DY + C Y − 6000 = 0, and summing moments about D, ΣM D = −(4)6000 + 3C Y = 0. Solving these equations simultaneously gives values for all of the forces in the frame. The values are A X = 15,000 N, AY = −8,000 N, B X = 15,000 N, BY = −16,000 N, C Y = 8,000 N, D X = 0, and DY = 2,000 N.

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Problem 6.86 The frame supports a 30-lb suspended weight. Determine the reactions on member BCDE at C and E.

Solution: Member AB is a two-force member. The free-body diagram of the entire frame is T

B 6 in

6 in

30 lb

6 in

30 lb

D A

458

From the equilibrium equations

ΣFx = E x − T sin 45 ° = 0, ΣFy = E y − T cos 45 ° − 30 lb = 0,

18 in

6 in

ΣM point E = (18 in)T sin 45 ° − (12 in)(30 lb) = 0, we obtain T = 28.3 lb, E x = 20 lb, E y = 50 lb. Member DF is also a two-force member. The free-body diagram of member BCDE is

B 28.3 lb

Cy 458 P D

20 lb

458 y

50 lb We write the equilibrium equations ΣFx = C x + P cos 45 ° + 20 lb − (28.3 lb)sin 45 ° = 0, ΣFy = C y + P sin 45 ° + 50 lb − (28.3 lb) cos 45 ° = 0, ΣM point D = −(6 in)C x + (6 in)(20 lb) + (12 in)(28.3 lb)sin 45 ° = 0, and solve, obtaining P = 84.9 lb, C x = 60 lb, C y = 30 lb. C x = 60 lb, C y = 30 lb, E x = 20 lb, E y = 50 lb.

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Problem 6.87 Determine the reactions on the horizontal bar at B, C, and D. 400 lb A 12 in 6 in D

B 8 in

8 in 16 in

Solution:

Consider the free-body diagram of the complete

structure:

ΣFy = −200 lb − D y = 0,

400 lb

ΣM point D = (6 in)T − (6 in)(200 lb) = 0.

From the equilibrium equations ΣFx = −T − D x = 0,

200 lb

we obtain T = 200 lb, D x = −200 lb, and D y = −200 lb. Finally, from the free-body diagram of member BCD, y

200 lb we obtain the equations From the equilibrium equations

ΣFx = B x + C x + D x = 0,

ΣFx = A x + B x = 0,

ΣFy = B y + C y + D y = 0,

ΣM point B = −(12 in) A x − (8 in)(400 lb) − (38 in)(200 lb) = 0,

ΣM point B = (16 in)C y + (32 in) D y = 0.

we obtain A x = −900 lb, B x = 900 lb. The free-body diagram of the pulley is y

Using the known values of B x , D x , and D y , these equations yield B y = −200 lb, C x = −700 lb, and C y = 400 lb.

B x = 900 lb, B y = −200 lb, C x = −700 lb, C y = 400 lb, D x = −200 lb, D y = −200 lb.

Dx Dy

200 lb

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Problem 6.88 The weight W = 80 lb. Determine the forces on member ABCD. 5 in

11 in

12 in

Dy Dx

By Cx

3 in B

8 in W F

Solution: The complete structure as a free body: The sum of the moments about A: ΣM A = −31W + 8E x = 0,

from which C x = 310 lb . The sum of the moments about E: ΣM E = 8F − 16C y + 8C x = 0. For frictionless pulleys, F = W, and thus C y = 195 lb . The sum of forces parallel to y: ΣFy = E y − C y + F = 0,

from which E x = 310 lb. The sum of the forces:

from which E y = 115 lb .

ΣFx = E x + A x = 0,

Eq. (1) above is now solvable: A y = −35 lb .

from which A x = −310 lb .

Element ABCD: The forces exerted by the pulleys on element ABCD are, by inspection: B x = W = 80 lb , B y = 80 lb , D x = 80 lb , and

ΣFy = E y + A y − W = 0,

D y = −80 lb , where the negative sign means that the force is reversed

from which (1) E y + A y = W.

from the direction of the arrows shown on the free body diagram.

Element CFE: The sum of the forces parallel to x: ΣFx = E x − C x = 0,

Problem 6.89 The athlete using the exercise machine holds the 80-lb weight stationary in the position shown. There are pinned connections at A, C, and F and a fixed support at E. What are the reactions on member ACD at A and C? (Neglect the size of the pulleys in writing equilibrium equations.)

2 ft 2 in 9 in

1 ft 6 in

2 ft B

D 608

6 ft 80 lb

Solution:

Member CF is a two-force member. We draw the freebody diagram of member ACD: D B

608

y x

Cy 80 lb

80 lb From the equilibrium equations ΣFx = A x + (80 lb) cos60 ° = 0, ΣFy = A y + C y − 80 lb − (80 lb)sin 60° = 0, ΣM point A = (50 in)C y − (26 in)(80 lb) − (9 in)(80 lb) cos60° − (68 in)(80 lb)sin 60 ° = 0, we obtain A x = −40 lb, A y = 6.26 lb, C x = 0, C y = 143 lb. A x = −40 lb, A y = 6.26 lb, C x = 0, C y = 143 lb.

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Problem 6.90 The weight of the 200-lb object effectively acts midway between points B and E. The hydraulic piston DF is pinned at D and F . Determine the reactions on member CDE at C and E.

Solution:

Members AB and DF are two-force members. Let us draw the free-body diagrams of members BE and CDE: 200 lb Ex

B TAB

y TDF x

200 lb

E 8 in

The angle α = arctan(8 in/6 in) = 53.1 °. For member BE we write the equilibrium equations

ΣFx = −E x − T AB cos α = 0, ΣFy = −E y − T AB sin α − 200 lb = 0,

8 in C

ΣM point E = −(12 in)T AB sin α − (6 in)(200 lb) = 0. x

Solving, we obtain T AB = −125 lb, E x = 75 lb, E y = −100 lb. For member CDE we write the equilibrium equations ΣFx = C x + E x + T DF cos α = 0,

12 in

ΣFy = C y + E y − T DF sin α = 0, 6 in

6 in

ΣM point C = −(16 in) E x + (12 in) E y − (8 in)T DF cos α − (6 in)T DF sin α = 0. Solving yields T DF = −250 lb, C x = 75 lb, C y = −100 lb. C x = 75 lb, C y = −100 lb, E x = 75 lb, E y = −100 lb.

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Problem 6.91 The mass of the suspended object is m = 50 kg. Determine the reactions on member ABC. 0.2 m

Solution:

Begin with an examination of the pulley at B.

1 (490.5 N) = 0 ⇒ B x = 347 N 2 1 (490.5 N) = 0 ⇒ B y = −837 N ΣFy : −B y − 490.5 N − 2 ΣFx : −B x +

B 1 By

0.6 m

1 Bx

0.8 m

C 0.2 m

490.5 N

0.6 m

490.5 N

Now examine the entire structure ΣM D : −(490.5 N)(1.6 m) − A x (0.6 m) = 0 ⇒ A x = −1308 N Ay Ax

Dy Dx

490.5 N

Finally look at member ABC ΣM C : − A x (0.6 m) − A y (1.4 m) − B x (0.6 m) − B y (0.6 m) = 0 ⇒ A y = 771 N ΣFx : A x + B x + C x = 0 ⇒ C x = 961 N ΣFy : A y + B y + C y = 0 ⇒ C y = 66.6 N

Ay Ax

Cx In Summary A x = −1308 N, A y = 771 N B x = 347 N, B y = −837 N C x = 961 N, C y = 66.6 N

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Problem 6.92 The unstretched length of the spring is L 0 . Show that when the system is in equilibrium the angle α satisfies the relation sin α = 2( L 0 − 2 F /k )/L.

F 1 L 4 1 L 4 1 L 2

Solution:

Since the action lines of the force F and the reaction E are co-parallel and coincident, the moment on the system is zero, and the system is always in equilibrium, for a non-zero force F. The object is to find an expression for the angle α for any non-zero force F.

F L

The complete structure as a free body: The sum of the moments about A ΣM A = −FL sin α + EL sin α = 0,

a Ax

from which E = F. The sum of forces: By

ΣFx = A x = 0,

ΣFy = A y + E − F = 0, from which A y = 0, which completes a demonstration that F does not exert a moment on the system. The spring C: The elongation of the L spring is ∆s = 2 sin α − L O , from which the force in the spring is 4 L T = k sin α − L O 2

(

from which. A x = 0.

L 4

a L 4

)

Element BE: The strategy is to determine C y , which is the spring force on BE. The moment about E is L L L ΣM E = − C y cos α − B y cos α − B x cos α = 0, 4 2 2 Cy from which + B y = −B x . The sum of forces: 2 ΣFx = B x = 0, from which B x = 0. ΣFy = C y + B y + E = 0, from which C y + B y = −E = −F. The two simultaneous equations are solved: C y = −2 F , and B y = F.

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Problem 6.93 The pin support B will safely support a force of 24-kN magnitude. Based on this criterion, what is the largest mass m that the frame will safely support?

Solution: Cx

By Bx

500 mm

100 mm

Ay E

Ey Ex Ey

Ex F

The weight is given by W = mg = 9.81 g The complete structure as a free body:

300 mm m

Sum the forces in the x-direction: ΣFx = A x = 0, from which A x = 0

300 mm

400 mm

Element ABC: The sum of the moments about A: ΣM A = +0.3B x + 0.9C x − 0.4W = 0, from which (1) 0.3B x + 0.9C x = 0.4W . The sum of the forces: ΣFx = −B x − C x + W + A x = 0, from which (2) B x + C x = W. Solve the simultaneous Eqs. (1) and 5 W 6 Element BE: The sum of the moments about E: (2) to obtain B x =

ΣM E = 0.4W − 0.7 B y = 0, 4 from which B y = W . The magnitude of the reaction at B is 7 B = W

( 65 ) + ( 47 ) = 1.0104W .

24 = 23.752 kN is the 1.0104 maximum load that can be carried. Thus, the largest mass that can be supported is m = W /g = 23752 N/9.81 m/s 2 = 2421 kg.

For a safe value of B = 24 kN, W =

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Problem 6.94 The hydraulic piston CD can be extended to adjust the height of the horizontal platform. The electric motor on the platform has a mass of 120 kg. When the platform is at the position shown, what are the magnitude of the axial force in the hydraulic piston and the reactions on member AEF at E?

0.36 m

F E 0.32 m

0.2 m

Solution:

The weight of the motor is W = (120 kg)(9.81 m/s 2 ) = 1180 N. Consider the free-body diagram of the entire system:

0.2 m

0.4 m

From the equation ΣM point B = (0.8 m) F − (0.36 m)W = 0, the reaction F = G = 530 N. Finally, we draw the free-body diagram of member AEF:

y y

x Ax

Ey Ay

ΣFx = A x = 0,

From the equations

ΣFy = A y + G − W = 0,

ΣFx = E x = 0,

ΣM point A = (0.8 m)G − (0.36 m)W = 0, we obtain A x = 0, A y = 647 N, G = 530 N. Now we draw the free-body diagram of the platform and motor:

ΣFy = A y + E y + TCD − F = 0, ΣM point A = −(0.16 m) E x + (0.4 m) E y + (0.2 m)TCD − (0.8 m) F = 0, we obtain E x = 0, E y = 2240 N, TCD = −2350 N.

Magnitude = 2.35 kN, E x = 0, E y = 2.24 kN.

By Bx

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From the equilibrium equations

TCD

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Problem 6.95 The electric motor on the horizontal platform has a mass of 120 kg. Suppose that the hydraulic piston CD is extended so that the platform is raised 0.2 m above the position shown. What are the magnitude of the axial force in the hydraulic piston and the reactions on member AEF at E?

y 0.36 m

F E 0.32 m

0.2 m

Solution:

The new height of point F above point G is 0.32 m + 0.2 m = 0.52 m. The length of member AEF is L = (0.8 m) 2 + (0.32 m) 2 = 0.862 m. Therefore the new distance from point A to point G (and from point B to point F ) is b = L2 − (0.52 m) 2 = 0.687 m.

0.2 m

0.4 m

From the equation ΣM point B = bF − (0.36 m)W = 0, the reaction F = G = 617 N. Finally, we draw the free-body diagram of member AEF:

The weight of the motor is W = (120 kg)(9.81 m/s 2 ) = 1180 N. Consider the free-body diagram of the entire system:

F y

TCD

Ex Ey

y Ay

From the equations

ΣFx = E x = 0, b

ΣFy = A y + E y + TCD − F = 0, 1 1 1 ΣM point A = − (0.52 m) E x + bE y + bTCD − bF = 0, 2 2 4

From the equilibrium equations ΣFx = A x = 0,

we obtain E x = 0, E y = 2240 N, TCD = −2350 N.

ΣFy = A y + G − W = 0, ΣM point A = bG − (0.36 m)W = 0,

Magnitude = 2.35 kN, E x = 0, E y = 2.41 kN.

we obtain A x = 0, A y = 560 N, G = 617 N. Now we draw the free-body diagram of the platform and motor:

A x = 0, A y = 560 N, E x = 0, E y = 2410 N.

W y x

430

By Bx

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Problem 6.96 The frame shown is used to support high-tension wires. If b = 3 ft, α = 30°, and W = 200 lb, what is the axial force in member HJ?

Ay h b G

Ax B

B a

Dy Dx E

H a

E G

W H a

Member BDEF: The sum of the forces:

These three equations have the solution: A x = 173.21 lb, A y = −100 lb, and B = −346.4 lb.

ΣFx = D x − B sin α − E sin α = 0, ΣFy = D y − W + B cos α − E cos α = 0. The sum of the moments about D:

ΣM D = −2bW − bE cos α − hE sin α − bB cos α + hB sin α = 0. b

Solution:

Joints B and E are sliding joints, so that the reactions are normal to AC and BF, respectively. Member HJ is supported by pins at each end, so that the reaction is an axial force. The distance h = b tan α = 1.732 ft Member ABC. The sum of the forces: ΣFx = A x + B sin α = 0, ΣFy = A y − W − B cos α = 0.

These three equations have the solution: D x = −259.8 lb, D y = 350 lb, E = −173.2 lb. Member EGHI : The sum of the forces: ΣFx = G x + E sin α − H cos α = 0, ΣFy = G y − W + E cos α + H sin α = 0. The sum of the moments about H: ΣM H = bG y − hG x + bW + 2bE cos α − 2hE sin α = 0. These three equations have the solution: G x = 346.4 lb, G y = 200 lb, and H = 300 lb. This is the axial force in HJ.

The sum of the moments about B: ΣM B = bA y − hA x + bW = 0.

Problem 6.97 Determine the force exerted on the ball by the bolt cutters and the magnitude of the axial force in the two-force member AB.

20 lb

Solution:

Free-body diagrams of the top head and the top handle

are shown. From the head we learn that ΣFx : C x = 0

20 in

3 in

6 in

4 in 20 lb

From the handle we have ΣM D : −(20 lb)(20 in) + C y (4 in) = 0 ⇒ C y = 100 lb Now we return to the head ΣM A : C y (6 in) − F (3 in) = 0 ΣFy : F − T AB + C y = 0 Solving yields Force on the ball = F = 200 lb, Axial force = T AB = 300 lb

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Problem 6.98 The person exerts 20-N forces to the pliers as shown. (a) What is the magnitude of the forces the pliers exert on the bolt at B? (b) Determine the magnitude of the force the members of the pliers exert on each other at the pinned connection C .

25 mm B

80 mm

50 mm 458 20 N

20 N

Solution:

Look at the piece that has the lower jaw of the pliers

(a) ΣM C : B(25 mm) − (20 N cos 45 °)(80 mm)

− (20 N sin 45 °)(50 mm) = 0 Cx

B = 73.5 N (b) ΣFx : C x − 20 N sin 45° = 0 ⇒ C x = 14.14 N ΣFy : C y − B − 20 N cos 45° = 0 ⇒ C y = 87.7 N Thus the magnitude is C =

432

C x 2 + C y 2 = 88.8 N

458 20 N

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Problem 6.99 Fig. a is a diagram of the bones and biceps muscle of a person’s arm supporting a mass. Tension in the biceps muscle holds the forearm in the horizontal position, as illustrated in the simple mechanical model in Fig. b. The weight of the forearm is 9 N, and the mass m = 2 kg. (a) Determine the tension in the biceps muscle AB. (b) Determine the magnitude of the force exerted on the upper arm by the forearm at the elbow joint C.

Solution: Make a cut through AB and BC just above the elbow joint C. The angle formed by the biceps muscle with respect to the 290 forearm is α = tan −1 = 80.2 °. The weight of the mass is 50 W = 2(9.81) = 19.62 N.

(

)

The section as a free body: The sum of the moments about C is ΣM C = −50T sin α + 150(9) + 350W = 0, from which T = 166.76 N is the tension exerted by the biceps muscle AB. The sum of the forces on the section is ΣFX = C x + T cos α = 0, from which C x = −28.33 N. ΣFY = C y + T sin α − 9 − W = 0, from which C y = −135.72. The magnitude of the force exerted by the forearm on the upper arm at joint C is F =

C x2 + C y2 = 138.65 N

T W

9N 200 mm

a Cy

50 150 mm mm

290 mm

(a)

A 50 mm

m 200 mm

150 mm (b)

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Problem 6.100 The bones and tendons in a horse’s rear leg are shown in Fig. a. A biomechanical model of the leg is shown in Fig. b. If the horse is stationary and the normal force exerted on its leg by the ground is N = 1200 N, determine the tensions in the superficial digital flexor BC and the patellar ligament DF.

Solution:

The free-body diagrams for AB and AE are shown. The

angle α = tan −1 (18/58) = 17.2 ° The equilibrium equations for AB are ΣM A : −(1200 N)(10 cm) + TBC cos α(8 cm) + TBC sin α(3 cm) = 0 ΣFx : A x − TBC sin α = 0 ΣFy : A y + TBC cos α + 1200 N = 0 One of the equilibrium equations for AE is ΣM E : −T DF (8 cm) − A x (49 cm) − A y (10 cm) = 0

6 cm 6 cm 6 cm

Solving these four equations yields

A x = 417 N, A y = −2540 N

TBC = 1410 N, T DF = 625 N

40 cm

3 cm

72 cm

(a)

434

8 10 8 cm cm cm (b)

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Problem 6.101 The pressure force exerted on the p iston is 2 kN toward the left. Determine the couple M necessary to keep the system in equilibrium.

B 300 mm

Solution: From the diagram, the coordinates of point B are (d , d ) where d = 0.3cos(45 °). The distance b can be determined from the Pythagorean Theorem as b = (0.35) 2 − d 2 . From the diagram, the angle θ = 37.3 °. From these calculations, the coordinates of points B and C are B (0.212, 0.212), and C (0.491, 0) with all distances being measured in meters. All forces will be measured in Newtons. The unit vector from C toward B is u CB = −0.795i + 0.606 j.

350 mm

The equations of force equilibrium at C are 458 A

ΣFX = FBC cos θ − 2000 = 0, C

and ΣFY = N − FBC sin θ = 0. 400 mm

Solving these equations, we get N = 1524 Newtons(N ), and FBC = 2514 N. The force acting at B due to member BC is FBC u BC = −2000 i + 1524 j N. The position vector from A to B is r AB = 0.212 i + 0.212 j m, and the moment of the force acting at B about A, calculated from the cross product, is given by M FBC = 747.6k N-m (counter - clockwise). The moment M about A which is necessary to hold the system in equilibrium, is equal and opposite to the moment just calculated. Thus, M = −747.6k N-m (clockwise). B 0.3 m 458 A

0.35 m d

FBC FBCY

FBCX

2000 N

N y M A

FBC uCB

r AB x

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Problem 6.102 In Problem 6.101, determine the forces on member AB at A and B. B 300 mm

350 mm

In the solution of Problem 6.101, we found that the force acting at point B of member AB was FBC u BC = −2000 i + 1524 j N, and that the moment acting on member BC about point A was given by M = −747.6k N-m (clockwise). Member AB must be in equilibrium, and we ensured moment equilibrium in solving Problem 6.101. From the free body diagram, the equations for force equilibrium are ΣFX = A X + FBC u BCX = A X − 2000 N = 0,

458 A

Solution:

and ΣFY = AY + FBC u BCY = AY + 1524 N = 0.

Thus, A X = 2000 N, and AY = −1524 N. 400 mm FBC uCB y

B M

Problem 6.103 In Example 6.8, suppose that the object being held by the pliers is moved to the left so that the horizontal distance from D to the object at E decreases from 30 mm to 20 mm. Draw a sketch of the pliers showing the new position of the object. What forces are exerted on the object at E as a result of the 150-N forces on the pliers?

Solution:

The analysis of the bottom grip of the pliers (member 3)

is unchanged. The reactions D x = −1517 N, D y = 500 N. From the free-body diagram of the lower jaw (member 2) we obtain ΣM C : −E (20 mm) − D x (30 mm) = 0 Therefore E = 2280 N 2280 N

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Problem 6.104 The shovel of the excavator is supported by a pin support at E and the two-force member BC. The 300-lb weight W of the shovel acts at the point shown. Determine the reactions on the shovel at E and the magnitude of the axial force in the two-force member BC.

Solution: α =

The angle

tan −1 (3/15) = 11.3 °

The equilibrium equations for the shovel are ΣFx : E x − TBC cos α = 0 ΣFy : E y + TBC sin α − 300 lb = 0 ΣM C : −(300 lb)(20 in) + E x (12 in) − E y (7 in) = 0 Solving yields E x = 604 lb, E y = 179 lb, TBC = 616 lb Thus E x = 604 lb, E y = 179 lb, axial force = 616 lb

Hydraulic cylinder Shovel

15 in

3 in E

C 12 in

12 in 7 in 20 in

Problem 6.105 The shovel of the excavator has a pin support at E. The position of the shovel is controlled by the horizontal hydraulic piston AB, which is attached to the shovel through a linkage of the two-force members BC and BD. The 300-lb weight W of the shovel acts at the point shown. What is the magnitude of the force the hydraulic piston must exert to hold the shovel in equilibrium?

Solution: From the solution to Problem 6.104 we know that the force in member BC is TBC = 616 lb We draw a free-body diagram of joint B and note that AB is h orizontal. The angles are α = tan −1 (3/15) = 11.3 ° β = tan −1 (4/15) = 14.9 ° The equilibrium equations for joint B are ΣFx : TBC cos α − T AB − TBD sin β = 0 ΣFy : −TBD cos β − TBC sin α = 0 Solving yields T AB = 637 lb, TBD = −125 lb 637 lb

Hydraulic cylinder Shovel

15 in

3 in E

C 12 in

12 in 7 in 20 in

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Problem 6.106 The person exerts 20-N forces on the handles of the shears. Determine the magnitude of the forces exerted on the branch at A.

Solution:

Assume that the shears are symmetrical.

Consider the 2 pieces CD and CE ΣFx = 0 ⇒ D x = E x ΣFy = 0 ⇒ D y = E y ΣM C = 0 ⇒ D x = E x = 0 20 N Dy

C 20 N

D B

20 N

Now examine CD by itself ΣM C = −(20 N)(90 mm) + D y (25 mm) = 0 ⇒ D y = 72 N

20 N

36 mm 25 mm 25 mm 65 mm

20 N

Cy Dx 5 0 Cx Finally examine DBA ΣM B : A(36 mm) − D y (50 mm) = 0 A

Dx5 0

Solving we find A = 100 N

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Problem 6.107 The person exerts 40-N forces on the handles of the locking wrench. Determine the magnitude of the forces the wrench exerts on the bolt at A.

Solution:

Recognize that DE is a 2-force member. Examine part CD

75 ΣFx : FDE + C x = 0 5689 8 ΣFy : FDE + C y + 40 N = 0 5689 8 ΣM C : FDE (30 mm) + (40 N)(105 mm) = 0 5689 Solving we find C x = 1312.5 N, C y = 100 N, FDE = −1320 N Now examine ABC ΣM B : A(50 mm) − C x (40 mm) = 0 ΣFx : B x − C x = 0 ΣFy : B y − C y − A = 0

B C

40 N

Solving: A = 1050 N, B x = 1312.5 N B y = 1150 N 8 mm 40 mm

Answer: A = 1050 N

E D

8 30 mm

75 mm

FDE

Cx 50 mm

40 N

40 N By A

Cx Cy

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Problem 6.108 Determine the magnitude of the force the members of the wrench exert on each other at B and the axial force in the two-force member DE.

40 N

B x2 + B y2 = 1745 N

FDE = 1320 N(C )

8 mm 40 mm

30 mm

75 mm

40 N

Problem 6.109 This device is designed to exert a large force on the horizontal bar at A for a stamping operation. If the hydraulic cylinder DE exerts an axial force of 800 N and α = 80°, what horizontal force is exerted on the horizontal bar at A?

D m

0 25

Solution: Define the x -y coordinate system with origin at C. The projection of the point D on the coordinate system is R y = 250 sin α = 246.2 mm, and R x = 250 cos α = 43.4 mm. The angle formed by member DE with the positive x-axis is Ry   = 145.38 °. The components of θ = 180 − tan −1   400 − R x 

908

B =

From the previous problem we have

50 mm

Solution:

C 400 mm

the force produced by DE are Fx = F cos θ = −658.3 N, and Fy = F sin θ = 454.5 N. The angle of the element AB with the positive x-axis is β = 180 − 90 − α = 10 °, and the components of the force for this member are Px = P cos β and Py = P sin β , where P is to be determined. The angle of the arm BC with the positive x-axis is γ = 90 + α = 170 °. The projection of point B is L x = 250 cos γ = −246.2 mm, and L y = 250 sin γ = 43.4 mm. Sum the moments about C: ΣM C = R x Fy − R y Fx + L x Py − L y Px = 0. Substitute and solve: P = 2126.36 N, and Px = P cos β = 2094 N is the horizontal force exerted at A.

Fy Fx Cx

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Problem 6.110 This device raises a load W by extending the hydraulic actuator DE. The bars AD and BC are 4 ft long, and the distances b = 2.5 ft and h = 1.5 ft. If W = 300 lb, what force must the actuator exert to hold the load in equilibrium? b

ΣFy = −W + C y + D y = 0. ΣFx = C x + D x = 0.

ΣM C = −bW + dD y = 0. These have the Solution:

B h

The complete structure as a free body: The sum of the forces:

The sum of the moments about C: W

( )

h The angle ADC is α = sin −1 = 22.02 °. The dis4 tance CD is d = 4 cos α.

Solution:

C y = 97.7 lb, D y = 202.3 lb, and C x = −D x . Divide the system into three elements: the platform carrying the weight, the member AB, and the member BC. The Platform: (See Free body diagram) The moments about the point A: ΣM A = −bW − dB = 0. The sum of the forces: ΣFy = A + B + W = 0. These have the solution: B = −202.3 lb, and A = −97.7 lb. Element BC : The sum of the moments about E is h d d ΣM C = − C + C + B = 0, from which 2 y 2 x 2

( )

(1) d C x − hC y − dB = 0. The sum of the forces: ΣFx = C x − E x = 0, from which (2) E x − C x = 0, ΣFy = C y − E y + B = 0. from which W A

B Ex

B A

Cx E

Ex Dx

(3) C y − E y + B = 0 Element AD: The sum of the moments about E: ΣM E =

( d2 ) D + ( h2 ) D − ( d2 ) A = 0, y

from which (4) dD y + hD x − dA = 0. These are four equations in the four unknowns: E X , E Y , D x , C X and D X Solving, we obtain D x = −742 lb.

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Problem 6.111 The four-bar linkage operates the forks of a fork lift truck. The force supported by the forks is W = 8 kN. Determine the reactions on member CDE.

Solution: W5 8 kN

0.7 m 0.15 m

0.2 m

W Cy

C 0.15 m 0.2 m

D E

Forks FAB 0.3 m

Cx 0.2 m

Ey Ex

D 3 2 FDF

Consider body BC. Note that AB is a 2-force body. ΣFx : −C x = 0 ΣFB : −C y (0.2 m) − (8 kN)(0.9 m) = 0 ⇒ C x = 0, C y = −36 kN Now examine CDE. Note that DF is a 2-force body. ΣM E : −C y (0.15 m) − C x (0.15 m) + ΣFx : C x + E x + ΣFy : C y + E y −

3 FDF (0.15 m) = 0 13

2 FDF = 0 13 3 FDF = 0 13

Solving we find FDF = −43.3 kN, E x = 24 kN, E y = 0 Note that D x =

2 3 FDF , D y = − FDF 13 13

Summary: C x = 0, C y = −36 kN D x = −24 kN, D y = 36 kN E x = 24 kN, E y = 0

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Problem 6.112 If the horizontal force on the scoop is F = 2000 lb, what is the magnitude of the axial force in the hydraulic actuator AC?

38 in B

28 in

Scoop

10 in

A F 10 in

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12 in

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6.112

(Continued)

Solution: We start with the free-body diagram of the scoop. Note that BC is a two-force body. The angle α = tan −1 (38/32) = 49.9 ° We have the following equilibrium equation ΣM A : −(2000 lb) (10 in) + TBC cos α(28 in) + TBC sin α(12 in) = 0 ⇒ TBC = 735 lb Now we work with the free-body diagram of joint C. The angles β = tan −1 (20/66) = 16.9 ° γ = tan −1 (10/38) = 14.7° The equilibrium equations are ΣFx : TBC cos α + T AC sin β − TCD sin γ = 0 ΣFy : −TBC sin α − T AC cos β − TCD cos γ = 0 Solving yields T AC = −1150 lb, TCD = 553 lb Thus 1150 lb

Problem 6.113 A car’s double-wishbone suspension is shown. The shock absorber BC is pinned at B and C. The bar BD is horizontal. Determine the axial force in the shock absorber if the vertical force exerted on the tire by the road is F = 4.0 kN.

Solution:

We draw the free-body diagram of the tire:

a TAC

TBC

b 42 mm C

TBD B 290 mm F

D 225 mm

192 mm

205 mm

The angles α = arctan(42 mm/225 mm) = 10.6 ° and β = arctan(290 mm/225 mm) = 52.2 °. equations

From

the

equilibrium

ΣFx = T AC cos α + TBC cos β + TBD = 0, ΣFy = −T AC sin α + TBC sin β + F = 0, ΣM point B = −(332 mm)T AC cos α − (192 mm)F = 0, we obtain T AC = −2.35 kN, TBC = −5.61 kN, TBD = 5.75 kN. 5.61 kN (C).

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Problem 6.114 The structure shown in the diagram (one of the two identical structures that support the scoop of the excavator) supports a downward force F = 1800 N at G. Members BC and DH can be treated as two-force members. Determine the reactions on member CDK at K .

320 mm C

Shaft 100 mm

Scoop

260 mm

D 160 mm

Solution: ΣM J :

Start with the scoop

B 180 mm

G F

K 1040 mm

380 mm

1120 mm

200 mm

Now examine CDK

4 1 FBC (0.44 m) − FBC (0.06 m) 17 17 − (1800 N)(0.2 m) = 0

ΣM K :

56 FDH (0.26 m) − 3161

⇒ FBC = 873 N

ΣFx : −

56 FDH + 3161

4 FBC + K x = 0 17

ΣFy : −

5 FDH − 3161

1 FBC + K y = 0 17

FBC 1 4

4 FBC (0.52 m) = 0 17

Solving we find K x = 847 N, K y = 363 N 1800 N

1 FBC

FDH Kx

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Problem 6.115 The loads F1 = 440 N and F2 = 160 N. Determine the axial forces in the members. Indicate whether they are in tension (T) or compression (C).

The sum of the moments about C is

M C = −0.7 BY + 0.7 F1 + 0.4 F2 = 0, from which B y =

0.7 F1 + 0.4 F2 = 531.43 N . 0.7

The axial loads at joint B are

F1 A

Solution:

AB = −B y = −531.4 N (C ) ,

and BC = 0 . Similarly, the sum of the forces at the joint A is

400 mm

ΣFxA = −F2 + AC cos β = 0,

from which

200 mm

AC =

F2 = 184.3 N (T ) cos β

700 mm

Problem 6.116 The truss supports a load F = 10 kN. Determine the axial forces in members AB, AC, and BC. B

3m C

F AX

Solution:

Find the support reactions at A and D. ΣFx : A x = 0

ΣFy : A y + D y − 10 = 0

a +ΣM A : (−4)(10) + 7 D y = 0

10 kN

Solving, A x = 0, FAB

A y = 4.29 kN D y = 5.71 kN

Joint A:

tan θ = 43 FAC

θ = 36.87 ° ( A y = 4.29 k Ν) ΣFx : FAB cos θ + FAC = 0

ΣFy : A y + FAB sin θ = 0 Solving, FAB = −7.14 kN (C ) FAC = 5.71 kN (T )

FBC

Joint C: ΣFx : FCD − FAC = 0 ΣFy : FBC − 10 kN = 0

FAC

FCD 10 kN

Solving FBC = 10 kN (T ) FCD = +5.71 kN (T )

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Problem 6.117 Each member of the truss will safely support a tensile force of 40 kN and a compressive force of 32 kN. Based on this criterion, what is the largest downward load F that can safely be applied at C? B

3m C

1 C

3m D

3m 4m

3m F

Solution: Assume a unit load F and find the magnitudes of the tensile and compressive loads in the truss. Then scale the load F up (along with the other loads) until either the tensile limit or the compressive limit is reached. External Support Loads: ΣFx : A x = 0

(1)

ΣFy : A y + D y − F = 0

(2)

ΣM A : − 4 F + 7 D y = 0

(3)

AX F

y FAB

Joint A: 3 tan θ = 2 θ = 36.87 °

ΣFx : FAC + FAB cos θ = 0

(4)

ΣFy : FAB sin θ + A y = 0

(5)

Joint C

x FAC

ΣFx : FCD − FAC = 0

(6)

ΣFy : FBC − F = 0

(7)

y FBD

Joint D 3 3 φ = 45 °

tan φ =

x ΣFx : −FCD − FBD cos φ = 0

(8)

ΣFy : FBD sin φ + D y = 0

(9)

FCD DY

Setting F = 1 and solving, we get the largest tensile load of 0.571 in AC and CD. The largest compressive load is 0.808 in member BD.

y FBC

Largest Tensile is in member BC . BC = F = 1 The compressive load will be the limit Fmax 32 = 1 0.808 Fmax = 40 kN

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FAC

FCD

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Problem 6.118 Determine the axial forces in members AB, BC, and BD. B

Solution:

The free-body diagram of the entire structure is

0.8 m Ax A

G C

F 2 kN

0.6 m

4 kN 0.6 m

0.6 m

2 kN

x 4 kN

From the equilibrium equations ΣFx = A x = 0, ΣFy = A y + G − 2 kN − 4 kN = 0, ΣM point A = (2.4 m)G − (0.6 m)(2 kN) − (1.8 m)(4 kN) = 0, The reactions A x = 0, A y = 2.5 kN, G = 3.5 kN. The free-body diagram of joint A is y

TAB

TAC

Ay The angle α = arctan(0.8 m/0.6 m) = 53.1 °. From the equilibrium equations ΣFx = T AB cos α + T AC = 0, ΣFy = T AB sin α + A y = 0, we obtain T AB = −3.13 kN, T AC = 1.88 kN. The free-body diagram of joint B is y B a

TAB

TBD

TBC

From the equilibrium equations ΣFx = −T AB cos α + TBD = 0, ΣFy = −T AB sin α − TBC = 0, we obtain TBC = 2.5 kN, TBD = −1.88 kN. AB: 3.13 kN (C); BC: 2.5 kN (T); BD: 1.88 kN (C).

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Problem 6.119 Use the method of sections to determine the axial forces in members CF , DF , and DE. B

Solution:

The free-body diagram of the entire structure is

0.8 m A

G C

2 kN

x 4 kN

F 2 kN

4 kN

From the equilibrium equations ΣFx = A x = 0,

0.6 m

ΣFy = A y + G − 2 kN − 4 kN = 0, ΣM point A = (2.4 m)G − (0.6 m)(2 kN) − (1.8 m)(4 kN) = 0, The reactions A x = 0, A y = 2.5 kN, G = 3.5 kN. We obtain a section by passing a plane through members CF , DE, and DF: E

TDE TDF

TCF

4 kN

The angle α = arctan(0.8 m/0.6 m) = 53.1 °. From the equilibrium equations ΣFx = −TCF − T DE − T DF cos α = 0, ΣFy = T DF sin α + G − 4 kN = 0, ΣM point F = (0.6 m)G + (0.8 m)T DE = 0, we obtain TCF = 2.25 kN, T DF = 0.625 kN, T DE = −2.63 kN. CF : 2.25 kN (T); DF : 0.625 kN (T); DE: 2.63 kN (C).

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Problem 6.120 The truss supports loads at F and H. Determine the axial forces in members AB, AC, BC, BD, CD, and CE.

Solution: 200 lb 100 lb

200 lb F 4 in 4 in

E C

4 in

100 lb

Ay Joint A

6 in

6 in BD a BC

AC a

I 12 in

AB Joint B

6 in CD

BC a

AC Joint C

6 in The complete structure as a free body: The sum of the moments about I: ΣM A = 100(6) + 200(12) − 24 AY = 0, from which AY = 125 lb. The sum of forces: ΣFx = A x = 0. The method of joints: The angles of the inclined members with the horizontal are α = tan −1 (0.6667) = 33.69 ° Joint A: ΣFx = AC cos α = 0, from which AC = 0. ΣFy = A y + AB + AC sin α = 0, from which AB = −125 lb (C ) Joint B: ΣFyt = − AB + BD sin α = 0, from which BD = −225.3 lb (C ) . ΣFx = BD cos α + BC = 0, from which BC = 187.5 lb (T ) Joint C: ΣFx = −BC − AC cos α + CE cos α = 0, from which CE = 225.3 lb (T ) ΣFy = − AC sin α + CD + CE sin α = 0, from which CD = −125 lb (C )

450

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Problem 6.121 Determine the axial forces in members EH and FH. 200 lb F 4 in

4 in

CE = 225.3 lb (T ), CD = −125 lb (C ), BD = −225.3 lb (C ).

100 lb

The method of joints: The angle of inclined members with the horizontal is α = 33.69 °.

Joint D: ΣFy = −BD sin α − CD + DF sin α = 0,

E C

4 in

Use the results from the solution to Problem 6.132:

from which DF = −450.7 lb (C ). ΣFx = −DE cos α + DE − BD cos α = 0, from which DE = 187.5 lb (T ) Joint F:

6 in

ΣFx = −DF cos α + FH cos α = 0, from which FH = −450.7 lb (C ) ΣFy = −200 − DF sin α − FH sin α − EF = 0,

Solution:

from which EF = 300 lb (T ) DF a DE BD CD Joint D

200 lb a DF

EF DE a CE

a FH EF

Joint F

Joint E:

EH a

ΣFy = −CE sin α + EF − EG sin α = 0,

from which EG = 315 lb (T )

Joint E

ΣFx = −DE + EH − CE cos α + EG cos α = 0, from which EH = 112.5 lb (T )

Problem 6.122 Determine the axial forces in members BD, CD, and CE. 10 kN

2m C B D

Use the method of sections y

10 kN

14 kN

Solution:

2m B

2m 14 kN

1.5 m ®

FBD

® x ®

FCD

FCE

2 1.5 θ = 53.13 °

tan θ = 6m

ΣFx : FCE cos θ − FCD cos θ + 24 = 0 ΣFy : −FBD − FCD sin θ − FCE sin θ = 0 ΣM B : −2(10) − 1.5 FCD sin θ − 1.5FCE sin θ = 0 3 eqns-3 unknowns. Solving FBD = 13.3 kN, FCD = 11.7 kN, FCE = −28.3 kN

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Problem 6.123 Determine the axial forces in members DF , EF , and EG.

10 kN

A 2m

14 kN

C B

2m E

2m G

2m I

Solution: tan θ =

Use method of sections

2 1.5

θ = 53.13 ° tan φ =

2 3

φ = 33.69 ° ΣFx : 24 + FEG cos θ − FEF cos φ = 0 ΣFy : −FDF − FEF sin φ − FEG sin φ = 0 ΣM E : 3FDF − 2(14) − 4(10) = 0 Solving, FEG = −32.2 kN (C ) FDF = 22.67 kN (T ) FEF = 5.61 kN (T ) A

10 kN

A 14 kN

10 kN

C B 2m

2m 14 kN

E 2m

2m D FDF

FEG

FEF

452

1.5

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Problem 6.124 The truss supports a 400-N load at G. Determine the axial forces in members AC , CD, and CF.

from which BD = −632.5 N (C ) ΣFy = AB + BD sin θ = 0, from which AB = 200 N (T )

400 N

Joint A:

ΣFy = A y − AD sin α AD − AB = 0, from which AD = 233.2 N (T )

300 mm 600 mm

ΣFx = A x + AC + AD cos α AD = 0, from which AC = 480 N (T )

F Ay

900 mm

400 N

600 mm

300 mm 300 mm 300 mm

Solution:

The complete structure as a free body: The sum of the moments about A:

AB B

ΣM A = −900(400) + 600 B = 0, from which B = 600 N. The sum of forces:

BD u

Joint B

ΣFx = A x + B = 0,

AD CD aAD DF u BD

from which A x = −600 N. ΣFy = A y − 400 = 0, from which A y = 400 N.

Joint D

AC aAD AD

AB Joint A AC

CE aCF CF CD Joint C

The method of joints: The angle from the horizontal of element BD is θ = tan −1

( 300 ) = 18.43°. 900

The angle from the horizontal of element AD is α AD = 90 − tan −1

( 600 −300 ) = 59.04°. 300 tan θ

The angle from the horizontal of element CF is αCF =

 90 − tan −1 

 300  600(1 − tan θ)  = 53.13 °.

Joint D: ΣFx = − AD cos α AD − BD cos θ + DF cos θ = 0, from which DF = −505.96 N (C ) ΣFy = AD sin α AD + CD − BD sin θ + DF sin θ = 0, from which CD = −240 N (C ) Joint C:

Joint B:

ΣFy = −CD − CF sin αCF = 0,

ΣFx = B + BD cos θ = 0,

from which CF = 300 N (T )

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Problem 6.125 Determine the axial forces in members CE, EF , and EH. 400 N A

G 300 mm

600 mm

H F

Solution:

Use the results of the solution of Problem 6.124:

AC = 480 N (T ), CF = 300 N (T ), DF = −505.96 N (C ), θ = 18.4 °, αCF = 53.1 °. The method of joints: The angle from the horizontal of element EH is 300 α EH = 90 − tan −1 = 45 ° 600 − 900 tan θ Joint C:

(

)

ΣFx = − AC + CE + CF cos αCF = 0,

from which CE = 300 N (T )

B 300 mm 300 mm 300 mm

CF aCF u DF

CE aCF CF

Joint C

EG a

Joint F

Joint E

Joint F: ∑ Fy = −CF cos αCF − DF cos θ + FH cos θ = 0, from which FH = −316.2 N (C ) ΣFy = EF + CF sin αCF − DF sin θ + FH sin θ = 0, from which EF = −300 N (C ) Joint E: ΣFy = −EH sin α EH − EF = 0, from which EH = 424.3 N (T )

Problem 6.126 Which members have the largest tensile and compressive forces, and what are their values?

Solution: The axial forces for all members have been obtained in Problems 6.124 and 6.125 except for members EG and GH. These are: Joint E:

400 N

ΣFx = −CE + EG + EH cos α EH = 0,

from which EG = 0

300 mm

Joint G: ΣFy = −GH − 400 = 0,

600 mm

H F D B 300 mm 300 mm 300 mm

from which GH = −400 N (C ). This completes the determination for all members. A comparison of tensile forces shows that AC = 480 N (T ) is the largest value, and a comparison of compressive forces shows that BD = −632.5 N (C ) is the largest value. CE EF Joint E

454

EG aEH EH

400 N GH

Joint G

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Problem 6.127 The Howe truss helps support a roof. Model the supports at A and G as roller supports. Use the method of joints to determine the axial forces in members BC , CD, CI, and CJ. 6 kN

4 kN D E 4 kN C 2 kN F 2 kN 4m B A G x H I J K L AY 12 m GY

4 kN

4 kN D

2 kN

6 kN

2 kN E

B A

4m G

H 2m

I 2m

J 2m

K 2m

L 2m

TDF F

TFH

E TEF TFG

TBX THI

TAH

TIJ

TBH

x TAH H THI

y 4 kN

y TCX I

TAB

TBC

Solution:

TCI

TCD x TCJ

The free body diagrams for the entire truss and the required joints are shown. The whole truss: The equations of equilibrium for the entire truss are:

Joint C:

ΣFX = 0,

ΣFX = −TBC u BCX + TCJ u CJX + TCD u CDX = 0,

ΣFY = AY + G Y − 18 kN = 0.

ΣFY = −TBC u BCY + TCJ u CJY + TCD u CDY − TCI − 4 = 0.

Instead of using the moment equation here (it would work), we see that the loading is symmetric. Thus, AY = G Y = 9 kN.

Solving these equations in sequence (we can solve at each joint before going to the next), we get

We need unit vectors along AB, BC , CD, (note that these are the same), and along BI, and CJ. We get u AB = u BC = u CD = 0.832 i + 0.555 j, u BI = 0.832 i − 0.555 j,

T AB = −16.2 kN, T AH = 13.5 kN, TBH = 0 kN, T HI = 13.5 kN, TBC = −14.4 kN, TBI = −1.80 kN, TIJ = 12.0 kN, TCI = 1.00 kN, TCJ = −4.17 kN, and TCD = −11.4 kN.

and u CJ = 0.6 i − 0.8 j. Joint A: The equations of equilibrium are ΣFX = T ABu ABX + T AH = 0 and ΣFY = T ABu ABY + AY = 0. Joint H: The equations of equilibrium are ΣFX = −T AH + T HI = 0, and ΣFY = TBH = 0. Joint B: ΣFX = −T ABu ABX + TBC u BCX + TBI u BIX = 0, ΣFY = −T ABu ABY + TBC u BCY + TBI u BIY − TBH − 2 = 0, Joint I: ΣFX = −T HI + TIJ − TBI u BIX = 0, and ΣFY = TCI − TBI u BIY = 0,

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Problem 6.128 Use the method of sections to determine the axial forces in members CD, CJ , and IJ .

Solution:

The free body diagram of the section is shown at the right. The support force at A is already known from the solution to Problem 6.139. The equations of equilibrium for the section are

6 kN

ΣFX = TCD u CDX + TCJ u CJX + TIJ = 0, ΣFY = TCD u CDY + TCJ u CJY + AY = 0,

4 kN

4 kN D

2 kN

and ΣM C = y C TIJ − 4 AY = 0.

2 kN

Solving, we get

B A

TIJ = 12.0 kN, TCJ = −4.17 kN,

G H 2m

I 2m

J 2m

K 2m

and TCD = −11.4 kN.

L 2m

Note that these values check with the values obtained in Problem 6.139. 2m 4 kN TCD 2 kN

TCJ

TIJ

Problem 6.129 The suspended 800-lb crate is supported by a truss. The y -axis is vertical. Determine the axial forces in members AB, AC, and AD.

Solution:

The position vectors from point A to points B, C , and D are

r AB = −10 i + 6k ft, r AC = −10 i − 4 k ft, r AD = −10 i + 8 j ft. Dividing each vector by its magnitude, we obtain unit vectors that point from point A toward points B, C , and D:

y D

e AB = −0.857 i + 0.514 k, e AC = −0.928i − 0.371k, e AD = −0.781i + 0.625 j. Let T AB , T AC , and T AD denote the tensile axial loads in the three bars: y

8 ft C 4 ft

TAD TAC

10 ft

6 ft

TAB

z z

800 lb

The equilibrium equation for the joint A is T AB e AB + T AC e AC + T AD e AD − (800 lb) j = 0: T AB (−0.857 i + 0.514 k) + T AC (−0.928i − 0.371k) + T AD (−0.781i + 0.625 j) − (800 lb) j = 0 . This yields the three equations −0.857T AB − 0.928T AC − 0.781T AD = 0, 0.625T AD = 800 lb, 0.514T AB − 0.371T AC = 0. Solving, we obtain T AB = −466 lb, T AC = −646 lb, T AD = 1280 lb. AB: 466 lb (C); AC: 646 lb (C); AD: 1280 lb (T).

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Problem 6.130 The mass of the suspended object is 900 kg. Determine the axial forces in the bars AB and AC. Strategy: Draw the free-body diagram of joint A.

Solution:

The free-body diagram of joint A is. TAD

y TAB

D (0, 4, 0) m A (3, 4, 4) m

TAC

(900) (9.81) N The position vectors from pt A to pts B, C , and D are

B (0, 0, 3) m

C (4, 0, 0) m x

r AB = −3i − 4 j − k (m), r AC = i − 4 j − 4 k (m), r AD = −3i − 4 k (m). Dividing these vectors by their magnitudes, we obtain the unit vectors e AB = −0.588i − 0.784 j − 0.196k, e AC = 0.174 i − 0.696 j − 0.696k, e AD = −0.6 i − 0.8k. From the equilibrium equation T AB e AB + T AC e AC + T AD e AC − (900)(9.81) j = O, We obtain the equations −0.588T AB + 0.174T AC − 0.6T AD = 0, −0.784T AB − 0.696T AC − (900)(9.81) = 0, −0.196T AB − 0.696T AC − 0.8T AD = 0. Solving, we obtain T AB = −7200 N, T AC = −4560 N, T AD = 5740 N.

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Problem 6.131 The suspended 360-kg crate is supported by a truss. The y -axis is vertical. Determine the axial forces in members AB, AC , and AD.

(0, 1 23) m

(0, 4, 2) m

A (4, 0, 0) m

B z

Solution:

The weight of the suspended crate is W = (360 kg) (9.81 m/s 2 ) = 3.53 kN. The position vectors from point A to points B, C , and D are r AB = −4 i − j + k (m),

(0, 21, 1) m

The equilibrium equation for the joint A is T AB e AB + T AC e AC + T AD e AD − Wj = 0: T AB (−0.943i − 0.236 j + 0.236k) + T AC (−0.784 i + 0.196 j − 0.588k) + T AD (−0.667 i + 0.667 j + 0.333k) − Wj = 0 .

r AC = −4 i + j − 3k (m), r AD = −4 i + 4 j + 2k (m).

This yields the three equations

Dividing each vector by its magnitude, we obtain unit vectors that point from point A toward points B, C , and D :

−0.236T AB + 0.196T AC + 0.667T AD = W ,

−0.943T AB − 0.784T AC − 0.667T AD = 0, 0.236T AB − 0.588T AC + 0.333T AD = 0.

e AB = −0.943i − 0.236 j + 0.236k, e AC = −0.784 i + 0.196 j − 0.588k,

Solving, we obtain T AB = −3.41 kN, T AC = 0.819 kN, T AD = 3.85 kN.

e AD = −0.667 i + 0.667 j + 0.333k. Let T AB , T AC , and T AD denote the tensile axial loads in the three bars:

AB: 3.41 kN (C); AC: 0.819 kN (T); AD: 3.85 kN (T).

TAC TAD TAB

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Problem 6.132 The mass m = 120 kg. Determine the forces on member ABC.

300 mm

200 mm

Solution:

The weight of the hanging mass is given by

(

m W = mg = 120 kg 9.81 2 s

) = 1177 N.

The complete structure as a free body: The equilibrium equations are: ΣFX = A X + E X = 0, ΣFY = AY − W = 0, and ΣM A = 0.3E X − 0.4W = 0.

Ay Ax B

200 mm

Cx B

W Cx

B Cy

B Ex

Solving, we get A X = −1570 N, AY = 1177 N, and E X = 1570 N. Element ABC: The equilibrium equations are ΣFX = A x + C X = 0, ΣFY = AY + C Y − BY − W = 0, and: ΣM A = −0.2 BY + 0.4C Y − 0.4W = 0. Solution gives BY = 2354 N (member BD is in tension), C X = 1570 N, and C Y = 2354 N.

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Problem 6.133 The mass of the pulley is 5 kg, the mass of the suspended object is 20 kg, and the mass of the horizontal bar is 8 kg. Determine the reactions on the bar at A and C.

0.4 m A

0.8 m

Solution:

0.8 m

Now consider the free-body diagram of the pulley,

Consider the free-body diagram of the entire system:

a Cy Cx

MA Ax

x Ay

WP WS

WS where C x and C y are the forces exerted on the pulley by the bar. From the equilibrium equations ΣFx = −WS cos α − C x = 0,

The angle α = arcsin(0.4 m/0.8 m) = 30 °. From the equilibrium equations ΣFx = A x = 0,

ΣFy = −WS sin α − WS − C y − W P = 0, we obtain C x = −170 N, C y = −343 N. A x = 0, A y = 324 N, M A = 534 N-m counterclockwise,

ΣFy = A y − W B − W P − WS = 0, ΣM point A = −(0.8 m)W B − (1.6 m)W P − (2.0 m)WS + M A = 0,

C x = −170 N, C y = −343 N.

we obtain A x = 0, A y = 324 N, M A = 534 N-m.

Problem 6.134 Determine the axial forces in the twoforce members AB, AC, and DE, and indicate whether they are in tension (T) or compression (C). B

4 kN

Solution: 4 kN TAB

We draw the free-body diagram of member BCD:

C TAC 6 in

D TDE

458 6 in

From the equilibrium equations 6 in

ΣFx = 4 kN − T AC sin 45 ° = 0, ΣFy = −T AB − T DE − T AC cos 45 ° = 0,

ΣM point B = −(6 in)T AC cos 45 ° − (12 in)T DE = 0, we obtain T AB = −2 kN, T AC = 5.66 kN, T DE = −2 kN.

6 in

460

6 in

AB: 2 kN (C); AC: 5.66 kN (T); DE: 2 kN (C).

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Problem 6.135 The 600-lb weight of the scoop acts at a point 1 ft 6 in to the right of the vertical line CE. The line ADE is horizontal. The hydraulic actuator AB can be treated as a two-force member. Determine the axial force in the hydraulic actuator AB and the forces exerted on the scoop at C and E.

Solution:

The free body diagrams are shown at the right. Place the coordinate origin at A with the x-axis horizontal. The coordinates (in ft) of the points necessary to write the needed unit vectors are A(0, 0), B(6, 2), C (8.5, 1.5), and D (5, 0). The unit vectors needed for this problem are u BA = −0.949 i − 0.316 j,

C 2 ft

D 5 ft

1 ft

TCB

Scoop

2 ft 6 in

1.5 ft

u BC = 0.981i − 0.196 j,

1 ft 6 in E

E EY

and u BD = −0.447 i − 0.894 j. 600 lb

The scoop: The equilibrium equations for the scoop are ΣFX = −TCBu BCX + E X = 0,

ΣFY = −TCBu BCY + E Y − 600 = 0, and ΣM C = 1.5E X − 1.5(600 lb) = 0.

TBA

x TCB

Solving, we get E X = 600 lb,

TBD

E Y = 480 lb, and TCB = 611.9 lb. Joint B: The equilibrium equations for the scoop are ΣFX = TBAu BAX + TBD u BDX + TCBu BCX = 0, and ΣFY = TBAu BAY + TBD u BDY + TCBu BCY = 0. Solving, we get TBA = 835 lb, and TBD = −429 lb.

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Problem 6.136 Determine the force exerted on the bolt by the bolt cutters.

100 N A

40 mm C 55 mm

75 mm B D

Solution:

The equations of equilibrium for each of the members will be developed.

90 mm 60 mm 65 mm

300 mm

Member AB: The equations of equilibrium are: ΣFX = A X + B X = 0, and ΣM B = 90 F − 75 A X − 425(100) = 0

75 mm

Member BD: The equations are

ΣFX = −B X + D X = 0,

100 N

ΣFY = AY + BY = 0,

100 N

40 mm

55 mm

ΣFY = −BY + DY + 100 = 0, and ΣM B = 15 D X + 60 DY + 425(100) = 0.

90 mm

60 mm 65 mm

300 mm

Member AC: The equations are ΣFX = − A X + C X = 0, ΣFY = − AY + C Y + F = 0, and ΣM A = −90 F + 125C Y + 40C X = 0. Member CD: The equations are:

CY 40 mm

75 mm

55 mm X

ΣFX = −C X − D X = 0, ΣFY = −C Y − DY = 0. 90 mm

Solving the equations simultaneously (we have extra (but compatible) equations, we get F = 1051 N, A X = 695 N, AY = 1586 N, B X = −695 N, BY = −435 N, C X = 695 N, C Y = 535 N, D X = −695 N, and D y = −535 N

60 mm 65 mm

300 mm

BY BX

B D DY

60 mm 65 mm

300 mm

100 N

CY C

DX D DY

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Problem 6.137 Determine the magnitude of the force the members of the bolt cutters exert on each other at the pin connection B and the axial force in the two-force member CD.

Solution:

From the solution to 6.136, we know B X = −695 N, and BY = −435 N. We also know that C X = 695 N, and C Y = 535 N, from which the axial load in member CD can be calculated. The load in CD is given by TCD = C X 2 + C Y 2 = 877 N

100 N A

40 mm C 55 mm

75 mm B D

90 mm 60 mm 65 mm

300 mm

100 N

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Chapter 7 Problem 7.1 The dimensions b = 12 in and h = 9 in. Using the method described in Example 7.1, determine the coordinates of the centroid of the triangle. y

The y coordinate of the midpoint (centroid) of the strip is y strip = f ( x )/2. The y coordinate of the centroid of the area is (see Example 7.1) b f ( x)

y =

∫ A y dA = ∫ A dA

f ( x ) dx ∫0 2 y strip

∫ f ( x) dx 0

x b

( (

) )

x 2 ∫0 2 1 − b dx = b x ∫0 h 1 − b dx 1 2 bh = 6 1 bh 2 = 3 in. b1

x = 4 in, y = 3 in.

Solution:

Consider the vertical strip element shown: y

f(x) h dA x x

dx b

The straight line defining the triangle’s hypotenuse can be expressed as a linear function f ( x ) = cx + d . When x = 0, f ( x ) = h, and when x = b, f ( x ) = 0. Using these two conditions to determine c and d yields c = −h /b and d = h, so

(

f ( x) = h 1 −

)

x . b

The area of the strip element is dA = f ( x ) dx, so the x coordinate of the centroid of the area is b

∫ x dA = ∫0 x f ( x) dx x = A b ∫ A dA ∫0 f ( x) dx ∫0 x h (1 − b ) dx x

∫0 h (1 − b ) dx b

1 2 b h = 6 1 bh 2 1 = b 3 = 4 in.

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Problem 7.2 In Example 7.2, suppose that the area is redefined as shown. Determine the x coordinate of the centroid.

Solution:

Consider the vertical strip element shown: y y51

y y 5 x2

y51 (1, 1)

x y 5 x2

The height of the strip element is 1 − x 2 , so the area of the element is dA = (1 − x 2 ) dx. The x coordinate of the centroid of the area is 1

∫ x dA = ∫0 x (1 − x 2 ) dx x = A 1 ∫ A dA ∫0 (1 − x 2 ) dx =

1  1 x2 − 1 x4   2 4  0

1  x − 1 x3   3  0 = 0.375.

x = 0.375.

Problem 7.3 In Example 7.2, suppose that the area is redefined as shown. Determine the y coordinate of the centroid.

The height of the strip element is 1 − x 2 , so the area of the element is dA = (1 − x 2 ) dx. The y coordinate of the centroid (midpoint) of the strip is x 2 + (1/2)(1 − x 2 ) = (1/2)(1 + x 2 ). The y coordinate of the centroid of the area is (see Example 7.2)

y y51

y =

(1, 1)

∫ A y dA ∫ A dA 1

(1 + x 2 )  [ (1 − x 2 ) dx ] ∫0  2  y strip 1

∫0 (1 − x 2 ) dx 1

y 5 x2

∫ (1 − x 4 ) dx = 01 ∫0 (1 − x 2 ) dx = 0.6. y = 0.6.

Solution:

Consider the vertical strip element shown: y y51

y 5 x2 dA

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Problem 7.4 Determine the centroid of the area.

Solution:

Consider the vertical strip element shown: y

y 5 x2 2 x 1 1

x x

The height of the strip is y = x 2 − x + 1, so dA = ( x 2 − x + 1) dx. The x coordinate of the centroid of the area is 3

∫ x dA = ∫0 x( x 2 − x + 1) dx x = A 3 ∫ A dA ∫0 ( x 2 − x + 1) dx 15.75 7.5 = 2.1. =

The y coordinate of the midpoint (centroid) of the strip is y =

1 2 ( x − x + 1). 2

The y coordinate of the centroid of the area is (see Example 7.1) 31

y =

∫ A y dA = ∫ A dA

( x 2 − x + 1)( x 2 − x + 1) dx ∫0 2 dA

y strip 3

∫0 ( x 2 − x + 1) dx

14.55 7.5 = 1.94. =

x = 2.1, y = 1.94.

Problem 7.5 Determine the coordinates of the centroid of the area. y

Solution:

∫ x( y dx) = ∫3 x ( 8 − 3 x ) dx = 11 x = 3 9 9 2 2 ∫3 y dx ∫3 ( 8 − 3 x ) dx 9

6 2 3

∫ y( y dx) = ∫3 2 ( 8 − 3 x ) dx = 13 y = 3 29 9 2 6 ∫3 y dx ∫3 ( 8 − 3 x ) dx 91

466

Use a vertical strip - The equation of the line is

2x y = 8− 3

x = 5.5 y = 2.17

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Problem 7.6 If a = 3 in, what is the x coordinate of the centroid of the area?

Solution:

Let dA be a vertical strip of width dx:

dA y 5 1 x2

y 5 9 x2 0

a dx

(

The area dA = 1 −

x =

)

1 2 x dx. The x coordinate of the centroid is 9

∫ A x dA ∫ A dA ∫0 x (1 − 9 x 2 ) dx 1

∫0 (1 − 9 x 2 ) dx 1

a  x2 − x4   2  36  0 = a  x − x3    27  0 = 1.125 in.

x = 1.125 in.

Problem 7.7 If a = 3 in, what is the y coordinate of the centroid of the area? Determine the answer using an element of area dA in the form of a horizontal strip of height dy.

Solution:

Let dA be a horizontal strip of the area of height dy:

y dy dA

y 5 1 x2 9

y 1

y 5 9 x2 0

Then dA = 3 y 2 dy. The y coordinate of the centroid is

y =

∫ A y dA ∫ A dA a2 /9

∫0

(

y 3 y 2 dy

a2 /9

∫0

)

1 3 y 2 dy

a2 /9

 6 25   y   5  0 =  a2 /9  32   2y   0 = 0.6 in. y = 0.6 in.

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Problem 7.8 Suppose that an art student wants to paint a panel of wood as shown, with the horizontal and vertical lines passing through the centroid of the area, and asks you to determine the coordinates of the centroid. What are they? y

The height of the strip is y = x + x 3, so dA = ( x + x 3 ) dx. The x coordinate of the centroid of the area is 2 ft

∫ x dA = ∫0 x ( x + x 3 ) dx x = A 2 ft ∫ A dA ∫0 ( x + x 3 ) dx 9.07 ft 3 6 ft 2 = 1.51 ft. =

y 5 x 1 x3

The y coordinate of the midpoint (centroid) of the strip is y =

1 ( x + x 3 ). 2

The y coordinate of the centroid of the area is (see Example 7.1) 2 ft 1

Solution:

y =

2 ft

∫ y dA A

∫ dA

∫0

( x + x 3 )( x + x 3 ) 2 dA y strip

2 ft

∫0

( x + x 3 ) dx

16.9 ft 3 6 ft 2 = 2.81 ft.

Consider the vertical strip element shown: y

x = 1.51 ft, y = 2.81 ft.

dA x

x dx

Problem 7.9 Determine the value of the constant c so that the y coordinate of the centroid of the area is y = 2. What is the x coordinate of the centroid? y

Solution: is dA =

The height of a vertical strip of width dx is cx 2 so the area

cx 2 dx.

1 The y coordinate of the midpoint of the vertical strip is cx 2 . We let this 2 be the value of y in Eq. (7.7): c  x5 4

∫ y dA = ∫2 ( 2 cx 2 ) cx 2 dx = 2  5  2 = 5.31c y= A 4 4  x3  ∫ A dA ∫2 cx 2 dx  3  4

= 2 ⇒ c = 0.376

y 5 cx2

The x coordinate of the centroid is 4

 x4 

∫ x dA = ∫2 x(cx 2 ) dx =  4  2 = 3.21 x = A 4 4  x3  ∫ dA ∫2 cx 2 dx  3  A

Notice that the value of x does not depend on the value of c. c = 0.376, x = 3.21

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Problem 7.10 If a = 1 m and b = 3 m, what is the x coordinate of the centroid of the area?

Solution:

Let dA be a vertical strip of width dx: y y52

b dx

y5 x

The area dA = (2/x ) dx. The x coordinate of the centroid is

x =

∫ A x dA ∫ A dA b

2

∫a x  x  dx b2

∫a x dx b

[ 2 x ]a b 2[ ln x ]a

= 1.82 m. x = 1.82 m.

Problem 7.11 If a = 1 m and b = 3 m, what is the y coordinate of the centroid of the area?

Solution:

Let dA be a vertical strip of the area of width dx. The y coordinate of the midpoint (centroid) of the strip is y strip = (1/2)(2/x ) = 1/x. y

y= 2

1 x 2 y5 x

dx The y coordinate of the centroid is (see Example 7.1)

y =

∫ A y dA ∫ A dA

)( x dx ) ∫a ( x b

y strip

∫a x

1 b 2  −   x a = 2[ ln x ]ba = 0.607 m. y = 0.607 m.

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Problem 7.12 Determine the centroid of the enclosed area. y

Solution:

We need to determine where the function intersects the x-axis. Setting 2 x − x 2 /5 = 0, we obtain x = 10. Consider the vertical strip element y

y 5 2x – 1 x2 5

y 5 2x 2 1 x2 5

The area of the strip is dA = (2 x − x 2 /5) dx. The x coordinate of the centroid is

∫ x dA = ∫0 x ( 2 x − 5 x 2 ) dx x = A 10 1 ∫ A dA ∫0 ( 2 x − 5 x 2 ) dx 1

= 5. (We could have anticipated this result by recognizing the symmetry of the parabola.) The height of the midpoint (centroid) of the strip element is y strip = (1/2)(2 x − x 2 /5). The y coordinate of the centroid is (see Example 7.1) 2

∫ y strip dA = ∫0 2 ( 2 x − 5 x 2 ) dx y = A 10 1 ∫ A dA ∫0 ( 2 x − 5 x 2 ) dx 10 1

= 2. x = 5, y = 2.

Problem 7.13 Determine the coordinates of the centroid of the area.

y y52

1 2 x 1 4x 2 7 4 y55

Solution:

Use a vertical strip. We first need to find the

x intercepts. 1 y = − x 2 + 4 x − 7 = 5 ⇒ x = 4, 12 4 12 12  1  ∫ x( y dx) = ∫4 x  − 4 x 2 + 4 x − 7 − 5  dx = 8 x = 4 12 12 1 ∫4 y dx ∫4  − 4 x 2 + 4 x − 7 − 5  dx

( (

) )

∫ y c ( y dx) y = 4 12 ∫4 y dx ∫  ( − x 2 + 4 x − 7 ) + 5   ( − 4 x 2 + 4 x − 7 ) − 5  dx 33 = 4 2 4 = 12 1 5 ∫4  ( − 4 x 2 + 4 x − 7 ) − 5  dx 12 1

x = 8, y = 6.6

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Problem 7.14 Determine the x coordinate of the centroid of the area.

Solution:

By setting x = x 3 , we see that the two curves intersect at x = 1. Consider the vertical strip element shown: y

y 5 x3

y 5 x3 y5x

y5x

The height of the strip element is x − x 3 , so the area of the element is dA = ( x − x 3 ) dx. The x coordinate of the centroid of the area is 1

∫ x dA = ∫0 x ( x − x 3 ) dx x = A 1 ∫ A dA ∫0 ( x − x 3 ) dx =

1  1 x3 − 1 x5   3 5  0

1  1 x2 − 1 x4   2 4  0 = 0.533.

x = 0.533.

Problem 7.15 Determine the y coordinate of the centroid of the area. y y 5 x3

The height of the strip element is x − x 3 , so the area of the element is dA = ( x − x 3 ) dx. The y coordinate of the centroid (midpoint) of the strip is x 3 + (1/2)( x − x 3 ) = (1/2)( x + x 3 ). The y coordinate of the centroid of the area is (see Example 7.2) y =

y5x

∫ A y dA ∫ A dA 1

( x + x 3 )  [( x − x 3 ) dx ] ∫0  2  y strip 1

∫0 ( x − x 3 ) dx

1 1 2 ∫ ( x − x 6 ) dx = 2 0 1

∫0 ( x − x 3 ) dx

= 0.381.

Solution:

By setting x = x 3 , we see that the two curves intersect at x = 1. Consider the vertical strip element shown:

y = 0.381.

y y 5 x3

y5x

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Problem 7.16 If h = 2, what is the centroid of the enclosed area?

Solution: 2x −

y y 5 2x 2

1 x2 5

We need to determine where the two functions intersect.

Setting 1 2 x = 2 5

and solving, we obtain x = 1.13 and x = 8.87. Consider the vertical strip element

y5h

y y 5 2x 2 1 x2 5

y5h x

The area of the strip is dA = (2 x − x 2 /5 − h) dx. The x coordinate of the centroid is

∫ x dA = ∫1.13 x ( 2 x − 5 x 2 − h ) dx x = A 8.87 1 ∫ A dA ∫1.13 ( 2 x − 5 x 2 − h ) dx 1

8.87

= 5. (We could have anticipated this result by recognizing the symmetry of the parabola.) The height of the midpoint (centroid) of the strip element is y strip = h + (1/2)(2 x − x 2 /5 − h). The y coordinate of the centroid is (see Example 7.1)

∫ y strip dA = ∫1.13 [h + (1/2)(2 x − x 2 /5 − h)]( 2 x − 5 x 2 − h ) dx y = A 8.87 1 ∫ A dA ∫1.13 ( 2 x − 5 x 2 − h ) dx 8.87

= 3.20 x = 5, y = 3.20.

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Problem 7.17 If the angle θ = 30 °, what is the x coordinate of the centroid of the enclosed area?

Solution:

The straight line is described by the function y = x tan θ. We need to determine where the two functions intersect. Setting 10 x − x 2 − 15 = x tan θ and solving, we obtain x = 2.03 and x = 7.39. Consider the vertical strip element

y y 5 10x 2 x2 2 15

u x

The area of the strip is dA = (10 x − x 2 − 15 − x tan θ) dx. The x coordinate of the centroid is 7.39

∫ x dA = ∫2.03 x(10 x − x 2 − 15 − x tan θ) dx x = A 7.39 ∫ A dA ∫2.03 (10 x − x 2 − 15 − x tan θ) dx = 4.71. x = 4.71.

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Problem 7.18 If the angle θ = 30 °, what is the y coordinate of the centroid of the enclosed area?

Solution: The straight line is described by the function y = x tan θ. We need to determine where the two functions intersect. Setting 10 x − x 2 − 15 = x tan θ

y y 5 10x 2 x2 2 15

and solving, we obtain x = 2.03 and x = 7.39. Consider the vertical strip element y y 5 10x 2 x2 2 15

u x

The area of the strip is dA = (10 x − x 2 − 15 − x tan θ) dx. The height of the midpoint (centroid) of the strip element is y strip = x tan θ + (1/2)(10 x − x 2 − 15 − x tan θ). The y coordinate of the centroid is (see Example 7.1) 7.39

∫ y strip dA = ∫2.03 [(1/2)(10 x − x 2 − 15 + x tan θ)](10 x − x 2 − 15 − x tan θ) dx y = A 7.39 ∫ A dA ∫2.03 (10 x − x 2 − 15 − x tan θ) dx = 5.60. y = 5.60.

Problem 7.19 What is the x coordinate of the centroid of the area?

Solution:

Use vertical strips, do an integral for the parabola then subtract the square First find the intercepts

y y52

1 2 x 1 2x 6

2 6

474

1 y = − x 2 + 2 x = 0 ⇒ x = 0, 12 6 12 1 x − x 2 + 2 x dx − 7(2)(2) ∫ 65 0 6 = x = 12 1 2 11 ∫0 − 6 x + 2 x dx − (2)(2) x = 5.91

( (

) )

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Problem 7.20 What is the y coordinate of the centroid of the area?

Solution:

Use vertical strips, do an integral for the parabola then subtract the square First find the intercepts

y y52

1 2 x 1 2x 6

1 y = − x 2 + 2 x = 0 ⇒ x = 0, 12 6 2 12 1 1 − x 2 + 2 x dx − 1(2)(2) ∫ 139 0 2 6 = y = 12 1 2 55 ∫0 − 6 x + 2 x dx − (2)(2) y = 2.53

( (

2 6

) )

Problem 7.21 An agronomist wants to measure the rainfall at the centroid of a plowed field between two roads. What are the coordinates of the point where the rain gauge should be placed?

Solution:

Consider the vertical strip element shown:

y yL y U dA

x dx 0.6 mi

The lower straight road is described by the linear function yL =

0.3 mi

x 0.5 mi

0.6 mi 0.2 mi

0.3 x mi, 1.3

and the upper straight road is described by yU = 0.3 +

0.6 x mi. 1.3

The height of the strip is y = yU − y L = 0.3 +

0.3 x mi, 1.3

so dA = [0.3 + (0.3/1.3) x ] dx. The x coordinate of the centroid of the area is

∫ x dA = ∫0.5 mi x ( 0.3 + 1.3 x ) dx x = A 1.1 mi 0.3 ∫ A dA ∫0.5 mi ( 0.3 + 1.3 x ) dx 0.3

1.1 mi

0.237 0.291 = 0.814 mi. =

The y coordinate of the midpoint (centroid) of the strip is 1 1 ( y − y L ) = ( yU + y L ) 2 U 2 1 0.9 0.3 + x mi. = 2 1.3

y = yL +

(

)

The y coordinate of the centroid of the area is (see Example 7.1) x )( 0.3 + x ) dx ( 0.3 + ∫0.5 mi 2 1.3 1.3

y =

∫ A y dA = ∫ A dA

0.3

0.9

1.1 mi 1

y strip

∫0.5 mi (

0.3 x dx 1.3

1.1 mi

0.3 +

)

0.126 = 0.291 = 0.432 mi. x = 0.814 mi, y = 0.432 mi.

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Problem 7.22 The cross section of an Earth-fill dam is shown. Determine the coefficients a and b so that the y coordinate of the centroid of the cross section is 10 m.

Solution:

The area: The elemental area is a vertical strip of length y and width dx, where y = ax − bx 3 . Note that y = 0 at x = 100, thus b = a × 10 −4. Thus A = ∫ dA = a ∫

100 0

( x − (10 −4 ) x 3 ) dx

= (0.5a)[ x 2 − (0.5 × 10 −4 ) x 4 ]100 0 = 0.5a × 10 4 − 0.25b × 10 8 ,

y 5 ax 2 bx

and the area is A = 0.25a × 10 4. The y -coordinate: The y-coordinate of the centroid of the elemental area is

100 m

y = (0.5)(ax − bx 3 ) = (0.5a)( x − (10 −4 ) x 3 ), from which y A = ∫ y dA A

= (0.5)a 2 ∫ = (0.5)a 2 ∫

100 0 100 0

( x − (10 −4 ) x 3 ) 2 dx ( x 2 − 2(10 −4 ) x 4 + (10 −8 ) x 6 ) dx

2x 5 x3 x 7 100 = (0.5a 2 )  − (10 −4 ) + (10 −8 )   3 5 7 0 = 3.81a 2 × 10 4. Divide by the area: y =

3.810 a 2 × 10 4 = 15.2381a. 0.25a × 10 4

For y = 10, a = 0.6562 , and b = 6.562 × 10 −5 m −2

Problem 7.23 The Spitfire used by Great Britain in World War II had a wing with an elliptical planform. The dimensions a = 17 ft 4 in and b = 3 ft 6 in. What is the x coordinate of the centroid of the wing’s area?

Solution:

Consider a vertical strip element of width dx: y

2 1/2

y 5 b(12 x2 ) a dA

y x2 y2 1 2 51 2 a b

The area dA = 2b(1 − x 2 /a 2 ) 1/ 2 dx. The x coordinate of the centroid is x 2 1/ 2

∫ x dA = ∫0 2bx (1 − a 2 ) dx x = A a x 2 1/ 2 ∫ A dA ∫0 2b (1 − a 2 ) dx a

a 1 2b  − (a 2 − x 2 ) 3/ 2   3a 0 = a 1 x  x 2 1/ 2 2b  x 1 − 2 + a arcsin  2  a  0 a 4a = 3π = 7.36 ft.

{ (

)

( )}

x = 7.36 ft.

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Problem 7.24 The radius of the circular part of the boundary is R = 4 m. What are the coordinates of the centroid of the area? Strategy: Write the equation for the circular part of the boundary in the form y = ( R 2 − x 2 ) 1/ 2 , and use a vertical “strip” of width dx as the element of area dA.

Solution:

Consider the vertical strip element shown:

dA x x

The height of the strip is y = ( R 2 − x 2 ) 1/ 2 , so dA = ( R 2 − x 2 ) 1/ 2 dx. The x coordinate of the centroid of the area is R 2m

∫ x dA = ∫2 m x( R 2 − x 2 )1/ 2 dx x = A R ∫ A dA ∫2 m ( R 2 − x 2 )1/ 2 dx

13.9 m 3 4.91 m 2 = 2.82 m. =

The y coordinate of the midpoint (centroid) of the strip is y =

1 2 ( R − x 2 ) 1/ 2 . 2

The y coordinate of the centroid of the area is (see Example 7.1) R

y = =

∫ A y dA = ∫ A dA

( R 2 − x 2 ) 1/ 2 ( R 2 − x 2 ) 1/ 2 dx ∫2 m 2 dA

y strip R

∫2 m ( R 2 − x 2 )1/ 2 dx

6.67 m 3

4.91 m 2 = 1.36 m. x = 2.82 m, y = 1.36 m.

Problem 7.25* If R = 6 and b = 3, what is the y coordinate of the centroid of the area?

Solution: α = cos −1 A = ∫

y =

( Rb ) = cos ( 63 ) = 60° = π3 −1

π /3

∫0 r dr d θ = ∫0 ∫0 r dr d θ = 6π

1 π /3 6 2 6 r sin θ dr d θ = = 1.910 A ∫0 ∫0 π

R b

α 0

We will use polar coordinates. First find the angle α

x a b

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Problem 7.26* What is the x coordinate of the centroid of the area in Problem 7.25?

Solution:

See the solution to 7.25

1 π /3 6 2 6 3 r cos θ dr d θ = = 3.31 A ∫0 ∫0 π

x = y

R x

Problem 7.27 In Practice Example 7.3, suppose that the area is placed as shown. Let the dimensions R = 6 in, c = 14 in, and b = 18 in. Use Eq. (7.9)1 to determine the x coordinate of the centroid. y

Solution: Let the semicircular area be area 1, let the rectangular area be area 2, and let the triangular area be area 3. The areas and the x coordinates of their centroids are 1 πR 2 , 2

A1 =

A2 = c (2 R), A3 =

R c

1 b (2 R), 2

4R , 3π 1 x 2 = c, 2 1 x3 = c + b 3

x1 = −

The x coordinate of the centroid of the composite area is x1 A1 + x 2 A2 + x 3 A3 A1 + A2 + A3 1 4R 1 1 c (2cR) + c + b (bR) − πR 2 + 3 3π 2 2 = 1 π R 2 + 2cR + bR 2

x =

(

)(

) ( )

(

)

Substituting the values for R, b, and c yields x = 9.60 in

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Problem 7.28 The dimensions b = 4 in and h = 3 in. Determine the coordinates of the centroid of the area.

Solution: y

h h

h 11 b

x b

We can treat this area as the two rectangles shown. The x coordinate of the centroid is

x b

x = =

x1 A1 + x 2 A2 A1 + A2 ( 12 b ) (2bh) + ( b + 12 b ) (bh)

2bh + bh

= 3.33 in. The y coordinate is y =

y1 A1 + y 2 A2 A1 + A2

(h)(2bh) + ( 12 h ) (bh) 2bh + bh = 2.50 in. =

x = 3.33 in, y = 2.50 in.

Problem 7.29 Determine the coordinates of the centroids. y

Break into a rectangle, a triangle and a circular hole

x =

5[(10)(8)] + 12[ 12 (8)(6) ] − 4[π (2) 2 ] = 6.97 in (10)(8) + 12 (8)(6) − π (2) 2

y =

4[(10)(8)] + 13 8[ 12 (8)(6) ] − 3[π(2) 2 ] = 3.79 in (10)(8) + 12 (8)(6) − π(2) 2

x = 6.97 in y = 3.79 in

2 in

8 in

Solution:

3 in x 4 in

6 in 10 in

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Problem 7.30 Determine the coordinates of the centroids.

Solution: y

R R

2 x

R 20 in

4R 3p

This area can be treated as a square area (area 1) with a quarter-circular “hole” (area 2). The location of the centroid of the quarter-circular area is from Appendix B. The x coordinate of the centroid of the area is x

x =

= =

x1 A1 + x 2 A2 A1 + A2

(

)

4R ( − 14 π R 2 ) 3π 1 2 2 R − 4 πR

( 12 R ) ( R 2 ) + R −

− 3π ( 10 )R 12 − 3π

= 4.47 in.

Because of the symmetry of the area, y = x. x = y = 4.47 in.

Problem 7.31 Determine the coordinates of the centroids. Solution:

Use a big triangle and a triangular hole

1.0[ 12 (1.0)(0.8) ] − ( 0.6 + 23 0.4 )[ 12 (0.4)(0.8) ] = 0.533 1 1 2 (1.0)(0.8) − 2 (0.4)(0.8)

0.8[ 12 (1.0)(0.8) ] − ( 13 0.8 )[ 12 (0.4)(0.8) ] = 0.267 1 1 2 (1.0)(0.8) − 2 (0.4)(0.8)

x = 3 y = 3

0.8 m

x = 0.533 m y = 0.267 m

x 0.6 m 1.0 m

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Problem 7.32 Determine the coordinates of the centroids.

Solution: y

y 1

22 R

b 0.3 m

We can treat this area as the two parts shown, where b = R = 0.3 m. Using the result from Appendix B for the quarter-circular area, the x coordinate of the centroid is

x 0.3 m

x =

x1 A1 + x 2 A2 A1 + A2

(

4R 3π 1 bR + π R 2 4

( 12 b)(bR) + b +

)( 14 π R ) 2

= 0.272 m. The y coordinate is y =

y1 A1 + y 2 A2 A1 + A2 ( 12 R ) (bR) +

bR +

( 43πR )( 14 π R ) 2

1 πR 2 4

= 0.140 m. x = 0.272 m, y = 0.140 m.

Problem 7.33 Determine the coordinates of the centroids.

Break into 4 pieces (2 rectangles, a quarter circle, and a

triangle)

(

)(

)

(

)(

) )

4[0.4] π[0.4] 2 3π 4 1  + 0.55(0.3)(0.7) + (0.8)  (0.3)(0.7)  2  x = 1 π[0.4] 2 (0.4)(0.3) + + (0.3)(0.7) + (0.3)(0.7) 2 4 4[0.4] π[0.4] 2 0.15[(0.4)(0.3)] + 0.3 + 3π 4 1 1 (0.3)(0.7) 0.7 + 0.35(0.3)(0.7) + 2 3 y = 1 π[0.4] 2 (0.4)(0.3) + + (0.3)(0.7) + (0.3)(0.7) 2 4 x = 0.450 m, y = 0.312 m 0.2[(0.4)(0.3)] + 0.4 −

400 mm

(

300 mm x 300 mm

300 mm

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Problem 7.34 Determine the x coordinate of the centroid of the area of the protractor. y

50 mm

608

20 mm

Source: Courtesy of Oleksii Arseniuk/Shutterstock.

Solution:

We can represent the area of the protractor as a composite consisting of (a) a solid circular area; (b) a semicircular “hole”; (c) a sector “hole”; (d) a triangular area to “fill in” part of the sector: y

The area and the x coordinate of the centroid of the solid circular area are Aa = π (70 mm) 2 , x a = 0. From Appendix B, the area and the x coordinate of the centroid of the semicircular “hole” are

4(50 mm) 1 Ab = − π (50 mm) 2 , x b = . 2 3π

50 mm

70 mm

The angle α = 60 ° = π (60 °/180 °) radians. The area and the x coordinate of the centroid of the sector “hole” are Ac = −α(50 mm) 2 , x c = −

(a)

(b)

608

(c)

482

The area and the x coordinate of the centroid of the triangular area are 2 Ad = (50 mm)sin 30 °(50 mm) cos30 °, x d = − (50 mm)sin 30 °. 3 The x coordinate of the centroid of the entire area is

2(50 mm)sin α . 3α

308 x

50 mm

50 mm (d)

x a Aa + x b A b + x c A c + x d A d Aa + A b + A c + A d = −2.94 mm.

x =

x = −2.94 mm.

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Problem 7.35 The dimension h = 40 mm. Determine the x coordinate of the centroid of the area.

Solution:

We consider the area as a composite consisting of two rectangles (areas 1 and 2), a half-circular area (area 3), and a circular hole (area 4).

y 40 mm

y y 20 mm

40 mm

20 mm

4 3

20 mm

2 h

20 mm

80 mm

120 mm Areas 1 and 2 and the x coordinates of their centroids are A1 = (120 mm)h = 4800 mm 2 , x1 = 20 mm, A2 = (80 mm)(20 mm) = 1600 mm 2 , x 2 = 0. Area 3 and the x coordinate of its centroid are 1 π (40 mm) 2 = 2510 mm 2 , 2 x 3 = 0. A3 =

Area 4 and the x coordinate of its centroid are A4 = −π (20 mm) 2 = −1260 mm 2 , x 4 = 0. The x coordinate of the centroid of the area is x1 A1 + x 2 A2 + x 3 A3 + x 4 A4 A1 + A2 + A3 + A4 = 12.5 mm.

x =

x = 12.5 mm.

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Problem 7.36 The dimension h = 40 mm. Determine the y coordinate of the centroid of the area.

Solution: We consider the area as a composite consisting of two rectangles (areas 1 and 2), a half-circular area (area 3), and a circular hole (area 4). y

20 mm

y 40 mm

20 mm

40 mm 3 20 mm

2 h

20 mm

x 120 mm

80 mm

Areas 1 and 2 and the y coordinates of their centroids are A1 = (120 mm)(40 mm) = 4800 mm 2 , h , 2 A2 = (80 mm)(20 mm) = 1600 mm 2 , y1 =

y 2 = h + 10 mm. Area 3 and the y coordinate of its centroid are A3 = 12 π (40 mm) 2 = 2510 mm 2 , 4(40 mm) . 3π Area 4 and the y coordinate of its centroid are y 3 = h + 20 mm +

A4 = −π (20 mm) 2 = −1260 mm 2 , y 4 = h + 20 mm. The y coordinate of the centroid of the area is y1 A1 + y 2 A2 + y 3 A3 + y 4 A4 A1 + A2 + A3 + A4 = 38.4 mm.

y =

y = 38.4 mm.

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Problem 7.37 The dimensions b = 42 mm and h = 22 mm. Determine the y coordinate of the centroid of the beam’s cross section.

Solution: y B

2 1

200 mm

x 120 mm

b We denote B = 200 mm, H = 120 mm. The areas of the parts shown are x

a1 = bH ,

a 2 = hB.

The y coordinate of the centroid is y = =

y1 A1 + y 2 A2 A1 + A2 ( 12 H ) (bH ) + ( H + 12 h ) (hB)

bH + hB

= 93.1 mm. y = 93.1 mm.

Problem 7.38 If the cross-sectional area of the beam is 12,000 mm 2 and the y coordinate of its centroid is y = 100 mm, what are the dimensions b and h?

Solution: y B

2 1

200 mm h

120 mm

b The total area is x

a = a1 + a 2 = bH + hB. The y coordinate of the centroid is y = =

y1 A1 + y 2 A2 A1 + A2 ( 12 H ) (bH ) + ( H + 12 h ) (hB)

bH + hB

Setting a = 12,000 mm 2 and y = 100 mm and solving, we obtain b = 47.2 mm, h = 31.7 mm. b = 47.2 mm, h = 31.7 mm.

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Problem 7.39 Determine the y coordinate of the centroid of the beam’s cross section.

Solution:

Take advantage of the symmetry. Work with only one half of the structure. Break into 2 rectangles, a quarter circle, and a quarter circular hole.

y y =

(

4(3)(8) + (11.5)(3)(3) + 8 +

4[5] 3π

(3)(8) + (3)(3) +

)( π[5]4 ) − ( 8 + 4[2] ) π[2] 3π ( 4 ) 2

π[5] 2 π[2] 2 − 4 4

y = 7.48 in 5 in

2 in

8 in

x 3 in

5 in

3 in

5 in

Problem 7.40 Determine the coordinates of the centroid of the plane’s vertical stabilizer.

Solution: y

7.813 m 1

2 10 m

10 m

10 m 528

10.763 m

808

10 m x

3 9m

1.763 m

We can treat the plane’s stabilizer as a rectangle and two triangular “holes.” The geometric information given leads to the dimensions shown. The x coordinate of the centroid is

Source: Courtesy of Oleksii Arseniuk/Shutterstock. x =

x1 A1 + x 2 A2 + x 3 A3 A1 + A2 + A3

( 12 10.763 )(10.763)(10) + ( 13 7.813 )  − 12 (7.813)(10)  + ( 9 + 23 1.763 )  − 12 (1.763)(10)  (10.763)(10) −

= 6.49 m.

1 1 (7.813)(10) − (1.763)(10) 2 2

The y coordinate of the centroid is y =

y1 A1 + y 2 A2 + y 3 A3 A1 + A2 + A3

( 12 10 )(10.763)(10) + ( 23 10 )  − 12 (7.813)(10)  + ( 13 10 )  − 12 (1.763)(10)  (10.763)(10) −

= 4.16 m.

1 1 (7.813)(10) − (1.763)(10) 2 2

x = 6.49 m, y = 4.16 m.

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Problem 7.41 The area has elliptical boundaries. If a = 30 mm, b = 15 mm, and ε = 6 mm, what is the x coordinate of the centroid of the area? y

Outer ellipse 4(a + ε) 3π π (a + ε)(b + ε) A1 = 4

x1 =

Inner Ellipse 4a 3π πab A2 = 4

x2 =

b x a

Solution:

The equation of the outer ellipse is

y2 x2 + = 1 ( a + ε) 2 ( b + ε) 2

and for the inner ellipse y2 x2 + 2 = 1 2 a b

–

We will handle the problem by considering two solid ellipses For any ellipse β

2 2 ∫ x dA = α ∫0 x α − x dx x = β α 2 − x 2 dx ∫ dA α∫

b y = a !a2 – x2

dA = y dx

b a

From integral tables (α 2 − x 2 ) 3/ 2 3 2 − x2 x α α2 x α 2 − x 2 dx = + sin −1 2 2 α

∫ x α 2 − x 2 dx = − ∫

( )

Substituting x =

[ −(α 2 − x 2 ) 3/ 2 ]0

 x α2 − x 2 α2 x  + sin   α  0 3 2  α 3 /3 [−0 + α 3 /3] = 2 x =  0 + α2 π − 0 − 0  α π /4   2 2 4α x = 3π

( )

For the composite x =

x 1 A1 − x 2 A2 A1 − A2

Substituting, we get x 1 = 15.28 mm

x 2 = 12.73 mm

A1 = 2375 mm 2

A2 = 1414 mm 2

and x = 19.0 mm

( )

Also Area = ∫ dA =

α β α 2 + x 2 dx ∫ 0 α α

β  x α2 − x 2 x  α2 + sin −1   2 2 α  α  0 β α2 π Area = = παβ /4 α 2 2

( )

( )( )

(The area of a full ellipse is παβ so this checks. Now for the composite area. For the outer ellipse, α = a + ε β = b + ε and for the inner ellipse α = a β = b

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Problem 7.42 By determining the x coordinate of the centroid of the area shown in Problem 7.41 in terms of a, b, and ε, and evaluating its limit as ε → 0, show that the x coordinate of the centroid of a quarter-elliptical line is

4 a(a + 2b) x = . 3π (a + b) b x a

Solution:

From the solution to 7.41, we have

4(a + ε) 3π 4a x2 = 3π x1 =

x1 A1 − x 2 A2 A1 − A2 1 [(2ab + a 2 ) + (2a + b)ε + ε 2 ] ε x = 3 π [(a + b) + ε] ε 4 4 a(a + 2b) 4(2a + b)ε 4 2 x = ε + + 3π (a + b) 3π 3π

Finally x =

π (a + ε)(b + ε) 4 πab A2 = 4

A1 =

( a + ε) 2 ( b + ε) 3 a 2b x 2 A2 = 3 π A1 − A2 = (ab + aε + bε + ε 2 − ab) 4 π A1 − A2 = (aε + bε + ε 2 ) 4 ( x1 A1 − x 2 A2 ) = 13 (a 2 b + 2abε + bε 2 + a 2ε + 2aε 2 + ε 3 − a 2 b)

so x 1 A1 =

Taking the limit as ε → 0 x =

4 a(a + 2b) 3π (a + b)

( x1 A1 − x 2 A2 ) = 13 ((2ab + a 2 )ε + (2a + b)ε 2 + ε 3 )

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Problem 7.43 The planforms of three sails of the gaffrigged sloop are shown. Determine the centroid of the topsail (a) in terms of the coordinate system shown. The coordinates of the points are expressed in meters and h = 7.0 m.

y y

(6.5, h)

(0, 8.5)

(0, 8.0)

(3.5, h)

(4.5, 1.5) x

x (a)

Solution:

We can treat the area of the sail as a composite consisting of a solid rectangle (area 1) and two triangular “holes” (areas 2 and 3). y

y 3

The areas and centroids of the triangles are 1 A2 = − (3.5 m)(7.0 m), 2

x2 =

2 (3.5 m), 3

y2 =

1 A3 = − (3.5 m)(1.5 m), 2

x3 =

2 (3.5 m), 3

y 3 = 7.0 m +

y = x

3.5 m

1 (7.0 m), 3 2 (1.5 m). 3

x1 A1 + x 2 A2 + x 3 A3 = 1.17 m, A1 + A2 + A3

and the y component is

2 x

(c)

The x component of the centroid of the sail is

1.5 m

7.0 m 1

(11.5, 1.5) x

(b)

x = 8.5 m

(3.5, h)

y1 A1 + y 2 A2 + y 3 A3 = 5.17 m. A1 + A2 + A3

x = 1.17 m, y = 5.17 m.

3.5 m

The area and centroid of the rectangle are A1 = (3.5 m)(8.5 m),

x1 =

1 (3.5 m), 2

y1 =

1 (8.5 m). 2

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Problem 7.44 Determine the centroid of the jib (b) in terms of the coordinate system shown. The coordinates of the points are expressed in meters and h = 10.0 m.

y y

(6.5, h)

(0, 8.5)

(0, 8.0)

(3.5, h)

(4.5, 1.5) x

x (a)

Solution:

We can treat the area of the sail as a composite consisting of a solid triangle (area 1), two triangular “holes” (areas 2 and 3), and a rectangular “hole” (area 4). y

(3.5, h)

(11.5, 1.5) x

(b)

(c)

The areas and centroids of the triangles are 1 A2 = − (4.5 m)(1.5 m), 2

x2 =

1 A3 = − (2.0 m)(8.5 m), 2

x 3 = 4.5 m +

y 3 = 1.5 m +

2 (4.5 m), 3

y2 =

1 (1.5 m), 3

2 (2.0 m), 3

1 (8.5 m), 3

and the area and centroid of the rectangle are 8.5 m

10 m

x 6.5 m

4.5 m

4 2.0 m

The area and centroid of area 1 are

490

1 (6.5 m)(10 m), 2

x1 =

2 (6.5 m), 3

y1 =

1 (10 m). 3

x 4 = 4.5 m +

1 (2.0 m), 2

y4 =

1 (1.5 m). 2

The x component of the centroid of the sail is

A1 =

A4 = −(2.0 m)(1.5 m),

1.5 m

x =

x1 A1 + x 2 A2 + x 3 A3 + x 4 A4 = 3.67 m, A1 + A2 + A3 + A4

and the y component is y1 A1 + y 2 A2 + y 3 A3 + y 4 A4 = 3.83 m. A1 + A2 + A3 + A4 x = 3.67 m, y = 3.83 m.

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Problem 7.45 Determine the centroid of the mainsail (c) in terms of the coordinate system shown. The coordinates of the points are expressed in meters and h = 14.5 m.

y y

(3.5, h)

(6.5, h)

(0, 8.5)

(0, 8.0)

(3.5, h)

(4.5, 1.5) x

x (a)

(b)

(11.5, 1.5) x (c)

Solution:

We treat the area of the sail as a composite consisting of a solid rectangle (area 1) and three triangular “holes” (areas 2-4). y

6.5 m

13 m 3.5 m 8.0 m

x 11.5 m

1.5 m

11.5 m

The area and centroid of the rectangle are A1 = (11.5 m)(14.5 m),

x1 =

1 (11.5 m), 2

y1 =

1 (14.5 m). 2

The areas and centroids of the triangles are 1 1 2 A2 = − (3.5 m)(6.5 m), x 2 = (3.5 m), y 2 = 8 m + (6.5 m), 2 3 3 1 2 2 A3 = − (8.0 m)(13 m), x 3 = 3.5 m + (8.0 m), y 3 = 1.5 m + (13 m), 2 3 3 2 1 1 A4 = − (11.5 m)(1.5 m), x 4 = (11.5 m), y 4 = (1.5 m). 2 3 3 The x component of the centroid of the sail is x =

x1 A1 + x 2 A2 + x 3 A3 + x 4 A4 = 4.43 m, A1 + A2 + A3 + A4

and the y component is y =

y1 A1 + y 2 A2 + y 3 A3 + y 4 A4 = 5.65 m. A1 + A2 + A3 + A4

x = 4.43 m, y = 5.65 m.

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Problem 7.46 The beam has pin and roller supports and is loaded by a triangular distributed load. Determine the reactions at A and B.

60 N/m A

B 4m

Solution:

We draw the free-body diagram of the beam, representing the triangular distributed load by an equivalent force equal to the “area” of the load. The load acts at the centroid of the distribution.

x B

12 m

ΣFx = A x = 0, ΣFy = A y + B −

1 (8 m)(60 N/m) 2

From the equilibrium equations

1 (8 m)(60 N/m) = 0, 2 1 1 ΣM point B = −(12 m) A y +  (8 m)   (8 m)(60 N/m)  = 0, 3 2 

1 (8 m) 3

we obtain A x = 0, A y = 53.3 N, B = 187 N. A x = 0, A y = 53.3 N, B = 187 N.

Problem 7.47 Determine the reactions at A and B.

6 ft

200 lb/ft

6 ft

4 ft

200 lb/ft

Solution:

From the free-body diagram of the bar (with the distributed loads represented by equivalent force), the equilibrium equations are

ΣFx : A x − 600 lb = 0 ΣFy : A y + B − 800 lb = 0 ΣM A : −(800 lb) (2 ft) − (600 lb) (4 ft) + B(10 ft) = 0 Solving yields A x = 600 lb, A y = 400 lb, B = 400 lb.

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Problem 7.48 In Example 7.6, suppose that the distributed loads are modified as shown. Determine the reactions on the beam at A and B.

400 N/m

6m 600 N/m

400 N/m B 6m

Solution:

The distributed loads are represented by three equivalent forces. The equilibrium equations are ΣFx : 1200 N + A x = 0 ΣFy : A y + B − 2400 N − 600 N = 0 ΣM A : −(1200 N) (4 m) = (600 N) (2 m) − (2400 N) (3 m) + B (6 m) = 0 Solving yields A x = −1200 N, A y = 800 N, B = 2200 N

Problem 7.49 In Example 7.7, suppose that the distributed load acting on the beam from x = 0 to x = 10 ft is given by w = 350 + 0.3 x 3 lb/ft. (a) Determine the downward force and the clockwise moment about A exerted by the distributed load. (b) Determine the reactions at the fixed support. Solution: (a) The force and moment are R = ∫

10 0

M = ∫

(350 + 0.3 x 3 ) dx = 4250 lb

10 0

x (350 + 0.3 x 3 ) dx = 23500 ft-lb

R = 4250 lb, M = 23,500 ft-lb (b) The equilibrium equations are ΣFx : A x = 0, ΣFy : A y − 4250 lb + 2000 lb = 0, ΣM A : M A − 23500 ft-lb + (2000 lb)(20 ft) + 10,000 ft-lb = 0. Solving yields A x = 0, A y = 2250 lb, M A = −26,500 ft-lb

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Problem 7.50 The beam is subjected to a distributed load that can be approximated by the function shown. The dimensions a = 1 m and b = 3 m. Determine the reactions at the built-in support A.

y w5 2 x kN/m

A a b

Solution:

Consider the free-body diagram of the beam:

The clockwise moment exerted by the distributed load about A is b

M DL = ∫ x

w 5 x2 kN/m

( 2x ) dx = 2(b − a) = 4 kN-m.

From the equilibrium equations ΣFx = A x = 0, ΣFy = A y − 2.20 kN = 0,

ΣM point A = M A − 4 kN-m = 0,

MA Ay a

we obtain A x = 0, A y = 2.20 kN, M A = 4 kN-m counterclockwise. b

A x = 0, A y = 2.20 kN, M A = 4 kN-m counterclockwise.

The downward force exerted by the distributed load is FDL = ∫

b2 a

dx = 2[ ln x ]a = 2.20 kN.

Problem 7.51 An engineer measures the forces exerted by the soil on a 10-m section of a building foundation and finds that they are described by the distributed load w = −10 x − x 2 + 0.2 x 3 kN/m. (a) Determine the magnitude of the total force exerted on the foundation by the distributed load. (b) Determine the magnitude of the moment about A due to the distributed load.

Solution: (a) The total force is F = −∫

12 0

(10 x + x 2 − 0.2 x 3 ) dx

x3 0.2 4  10 =  −5 x 2 − + x   0 3 4 F = 333.3 kN (b) The moment about the origin is

M = −∫

10 0

(10 x + x 2 − 0.2 x 3 ) x dx

10 1 0.2 5  10 =  − x 3 − x 4 + x  ,  3 0 4 5 M = 1833.33 kN.

10 m x

The distance from the origin to the equivalent force is d =

M = 5.5 m, F

from which M A = (d + 2) F = 2500 kN m.

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Problem 7.52 Determine the reactions on the beam at A and B.

6 kN/m

3 kN/m A B 4m

Solution:

First we express the left part of the distributed load by the sum of a uniformly distributed load and a triangular distributed load,

3 kN-m

6 kN-m

3 kN-m

From the equilibrium equations ΣFx = A x = 0, ΣFy = A y + B − 12 kN − 6 kN − 6 kN = 0, 2 ΣM point A = (4 m) B − (2 m)(12 kN) −  (4 m)  (6 kN) 3  2 −  6 m − (2 m)  (6 kN) = 0,   3 we obtain A x = 0, A y = 7 kN, B = 17 kN.

A x = 0, A y = 7 kN, B = 17 kN.

then we represent the three distributed loads by their equivalent forces:

y Ay

12 kN

6 kN

x 2 3 (2 m)

2m 2 (4 m) 3

Problem 7.53 The aerodynamic lift of the wing is described by the distributed load w = −300 1 − 0.04 x 2 N/m. The mass of the wing is 27 kg, and its center of mass is located 2 m from the wing root R. (a) Determine the magnitudes of the force and the moment about R exerted by the lift of the wing. (b) Determine the reactions on the wing at R.

x 2m 5m

Solution: w

(a) The force due to the lift is 5

F = −w = ∫ 300(1 − 0.04 x 2 ) 1/ 2 dx,

F =

300 5 (25 − x 2 ) 1/ 2 dx 5 ∫0 5

 x 25 − x 2 25 −1 x  sin F = 60  +  = 375π N,  2 2 5  0

( )

(b) The sum of the moments about the root:

F = 1178.1 N. The moment about the root due to the lift is 5

M = 300 ∫ (1 − 0.04 x 2 ) 1/ 2 x dx, 0

(25 − x 2 ) 3/ 2  60(25) 3/ 2 M = −60  = 2500  =  3 3 0 M = 2500 Nm.

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mg FR

ΣM = M R + 2500 − 27 g(2) = 0, from which M R = −1970 N-m. The sum of forces ΣFy = FR + 1178.1 − 27 g = 0, from which FR = −1178.1 + 27 g = −913.2 N

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Problem 7.54 Determine the reactions on the bar at A and B.

400 lb/ft y

B 2 ft

600 lb/ft

2 ft

400 lb/ft

x 4 ft

Solution:

4 ft

First replace the distributed loads with three equivalent

forces. 800 lb

The equilibrium equations ΣFx : B x + 800 lb = 0 ΣM B : (800 lb)(1 ft) − A(4 ft) + (1600 lb)(6 ft)

400 lb

+ (400 lb)(6.67 ft) = 0 ΣFy : A + B y − 1600 lb − 400 lb = 0 Solving:

1600 lb

A = 3267 lb, B x = −800 lb, B y = −1267 lb

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Problem 7.55 Determine the reactions on member AB at A and B. 300 lb/ft A

6 ft

300 lb/ft

Solution:

From the free-body diagram of the entire structure (with the distributed loads represented by equivalent forces), one of the equilibrium equations is ΣM C : − A x (6 ft) − (3600 lb) (6 ft) − (900 lb) (4 ft) − (1800 lb) (9 ft) = 0

From the free-body diagram of member AB we have the equilibrium equations ΣFx : A x + B x = 0 ΣFy : A y + B y − 3600 lb = 0 ΣM A : B y (12 ft) − (3600 lb) (6 ft) = 0 Solving yields A x = −6900 lb,

A y = 1800 lb,

B x = 6900 lb, B y = 1800 lb.

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Problem 7.56 Determine the axial forces in members BD, CD, and CE of the truss and indicate whether they are in tension (T) or compression (C).

2m C

4 kN/m

Solution:

We start by analyzing the horizontal bar with the dis-

8 kN/m

tributed load

ΣM G : (16 kN)(0.667 m) + (32 kN)(2 m) − FC (4 m) = 0 ⇒ FC = 18.67 kN ΣFy = FC + FG − 32 kN − 16 kN = 0 32 kN

⇒ FG = 29.33 kN Now work with the whole structure in order to find the reactions at A ΣFx : A x = 0 ⇒ A x = 0 ΣM H : FG (2 m) + FC (6 m) − A y (8 m) = 0 ⇒ A y = 21.3 kN

16 kN Ax

Finally, cut through the truss and look at the left section ΣM C : − A x (2 m) − A y (2 m) − BD(2 m) = 0 Ay

ΣM D : − A y (4 m) + FC (2 m) + CE (2 m) = 0 ΣFy : A y − FC +

1 CD = 0 2

Solving we find FC

BD = −21.3 kN, CD = −3.77 kN, CE = 24 kN In summary:

B BD

BD = 21.3 kN(C ), CD = 3.77 kN(C ), CE = 24 kN(T )

CD Ay C

CE FC

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Problem 7.57 Determine the reactions on member ABC at A and B.

400 N/m 200 N/m C 160 mm B

240 mm E

400 N/m

160 mm

Solution:

Work on the entire structure first to find the reactions at A. Replace the distributed forces with equivalent concentrated forces

160 mm

48 N 96 N

ΣFx : A x + 160 N = 0 ΣM E : (96 N)(0.08 m) + (48 N)(0.16 m) − (160 N)(0.2 m) − A y (0.32 m) = 0 Solving: A x = −160 N, A y = −52 N Now look at body ABC. Take advantage of the two-force body CD. ΣM B : A x (0.24 m) + (160 N)(0.04 m) − (48 N)(0.16 m) − (96 N)(0.24 m) − ΣFx : A x + B x + 160 N −

7 CD(0.32 m) + 65

4 CD(0.16 m) = 0 65

4 CD = 0 65

ΣFy : A y + B y − 48 N − 96 N −

160 N

7 CD = 0 65

Solving:

A x = −160 N, A y = −52 N

B x = −157 N, B y = −78.4 N

48 N 96 N

C 7

By 4 Bx

160 N

Ax Ay

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Problem 7.58 Determine the forces on member ABC of the frame.

A 1m

3 kN/m B

C 2m

Solution:

The free body diagram of the member on which the distributed load acts is

(4 m)(3 kN/m) 5 12 kN BX

From the equilibrium equations ΣFx = B x = 0,

ΣFy = B y + E − 12 = 0, Σm (leftend) = 3E − (2)(12) = 0,

4 kN AY

Σm (leftend) = −2C y − (4)(8) = 0,

we obtain C y = −16 kN. Then from the middle free body diagram, we write the equilibrium equations ΣFy = A y − 4 − 16 = 0,

Σm (rightend) = −2 A x − 2 A y + (1)(4) = 0

obtaining A x = −18 kN, A y = 20 kN, C x = 18 kN.

8 kN

ΣFx = A x + C x = 0,

Problem 7.59 Use the method described in Practice Example 7.8 to determine the centroid of the truncated cone.

we find that B x = 0, B y = 4 kN, and E = 8 kN. From the lower fbd, writing the equilibrium equation

Solution:

Just as in Active Example 7.8, the volume of the disk

element is dV = π

( Rh x ) dx 2

the x coordinate of the centroid is

∫ x dV = ∫h / 2 xπ ( h x ) x = 45 h x = V h R 2 56 ∫V dV ∫h / 2 π ( h x ) x h

x =

45 h 56

R h 2

500

x h 2

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Problem 7.60 A grain storage tank has the form of a surface of revolution with the profile shown. The height of the tank is 7 m and its diameter at ground level is 10 m. Determine the volume of the tank and the height above ground level of the centroid of its volume.

y y 5 ax1/2 7m

10 m x

Solution:

dV = π y 2 dx 7

x =

y 5 ax1/2

∫0 ∫ = 07 7 2 ∫0 π y dx ∫0 πa 2 x dx χπ y 2 dx

χπa 2 x dx

y dx

x =

[ x 3 /3 ]0

[ x 2 /2 ]0

= 4.67 m

dV 5 py 2 dx

The height of the centroid above the ground is 7 m − x h = 2.33 m

The volume is 7

V = ∫ πa 2 x dx = πa 2 0

( 492 ) m

To determine a, y = 5, m when x = 7 m. y = ax 1/ 2 , 5 = a 7 a = 5/ 7a 2 = 25/7 V = π

( 257 )( 492 ) = 275 m

V = 275 m 3

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Problem 7.61 The object shown, designed to serve as a pedestal for a speaker, has a profile obtained by revolving the curve y = 0.167 x 2 about the x-axis. What is the x coordinate of the centroid of the object?

z x 0.75 m 0.75 m

Solution: y 5 0.167 x 2 dV 5 p y 2dx x

dv 1.50

∫ x dV = ∫0.75 xπ(0.167 x 2 ) 2 dx x = V 1.50 ∫V dV ∫0.75 π(0.167 x 2 ) 2 dx 1.5

5 1.5 [ x 6 /6 ]0.75 π (0.167) 2 ∫0.75 x dx x = ⋅ 1.5 = 1.5 2 π (0.167) [ x 5 /5 ]0.75 x 4 dx

∫0.75

x = 1.27 m

Problem 7.62 The volume of a nose cone is generated by rotating the function y = x − 0.2 x 2 about the xaxis. (a) What is the volume of the nose cone? (b) What is the x coordinate of the centroid of the volume?

z x 2m

Solution: (a)

V = ∫

2m 0 2m

(b)

502

∫ x = 0

π y 2 dx = ∫ x (π y 2 dx ) V

2m 0

π ( x − 0.2 x 2 ) 2 dx = 4.16 m 3 2

∫ π x( x − 0.2 x 2 ) 2 dx = 1.411 m = 0 4.155 m 3

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Problem 7.63 Determine the centroid of the hemispherical volume.

Solution:

The equation of the surface of a sphere is x 2 + y 2 + z 2 = R 2.

The volume: The element of volume is a disk of radius ρ and thickness dx. The radius of the disk at any point within the hemisphere is ρ 2 = y 2 + z 2 . From the equation of the surface of the sphere, ρ 2 = ( R 2 − x 2 ). The area is πρ 2 , and the element of volume is dV = π ( R 2 − x 2 ) dx, from which V =

Vsphere 2π 3 = R . 2 3

The x -coordinate is: R

∫V x dV = π ∫0 ( R 2 − x 2 ) x dx R 2x 2 x4  R = π  −   2 4 0 π 4 = R . 4 Divide by the volume: x =

( πR4 )( 2π3R ) = 83 R. 4

By symmetry, the y- and z-coordinates of the centroid are zero.

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Problem 7.64 The volume consists of a segment of a sphere of radius R. Determine its centroid.

R 2

Solution:

The volume: The element of volume is a disk of radius ρ and thickness dx. The area of the disk is πρ 2 , and the element of volume is πρ 2 dx. From the equation of the surface of a sphere (see solution to Problem 7.63) ρ 2 = R 2 − x 2 , from which the element of volume is dV = π ( R 2 − x 2 ) dx. Thus

V = ∫ dV = π ∫ V

R R/2

x3 R

( R 2 − x 2 ) dx

R – 2 x

( )

5π 3  = π  R 2 x − = R .   3  R/2 24 R

The x -coordinate : R

∫V x dV = π ∫ R / 2 ( R 2 − x 2 ) x dx R 2x 2 x4  R 9π 4 = π  − = R .  2 4  R / 2 64 Divide by the volume: x =

27 R = 0.675 R. ( 9π64R )( 5π24R ) = 40 4

By symmetry the y- and z-coordinates are zero.

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Problem 7.65 A volume of revolution is obtained by revolving the curve x 2 /a 2 + y 2 / b 2 = 1 about the x-axis. Determine its centroid.

x2 y2 1 2 51 a2 b

z x

Solution:

The volume: The element of volume is a disk of radius y and thickness dx. The area of the disk is π y 2 . From the equation for the surface of the ellipse,

(

π y 2 = πb 2 1 −

x2 a2

y x2 1 y2 5 1 a2 b2

) (

and dV = π y 2 dx = πb 2 1 −

)

x2 dx, a2

from which V = ∫ dV = πb 2 ∫ V

a 0

(1 − ax ) dx 2 2

x3 a 2πb 2 a = πb 2  x − . =  3a 2  0 3 The x -coordinate :

∫V x dV = πb 2 ∫0 (1 − a 2 ) x dx a

x2 x4 a πb 2 a 2 . = πb 2  − =  2  2 4a  0 4 Divide by volume: x =

( πb4a )( 2π3b a ) = ( 83 )a. 2

By symmetry, the y- and z-coordinates of the centroid are zero.

Problem 7.66 In Example 7.9, determine the y coordinate of the centroid of the line.

Solution:

The expression derived in Example 7.9 for the element dL of the line in terms of x is dL =

1 + 4 x 2 dx

The y coordinate of the centroid is 1

∫ y dL = ∫0 x 2 1 + 4 x 2 dx = 0.410 y = L 1 ∫ L dL ∫0 1 + 4 x 2 dx y = 0.410

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Problem 7.67 Determine the coordinates of the centroid of the line.

y 5 x2

Solution: dy 2

∫ x ds = ∫−1 x 1 + ( dx ) dx = ∫−1 x 1 + (2 x) 2 dx x = −12 2 2 dy 2 1 + (2 x ) 2 dx ∫ ∫−1 ds + 1 dx ( ) − 1 ∫−1 2

dy 2

∫ y ds = ∫−1 y 1 + ( dx ) dx = ∫−1 x 2 1 + (2 x) 2 dx y = −12 2 2 dy 2 ∫−1 1 + (2 x) 2 dx ∫−1 ds ∫−1 1 + ( ) dx 2

x = 0.801 y = 1.482

Problem 7.68 Determine the x coordinate of the centroid of the line.

Solution: length is dL =

The length: Noting that

dy = ( x − 1) 1/ 2 , the element of dx

( dydx ) dx = x dx

from which 5

5 2 L = ∫ dL = ∫ ( x ) 1/ 2 dx =  ( x ) 3/ 2  = 6.7869. L 1 3 1

2 y 5 (x 2 1)3/2 3

The x -coordinate : 2



∫ L x dL = ∫0 x 3/ 2 dx =  5 x 5/ 2  1 = 21.961. Divide by the length: x = 0

21.961 = 3.2357 6.7869

Problem 7.69 Determine the x coordinate of the centroid of the line. y

Solution:

The length: Noting that

dL =

( dydx ) dx = 1 + x dx

dy = x 1/ 2 the element of length is dx

from which 2

2 2 L = ∫ dL = ∫ (1 + x ) 1/ 2 dx =  (1 + x ) 3/ 2  = 2.7974 L 0 3 0

2 y 5 x3/2 3

The x -coordinate : 2

 (1 + x ) 5/ 2 − (1 + x ) 3/ 2   5 3 0

∫ L x dL = ∫0 x(1 + x)1/ 2 dx = 2 

( ) ( )

3 5/ 2 3 3/ 2 1 1  = 2  − − +  = 3.0379.  5 3 5 3  0

506

Divide by the length: x = 1.086

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Problem 7.70 Use the method described in Example 7.10 to determine the centroid of the circular arc.

Solution:

The length of the differential line element of the circular arc is dL = Rd θ. The coordinates of the centroid are α

x =

∫ L x dL = ∫0 ( R cos θ) R d θ = R sin α α α ∫ L dL ∫0 R d θ α

∫ y dL = ∫0 ( R sin θ) R d θ = R (1 − cos α) y = L α α ∫ L dL ∫0 R d θ a

Thus x =

R(1 − cos α) R sin α ,y = α α

x R

Problem 7.71 In Practice Example 7.11, suppose that the cylinder is hollow with inner radius R/2 as shown. If the dimensions R = 6 in, h = 12 in, and b = 10 in, what is the x coordinate of the centroid of the volume?

Solution:

Let the cone be volume 1, let the solid cylinder be volume 2, and let the cylindrical hole be volume 3. The volumes and the x coordinates of the their centroids are

V1 = 13 π R 2 h,

x1 = 43 h,

V2 = π R 2 b,

x 2 = h + 12 b, 2

V3 = −π ( 12 R ) b,

h b R 2

x 3 = h + 12 b

The x coordinate of the centroid of the composite volume is x =

x1V1 + x 2V3 + x 3V3 V1 + V2 + V3

( 3 h )( 13 πR h ) + ( h + 12 b )(πR b) + ( h + 12 b )( −π ( 12 R ) b ) = 4 1 1 πR h + πR b − π( R ) b 3 2 2

Substituting the values for R, h, and b, we have x = 14.2 in.

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Problem 7.72 Use the procedure described in Example 7.12 to determine the x component of the centroid of the volume.

25 mm

x z

Solution:

Let the rectangular part without the cutout be volume 1, let the semicylindrical part be volume 2, and let the cylindrical hole be volume 3. The volumes and the x coordinates of their centroids are V1 = (60) (50) (20) mm 3 ,

x1 = 30 mm,

1 V2 = π (25) 2 (20) mm 3 , 2

4(25) x 2 = 60 + mm, 3π

V3 = −π (10) 2 (20) mm 3 ,

x 3 = 60 mm.

10 mm 20 mm

60 mm

The x coordinate of the centroid of the composite volume is x =

x1V1 + x 2V2 + x 3V3 V1 + V2 + V3 4(25)   1 (30) [(60) (50) (20)] +  60 + π(25) 2 (20)  + (60) [−π(10) 2 (20)]   3π   2 1 (60) (50) (20) + π(25) 2 (20) − π(10) 2 (20) 2

= 38.3 mm. x = 38.3 mm

Problem 7.73 Determine the centroids of the volumes.

Solution: The object will be divided into a cone and a hemisphere. From symmetry y = z = 0 Using tables we have in the x direction

(

) (

3R 1 3 (4 R) π R 2 [4 R] + 4 R + 8 3 x = 4 1 2π R 3 π R 2 [4 R] + 3 3

)( 2π3R ) = 83R 3

508

In summary x =

83 R , y = 0, z = 0 24

x 4R

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Problem 7.74 Determine the centroids of the volumes.

Solution:

We have a hemisphere and a hemispherical hole. From symmetry y = z = 0

200 mm

In the x direction we have

( − 3[3008 mm] )( 2π[3003 mm] ) 3[200 mm] 2π[200 mm] − (− )( ) 8 3 x = 3

300 mm

2π[300 mm] 3 3

−

2π[200 mm] 3 3

We have

x = −128 mm, y = 0, z = 0

Problem 7.75 Determine the centroids of the volumes.

Solution:

This is a composite shape. Let us consider a solid cylinder and then subtract the cone. Use information from the appendix Volume

Volume (mm 3 )

x (mm)

πR 2 L 1 2 3 πr h

1.1706 × 10 7 1.3572 × 10 6

L/2 L -h /4

230 370

Cylinder Cone R = 90 mm

L = 460 mm

z 60 mm 90 mm 360 mm

460 mm

r = 60 mm h = 360 mm x =

X CyLVCyL − X CONE VCONE VCyL − VCONE

x = 211.6 mm y = z = 0 mm

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Problem 7.76 Determine the centroids of the volumes.

20 mm 25 mm

Solution:

75 mm

Break the composite object into simple shapes, find the volumes and centroids of each, and combine to find the required centroid. Object

Volume (V)

LWH

H /2

L/2

hWD

( H + h /2)

D/2

π R 2 D/2

D/2

πr 2 D

( H + h + 43πR ) ( H + h)

D/2

120 mm

where R = W /2. For the composite,

x 1V1 + x 2V2 + x 3V3 − x 4V4 V1 + V2 + V3 − V4 (H) 25 mm

with similar eqns for y and z The dimensions, from the figure, are L = 120 mm

+ (L)

W = 100 mm

100

mm (W)

H = 25 mm

r = 20 mm h = 75 mm

1 +

D = 25 mm

) m (h

R = 50 mm Object

V mm 3

x (mm)

y (mm)

z (mm)

300000

12.5

187500

62.5

12.5

98175

121.2

12.5

−4

31416

100

12.5

–

x =

25 mm 100 mm

100

D) m(

m 25

m 75

Substituting into the formulas for the composite, we get x = 0 y = 43.7 mm z = 38.2 mm

2 H

y z r = 20 mm x 4 z

510

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Problem 7.77 Determine the centroids of the volumes.

y 1 in

1.75 in

Solution:

Divide the object into six volumes: (1) A cylinder 5 in long of radius 1.75 in, (2) a cylinder 5 in long of radius 1 in, (3) a block 4 in long, 1 in thick, and 2(1.75) = 3.5 in wide. (4) Semicylinder 1 in long with a radius of 1.75 in, (5) a semi-cylinder 1 in long with a radius of 1.75 in. (6) The composite object. The volumes and centroids are: Volume V1

Vol, cu in 48.1

x, in

y, in

z, in

2.5

15.7

2.5

0.5

4.81

0.743

0.5

4.81

4.743

The composite volume is V = V1 − V2 + V3 − V4 + V5 = The composite centroid: x =

V1x 1 − V2 x 2 + V3x 3 − V4 x 4 + V5x 5 = 1.02 in, V

y =

V1 y1 − V2 y 2 + V3 y 3 − V4 y 4 + V5y 5 = 1.9 in, V

5 in

z 4 in

1 in x

1 in

1.75 in

46.4 in 3.

z z

x 5 in

4 in y 1 in x

z = 0

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Problem 7.78 Determine the centroids of the volumes.

30 mm 60 mm

x 180 mm

Solution:

Consider the composite volume as being made up of three volumes, a cylinder, a large cone, and a smaller cone which is removed

Object

Cylinder

πr 2 L/2

L/4

1 πR 2 L 3 1 2 L πr 3 2

Cone 1

( )

Cone 2

180 mm

3 L /4

L/2

3( L /2)/4 y

(mm 3 )

(mm)

Cylinder

5.089 × 10 5

Cone 1

1.357 × 10 6

270

Cylinder

Cone 2

1.696 × 10 5

135

60 mm x

L = 360 mm y

r = 30 mm

360 mm cone

120 mm

R = 60 mm

For the composite shape x CylVCyL + x 1V1 − x 2V2 x = VCyL + V1 − V2

x = 229.5 mm

cone –

60 mm 3

180 mm

512

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Problem 7.79 The dimensions of the Gemini spacecraft (in meters) were a = 0.70, b = 0.88, c = 0.74, d = 0.98, e = 1.82, f = 2.20, g = 2.24, and h = 2.98. Determine the centroid of its volume.

g e

Solution:

The spacecraft volume consists of three truncated cones and a cylinder. Consider the truncated cone of length L with radii at the ends R1 and R 2 , where R 2 > R1 . Choose the origin of the x – y coordinate system at smaller end. The radius of the cone is a linear function of the length; from geometry, the length of the cone before truncations was (1) H = (2)

πR 22 H . The length of the truncated portion is 3

π R12 η . The volume of the truncated cone is the difference of the 3 two volumes, π L  R 23 − R13  . The centroid of the removed part of the  3  R 2 − R1  cone is

(5) V =

( )

3 (6) x η = η, and the centroid of the complete cone is 4 (7) x h =

( 43 ) H , measured from the pointed end. From the compos-

ite theorem, the centroid of the truncated cone is Vh x h − Vη x η − η + x, where x is the x-coordinate of the V left hand edge of the truncated cone in the specific coordinate system. These eight equations are the algorithm for the determination of the volumes and centroids of the truncated cones forming the spacecraft.

(8) x =

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R2 L with volume ( R 2 − R1 )

R1 L (3) η = with volume ( R 2 − R1 ) (4)

Beginning from the left, the volumes are (1) a truncated cone, (2) a cylinder, (3) a truncated cone, and (4) a truncated cone. The algorithm and the data for these volumes were entered into TK Solver Plus and the volumes and centroids determined. The volumes and x-coordinates of the centroids are: Volume

Vol, cu m

x, m

0.4922

0.4884

0.5582

1.25

3.7910

2.752

11.8907

4.8716

Composite

16.732

3.999

The last row is the composite volume and x-coordinate of the centroid of the composite volume. The total length of the spacecraft is 5.68 m, so the centroid of the volume lies at about 69% of the length as measured from the left end of the spacecraft. Discussion: The algorithm for determining the centroid of a system of truncated cones may be readily understood if it is implemented for a cone of known dimensions divided into sections, and the results compared with the known answer. Alternate algorithms (e.g. a Pappus-Guldinus algorithm) are useful for checking but arguably do not simplify the computations End discussion.

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Problem 7.80 Two views of a machine element are shown. Determine the centroid of its volume.

y 24 mm

8 mm

Solution:

We divide the volume into six parts as shown. Parts 3 and 6 are the “holes”, which each have a radius of 8 mm. The volumes are

18 mm 8 mm x z

V1 = (60)(48)(50) = 144,000 mm 3 , 20 mm 50 mm

V2 = 12 Π(24) 2 (50) = 45,239 mm 3 , V3 = Π(8) 2 (50) = 10,053 mm 3 , V4 = (16)(36)(20) = 11,520 mm 3 ,

16 mm

V5 = 12 Π(18) 2 (20) = 10,179 mm 3 , V6 = Π(8) 2 (20) = 4021 mm 3 .

The coordinates of the centroids are

x 1 = 25 mm, y 1 = 30 mm,

z 1 = 0,

x 2 = 25 mm, y 2 = 60 +

60 mm

4(24) = 70.2 mm, 3Π

4(18) = 47.6 mm, 3Π

z 2 = 0,

z 5 = 24 + 16 +

x 3 = 25 mm,

x 6 = 10 mm,

y 3 = 60 mm,

y 6 = 18 mm,

z 3 = 0,

z 6 = 24 + 16 = 40 mm.

x 4 = 10 mm,

The x coordinate of the centroid is

y 4 = 18 mm, x =

z 4 = 24 + 8 = 32 mm, x 5 = 10 mm,

x 1V1 + x 2V2 − x 3V3 + x 4V4 + x 5V5 − x 6V6 = 23.65 mm. V1 + V2 − V3 + V4 + V5 − V6

Calculating the y and z coordinates in the same way, we obtain y = 36.63 mm and z = 3.52 mm

y 5 = 18 mm,

Problem 7.81 In Example 7.13, suppose that the circular arc is replaced by a straight line as shown. Determine the centroid of the three-segment line. Solution:

Let the new straight-line segment be line 1 and let the segment in the x -z plane be line 2. Let the other line segment be line 3. The centroid locations of the parts and their lengths are x1 = 0,

y1 = 1 m,

z 1 = 1 m,

L1 = 2.83 m,

x 2 = 2 m,

y 2 = 1 m,

z 2 = 2 m,

L 2 = 4 m,

x 3 = 2 m,

y 3 = 1 m,

z 3 = 1 m,

L 2 = 4.90 m.

y (0, 2, 0) m

(0, 0, 2) m z

x (4, 0, 2) m

Applying Eqs. (7.18) yields x = 1.52 m,

514

y = 0.659 m, z = 1.34 m

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Problem 7.82 Determine the centroids of the line.

Solution:

The object is divided into two lines and a composite.

(1) L1 = 6 m, x 1 = 3 m, y 1 = 0. (2) L 2 = 3π m, x 2 = 6 +

6 m (Note: See Example 7.13) y 2 = 3. π

(3) The composite length: L = 6 + 3π m. The composite centroid: x =

L1x 1 + L 2 x 2 = 6 m, L

y =

3π = 1.83 m 2+ π

Problem 7.83 Determine the centroids of the line.

Solution:

Break the composite line into three parts (the quarter circle and two straight line segments) (see Appendix B). xi

Part 1

2 R /π

π R/2

Part 2

Part 3

( R = 2 m)

x 1 L1 + x 2 L 2 + x 3 L 3 = 1.4 m L1 + L 2 + L 3 y L + y 2L2 + y3L3 y = 1 1 = 1.4 m L1 + L 2 + L 3 x =

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x 2m

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Problem 7.84 The semicircular part of the line lies in the x–z plane. Determine the centroid of the line.

Solution: y

y 100 mm

100 mm

160 mm

120 mm

160 mm x

120 mm

The bar is divided into three segments plus the composite. The lengths and the centroids are given in the table: The composite length is:

L = ∑L i . i =1

The composite coordinates are: 3

∑L i x i

x = i =1

∑L i y i

and y = i = 1

Segment

Length, mm

x, mm

y, mm

z, mm

120π

240 π

120

100

188.7

Composite

665.7

65.9

21.7

68.0

Problem 7.85 Determine the centroid of the line. y

200 mm 608

Solution:

x =

y =

Break into a straight line and an arc.

1 2 2 (200 mm tan 60°) cos30 ° +

2 π /3

∫0

200 mm tan 60° + ∫

(200 mm) 2 2 π /3

1 2 (200 mm tan 60°)(200 mm sin 60°) +

200 mm tan 60° + ∫

2 π /3 0

1 + cos θ ) d θ ( cos60 ° = 332 mm

(200 mm) d θ 2 π /3

∫0

(200 mm) 2 (sin θ) d θ

(200 mm) d θ

= 118 mm

x = 332 mm, y = 118 mm

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Problem 7.86 Use the method described in Practice Example 7.14 to determine the area of the curved part of the surface of the truncated cone.

Solution:

Work with the solid line shown. The surface area is

given by A = 2π yL = 2π

y A =

( 34R ) ( R2 ) + ( h2 ) 2

3π R h2 + R2 4

R h 2

x h 2

Problem 7.87 Use the second Pappus–Guldinus theorem to determine the volume of the truncated cone.

Solution:

Work with the trapezoidal area

3 Rh 8 ( R /2)(h /2)( R /4) + (1/2)( R /2)(h /2)[(1/3)( R /2) + ( R /2)] 7R y = = 18 A 7 R 3 Rh 7π R 2 h = V = 2π yA = 2π 18 8 24 7π R 2 h V = 24 A = ( R /2)(h /2) + (1/2)( R /2)(h /2) =

( )(

)

R h 2

x R/2 h 2

h/2

Problem 7.88 The area of the shaded semicircle is 12 πR 2 . The volume of a sphere is 43 πR 3 . Extend the approach described in Example 7.15 to the second Pappus–Guldinus theorem and determine the centroid y S of the semicircular area.

The semicrcular area is A = 12 π R 2 , and y s is the y coordinate of its centroid. Rotating the area about the x-axis generates the volume of a sphere. The second Pappus–Guldinus theorem states that the volume of the sphere is V = 2π y s A = 2π y s ( 12 π R 2 )

Solving for y s yields y s =

_ yS

h/4

Solution:

4 3 3 πR

h/4

4R 3π

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Problem 7.89 Use the second Pappus–Guldinus theorem to determine the volume generated by revolving the curve about the y -axis. y

Solution:

The x coordinate of the centroid:. The element of area is the vertical strip of height (1 − y ) and width dx. Thus 1

A = ∫ (1 − y ) dx = ∫ (1 − x 2 ) dx. Integrating,

(1, 1)

x3 1 2 A =  x −  = .  3 0 3  x2

x 4 1

∫ A x dA = ∫0 ( x − x 3 ) dx =  2 − 4  0 = 4 , y 5 x2

divide by the area: x =

3 π . The volume is V = 2πx A = 2 8

Problem 7.90 The length of the curve is L = 1.479, and the area generated by rotating it about the x-axis is A = 3.810. Use the first Pappus–Guldinus theorem to determine the y coordinate of the centroid of the curve.

Solution:

The surface area is A = 2πy L, from which

A y = = 0.41 2π L

y (1, 1)

y 5 x2

Problem 7.91 Use the first Pappus–Guldinus theorem to determine the area of the surface generated by revolving the curve about the y -axis.

Solution: The length of the line is given in Problem 7.90. L = 1.479. The elementary length of the curve is 1+

Noting

dy = 2 x, the element of line is dL = (1 + 4 x 2 ) 1/2 . dx

y (1, 1)

( dydx ) dx.

dL =

The x -coordinate: 1

∫ L x dL = ∫0 x(1 + 4 x 2 )1/ 2 dx y5x

518

1 1  5 3/ 2 − 1 = 0.8484. (1 + 4 x 2 ) 3/ 2  = 0 12  12

Divide by the length to obtain x = 0.5736. The surface area is A = 2πx L = 5.33

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Problem 7.92 The equation describing the profile of the rocket nose cone is shown. Use the second PappusGuldinus theorem to determine the nose cone’s volume.

Solution:

The area A that can be rotated about the axis of symmetry of the nose cone to generate its volume is shown. y

dA y 5 19 x2

The width of the horizontal strip is 3 y 1/ 2 , so its area is dA = 3 y 1/ 2 dy. Integrating to determine the area A, we obtain 1m

y 5 1 x2 9 0

∫ A dA = ∫0

1  3 3 y 2 dy =  2 y 2   0

= 2 m 2.

The y coordinate of the centroid of the area A is

5 1m

∫ y dA = ∫0 y ( 3y 2 dy ) =  5 y 2  0 y = A 2 m2 2 m2 ∫ A dA 1m

6



= 0.6 m.

1.0 m 0.6 m x

The distance from the axis of symmetry to the centroid is 1 m − 0.6 m = 0.4 m. Applying the second Pappus-Guldinus theorem, the volume of the nose cone is Volume = 2π (0.4 m) A = 2π (0.4 m)(2 m 2 ) = 5.03 m 3 . Volume = 5.03 m 3 .

Problem 7.93 The coordinates of the centroid of the line are x = 332 mm and y = 118 mm. Use the first Pappus–Guldinus theorem to determine the area of the surface of revolution obtained by revolving the line about the x-axis.

Solution: L = 200 mm tan 60 ° + 200 mm

° π = 765 mm ( 120 180 ° )

A = 2π yL = 2π (0.118 m)(0.765 m) = 0.567 m 2

200 mm 608

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Problem 7.94 The coordinates of the centroid of the area between the x-axis and the line are x = 355 mm and y = 78.4 mm. Use the second Pappus–Guldinus theorem to determine the volume obtained by revolving the area about the x-axis.

200 mm

Solution: A =

608

The area is

(

)

1 120 ° (0.2 m)(0.2 m tan 60 °) + π (0.2 m) 2 = .0765 m 2 2 360 °

V = 2π yA = 2π (0.0784 m)(0.0765 m 2 ) = 0.0377 m 3

Problem 7.95 The volume of revolution contains a hole of radius R. (a) Use integration to determine its volume. (b) Use the second Pappus–Guldinus theorem to determine its volume.

R1a R

Solution: (a) The element of volume is a disk of radius y and thickness dx. The area of the disk is π ( y 2 − R 2 ). The radius is y =

( ah ) x + R,

from which dV = π Denote m =

( ah x + R ) dx − πR dx. 2

( ah ), dV = π(m x + 2mRx) dx, 2

from which h

V = ∫ dV = πm ∫ (mx 2 + 2 Rx ) dx V

h x3 mh = πm  m + Rx 2  = πmh 2 + R  3 0 3

(

= πah

)

( a3 + R ).

(b) The area of the triangle is A =

( 12 )ah. The y-coordinate of the

centroid is y = R + ( 13 ) a. The volume is V = 2πy A = πah ( R + ( 13 ) a )

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Problem 7.96 Determine the volume of the volume of revolution.

Solution:

The area of the semicircle is A =

4r . The volume is 3π πr 2 4r 4r V = 2π = π 2r 2 R + . R+ 2 3π 3π

πr 2 . The centroid is 2

y = R+

( )(

)

(

)

For r = 40 mm and R = 140 mm, V = 2.48 × 10 −3 m 3

140 mm

80 mm

Problem 7.97 Determine the surface area of the volume of revolution. Solution:

The length and centroid of the semicircle is L o = πr, 2r . The length and centroid of the inner line is L i = 2r , y = R+ π and y = R.

(

140 mm

)

2r A = 2π (πr ) R + + 2π (2r )( R) = 2πr (π R + 2r + 2 R). π For r = 40 mm and R = 140 mm, A = 0.201 m 2 80 mm

Problem 7.98 The volume of revolution has an elliptical cross section. Determine its volume.

Solution:

Use the second theorem of Pappus-Guldinus. The centroid of the ellipse is 180 mm from the axis of rotation. The area of the ellipse is πab where a = 115 mm, b = 65 mm.

The centroid moves through a distance d = 2π R = 2π (180 mm) as the ellipse is rotated about the axis.

230 mm

V = Ad = πabd = 2.66 × 10 7 mm 3 130 mm

v = 0.0266 m 3

2b 180 mm 2a

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Problem 7.99 Suppose that the bar in Practice Example 7.16 is replaced with this 100-kg homogeneous bar. (a) What is the x coordinate of the bar’s center of mass? (b) Determine the reactions at A and B.

0.5 m

1m A

x 1m

Solution: (a) Let the new horizontal segment of the bar part 3. The x coordinate of the centroid of the bar’s axis, which coincides with its center of mass is x =

x 1 L1 + x 2 L 2 + x 3 L 3 (0.5)(1) + (1)(1) + (0.75)(0.5) = = 0.75 m L1 + L 2 + L 3 0.5 + 1 + 0.5

x = 0.75 m (b) The equilibrium equations are ΣM A : B(1 m) − (981 N) (0.75 m) = 0, ΣFx : A x − B = 0, ΣFy : A y − 981 N = 0. Solving yields A x = 736 N, A y = 981 N, B = 736 N.

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Problem 7.100 The mass of the homogeneous flat plate is 50 kg. Determine the reactions at the supports A and B.

100 mm

400 mm 200 mm

A 600 mm 800 mm

600 mm

Solution:

Divide the object into three areas and the composite. Since the distance to the action line of the weight is the only item of importance, and since there is no horizontal component of the weight, it is unnecessary to determine any centroid coordinate other than the x-coordinate. The areas and the x-coordinate of the centroid are tabulated. The last row is the composite area and x-coordinate of the centroid. Area

A, sq mm

Rectangle

3.2 × 10 5

400

Circle

3.14 × 10 4

600

Triangle

1.2 × 10 5

1000

Composite

4.09 × 10 5

561

AX AY

500 N X

B 1400 mm

The composite area is A = Arect − Acirc + Atriang . The composite x-coordinate of the centroid is x =

Arect x rect − Acirc x circ + Atriang x triang . A

The sum of the moments about A: ΣM A = −500(561) + 1400 B = 0, from which B = 200 N. The sum of the forces: ΣFy = A y + B − 500 = 0, from which A y = 300 N. ΣFx = A x = 0

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Problem 7.101 The suspended sign is a homogeneous flat plate that has a mass of 130 kg. Determine the axial forces in members AD and CE. (Notice that the y -axis is positive downward.) A

The area: The element of area is the vertical strip of length y and width dx. The element of area dA = y dx = (1 + ax 2 ) dx, where a = 0.0625. Thus 4

4 ax 3  A = ∫ dA = ∫ (1 + ax 2 ) dx =  x + = 5.3333 sq ft. A 0  3  0

E B

Solution: The strategy is to determine the distance to the action line of the weight (x-coordinate of the centroid) from which to apply the equilibrium conditions to the method of sections.

The x -coordinate: 4

 x2

ax 4  4

∫ A x dA = ∫0 x(1 + ax 2 ) dx =  2 + 4  0 = 12. 12 = 2.25 ft. 5.3333 The equilibrium conditions: The angle of the member CE is

Divide A: x = y

y 5 1 1 0.0625x2

α = tan −1 ( 14 ) = 14.04 °. The weight of the sign is W = 130(9.81) = 1275.3 N. The sum of the moments about D is ΣM D = −2.25W + 4CE sin α = 0, from which CE = 2957.7 N (T ) . Method of sections: Make a cut through members AC , AD and BD and consider the section to the right. The angle of member AD is β = tan −1 ( 12 ) = 26.57 °. The section as a free body: The sum of the vertical forces: ΣFY = AD sin β − W = 0 from which AD = 2851.7 N (T )

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Problem 7.102 The bar has a mass of 80 kg. What are the reactions at A and B?

Solution:

Break the bar into two parts and find the masses and centers of masses of the two parts. The length of the bar is L = L1 + L 2 = 2 m + 2π R /4( R = 2 m) L = 2+ πm

Length i (m)

Part 2m 2m B

x i (m)

Mass i (kg)

( ) ( 2 +π π ) 80

2 80 2+ π

m1 = 31.12 kg

x1 = 1 m

m 2 = 48.88 kg

x 2 = 3.27 m

ΣFx :

Ax = 0

ΣFy :

A y + B y − m1 g − m 2 g = 0

( 2 + 2πR )

ΣM A : − x 1m1g − x 2 m 2 g + 4 B y = 0 Solving A x = 0, A y = 316 N, B = 469 N y

m1g m2g

AX X2 AY 4m

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Problem 7.103 The mass of the bar per unit length is 2 kg/m. Choose the dimension b so that part BC of the suspended bar is horizontal. What is the dimension b, and what are the resulting reactions on the bar at A?

Solution:

We must have

∑ M A : ρ g(1.0 m)(0.5 m cos30 °) − ρ gb

( b2 − 1.0 m cos30° ) = 0

⇒ b = 2.14 m Then ∑ Fx : A x = 0 ∑ Fy : A y − ρ g(1.0 m) − ρ g(b) = 0 ⇒

A x = 0, A y = 61.6 N, b = 2.14 m Ay

308

Ax b

rg(1.0 m) rgb

Problem 7.104 The semicircular part of the homogeneous slender bar lies in the x–z plane. Determine the center of mass of the bar.

Solution:

L = ∑L i .

The bar is divided into three segments plus the composite. The lengths and the centroids are given in the table: The composite length is: 3

i =1

The composite coordinates are: 3

∑L i x i

x = i =1

10 in 16 in

∑L i y i

12 in z

and y = i = 1 x

Length, in

x, in

y, in

12π

24 π

18.868

Composite

66.567

6.594

2.168

6.796

Segment

526

L z, in

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Problem 7.105 The density of the cone is given by the equation ρ = ρ 0 (1 + x /h), where ρ 0 is a constant. Use the procedure described in Example 7.17 to show that the mass of the cone is given by m = (7/4)ρ 0V , where V is the volume of the cone, and that the x coordinate of the center of mass of the cone is x = (27/35)h.

Solution: Consider an element of volume dV of the cone in the form of a “disk” of width dx. The radius of such a disk at position x is ( R /h) x, so dV = π[( R /h) x ] 2 dx. The mass of the cone is h

m = ∫ ρ dV = ∫ ρ 0 (1 + x /h) π[( R /h) x ] 2 dx = V

7 7 ρ π R 2 h = ρ 0V . 12 0 3

The x coordinate of the center of mass is y

∫ xρ dV = ∫0 xρ 0 (1 + x /h) π [( R /h) x] 2 dx = 27 h. x = V h 35 ∫V ρ dV ∫0 ρ 0 (1 + x /h) π [( R /h) x] 2 dx

R x h

Problem 7.106 A horizontal cone with 800-mm length and 200-mm radius has a fixed support at A. Its density is ρ = 6000(1 + 0.4 x 2 ) kg/m 3 , where x is in meters. What are the reactions at A?

y 200 mm x

y 800 mm 200 mm x

The x -coordinate of the mass center: 0.8

 x4

x 6  0.8

∫m x dm = 375π ∫0 (1 + ax 2 ) x 3 dx = 375π  4 + a 6  0 800 mm

= 141.23. Divide by the mass: x = 0.6089 m The equilibrium conditions: The sum of the moments about A:

Solution:

The strategy is to determine the distance to the line of action of the weight, from which to apply the equilibrium conditions. The mass: The element of volume is a disk of radius y and thickness dx. y varies linearly with x: y = 0.25 x. Denote a = 0.4. The mass of the disk is dm = ρπ y 2 dx = 6000 π (1 + ax 2 )(0.25 x ) 2 dx = 375π(1 + ax 2 ) x 2 dx,

from which x3 x 5 0.8 (1 + ax 2 ) x 2 dx = 375π  + a  0  3 5 0 = 231.95 kg

m = 375π ∫

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∑ M = M A − mgx = 0, from which M A = mgx = 231.94(9.81)(0.6089) = 1385.4 N-m . The sum of the vertical forces: ΣFY = AY − mg = 0 from which AY = 2275.4 N . The horizontal component of the reaction is zero, ΣFX = 0.

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Problem 7.107 In Practice Example 7.18, suppose that bar 1 is replaced by a bar with the same dimensions that consists of aluminum alloy with a density of 2600 kg/m 3 . Determine the x coordinate of the center of mass of the machine part.

Solution:

The mass of bar 1 is

m = (7.68 × 10 −4 m 3 ) (2600 kg/m 3 ) = 2.00 kg The x coordinate of the center of mass is x =

x1m1 + x 2 m 2 (40 mm) (2.00 kg) + (200 mm) (5.99 kg) = 2.00 kg + 5.99 kg m1 + m 2

= 160 mm x = 160 mm

Problem 7.108 The cylindrical tube is made of aluminum with density 2700 kg/m 3 . The cylindrical plug is made of steel with density 7800 kg/m 3 . Determine the coordinates of the center of mass of the composite object. y

Solution:

The volume of the aluminum tube is

V Al = π (0.035 2 − 0.02 2 )(0.2) = 5.18 × 10 −4 m 3 . The mass of the aluminum tube is m Al = (2700)V Al = 1.4 kg. The centroid of the aluminum tube is x AL = 0.1 m, y Al = z Al = 0. The volume of the steel plug is VFe = π (0.02) 2 (0.1) = 1.26 × 10 −4 m 3 . The mass of the steel plug is m Fe = (7800)VFe = 0.9802 kg. The centroid of the steel plug is x Fe = 0.15 m, y Fe = z Fe = 0. The composite mass is m = 2.38 kg. The composite centroid is

m Al (0.1) + m Fe (0.15) = 0.121 m m y = z = 0 x =

x y

Tube A

Plug

20 mm x

z 35 mm

100 mm

Section A-A

Problem 7.109 In Example 7.19, suppose that the object is redesigned so that the radius of the hole in the hollow cylinder is increased from 2 in to 3 in. What is the x coordinate of the center of mass of the object?

Solution:

The volume of the cylinder is

Vcylinder = (12 in)π [(4 in) 2 − (3 in) 2 ] = 264 in 3 = 0.153 ft 3 Its weight is Wcylinder = (0.153 ft 3 ) (530 lb/ft 3 ) = 80.9 lb. The x coordinate of the center of mass (same as center of weight) is x = =

x barW bar + x cylinderWcylinder W bar + Wcylinder (1.86 in) (15.6 lb) + (10 in) (80.9 lb) = 8.68 in. 15.6 lb + 80.9 lb

x = 8.68 in.

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Problem 7.110 A machine consists of three parts. The masses and the locations of the centers of mass of two of the parts are shown in the following table: Part

Mass (kg)

x (mm)

y (mm)

z (mm)

2.0

100

−20

4.5

150

Solution:

The composite mass is m = 2.0 + 4.5 + 2.5 = 9 kg. The location of the third part is 120(9) − 2(100) − 4.5(150) = 82 mm 2.5 80(9) − 2(50) − 4.5(70) y3 = = 122 mm 2.5 2(20) z3 = = 16 mm 2.5

x3 =

The mass of part 3 is 2.5 kg. The design engineer wants to position part 3 so that the center of mass location of the machine is x = 120 mm, y = 80 mm, z = 0. Determine the necessary position of the center of mass of part 3. Problem 7.111 Two views of a machine element are shown. Part 1 is aluminum alloy with density 2800 kg/m 3 , and part 2 is steel with density 7800 kg/m 3 . Determine the coordinates of its center of mass.

24 mm

= 179, 186 mm 3 = 17.92 × 10 −5 m 3 , V2 = [ (16)(36) + 12 π (18) 2 − π (8) 2 ] (20) = 17, 678 mm 3

m1 = S1V1 = (2800)(17.92 × 10 −5 ) = 0.502 kg, m 2 = S 2V2 = (7800)(1.77 × 10 −5 ) = 0.138 kg.

8 mm

18 mm

20 mm

V1 = [ (60)(48) + 12 π (24) 2 − π (8) 2 ] (50)

so their masses are

The volumes of the parts are

= 1.77 × 10 −5 m 3 ,

Solution:

60 mm

The x coordinates of the centers of mass of the parts are x 1 = 25 mm, x 2 = 10 mm, so x =

8 mm z

x 1m1 + x 2 m 2 = 21.8 mm m1 + m 2

16 mm

50 mm

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Problem 7.112 The loads F1 = F2 = 25 kN. The mass of the truss is 900 kg. The members of the truss are homogeneous bars with the same uniform cross section. (a) What is the x coordinate of the center of mass of the truss? (b) Determine the reactions at A and G.

F1 3m F2

B E

3m G

C 4m

Solution: (a) The center of mass of the truss is located at the centroid of the composite line of the axes of the members. The lengths of the diagonal members are (4 m) 2 + (3 m) 2 = 5 m. The lengths and x coordinates of the centroids of the axes of the members are Member

Length

x coordinate

The x coordinate of the centroid of the composite line, which is coincident with the center of mass of the truss, is x =

Σx i L i (2) (5 + 4) + (4) (3) + (6) (5 + 4 + 5 + 4) + (8) (3 + 3) = = 5.17 m ΣL i 5+ 4+3+ 5+ 4+ 5+ 4+3+3

x = 5.17 m (b) The equilibrium equations for the truss are ΣFx : A x + 25 kN + 25 kN, ΣFy : A y + G − (900)(9.81)N = 0, ΣM A : −(25 kN) (3 m) − (25 kN) (6 m) − (900) (9.81) N(5.17 m) + G (8 m) = 0. Solving yields A xc = −50 kN, A y = −25.0 kN, G = 33.8 kN.

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Problem 7.113 With its engine removed, the mass of the car is 1100 kg and its center of mass is at C. The mass of the engine is 220 kg. (a) Suppose that you want to place the center of mass E of the engine so that the center of mass of the car is midway between the front wheels A and the rear wheels B. What is the distance b? (b) If the car is parked on a 15 ° slope facing up the slope, what total normal force is exerted by the road on the rear wheels B?

Solution: (a) The composite mass is m = mC + m E = 1320 kg. The x-coordinate of the composite center of mass is given: x =

2.6 = 1.3 m, 2

from which the x-coordinate of the center of mass of the engine is xE = b =

(1.3 m − 1.14 mC ) = 2.1 m. mE

The y-coordinate of the composite center of mass is y =

(b) Assume that the engine has been placed in the new position, as given in Part (a). The sum of the moments about B is

E C 0.6 m

0.45 m A

1.14 m

0.45 mC + 0.6 m E = 0.475 m. m

ΣM A = 2.6 A + y mg sin(15 °) − (2.6 − x )mg cos(15 °) = 0, from which A = 5641.7 N. This is the normal force exerted by the road on A. The normal force exerted on B is obtained from;

b 2.60 m

ΣFN = A − mg cos(15 °) + B = 0, from which B = 6866 N

Problem 7.114 The airplane is parked with its landing gear resting on scales. The weights measured at A, B, and C are 30 kN, 140 kN, and 146 kN, respectively. After a crate is loaded onto the plane, the weights measured at A, B, and C are 31 kN, 142 kN, and 147 kN, respectively. Determine the mass and the x and y coordinates of the center of mass of the crate.

Solution:

The weight of the airplane is W A = 30 + 140 + 146 = 316 kN. The center of mass of the airplane: ΣM y -axis = 30(10) − x AW A = 0,

from which x A = 0.949 m. ΣM x -axis = (140 − 146)(6) + y AW A = 0, from which y A = 0.114 m. The weight of the loaded plane: W = 31 + 142 + 147 = 320 kN. The center of mass of the loaded plane: ΣM y-axis = (31)10 − x W = 0,

from which x = 0.969 m.

A x

from which y = 0.0938 m. The weight of the Wc = W − W A = 4 kN. The center of mass of the crate:

ΣM x-axis = (142 − 147)(6) + y W = 0,

10 m y

crate

Wx − W A x A = 2.5 m, Wc Wy − W A y A yc = = −1.5 m. Wc xc =

The mass of the crate: mc =

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Wc × 10 3 = 407.75 kg 9.81

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Problem 7.115 A suitcase with a mass of 90 kg is placed in the trunk of the car described in Example 7.20. The position of the center of mass of the suitcase is x s = −0.533 m, y s = 0.762 m, z s = −0.305 m. If the suitcase is regarded as part of the car, what is the new position of the car’s center of mass?

Solution: In Example 7.20, the following results were obtained for the car without the suitcase Wc = 17303 N x c = 1.651 m y c = 0.584 m z c = 0.769 m For the suitcase Ws = (90) g, y = 0.762 m,

x s = −0.533 m, z = −0.305 m.

The new center of mass is at xN =

x cW c + x s W s (Wc + Ws )

with similar eqns for y N and z N Solving, we get x N = 1.545 m, y N = 0.593 m, z N = 0.717 m

Problem 7.116 A group of engineering students constructs a miniature device of the kind described in Example 7.20 and uses it to determine the center of mass of a miniature vehicle. The data they obtain are shown in the following table:

Measured Loads (lb) α = 0

Left front wheel, N LF Right front wheel, N RF Left rear wheel, N LR Right rear wheel, N RR

ΣM z -axis = (Wheelbase)( N LF + N RF ) − x W = 0, from which x =

Wheelbase = 36 in Track = 30 in

Solution: The weight of the go-cart: W = 35 + 36 + 27 + 29 = 127 lb. The sum of the moments about the z-axis

35 36 27 29

α = 10 °

32 33 34 30

Determine the center of mass of the vehicle. Use the same coordinate system as in Example 7.20.

36(35 + 36) = 20.125 in. W

The sum of the moments about the x-axis: ΣM x -axis = zW − (Track )( N RF + N RR ) = 0, from which z =

(30)(36 + 29) = 15.354 in. W

With the go-cart in the tilted position, the sum of the moments about the z-axis ΣM z -axis = (Wheelbase)( N LF + N RF ) + y W sin(10 °) − x W cos(10 °) = 0, from which x W cos(10 °) − (36)(32 + 33) W sin(10 °) = 8.034 in.

y =

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Problem 7.117 Determine the centroid of the area by letting dA be a vertical strip of width dx.

Solution: The area: The length of the vertical strip is (1 − y), so that the elemental area is dA = (1 − y ) dx = (1 − x 2 ) dx. The area: 1



x3 1

∫ A dA = ∫0 (1 − x 2 ) dx =  x − 3  0 = 1 − 3 = 3 .

The x -coordinate :

(1, 1)

1 x2 x 4 1 1 3 x A = ∫ x dA = ∫ x (1 − x 2 ) dx =  − = : x = A 0  2 4  0 4 8

The y -coordinate: The y-coordinate of the centroid of each element of area is located at the midpoint of the vertical dimension of the area element.

y 5 x2 x

y = y + 12 (1 − x 2 ). Thus

∫ A y dA = ∫0 ( x 2 + ( 2 )(1 − x 2 ) ) (1 − x 2 ) dx 1

= y =

Problem 7.118 Determine the centroid of the area by letting dA be a horizontal strip of height dy.

5 1

( 12 )  x − x5  = 25 . 0

3 5

Solution: The area: The length of the horizontal strip is x, hence the element of area is dA = x dy = y 1/ 2 dy.

Thus (1, 1)

1 2 y 3/ 2  2 A = ∫ y 1/ 2 dy =   = 0 3  3 0

Check: The x -coordinate: The x-coordinate of the centroid of each element of area is x = 12 x = 12 y 1/ 2 . Thus

y 5 x2 x

∫ A( 2 ) y 1/ 2 dA = ( 2 ) ∫0 y dy = ( 2 )  2  0 = 4 . 1

1  y2 

Divide by the area: x =

3 8

The y -coordinate: 1

y A = ∫ y dA = ∫ y ( y 1/ 2 dy ) A

1 2  2 y 5/ 2  = ∫ y 3/ 2 dy =   = . 0 5  5 0 3 Divide by the area: y = 5

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Problem 7.119 Determine the centroid of the area.

Solution: The strategy is to develop useful general results for the triangle and the rectangle.

The rectangle: The area of the rectangle of height h and width w is w

A = ∫ h dx = hw = 4800 cm 2 . 0

The x -coordinate: 60 cm

∫0 hx dx = h  2  0 = ( 2 ) hw 2 .  x2 

x 80 cm

60 cm

w = 40 cm 2

Divide by the area: x = The y -coordinate:

( 12 ) ∫ h dx = ( 12 ) h w. w

Divide by the area: y =

( 12 ) h = 30 cm

The triangle: The area of the triangle of altitude a and base b is (assuming that the two sides a and b meet at the origin) b

A = ∫ y ( x ) dx = ∫

( − ab x + a ) dx =  − ax2b − ax  2

b 0

ab ab =  − + ab  = = 1800 cm 2  2  2 Check: This is the familiar result. check. The x -coordinate: ax 2  b

∫0 ( − b x + a ) x dx =  − 3b + 2  0 = b

 ax 3

Divide by the area: x =

ab 2 . 6

b = 20 cm 3

The y -coordinate:

∫ A y dA = ( 2 ) ∫0 ( − b x + a ) dx 1

= −

(

)

3  b  a ba 2 .  − x + a  =  0 6a  b 6

Divide by the area: y = x =

a 20 cm. The composite: 3

x R A R + x T AT 40(4800) + 100(1800) = A R + AT 4800 + 1800

= 56.36 cm (30)(4800) + (20)(1800) y = 4800 + 1800 = 27.27 cm

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Problem 7.120 Determine the centroid of the area.

Solution:

20 mm

40 mm

Divide the object into five areas:

(1)

The rectangle 80 mm by 80 mm,

(2)

The rectangle 120 mm by 80 mm,

(3)

the semicircle of radius 40 mm,

(4)

The circle of 20 mm radius, and

(5)

the composite object. The areas and centroids:

(1)

A1 = 6400 mm 2 , x 1 = 40 mm, y 1 = 40 mm,

(2)

A2 = 9600 mm 2 ,

40 mm

80 mm

x 2 = 120 mm, y 2 = 60 mm, x

(3)

120 mm

A3 = 2513.3 mm 2 , x 3 = 120 mm, y 3 = 136.98 mm,

160 mm

(4)

A4 = 1256.6 mm 2 , x 4 = 120 mm, y 4 = 120 mm.

(5)

he composite area: T A = A1 + A2 + A3 − A4 = 17256.6 mm 2 . The composite centroid:

Problem 7.121 The cantilever beam is subjected to a triangular distributed load. What are the reactions at A?

x =

A1x 1 + A2 x 2 + A3x 3 − A4 x 4 = 90.3 mm . A

y =

A1y 1 + A2 y 2 + A3y 3 − A4 y 4 = 59.4 mm A

Solution: 200 N/m

AX AY

10 m

200 N/m

x 10 m

The load distribution is a straight line with intercept w = 200 N/m at 200 x = 0, and slope − = −20 N/m 2 . The sum of the moments is 10

(

ΣM = M A − ∫

10 0

)

(−20 x + 200) x dx = 0,

from which 10 20 M A =  − x 3 + 100 x 2  = 3333.3 N-m.  3 0

The sum of the forces: ΣFy = A y − ∫

10 0

(−20 x + 200) dx = 0,

from which A y = [−10 x 2 + 200 x ]10 0 = 1000 N, and ΣFx = A x = 0

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Problem 7.122 What is the axial load in member BD of the frame?

Solution:

The distributed load is two straight lines: Over the interval 0 ≤ y ≤ 5 the intercept is w = 0 at y = 0 and the slope is 100 = 20. 5 Over the interval 5 ≤ y ≤ 10, the load is a constant w = 100 N/m. The moment about the origin E due to the load is +

100 N/m

M E = ∫ (20 y ) y dy + ∫ 100 y dy,

from which 5

20 100 2  M E =  y 3  +  y  = 4583.33 N-m.  3 0  2 5 Check: The area of the triangle is

5m E

A 10 m

F1 = ( 12 ) (5)(100) = 250 N. The area of the rectangle: F2 = 500 N. The centroid distance for the triangle is d 1 = ( 23 ) 5 = 3.333 m. The centroid distance of the rectangle is d 2 = 7.5 m. The moment about E is M E = d 1F1 + d 2 F2 = 4583.33 Nm check. The Complete Structure: The sum of the moments about E is ΣM = −10 A R + M E = 0, where A R is the reaction at A, from which A R = 458.33 N. The element ABC: Element BD is a two force member, hence B y = 0. The sum of the moments about C: ΣM C = −5B x − 10 A y = 0, where A y is equal and opposite to the reaction of the support, from which B x = −2 A y = 2 A R = 916.67 N. Since the reaction in element BD is equal and opposite, B x = −916.67 N, which is a tension in BD. Cy

By Bx Ay

Cy Bx By

Dx Dy

Dy Dx Ey Ex

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Problem 7.123 An engineer estimates that the maximum wind load on the 40-m tower in Fig. a is described by the distributed load in Fig. b. The tower is supported by three cables, A, B, and C, from the top of the tower to equally spaced points 15 m from the bottom of the tower (Fig. c). If the wind blows from the west and cables B and C are slack, what is the tension in cable A? (Model the base of the tower as a ball-and-socket support.) 200 N/m

(a)

400 N/m (b)

40 0

(−5 y + 400) y dy,

40 5 M W =  − y 3 + 200 y 2  = 213.33 kN-m,  3 0

clockwise about the base, looking North. The angle formed by the cable with the horizontal at the top of the tower is

( 15 ) = 69.44°. 40

ΣM = −M W + 40T A cos θ = 0, from which

15 m

B, C

MW = ∫

The sum of the moments about the base of the tower is N

The load distribution is a straight line with the intercept w = 400 N/m, and slope −5. The moment about the base of the tower due to the wind load is

θ = 90 ° − tan −1 B

40 m

Solution:

TA =

1 ( 40 cos ) M = 15.19 kN θ W

200 N/m

(c)

u 40 m

Fx Fy

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Problem 7.124 Determine the reactions on member ABCD at A and D. 2 kN/m

2 kN/m D

ΣFx : − D x − FCE cos 45 ° = 0 ΣFy :

− D y − FCE sin 45 ° − 4 = 0

ΣM E : (1) D y = 0 Solving, we get

For DE:

A x = 7 kN A y = −6 kN

D x = 4 kN Dy = 0

also FBF = −14.14 kN(c)

FCE = −5.66 kN(c)

1m A

DX 1m FCE

Solution:

458

3 kN

2 kN / m y

FBF

458

AX AY

x 1

4 kN 2 kN / m 1m 2m First, replace the distributed forces with equivalent concentrated forces, then solve for the loads. Note that BF and CE are two force members. Distributed Load on ABCD, F1 By area analogy, concentrated load is applied at y = ±m. The load is 1 2 (2)(3) kN F1 = 3 kN

DX E DY

458 FCE 2

By the area analogy, F2 = 4 kN applied at x = 1 m Assume FCE and FBF are tensions For ABCD:

538

ΣFx :

A x + FBF cos 45 ° + FCE cos 45 ° + D x + 3 kN = 0

ΣFy :

A y + D y − FBF sin 45 ° + FCE sin 45 ° = 0

ΣM A :

−1( FBF cos 45°) − 2( FCE cos 45°) − 2(3) − 3 D x = 0

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Problem 7.125 Estimate the centroid of the volume of the Apollo lunar return configuration (not including its rocket nozzle) by treating it as a cone and a cylinder.

Solution: V1 =

The volume of the cone is

π R 2h = 428.93 ft 3 . 3

x-coordinate of the centroid from the nose of the 3h cone is x 1 = = 7.5 ft. The volume of the cylinder is 4 2 V2 = π R L = 1801.5 ft 3 . The x-coordinate of the centroid from the L nose of the cone is x 2 = h + = 17 ft. The composite volume is 2 V = V1 + V2 = 2230.4 ft 3 . The x-coordinate of the composite centroid is The

12.8 ft

x = Nozzle 10 ft

V1 x1 + V2 x 2 = 15.2 ft . V

The y- and z-coordinates are zero, from symmetry.

14 ft

Problem 7.126 The shape of the rocket nozzle of the Apollo lunar return configuration is approximated by revolving the curve shown around the x-axis. In terms of the coordinate system shown, determine the centroid of the volume of the nozzle.

Solution: x =

∫ x dV . ∫ dV

Let dV be a disk of radius y and thickness dx Thus, dV = π y 2 dx, where

y y 5 0.350 1 0.435x 2 0.035x2

y 2 = (0.350 + 0.435 x − 0.035 x 2 ) 2 y 2 = (a + bx + cx 2 ) 2 y 2 = a 2 + 2abx + (2ac + b 2 ) x 2 + 2bcx 3 + c 2 x 4 a = 0.350 b = 0.435 x

c = −0.035 2.83

∫ (a 2 x + 2abx 2 + (2ac + b 2 ) x 3 + 2bcx 4 + c 2 x 5 ) dx x = 0 2.83 ∫0 (a 2 + 2abx + (2ac + b 2 ) x 2 + 2bcx 3 + c 2 x 4 ) dx 2.83 m

2.83  a 2 x 2 + 2ab x 3 + (2ac + b 2 ) x 4   2 3 3     5 6   + 2bc x + c 2 x  0  5 6 x = 2.83  a 2 x + 2ab x 2 + (2ac + b 2 ) x 3   2 3     4 5 x x   + 2bc + c2  0  4 5

( )

( ) ( ) ( ) ( ) ( ) ( )

( )

Evaluating, x =

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4.43 = 1.87 m 3.37

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Problem 7.127 Determine the coordinates of the centroid of the volume. y

Solution:

From symmetry y = z = 0

π (0.04 2 − 0.03 2 )(0.12)(0.06) + π (0.03 2 − 0.02 2 )(0.22)(0.11) π (0.04 2 − 0.03 2 )(0.12) + π (0.03 2 − 0.02 2 )(0.22) x = 0.0884 m = 88.4 mm, y = z = 0 x =

120 mm 100 mm

40 mm

z 30 mm

20 mm

Problem 7.128 Determine the surface area of the volume of revolution.

Solution:

The outer surface: The length of the line is L1 = 4 2 + 6 2 = 7.2111 in. The y-coordinate of the centroid is y = 5 + 2 = 7 in. The surface area is A1 = 2π L1y 1 = 317.16 in 2 .

The side surface: The length of the line is L 2 = 4 in. The y-coordinate of the centroid is y 2 = 5 + 2 = 7 in. The surface area is A2 = 2π L 2 y 2 = 87.96 in 2 .

5 in

The inner surface: The length of the line is L 3 = 6 in. The y-coordinate is y 3 = 5 in. The surface area is A3 = 2π L 3y 3 = 188.5 in 2 . The total surface: A = A1 + A2 + A3 = 681.6 in 2

9 in

6 in

Problem 7.129 Determine the y coordinate of the center of mass of the homogeneous steel plate. y 10 mm

20 mm

Solution:

Divide the object into five areas: (1) The lower rectangle 20 by 80 mm, (2) an upper rectangle, 20 by 40 mm, (3) the semicircle of radius 20 mm, (4) the circle of radius 10 mm, and (5) the composite part. The areas and the centroids are tabulated. The last row is the composite and the centroid of the composite. The composite area is 3

A = ∑ Ai − A 4 . 1

20 mm

The centroid: 3

20 mm 80 mm

x =

∑ Ai x i − A 4 x 4 1

and y =

∑ Ai y i − A 4 y 4 1

The following relationships were used for the centroids: For a rectangle: the centroid is at half the side and half the base. For a semicircle, the cen4R troid is on the centerline and at from the base. For a circle, the centroid 3π is at the center.

540

Area

A, sq mm

x, mm

y, mm

1600

800

628.3

48.5

314.2

Composite

2714

48.2

21.3

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Problem 7.130 Determine the x coordinate of the center of mass of the homogeneous steel plate. y

Solution:

The quarter circle: The equation of the circle is x 2 + y 2 = R 2 . Take the elemental area to be a vertical strip of height y = R 2 − x 2 and width dx, hence the element of area is dA =

R 2 − x 2 dx, and the area is

A = ∫

220 mm

R 0

 x R2 − x2 R2 x  πR 2 R 2 − x 2 dx =  + sin −1  =  2 2 R  0 4

( )

The x -coordinate: R

R ( R 2 − x 2 ) 3/ 2  R3 x C A = ∫ x dA = ∫ x R 2 − x 2 dx =  −  = A 0 3 3  0 4R xC = 3π

150 mm x

The rectangle: The area is A = 50(150) = 7500 mm 2 . x-coordinate of the centroid is x R = 25 mm. The composite: The area of the quarter circle is

50 mm

AC =

The

π (220) 2 = 3.8013 × 10 4 mm. 4

The area of the rectangle is A R = 50(150) = 0.75 × 10 4 mm 2 . The composite area is A = AC − A R = 3.0513 × 10 4 mm 2 . The centroid: AC x C − A R x R = 110 mm A

x =

Problem 7.131 The area of the homogeneous plate is 10 ft 2 . The vertical reactions on the plate at A and B are 80 lb and 84 lb, respectively. Suppose that you want to equalize the reactions at A and B by drilling a 1-ft-diameter hole in the plate. What horizontal distance from A should the center of the hole be? What are the resulting reactions at A and B?

5 ft

Solution:

The weight of the plate is W = 80 + 84 = 164 lb. From the sum of moments about A, the centroid is x =

84(5) = 2.56 ft. W

The weight density is w =

W = 16.4 lb/ft 2 . 10

The weight of the cutout is WC = π (0.5 2 ) w = 12.88 lb. The new weight of the plate is W2 = W − WC = 151.12 lb. The new centroid must be at 5 x2 = = 2.5 ft for the reactions to be equal. 2 Therefore the x-coordinate of the center of the circle will be xC =

Wx − W 2 x 2 = 3.26 ft. WC

The reactions at A and B will be A = B =

W2 = 75.56 lb 2

X 5 ft

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Problem 7.132 The plate is of uniform thickness and is made of homogeneous material whose mass per unit area of the plate is 2 kg/m 2 . The vertical reactions at A and B are 6 N and 10 N, respectively. What is the x coordinate of the centroid of the hole?

Solution:

Choose an origin at A. The basic relation is WC x C = Wx − W H x H , where WC is the weight of the composite plate (the one with the hole), W is the weight of the plate without the hole, W H is the weight of the material removed from the hole, and x C , x, and x H are the x-coordinates of the centroids of the composite plate, the plate without the hole, and the hole, respectively. The composite weight: ΣFY = A + B − WC = 0,

1m B

from which WC = 16 N. The x-coordinate of the centroid: ΣM A = −WC x C + 2 B = 0, from which x C = 1.25 m. The weight of the plate without the hole and the x-coordinate of the centroid: W = ρ Ag = ( 12 ) (2)(1)(2)(9.81) = 19.62 N, and x = ( 23 ) 2 = 1.3333 m. The weight of the material removed from the hole: W H = W − WC = 3.62 N. Solve: x H =

Wx − WC x C = 1.702 m WH

XC XH 2m

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Problem 7.133 Determine the center of mass of the homogeneous sheet of metal. y

Solution:

Divide the object into four parts: (1) The lower plate, (2) the left hand plate, (3) the semicircular plate, and (4) the composite plate. The areas and centroids are found by inspection: (1)

Area: A1 = 9(12) = 108 in 2 , x 1 = 0.5 in, y 1 = −8 in, z 1 = 6 in.

(2)

A2 = 8(12) = 96 in 2 , x 2 = −4 in,

4 in 8 in

y 2 = −4 in, z 2 = 6 in. A3 = π 4(12) = 150.8 in 2 , 2(4) x 3 = 0, y = = 2.546 in, z = 6 in. π The composite area is

(3)

A = ∑ Ai = 354.796 in 2 .

12 in

The centroid for the composite: 3

9 in x =

∑ Ai x i 1

= −0.930 in

y =

∑ Ai y i 1

= −2.435 in

z =

Problem 7.134 Determine the center of mass of the homogeneous object. 60 mm

10 mm

30 mm

z x

= 6 in

Divide the object into three parts and the composite: (1) A triangular solid 30 mm altitude, 60 mm base, and 10 mm thick. (2) A rectangle 60 by 70 mm by 10 mm. (3) A semicircle with radius 20 mm and 10 mm thick. The volumes and their centroids are determined by inspection:

60 mm

20 mm

Solution:

(1)

∑ Ai z i

V1 =

( 12 ) (30)(60)(10) = 9000 mm , 3

x 1 = 5 mm, y 1 = 10 + z1 =

x y

(2)

30 = 20 mm, 3

60 = 20 mm 3

V2 = (60)(70)(10) = 42000 mm 3 , x 2 = 35 mm, y 2 = 5 mm,

30 mm z

z 2 = 30 mm

10 mm (3)

π 20 2 (10) = 6283.2 mm 3 , 2 4(20) = 61.51 mm, x 3 = 70 − 3π y 3 = 5 mm, z 3 = 30 mm. V3 =

The composite volume is V = V1 + V2 − V3 = 44716.8 mm 3 . The centroid is V1x 1 + V2 x 2 − V3x 3 = 25.237 mm V V y + V 2 y 2 − V3 y 3 = 8.019 mm y = 1 1 V V z + V 2 z 2 − V3 z 3 = 27.99 mm z = 1 1 V

x =

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Problem 7.135 Determine the center of mass of the homogeneous object. y

5 in

1.5 in

Solution: Divide the object into five parts plus the composite. (1) A solid cylinder with 1.5 in radius, 3 in long. (2) A rectangle 3 by 5 by 1 in (3) A solid cylinder with radius 1.5 in, 2 in long. (4) A semicircle with radius 1.5 in, 1 inch thick, (5) a semicircle with radius 1.5 in, 1 inch thick. The volumes and centroids are determined by inspection. These are tabulated:

z Top View

y 1 in

3 in

2 in

Side View

Part No

Vol, cu in

x, in

y, in

z, in

21.205

2.5

14.137

−0.5

3.534

0.6366

3.534

4.363

Composite

43.27

2.09

0.3267

The composite is 3

V = ∑Vi − ∑Vi . The centroid: 3

x =

∑Vi x i − ∑Vi x i 1

with a corresponding expression for y. The z-coordinate is zero because of symmetry.

Problem 7.136 The arrangement shown can be used to determine the location of the center of mass of a person. A horizontal board has a pin support at A and rests on a scale that measures weight at B. The distance from A to B is 2.3 m. When the person is not on the board, the scale at B measures 90 N. (a) When a 63-kg person is in position (1), the scale at B measures 496 N. What is the x coordinate of the person’s center of mass? (b) When the same person is in position (2), the scale measures 523 N. What is the x coordinate of their center of mass?

Solution: WB

1.15 m

2.3 m

By 5 90 N

W WB

1.15 m

y Ay

2.3 m

W = mg = 63 g = 618 N (a) Unloaded Beam (assume uniform beam) B A

ΣFy :

Ay + B y − WB = 0

ΣM A :

(−1.15)W B + 2.3B y = 0

Solving, A y = 90 N, W B = 180 N

(1)

(b) ΣFy :

y B A

Ay + B y − WB − W = 0

ΣM A : 2.3B y − 1.15W B − x W = 0 x

(2)

W = 618 N, W B = 180 N For (a), B y = 496 N. Solving the equations for this case yields x = 1.511 m For (b), B y = 523 N. Solving the equations for this case yields x = 1.611 m

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Problem 7.137 If a string is tied to the slender bar at A and the bar is allowed to hang freely, what will be the angle between AB and the vertical? B

Solution:

When the bar hangs freely, the action line of the weight will pass through the mass center. With a homogenous, slender bar, the mass center corresponds to the centroid of the lines making up the bar. Choose the origin at A, with the x-axis parallel to the lower bar. Divide the bar into three segments plus the composite: (1) The segment from A to the semi circle, (2) the segment AB, and (3) the semicircle. (1)

8 2 + 8 2 = 11.314 in, x 2 = 4, y 2 = 4 2(4) (3) L 3 = 4 π = 12.566 in, x 3 = 8 + = 10.546 in, y 3 = 4. π The composite length

(2) 4 in

L1 = 8 in, x 1 = 4 in, y 1 = 0. L2 =

A 8 in

L = ∑L i = 31.88 in. 1

The composite centroid: L1x 1 + L 2 x 2 + L 3x 3 = 6.58 in, L L y + L 2 y 2 + L 3y 3 y = 1 1 = 2.996 in L The angle from the point A to the centroid relative to the lower bar is x =

α = tan −1

( xy ) = 24.48°.

The angle between AB and the lower bar is 45 °, hence the angle between the line from A to the centroid and AB is β = 45 − α = 20.52 ° Since the line from A to the centroid will be vertical, this is the angle between AB and the vertical.

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Problem 7.138 When the truck is unloaded, the total reactions at the front and rear wheels are A = 54 kN and B = 36 kN. The density of the load of gravel is ρ = 1600 kg/m3. The dimension of the load in the z direction is 3 m, and its surface profile, given by the function shown, does not depend on z. What are the total reactions at the front and rear wheels of the loaded truck?

The horizontal distance from A to the center of mass of the load is d L = x L + 2.8 m = 4.344 m Now we can find the wheel loads on the loaded truck ΣFx : no forces ΣFy : A y + B y − mT g − m L g = 0 ΣM A : 5.2 B y − x T mT g − d L m L g = 0 Solving A y = 80.7 kN, B y = 171.6 kN

y y 5 1.5 2 0.45x 1 0.062 x 2

mTg

x 5.2 m

A 2.8 m

36 kN

54 kN

3.6 m 5.2 m

mL g

y = 1.5 − 0.45x + 0.062x2 dm = p(3)y dx

Solution:

First, find the location of the center of mass of the unloaded truck (and its mass). Then find the center of mass and mass of the load. Combine to find the wheel loads on the loaded truck. Unloaded Truck

ΣFx : no forces ΣFy : 54000 + 36000 − mT g = 0( N )

3.6 m

ΣM A : − x T mT g + 5.2(36) = 0

Solving x T = 2.08 m, mT = 9174 kg

Next, find x L and m L (for the load) xL =

mL g mT g

∫m x dm Num = mL ∫m dm L

where m L = ∫ mL = ∫

3.6 0

dm, Num = ∫

3ρ y dx = 3ρ ∫

3.6 0

x dm

5.2 m

(1.5 − 0.45 x + 0.062 x 2 ) dx

( )

3.6

x2 x3  m L = 3ρ  1.5 x − 0.45 + 0.062   2 3 0 m L = 16551 kg Num = 3ρ ∫

3.6 0

(1.5 x − 0.45 x 2 + 0.062 x 3 ) dx

( )

3.6

x2 x3 x4  − 0.45 + 0.062 Num = 3ρ  1.5   2 3 4 0 Num = 25560 kg ⋅ m xL =

Num = 1.544 m mL

measured from the front of the load

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Problem 7.139 The mass of the Moon is 0.0123 times the mass of the Earth. If the Moon’s center of mass is 383,000 km from the center of mass of the Earth, what is the distance from the center of mass of the Earth to the center of mass of the Earth–Moon system?

Solution: mE mM

X 383,000 km x(m E + m M ) = (383,000 m M ) mM (383,000) mE + mM m M /m E = (383,000) 1 + m M /m E 0.0123 = (383,000) 1 + 0.0123 = 4650 km.

so x =

(The Earth’s radius is 6370 km, so the center of mass of the Earth-Moon system is within the Earth.)

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Chapter 8 Problem 8.1 Use the method described in Practice Example 8.1 to determine I y and k y for the rectangular area.

Solution:

Consider the vertical strip element shown: y 0.2 m

0.4 m

y dA 0.6 m

0.6 m

x x

0.2 m

0.4 m

The area dA = (0.6 m) dx. The moment of inertia of the area about the y-axis is I y = ∫ x 2 dA = ∫ A

0.6 m 0.2 m

x 2 (0.6 m) dx

= 0.0416 m 4 . The radius of gyration about the y-axis is ky =

Iy = A

0.0416 m 4 = 0.416 m. (0.4 m)(0.6 m)

I y = 0.0416 m 4 , k y = 0.416 m.

Problem 8.2 Use the method described in Practice Example 8.1 to determine I x and k x for the rectangular area.

Solution:

Consider the vertical strip element shown: y 0.2 m

0.4 m

dA 0.6 m

0.6 m

x x

0.2 m

0.4 m

From Practice Example 8.1, the moment of inertia of the strip element about the x-axis is ( I x ) strip =

1 (0.6 m) 3 dx. 3

We obtain the moment of inertia of the area about the x-axis by integrating this expression: Ix = ∫

0.6 m 1

(0.6 m) 3 dx 3 = 0.0288 m 4 . 0.2 m

The radius of gyration about the x-axis is kx =

Ix = A

0.0288 m 4 = 0.346 m. (0.4 m)(0.6 m)

I x = 0.0288 m 4 , k x = 0.346 m.

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Problem 8.3 The dimensions b = 4 in and h = 3 in. Use the method described in Practice Example 8.1 to determine I y and k y for the area shown.

Solution:

2h dA h

x x b

From x = 0 to x = b the area of the strip element shown is dA = 2h dx, and from x = b to x = 2b it is dA = h dx. The moment of inertia of the area about the y-axis is

x b

I y = ∫ x 2 dA = ∫ x 2 (2h dx ) + ∫ A

x 2 (h dx )

x3 b x 3 2b = 2h   + h    3 0  3 b = 2h

(

b3 8b 3 b3 +h − 3 3 3

)

= 3hb 3 = 3(3 in)(4 in) 3 = 576 in 4 . The area is A = b(2h) + bh = 3bh, So the radius of gyration about the y-axis is Iy = A

ky =

3hb 3 3bh

= b = 4 in. I y = 576 in 4 , k y = 4 in.

Problem 8.4 (a) Determine the moment of inertia I y of the beam’s rectangular cross section about the y-axis. (b) Determine the moment of inertia I y ′ of the beam’s cross section about the y ′-axis . Using your numerical values, show that I y = I y ′ + d x 2 A, where A is the area of the cross section.

Solution: 40 mm

(a)

Iy = ∫

(b)

I y' = ∫

60 mm

∫0

20 mm

−20 mm

x 2 dy dx = 1.28 × 10 6 mm 4

30 mm

∫−30 mm x 2dy dx = 3.2 × 10 5 mm 4

y I y = I y' + d x 2 A

y9 dx

1.28 × 10 6 mm 4 = 3.2 × 10 5 mm 4 + (20 mm ) 2 [(40 mm)(60 mm)]

60 mm x9

O9 dy O

x 40 mm

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Problem 8.5 (a) Determine the polar moment of inertia J O of the beam’s rectangular cross section about the origin O. (b) Determine the polar moment of inertia J O ′ of the beam’s cross section about the origin O′. Using your numerical values, show that J O = J O ′ + (d x 2 + d y 2 ) A, where A is the area of the cross section.

Solution: 40 mm

(a)

JO = ∫

(b)

J O' = ∫

60 mm

∫0

20 mm

−20 mm

( x 2 + y 2 ) dy dx = 4.16 × 10 6 mm 4

30 mm

∫−30 mm ( x 2 + y 2 ) dy dx = 1.04 × 10 6 mm 4

y J O = J O' + (d x 2 + d y 2 ) A

(c)

4.16 × 10 6 mm 4 = 1.04 × 10 6 mm 4 + [(20 mm) 2 + (30 mm) 2 ][(40 mm)(60 mm)]

60 mm x9

O9 dy O

x 40 mm

Solution:

Problem 8.6 Determine I y and k y .

A = (0.3 m) (1 m) +

Iy = ∫ 0.6 m

0.3 m

ky = x

1m 0

0.3 m + 0.3 x

∫0

Iy = A

1 (0.3 m)(1 m) = 0.45 m 2 2 x 2 dy dx = 0.175 m 4

0.175 m 4 = 0.624 m 0.45 m 2

Solution:

Problem 8.7 Determine J O and k O .

A = (0.3 m) (1 m) +

JO = ∫

1m 0

0.6 m

0.3 m

kO = x

1 (0.3 m)(1 m) = 0.45 m 2 2

0.3 m + 0.3 x

∫0

( x 2 + y 2 ) dy dx = 0.209 m 4

0.209 m 4 = 0.681 m 0.45 m 2

Solution:

Problem 8.8 Determine I xy . y

I xy = ∫

1m 0

0.3 m + 0.3 x

∫0

x yd ydx = 0.0638 m 4

0.6 m

0.3 m

x 1m

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Problem 8.9 The coordinates a = 1 m and b = 3 m. Determine I y .

Solution:

Let dA be a vertical strip of width dx: y

y 5 x2 dA a

y 5 2x

The area dA = (2/x )dx. The moment of inertia about the y-axis is

I y = ∫ x 2 dA A

= ∫ x2 a

( 2x ) dx b

x2 = 2    2 a = 8 m 4. I y = 8 m 4.

Problem 8.10 The coordinates a = 1 m and b = 3 m. Determine I x .

Solution:

Let dA be a vertical strip of width dx: y

y 5 x2 dA a

y 5 2x

dx The height of the strip is 2/x. From Example 8.1, the moment of inertia of the strip element about the x-axis is

( I x ) strip =

( )

1 2 3 dx. 3 x

We integrate this expression from x = a to x = b to determine the moment of inertia of the area about the x-axis: Ix = ∫ =

( )

2 3 dx 3 x

b1 a

8  x −2  b   3  −2  a

= 1.19 m 4 . I x = 1.19 m 4 .

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Problem 8.11 The coordinates a = 1 m and b = 3 m. Determine J O .

Solution:

We will determine I x and I y and use the relation J O = I x + I y . Let dA be a vertical strip of width dx: y

y 5 x2 dA

y 5 2x

The area dA = (2/x )dx. The moment of inertia about the y-axis is I y = ∫ x 2 dA A

= ∫ x2 a

( 2x ) dx b

x2 = 2    2 a = 8 m 4. The height of the strip is 2/x. From Example 8.1, the moment of inertia of the strip element about the x-axis is ( I x ) strip =

( )

1 2 3 dx. 3 x

We integrate this expression from x = a to x = b to determine the moment of inertia of the area about the x-axis: Ix = ∫ =

( 2 ) dx 3 x

b1 a

8  x −2  b 3  −2  a

= 1.19 m 4 . The polar moment of inertia is JO = I x + I y = 9.19 m 4 . J O = 9.19 m 4 .

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Problem 8.12 The coordinates a = 1 m and b = 3 m. Determine I xy .

Solution:

Let dA be a vertical strip of width dx: y

y 5 x2 dA a

y 5 2x

dx The height of the strip is 2/x. From Example 8.1, the product of inertia of the strip element is

( I xy ) strip =

( )

1 2 2 x dx. 2 x

We integrate this expression from x = a to x = b to determine the product of inertia of the area: I xy = ∫

( )

2 2 x dx 2 x

b1 a

= 2[ ln x ]ba = 2.20 m 4 . I xy = 2.20 m 4 .

Solution: Setting the given equation equal to zero, the point where the upper profile intersects the x-axis is x = 10.

Problem 8.13 Determine I y and k y . y y 5 2x – 1 x2 5

Let dA be a vertical strip of width dx: y

y 5 2x 2

dA x

(

The area dA = 2 x − A = ∫ dA = ∫ A

10 0

1 x2 5

)

1 2 x dx. The total area is 5

( 2x − 15 x ) dx 2

= 33.3. The moment of inertia about the y-axis is I y = ∫ x 2 dA = ∫ A

10 0

(

x 2 2x −

)

1 2 x dx 5

= 1000. The radius of gyration about the y-axis is ky =

Iy = 5.48. A

I y = 1000, k y = 5.48.

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Problem 8.14 Determine I x and k x .

Solution:

Setting the given equation equal to zero, the point where the upper profile intersects the x-axis is x = 10.

y y 5 2x – 1 x2 5

Let dA be a vertical strip of width dx: y

y 5 2x2

1 x2 5

dA x x

(

The area dA = 2 x − A = ∫ dA = ∫ A

10 0

)

1 2 x dx. The total area is 5

( 2x − 15 x ) dx 2

= 33.3. 1 2 x . From Example 8.1, the moment of 5 inertia of the strip element about the x-axis is The height of the strip is 2 x −

( I x ) strip =

3 1 1 2 x − x 2 dx. 3 5

(

)

We integrate this expression from x = 0 to x = 10 to determine the moment of inertia of the area about the x-axis: Ix = ∫

10 1 0

( 2x − 15 x ) dx 2

= 190. The radius of gyration about the x-axis is kx =

Ix = 2.39. A

I x = 190, k x = 2.39.

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Solution: Setting the given equation equal to zero, the point where the upper profile intersects the x-axis is x = 10.

Problem 8.15 Determine J O and k O . y y 5 2x – 1 x2 5

Let dA be a vertical strip of width dx: y

y 5 2x 2 1 x2 5

dA x x

(

The area dA = 2 x − A = ∫ dA = ∫ A

10 0

)

1 2 x dx. The total area is 5

(2 x −

1 2 x ) dx 5

= 33.3. 1 2 x . From Example 8.1, the moment of 5 inertia of the strip element about the x-axis is The height of the strip is 2 x −

( I x ) strip =

3 1 1 2 x − x 2 dx. 3 5

(

)

We integrate this expression from x = 0 to x = 10 to determine the moment of inertia of the area about the x-axis: Ix = ∫

10 1 0

( 2x − 15 x ) dx 2

= 190. The moment of inertia about the y-axis is I y = ∫ x 2 dA = ∫ A

10 0

(

x 2 2x −

)

1 2 x dx 5

= 1000. The polar moment of inertia about the origin is JO = I x + I y = 1190. The radius of gyration about the origin is kO =

JO = 5.98. A

J O = 1190, k O = 5.98.

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Problem 8.16 Determine I xy .

Solution:

Let dA be a vertical strip of width dx: y

y 5 2x 2 1 x2 5

y 5 2x – 1 x2 5

dA x x

The height of the strip is 2 x −

1 2 x . From Example 8.1, the product of 5

inertia of the strip element is 2 1 1 2 x − x 2 x dx. 2 5

(

( I xy ) strip =

)

We integrate this expression from x = 0 to x = 10 to determine the product of inertia of the area: I xy = ∫

( 2 x − 15 x ) x dx 2

b1 a

= 333. I xy = 333.

Solution:

Problem 8.17 Determine I y and k y . y 1 y 5 2 x 2 1 4x 2 7 4 y55

1 −4 ± − x 2 + 4x − 7 = 5 ⇒ x = 4 A = ∫

12 4

−x 2 / 4 + 4 x − 7

∫5

4 ∫5

ky =

Iy = A

Ix = ∫

y 1 2 x 1 4x 2 7 4

kx =

12 4

16 − 4(−1/4)(−12) = 4,12 2(−1/4)

dy dx = 21.33

−x 2 / 4 + 4 x − 7

Iy = ∫

Solution:

Problem 8.18 Determine I x and k x .

y52

First we need to locate the points where the curve inter-

sects the line.

x 2 dy dx = 1434

1434 = 8.20 21.33

See Solution to Problem 8.17 −x 2 / 4 + 4 x − 7

∫5

Ix = A

y 2 dy dx = 953

953 = 6.68 21.33

y55

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Problem 8.19 (a) Determine I y and k y by letting dA be a vertical strip of width dx. (b) The polar moment of inertia of a circular area with its center at the origin is J O = 12 π R 4 . Explain how you can use this information to confirm your answer to (a).

Solution: The equation of the circle is x 2 + y 2 = R 2 , from which y = ± R 2 − x 2 . The strip dx wide and y long has the elemental area dA = 2 R 2 − x 2 dx. The area of the semicircle is A =

R πR 2 I = ∫ x 2 dA = 2 ∫ x 2 R 2 − x 2 dx 0 A 2 y

x ( R 2 − x 2 ) 3/ 2 R 2 x ( R 2 − x 2 ) 1/ 2 R4 x R = 2  − + + sin −1  R 0 4 8 8 

( )

= x R

πR 4 8 Iy R = A 2

ky =

(b) If the integration were done for a circular area with the center at the origin, the limits of integration for the variable x would be from –R to R, doubling the result. Hence, doubling the answer above, Iy =

πR 4 . 4

By symmetry, I x = I y , and the polar moment would be J O = 2I y =

πR 4 , 2

which is indeed the case. Also, since k x = k y by symmetry for the full circular area, kO =

Iy Ix + = A A

Iy = A

JO A

as required by the definition. Thus the result checks.

Problem 8.20 (a) Determine I x and k x by letting dA be a horizontal strip of height dy. (b) The polar moment of inertia of a circular area with its center at the origin is J O = 12 π R 4 . Explain how you can use this information to confirm your answer to (a).

Solution:

dA =

Use the results of the solution to Problem 8.19,

πR 2 A = . The equation for the circle is x 2 + y 2 = R 2 , from which 2 x = ± R 2 − y 2 . The horizontal strip is from 0 to R, hence the element of area is R 2 − y 2 dy.

I x = ∫ y 2 dA = ∫ =  −  x R

−R

y 2 R 2 − y 2 dy

y ( R 2 − y 2 ) 3/ 2 R 2 y ( R 2 − y 2 ) 1/ 2 y R R4 sin −1 + +  4 8 8 R  −R

( )

R4 π R4 π  πR 4 =  + =  8 2 8 2  8 kx =

Ix R = . A 2

(b) If the area were circular, the strip would be twice as long, and the moment of inertia would be doubled: Iy =

πR 4 . 4

By symmetry I y = I x , and J O = 2 I x =

πR 4 , 2

which is indeed the result. Since k x = k y by symmetry for the full circular area, the kO =

Iy Ix + = A A

I 2 x = A

JO A

as required by the definition. This checks the answer.

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Problem 8.21 Use the procedure described in Example 8.2 to determine the moments of inertia I x and I y for the annular ring. y

Solution: We first determine the polar moment of inertia J O by integrating in terms of polar coordinates. Because of symmetry and the relation J O = I x + I y , we know that I x and I y each equal 12 J O . Integrating as in Example 8.2, the polar moment of inertia for the annular ring is J O = ∫ r 2 dA = ∫ A

Ro Ri

Therfore I x = I y =

r 2 (2πr ) dr =

1 π ( R o4 − Ri4 ) 2

1 π ( R o4 − Ri4 ) 4

x Ri

Problem 8.22 What are the values of I y and k y for the elliptical area of the airplane’s wing? y x2 y2 1 2 51 a2 b

Iy =

2b a4 π a 8 2

Iy =

2 a 3 bπ 8

Evaluating, we get I y = 49.09 m 4 The area of the ellipse (half ellipse) is

A = 2∫

a 0 a

Solution: I y = ∫ x 2 dA = ∫

−0

I y = 2∫ I y = 2∫

[ x 2 y]

(

I y = 2b ∫ x 2 1 − 0

)

2b (a 2 − x 2 ) 1/ 2 dx a ∫0

2b  x a 2 − x 2 a2 x  sin −1 +   2 2 a  a  0

2b   a 0 0   a2 a   0 a a2 sin −1 sin −1 + +   − 2 2 a   2 a   2 a  

( )

a 2x a 2 − x 2 2b  x (a 2 − x 2 ) 3/ 2 + − a  4 8

( )

2b a2 π πab = 2 a 2 2

Finally ky =

Rewriting

Iy =

)

x2 dx a

A = 7.85 m 2

x2 dx a2

2b a 2 2 x a − x 2 dx a ∫0

dy dx

Evaluating, we get

x 2 1/ 2 dx a2

Iy =

1/ 2

)

A =

x2 b (1− 2 )1/ 2 a dx 0

I y = 2∫ x 2b 1 −

(

x2 a

∫−y x 2 dy dx

x 2 dy dx 0 ∫0 a

∫0

A = 2∫ b 1 −

(

b 1−

Iy = A

49.09 7.85

k y = 2.5 m

a4 x  sin −1  a  0 8

( )

x2 + y2 5 1 a 2 b2

(from the integral tables)    0 0  a3 a2 − a2 a4 a  2b   a(a 2 − a 2 ) 3 / 2 Iy = − + + sin −1  a   4 8 8 a         0 0    0 a2 ⋅0 a2  0(a 2 ) 3 / 2 0   a4 1 − − + + sin   4 8 8 a       

( )

558

a y 5 b 1– x2 a2

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Problem 8.23 What are the values of I x and k x for the elliptical area of the airplane’s wing?

Solution: I x = ∫ y 2 dA = 2 ∫ A

a y3 a  I x = 2 ∫   0  3 0

x2 y2 1 2 51 2 a b

I x = 2∫

Ix =

a b3 0

3a 3

a 0 ∫0

a2 −x2

y 2 dy dx

( a 2 − x 2 ) 3 / 2 dx

2b 3  x ( a 2 − x 2 ) 3 / 2 3a 2 x a 2 − x 2 3 x  + + a 4 sin −1   3a 3  4 8 8 a 

( )

a 0

 0(a 2 ) 2b 3  a(0) 3a 3 0 3 π 3a 2 ⋅ 0 a 2 Ix = + + a4 − − + 0   3a 3  4 8 8 2 4 8

( ) ( π2 )

Ix =

2b3 3 . a4 3 a3 8

Ix =

3 ab 3π ab 3π = 3 .8 8

Evaluating (a = 5, b = 1) Ix =

5π = 1.96 m 4 8

From Problem 8.22, the area of the wing is A = 7.85 m 2 kx =

Ix = A

1.96 k = 0.500 m 7.85 x

y2 x2 — — + 2 51 2 a b

x y 5 b a2 – x2 a a

Problem 8.24 The radius R = 50 mm. Determine I y by letting dA be a vertical strip of width dx. y

The equation for the circular arc is x 2 + y 2 = R 2, so the function f ( x ) = ( R 2 − x 2 ) 1/ 2 . The moment of inertia of the area about the y-axis is I y = ∫ x 2 dA A

R x

= ∫ x 2 f ( x ) dx 0

Solution:

= ∫ x 2 ( R 2 − x 2 ) 1/ 2 dx 0

Consider the strip element y f(x)

1 1 1 x R =  − x ( R 2 − x 2 ) 3/ 2 + R 2 x ( R 2 − x 2 ) 1/ 2 + R 4 arcsin  4 8 8 R  0

( )

= R

= 1.23E6 mm 4 . x

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1 πR 4 16

I y = 1.23E6 mm 4 .

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Problem 8.25 The radius R = 50 mm. Determine I xy by letting dA be a vertical strip of width dx (see Practice Example 8.1). y

The equation for the circular arc is x 2 + y 2 = R 2, so the function f ( x ) = ( R 2 − x 2 ) 1/ 2 . From Example 8.1, the product of inertia of the vertical strip element is ( I xy ) strip =

We integrate this expression from x = 0 to x = R to obtain the product of inertia of the entire area:

R x

Solution:

= ∫

f(x) dA

I xy = ∫

Consider the strip element y

1[ f ( x ) ]2 x dx. 2

[ f ( x ) ]2 x dx

( R 2 − x 2 ) x dx

1  2 x2 x4  R − R  2 2 4  0

1 4 R 8

= 7.81E5 mm 4 . x

Problem 8.26 The radius R = 50 mm. Use the method described in Example 8.2 to determine I x and I y for the quarter-circular area.

I xy = 7.81E5 mm 4 .

Solution:

Consider the element of area y dA

dr r x

The area dA = (π /2)r dr . The polar moment of inertia of the entire area is

R x

J O = ∫ r 2 dA A

= ∫ r 2 (π /2)r dr 0

π r4 R 2  4  0

1 4 πR . 8

The moments of inertia I x = I y due to symmetry, and J O = I x + I y , so Ix = Iy =

1 1 J = πR 4 2 O 16

= 1.23E6 mm 4 . I x = I y = 1.23E6 mm 4 .

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Problem 8.27 The dimensions b = 4 in and h = 3 in. Determine I x and k x for the area by dividing it into two rectangles as shown. y

Solution: From Appendix B, the moments of inertia of the two rectangles about the x-axis are (I x )1 =

1 1 b(2h) 3 = 288 in 4 , ( I x ) 2 = bh 3 = 36 in 4 . 3 3

The moment of inertia of the area about the x-axis is I x = ( I x ) 1 + ( I x ) 2 = 324 in 4 . The total area is A = b(2h) + bh = 36 in 2 , so

kx =

Ix = 3 in. A

I x = 324 in 4 , k x = 3 in.

2 x b

Problem 8.28 The dimensions b = 4 in and h = 3 in. Determine I y and k y for the area by dividing it into two rectangles as shown.

Solution: y

h y9 x9

1 b

x b

From Appendix B, the moment of inertia of rectangle 1 about the y-axis is

2 x

(I y )1 =

1 (2h)b 3 = 128 in 4 . 3

The moment of inertia of rectangle 2 about the y'-axis is ( I y' ) 2 =

1 hb 3 = 16 in 4 . 12

Applying the parallel-axis theorem, the moment of inertia of rectangle 2 about the y-axis is

(

( I y ) 2 = ( I y' ) 2 + b +

1 2 b (bh) = 448 in 4 . 2

)

The moment of inertia of the area about the y-axis is I y = ( I y ) 1 + ( I y ) 2 = 576 in 4 . The total area is A = b(2h) + bh = 36 in 2 , so ky =

Iy = 4 in. A

I y = 576 in 4 , k y = 4 in.

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Problem 8.29 The dimensions d = 300 mm, b f = 120 mm, t f = 20 mm, and t w = 15 mm. Determine I x and k x for the beam’s cross section.

Solution: y

d tw

Source: Courtesy of aliaksab/123RF. tf

We divide the cross section into five rectangles as shown. From Appendix B, the moment of inertia of the large rectangle about the x-axis is

1 t d 3 = 3.38E7 mm 4 . 12 w

( I x ) large =

The area of each small rectangle is Asmall = (

bf t − w )t f = 1050 mm 2 , 2 2

and the distance from the x-axis to the centroid of each small rectangle is h = x

tf d − = 140 mm. 2 2

Applying the parallel axis theorem, the moment of inertia of each small rectangle about the x-axis is ( I x ) small =

bf t  t  1  b f − w  t f = 2.06E7 mm 4 . − w  t f 3 + h 2   2 12  2 2 2

The moment of inertia of the complete cross section is I x = ( I x ) large + 4( I x ) small = 1.16E8 mm 4 . bf

The area of the complete cross section is A = t w d + 4 Asmall = 8700 mm 2 , and the radius of gyration is kx =

Ix = 116 mm. A

I x = 1.16E8 mm 4 , k x = 116 mm.

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Problem 8.30 In Example 8.4, determine I x and k x for the composite area.

Solution:

The area is divided into a rectangular area without the cutout (part 1), a semicircular areas without the cutout (part 2), and the circular cutout (part 3). Using the results from Appendix B, the moment of inertia of part 1 about the x-axis is (I x )1 =

1 (120 mm)(80 mm) 3 = 5.12 × 10 6 mm 4 , 12

the moment of inertia of part 2 is (I x ) 2 =

1 π (40 mm) 4 = 1.01 × 10 6 mm 4 , 8

and the moment of inertia of part 3 is (I x ) 3 =

1 π (20 mm) 4 = 1.26 × 10 5 mm 4 . 4

The moment of inertia of the composite area is I x = ( I x ) 1 + ( I x ) 2 + ( I x ) 3 = 6.00 × 10 6 mm 4 . From Example 8.4, the composite area is A = 1.086 × 10 4 mm 4 , so the radius of gyration about the x-axis is Ix = A

kx =

6.00 × 10 6 mm 4 = 23.5 mm. 1.086 × 10 4 mm 2

I x = 6.00 × 10 6 mm 4 , k x = 23.5 mm.

Solution:

Problem 8.31 Determine I x and k x .

Break into 3 rectangles—See 8.29

First locate the centroid

d =

0.8 m 0.2 m

(0.6)(0.2)(0.1) + (0.2)(0.6)(0.5) + (0.8)(0.2)(0.9) = 0.54 m (0.6)(0.2)+ (0.2)(0.6) + (0.8)(0.2)

1 I x =  (0.2)(0.6) 3 + (0.2)(0.6)(d – 0.5) 2   12  1 +  (0.6)(0.2) 3 + (0.6)(0.2)(d – 0.1) 2   12 

0.6 m

1 +  (0.8)(0.2) 3 + (0.8)(0.2)(0.9 – d ) 2  = 0.0487 m 4  12 

0.2 m

kx =

Ix = A

0.0487 m 4 = 0.349 m 0.4 m 4

0.2 m 0.6 m

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Problem 8.32 Determine I y and k y .

Solution:

Break into 3 rectangles—See 8.29

1 1 1 (0.6)(0.2) 3 + (0.2)(0.6) 3 + (0.2)(0.8) 3 12 12 12

Iy =

= 0.01253 m 4

0.8 m 0.2 m

Iy = A

ky =

0.01253 m 4 = 0.1770 m 0.4 m 2

0.6 m

0.2 m 0.2 m 0.6 m

Solution:

Problem 8.33 Determine J O and k O .

See 8.29, 8.31 and 8.32

J O = I x + I y = 0.0612 m 4

KO =

0.8 m

JO = A

0.0612 m 4 = 0.391 m 0.4 m 4

0.2 m

0.6 m

0.2 m 0.2 m 0.6 m

Problem 8.34 If you design the beam cross section so that I x = 6.4 × 10 5 mm 4 , what are the resulting values of I y and J O ?

Solution:

The area moment of inertia for a triangle about the base is

( )

1 Ix = bh 3 , 12 from which

Ix = 2

( 121 ) (60)h = 10h mm , 3

I x = 10 h 3 = 6.4 × 10 5 mm 4 , h

from which h = 40 mm. x

Iy = 2

30 mm

from which Iy =

30 mm

( 121 ) (2h)(30 ) = ( 13 ) h(30 )

( 13 ) (40)(30 ) = 3.6 × 10 mm 3

and J O = I x + I y = 3.6 × 10 5 + 6.4 × 10 5 = 1 × 10 6 mm 4

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Problem 8.35 Determine I y and k y .

Part (2): The middle rectangle: A2 = (200 − 80)(40) = 4.8 × 10 3 mm 2 ,

d x 2 = 20 mm,

160 mm

I yy 2 =

40 mm

( 121 )(120)(40 ) = 6.4 × 10 mm . 3

From which,

200 mm

I y 2 = d x 2 2 A2 + I yy 2 = 2.56 × 10 6 mm 4 .

40 mm 40 mm

120 mm

Part (3) The bottom rectangle: A3 = 120(40) = 4.8 × 10 3 mm 2 , dx 3 =

120 = 60 mm, 2

Part (1): The top rectangle.

I yy3 =

( 121 )40(120 ) = 5.76 × 10 mm

A1 = 160(40) = 6.4 × 10 3 mm 2 ,

From which

160 d x1 = = 80 mm, 2

I y 3 = d x 3 2 A3 + I yy 3 = 2.304 × 10 7 mm 4

Solution:

I yy1 =

Divide the area into three parts:

ky =

I y1 = d x1 2 A1 + I yy1 = 5.4613 × 10 7 mm 4 .

Problem 8.36 Determine I x and k x .

Iy = 70.8 mm. ( A1 + A2 + A3 )

From which I x1 = d y1 2 A1 + I xx1 = 2.082 × 10 8 mm 4

Part (2): The middle rectangle: 160 mm

A2 = 4.8 × 10 3 mm 2 , 40 mm

40 mm 40 mm

120 mm

Solution:

I y = I y1 + I y 2 + I y 3 = 8.0213 × 10 7 mm 4

From which

200 mm

The composite:

( 121 )(40)(160 ) = 1.3653 × 10 mm . 3

Use the solution to Problem 8.35. Divide the area into

d y2 =

120 + 40 = 100 mm, 2

I xx 2 =

( 121 )(40)(120 ) = 5.76 × 10 mm 3

from which I x 2 = d y 2 2 A2 + I xx 2 = 5.376 × 10 7 mm 4 Part (3) The bottom rectangle: A3 = 4.8 × 10 3 mm 2 ,

three parts:

d y 3 = 20 mm,

Part (1): The top rectangle.

I xx3 =

A1 = 6.4 × 10 3 mm 2 ,

and I x 3 = d y 3 2 A3 + I xx 3 = 2.56 × 10 6 mm 4 .

d y1 = 200 − 20 = 180 mm,

The composite:

( )

1 I xx1 = (160)(40 3 ) = 8.533 × 10 5 mm 4 . 12

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I x = I x1 + I x 2 + I x 3 = 2.645 × 10 8 mm 4 kx =

( 121 )120(40 ) = 6.4 × 10 mm

Ix = 128.6 mm ( A1 + A2 + A3 )

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from which

Problem 8.37 Determine I xy .

I xy1 = d x1d y1 A1 + I xxyy1 = 9.216 × 10 7 mm 4 .

Part (2) A2 = (200 − 80)(40) = 4.8 × 10 3 mm 2 ,

160 mm

d x2 = 20 mm,

40 mm 200 mm

d y2 = 40 mm

120 + 40 = 100 mm, 2

from which 40 mm

120 mm

I xy 2 = d x 2 d y 2 A2 = 9.6 × 10 6 mm 4 . Part (3): A3 = 120(40) = 4.8 × 10 3 mm 2 , 120 = 60 mm, 2 d y 3 = 20 mm, d x3 =

Solution:

(See figure in Problem 8.35). Use the solutions in Problems 8.35 and 8.36. Divide the area into three parts: Part (1): A1 = 160(40) = d x1 =

6.4 × 10 3 mm 2 ,

from which I xy 3 = d x 3d y 3 A3 = 5.76 × 10 6.

160 = 80 mm, 2

The composite: I xy = I xy1 + I xy 2 + I xy 3 = 1.0752 × 10 8 mm 4

d y1 = 200 − 20 = 180 mm, I xxyy1 = 0,

Problem 8.38 Determine I x and k x .

Part (2) A2 = (200 − 80)(40) = 4.8 × 10 3 mm 2 , 160 = 80 mm, d x 2 = 20 mm, 2 120 d y2 = + 40 = 100 mm, 2 d x1 =

y 160 mm

Part (3) A3 = 120(40) = 4.8 × 10 3 mm 2 ,

40 mm 200 mm

x 40 mm

d x3 =

120 = 60 mm, d y 3 = 20 mm. 2

The total area is 40 mm

120 mm

A = A1 + A2 + A3 = 1.6 × 10 4 mm 2 . The centroid coordinates are A1d x1 + A2 d x 2 + A3d x 3 = 56 mm, A A1d y1 + A2 d y 2 + A3d y 3 y = = 108 mm A from which x =

Solution:

The strategy is to use the relationship I x = d 2 A + I xc , where I xc is the area moment of inertia about the centroid. From this I xc = −d 2 A + I x . Use the solutions to Problems 8.35, 8.36, and 8.37. Divide the area into three parts and locate the centroid relative to the coordinate system in the Problems 8.35, 8.36, and 8.37.

I xc = −y 2 A + I x = −1.866 × 10 8 + 2.645 × 10 8 = 7.788 × 10 7 mm 4

Part (1) A1 = 6.4 × 10 3 mm 2 , d y1 = 200 − 20 = 180 mm.

566

k xc =

I xc = 69.77 mm A

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Solution: The strategy is to use the relationship I y = d 2 A + I yc , where I yc is the area moment of inertia about the centroid. From this I yc = −d 2 A + I y . Use the solution to Problem 8.38. The centroid coordinates are x = 56 mm, y = 108 mm, from which

Problem 8.39 Determine I y and k y . y 160 mm

I yc = −x 2 A + I y = −5.0176 × 10 7 + 8.0213 × 10 7 = 3.0 × 10 7 mm 4 ,

40 mm 200 mm

k yc =

40 mm

I yc = 43.33 mm A

40 mm 120 mm

Problem 8.40 Determine I xy .

Solution:

Use the solution to Problem 8.37. The centroid coordinates

are y

x = 56 mm, y = 108 mm,

160 mm

from which I xyc = −xy A + I xy = −9.6768 × 10 7 + 1.0752 × 10 8

40 mm 200 mm

= 1.0752 × 10 7 mm 4 x

40 mm 40 mm 120 mm

Solution:

Problem 8.41 Determine I x and k x .

Divide the area into two parts:

Part (1): a triangle and Part (2): a rectangle. The area moment of inertia for a triangle about the base is

Ix = 4 ft

3 ft

( 121 ) bh . 3

The area moment of inertia about the base for a rectangle is Ix =

3 ft x

( 13 ) bh . 3

Part (1) I x1 =

( 121 )4(3 ) = 9 ft .

Part (2) I x2 =

( 13 )3(3 ) = 27.

The composite: I x = I x1 + I x 2 = 36 ft 4 . The area: A = kx =

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( 12 )4(3) + 3(3) = 15 ft . 2

Ix = 1.549 ft. A

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Solution:

Problem 8.42 Determine J O and k O .

(See Figure in Problem 8.41.) Use the solution to

Problem 8.41.

Part (1): The area moment of inertia about the centroidal axis parallel to the base for a triangle is I yc =

3 ft

4 ft

( 361 )bh = ( 361 )3(4 ) = 5.3333 ft , 3

from which

3 ft

I y1 =

( 83 ) A + I 1

= 48 ft 4 .

where A1 = 6 ft 2 . Part (2): The area moment of inertia about a centroid parallel to the base for a rectangle is I yc =

( 121 ) bh = ( 121 )3(3 ) = 6.75 ft , 3

from which I y 2 = (5.5 2 ) A2 + I yc = 279 ft 4 , where A2 = 9 ft 2 . The composite: I y = I y1 + I y 2 = 327 ft 4 , from which, using a result from Problem 8.41, J O = I x + I y = 327 + 36 = 363 ft 4 and k O =

Solution:

Problem 8.43 Determine I xy .

(See Figure in Problems 8.41.) Use the results of the solutions to Problems 8.41 and 8.42. The area cross product of the moment of inertia about centroidal axes parallel to the bases for a triangle is 1 2 2 I x'y' = b h , and for a rectangle it is zero. Therefore: 72

3 ft

4 ft

I xy1 =

( 721 ) (4 )(3 ) + ( 83 )( 33 ) A = 18 ft 2

I xy = I x'y'1 + I xy 2 = 92.25 ft 4

Solution:

Problem 8.44 Determine I x and k x .

Use the results of Problems 8.41, 8.42, and 8.43. The strategy is to use the parallel axis theorem and solve for the area moment of inertia about the centroidal axis. The centroidal coordinate

y = 3 ft

A1 (1) + A2 (1.5) = 1.3 ft. A

from which 3 ft

I xc = −y 2 A + I x = 10.65 ft 4

and k xc =

568

and I xy 2 = (1.5)(5.5) A2 = 74.25 ft 4 ,

3 ft

4 ft

JO = 4.92 ft A

I xc = 0.843 ft A

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Solution: Use the results of Problems 8.41, 8.42, and 8.43. The strategy is to use the parallel axis theorem and solve for the area moment of inertia about the centroidal axis. The centroidal coordinate:

Problem 8.45 Determine J O and k O . y

4 ft

x =

3 ft

( 83 ) + A (5.5) = 4.3667 ft, 2

from which 3 ft

I YC = −x 2 A + I Y = 40.98 ft 4 . Using a result from Problem 8.44, J O = I XC + I YC = 10.65 + 40.98 = 51.63 ft 4 JO = 1.855 ft A

and k O =

Problem 8.46 Determine I xy .

Solution: Use the results of Problems 8.41–8.45. The strategy is to use the parallel axis theorem and solve for the area moment of inertia about the centroidal axis. Using the centroidal coordinates determined in Problems 8.44 and 8.45,

4 ft

I xy = −xy A + I xy = −85.15 + 92.25 = 7.1 ft 4

3 ft 3 ft

I x 2 = 1.257 × 10 5 + π (20) 2 (80) 2

Problem 8.47 Determine I x and k x .

I x 2 = 0.126 × 10 6 + 8.042 × 10 6 mm 4

= 8.168 × 10 6 mm 4 = 0.817 × 10 7 mm 4 I x = I x1 − I x 2 = 3.79 × 10 7 mm 4 Area = 8343 mm 2 120 mm 80 mm

kx =

20 mm

Ix = 97.4 mm Area

20 m

40 mm 80 mm

40 mm y9

Solution:

Let Part 1 be the entire rectangular solid without the hole and let part 2 be the hole.

40 mm

Area = hb − π R 2 = (80)(120) − π R 2 I x1 =

1 3 bh 3

where b = 80 mm h = 120 mm

I x1 =

120 mm

Part 2

1 (80)(120) 3 = 4.608 × 10 7 mm 4 3

dy = 80 mm

For Part 2, I x ′ 2 = 14 π R 4 = 14 π (20) 4 mm 4 I x ′ 2 = 1.257 × 10 5 mm 4 I x 2 = I x ′2 + d y

Part 1

where A = π R 2 = 1257 mm 2

80 mm

d = 80 mm

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Problem 8.48 Determine J O and k O .

120 mm 80 mm

20 mm x 40 mm 80 mm

Solution:

For the rectangle,

J O1 = I x 1 + I y 1 =

40 mm

1 1 3 3 3 bh + 3 hb

J O1 = 4.608 × 10 7 + 2.048 × 10 7 mm 4 J O1 = 6.656 × 10 7 mm 4

y9 R 5 20 mm

A1 = bh = 9600 mm 2 x9

For the circular cutout about x'y' A2

J'O 2 = I x' 2 + I y'2 = 14 π R 4 + 14 π R 4 J'O 2 = 1.257 × 10 5 + 1.257 × 10 5 mm 4 J'O 2 = 2.513 × 10 5 mm 2

(h) 80 mm

120 mm

Using the parallel axis theorem to determine J O2 (about x, y ) J O 2 = J ′02 + (d x 2 + d y 2 ) A2 A2 = π R 2 = 1257 mm 2 J O 2 = 1.030 × 10 7 mm 4

x 80 mm (b)

J O = J O1 − J O 2 J O = 6.656 × 10 7 − 1.030 × 10 7 mm 4 J O = 5.63 × 10 7 mm 4 kO =

JO = Area

JO A1 − A2

k O = 82.1 mm

570

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Problem 8.49 Determine I xy .

120 mm 80 mm

20 mm x 40 mm 80 mm

Solution:

y 80 mm

A1 = (80)(120) = 9600 mm 2 A2 =

πR 2

π (20) 2

= 1257 mm 2 R 5 20 mm

For the rectangle ( A1 ) 1 1 2 2 b h = (80) 2 (120) 2 4 4 I xy1 = 2.304 × 10 7 mm 2

I xy1 =

For the cutout I x'y'2 = 0

x9 A2 A1

120 mm

A2 dy 5 80 mm

and by the parallel axis theorem I xy 2 = I x'y' 2 + A2 (d x )(d y ) I xy 2 = 0 + (1257)(40)(80)

dx 5 40 mm

I xy 2 = 4.021 × 10 6 mm 4 I xy = I xy1 − I xy 2 I xy = 2.304 × 10 7 − 0.402 × 10 7 mm 4 I xy = 1.90 × 10 7 mm 4

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Problem 8.50 Determine I x and k x .

20 mm 120 mm 80 mm

40 mm 80 mm

Solution:

We must first find the location of the centroid of the total area. Let us use the coordinates XY to do this. Let A1 be the rectangle and A2 be the circular cutout. Note that by symmetry X c = 40 mm

Y 80 mm

Area

Rectangle1

9600 mm2

40 mm

60 mm

Circle2

1257 mm2

40 mm

80 mm

R 5 20 mm 120 mm

A1 = 9600 mm 2 A2 = 1257 mm 2

80 mm

For the composite, A1 X c1 − A2 X c 2 = 40 mm A1 − A2 A X − A2 X c 2 Yc = 1 c1 = 57.0 mm A1 − A2

Xc =

X 40 mm

Now let us determine I x and k x about the centroid of the composite body. Rectangle about its centroid (40, 60) mm 1 1 bh 3 = (80)(120) 3 12 12 I x1 = 1.152 × 10 7 mm 3 , I x1 =

Now to C → d y 2 = 80 − 57 = 23 mm I xc 2 = I x 2 + (d y 2 ) 2 A2 I xc 2 = 7.91 × 10 5 mm 4

Now to C

For the composite about the centroid

I xc1 = 1 x1 + (60 − Yc ) 2 A1

I x = I xc1 − I xc 2

I xc1 = 1.161 × 10 7 mm 4

I x = 1.08 × 10 7 mm 4

Circular cut out about its centroid

The composite Area = 9600 − 1257 mm 2

A2 = π R 2 = (20) 2 π = 1257 mm 2 I x 2 = 14 π R 4 = π (20) 4 /4 I x 2 = 1.26 × 10 5 mm 4

572

40 mm

= 8343 mm 2 kx =

Ix = 36.0 mm A

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Problem 8.51 Determine I y and k y .

20 mm 120 mm 80 mm

40 mm 80 mm

Solution:

From the solution to Problem 8.50, the centroid of the composite area is located at (40, 57.0) mm. The area of the rectangle, A1 , is 9600 mm 2 .

The area of the cutout, A2 , is 1257 mm 2 . The area of the composite is 8343 mm 2 . (1) Rectangle about its centroid (40, 60) mm. I y1 =

1 1 hb 3 = (120)(80) 3 12 12

I y1 = 5.12 × 10 6 mm 4

80 mm

d x1 = 0 (2) Circular cutout about its centroid (40, 80) I y 2 = π R 4 /4 = 1.26 × 10 5 mm 4

40 mm

d x2 = 0 Since d x1 and d x2 are zero. (no translation of axes in the x-direction), we get

80 mm

I y = I y1 − I y 2 I y = 4.99 × 10 6 mm 4 Finally, ky =

Iy = A1 − A2

4.99 × 10 6 8343

k y = 24.5 mm

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Problem 8.52 Determine J O and k O .

20 mm 120 mm 80 mm

40 mm 80 mm

Solution:

From the solutions to Problems 8.51 and 8.52,

I x = 1.07 × 10 7 mm 4 I y = 4.99 × 10 6 mm 4 and A = 8343 mm 2 JO = I x + I y kO =

= 1.57 × 10 7 mm 4

JO = 43.4 mm A

120 mm

80 mm

574

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Problem 8.53 Determine I y and k y .

12 in x 20 in

Solution:

Treat the area as a circular area with a half-circular cutout: From Appendix B,

y 1 12 in.

( I y ) 1 = 14 π (20) 4 in 4 and ( I y ) 2 = 18 π (12) 4 in 4 ,

2 x

20 in.

2 in. x

20 in.

So I y = 14 π (20) 4 − 18 π (12) 4 = 1.18 × 10 5 in 4 . The area is A = π (20) 2 − so, k y =

Iy = A

1 π (12) 2 = 1030 in 2 2

1.18 × 10 5 1.03 × 10 3

= 10.7 in

Solution:

Problem 8.54 Determine J O and k O .

Treating the area as a circular area with a half-circular cutout as shown in the solution of Problem 8.53, from Appendix B, ( J O ) 1 = ( I x ) 1 + ( I y ) 1 = 12 π (20) 4 in 4

and ( J O ) 2 = ( I x ) 2 + ( I y ) 2 = 14 π (12) 4 in 4 . Therefore J O = 12 π (20) 4 − 14 π (12) 4 = 2.35 × 10 5 in 4 .

12 in x 20 in

From the solution of Problem 8.53, A = 1030 in 2 RO = =

BandF_6e_ISM_CH08.indd 575

JO A

2.35 × 10 5 = 15.1 in. 1.03 × 10 3

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Problem 8.55 Determine I y and k y if h = 3 m. Solution:

Break the composite into two parts, a rectangle and a 1.2 m

semi-circle. For the semi-circle I x ′c =

( π8 − 89π ) R

I y ′c =

1 4 4R πR d = 8 3π

h y9

y d

x9 1.2 m AC

d 5 4R 3p To get moments about the x and y axes, the (d xc , d yc ) for the semicircle are d xc = 0,

d yc = 3 m +

AR 3m 5 h

4R 3π

and Ac = π R 2 /2 = 2.26 m 2 I y ′c =

1 4 πR 8

2 A (d and I yc = I y ′c + d xc x = 0)

To get moments of area about the x, y axes, d xR = 0, d yR = 1.5 m

I yc = I y'c = π (1.2) 4 /8

I yR = I y ′R + ( d xR ) 2 (bh)

I yc = 0.814 m 4

I yR = I y ′R =

For the Rectangle

1 (3)(2, 4) 3 m 4 12

I x ′R =

1 bh 3 12

I yR = 3.456 m 2

I y ′R =

1 hb 3 12

A R = bh = 7.2 m 2 I y = I yc + I yR

A R = bh

I y = 4.27 m 2 To find k y , we need the total area, A = A R + Ac y9 y

A = 7.20 + 2.26 m 2 A = 9.46 m 2

2.4 m

ky = h

576

Iy = 0.672 m A

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Problem 8.56 Determine I x and k x if h = 3 m.

1.2 m

Solution:

Break the composite into two parts, the semi-circle and the rectangle. From the solution to Problem 8.55, I x ′c =

( π8 − 89π ) R (

d yc = 3 +

)

4R m 3π

Ac = 2.26 m 2 I xc = I x ′c + Ac d yc 2

R 5 1.2 m h53m b 5 2.4 m

Substituting in numbers, we get I x ′c = 0.0717 m 4 d yc = 3.509 m

I xc = 27.928 m 2

2.4 m

and I xc = I x ′c + Ac d y 2 yc9

For the Rectangle h = 3 m, b = 2.4 m Area: A R = bh = 7.20 m 2 1 I x ′R = bh 3 , d yR = 1.5 m 12 I xR = I x ′R + d yR 2 A R

xc9

Substituting, we get I x ′R = 5.40 m 4 I xR = 21.6 m 4

4R 3p

For the composite, I x = I xR + I xc I x = 49.5 m 4 Also k x =

Ix = 2.29 m A R + Ac

k x = 2.29 m

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Problem 8.57 If I y = 5 m 4 , what is the dimension h?

1.2 m

Solution:

From the solution to Problem 8.55, we have:

y9c

For the semicircle

xc9

I y ′c = I y = π (1.2) 4/8 = 0.814 m 2 For the rectangle 1 I y ′R = I yR = (h)(2.4) 3 m 4 12 Also, we know I yR + I yc = 5 m 4 . Hence 0.814 +

1.2

y9R h

x9R b 2.4 m

1 (h)(2.4) 3 = 5 12

Solving, h = 3.63 m

578

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Problem 8.58 Determine the moment of inertia about the x-axis of the protractor’s area. y9 y

x, x9

608

50 mm

20 mm

Source: Courtesy of narahmommy/123RF.

Solution:

(I x ) a =

From Appendix B, the moment of inertia of the circular area about the x-axis is 1 π (70 mm) 4 . 4

The moment of inertia of the semicircular “hole” about the x-axis is (I x ) b =

50 mm

70 mm

(a)

y 608

(c)

308 x

50 mm

1 1 (50 mm) 4 (α − sin 2α). 4 2

Consider the upper half of the triangular area in Fig. (d). Its horizontal dimension is b = (50 mm)sin 30 ° and its vertical dimension is h = (50 mm) cos30 °. Its moment of inertia about the x-axis is

(d)

BandF_6e_ISM_CH08.indd 579

Let the angle α = 60 ° = (60 °/180 °)π radians. The moment of inertia of the sector “hole” about the x-axis is (I x ) c =

(b)

1 π (50 mm) 4 . 8

( I x ) upper half =

1 bh 3 , 12

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Problem 8.58

(Continued)

so the moment of inertia of the entire triangular area is ( I x ) d = 2( I x ) upper half =

1 3 1 bh = [ (50 mm)sin 30 ° ][ (50 mm) cos30 ° ]3 . 6 6

The moment of inertia about the x-axis of the composite area is I x = (I x ) a − (I x ) b − (I x ) c + (I x ) d = 1.58E7 mm 4 . I x = 1.58E7 mm 4 .

Problem 8.59 Determine the moment of inertia about the y-axis of the protractor’s area. y9 y

x, x9

608

50 mm

20 mm

580

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Problem 8.59

(Continued)

Let the angle α = 60 ° = (60 °/180 °)π radians. The moment of inertia of the sector “hole” about the y-axis is

Solution:

(I y ) c =

(

)

1 1 (50 mm) 4 α + sin 2α . 4 2

50 mm

70 mm

( I y ) upper half =

1 3 hb , 4

so the moment of inertia of the entire triangular area is ( I y ) d = 2( I y ) upper half =

(a)

(b) y

y 608

(c)

1 3 1 hb = [ (50 mm) cos30 ° ] 2 2 3 [ (50 mm)sin 30 ° ] .

The moment of inertia about the y-axis of the composite area is

308 x

50 mm

50 mm (d)

I y = (I y ) a − (I y ) b − (I y ) c + (I y ) d = 1.44E7 mm 4 . I y = 1.44E7 mm 4 .

From Appendix B, the moment of inertia of the circular area about the y-axis is (I y ) a =

1 π (70 mm) 4 . 4

The moment of inertia of the semicircular “hole” about the y-axis is (I y ) b =

1 π (50 mm) 4 . 8

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Problem 8.60 Determine the moment of inertia about the y ′-axis through the centroid of the area of the protractor. Strategy: Determine the moment of inertia about the y-axis and then apply the parallel-axis theorem to the entire area. y9 y

x, x9

608

50 mm

20 mm

582

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Problem 8.60

The x coordinate of the centroid of the entire area is

(Continued)

Solution:

We first need to locate the centroid. We can represent the area of the protractor as a composite consisting of (a) a solid circular area; (b) a semicircular “hole”; (c) a sector “hole”; (d) a triangular area to “fill in” part of the sector: y

x a Aa − x b A b − x c A c + x d A d A = −2.94 mm.

x =

We will next determine the moment of inertia of the entire area about the y-axis. From Appendix B, the moment of inertia of the circular area about the y-axis is (I y ) a =

50 mm

70 mm

(a)

(I y ) c =

308 x

50 mm

(c)

(d)

4(50 mm) 1 π (50 mm) 2 , x b = . 2 3π

The angle α = 60 ° = π (60 °/180 °) radians. The area and the x coordinate of the centroid of the sector “hole” are Ac = α(50 mm) 2 , x c = −

1 π (50 mm) 4 . 8

Let the angle α = 60 ° = (60 °/180 °)π radians. The moment of inertia of the sector “hole” about the y-axis is

(b) y

608

The moment of inertia of the semicircular “hole” about the y-axis is (I y ) b =

1 π (70 mm) 4 . 4

2(50 mm)sin α . 3α

The area and the x coordinate of the centroid of the triangular area are 2 Ad = (50 mm)sin 30 °(50 mm) cos30°, x d = − (50 mm)sin 30 °. 3 The entire area of the protractor is

(

)

1 1 (50 mm) 4 α + sin 2α . 4 2

1 3 hb , 4

so the moment of inertia of the entire triangular area is ( I y ) d = 2( I y ) upper half =

1 3 1 hb = [ (50 mm) cos30 ° ] 2 2 3 [ (50 mm)sin 30 ° ] .

The moment of inertia about the y-axis of the composite area is I y = (I y ) a − (I y ) b − (I y ) c + (I y ) d = 1.44E7 mm 4 . We can now determine the moment of inertia of the entire area about the y′ axis by applying the parallel axis theorem: I y′ = I y − d x 2 A = 1.44E7 mm 4 − (2.94 mm) 2 (9931 mm 2 ) = 1.43E7 mm 4 . I y ′ = 1.43E7 mm 4 .

A = Aa − Ab − Ac + Ad = 9931 mm 2 .

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Problem 8.61 Determine I y and k y .

y 30 in

x9 3

40 in

30 in

x 40 in

20 in x

x = 20 in +

20 in

Solution:

The x coordinate of the centroid of the quarter-circular area is

We’ll divide the area into the parts shown:

4R = 32.7 in. 3π

Using the result given in Appendix B, its moment of inertia about the y-axis is (I y ) 3 =

( 16π − 94π ) R + x ( 14 πR ) 4

= 8.02E5 in 4 . 30 in

40 in

The moment of inertia of the entire area about the y-axis is

I y = ( I y ) 1 + ( I y ) 2 + ( I y ) 3 = 2.55E6 in 4 . The total area is A = A1 + A2 + A3 = 3310 in 2 , and the radius of gyration about the y-axis is

ky =

x 20 in

Iy = 27.8 in. A

I y = 2.55E6 in 4 , k y = 27.8 in.

From Appendix B, the moment of inertia of rectangle 1 about the y-axis is (I y )1 =

1 (40 in)(50 in) 3 = 1.67E6 in 4 . 3

and the moment of inertia of rectangle 2 is (I y ) 2 =

584

1 (30 in)(20 in) 3 = 8.00E4 in 4 . 3

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Problem 8.62 Determine I x and k x .

y 30 in

x9 3

40 in

30 in

x 40 in

20 in x

y = 40 in +

20 in

Solution:

The y coordinate of the centroid of the quarter-circular area is

We’ll divide the area into the parts shown:

4R = 52.7 in. 3π

Using the result given in Appendix B, its moment of inertia about the y-axis is (I x ) 3 =

( 16π − 94π ) R + y ( 14 πR ) 4

= 2.01E6 in 4 . 30 in

The moment of inertia of the entire area about the y-axis is I x = ( I x ) 1 + ( I x ) 2 + ( I x ) 3 = 4.94E6 in 4 .

The total area is A = A1 + A2 + A3 = 3310 in 2 , and the radius of gyration about the y-axis is

40 in

x 20 in

kx =

Ix = 38.6 in. A

I x = 4.94E6 in 4 , k x = 38.6 in.

From Appendix B, the moment of inertia of rectangle 1 about the x-axis is (I x )1 =

1 (30 in)(40 in) 3 = 6.40E5 in 4 . 3

and the moment of inertia of rectangle 2 is (I x ) 2 =

1 (20 in)(70 in) 3 = 2.29E6 in 4 . 3

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Problem 8.63 Determine I xy .

Applying the parallel-axis theorem, we have ( I xy ) 1 = I x ′y ′ + xyA

= 0 + xyA = 8.40E5 in 4 . The area of part 2 is A = (20 in)(70 in) = 1400 in 2 . The coordinates of its centroid are 30 in

x = 10 in, y = 35 in. Applying the parallel-axis theorem, we have

40 in

( I xy ) 2 = I x ′y ′ + xyA = 0 + xyA = 4.90E5 in 4 .

The area of part 3 is

20 in

Solution:

We’ll divide the area into the parts shown:

A =

1 π (30 in) 2 = 707 in 2 . 4

From Appendix B, the coordinates of its centroid are 4 ( 30 in ) = 32.7 in, 3π 4 ( 30 in ) = 52.7 in. y = 40 in + 3π

x = 20 in + 30 in

Applying the parallel-axis theorem, we have ( I xy ) 3 = I x ′y ′ + xyA

40 in

( 81 − 94π )(30 in) + xyA 4

= 1.21E6 in 4 .

x 20 in

We obtain I xy = ( I xy ) 1 + ( I xy ) 2 + ( I xy ) 3 = 2.54E6 in 4 .

Let us analyze the three areas in sequence. The area of part 1 is

I xy = 2.54E6 in 4 .

A = (30 in)(40 in) = 1200 in 2 . The coordinates of its centroid are x = 35 in, y = 20 in.

586

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Problem 8.64 The dimensions b = 16 in and h = 4 in. Determine I y .

2h h x b

Solution: y

The area consists of a rectangle and a triangle: y9

Applying the parallel-axis theorem, the moment of inertia of the triangle about the y-axis is ( I y ) tri = ( I y ′ ) tri +

h h

x b/3

1 1 hb 3 + hb 3 36 18

1 hb 3 . 12

I y = ( I y ) rect + ( I y ) tri

From Appendix B, the moment of inertia of the rectangle about the y-axis is 1 3 hb , 3

and the moment of inertia of the triangle about the y ′ -axis through its centroid is ( I y ′ ) tri =

The moment of inertia of the entire area about the y-axis is b

( I y ) rect =

( b3 ) 12 bh

1 3 1 hb + hb 3 3 12 5 = hb 3 12 = 6827 in 4 . =

I y = 6827 in 4 .

1 hb 3 . 36

Problem 8.65 The dimensions b = 16 in and h = 4 in. Determine J O . 2h

Solution:

The area consists of a rectangle and a triangle:

Applying the parallel-axis theorem, the polar moment of inertia of the triangle about the origin of the x -y coordinate system is x'

h/3 h

x b/3 b

 h 2 b 2 1 ( J O ) tri = ( J O ′ ) tri +  h + +  bh  3 3  2 1 1 16 3 1 bh 3 + hb 3 + bh + hb 3 = 36 36 18 18 11 3 1 bh + hb 3 . = 12 12

(

)

( ) (

From Appendix B, the polar moment of inertia of the rectangle is

The polar moment of inertia of the entire area is

( J O ) rect = ( I x ) rect + ( I y ) rect

J O = ( J O ) rect + ( J O ) tri 1 1 11 3 1 = bh 3 + hb 3 + bh + hb 3 3 3 12 12 15 3 5 bh + hb 3 = 12 12 = 8107 in 4 .

1 3 1 bh + hb 3 . 3 3

The polar moment of inertia of the triangle about its centroid is ( J O ′ ) tri = ( I x ′ ) tri + ( I y ′ ) tri =

1 1 bh 3 + hb 3 . 36 36

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)

J O = 8107 in 4 .

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Problem 8.66 The dimensions b = 16 in and h = 4 in. Determine I xy . y

2h h x b

Solution:

The area consists of a rectangle and a triangle:

changes the sign of the product of inertia: ( I x ′y ′ ) tri = −

1 2 2 b h . 72

Applying the parallel-axis theorem, the product of inertia of the triangle is x9

h/3

( I xy ) tri = ( I x ′y ′ ) tri +

= −

x b/3

= b

From Appendix B, the product of inertia of the rectangle is ( I xy ) rect =

1 2 2 b h . 4

( b3 )( h + h3 )( 12 bh )

1 2 2 2 b h + b 2h 2 72 9

15 2 2 b h . 72

The product of inertia of the entire area is I xy = ( I xy ) rect + ( I xy ) tri

For the triangle in the orientation y9

1 2 2 15 2 2 b h + b h 4 72

11 2 2 b h 24

= 1877 in 4 . x9

I xy = 1877 in 4 .

Appendix B gives the product of inertia ( I x ′y ′ ) tri =

1 2 2 b h . 72

Reflecting the triangle into the orientation y9

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Problem 8.67 Determine I y and k y .

y 6 in 2 in

8 in

Solution:

We divide the composite area into a triangle (1), rectangle (2), half-circle (3), and circular cutout (4):

Triangle: (I y )1 =

1 (12)(8) 3 = 1536 in 4 4

3 4

Rectangle: (I y ) 2 =

8 in

1 (12)(8) 3 + (12) 2 (8)(12) = 14,336 in 4 . 12

Half-Circle:

(

)

Circular cutout: (I y ) 4 =

8 in.

4(6)  2 1 π 8 (I y ) 3 = − π (6) 2 = 19, 593 in 4 (6) 4 +  16 +  8 9π 3π  2

12 in.

1 π (2) 4 + (16) 2 π (2) 2 = 3230 in 4 . 4

Therefore

8 in.

12 in.

I y = ( I y ) 1 + ( I y ) 2 + ( I y ) 3 − ( I y ) 4 = 3.224 × 10 4 in 4 . y

The area is

A = A1 + A2 + A3 − A4 1 1 = (12)(8) + (8)(12) + π (6) 2 − π (2) 2 = 188 in 2 , 2 2 Iy 3.224 × 10 4 so k y = = = 13.1 in. A 188

6 in.

x 16 1 4(6) in. 3p

2 in.

16 in.

Rectangle:

Problem 8.68 Determine J O and k O .

( I x ) 2 = 13 (8)(12) 3 = 4608 in 4 .

Half Circle: ( I x ) 3 = 18 π (6) 4 + (6) 2 12 π (6) 2 = 2545 in 4 .

6 in

Circular Cutout:

2 in

( I x ) 4 = 14 π (2) 4 + (6) 2 π (2) 2 = 465 in 4 . Therefore 8 in

8 in

Solution:

I y is determined in the solution to Problem 8.67. We will determine I x and use the relation J O = I x + I y . Using the figures in the solution to Problem 8.67, Triangle: (I x )1 =

I x = ( I x ) 1 + ( I x ) 2 + ( I x ) 3 − ( I x ) 4 = 7840 in 4 . Using the solution of Problem 8.67, J O = I x + I y = 0.784 × 10 4 + 3.224 × 10 4 = 4.01 × 10 4 in 4 . From the solution of Problem 8.67, A = 188 in 2 , so R0 =

JO = A

4.01 × 10 4 = 14.6 in. 188

1 (8)(12) 3 = 1152 in 4 . 12

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Problem 8.69 Determine I y and k y .

y 4 in 2 in 4 in

8 in x

12 in 16 in

Solution: Divide the area into four parts: Part (1) The rectangle 8 in by 16 in. Part (2): The rectangle 4 in by 8 in. Part (3) The semicircle of radius 4 in, and Part (4) The circle of radius 2 in. Part (1): A1 = 16(8) = 128 in 2 ,

The area moments of inertia about the centroid of the semicircle are I yy 3 =

( 81 ) π(4 ) = 100.53 in ,

x 1 = 8 in,

I xx 3 =

( 81 ) π(4 ) − ( 4(4) ) A = 28.1 in . 3π

y 1 = 4 in,

Check:

I xx1 =

I xx3 = 0.1098( R 4 ) = 28.1 in 4 .

( 121 ) 16(8 ) = 682.67 in ,

check.

( )

Part (4): A4 = π (2 2 ) = 12.566 in 2 ,

1 I yy1 = 8(16 3 ) = 2730.7 in 4 . 12

x 4 = 12 in,

Part (2): A2 = 4(8) = 32 in 2 ,

y 4 = 12 in,

x 2 = 12 in,

I xx4 = ( 14 ) π (2 4 ) = 12.566 in 4 ,

y 2 = 10 in, I xx2 =

( 121 ) 8(4 ) = 42.667 in ,

I yy2 =

( )

I yy 4 = I xx 4 = 12.566 in 4 .

The composite area:

1 4(8 3 ) = 170.667 in 4 . 12

Part (3): A3 =

A = ∑ Ai − A4 = 172.566 in 2 . 1

π (4 2 ) = 25.133 in 2 , 2

The area moment of inertia:

x 3 = 12 in. y 3 = 12 +

I y = x 1 2 A1 + I yy1 + x 2 2 A2 + I yy 2 + x 3 2 A3 + I yy 3 − x 4 2 A4 − I yy 4 I y = 1.76 × 10 4 in 4 ,

( 4(4) ) = 13.698 in. 3π

ky =

Iy = 10.1 in A

Solution:

Problem 8.70 Determine I x and k x .

Ix =

Use the results in the solution to Problem 8.69.

y 1 2 A1 + I xx1 + y 2 2 A2

+ I xx 2 + y 3 2 A3 + I xx 3 − y 4 2 A4 − I xx 4

I x = 8.89 × 10 3 in 4

4 in 2 in

kx =

4 in

Ix = 7.18 in A

8 in

12 in

16 in

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Problem 8.71 Determine I xy .

Solution:

Use the results in the solution to Problem 8.69.

I xy = x 1y 1 A1 + x 2 y 2 A2 + x 3y 3 A3 − x 4 y 4 A4 ,

I xy = 1.0257 × 10 4 in 4 4 in 2 in 4 in

8 in x

12 in 16 in

Solution:

Problem 8.72 Determine I y and k y .

Use the results in the solutions to Problems 8.69 to 8.71. The centroid is

x = 4 in =

1024 + 384 + 301.6 − 150.8 = 9.033 in, 172.567

from which

2 in

4 in

x 1 A1 + x 2 A2 + x 3 A3 − x 4 A4 A

I yc = −x 2 A + I y = −1.408 × 10 4 + 1.7598 × 10 4 = 3518.2 in 4 x 8 in

k yc =

I yc = 4.52 in A

12 in 16 in

Problem 8.73 Determine I x and k x .

Solution: Use the results in the solutions to Problems 8.69 to 8.71. The centroid is

y = 4 in

y 1 A1 + y 2 A2 + y 3 A3 − y 4 A4 = 5.942 in, A

from which I xc = −y 2 A + I x = −6092.9 + 8894 = 2801 in 4

2 in

4 in

k xc =

I xc = 4.03 in A

8 in

12 in 16 in

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Solution:

Problem 8.74 Determine I xy . y

Use the results in the solutions to Problems 8.69–8.71.

I xyc = −xy A + I xy = −9.263 × 10 3 + 1.0257 × 10 4 = 994.5 in 4 4 in

2 in

4 in

x 8 in

12 in 16 in

Problem 8.75 Determine I y and k y .

5 mm 15 mm 50 mm

Solution: (I y )1 =

15 mm

1 (50 + 15 + 15)(30) 3 = 180,000 mm 4 12

(I y ) 2 = (I y ) 3 = (I y ) 4 =

5 mm

We divide the area into parts as shown:

15 mm 10 15 15 10 mm mm mm mm

1 (30)(10) 3 + (20) 2 (10)(30) 12

= 122, 500 mm 4 (I y ) 5 = (I y ) 6 = (I y ) 7 =

 1 π (15) ( π8 − 98π ) (15) +  25 + 4(15) 3π  2 4

y 1

= 353, 274 mm 4 ( I y ) 8 = ( I y ) 9 = ( I y ) 10 =

1 π (5) 4 + (25) 2 π (5) 2 = 49, 578 mm 4 . 4

Therefore, I y = ( I y ) 1 + 3( I y ) 2 + 3( I y ) 5 − 3( I y ) 8 = 1.46 × 10 6 mm 4 .

2 5 8 50 mm

7 4 10

3 6 9

10 15 mm mm

The area is A = A1 + 3 A2 + 3 A5 − 3 A8 = (30)(80) + 3(10)(30) + 3

( 12 )π(15) − 3π(5) 2

= 4125 mm 2 so k y =

592

Iy = A

1.46 × 10 6 = 18.8 mm 4125

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Problem 8.76 Determine J O and k O .

5 mm 15 mm 50 mm 5 mm

5 mm

Solution:

I y is determined in the solution to Problem 8.75. We will determine I x and use the relation J O = I x + I y . Dividing the area as shown in the solution to Problem 8.75, we obtain (I x )1 =

1 (30)(80) 3 + (25) 2 (30)(80) = 2, 780, 000 mm 4 12

(I x ) 2 =

1 (10)(30) 3 + (50) 2 (10)(30) = 772, 500 mm 4 12

(I x ) 8 =

= 4.34 × 10 6 mm 4 and J O = I x + I y = 5.80 × 10 6 mm 4 . From the solution to Problem 8.75, A = 4125 mm 2

1 π (15) 4 = 19,880 mm 4 , 8

so k O =

1 π (5) 4 + π (5) 2 (50) 2 , 4

( I x ) 9 = ( I x ) 10 =

10 15 15 10 mm mm mm mm

I x = ( I x ) 1 + ( I x ) 2 + 2( I x ) 3 + ( I x ) 5 + 2( I x ) 6 − ( I x ) 8 − 2( I x ) 9

1 1 π (15) 4 + (50) 2 π (15) 2 = 903, 453 mm 4 8 2

(I x ) 6 = (I x ) 7 =

15 mm

Therefore

1 (I x ) 3 = (I x ) 4 = (10)(30) 3 = 22, 500 mm 4 12 (I x ) 5 =

15 mm

1 π (5) 4 = 491 mm 4 . 4

JO A 5.80 × 10 6 4125

= 37.5 mm.

Problem 8.77 Determine I x and I y for the beam’s cross section. y

Solution:

Use the symmetry of the object

Ix 1 1 = (3 in.)(8 in) 3 +  (3 in)(3 in) 3 + (3 in) 2 (11.5 in) 2   12  2 3  π (5 in) 4 π (5 in) 2 4[5 in] 2 +  −  16 4 3π π(5 in) 2 4[5 in] 2  + 8 in +   4 3π  π(2 in) 4 π(2 in) 2 4[2 in] 2 − −  16 4 3π π(2 in) 2 4[2 in] 2  + 8 in +   4 3π

(

5 in

2 in

)

(

)

(

)

Solving we find I x = 7016 in 4

8 in

x 3 in

5 in

3 in

Iy 1 1 = (3 in)(3 in) 3 +  (8 in)(3 in) 3 + (8 in)(3in)(6.5 in) 2   12  2 3  π (5 in) 4 π (5 in) 2 4[5 in] 2 +  −  16 4 3π π(5 in) 2 4[5 in] 2  + 3 in +   4 3π 2  π(2 in) 4 π(2 in) 2 4[2 in] − −  16 4 3π π(2 in) 2 4[2 in] 2  + 3 in +   4 3π

(

)

(

)

(

)

Solving we find I y = 3122 in 4

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Problem 8.78 Determine I x and I y for the beam’s cross section.

5 in

2 in

Solution: Use Solution 8.77 and 7.39. From Problem 7.39 we know

8 in

that y = 7.48 in, A = 98.987 in 2 I x = 7016 in 4 − Ay 2 = 1471 in 4 I y = 3122 in 4 − A(0) 2 = 3122 in 4

3 in

Problem 8.79 The area A = 2 ×10 4 mm 2 . Its moment of inertia about the y-axis is I y = 3.2 × 10 8 mm 4 . Determine its moment of inertia about the ŷ-axis.

5 in

ŷ

5 in

3 in

Solution:

Use the parallel axis theorem. The moment of inertia about the centroid of the figure is I yc = −x 2 A + I y = −(120 2 )(2 × 10 4 ) + 3.2 × 10 8 = 3.20 × 10 7 mm 4 . The moment of inertia about the ŷ axis is I yˆ =

x2A + I

x, x̂ 100 mm

120 mm

I yˆ = (220 2 )(2 × 10 4 ) + 3.2 × 10 7 = 1 × 10 9 mm 4 y

Problem 8.80 The area A = 100 in 2 and it is symmetric about the x ′-axis. The moments of inertia I x ′ = 420 in 4, I y ′ = 580 in 4, J O = 11, 000 in 4, and I xy = 4800 in 4 . What are I x and I y ?

A x9 O9

Solution: (1) (2) (3) (4) (5) (6)

I x = y 2 A + I xc , I y = x 2 A + I yc , J O = Ar 2 + J c , JO = I x + I y, J c = I xc + I yc , and I xy = Axy + I xyc ,

where the subscript c applies to the primed axes, and the others to the unprimed axes. The x, y values are the displacement of the primed axes from the unprimed axes. The steps in the demonstration are: (i)

The basic relationships:

(iv) The roots: x1 2 = 64, and x 2 2 = 36. The corresponding values of y are found from y = r 2 − x 2 from which ( x1 , y1 ) = (8, 6), and ( x 2 , y 2 ) = (6, 8). (v) Substitute these pairs to obtain the possible values of the area moments of inertia:

From symmetry about the x c axis, the product of inertia I xyc = 0.

J − JC (ii) From (3): r 2 = O = 100 in 2 , from which r 2 = x 2 + A 2 2 y = 100 in I xy (iii) From (6) and I xyc = 0, y = , from which x 2r 2 = x 4 + Ax 2  I xy  4 2  A  . From which: x − 100 x + 2304 = 0.

594

I x1 = Ay1 2 + I xc = 4020 in 4 , I y1 = Ax1 2 + I yc = 6980 in 4 I x 2 = Ay 2 2 + I xc = 6820 in 4 , I y 2 = Ax 2 2 + I yc = 4180 in 4

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Problem 8.81 Determine the moment of inertia of the beam cross section about the x-axis. Compare your result with the moment of inertia of a solid square cross section of equal area. (See Example 8.5.)

y 20 mm x 160 mm

Solution:

We first need to find the location of the centroid of the composite. Break the area into two parts. Use X , Y coords.

20 mm 100 mm

100 20

A1 5 2000 mm2 XC 5 0

160 mm

100 mm y Y

A2 5 3200 mm2 XC25 0

20 mm

YC2 5 80 mm

YC1 5 170 mm

20 mm 160 mm

For the composite Xc =

X c1 A1 + X c 2 A2 = 0 A1 + A2

Yc =

Yc1 A1 + Yc 2 A2 A1 + A2

Substituting, we get X c = 0 mm

20 mm

Yc = 114.6 mm We now find I x for each part about its center and use the parallel axis theorem to find I x about C.

Finally,

Part (1): b1 = 100 mm, h1 = 20 mm

I x = 1.686 × 10 7 mm 4

1 1 b h3 = (100)(20) 3 mm 4 12 1 1 12

for our composite shape.

I x ′1 =

I x = I x1 + I x 2

I x ′1 = 6.667 × 10 4 mm 4

Now for the comparison. For the solid square with the same total area A1 + A2 = 5200 mm 2 , we get a side of length

dy1 = Yc1 − Yc = 55.38 mm

l 2 = 5200: l = 72.11 mm

I x1 =

I x ′1 + (dy1 ) 2 ( A1 )

I x1 = 6.20 × 10 6 mm 4 Part (2) b 2 = 20 mm, h 2 = 160 mm 1 1 I x ′2 = (b )(h ) 3 = (20)(160) 3 mm 4 12 2 2 12 I x ′ 2 = 6.827 × 10 6 mm 4 dy 2 = Yc 2 − Yc = −34.61 mm

And for this solid section I xSQ =

1 1 4 bh 3 = l 12 12

I xSQ = 2.253 × 10 6 mm 4 Ratio = I x /I xSQ =

1.686 × 10 7 2.253 × 10 6

Ratio = 7.48 This matches the value in Example 8.5.

I x 2 = I x ′ 2 + (dy ) 2 A2 I x 2 = 1.066 × 10 7 mm 4

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Problem 8.82 In terms of the coordinate system shown, the coordinates of the points of the airliner’s wing are A: (19.0, 35.0) m, B: (22.6, 35.0) m, C : (13.2, 9.4) m, D: (10.0, 0) m, E : (13.2, 0) m. Determine the moment of inertia of the wing’s area about the x-axis. (Disregard the area of the “winglet,” the part of the wing above points A and B.)

Solution: To model the wing’s area with a composite area, we begin with the triangle and rectangle in Fig. a. In Fig. b we subtract a triangular “hole”. We then add the triangle in Fig. c. That sequence gives us the wing’s area. y

y A

(a)

(b)

D E

(c)

From the given coordinates of the points and using results from Appendix B, the moment of inertia about the x-axis of the triangle in Fig. a is ( I x ) triangle a =

1 (19 m)(35 m) 3 = 6.79E4 m 4 . 12

The moment of inertia of the rectangle in Fig. a is ( I x ) rectangle a =

1 (22.6 m − 19.0 m)(35 m) 3 = 5.15E4 m 4 . 3

The moment of inertia of the triangular “hole” in Fig. b is ( I x ) triangle b =

1 (22.6 m − 10.0)(35 m) 3 = 4.50E4 m 4 . 12

The moment of inertia of the triangle in Fig. c is ( I x ) triangle c =

1 (13.2 m − 10.0 m)(9.4 m) 3 = 221 m 4 . 12

The moment of inertia of the wing about the x-axis is I x = ( I x ) triangle a + ( I x ) rectangle a − ( I x ) triangle b + ( I x ) triangle c = 7.45E4 m 4 . I x = 7.45E4 m 4 .

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Problem 8.83 If the beam in Fig. a is subjected to couples of magnitude M about the x-axis (Fig. b), the beam’s longitudinal axis bends into a circular arc whose radius R is given by

y z (a) Unloaded.

EI x , R = M where I x is the moment of inertia of the beam’s cross section about the x - and y -axes. The value of the term E, which is called the modulus of elasticity, depends on the material of which the beam is constructed. Suppose that a beam with the cross section shown in Fig. c is subjected to couples of magnitude M = 180 N-m. As a result, the beam’s axis bends into a circular arc with radius R = 3 m. What is the modulus of elasticity of the beam’s material? (See Example 8.5.)

(b) Subjected to couples at the ends. y 3 mm

Solution:

The moment of inertia of the beam’s cross section about

the x-axis is Ix =

{

1 1 (3)(9) 3 + 2  (9)(3) 3 + (6) 2 (9)(3)   12  12

}

9 mm mm 4

= 2170 mm 4 = 2.17 × 10 −9 m 4 . The modulus of elasticity is (3 m)(180 N-m) RM E = = = 2.49 × 10 11 N/m 2 Ix 2.17 × 10 −9 m 4

3 mm 3 mm 9 mm (c) Beam cross section.

E = 2.49 × 10 11 N/m 2 .

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Problem 8.84 Suppose that you want to design a beam made of material whose density is 8000 kg/m 3 . The beam is to be 4 m in length and have a mass of 320 kg. Design a cross section for the beam so that I x = 3 × 10 −5 m 4 . (See Example 8.5.) Solution: The strategy is to determine the cross sectional area, and then use the ratios given in Figure 8.14 to design a beam. The volume of the beam is V = AL = 4 A m 3 . The mass of the beam is m = V (8000) = 32000 A = 320 kg, from which A = 0.01 m 2 . The moment of inertia for a beam of square cross section with this area is

( 121 )(0.1)(0.1 ) = 8.333 × 10

I xxb =

h b

−6 m 4 .

The ratio is

I - beam Flange dimension 1 .8 f .6 ( .4 b ) .2 * 0 E –.2 + –.4 5 –.6 –.8 –1 .03 .04 .05 .06 .07 .08 .09 .1 b

3 × 10 −5 R = = 3.6. 8.333 × 10 −6 From Figure 8.6, this ratio suggests an I -beam of the form shown in the sketch. Choose an I -beam made up of three equal area rectangles, of dimensions b by hm in section. The moment of inertia about the centroid is I x = y 1 2 A1 + I xx1 + y 2 2 A2 + I xx 2 + y 3 2 A3 + I xx 3 . Since all areas are equal, A1 = A2 = A3 = bh, and y 1 =

b+ h , 2

y 2 = 0, and y 3 = −y 1 , this reduces to Ix =

( 16 )bh + 2( b +2 h ) hb + ( 121 )hb . 3

A , where A is the known total cross section area. 3 These are two equations in two unknowns. Plot the function

Note that bh =

f (b) =

( 61 )bh + 2( b +2 h ) bh + ( 121 )hb − I 3

A subject to the condition that hb = . The function was graphed using 3 TK Solver Plus. The graph crosses the zero axis at approximately b = 0.0395 m. and b = 0.09 m. The lower value is an allowable value for h and the greater value corresponds to an allowable value of b. Thus the I beam design has the flange dimensions, b = 90 mm and h = 39.5 mm.

Problem 8.85 The area in Fig. a is a C230 × 30 American Standard Channel beam cross section. Its cross-sectional area is A = 3790 mm 2 and its moments of inertia about the x-axis are I x = 25.3 × 10 6 mm 4 and I y = 1 × 10 6 mm 4 . Suppose that two beams with C230 × 30 cross sections are riveted together to obtain a composite beam with the cross section shown in Fig. b. What are the moments of inertia about the x - and y -axes of the composite beam?

14.8 mm

(a)

(b)

Solution: I x = 2(25.3 × 10 6 mm 4 ) = 50.6 × 10 6 mm 4 I y = 2(10 6 mm 4 + [3790 mm 2 ][14.8 mm] 2 ) = 3.66 × 10 6 mm 4

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Problem 8.86 The area in Fig. a is an L152 × 102 ×12.7 Angle beam cross section. Its cross-sectional area is A = 3060 mm 2 and its moments of inertia about the x - and y -axes are I x = 7.24 × 10 6 mm 4 and I y = 2.61 × 10 6 mm 4 . Suppose that four beams with L152 × 102 × 12.7 cross sections are riveted together to obtain a composite beam with the cross section shown in Fig. b. What are the moments of inertia about the x - and y -axes of the composite beam?

y y 24.9 mm

50.2 mm

Solution: (a) I x = 4(7.24 × 10 6 mm 4 + [3060 mm 2 ][50.2 mm] 2 ) = 59.8 × 10 6 mm 4 I y = 4(2.61 × 10 6 mm 4 + [3060 mm 2 ][24.9 mm] 2 ) = 18.0 × 10 6 mm 4

Problem 8.87 The dimensions b = 4 in and h = 3 in. The moments of inertia of the area in terms of the xy coordinate system are I x = 324 in 4 , I y = 576 in 4 , and I xy = 252 in 4 . Determine I x ′ , I y ′ , and I x ′ y ′ for θ = 50 °.

(b)

y y9

h x b

Solution: I x′ =

From Eqs. 8.23–8.25, we obtain

324 + 576 324 − 576 cos2(50 °) − 252sin 2(50 °) + 2 2

= 224 in 4 , I y′ =

324 + 576 324 − 576 cos2(50 °) + 252sin 2(50 °) − 2 2

= 676 in 4 , I x ′y ′ =

324 − 576 sin 2(50 °) + 252cos2(50 °) 2

= −168 in 4 . I x ′ = 224 in 4 , I y ′ = 676 in 4 , I x ′y ′ = −168 in 4 .

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Problem 8.88 The dimensions b = 4 in and h = 3 in. The moments of inertia of the area in terms of the xy coordinate system are I x = 324 in 4 , I y = 576 in 4 , and I xy = 252 in 4 . Determine a set of principal axes and the corresponding principal moments of inertia.

y y9

h x b

Solution: tan 2θ p =

From Eq. 8.26,

2 I xy 2(252 in 4 ) = , Iy − Ix (576 in 4 ) − (324 in 4 )

y9 x9

we obtain θ p = 31.7 °. A set of principal axes is shown:

31.78

From Eqs. 8.23 and 8.24, the corresponding principal stresses are I x′ =

324 + 576 324 − 576 cos2(31.7 °) − 252sin 2(31.7 °) + 2 2

= 168 in 4 , I y′ =

324 + 576 324 − 576 cos2(31.7 °) + 252sin 2(31.7 °) − 2 2

= 732 in 4 . θ p = 31.7 °, I x ′ = 168 in 4 , I y ′ = 732 in 4 .

Problem 8.89 In Example 8.7, suppose that the area is reoriented as shown. Determine a set of principal axes and the corresponding principal moments of inertia. Based on the results of Example 8.7, can you predict a value of θ p without using Eq. (8.26)?

y 1 ft

3 ft

Solution:

Based on Example 8.7, the moments and product of inertia of the reoriented area are

1 ft

I x = 10 ft 4 , I y = 22 ft 4 , I xy = 6 ft 4 . From Eq. (8.26), tan 2θ p =

2 I xy 2(6) = = 1 ⇒ θ p = 22.5 ° Ix − Iy 22 − 10

4 ft

This value could have been anticipated fromExample 8.7 by reorienting the axes. Substituting the angle into Eqs. (8.23) and (8.24), the principal moments of inertia are Ix + Iy Ix − Iy + cos 2θ p − I xy sin 2θ p 2 2 10 + 22 10 − 22 = + cos 45 ° − 6sin 45 ° = 7.51 ft 4 , 2 2

I x′ =

I y′ = =

Ix + Iy Ix − Iy − cos 2θ p + I xy sin 2θ p 2 2 10 + 22 10 − 22 − cos 45 ° + 6sin 45 ° = 24.5 ft 4 , 2 2

θ p = 22.5 °, principal moments of inertia are 7.51 ft 4 , 24.5 ft 4 .

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Problem 8.90 The moments of inertia of the area are I x = 1.26 × 10 6 in 4 , I y = 6.55 × 10 5 in 4 , and I xy = −1.02 × 10 5 in 4 . Determine the moments of inertia of the area I x ′ , I y ′ , and I x ′ y ′ if θ = 30 °.

x9 u

Solution: I x′ =

Applying Eqs. (8.23)–(8.25),

Ix + Iy Ix − Iy + cos 2θ − I xy sin 2θ. 2 2

=  

1.26 + 0.655 1.26 − 0.655 + cos60° − (−0.102)sin 60°  × 10 6 in 4  2 2

= 1.20 × 10 6 in 4 I y′ =

Ix + Iy Ix − Iy − cos 2θ + I xy sin 2θ 2 2

=  

1.26 + 0.655 1.26 − 0.655 − cos60° + (−0.102)sin 60°  × 10 6 in 4  2 2

= 7.18 × 10 5 in 4 I x ′y ′ =

Ix − Iy sin 2θ + I xy cos 2θ 2

=  

1.26 − 0.655 sin 60° + (−0.102) cos60°  × 10 6 in 4  2

= 2.11 × 10 5 in 4 I x ′ = 1.20 × 10 6 in 4 , I y ′ = 7.18 × 10 5 in 4 , I x ′y ′ = 2.11 × 10 5 in 4 .

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Problem 8.91 The moments of inertia of the area are I x = 1.26 × 10 6 in 4 , I y = 6.55 × 10 5 in 4 , and I xy = −1.02 × 10 5 in 4 . Determine a set of principal axes and the corresponding principal moments of inertia.

x9 u x

Solution: tan 2θ p =

From Eq. (8.26),

2 I xy 2(−.102) = = 0.337 Iy − Ix 0.655 − 1.26

⇒ θ p = 9.32º Substituting this angle into Eqs. (8.23) and (8.24), the principal moments of inertia are I x′ =

Ix + Iy Ix − Iy + cos 2θ p − I xy sin 2θ p 2 2

=  

1.26 + 0.655 1.26 − 0.655 + cos18.63 ° − (−0.102)sin18.63 °   2 2

× 10 6 in 4 = 1.28 × 10 6 in 4 I y′ =

Ix + Iy Ix − Iy − cos 2θ p + I xy sin 2θ p 2 2

=  

1.26 + 0.655 1.26 − 0.655 − cos18.63 ° + (−0.102)sin18.63 °   2 2

× 10 6 in 4 = 6.38 × 10 5 in 4 θ p = 9.32 °, principal moments of inertia are 1.28 × 10 6 in 4 , 6.38 × 10 5 in 4 .

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Problem 8.92* Determine a set of principal axes and the corresponding principal moments of inertia.

y 160 mm 40 mm

200 mm

40 mm 40 mm 120 mm

Solution:

We divide the area into 3 rectangles as shown: In terms of the xˆ , yˆ coordinate system, the position of the centroid is

from Eq. (8.26),

xˆ A + xˆ 2 A1 + xˆ 3 A3 xˆ = 1 1 A1 + A2 + A3

tan 2θ p =

(20)(40)(200) + (100)(120)(40) + (80)(80)(40) = 56 mm, (40)(200) + (120)(40) + (80)(40)

yˆ A + yˆ 2 A1 + yˆ 3 A3 yˆ = 1 1 A1 + A2 + A3 =

(100)(40)(200) + (180)(120)(40) + (20)(80)(40) = 108 mm. (40)(200) + (120)(40) + (80)(40)

2 I xy 2(10.75 × 10 6 ) = , Iy − Ix (30.04 × 10 6 ) − (77.91 × 10 6 )

we obtain θ p = −12.1 °. We can orient the principal axes as shown: Substituting the values of I x , I y and I xy into Eqs. (8.23) and (8.24) and setting θ = −12.1 °, we obtain I x 1 = 80.2 × 10 6 mm 4 I y 1 = 27.7 × 10 6 mm 4 . y

^ y

The moments and products of inertia in terms of the xˆ , yˆ system are

160 mm

Iˆ x = ( Iˆ x ) 1 + ( Iˆ x ) 2 + ( Iˆ x ) 3 =

1 1 1 (40)(200) 3 + (120)(40) 3 + (180) 2 (120)(40) + (80)(40) 3 3 12 3

= 26.5 × 10 7 mm 4 , Iˆ y = ( Iˆ y ) 1 + ( Iˆ y ) 2 + ( Iˆ y ) 3 =

1 1 1 (200)(40) 3 + (40)(120) 3 + (100) 2 (120)(40) + (40)(80) 3 3 12 12 + (80) 2 (80)(40) = 8.02 × 10 7 mm 4 ,

1 200 mm

40 mm x

40 mm 40 mm

3 120 mm

y y9

Iˆ xy = ( Iˆ xy ) 1 + ( Iˆ xy ) 2 + ( Iˆ xy ) 3 = (20)(100)(40)(200) + (100)(180)(40)(120) + (20)(80)(40)(80) = 10.75 × 10 7 mm. The moments and product of inertia in terms of the xˆ , yˆ system are

x 12.18 x9

I x = Iˆ x − (yˆ ) 2 A = 77.91 × 10 6 mm 4 , I = Iˆ − (xˆ ) 2 A = 30.04 × 10 6 mm 4 , y

ˆ ˆ A = 10.75 × 10 6 mm 4 , I xy = Iˆ xy − xy

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Problem 8.93 In Practice Example 8.6, suppose that the vertical 3-m dimension of the triangular area is increased to 4 m. Use Mohr’s circle to determine a set of principle axes and the corresponding principle moments of inertia.

The angle and principal moments are now

Solution:

θ p = 28.2 °, principal moments of inertia are 4.21 m 4 , 81.1 m 4

The vertical 3-m dimension is increased to 4 m. From Problem 8.87, the moments and product of inertia for the unrotated system are Ix =

1 (4 m)(4 m) 3 = 21.3 m 4 , 12

Iy =

1 (4 m) 3 (4 m) = 64 m 4 , 4

I xy =

1 (4 m) 2 (4 m) 2 = 32 m 4 . 8

tan (2θ p ) =

32 ⇒ θ p = 28.2º , 64 − 42.7

I 1 = C + R = 81.1 m 4 , I 2 = C − R = 4.21 m 4 .

Mohr’s circle (shown) has a center and radius given by C =

21.3 + 64 = 42.7 m 4 2

R =

( 21.3 2− 64 ) + (32) = 38.5 m

Problem 8.94 For the area shown in Problem 8.89, use Mohr’s circle to determine the moments of inertia I x ′ , I y ′ , and I x ′ y ′ if θ = 30 °. (See Example 8.7.)

Now we can calculate the new inertias I x = C − R cos15 ° = 7.80 ft 4 I y = C + R cos15 ° = 24.2 ft 4 I xy = −R sin15 ° = −2.20 ft 4

Solution: Based on Example 8.7, the moments and product of inertia of the reoriented area are

I x ′ = 7.80 ft 4 , I y ′ = 24.2 ft 4 , I x ′y ′ = −2.20 ft 4 .

I x = 10 ft 4 , I y = 22 ft 4 , I xy = 6 ft 4 . For Mohr’s circle we have the center, radius, and angle

604

C =

10 + 22 = 16 ft 4 , 2

R =

( 22 −2 10 ) + (6) = 8.49 ft ,

θp =

1 6 tan −1 = 22.5º 22 − 16 2

(

)

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Problem 8.95 Solve Problem 8.89 by using Mohr’s circle. Solution: Based on Example 8.7, the moments and product of inertia of the reoriented area are I x = 10 ft 4 , I y = 22 ft 4 , I xy = 6 ft 4 . For Mohr’s circle we have the center, radius, and angle C =

10 + 22 = 16 ft 4 , 2

R =

( 22 −2 10 ) + (6) = 8.49 ft ,

θp =

1 6 tan −1 = 22.5 ° 2 22 − 16

(

)

Now we can calculate the principal moments of inertias I 1 = C + R = 24.5 ft 4 I 2 = C − R = 7.51 ft 4 θ p = 22.5º , principal moments of inertia are 7.51 ft 4 , 24.5 ft 4 .

Problem 8.96 Solve Problem 8.90 by using Mohr’s circle. Solution: C =

For Mohr’s circle we have the center, radius, and angle

( 12.6 +2 6.55 ) × 10 = 9.58 × 10 in , 5

9 9

12.6 − 6.55 2 + (−1.02) 2 × 10 5 = 3.19 in 4 , 2

R =

(

)

θp =

1 1.02 = 9.32 ° tan −1 2 12.6 − 9.58

(

)

Now we can calculate the new inertias I x′ = C + R cos22.7 ° = 12.0 × 10 5 in 4 ,

9 9

I y′ = C − R cos22.7° = 7.18 × 10 5 in 4 , I x′y′ = R sin 22.7 ° = 2.11 × 10 5 in 4 . I x′ = 1.20 × 10 6 in 4 , I y′ = 7.18 × 10 5 in 4 , I x′y′ = 2.11 × 10 5 in 4 .

Problem 8.97 Solve Problem 8.91 by using Mohr’s circle. Solution: C =

For Mohr’s circle we have the center, radius, and angle

( 12.6 +2 6.55 ) × 10 = 9.58 × 10 in , 5

R =

( 12.6 −2 6.55 ) + (−1.02) × 10 = 3.19 in ,

θp =

1 1.02 = 9.32 ° tan −1 2 12.6 − 9.58

(

)

Now we can calculate the principal inertias I 1 = C + R = 12.8 × 10 5 in 4 , I y ′ = C − R = 6.38 × 10 5 in 4 , θ p = 9.32 °, principal moments of inertia are 1.28 × 10 6 in 4 , 6.38 × 10 5 in 4 .

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Problem 8.98* Solve Problem 8.92 by using Mohr’s circle.

50 3106

(+)

Solution:

The moments and product of inertia are derived in terms of the xy coordinate system in the solution of Problem 8.92: 1

I x = 77.91 × 10 6 mm 4

2 up

I y = 30.04 × 10 6 mm 4

2 (30.0, 210.8)

I xy = 10.75 × 10 6 mm 4 . The Mohr’s circle is: Measuring the 2θ p, angle we estimate that θ p = −12º , and the principle moments of inertia are approximately 81 × 10 6 mm 4 and 28 × 10 6 mm 4 the orientation of the principal axes is shown in the solution of Problem 8.92.

Problem 8.99 Derive Eq. (8.22) for the product of inertia by using the same procedure we used to derive Eqs. (8.20) and (8.21).

(77.9, 10.8) (+) 100 3106

2 50 3106

Solution: Suppose that the area moments of inertia of the area A are known in the coordinate system ( x, y ), I x = ∫ y 2 dA, A

I y = ∫ x 2 dA, A

and I xy = ∫ xyA. A

The objective is to find the product of inertia in the new coordinate system ( x ′, y ′) in terms of the known moments of inertia. The new ( x ′, y ′) system is formed from the old ( x, y ) system by rotation about the origin through a counterclockwise angle θ. By definition, I x ′y ′ = ∫ x ′y ′ dA. A

From geometry, x ′ = x cos θ + y sin θ, and y ′ = −x sin θ + y cos θ. The product is x ′y ′ = xy cos 2 θ − xy sin 2 θ + y 2 cos θ sin θ − x 2 cos θ sin θ. Substitute into the definition: I x ′y ′ = (cos 2 θ − sin 2 θ) ∫ xy dA A

(

)

+ (cos θ sin θ) ∫ y 2 dA − ∫ x 2 dA , A

from which I x ′y ′ = (cos 2 θ − sin 2 θ) I xy + ( I x − I y )sin θ cos θ, which is the expression required. x9 y y9

A 0

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Problem 8.100 The axis L O is perpendicular to both segments of the L-shaped slender bar. The mass of the bar is 6 kg and the material is homogeneous. Use the method described in Example 8.10 to determine the moment of inertia of the bar about L O .

Solution:

Use Example 8.10 as a model for this solution.

Introduce the coordinate system shown and divide the bar into two parts as shown y51

y dy

2 r

1 0

x 2

( I 0 ) 1 = ∫ r 2 dm = ∫ ρ Ax 2 dx = ρ A 0

x3 2 3 0

8 ρA 3

(I 0 )1 = However

m1 = ρ Al1 = (2ρ A). Since part 1 is 2/3 of the length, its mass is 2/3(6 kg) = 4 kg. Part 2 has mass 2 kg. For part 2, dm = ρ A dy and r =

22 + y 2 1

( I 0 ) 2 = ∫ r 2 dm = ∫ ρ A(2 2 + y 2 ) dy m2

(

(I 0 ) 2 = ρ A 4 y +

y3 3

) = ρ A 133 1

13 8 21 I 0 TOTAL = ρA + ρA = ρA 3 3 3 ( I 0 ) TOTAL = 7ρ A The total mass = 3ρ A = 6 kg I 0 TOTAL =

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7 (6 kg) ⋅ m 2 = 7 kg m 2 6

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Problem 8.101 Two homogeneous slender bars, each of mass m and length l, are welded together to form the T-shaped object. Use integration to determine the moment of inertia of the object about the axis through point O that is perpendicular to the bars.

Solution:

Divide the object into two pieces, each corresponding to a slender bar of mass m; the first parallel to the y-axis, the second to the x-axis. By definition l

I = ∫ r 2 dm + ∫ r 2 dm. 0

For the first bar, the differential mass is dm = ρ A dr . Assume that the second bar is very slender, so that the mass is concentrated at a distance l from O. Thus dm = ρ A dx, where x lies between the limits −

O l

l l ≤ x ≤ . 2 2

The distance to a differential dx is r = becomes l

l 2

−2

l 2 + x 2 . Thus the definition

I = ρ A ∫ r 2 dr + ρ A ∫ l (l 2 + x 2 ) dx I l

l/2

r3 x3  = ρ A   + ρ A  l 2 x +  3 0  3  −l / 2 = ml 2

Problem 8.102 The slender bar lies in the x – y plane. Its mass is 6 kg and the material is homogeneous. Use integration to determine its moment of inertia about the z-axis.

ml ( 13 + 1 + 121 ) = 17 12

Solution: Iz =

The density is ρ =

∫0 ρ x 2 dx + ∫0

6 kg = 2 kg/m 3m

ρ[(1 m + s cos50 °) 2 + (s sin 50 °) 2 ] ds

I z = 15.14 kg m 2 y

2m 2m

508

Problem 8.103 Use integration to determine the moment of inertia of the slender 6-kg bar about the y-axis.

Solution: Iy = ∫

1m 0

See solution for 8.102 ρ x 2 dx + ∫

2m 0

ρ (1 m + s cos 50 °) 2 ds = 12.01 kg m 2

508

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Problem 8.104 The homogeneous thin plate has mass m = 12 kg and dimensions b = 2 m and h = 1 m. Use the procedure described in Practice Example 8.9 to determine the moments of inertia of the plate about the x - and y -axes.

h x

Solution:

From Appendix B, the moments of inertia about the x- and

y-axes are

1 1 Ix = bh 3 , I y = hb 3 . 36 36 Therefore the moments of inertia of the plate about the x- and y-axes are I x -axis =

I y -axis =

(

)

m m 1 3 1 1 I − b = mh 2 = (12 kg)(1 m) 2 = 0.667 kg-m 2 1 A x 18 8 bh 36 2

(

)

m m 1 1 1 I = hb 3 = mb 2 = (12 kg)(2 m) 2 = 2.67 kg-m 2 1 A y 18 18 bh 36 2

I x -axis = 0.667 kg-m 2 , I y -axis = 2.67 kg-m 2 .

Problem 8.105 The homogeneous thin plate is of uniform thickness and mass m. (a) Determine its moments of inertia about the x - and z -axes. (b) Let R i = 0 and compare your results with the values given in Appendix C for a thin circular plate. y Ro

Solution: (a)

πR 4 . 4 π 4 4 For the plate with a circular cutout, I x = ( R o − Ri ). The 4 m area mass density is , thus for the plate with a circular cut, A m m , from which the moments of inertia = 2 2 A π( R o − Ri ) The area moments of inertia for a circular area are I x = I y =

I x -axis =

Ri x

m( R o 4 − R i 4 ) m = ( R o 2 + Ri 2 ) 4( R o 2 − Ri 2 ) 4

I z -axis = 2 I x -axis =

m ( R 2 + Ri 2 ). 2 o

(b) Let Ri = 0, to obtain m 2 R , 4 o m = Ro 2 , 2

I x -axis = I z -axis

which agrees with table entries.

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Problem 8.106 The homogeneous thin plate is of uniform thickness and weighs 20 lb. Determine its moment of inertia about the y-axis.

Solution: 1 2 x ft 4

y = 4−

The plate’s area is

1 y 5 4 2 x 2 ft 4

A = ∫

−4

( 4 − 14 x ) dx = 21.3 ft . 2

The plate’s density per unit area is δ = (20/32.2)/21.3 = 0.0291 slug/ft 2 .

The moment of inertia about the y-axis is 4 1 I y-axis = ∫ x 2δ 4 − x 2 dx −4 4

(

)

= 1.99 slug-ft 2 . y

y542

24 ft

Problem 8.107 Determine the moment of inertia of the plate about the x-axis. y

y542

Solution:

See the solution of Problem 8.106. The mass of the strip

element is

(

m (strip) = δ 4 −

1 2 x ft 4

4 ft

1 2 x ft 4

)

1 2 x dx. 4

The moment of inertia of the strip about the x-axis is I (strip) = =

2 1 1 m 4 − x2 3 (strip) 4

(

)

3 1 1 δ 4 − x 2 dx, 3 4

(

)

so the moment of inertia of the plate about the x-axis is I ( x -axis) = ∫

−4

Problem 8.108 The mass of the object is 10 kg. Its moment of inertia about L1 is 10 kg-m 2 . What is its moment of inertia about L 2? (The three axes lie in the same plane.)

610

0.6 m

( 3

4 1

δ 4−

1 2 3 x dx = 2.27 slug-ft 2 . 4

)

Solution: The strategy is to use the data to find the moment of inertia about L, from which the moment of inertia about L 2 can be determined. I L = −(0.6) 2 (10) + 10 = 6.4 m 2 , from which I L 2 = (1.2) 2 (10) + 6.4 = 20.8 m 2

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Problem 8.109 An engineer gathering data for the design of a maneuvering unit determines that the astronaut’s center of mass is at x = 1.01 m, y = 0.16 m and that her moment of inertia about the z-axis is 105.6 kg-m 2 . Her mass is 81.6 kg. What is her moment of inertia about the z ′-axis through her center of mass?

d =

The distance d from the z-axis to the z ′-axis is

(1.01) 2 + (0.16) 2

= 1.0226 m. From the parallel-axis theorem, I ( z -axis) = I ( z ′ -axis) + d 2 m : 105.6 = I ( z ′ -axis) + (1.0226) 2 (81.6).

Solution:

Solving, we obtain

I ( z ′ -axis) = 20.27 kg-m 2 .

Problem 8.110 Two homogeneous slender bars, each of mass m and length l, are welded together to form the T-shaped object. Use the parallel-axis theorem to determine the moment of inertia of the object about the axis through point O that is perpendicular to the bars.

Solution: Divide the object into two pieces, each corresponding to a bar of mass m. By definition l

I = ∫ r 2 dm. 0

For the first bar, the differential mass is dm = ρ A dr , from which the moment of inertia about one end is l

l r3 ml 2 I 1 = ρ A∫ r 2 dr = ρ A   = . 0  3 0 3

For the second bar l

O l

2 r3 2 ml 2 I 2 = ρ A∫ l r 2 dr = ρ A   = −2  3  − 2l 12

is the moment of inertia about the center of the bar. From the parallel axis theorem, the moment of inertia about O is Io =

Problem 8.111 Use the parallel-axis theorem to determine the moment of inertia of the T-shaped object about the axis through the center of mass of the object that is perpendicular to the two bars. (See Practice Example 8.11.)

ml 2 ml 2 17 2 ml . + l 2m + = 3 12 12

Solution:

The location of the center of mass of the object is m ( 2l ) + lm 3 x = = l. Use the results of Problem 8.110 for the 2m 4 moment of inertia of a bar about its center. For the first bar, I1 =

( 4l ) m + ml12 = 487 ml . 2

For the second bar, I2 = l

( 4l ) m + ml12 = 487 ml .

I c = I1 + I 2 =

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The composite:

7 ml 2 24

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Problem 8.112 The length of the slender bar is L = 2 m. It swings from a pin support at O. If the bar is displaced a small amount from the vertical and released, the time in seconds required for it to go through one complete swing is T = 2π

Iz , mgd

Solution:

From Appendix C, the bar’s moment of inertia about a perpendicular axis through the center of mass is I z′ =

1 mL2 . 12

Applying the parallel axis theorem, the bar’s moment of inertia about the z -axis through point O is I z = I z ′ + d 2m

where g = 9.81 m/s 2 , m is the bar’s mass, and I z is the moment of inertia of the bar about the z-axis perpendicular to the bar through point O. If the distance d = 0.3 m, what is the time T ?

1 mL2 + d 2 m. 12

The time T is T = 2π

Iz mgd

1 mL2 + d 2 m = 2π 12 mgd

y, y9

1 (2 m) 2 + (0.3 m) 2 = 2π 12 (9.81 m/s 2 )(0.3 m)

x O d

= 2.38 s. x9

2.38 s.

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Problem 8.113 In Problem 8.112, determine the value of the distance d for which the time T required for the bar to complete one swing is a minimum. What is the time required?

Solution:

From Appendix C, the bar’s moment of inertia about a perpendicular axis through the center of mass is I z′ =

1 mL2 . 12

Applying the parallel axis theorem, the bar’s moment of inertia about the z -axis through point O is

y, y9

I z = I z ′ + d 2m =

x O

1 mL2 + d 2 m. 12

The time T is

d x9 L

Iz mgd

T = 2π

1 mL2 + d 2 m = 2π 12 mgd 1 2 L + d2 . = 2π 12 gd We want to find the value of d that minimizes the quantity 1 f (d ) = 12

L2 + d 2 d

1 2 −1 = L d + d. 12 Equating the derivative of this function to zero, f ′(d ) = −

1 2 −2 L d + 1 = 0, 12

we obtain L 12 2m = 12

d =

= 0.577 m. The time T is 1 2 L + d2 T = 2π 12 gd 1 (2 m) 2 + (0.577 m) 2 = 2π 12 (9.81 m/s 2 )(0.577 m) = 2.16 s. d = 0.577 m, T = 2.16 s.

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Problem 8.114 The homogeneous slender bar weighs 5 lb. Determine its moment of inertia about the z-axis. y

The center of mass of part 3 is located to the right of its center C a distance 2 R /π = 2(4)/π = 2.55 in. The moment of inertia of part 3 about C is

∫m r 2 dm = m3r 2 = (0.0612)(4) 2 = 0.979 slug-in 2 .

The moment of inertia of part 3 about the center of mass of part 3 is therefore I 3 = 0.979 − m 3 (2.55) 2 = 0.582 slug-in 2 . 4 in

The moment of inertia of the bar about the z-axis is I ( z -axis) = 13 m1L1 2 + 13 m 2 L 2 2 + I 3 + m 3 [(8 + 2.55) 2 + (4) 2 ] = 11.6 slug-in 2 = 0.0802 slug-ft 2 .

y x 8 in 2

Solution:

The Bar’s mass is m = 5/32.2 slugs. Its length is L = L1 + L 2 + L 3 = 8 + 8 2 + 8 2 + π (4) = 31.9 in.

(

m2 =

 2(64)  5 L2 = 0.0551 slugs, m =    31.9  32.2 L

m3 =

L3 4π m = L 31.9

(

)

5 ) = 0.0612 slugs. )( 32.2

Problem 8.115 Determine the moment of inertia of the 5-lb bar about the z ′-axis through its center of mass. y

Solution:

In the solution of Problem 8.114, it is shown that the moment of inertia of the bar about the z-axis is I ( z -axis) = 11.6 slug-in 2 . The x and y coordinates of the center of mass coincide with the centroid of the axis: x =

4 in

y = x9 =

x 8 in

614

8 in

5 ) = 0.0390 slugs, )( 32.2

L1 8 m = L 31.9

x9 3

The masses of the parts are therefore, m1 =

4 in

x 1 L1 + x 2 L 2 + x 3 L 3 L1 + L 2 + L 3 2(4)  π (4) (4)(8) + (4) 8 2 + 8 2 +  8 +  π  = 6.58 in, 2 2 8 + 8 + 8 + π (4) y 1 L1 + y 2 L 2 + y 3 L 3 L1 + L 2 + L 3 0 + (4) 8 2 + 8 2 + (4)π (4) = 3.00 in. 8 + 8 2 + 8 2 + π (4)

The moment of inertia about the z-axis is I ( z ′ -axis) = I ( z -axis) − (x 2 + y 2 )

5 ( 32.2 ) = 3.43 slug-in . 2

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Problem 8.116 The rocket is used for atmospheric research. Its weight and its moment of inertia about the z-axis through its center of mass (including its fuel) are 10 kip and 10,200 slug-ft 2 , respectively. The rocket’s fuel weighs 6000 lb, its center of mass is located at x = −3 ft, y = 0, z = 0, and the moment of inertia of the fuel about the axis through the fuel’s center of mass parallel to z is 2200 slug-ft 2 . When the fuel is exhausted, what is the rocket’s moment of inertia about the axis through its new center of mass parallel to z? y

Solution:

Denote the moment of inertia of the empty rocket as I E about a center of mass x E , and the moment of inertia of the fuel as I F about a mass center x F . Using the parallel axis theorem, the moment of inertia of the filled rocket is I R = I E + x E 2m E + I F + x F 2m F , about a mass center at the origin ( x R = 0). Solve: I E = I R − x E 2 m E − I F − x F 2m F . The objective is to determine values for the terms on the right from the data given. Since the filled rocket has a mass center at the origin, the mass center of the empty rocket is found from 0 = m E x E + m F x F , from which m  x E = − F  x F .  mE 

Using a value of g = 32.2 ft/s 2 , mF =

WF 6000 = = 186.34 slug, g 32.2

mE =

(W R − W F ) 10000 − 6000 = = 124.23 slug. g 32.2

From which 186.335 ( 124.224 )(−3) = 4.5 ft

xE = −

is the new location of the center of mass. Substitute: I E = I R − x E 2m E − I F − x F 2m F = 10200 − 2515.5 − 2200 − 1677.01 = 3807.5 slug-ft 2

Problem 8.117 The mass of the homogeneous thin plate is 36 kg. Determine its moment of inertia about the x-axis.

Solution:

Divide the plate into two areas: the rectangle 0.4 m by 0.6 m on the left, and the rectangle 0.4 m by 0.3 m on the right. The m mass density is ρ = . The area is A A = (0.4)(0.6) + (0.4)(0.3) = 0.36 m 2 ,

from which 0.4 m

0.4 m

ρ =

36 = 100 kg/m 2 . 0.36

The moment of inertia about the x-axis is

0.3 m

I x-axis = ρ ( 13 )(0.4)(0.6 3 ) + ρ ( 13 )(0.4)(0.3) 3 = 3.24 kg-m 2 0.3 m x

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Problem 8.118 Determine the moment of inertia of the 36-kg plate about the z-axis.

The basic relation to use is

I z -axis = I x -axis + I y -axis . The value of I x- axis is given in the solution of Problem 8.117. The moment of inertia about the y-axis using the same divisions as in Problem 8.117 and the parallel axis theorem is

y 0.4 m

Solution:

0.4 m

I y-axis = ρ

( 13 )(0.6)(0.4) + ρ ( 121 ) (0.3)(0.4) 3

+ (0.6) 2 ρ (0.3)(0.4) = 5.76 kg-m 2 ,

0.3 m

from which I z -axis = I x -axis + I y -axis = 3.24 + 5.76 = 9 kg-m 2 .

0.3 m x

Problem 8.119 The homogeneous thin plate weighs 10 lb. Determine its moment of inertia about the x-axis. y 5 in

Solution:

Divide the area into two parts: the lower rectangle 5 in by 10 in and the upper triangle 5 in base and 5 in altitude. The mass density W is ρ = . The area is gA A = 5(10) + ( 12 ) 5(5) = 62.5 in 2 .

5 in

Using g = 32 ft/s 2 , the mass density is ρ =

W = 0.005 slug/in 2 . gA

Using the parallel axis theorem, the moment of inertia about the x-axis is 10 in I x -axis = ρ

( 13 )(10)(5) + ρ ( 361 )(5)(5 ) 3

5 in

(

+ρ 5+ x

Problem 8.120 Determine the moment of inertia of the 10-lb plate about the y-axis.

5 in

5 2 1 (5)(5) = 4.948 slug-in 2 3 2

)( )

I x -axis = 0.03436 slug-ft 2

Solution:

Use the results of the solution in Problem 8.119 for the area and the mass density. I y-axis = ρ

( 13 )5(10 ) + ρ ( 361 )5(5 ) + ρ ( 5 + 103 ) ( 12 )5(5) 3

= 12.76 slug-in 2 = 0.0886 slug-ft 2

5 in

10 in 5 in x

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Problem 8.121 The thermal radiator (used to eliminate excess heat from a satellite) can be modeled as a homogeneous thin rectangular plate. Its mass is 5 slugs. Determine its moments of inertia about the x-, y-, and z-axes. y 3 ft

6 ft

Solution: ρ =

The area is A = 9(3) = 27 ft 2 . The mass density is

m 5 = = 0.1852 slugs/ft 2 . A 27

The moment of inertia about the centroid of the rectangle is I xc = ρ

( 121 ) 9(3 ) = 3.75 slug-ft ,

I yc = ρ

( 121 ) 3(9 ) = 33.75 slug-ft .

Use the parallel axis theorem: I x -axis = ρ A(2 + 1.5) 2 + I xc = 65 slug-ft 2 ,

3 ft

I y -axis = ρ A(4.5 − 3) 2 + I yc = 45 slug-ft 2 . I z -axis = I x -axis + I y -axis = 110 slug-ft 2

2 ft x

Problem 8.122 The homogeneous cylinder has mass m, length l, and radius R. Use integration as described in Example 8.13 to determine its moment of inertia about the x-axis. y

Solution:

The volume of the disk element is π R 2 dz and its mass is dm = ρπ R 2 dz, where ρ is the density of the cylinder. From Appendix C, the moment of inertia of the disk element about the x′-axis is 1 1 dmR 2 = (ρπ R 2 dz ) R 2 . 4 4

dI x ′ -axis =

Applying the parallel-axis theorem, the moment of inertia of the disk element about the x-axis is

dI x -axis = dI x ′ -axis + z 2 dm =

1 (ρπ R 2 dz ) R 2 + z 2 (ρπ R 2 dz ) 4

Integrating this expression from z = 0 to z = l gives the moment of inertia of the cylinder about the x-axis. I x -axis = ∫ R

l 0

( 41 ρπ R + ρπ R z ) dz = 14 ρπ R l + 13 ρπ R l . 4

2 2

2 3

In terms of the mass of the cylinder m = ρπ R 2l,

l I x-axis = z

Problem 8.123 The homogeneous cone is of mass m. Determine its moment of inertia about the z-axis, and compare your result with the value given in Appendix C. (See Example 8.13.)

1 1 mR 2 + ml 2 4 3

Solution:

The differential mass

( )

m 3m dm = πr 2 dz = 2 r 2 dz. V R h The moment of inertia of this disk about the z-axis is 12 mr 2 . The radius varies with z,

y r =

( Rh ) z,

from which

I z -axis =

3mR 2 h 4 3mR 2  z 5  h 3mR 2 z dz =   = 2h 5 ∫ 0 2h 5  5  0 10

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Problem 8.124 Determine the moments of inertia of the homogeneous cone of mass m about the x- and y-axes, and compare your results with the values given in Appendix C.

m 3m = . The differential V π R 2h 2 element of mass is dm = ρπr dz. The moment of inertia of this elemental disk about an axis through its center of mass, parallel to the x- and y-axes, is

Solution:

The mass density is ρ =

dI x = ( 14 ) r 2 dm.

Use the parallel axis theorem, I x = ∫ ( 14 )r 2 dm + ∫ z 2 dm. m

Noting that r = R

R z, then h

r 2 dm = ρ

( πhR ) z dz,

and z 2 dm = ρ

( πhR ) z dz.

Substitute: Ix = ρ

( π4hR )∫ z dz + ρ( πhR )∫ z dz. 4

Integrating and collecting terms: Ix =

5 h

( 34mRh + 3hm )  z5  = m( 203 R + 35 h ). 2

By symmetry, I y = I x

Problem 8.125 The mass of the homogeneous wedge is m. Use integration as described in Example 8.13 to determine its moment of inertia about the z-axis. (Your answer should be in terms of m, a, b, and h.)

Solution:

Consider a triangular element of the wedge of thickness dz. The mass of the element is the product of the density ρ of the material and the volume of the element, dm = ρ 12 bhdz. The moments of inertia of the triangular element about the x ′ - and y ′ -axes are given by Eqs. (8.30) and (8.31) in terms of the mass of the element, its triangular area, and the moments of inertia of the triangular area:

dI x ′ -axis =

dm I′ = A x

1 ρ bhdz 1 1 2 bh 3 = rbh 3dz, 1 12 bh 12 2

dI y ′ -axis =

dm I′ = A y

1 ρ bhdz 1 3 1 2 hb = rhb 3dz, 1 4 bh 4 2

(

)

(

)

The moment of inertia of this thin plate about the z-axis is dI z -axis = dI x ′ -axis + dI y ′ -axis =

1 1 ρbh 3dz + ρhb 3dz. 12 4

Integrating this expression from z = 0 to z = a gives the moment of inertia of the wedge about the z-axis: I z -axis = ∫

a 0

( 121 ρbh + 14 ρhb ) dz = 121 ρbh a + 14 ρhb a. 3

In terms of the mass m = ρ 12 bha, I z -axis =

618

1 1 mh 2 + mb 2 . 6 2

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Problem 8.126 The mass of the homogeneous wedge is m. Use integration as described in Example 8.13 to determine its moment of inertia about the x-axis. (Your answer should be in terms of m, a, b, and h.) y

Solution:

Consider a triangular element of the wedge of thickness dz. The mass of the element is the product of the density ρ of the material and the volume of the element, dm = ρ 12 bhdz. The moments of inertia of the triangular element about the x ′ axis is given by Eq. (8.30) in terms of the mass of the element, its triangular area, and the moments of inertia of the triangular area: dI x ′ -axis =

dm I = A x′

1 ρ bhdz 1 1 2 bh 3 = rbh 3dz, 1 36 bh 36 2

(

)

Applying the parallel-axis theorem, the moment of inertia of the triangular element about the x-axis is  1 2 dI x -axis = dI x ′ -axis +  z 2 + h  dm 3  

( )

 1 1 2 1 ρbh 3dz +  z 2 + h  ρ bhdz 36 3   2

1 1 ρbh 3dz + ρbhz 2 dz 12 2

( )

Integrating this expression from z = 0 to z = a gives the moment of inertia of the wedge about the x-axis: I x -axis = ∫

a 0

( 121 ρbh + 12 ρbhz ) dz = 121 ρbh a + 16 ρbha . 3

In terms of the mass m = ρ 12 bha, I x-axis =

Problem 8.127 In Example 8.12, suppose that part of the 3-kg bar is sawed off so that the bar is 0.4 m long and its mass is 2 kg. Determine the moment of inertia of the composite object about the perpendicular axis L through the center of mass of the modified object.

1 1 mh 2 + ma 2 . 6 3

Solution: The mass of the disk is 2 kg. Measuring from the left end of the rod, we locate the center of mass x =

(2 kg)(0.2 m) + (2 kg)(0.6 m) = 0.4 m. (2 kg) + (2 kg)

The center of mass is located at the point where the rod and disk are connected. The moment of inertia is I =

1 (2 kg)(0.4 m) 2 + 3

{ 12 (2 kg)(0.2 m) + (2 kg)(0.2 m) } 2

I = 0.227 kg-m 2 .

Problem 8.128 The L-shaped machine part is composed of two homogeneous bars. Bar 1 is tungsten alloy with density 14,000 kg/m 3 , and bar 2 is steel with density 7800 kg/m 3 . Determine its moment of inertia about the x-axis. y

240 mm

1 40 mm 2

80 mm

Solution:

The masses of the bars are

m1 = (14, 000)(0.24)(0.08)(0.04) = 10.75 kg m 2 = (7800)(0.24)(0.08)(0.04) = 5.99 kg. Using Appendix C and the parallel axis theorem the moments of inertia of the parts about the x-axis are I ( x -axis)1 =

1 m [(0.04) 2 + (0.24) 2 ] + m1 (0.12) 2 = 0.2079 kg-m 2 , 12 1

I ( x -axis)2 =

1 m [(0.04) 2 + (0.08) 2 ] + m 2 (0.04) 2 = 0.0136 kg-m 2 . 12 2

Therefore I ( x -axis) = I ( x -axis)1 + I ( x -axis)2 = 0.221 kg-m 2

80 mm

z 240 mm

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Problem 8.129 The homogeneous object is a cone with a conical hole. The dimensions R1 = 2 in, R 2 = 1 in, h1 = 6 in, and h 2 = 3 in. It consists of aluminum alloy with a density of 5 slug/ft 3 . Determine its moment of inertia about the x-axis.

Solution:

ρ = 5 slug/ft 3

1 1 π R 2 h = π (2 in) 2 (6 in) = 25.1 in 3 . 3 1 1 3

The mass of the conical object without the conical hole is m1 = πV1 = 0.0727 slug. From Appendix C, the moment of inertia of the conical object without the conical hole about the x-axis is

( I x ) 1 = m1 R1

( 35 h + 203 R ) 1

3 3 = (0.0727 slug)  (6 in) 2 + (2 in) 2  = 1.61 slug-in 2 5  20 The volume of the conical hole is

R2 h1

( 121 ftin ) = 0.00289 slug/in .

The volume of the conical object without the conical hole is V1 =

The density of the material is

V2 =

1 1 π R 2 h = π (1 in) 2 (3 in) = 3.14 in 3 . 3 2 2 3

The mass of the material that would occupy the conical hole is m 2 = ρV2 = 0.00909 slug. slug. The z coordinate of the center of mass of the material that would occupy the conical hole is z = h1 − h 2 +

3 3 h = 6 in − 3 in + (3 in) = 5.25 in. 4 2 4

Using Appendix C and applying the parallel-axis theorem, the moment of inertia about the x-axis of the material that would occupy the conical hole is (I x ) 2 = m 2

( 803 h + 203 R ) + z m = 0.255 slug-in . 2

The moment of inertia of the conical object with the conical hole is I x = ( I x ) 1 − ( I x ) 2 = 1.36 slug-in 2 . I x = 1.36 slug-in 2 .

Problem 8.130 The circular cylinder is made of aluminum (Al) with density 2700 kg/m 3 and iron (Fe) with density 7860 kg/m 3 . Determine its moments of inertia about the x ′- and y ′-axes.

Solution:

We have ρ A1 = 2700 kg/m 3 , ρ Fe = 7860 kg/m 3

We first locate the center of mass ρ Al (π[0.1 m] 2 [0.6 m])(0.3 m) + ρ Fe (π[0.1 m] 2 [0.6 m])(0.9 m) ρ Al (π[0.1 m] 2 [0.6 m]) + ρ Fe (π[0.1 m] 2 [0.6 m]) = 0.747 m

x =

We also have the masses y9

m Al = ρ Al π (0.1 m) 2 (0.6 m), m Fe = ρ Fe π (0.1 m) 2 (0.6 m)

Al z

I x′ =

600 mm z9

Now find the moments of inertia

200 mm 600 mm x, x9

1 1 m (0.1 m) 2 + m Fe (0.1 m) 2 = 0.995 kg m 2 2 Al 2

I y ′ = m Al

( [0.612m] + [0.14m] ) + m (x − 0.3 m)

+ m Fe

( [0.612m] + [0.14m] ) + m (0.9 m − x ) 2

= 20.1 kg m 2

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Problem 8.131 The homogeneous half-cylinder is of mass m. Determine its moment of inertia about the axis L through its center of mass.

Solution:

The centroid of the half cylinder is located a distance of 4R from the edge diameter. The strategy is to use the parallel axis the3π orem to treat the moment of inertia of a complete cylinder as the sum of the moments of inertia for the two half cylinders. From Problem 8.118, the moment of inertia about the geometric axis for a cylinder is I cL = mR 2 , where m is one half the mass of the cylinder.

( )

By the parallel axis theorem, I cL = 2

(( 43πR ) m + I ). 2

Solve I hL =

Problem 8.132 The homogeneous machine part is made of aluminum alloy with density ρ = 2800 kg/m 3 . Determine its moment of inertia about the z-axis. y

( I2 − ( 43πR ) m ) = ( mR2 − ( 916π ) mR ) 2

= mR 2

( 12 − 916π )

= mR 2

( 12 − 916π ) = 0.31987 mR = 0.32 mR

Therefore, the moment of inertia of part 2 about the z-axis through its center of mass that is parallel to the axis is 1 2 2 2 m 2 (0.04) − m 2 (0.0170)

= 0.000144 kg-m 2 .

Using this result, the moment of inertia of part 2 about the z-axis is I ( z -axis)2 = 0.000144 + m 2 (0.12 + 0.017) 2 = 0.00543 kg-m 2 .

20 mm x

The moment of inertia of the material that would occupy the hole 3 about the z-axis is

40 mm 120 mm

I ( z -axis)3 = 12 m 3 (0.02) 2 + m 3 (0.12) 2 = 0.00205 kg-m 2 . 40 mm

Therefore I ( z -axis) = I ( z -axis)1 + I ( z -axis)2 − I ( z -axis)3 = 0.00911 kg-m 2 .

Solution:

We divide the machine part into the 3 parts shown: (The dimension into the page is 0.04 m The masses of the parts are m1 = (2800)(0.12)(0.08)(0.04) = 1.075 kg, m 2 = (2800) 12 π (0.04) 2 (0.04) = 0.281 kg, m 3 = (2800)π (0.02) 2 (0.04) = 0.141 kg.

0.12 m 1

0.08 m y

Using Appendix C and the parallel axis theorem the moment of inertia of part 1 about the z-axis is 1 I ( z -axis)1 = m [(0.08) 2 + (0.12) 2 ] + m1 (0.06) 2 12 1

C 2 0.12 m

0.12 m x – 0.04 m

x 0.02 m

= 0.00573 kg-m 2 . The moment of inertia of part 2 about the axis through the center C that is parallel to the z-axis is 1 2 2 m2 R

= 12 m 2 (0.04) 2 .

The distance along the x-axis from C to the center of mass of part 2 is 4(0.04)/(3π ) = 0.0170 m.

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Problem 8.133 Determine the moment of inertia of the machine part described in Problem 8.132 about the x-axis. y

Solution:

We divide the machine part into the 3 parts shown in the solution to Problem 8.132. Using Appendix C and the parallel axis theorem, the moments of inertia of the parts about the x-axis are: I ( x -axis)1 =

1 1 I ( x -axis)2 = m 2  (0.04) 2 + (0.04) 2   12  4

20 mm x

= 0.0001501 kg-m 2 1 1 I ( x -axis)3 = m 3  (0.04) 2 + (0.02) 2   12  4

40 mm 120 mm

1 m [(0.08) 2 + (0.04) 2 ] = 0.0007168 kg-m 2 12 1

40 mm

= 0.0000328 kg-m 2 . Therefore, I ( x -axis) = I ( x -axis)1 + I ( x -axis)2 − I ( x -axis)3 = 0.000834 kg-m 2 .

Problem 8.134 The object consists of steel of density ρ = 7800 kg/m 3 . Determine its moment of inertia about the axis L O . 20 mm

Solution:

Divide the object into four parts: Part (1) The semicylinder of radius R = 0.02 m, height h1 = 0.01 m. Part (2): The rectangular solid L = 0.1 m by h 2 = 0.01 m by w = 0.04 m. Part (3): The semi-cylinder of radius R = 0.02 m, h1 = 0.01 m. Part (4) The cylinder of radius R = 0.02 m, height h = 0.03 m. Part (1)

O 100 mm

10 mm

30 mm LO

m1 =

ρπ R 2 h1 = 0.049 kg, 2

I1 =

m1 R 2 = 4.9 × 10 −6 kg-m 2 , 4

Part (2): m 2 = ρ wLh 2 = 0.312 kg, I2 =

( 121 )m (L + w ) + m ( L2 ) = 0.00108 kg-m . 2

Part (3) m 3 = m1 = 0.049 kg, 2

( 43πR ) m + I + m ( L − 43πR ) = 0.00041179 kg m .

I3 = −

Part (4) m 4 = ρπ R 2 h = 0.294 kg, I 4 = ( 12 ) m 4 ( R 2 ) + m 4 L2 = 0.003 kg m 2 . The composite: I Lo = I 1 + I 2 − I 3 + I 4 = 0.003674 kg m 2

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Problem 8.135 Determine the moment of inertia of the object in Problem 8.134 about the axis through the center of mass of the object parallel to L O . 20 mm

Solution: x =

The center of mass is located relative to L O

( 43πR ) + m (0.05) − m ( 0.1 − 43πR ) + m (0.1)

m1 −

m1 + m 2 − m 3 + m 4

= 0.066 m, I c = −x 2 m + I Lo = −0.00265 + 0.00367 = 0.00102 kg m 2

O 100 mm

10 mm

30 mm LO

Problem 8.136 The thick plate consists of steel of density ρ = 15 slug/ft 3 . Determine its moment of inertia about the z-axis. y

Solution: Divide the object into three parts: Part (1) the rectangle 8 in by 16 in, Parts (2) & (3) the cylindrical cut outs. Part (1): m1 = ρ8(16)(4) = 4.444 slugs. I1 =

( 121 )m (16 + 8 ) = 118.52 slug in . 2

Part (2): 4 in

2 in

m 2 = ρπ (2 2 )(4) = 0.4363 slug, x

z I2 =

4 in 4 in

8 in

4 in

m 2 (2 2 ) + m 2 (4 2 ) = 7.854 slug in 2 . 2

Part (3): m 3 = m 2 = 0.4363 slugs, I 3 = I 2 = 7.854 slug-in 2 . The composite: I z -axis = I 1 − 2 I 2 = 102.81 slug-in 2 I z -axis = 0.714 slug-ft 2

Problem 8.137 Determine the moment of inertia of the plate in Problem 8.136 about the x-axis. y

Solution:

Use the same divisions of the object as in Problem 8.136.

Part (1): I 1x -axis =

( 121 )m (8 + 4 ) = 29.63 slug-in , 1

Part (2): 4 in

2 in

2 in x

I 2 x -axis =

4 in

( 121 )m (3(2 ) + 4 ) = 1.018 slug-in . 2

The composite: 4 in

8 in

4 in

I x -axis = I 1x -axis − 2 I 2 x -axis = 27.59 slug-in 2 = 0.1916 slug-ft 2

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Problem 8.138 The dimensions b = 40 mm and h = 30 mm. Use the method described in Practice Example 8.1 to determine I y and k y for the area shown.

Solution:

Consider the vertical strip element shown: y

y f(x) h dA h

x x x b

dx b

(

f ( x) = h 1 −

)

x . b

The area of the strip element is dA = f ( x ) dx, so the moment of inertia of the triangle about the y-axis is I y = ∫ x 2 dA A

= ∫ x 2 f ( x ) dx 0

(

= ∫ x 2h 1 − 0

)

x dx b

x3 x4 b = h  −  3 4 b  0 =

1 hb 3 12

= 1.60E5 mm 4 . The product of inertia about the y-axis is

ky =

Iy = A

1 hb 3 1 12 = b = 16.3 mm. 1 6 hb 2

I y = 1.60E5 mm 4 , k y = 16.3 mm.

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Problem 8.139 The dimensions b = 40 mm and h = 30 mm. Use the method described in Practice Example 8.1 to determine the cross moment of inertia I xy for the area shown.

Solution:

Consider the vertical strip element shown: y

f(x)

y h dA

x h

dx b

x b

(

)

x . b

f ( x) = h 1 −

From Example 8.1, the product of inertia of the vertical strip element is 1 [ f ( x )] 2 x dx. 2 We integrate this expression from x = 0 to x = b to obtain the product of inertia of the triangle: ( I xy ) strip =

I xy = ∫

= ∫

[ f ( x )] 2 x dx

(

)

2  h 1 − x  x dx b  2 

(

)

1 2 b x x2 h 1 − 2 + 2 x dx 2 ∫0 b b

1 2  x2 2x 3 x4 b − + h  2  2 3b 4 b 2  0

1 2 2 h b 24

= 60,000 mm 4 . I xy = 60, 000 mm 4 .

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Problem 8.140 The dimensions b = 16 in and h = 4 in. Use the method described in Practice Example 8.1 to determine I y .

Solution:

Consider the vertical strip element shown:

f(x)

h x

h x b

The straight line defining the triangle’s hypotenuse can be expressed as a linear function f ( x ) = cx + d . When x = 0, f ( x ) = 2h, and when x = b, f ( x ) = h. Using these two conditions to determine c and d yields c = −h /b and d = 2h, so

(

f ( x) = h 2 −

)

x . b

The area of the strip element is dA = f ( x ) dx, so the moment of inertia of the triangle about the y-axis is I y = ∫ x 2 dA A

= ∫ x 2 f ( x ) dx 0

(

= ∫ x 2h 2 − 0

)

x dx b

x4 b

2x 3  = h  −  3 4 b  0 5 = hb 3 12 = 6827 in 4 . I y = 6827 in 4 .

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Problem 8.141 The dimensions b = 16 in and h = 4 in. Use the method described in Practice Example 8.1 to determine J O .

The area of the strip element is dA = f ( x ) dx, so the moment of inertia of the triangle about the y-axis is I y = ∫ x 2 dA A

= ∫ x 2 f ( x ) dx 0

(

)

x dx b

= ∫ x 2h 2 − 0

2x 3 x4 b = h  −  3 4 b  0

h x

= 6827 in 4 .

Solution:

We will determine I x and I y for the area and use the relation J O = I x + I y . Consider the vertical strip element shown: y

dA dx b

(

f ( x) = h 2 −

( I x ) strip =

)

x . b

Ix = ∫

= ∫

h x

From Example 8.1, the moment of inertia of the strip element about the x-axis is 1 [ f ( x )] 3 dx. 3

Therefore the moment of inertia of the area about the x-axis is

f(x) 2h

5 hb 3 12

[ f ( x )] 3 dx

(

)

3  h 2 − x  dx 3  b 

(

)

h3 b 12 x 6x 2 x3 8− + 2 − 3 dx 3 ∫0 b b b

h3  x4 b 6x 2 2x 3 8x − + 2 −  b b 3  4 b 3  0

= 1.25bh 3 = 1280 in 4 . We obtain J O = I x + I y = 8107 in 4 . J O = 8107 in 4 .

Problem 8.142 Determine I y and k y .

Solution:

By definition,

I y = ∫ x 2 dA.

The element of area is dA = dx dy. The limits on the variable x are 0 ≤ x ≤ 4. The area is

1 y 5 x 2 x2 4

x−x2 /4

A = ∫ dx ∫ x

x2 x3 4 dy =  − = 2.6667  2 12  0

x−x2 /4

I y = ∫ x 2 dx ∫

dy = ∫

4 0

( x − x4 ) x dx 2

x4 x5 4 =  −  = 12.8  4 20  0 from which ky =

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Iy = 2.19 A

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Problem 8.143 Determine I x and k x .

Solution:

By definition,

I x = ∫ y 2 dA,

from which

y5x 2

1 2 x 4

x−x2 /4

I x = ∫ dx ∫ Ix = x

y 2 dy =

x  ( 13 )  x4 − 203 x + 963 x − 448  = 0.6095. 4

From Problem 8.142, Ix = 0.4781 A

A = 2.667, k x =

Problem 8.144 Determine I xy .

( 13 ) ∫ ( x − x4 ) dx

Solution: I xy = ∫ xy dA,

x−x 2 /4

= ∫ x dx ∫

1 y 5 x 2 x2 4

Problem 8.145 Determine I y ′ and k y ′ .

( 12 ) ∫ ( x − x4 ) x dx

x x  + ( 12 )  x4 − 10  = 2.1333 96 

6 4

Solution:

y dy

The limits on the variable x are 0 ≤ x ≤ 4. By definition, 4

x−x 2 /4

Ay = ∫ y dA = ∫ dx ∫ A

y5x 2

1 2 x 4 x9 x

y dy

( 12 ) ∫ ( x − x4 ) dx

x  ( 12 )  x3 − x8 + 80  = 1.06667.

5 4

From Problem 8.142 the area y = 0.3999 = 0.4. Similarly, X−X 2 / 4

Ax = ∫ x dx ∫ 0

(

= ∫ x x− 0

A = 2.667,

from

which

x2 x3 x4 4 dx =  − = 5.3333,  3 4 16  0

)

from which x = 1.9999 = 2. The area moment of inertia is I yy = −x 2 A + I y . Using the result of Problem 8.142, I y = 12.8, from which the area moment of inertia about the centroid is I y ′ = −10.6666 + 12.8 = 2.133 and k y ′ =

628

I y′ = 0.8944 A

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Solution:

Problem 8.146 Determine I x ′ and k x ′ .

Using the results of Problems 8.143 and 8.145, I x = 0.6095 and y = 0.4. The area moment of inertia about the centroid is

I x ′ = −y 2 A + I x = 0.1828

y5x 2

and k x ′ =

1 2 x 4

I x′ = 0.2618 A

x9 x

Solution:

Problem 8.147 Determine I x ′ y ′ .

From Problems 8.143 and 8.144, I xy = 2.133 and x = 2, y = 0.4. The product of the moment of inertia about the centroid is

I x ′y ′ = −xy A + I xy = −2.133 + 2.133 = 0 y9

y5x 2

1 2 x 4 x9 x

Solution:

Problem 8.148 Determine I y and k y .

Divide the section into two parts: Part (1) is the upper rectangle 40 mm by 200 mm, Part (2) is the lower rectangle, 160 mm by 40 mm.

Part (1) A1 = 0.040(0.200) = 0.008 m 2 , y 1 = 0.180 m x 1 = 0, 1 0.04(0.2) 3 = 2.6667 × 10 −5 m 4 . I y1 = 12

40 mm

( )

Part (2): A2 = (0.04)(0.16) = 0.0064 m 2 ,

160 mm

80 mm

40 mm

80 mm

y 2 = 0.08 m, x 2 = 0, 1 (0.16)(0.04) 3 = 8.5 × 10 −7 m 4 . I y2 = 12

( )

The composite: A = A1 + A2 = 0.0144 m 2 , I y = I y1 + I y 2 , I y = 2.752 × 10 −5 m 4 = 2.752 × 10 7 mm 4 , Iy = 0.0437 m = 43.7 mm and k y = A

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Problem 8.149 Determine I x and k x .

Solution:

Use the results in the solution to Problem 8.148. Part (1)

A1 = 0.040(0.200) = 0.008 m 2 , y 1 = 0.180 m, 1 0.2(0.04 3 ) + (0.18) 2 A1 = 2.603 × 10 −4 m 4 . I x1 = 12 Part (2):

( )

40 mm

A2 = (0.04)(0.16) = 0.0064 m 2 , y 2 = 0.08 m, 1 (0.04)(0.16) 3 + (0.08) 2 A2 = 5.461 × 10 −5 m 4 . I x2 = 12 The composite: A = A1 + A2 = 0.0144 m 2 , The area moment of inertia about the x-axis is

( )

160 mm

80 mm

40 mm

80 mm

I x = I x1 + I x 2 = 3.15 × 10 −4 m 4 = 3.15 × 10 8 mm 4 , and k x =

Solution:

Use the results of the solutions to Problems 8.148–8.149. The centroid is located relative to the base at

Problem 8.150 Determine I x and k x .

x 1 A1 + x 2 A2 = 0, A y A + y 2 A2 yc = 1 1 = 0.1356 m. A The moment of inertia about the x-axis is

xc =

40 mm x

I xc = −y C 2 A + I X = 5.028 × 10 7 mm 4 and k xc =

160 mm

80 mm

40 mm

Ix = 0.1479 m = 147.9 mm A

I xc = 59.1 mm A

80 mm

Solution:

Use the results of the solutions to Problems 8.148–8.149. The area moments of inertia about the centroid are

Problem 8.151 Determine J O and k O .

I xc = 5.028 × 10 −5 m 4

and I yc = I y = 2.752 × 10 −5 m 4 , from which

40 mm x

and k O =

160 mm

JO = 0.0735 m A

= 73.5 mm

80 mm

630

J O = I xc + I yc = 7.78 × 10 −5 m 4 = 7.78 × 10 7 mm 4

40 mm

80 mm

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Problem 8.152 Determine I y and k y . y

Solution:

For a semicircle about a diameter:

I yy = I xx =

( 81 )π R ,

Iy = 2 ft

ky =

( 81 )π(4) − ( 81 )π(2) = π8 (4 − 2 ) = 94.25 ft , 4

2I y = 2.236 ft π (4 2 − 2 2 )

4 ft

Problem 8.153 Determine J O and k O . y

2 ft x

Solution:

For a semicircle:

I yy = I xx =

( 81 ) π R .

Ix =

π 4 (4 − 2 4 ) = 94.248 ft 4 . 8

kx =

2I x = 2.236 ft. π (4 2 − 2 2 )

Also use the solution to Problem 8.152. J O = I x + I y = 2(94.248) = 188.5 ft 4

4 ft

kO =

Problem 8.154 Determine I x and k x .

2J O = 3.16 ft π (4 2 − 2 2 )

Part (2): A2 = x2 = a +

y 3 ft

3 ft

( 12 )h(b − a) = 3 ft , 2

b−a = 2.3333 ft, 3

y2 =

( 23 )h = 4 ft,

I xc 2 =

( 361 ) (b − a)h = 6 ft . 3

Part (3): A3 = A2 ,

6 ft

x 3 = −x 2 , y 3 = y 2 , I xc 3 = I xc 2 . The composite area is x

2 ft

A = A1 + A2 + A2 = 30 ft 2 . The composite moment of inertia I x = (y 1 ) 2 A1 + I xc1 + (y 2 ) 2 A2 + I xc 2 + (y 3 ) 2 A3 + I xc 3 ,

Solution:

Break the area into three parts: Part (1) The rectangle with base 2a and altitude h; Part (2) The triangle on the right with base (b − a) and altitude h, and Part (3) The triangle on the left with base (b − a) and altitude h. Part (1) The area is

I x = 396 ft 4 kx =

Ix = A

396 = 3.633 ft 30 y

A1 = 2ah = 24 ft 2 .

The centroid is x1 = 0 and y 1 =

h = 3 ft. 2

The area moment of inertia about the centroid is I xc1 =

( 121 ) (2a)h = ( 16 ) ah = 72 ft . 3

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Problem 8.155 Determine I y and k y .

Solution:

Divide the area as in the solution to Problem 8.154. Part (1) The area is A1 = 2ah = 24 ft 2 . The centroid is x 1 = 0 and h y1 = = 3 ft. The area moment of inertia about the centroid is 2 1 2 I yc1 = h(2a) 3 = ha 3 = 32 ft 4 12 3

y 3 ft

( )

3 ft

( )

( 12 )h(b − a) = 3 ft ,

Part (2): A2 = 6 ft

x2 = a +

2 ft

b−a = 2.3333 ft, 3

y2 =

( 23 )h = 4 ft,

I yc 2 =

( 361 )h(b − a) = 0.1667 ft .

x 2 ft

Part (3): A3 = A2 , x 3 = −x 2 , y 3 = y 2 , I yc 3 = I yc 2 . The composite area is A = A1 + A2 + A2 = 30 ft 2 . The composite moment of inertia, I y = x 1 2 A1 + I yc1 + x 2 2 A2 + I yc 2 + x 3 2 A3 + I yc 3 , I y = 65 ft 4 and k y =

Problem 8.156 The moments of inertia of the area are I x = 36 m 4 , I y = 145 m 4 , and I xy = 44.25 m 4 . Determine a set of principal axes and the principal moments of inertia.

Solution: θ =

Iy = 1.472 ft A

The principal angle is 

 2 I xy

( 12 ) tan  I − I  = 19.54°. −1

The principal moments of inertia are I xP = I x cos 2 θ − 2 I xy sin θ cos θ + I y sin 2 θ = 20.298 = 20.3 m 4

I yP = I x sin 2 θ + 2 I xy sin θ cos θ + I y cos 2 θ = 160.70 m 4 3m

3m x

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Problem 8.157 The moment of inertia of the 31-oz bat about a perpendicular axis through point B is 0.093 slug-ft 2 . What is the bat’s moment of inertia about a perpendicular axis through point A? (Point A is the bat’s “instantaneous center,” or center of rotation, at the instant shown.)

31 = 0.06023 slugs. 16(32.17) Use the parallel axis theorem to obtain the moment of inertia about the center of mass C, and then use the parallel axis theorem to translate to the point A.

Solution:

The mass of the bat is m =

( 12 ) m + 0.093 = 0.0328 slug-ft 12

IC = − IA =

( 12 12+ 14 ) m + 0.0328 = 0.3155 slug-ft

12 in

14 in

Problem 8.158 The mass of the thin homogeneous plate is 4 kg. Determine its moment of inertia about the y-axis.

Solution:

Divide the object into two parts: Part (1) is the semicircle of radius 100 mm, and Part (2) is the rectangle 200 mm by 280 mm. The area of Part (1) A1 =

πR 2 = 15708 mm 2 . 2

The area of Part (2) is A2 = 280(200) = 56000 mm 2 . 100 mm

The composite area is A = A2 − A1 = 40292 mm 2 . The area mass density is

140 mm x 140 mm

200 mm

ρ =

4 = 9.9275 × 10 −5 kg/mm 2 . A

For Part (1) x 1 = y 1 = 0, I y1 = ρ

( 81 )π R = 3898.5 kg-mm . 4

For Part (2) x 2 = 100 mm. I y2 = x 2 2 ρ A2 + ρ

( 121 )(280)(200 ) = 74125.5 kg-mm . 3

The composite: I y = I y 2 − I y1 = 70226 kg-mm 2 = 0.070226 kg-m 2

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Problem 8.159 Determine the moment of inertia of the 4-kg plate about the z-axis. y

Solution:

Use the same division of the parts and the results of the solution to Problem 8.158. For Part (1), 1 I x1 = ρ π R 4 = 3898.5 kg-mm 2 . 8

( )

For Part (2) I x2 = ρ 100 mm

( 121 )(200)(280 ) = 36321.5 kg-mm . 3

The composite: I x = I x 2 − I x1 = 32423 kg-mm 2 , from which, using the result of the solution to Problem 8.158

140 mm x

I z = I x + I y = 32422 + 70226 = 102649 kg-mm 2 = 0.10265 kg-m 2

140 mm

200 mm

Problem 8.160 The homogeneous pyramid is of mass m. Determine its moment of inertia about the z-axis. y

Solution: ρ =

The mass density is

m 3m = 2 . V w h

The differential mass is dm = ρω 2 dz. The moment of inertia of this element about the z-axis is x

dI Z =

( 16 )ω dm. 2

Noting that ω = h

w dI z = ρ z w

wz , then h

z dz. ( 6wh ) z dz = mw 2h 4

Integrating: I z -axis =

( mw ) z dz = 101 mw 2h ∫ 2

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Problem 8.161 Determine the moments of inertia of the homogeneous pyramid of mass m about the x - and y -axes.

Solution: Use the results of the solution of Problem 8.160 for the mass density. The elemental disk is dm = ρω 2 dz. The moment of inertia about an axis through its center of mass parallel to the x-axis is dI X =

( 121 )ω dm. 2

Use the parallel axis theorem: I x -axis =

( 121 ) ∫ ω dm + ∫ z dm. 2

Noting that ω = h

I x -axis =

w z, the integral is h

ρw 4 h 4 ρw 2 h z dz + 2 ∫ z 4 dz. 0 12h 4 ∫0 h

Integrating and collecting terms

z w

I x-axis = m

( 201 w + 35 h ). 2

By symmetry, I y -axis = I x -axis

Problem 8.162 The homogeneous object weighs 400 lb. Determine its moment of inertia about the x-axis. y

Solution:

The volumes are

Vcyl = (46)π (9) 2 = 11, 706 in 3 , Vcone = 13 π (6) 2 (36) = 1357 in 3 , so V = Vcyl − Vcone = 10,348 in 3 .

The masses of the solid cylinder and the material that would occupy the conical hole are

6 in 9 in 36 in 46 in

x 36 in 46 in Side View

 Vcyl  400 m cyl =  = 14.052 slug,  V  32.2

(

m cone =

)

400 ) = 1.629 slug. ( V V )( 32.2 cone

Using results from Appendix C, I ( x -axis) =

1 3 m (9) 2 − m (6) 2 2 cyl 10 cone

= 551 slug-in 2 = 3.83 slug-ft 2

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Problem 8.163 Determine the moments of inertia of the 400-lb object about the y- and z-axes. y

Solution:

See the solution of Problem 8.162. The position of the center of mass of the material that would occupy the conical hole is x = (46 − 36) +

From Appendix C, 3 3 I (y ′ -axis)cone = m cone  (36) 2 + (6) 2   80  20

6 in 9 in 36 in

= 87.97 slug-in 2 . The moment of inertia about the y-axis for the composite object is

36 in 46 in

I ( y -axis) = m cyl [ 13 (46) 2 + 14 (9) 2 ] − ( I ( y′ -axis)cone + x 2 m cone )

Side View

46 in

3 (36) = 37 in. 4

= 7877 slug-in 2 = 54.7 slug-ft 2 . y9

6 in x , x9 X

36 in

Problem 8.164 Determine the moment of inertia of the 14-kg flywheel about the axis L.

Solution:

The flywheel can be treated as a composite of the objects

shown: The volumes are V1 = (150)π (250) 2 = 294.5 × 10 5 mm 3 , V2 = (150)π (220) 2 = 228.08 × 10 5 mm 3 , V3 = (50)π (220) 2 = 76.03 × 10 5 mm 3 ,

50 mm

V4 = (50)π (60) 2 = 5.65 × 10 5 mm 3 , V5 = (100)π (60) 2 = 11.31 × 10 5 mm 3 ,

70 mm

120 mm

L 100 mm

V6 = (100)π (35) 2 = 3.85 × 10 5 mm 3 . The volume V = V1 − V2 + V3 − V4 + V5 − V6 = 144.3 × 10 5 mm 3 ,

440 mm 500 mm

so the density is 150 mm

δ =

14 = 9.704 × 10 −7 kg/mm 3 . V

The moment of inertia is I L = 12 δV1 (250) 2 − 12 δV2 (220) 2 + 12 δV3 (220) 2 − 12 δV4 (60) 2 + 12 δV5 (60) 2 − 12 δV6 (35) 2 = 536,800 kg-mm 2 = 0.5368 kg-m 2 .

1 4

636

2 5

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Chapter 9 Problem 9.1 In Practice Example 9.1, suppose that the coefficient of static friction between the 180-lb crate and the ramp is µ s = 0.3. What is the magnitude of the smallest horizontal force the rope must exert on the crate to prevent it from sliding down the ramp?

Solution:

The free-body diagram is shown.

Assume that the crate is on the verge of slipping down the ramp. Then the friction force is f = µ s N and points up the ramp. The equilibrium equations are ΣFx : µ s N + T cos 20 ° − W sin 20 ° = 0, ΣFy : N − T sin 20 ° − W cos 20 ° = 0. Setting W = 180 lb and µ s = 0.3 and solving yields N = 173 lb, T = 10.4 lb. 10.4 lb. y x T 208 W

f N

Problem 9.2 A person places a 2-lb book on a table that is tilted at 15 °. The coefficients of friction between the book and the polished surface of the table are µ s = 0.22 and µ k = 0.18. (a) What is the smallest force she must apply to the stationary book (parallel to the table surface) to prevent it from sliding down? (b) What force must she apply to cause the book to start sliding up?

She can continue decreasing the force she applies until the friction force reaches its maximum value f = µ s N = µ sW cos15 °, where we have used the second equilibrium equation. From the first equilibrium equation, the minimum force she must apply is F = W sin15 ° − f = W sin15 ° − µ sW cos15 ° = 0.0926 lb (1.48 oz). (b) If she begins increasing the force she applies, the friction force decreases to zero and then begins increasing toward the left: y

x 158

158 F

N W

The equilibrium equations are ΣFx = F − f − W sin15 ° = 0, ΣFy = N − W cos15 ° = 0.

Solution: (a) If she pushes with the proper force, the book will be in equilibrium on the table with no friction force acting on it. If she begins decreasing the force she applies, an increasing friction force toward the right is necessary to keep the book in equilibrium: y

x 158

As she pushes harder, the resisting friction force will increase until it reaches its maximum value f = µ s N = µ sW cos15 °. The force she must exert is F = W sin15 ° + f = W sin15 ° + µ sW cos15 ° = 0.943 lb (15.1 oz). (a) 0.0926 lb. (b) 0.943 lb.

N W

We have the equilibrium equations ΣFx = F + f − W sin15 ° = 0, ΣFy = N − W cos15 ° = 0.

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Problem 9.3 A student wishes to push a 44-kg crate of books across the floor. The static coefficient of friction between the crate and the carpet is µ s = 0.6. (a) If he exerts a force F at an angle α = 20 °, what force is necessary to start the crate moving? (b) If he bends his knees more and exerts the force at α = 5 °, what force is necessary?

Solution:

The weight of the crate is W = (44 kg)(9.81 m/s 2 ) = 432 N. (a) The free-body diagram of the crate is y F a x W f

a F

The equilibrium equations are ΣFx = F cos α − f = 0, ΣFy = −F sin α + N − W = 0. The crate will slip and begin moving when f = µ s N. Solving these three equations with α = 20 °, we obtain F = 353 N (79.3 lb), f = 331 N, N = 552 N. (b) Solving the three equations with α = 5 °, we obtain F = 274 N (61.7 lb), f = 273 N, N = 456 N. (a) 353 N (79.3 lb), (b) 274 N (79.3 lb) (61.7 lb).

Problem 9.4 The 2975-lb car is parked on a sloped street. The brakes are applied to both its front and rear wheels. (a) If the coefficient of static friction between the car’s tires and the road is µ s = 0.8, what is the steepest slope (in degrees relative to the horizontal) on which the car could remain in equilibrium? (b) If the street were icy and the coefficient of static friction between the car’s tires and the road was µ s = 0.2, what is the steepest slope on which the car could remain in equilibrium?

Solution: Let α be the slope of the street in degrees. The equilibrium equations and impending slip friction equation are ΣFx : W sin α − f = 0, ΣFy : N − W cos α = 0, f = µs N Solving, we find that f = W sin α, N = W cos α, µ s = tan α. α = tan −1 (µ s ). (a) α = tan −1 (0.8) = 38.7 °. (b) α = tan −1 (0.2) = 11.3 °. (a) α = 38.7 °, (b) α = 11.3 °.

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Problem 9.5 A 1080-kg car is parked on a sloped street. The figure shows its wheels and the position of its center of mass. The street is icy, and as a result the coefficient of static friction between the car’s tires and the street surface is µ s = 0.2. Determine the steepest slope (in degrees relative to the horizontal) at which the car could remain in equilibrium if (a) the brakes are applied to both its front and rear wheels; (b) the brakes are applied to the front (lower) wheels only.

Solution:

The car’s weight is W = (1080 kg)(9.81 m/s 2 ) = 10.6 kN. (a) The free-body diagram of the car is

x N

where N is the total normal force exerted by the street on the car’s wheels and f is the total friction force exerted on the wheels. We have the equilibrium equations ΣFx = − f + W sin α = 0, ΣFy = N − W cos α = 0.

532 mm 1380 mm 2370 mm

When slip is impending, the friction force f = µ s N. From the equilibrium equations, f /N = sin α / cos α = tan α. Therefore slip is impending when tan α = µ s = 0.2. We obtain α = 11.3 °. (b) Let N F and N R be the total normal forces exerted by the front and rear wheels. Only the rear wheels can exerted a friction force:

a NR

W NF The equilibrium equations are ΣFx = − f F + W sin α = 0, ΣFy = N F + N R − W cos α = 0,

ΣM rear wheel contact = (2370 mm) N F − (532 mm)W sin α − (1380 mm)W cos α = 0. When slip of the rear wheels is impending, f F = µ s N F . Substituting this into the first equilibrium equation, we have N F = (W sin α)/µ s . Then substituting this expression into the third equilibrium equation and dividing the resulting equation by cos α, we obtain the equation tan α =

1380 . 1 − 532 µs

(2370)

This yields α = 6.95 °. (a) α = 11.3 °. (b) α = 6.95 °.

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Problem 9.6 The coefficient of kinetic friction between the 600-lb crate and the ramp is µ k = 0.18. The unstretched length of the spring is 4 ft, and the spring constant is k = 250 lb/ft. The crate is placed on the ramp and slowly lowered until it is held in equilibrium by the spring. What is the resulting distance x?

Solution:

As the crate slides downward, the friction between it and the ramp is determined by the kinetic coefficient of friction. The spring force will increase until the box is in equilibrium. The free-body diagram is

f W

N 508

The equilibrium equations are

ΣFx = F − W sin 50 ° + f = 0, k

ΣFy = −W cos 50 ° + N = 0. 508

Setting f = µ k N and solving, the spring force required for equilibrium is F = W sin 50 ° − µ kW cos 50 ° = (600 lb)[ sin 50 ° − (0.18) cos 50 ° ] = 390 lb. In terms of the spring constant, the spring force is F = k (4 ft − x ), so the crate is in equilibrium when F k 390 lb = 4 ft − 250 lb/ft = 2.44 ft.

x = 4 ft −

2.44 ft.

Problem 9.7 The coefficient of static friction between the 50-kg crate and the ramp is µ s = 0.35. The unstretched length of the spring is 800 mm, and the spring constant is k = 660 N/m. What is the minimum value of x at which the crate can remain stationary on the ramp?

Solution: Fs = (660 N/m)(0.8 m − x ) ΣF : FS − 490.5 N sin 50 ° + f = 0 ΣF : N − 490.5 N cos50 ° = 0 f = 0.35 N Solving: x = 0.398 m = 398 mm

490.5 N x

508 k 508 N f

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Problem 9.8 The coefficients of friction between the 30-kg crate and the ramp are µ s = 0.6 and µ k = 0.45. If the angle α = 20 °, what tension must the person exert to start the stationary crate moving up the ramp?

Solution:

The weight of the crate is W = (30 kg)(9.81 m/s 2 ) = 294 N. The free-body diagram is T

a W

x 108

f The equilibrium equations are

108

ΣFx = T cos α − W sin10 ° − f = 0, ΣFy = T sin α − W cos10 ° + N = 0. T cos α − W sin10 ° − µ k N = 0, T sin α − W cos10 ° = −N . T =

W (sin10 ° + µ k cos10 °) cos α + µ k sin α

d (cos α + µ k sin α) = −sin α + µ k cos α = 0 dα tan α = µ k . The crate will slip when the friction force exceeds f = µ s N. Substituting this expression into the equilibrium equations and solving them, we obtain T = 197 N, N = 223 N. Greater than 197 N (44.2 lb).

Problem 9.9 The coefficients of friction between the 30-kg crate and the ramp are µ s = 0.6 and µ k = 0.45. What value of the angle α would minimize the tension the person must exert to pull the crate up the ramp at a constant speed? What is the minimum tension?

Solution: The weight of the crate is W = (30 kg)(9.81 m/s 2 ) = 294 N. The free-body diagram is T

a W

x 108

f a

The equilibrium equations are ΣFx = T cos α − W sin10 ° − f = 0,

108

ΣFy = T sin α − W cos10 ° + N = 0. The friction force equals f = µ k N. Substituting this expression into the equilibrium equations and solving them for T, we obtain T =

W (sin10 ° + µ k cos10 °) . cos α + µ k sin α

To minimize the tension, we want to determine the value of α that makes the denominator cos α + µ k sin α a maximum. We set d (cos α + µ k sin α) = − sin α + µ k cos α = 0, dα which yields tan α = µ k . The angle is α = 24.2 °. Using this value and solving the equilibrium equations, we obtain T = 166 N, N = 222 N. α = 24.2 °, Tension is 166 N (37.2 lb).

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Problem 9.10 The mass of box A is 30 kg, and the mass of box B is 10 kg. The coefficients of friction between box A and the ramp are µ s = 0.30 and µ k = 0.28. Can the boxes remain in equilibrium in the positions shown? If so, what is the magnitude of the friction force exerted on box A by the ramp?

Solution: Let us draw individual free-body diagrams of the two boxes. We don’t know the direction of the friction force, so we choose it arbitrarily:

mAg

308

mBg

If the boxes are in equilibrium, their equilibrium equations are B

ΣFx = T − f − m A g sin 30 ° = 0, ΣFx = T − f − m A g sin 30 ° = 0,

308

ΣFy = T − m B g = 0. Solving, we obtain T = 98.1 N, N = 255 N, f = −49.1 N. The minus sign means the friction force necessary for equilibrium is in the direction up the incline, not down. The question is, can the surfaces support a friction force of that magnitude? The largest force that can be supported is µ s N = (0.3)(255 N) = 76.5 N. Yes, it can be supported and the magnitude of the friction force is 49.1 N. Yes, 49.1 N.

Problem 9.11 Box A weighs 100 lb, and the coefficients of friction between box A and the ramp are µ s = 0.30 and µ k = 0.28. For what range of weights of the box B will the system remain stationary?

Solution: First let’s find out how light box B can be by assuming that box A is on the verge of slipping down the inclined surface:

y A

308

μs N

B 308

The equilibrium equations are ΣFx = T + µ s N − W A sin30 ° = 0, ΣFy = N − W A cos30 ° = 0, ΣFy = T − W B = 0. Solving, we obtain T = 24.0 lb, N = 86.6 lb, W B = 24.0 lb. Now we determine how heavy B can be by assuming that box A is on the verge of slipping up the inclined surface:

T T 308

WA y

y x

N μsN

The equilibrium equations are ΣFx = T − µ s N − W A sin 30 ° = 0, ΣFy = N − W A cos30 ° = 0, ΣFy = T − W B = 0. Solving, we obtain T = 76.0 lb, N = 86.6 lb, W B = 76.0 lb. We see that the system will remain stationary if 24.0 ≤ W B ≤ 76.0 lb. 24.0 ≤ W B ≤ 76.0 lb.

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Problem 9.12 The truck weighs 4000 lb and has fourwheel drive. The coefficient of static friction between its tires and the muddy street surface is µ s = 0.4. The block of steel on the sloping ramp weighs 9000 lb, and the coefficient of static friction between its surface and the ramp is µ s = 0.6. The cable is horizontal. When the driver attempts to move the truck forward and slide the block down the ramp, will the block or the truck’s tires slip first?

Solution:

The free-body diagrams are

y WT

WB fB

NB NT

where T is the tension in the cable. The equilibrium equations for the truck are ΣFx = f T − T = 0, 208

ΣFy = N T − WT = 0. What is the largest tension the truck can exert without its wheels slipping? The maximum the friction force between its wheels and the road is f T = µ T N T = µ TWT , where µ T = 0.4. Therefore the largest tension the truck can exert is T = f T = µ TWT = 1600 lb. The equilibrium equations for the block are ΣFx = − f B cos20 ° + N B sin 20 ° + T = 0, ΣFy = f B sin 20 ° + N B cos20° − W B = 0. What is the largest tension that can be exerted on the block before it slips? Setting f B = µ B N B , where µ B = 0.6, and solving, we obtain  µ cos 20 ° − sin 20 °  T =  B W = 1740 lb.  µ B sin 20 ° + cos 20 °  B The truck’s tires slip first.

Problem 9.13 The coefficient of kinetic friction between the 100-kg box and the inclined surface is 0.35. Determine the tension T necessary to pull the box up the surface at a constant rate.

Solution: 3T

981 N T

608

N f

608

ΣF : 3T − (981 N)sin 60 ° − f = 0 ΣF : N − (981 N) cos60 ° = 0 f = 0.35 N Solving: T = 340 N

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Problem 9.14 The angles α = 15 °, β = 35 °. The mass of the box is 15 kg. The coefficients of friction between the box and the ramp are µ s = 0.30 and µ k = 0.28. The box is sliding down the ramp at a constant speed. What is the force T?

Solution:

The free-body diagram is T b mg

x μkN

The equilibrium equations are ΣFx = −T cos β + mg sin α + µ k N = 0, ΣFy = N + T sin β − mg cos α = 0. a

Solving, we obtain T = 79.5 N, N = 96.5 N. T = 79.5 N.

Problem 9.15 In explaining observations of ship launchings at the port of Rochefort in 1779, Coulomb analyzed the system shown to determine the minimum force T necessary to hold the box stationary on the inclined surface. Show that the result is T =

Solution:

T b mg

(sin α − µ s cos α)mg . cos β − µ s sin β

We assume that the box is on the verge of slipping down

the surface:

μsN

x N

The equilibrium equations are

ΣFx = −T cos β + mg sin α − µ k N = 0, ΣFy = N + T sin β − mg cos α = 0. Solving the first equation for N and substituting it into the second one, we can solve it for T, obtaining a

644

T =

(sin α − µ k cos α)mg . cos β − µ k sin β

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Problem 9.16 Two sheets of plywood A and B, each with weight W = 40 lb, lie on the bed of the truck. The coefficients of friction between the two sheets are µ s = 0.30 and µ k = 0.28, and the coefficients of friction between sheet B and the bed of the truck are µ s = 0.26 and µ k = 0.22. (a) If you apply a horizontal force to sheet A and apply no force to sheet B, can you slide A off the truck without causing B to move? What force is necessary to cause A to start moving? (b) If you prevent sheet A from moving by exerting a horizontal force on it, what horizontal force on sheet B is necessary to start it moving?

Solution: (a) Suppose we apply a small horizontal force F to sheet A. The resulting free-body diagrams of the two sheets are shown: W

A fA

NA W

NA f A

We can see that the normal forces N A = W = 40 lb and N B = N A + W = 80 lb. Also, the friction forces f A = F and f B = f A = F. If we increase the force F, sheet A will be on the verge of slipping when f A = 0.3 N A = 12 lb.

A B

At that point the friction force f B = f A = 12 lb. In order for sheet B to be on the verge of slipping, the friction force f B must reach f B = 0.26 N B = 20.8 lb. So there’s no problem moving sheet A without causing sheet B to slip, and the force required is 12 lb. (b) The free-body diagrams for this case are shown: W

A fA B

NA W

NA f A

Again, the normal forces N A = W = 40 lb and N B = N A + W = 80 lb. Sheet B will be on the verge of slipping at its upper and lower surfaces when FB = f A + f B = 0.3 N A + 0.26 N B = 32.8 lb. (a) Yes, you can. Force is 12 lb. (b) 32.8 lb.

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Problem 9.17 The weights of the two boxes are W1 = 100 lb and W2 = 50 lb. The coefficients of friction between the left box and the inclined surface are µ s = 0.12 and µ k = 0.10. Determine the tension the man must exert on the rope to pull the boxes upward at a constant rate.

308

Solution: 100 lb

308

50 lb

308 N W2

ΣF : T − 100 lb sin30 ° − f − 50 lb = 0 ΣF : N − 100 lb cos30 ° = 0 f = 0.10 N Solving: T = 109 lb

Problem 9.18 In Problem 9.17, for what range of tensions exerted on the rope by the man will the boxes remain stationary?

Solution:

See the figure in 9.17.

First solve for the largest force Tmax ΣF : Tmax − 100 lb sin30 ° − f − 50 lb = 0 ΣF : N − 100 lb cos30 ° = 0

308

⇒ Tmax = 110.4 lb

f = 0.12 N

Next solve for the smallest force Tmin . We need to turn the friction force in the opposite direction. ΣF : Tmin − 100 lbsin30 ° + f − 50 lb = 0

308

ΣF : N − 100 lbcos30 ° = 0

⇒ Tmin = 89.6 lb

f = 0.12 N W2

Thus for the boxes to remain stationary we must have 89.6 lb < T < 110.4 lb

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Problem 9.19 Each box weighs 10 lb. The coefficient of static friction between box A and box B is 0.24, and the coefficient of static friction between box B and the inclined surface is 0.3. What is the largest angle α for which box B will not slip? Strategy: Draw individual free-body diagrams of the two boxes and write their equilibrium equations assuming that slip of box B is impending.

Solution:

We have 6 unknowns, 4 equilibrium equations and 2 fric-

tion equations ΣFA : T − 10 lbsin α − f 2 = 0 ΣFA : N 2 − 10 lb cos α = 0 ΣFB : f 2 + f1 − 10 lbsin α = 0 ΣFB : N 1 − N 2 − 10 lb cos α = 0 f1 = 0.3 N 1 , f 2 = 0.24 N 2 Solving we find α = 40.0 ° 10 lb

A B A

a N2 10 lb

a N1

Problem 9.20 The masses of the boxes are m A = 15 kg and m B = 60 kg. The coefficient of static friction between boxes A and B and between box B and the inclined surface is 0.12. What is the largest force F for which the boxes will not slip?

Solution:

We have 6 unknowns, 4 equilibrium equations and 2 fric-

tion equations. ΣFA : T − F − (147.15 N)sin 20 ° + f 2 = 0 ΣFA : N 2 − (147.15 N) cos 20 ° = 0 ΣFB : T − (588.6 N)sin 20 ° − f1 − f 2 = 0 ΣFB : N 1 − N 2 − (588.6 N) cos 20 ° = 0 f1 = 0.12 N 1 , f 2 = 0.12 N 2

Solving we find F = 267 N

147.15 N

208

A f2

N2 B 588.6 N f1 208 N1

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Problem 9.21 In Problem 9.20, what is the smallest force F for which the boxes will not slip?

Solution:

See the solution for 9.20—change the directions of all of the friction forces. ΣFA : T − F − (147.15 N)sin 20 ° − f 2 = 0 ΣFA : N 2 − (147.15 N) cos 20 ° = 0 ΣFB : T − (588.6 N)sin 20 ° + f1 + f 2 = 0 ⇒ F = 34.8 N ΣFB : N 1 − N 2 − (588.6 N) cos 20 ° = 0

f1 = 0.12 N 1, f 2 = 0.12 N 2

B 208

Problem 9.22 The angle α = 30 °. The homogeneous cylinder weighs 50 lb, and its radius is 1 ft. The left surface is smooth. The coefficient of static friction between the surface of the cylinder and the right surface is µ s = 0.3. Determine the largest couple M that could be applied to the cylinder without causing it to slip if the couple is (a) counterclockwise; (b) clockwise.

Solution: y

M x 50 lb f

(a) We have the equilibrium equations

ΣFx = N L sin α − N R cos 45 ° − f cos 45 ° = 0,

458

ΣFy = N L cos α + N R cos 45 ° − f cos 45 ° = 50 lb, ΣM origin = M − Rf = 0. Slipping impends when f = µ k N R , so we have four equations in terms of N L , N R , f , M . Solving, we obtain N L = 44.0 lb, N R = 24.0 lb, f = 7.19 lb, M = 7.19 ft-lb. y

M x 50 lb NL

f NR

(b) The equilibrium equations are ΣFx = N L sin α − N R cos 45 ° + f cos 45 ° = 0, ΣFy = N L cos α + N R cos 45 ° + f cos 45 ° = 50 lb, ΣM origin = −M + Rf = 0. Solving together with f = µ k N R , we obtain N L = 27.9 lb, N R = 28.1 lb, f = 8.44 lb, M = 8.44 ft-lb. (a) 7.19 ft-lb. (b) 8.44 ft-lb.

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Problem 9.23 The homogeneous horizontal bar AB weighs 20 lb. The homogeneous disk weighs 30 lb. The coefficient of kinetic friction between the disk and the sloping surface is µ k = 0.24. What is the magnitude of the couple that would need to be applied to the disk to cause it to rotate at a constant rate in the clockwise direction? 5 ft 1 ft A

208

Solution:

From the free-body diagram of the bar,

ΣM B : (20 lb)(2.5 ft) + A y (5 ft) = 0 ⇒ A y = −10 lb. From the free-body diagram of the disk. ΣFx : A x + N sin 20 ° + µ k N cos 20 ° = 0, ΣFy : N cos 20 ° − µ k N sin 20 ° − 30 lb = 0, ΣM A : −M + µ k N (1 ft) = 0. Solving yields A x = −26.5 lb, N = 46.6 lb, M = 11.2 ft-lb. M = 11.2 ft-lb.

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Problem 9.24 The homogeneous horizontal bar AB weighs 20 lb. The homogeneous disk weighs 30 lb. The coefficient of kinetic friction between the disk and the sloping surface is µ k = 0.24. What is the magnitude of the couple that would need to be applied to the disk to cause it to rotate at a constant rate in the counterclockwise direction? 5 ft 1 ft A

208

Solution:

From the free-body diagram of the bar,

ΣM B : (20 lb)(2.5 ft) + A y (5 ft) = 0 ⇒ A y = −10 lb. From the free-body diagram of the disk, ΣFx : A x + N sin 20 ° − µ k N cos 20 ° = 0, ΣFy : N cos 20 ° + µ k N sin 20 ° − 30 lb = 0, ΣM A : M − µ k N (1 ft) = 0. Solving yields A x = −4.56 lb, N = 39.1 lb, M = 9.40 ft-lb. M = 9.40 ft-lb.

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Problem 9.25 The mass of the bar is 4 kg. The coefficient of static friction between the bar and the floor is 0.3. Neglect friction between the bar and the wall. (a) If α = 20 °, what is the magnitude of the friction force exerted on the bar by the floor? (b) What is the maximum angle α for which the bar will not slip?

Solution: (a) α = 20 ° f max = 0.3 N B ΣFx : N A − f B = 0 ΣFy : N B − 39.24 N = 0 ΣM B : (39.24 N)(0.5 m)sin α − N A (1.0 m) cos α = 0 Solving: f B = 7.14 N, f max = 11.77 N Since f B < f max , f B = 7.14 N (b)

f B = 0.3 N B ΣFx : N A − f B = 0

1m a

ΣFy : N B − 39.24 N = 0 ΣM B : (39.24 N)(0.5 m)sin α − N A (1.0 m) cos α = 0 ⇒ α = 31.0 °

a 1.0 m

39.24 N

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Problem 9.26 The coefficient of static friction between the bar and the floor and between the 4-kg bar and the wall is 0.3. What is the maximum angle α for which the bar will not slip?

Solution: f B = 0.3 N B , f A = 0.3 N B ΣFx : N A − f B = 0 ΣFy : N B + f A − 39.24 N = 0 ΣM B : (39.24 N)(0.5 m)sin α − N A (1.0 m) cos α − f A (1.0 m)sin α = 0 Solving α = 33.4 °

1m a

a 1.0 m

39.24 N

Problem 9.27 The ladder and the person weigh 30 lb and 180 lb, respectively. The center of mass of the 12-ft ladder is at its midpoint. The angle α = 30 °. Assume that the wall exerts a negligible friction force on the ladder. (a) If x = 4 ft, what is the magnitude of the friction force exerted on the ladder by the floor? (b) What minimum coefficient of static friction between the ladder and the floor is necessary for the person to be able to climb to the top of the ladder without slipping?

Solution: (a) Assume no slipping occurs α = 30 °, x = 4 ft ΣFx : f B − N A = 0 ΣFy : N B − 210 lb = 0 ΣM B : N A (12 ft cos α) − 30 lb (6 ft sin α) − 180 lb x = 0 Solving f B = 77.9 lb, N B = 210 lb (b) At the top of the ladder α = 30 °, x = 6 ft ΣFx : f B − N A = 0 ΣFy : N B − 210 lb = 0 ΣM B : N A (12 ft cos α) − 30 lb (6 ft sin α) − 180 lb x = 0 f B = µs N B ⇒ µ s = 0.536

NA a

180 lb 30 lb x

fB x

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Problem 9.28 The 30-lb ladder’s length L = 12 ft. Friction at the wall is negligible. Suppose that in designing a high-friction footing for the ladder, it is conservatively assumed that the angle α = 40 ° and that a 300-lb person is at the top of the ladder (that is, his center of mass is at x = L sin α ). What minimum coefficient of friction would be necessary to avoid slipping?

Solution:

The free-body diagram of the ladder is Wp P

408

N a

The equilibrium equations are ΣFx = f − P = 0, ΣFy = N − W L − W P = 0, ΣM foot of ladder = ( L cos 40 °) P −

( 12 L sin 40° )W − (L sin 40°)W = 0. L

Solving yields P = f = 264 lb, N = 330 lb. The necessary friction coefficient is µ s = f /N = 264/330 = 0.801. µ s = 0.801.

Problem 9.29 The ladder and the person weigh 30 lb and 200 lb, respectively. The ladder’s length L = 12 ft. The coefficient of static friction between the foot of the ladder and the floor is µ s = 0.4. Friction at the wall is negligible. If the angle α = 30 °, how high up the ladder (that is, to what horizontal position x) can the person go without causing the ladder to slip? Assume that the system is in equilibrium.

Solution:

The free-body diagram of the ladder is WP

308

f x

The equilibrium equations are ΣFx = f − P = 0, ΣFy = N − W L − W P = 0, ΣM foot of ladder = ( L cos30 °) P −

( 12 L sin 30° )W − xW = 0. L

Slip is impending when f = µ s N. Substituting this expression into the equilibrium equations and solving them, we obtain P = f = 92 lb, N = 230 lb, x = 4.33 ft. x = 4.33 ft.

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Problem 9.30 The disk weighs 50 lb and the bar weighs 25 lb. The coefficients of friction between the disk and the inclined surface are µ s = 0.6 and µ k = 0.5. (a) What is the largest couple M that can be applied to the stationary disk without causing it to start rotating? (b) What couple M is necessary to rotate the disk at a constant rate?

Solution: The Problem has 7 unknowns, 6 equilibrium equations and one friction equation. (a) No slip ΣM A : (25 lb)(10 in cos30°) − B 2 (20 in) = 0 ΣF : N − B 2 − 50 lb cos30° = 0 ΣM B : M − f (5 in) = 0 f = 0.6 N ⇒ M = 162.4 lb in

5 in

(b) Steady rotation. Replace the last equation with f = 0.5 N ⇒ M = 135.3 lb in

20 in

50 lb

308 M

25 lb

308

f N

Problem 9.31 The radius of the 40-kg homogeneous cylinder is R = 0.15 m. The slanted wall is smooth and the angle α = 30 °. The coefficient of static friction between the cylinder and the floor is µ s = 0.2. What is the largest couple M that can be applied to the cylinder without causing it to slip? R

Solution: Assume that slip of the wheel is impending. The equilibrium equations are ΣFx : P sin α − µ s N = 0, ΣFy : N + P cos α − W = 0, ΣM center : M − µ s NR = 0 where W = (40 kg)(9.81 m/s 2 ). Putting in the values for W , α, R, µ s and solving yields P = 117 N, N = 291 N, M = 8.74 N M = 8.74 N-m

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Problem 9.32 The homogeneous cylinder has weight W. The coefficient of static friction between the cylinder and both surfaces is µ s . What is the largest couple M that can be applied to the cylinder without causing it to slip? (Assume that the cylinder slips before rolling up the inclined surface.) R

Solution:

Assume that slip of the wheel is impending. The equilibrium equations are ΣFx : P sin α − µ s P cos α − µ s N = 0, ΣFy : N + P cos α + P sin α − W = 0, ΣM center : M − µ s NR − µ s PR = 0 Solving these equations for P, N , and M, we obtain M =

µ s RW[sin α + µ s (1 − cos α)] (1 + µ s 2 )sin α

M a

Problem 9.33 The homogeneous cylinder has weight W. The coefficient of static friction between the cylinder and both surfaces is µ s . What is the minimum value of µ s for which the couple M will cause the cylinder to roll up the inclined surface without slipping?

Solution:

Assume that slip of the cylinder is impending and the cylinder is on the verge of rolling up the inclined surface (the normal force exerted by the floor is zero). The equilibrium equations are ΣFx : P sin α − µ s P cos α = 0, ΣFy : P cos α + µ s P sin α − W = 0, ΣM center : M − µ s PR = 0

From the first equation we see that µ s = tan α.

M a

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Problem 9.34 The coefficient of static friction between the blades of the shears and the object they are gripping is 0.36. What is the largest value of the angle α for which the object will not slip out? Neglect the object’s weight. Strategy: Draw the free-body diagram of the object and assume that slip is impending.

Solution: ΣFx : 2 f cos(α /2) − 2 N sin (α /2) = 0 f = 0.36 N Solving tan (α /2) = 0.36 ⇒ α = 39.6 °

N f

f N

Problem 9.35 A stationary disk of 300-mm radius is attached to a pin support at D. The disk is held in place by the brake ABC in contact with the disk at C. The hydraulic actuator BE exerts a horizontal 400-N force on the brake at B. The coefficients of friction between the disk and the brake are µ s = 0.6 and µ k = 0.5. What couple must be applied to the stationary disk to cause it to slip in the counterclockwise direction?

Solution:

Assume impending slip. For counterclockwise motion the friction force f = µ s FN opposes the impending slip, so that it acts on the brake in a downward direction, producing a negative moment (clockwise) about A. The sum of the moments about A: ΣM A = −0.2(400) + (0.4 − 0.2µ s ) FN , from which FN = 285.7 N. The sum of the moments about the center of the disk: ΣM D = M − 0.3(µ s ) FN = 0, from which M = 51.43 N m.

C 200 mm 200 mm

300 mm

200 mm 400 N 200 mm

200 mm

300 mm

200 mm

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21/04/23 12:32 PM

Problem 9.36 The figure shows a preliminary conceptual idea for a device to exert a braking force on a rope. The coefficient of kinetic friction between the rope and the two bars is µ k = 0.22. The force F = 10 lb and the angle α = 30 °. (a) Determine the force T necessary to pull the rope downward at a constant rate. (b) If the force T acts upward at the top of the rope instead, determine the force necessary to pull the rope upward at a constant rate.

Solution: (a) Consider the free-body diagrams of one of the bars and the rope:

N F Ay

f 308

y T

For the bar we write the equilibrium equation ΣM bottom of bar = −[ (3 in) cos30 ° ] F + [ (6 in)sin 30 ° ] N

6 in

F F

− [ (6 in) cos30 ° ] f = 0. F

3 in a

And from the free-body diagram for the rope, ΣFy = 2 f − T = 0. The friction force f = µ k N. Solving, we obtain N = 14.0 lb, f = 3.08 lb, T = 6.16 lb.

(b) If the is pulled upward the free-body diagrams are T T f

F Ay

308

For the bar we write the equation ΣM bottom of bar = −[ (3 in) cos30 ° ] F + [ (6 in)sin 30 ° ] N + [ (6 in) cos30 ° ] f = 0. From the free-body diagram for the rope, ΣFy = −2 f + T = 0. The friction force f = µ k N. Solving, we obtain N = 6.27 lb, f = 1.38 lb, T = 2.76 lb. (a) T = 6.16 lb. (b) T = 2.76 lb.

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Problem 9.37 The mass of block B is 8 kg. The coefficient of static friction between the surfaces of the clamp and the block is µ s = 0.2. When the clamp is aligned as shown, what minimum force must the spring exert to prevent the block from slipping out?

Solution:

The free-body diagram of the block when slip is impending is shown. From the equilibrium equation µ s FT + µ s ( FT + W cos α) − W cos α = 0, we obtain FT =

(8)(9.81)(1 − 0.2) cos 45 ° 2(0.2) = 111 N.

458 160 mm

w(1 − µ s ) cos α 2µ s

The free-body diagram of the upper arm of the clamp is shown. Summing moments about the upper end,

200 mm

0.16 Fs + 0.1µ s FT − 0.36 FT = 0, the force exerted by the spring is 0.36 FT − 0.1µ s FT 0.16 [0.36 − 0.1(0.2)]111 = 0.16 = 236 N.

Fs = B

100 mm

msFT

ms (F 1Wcosa) T FT 1Wcosa 100 mm

FS msF T

Problem 9.38 By altering its dimensions, redesign the clamp in Problem 9.37 so that the minimum force the spring must exert to prevent the block from slipping out is 180 N. Draw a sketch of your new design.

Solution:

160 mm 200 mm

This problem does not have a unique solution.

458 160 mm 200 mm

100 mm

658

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Problem 9.39 The horizontal bar is attached to a collar that slides on the smooth vertical bar. The collar at P slides on the smooth horizontal bar. The total mass of the horizontal bar and the two collars is 12 kg. The system is held in place by the pin in the circular slot. The pin contacts only the lower surface of the slot, and the coefficient of static friction between the pin and the slot is 0.8. If the system is in equilibrium and y = 260 mm, what is the magnitude of the friction force exerted on the pin by the slot?

Solution:

The free-body diagram of the horizontal bar and right collar is as shown, where m1 is the mass of the horizontal bar and right collar, N 1 is the normal force exerted by the vertical bar, and N 2 is the force exerted by the left collar. From the equilibrium equations ΣFx = −N 1 = 0, ΣFx = N 2 − m1g = 0, we see that N 2 = m1g. The free body diagram of the left collar is as shown, where m 2 is the mass of the left collar and N , f are the normal and friction forces exerted by the curved slot. y = 260 mm = (300 mm)sin θ, so the angle θ = 60.1 °. From the equilibrium equations, ΣFx = − f + m 2 g cos θ + N 2 cos θ = 0, ΣFy = N − m 2 g sin θ − N 2 sin θ = 0, we obtain

f = (m 2 g + N 2 ) cos θ = (m 2 + m1 ) cos θ = (12)(9.81) cos60.1 ° = 58.7 N. y

300 mm

N2 N1

m1g

u x m2g

Problem 9.40 In Problem 9.39, what is the minimum height y at which the system can be in equilibrium?

Solution:

From the solution of Problem 9.39, the friction and normal forces exerted on the pin by the circular slot are f = (m 2 g + N 2 ) cos θ, N = (m 2 g + N 2 )sin θ, so

f = cot θ. When slip impends, N

f = µ s N = 0.8 N, P

so 0.8 = cot θ and θ = 51.3 °. The height y = 300 sin θ = 234 mm.

y 300 mm

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Problem 9.41 The rectangular 100-lb plate is supported by the pins A and B. If friction can be neglected at A and the coefficient of static friction between the pin at B and the slot is µ s = 0.4, what is the largest angle α for which the plate will not slip?

Solution: Choose a coordinate system with the x-axis parallel to the rail. The sum of the moments about A is ΣM A = −2W cos α − 2.25W sin α + 4 B = 0, from which B =

a 2 ft 3 in B

W (2 cos α + 2.25sin α). 4

The component of weight causing the plate to slide is F = W sin α. This must be balanced by the friction force: 0 = −W sin α + µ s B, from which W sin α W = (2 cos α + 2.25sin α). 4 µs

Reduce algebraically to obtain   µs α = tan −1   = 14.47 °  2 − 1.125µ s  2 ft 2 ft

2 ft

2 ft msB

2.25 ft B

A W

Problem 9.42 If you can neglect friction at B and the coefficient of static friction between the pin at A and the slot is µ s = 0.4, what is the largest angle α for which the 100-lb plate will not slip?

The force tending to make the plate slide is F = −W sin α. This is balanced by the friction force at A, 0 = −W sin α + µ s A N , from which W sin α W = (−2.25sin α + 2 cos α). 4 µs

a 2 ft 3 in B

Reduce algebraically to obtain   µs α = tan −1   = 9.27 °  2 + 1.125µ s 

Check: The normal reactions at A and B are unequal: as the slots are inclined from the horizontal, the parallel component of the gravity force reduces the normal force at A, and increases the normal force at B. check. 2 ft 2 ft

Check: The sum of the reactions at A and B are A N + B N = W cos α. check. The magnitude ( A N + B N ) 2 + (µ s A N ) 2 = W , hence the system is in equilibrium at impending slip. check.

Solution: The normal force acts normally to the slots, and the friction force acts parallel to the slot. Choose a coordinate system with the x-axis parallel to the slots. The normal component of the reaction at A is found from the sum of the moments about B: ΣM B = −2.25W sin α + 2W cos α − 4 A N = 0, from which

2 ft 2 ft ms AN

2.25 ft BN

AN W

W AN = (−2.25sin α + 2 cos α). 4

660

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Problem 9.43 The airplane’s weight is W = 2400 lb. Its brakes keep the rear wheels locked, and the coefficient of static friction between the wheels and the runway is µ s = 0.6. The front (nose) wheel can turn freely and so exerts a negligible friction force on the runway. Determine the largest horizontal thrust force T the plane’s propeller can generate without causing the rear wheels to slip.

Solution:

The free body diagram when slip of the rear wheels impends is shown. From the equilibrium equations Σf x = −T + µ s B = 0, Σf y = A + B − W = 0, ΣM ptA = 4T − 5W + 7 B = 0, we obtain A = 1120 lb, B = 1280 lb, and T = 766 lb.

T y

4 ft

W A

5 ft

2 ft

4 ft

msB A 5 ft

Problem 9.44 The refrigerator weighs 220 lb. It is supported at A and B. The coefficient of static friction between the supports and the floor is µ s = 0.2. If you assume that the refrigerator does not tip over before it slips, what force F is necessary for impending slip?

Solution:

B 2 ft

Assume that slip is impending. From the equilibrium

equations ΣFx : F − µ s A − µ s B = 0, ΣFy : A + B − W = 0. Solving we find F = µ s ( A + B) = µ sW = (0.2)(220 lb). F = 44 lb.

B b 2

b 2

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Problem 9.45 The refrigerator weighs 220 lb. It is supported at A and B. The coefficient of static friction between the supports and the floor is µ s = 0.2. The distance h = 60 in and the dimension b = 30 in. When the force F is applied to push the refrigerator across the floor, will it tip over before it slips? (See Example 9.3.)

Solution:

See the analysis in Example 9.3. The refrigerator will tip over before it slips if h >

b 30 in = = 75 in. 2µ s 2(0.2)

No.

B b 2

b 2

Problem 9.46 To obtain a preliminary evaluation of the stability of a turning car, imagine subjecting the stationary car to an increasing lateral force F at the height of its center of mass, and determine whether the car will slip (skid) laterally before it tips over. Show that this will be the case if b /h > 2µ s . (Notice the importance of the height of the center of mass relative to the width of the car. This reflects on recent discussions of the stability of sport utility vehicles and vans that have relatively high centers of mass.)

Equilibrium Eqs. ΣFx :

F − fL − fR = 0

ΣFy :

N L + N R − mg = 0

b mg = 0 2 Assume skid and tip simultaneously. ΣM A :

−hF + bN R −

f L = µs N L , f R = µ s N R (skid ) and N L = 0 (tip), ∴ f L = 0. f R = µ s mg. The equilibrium equations become F = f R = µ s N R = µ s mg

and the moment equation uses b −h(µ s mg) + b(mg) − mg = 0 2

h b 2

b 2

Solution:

For y

b > 2µ s , slip before tip h

b < 2µ s , tip before slip h

b big N low cm, relative to track width h

F h fL

662

b = 2µ s h

A NL

b small N high cm, relative to track width h

B b 2

b 2

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Problem 9.47 The motorist who stopped to help exerts a force P on the car at an angle α = 20 °. The 1760-kg car has front wheel drive. The driver spins the front wheels, and the coefficient of kinetic friction is µ k = 0.02. Snow behind the rear tires exerts a horizontal resisting force S. Getting the car to move requires overcoming a resisting force S = 420 N. What force P must the man exert?

Solution: ΣFx : S − µ k N F − P cos α = 0 ΣFy : N R + N F − mg − P sin α = 0 ΣM A : −(1.62)mg + 2.55 N F + (0.90) P cos α − (3.40) P sin α = 0 α = 20 °, m = 1760 kg, g = 9.81 m/s 2 S = 420 N, µ k = 0.02 3 equations in 3 unknowns ( N R , N F , and P) Solving the equations, we get P = 213 N N R = 6.34 kN N F = 11.00 kN y mg

Source: Courtesy of T.Den_Team/Shutterstock.

F a

1.62 m P a

A S

0.90 m S

1.62 m 2.55 m 3.40 m

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2.55 m NR

0.90 m

B mkNF

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Problem 9.48 In Problem 9.47, what value of the angle α minimizes the magnitude of the force P the man must exert to overcome the resisting force S = 420 N exerted on the rear tires by the snow? What force must he exert?

From Eq. (2), N R = −N F + mg + P sin α or N R = −

1 (S − P cos α) + mg + P sin α ( b) µk

Substitute (a) and (b) into (3) We get  1  −1.62 mg + 2.55  (S − P cos α)  µk  + 0.90 P cos α − 3.40 P sin α = 0 Use this equation to find

dP 2.55  dP cos α − 0.90 P sin α − cos α + P sin α  + 0.90  dα µ k  d α dP − 3.40 sin α − 3.40 P cos α = 0 dα or

Source: Courtesy of T.Den_Team/Shutterstock. P a

 dP  2.55 cos α   0.90 cos α − 3.40 sin α −  d α  µk

0.90 m S

Solution:

dP and set it to zero. dα

1.62 m 2.55 m 3.40 m

From the solution to Problem 9.47, we have

S − µ k N F − P cos α = 0

(1)

N R + N F − mg − P sin α = 0

(2)

−1.62 mg + 2.55 N F + 0.90 P cos α − 3.40 P sin α = 0

(3)

 2.55  + P sin α − 0.90 sin α − 3.40 cos α  = 0  µ k    2.55   −P   − 0.90  sin α − 3.40 cos α    µk dP   = 0 =   dα 2.55 cos α   0.90 cos α − 3.40 sin α − µk   3.40 tan α =  2.55   µ − 0.90  k Solving, α = 1.54 ° Substituting this back into Eqs. (1), (2), and (3), and solving, we get

where µ k = 0.02,

P = 202 N

S = 420 N, m = 1760 kg, and g = 9.81 m/s 2 . From Eq. (1), NF =

664

1 (S − P cos α) (a ) µk

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Problem 9.49 The coefficient of static friction between the 3000-lb car’s tires and the road is µ s = 0.5. Determine the steepest grade (the largest value of the angle α) the car can drive up at constant speed if the car has (a) rear-wheel drive; (b) front-wheel drive; (c) fourwheel drive. n 19 i

The friction force balances the component of gravity parallel to the incline: 0 = −W sin α + µ s F , from which W sin α W = (72 cos α − 19sin α). µs 107 Reduce and solve:     72 α = tan −1   = 17.17 °  107 + 19    µ s

(c) For four wheel drive: Use the reactions of the front and rear wheels obtained in Parts (a) and (b). The sum of the forces parallel to the incline is

35 i n

72 i

Solution:

The friction force acts parallel to the incline, and the normal force is normal to the incline. Choose a coordinate system with the x-axis parallel to the incline. The component of the weight that acts parallel to the incline is W sin α, and the component acting normally to the incline is W cos α.

Σ FX = −W sin α + µ s R + µ s F = 0, from which W sin α W = (35cos α + 19sin α + 72 cos α − 19sin α). µs 107 Reduce and solve: α = tan −1 (µ s ) = 26.57 ° Check: This result is the same as if the Mercedes with four wheel drive were a box on an incline, as it should be.

(a) For rear wheel drive: The moment about the point of contact of the front wheels: ΣM FW = 35W cos α + 19W sin α − 107 R = 0,

19 in

from which the normal reaction of the two rear wheels is R =

W (35cos α + 19sin α). 107

The force causing impending slip is W sin α, which is balanced by the friction force: 0 = W sin α − µ s R, from which

35 in 72 in a

W sin α W = (35cos α + 19sin α). µs 107

Reduce and solve: α =

35  = 10.18 °  107    − 19  µ  s

msF

tan −1 

is the maximum angle at impending slip.

W msR

F 35 in

72 in

(b) For front wheel drive: The moments about the point of contact of the rear wheels is ΣM RW = −72W cos α + 19W sin α + 107 F = 0, from which the normal reaction of the two front wheels is F =

W (72 cos α − 19sin α). 107

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Problem 9.50 The stationary cabinet has weight W. Determine the force F that must be exerted to cause it to move if (a) the coefficient of static friction at A and at B is µ s ; (b) the coefficient of static friction at A is µ sA and the coefficient of static friction at B is µ sB .

Solution: (a) The sum of the moments about B is ΣM B = −hF + from which A =

( b2 )W − bA = 0,

( )

W h − F. b 2

The sum of forces: ΣFy = −W + A + B = 0,

from which

( )

W h + F. b 2 ΣFx = F − µ s A − µ s B = 0, B = W − A =

b 2

from which F = µs

( W2 + ( bh ) F + W2 − ( bh ) F ) = µ W s

(b) Use the normal reactions found in Part (a). From the sum of forces parallel to the floor, F = µ sA A + µ sB B = µ sA

( W2 − ( bh ) F ) + µ ( W2 + ( hb )). sB

Reduce and solve: W (µ + µ sB ) 2 sA F = h 1 + (µ sA − µ sB ) b

(

)

F h

H A

b — 2

F h mSA A A

666

W b 2

b 2

B mSBB

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Problem 9.51 The table weighs 50 lb and the coefficient of static friction between its legs and the inclined surface is 0.7. (a) If you apply a force at A parallel to the inclined surface to push the table up the inclined surface, will the table tip over before it slips? If not, what force is required to start the table moving up the surface? (b) If you apply a force at B parallel to the inclined surface to push the table down the inclined surface, will the table tip over before it slips? If not, what force is required to start the table moving down the surface?

Solution: (a) Assume the table does not tip ΣF : F − f1 − f 2 − 50 lbsin 20 ° = 0 ΣF : N 1 + N 2 − 50 lb cos 20 ° = 0 ΣM 2 : −F (32 in) − N 1 (46 in) + (50 lb cos 20 °)(23 in) + (50 lbsin 20 °)(28 in) = 0 f1 = 0.7 N 1, f 2 = 0.7 N 2 Solving we find F = 50 lb, N 1 = −0.874 lb, N 2 = 47.9 lb Since N 1 < 0 we conclude that the table tips before it slips

(b) Assume the table does not tip 32 in

ΣF : −F + f1 + f 2 − 50 lbsin 20 ° = 0

28 in

ΣF : N 1 + N 2 − 50 lb cos 20 ° = 0 ΣM 2 : F (32 in) − N 1 (46 in) + (50 lb cos 20 °)(23 in) + (50 lbsin 20 °)(28 in) = 0

23 in

f1 = 0.7 N 1 ,

23 in 208

f 2 = 0.7 N 2

Solving we find F = 15.79 lb,

N 1 = 44.8 lb,

N 2 = 2.10 lb

Since N 2 > 0 we conclude that the table does not tip, and F = 15.79 lb

208 F

f2 50 lb f1

N1 F

208

f2 50 lb

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Problem 9.52 The coefficient of static friction between the right bar and the surface at A is µ s = 0.6. Neglect the weights of the bars. If α = 20 °, what is the magnitude of the friction force exerted at A?

Solution:

Note that the condition of impending slip does not necessarily apply. The moments about the left pin support: ΣM = −F cos α + 2 A sin α = 0, from which A =

F . 2 tan α

Isolate the right bar and take moments about the upper pin joint: ΣM = AL sin α − fL cos α = 0, from which f = A tan α. Substitute for A: a

f = A tan α =

F tan α F = 2 tan α 2

A F

a a f

Problem 9.53 The coefficient of static friction between the right bar and the surface at A is µ s = 0.6. Neglect the weights of the bars. What is the largest angle α at which the truss will remain stationary without slipping?

a A

Solution: From the solution to Problem 9.52, f = A tan α. Since f = µ s A at impending slip, µ s = tan α, from which α = tan −1 (0.6) = 30.96 °

a A

668

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Problem 9.54 The bar BC is supported by a rough floor at C. If F = 2 kN and bar BC does not slip at C, what is the magnitude of the friction force exerted on the bar at C?

Solution:

ΣM A : B y (1 m) − F (0.5 m) = 0, ⇒ By =

1 F. 2

From the free-body diagram of bar BC,

From the free-body diagram of bar AB,

ΣFy : N − B y = 0, ΣM B : N (0.2 m) − f (0.6 m) = 0 600 mm C

500 mm

200 mm

BandF_6e_ISM_CH09.indd 669

We obtain 1 1 1 1 N = B y = F = (2000 N). 3 3 6 6 f = 333 N.

f =

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Problem 9.55 The bar BC is supported by a rough floor at C. If F = 2 kN, what is the minimum coefficient of static friction for which bar BC will not slip at C?

Solution:

See the solution to Problem 9.54. Equilibrium requires that f = 13 N independently of the value of F. If µ s is less than 1/3, the bar will slip for any value of F, and if µ s is equal to or greater than 1/3 the bar will not slip for any value of F. µ s = 1/3.

F B

600 mm C 500 mm

500 mm

200 mm

Problem 9.56 The weight of the box is 20 lb and the coefficient of static friction between the box and the floor is µ s = 0.65. Neglect the weights of the bars. What is the largest value of the force F that will not cause the box to slip? 4 in

4 in

8 in

Solution:

Note that BC is a two force member.

Member AB ΣFx : A x − FBC cos 45 ° = 0

(1)

ΣFy : A y − F + FBC sin 45 ° = 0

(2)

ΣM A :

(3)

−4 F + 8FBC sin 45 ° = 0

Unknowns A x , A y , F , FBC W = 20 lb

µ s = 0.65 f = µ s N for impending slip. 8 in

ΣFx :

−µ s N + FBC cos 45 ° = 0

(4)

ΣFy : N − W − FBC sin 45 ° = 0

(5)

Unknowns N , FBC , A x , A y , F 5 equations, 5 unknowns Solving, we get F = 74.3 lb. F AY

FBC

458

AX 499

499

y FBC 458

W W 5 20 lb ms 5 0.65 x

f 5 ms N N

670

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Problem 9.57 The bar BC is supported by a rough floor at C. If F = 4 kN and bar BC does not slip at C, what is the magnitude of the friction force exerted on the bar?

Solution:

We draw the free-body diagrams of the members, showing the normal and friction forces at C: By

By B Bx 0.4 m

Bx F

F 6 KN

6 kN 0.4 m A

C Ax 0.4 m

0.4 m

The equilibrium equations for member AB are ΣFx = A x + B x + 6 kN = 0, ΣFy = A y + B y = 0, ΣM point A = −(0.8 m) B x − (0.4 m)(6 kN) = 0, and the equilibrium equations for member BC are ΣFx = −B x − f = 0, ΣFy = −B y + N − F = 0, ΣM point C = (0.8 m) B x + (0.8 m) B y + (0.4 m) F = 0. Setting F = 4 kN and solving, we obtain A x = −3 kN, A y = −1 kN, B x = −3 kN, B y = 1 kN, N = 5 kN, f = 3 kN. 3 kN.

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Problem 9.58 The bar BC is supported by a rough floor at C. If F = 2 kN, what is the minimum value of the coefficient of static friction necessary for the system to remain in equilibrium?

Solution:

We draw the free-body diagrams of the members, showing the normal and friction forces at C: By

By Bx

Bx F

6 KN

0.4 m

F 6 kN

0.4 m A

The equilibrium equations for member AB are 0.4 m

0.4 m

ΣFx = A x + B x + 6 kN = 0, ΣFy = A y + B y = 0, ΣM point A = −(0.8 m) B x − (0.4 m)(6 kN) = 0, and the equilibrium equations for member BC are ΣFx = −B x − f = 0, ΣFy = −B y + N − F = 0, ΣM point C = (0.8 m) B x + (0.8 m) B y + (0.4 m) F = 0. Setting F = 2 kN and solving, we obtain A x = −3 kN, A y = −2 kN, B x = −3 kN, B y = 2 kN, N = 4 kN, f = 3 kN. The required friction coefficient is µ s = f /N = 0.75. µ s = 0.75.

672

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Problem 9.59 The frame is supported by the normal and friction forces exerted on the plates at A and G by the fixed surfaces. The coefficient of static friction at A is µ s = 0.6. Will the frame slip at A when it is subjected to the loads shown?

These twelve equations in ten unknowns can be solved by iteration or by back substitution. The results in detail: A = −24 kN, f A = 13 kN, C y = 18 kN, C x = 13 kN,

B = 36 kN, D x = 5 kN,

6 kN A

D y = 18 kN, E = 36 kN,

G = −18 kN, 1m

f G = 5 kN.

8 kN

Solution:

(1) ΣFx = − f A + f G + 8 = 0, (2) ΣFy = − A + G − 6 = 0.

(4) ΣFx = − f A + C x = 0,

CX CY

DX DY

8 kN DX

The assumed directions are shown in the Figure; a negative sign means that the result is opposite to the assumed direction. The magnitude of the coefficient of static friction for the reaction at A required to hold the frame in equilibrium is

The elements as free bodies: Element ABC: (See Figure) (3) ΣM B = +1 A + 2C y − 12 = 0.

B E

fG The strategy is to write the equilibrium equations and solve for the unknown. The complete structure as a free body: Denote the normal forces as A and G, and the friction forces as f A and f G . The sum of forces:

6 kN C X

µ SA =

fA = 0.5417. A

Since this is less than the known value, µ s = 0.6, the frame will not slip at A.

(5) ΣFy = − A − B + C y − 6 = 0. Element BE: (6) ΣFy = B − E = 0. Element CD: (7) ΣM C = +2 D x − D y + 8 = 0, (8) ΣFx = −C x + D x + 8 = 0 (9) ΣFy = D y − C y = 0. Element DEG: (10) ΣM D = −2G − E = 0 (11) ΣFx = f G − D x = 0 (12) ΣFy = G − D y + E = 0.

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Problem 9.60 The frame is supported by the normal and friction forces exerted on the plate at A by the wall. (a) What is the magnitude of the friction force exerted on the plate at A? (b) What is the minimum coefficient of static friction at A necessary for the structure to remain in equilibrium?

We have 9 equations in 9 unknowns. Solving, we get

Also N = 15 kN B x = −15 kN, B y = 16 kN C y = −8 kN (opposite the direction assumed) D x = 0, D y = 2 kN

E x = −15 kN, E y = 14 kN 6 kN

f = −8 kN (friction acts down)

(a)

(b) µ MIN =

f = 0.533 N y

B 2m

(1)

Σ Fy : f + B y + C y = 0

(2)

Σ M A : 2 B y + 4C y = 0

(3)

Σ Fx : E x + D x − B x = 0

(4)

(5)

Σ M E : 1D y + 1D x − 2 B y − 2 B x = 0

(6)

Σ Fx :

−D x = 0

(7)

Σ Fy :

−D y − C y − 6 = 0

(8)

ΣMD :

−3C y − 4(6) = 0

(9)

1m EX

E Σ Fy : E y + D y − B y = 0

Solution: Draw a free body diagram of each member and write the corresponding equilibrium equations Σ Fx : N + B x = 0

DY DX

1m B BX

Unknowns: N, f , Bx , By , C y , Dx , Dy , E x , E y .

6 kN 3m

1m x

DX DY

674

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Problem 9.61 The direction cosines of the crane’s cable are cos θ x = 0.588, cos θ y = 0.766, cos θ z = 0.260. The y -axis is vertical. The stationary caisson to which the cable is attached weighs 2000 lb and rests on horizontal ground. If the coefficient of static friction between the caisson and the ground is µ s = 0.4, what tension in the cable is necessary to cause the caisson to slip? y

Solution: Let T be the force exerted on the caisson by the cable. We can express it in terms of the direction cosines as T = T (0.588i + 0.766 j + 0.260 k). If N = Nj is the normal force exerted on the caisson by the ground, the sum of the vertical forces on the caisson is ΣFy = 0.766T + N − 2000 = 0. (1) The magnitude of the horizontal force exerted by the cable is (0.588T ) 2 + (0.260T ) 2 = 0.643T . From the free-body diagram of the caisson, viewed perpendicular to the vertical plane containing the cable, we see that 0.643T − f = 0. (2) slip impends when f = µ s N. (3) Solving Eq. (1) for N and solving Eq. (2) for f and substituting the results into Eq. (3), we obtain 0.643T = µ s (2000 − 0.766T ). The solution of this equation is T = 843 lb.

x z

Problem 9.62* The 10-lb metal disk A is at the center of the inclined surface. The tension in the string AB is 5 lb. What minimum coefficient of static friction between the disk and the surface is necessary to keep the disk from slipping?

T = (5 lb)

r AB = −3.08i + 3.08 j − 2.46k (lb). r AB

The sum of the forces exerted on the disk by the string and the weight of the disk is

y B

Solution: The coordinates of the disk A are (5, 1, 4) ft, so the position vector of pt B relative to A is r AB = (0 − 5) i + (6 − 1) j + (0 − 4)k ft. We can express the force exerted on the disk by the string as

T − 10 j = −3.08i − 6.92 j − 2.46k (lb).

(0, 6, 0) ft

2 ft A

8 ft

10 ft z

The components of this force normal and parallel to the surface are balanced by the normal force and friction force, respectively. To determine these components, we need a unit vector e perpendicular to the surface: The angle β = arctan(2/8) = 14.0 °, so e = cos β j + sin β k = 0.970 j + 0.243k. The component of T − 10 j normal to the surface is [(T − 10 j ) ⋅ e]e = −7.09 j − 1.77k (lb). The magnitude of the normal force equals the magnitude of this vector: N = −7.09 j − 1.77k = 7.31 lb. The component of T − 10 j parallel to the surface is (T − 10 j) − [(T − 10 j) ⋅ e]e = −3.08i + 0.17 j − 0.69k (lb). The magnitude of the friction force is f = −3.08i + 0.17 j − 0.69k = 3.16 lb. Slip impends when f = µ s N so the minimum friction coefficient is µs =

f 3.16 = = 0.432. N 7.31 y

2 ft z

b 8 ft

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Problem 9.63* The 5-kg box is at rest on the sloping surface. The y -axis points upward. The unit vector 0.557 i + 0.743 j + 0.371k is perpendicular to the sloping surface. What is the magnitude of the friction force exerted on the box by the surface?

Solution: N = N mag (0.577 i + 0.743 j + 0.371k) W = −(49.05 N) j ΣF : f + N + W = 0 ⇒ f = −W − N We also know that the friction force is parallel to the surface:

f ⋅ (0.557 i + 0.743 j + 0.371k) = 0 ⇒ N mag = 36.45 N Then the friction force is f = (−20.3i + 22.0 j − 13.52k) N ⇒ f = 32.83 N

Problem 9.64* In Problem 9.63, what is the minimum coefficient of static friction necessary for the box to remain at rest on the sloping surface?

Solution: µs =

See 9.63

f 32.83 N = = 0.901 N mag 36.45 N

Problem 9.65 In Practice Example 9.4, the coefficients of friction between the wedge and the log are µ s = 0.22 and µ k = 0.20. What is the largest value of the wedge angle α for which the wedge would remain in place in the log when the force F is removed?

Solution:

Assume that F = 0 and the wedge is on the verge of slipping out of the log. The sum of the forces in the vertical direction is ΣFy : 2 N sin

( α2 ) − 2µ N cos( α2 ) = 0. s

The wedge will not slip out if µ s ≥ tan

( α2 ) ⇒ α ≤ 2 tan (µ ). −1

If µ s = 0.22, the largest value of α for which the wedge will remain in place is α = 24.8 °.

676

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Problem 9.66 The wedge shown is being used to split the log. The wedge weighs 20 lb and the angle α equals 30°. The coefficient of kinetic friction between the faces of the wedge and the log is 0.28. If the normal force exerted by each face of the wedge must equal 150 lb to split the log, what vertical force F is necessary to drive the wedge into the log at a constant rate? (See Practice Example 9.4) F

Solution:

The free-body diagram is shown. Summing forces in the vertical direction gives ΣFy : 2 N sin

( α2 ) + 2µ N cos( α2 ) − W − F = 0 k

Thus

( )

α α  F = 2 N  sin + µ k cos −W  2 2  = 2(150 lb)[sin15 ° + (0.28) cos15 °] − 20 lb = 139 lb F = 139 lb. F

mkN

Problem 9.67 The coefficient of static friction between the faces of the wedge and the log in Problem 9.66 is 0.30. Will the wedge remain in place in the log when the vertical force F is removed? (See Practice Example 9.4)

m kN N

Solution:

Assume that F = 0 and the wedge is on the verge of slipping out of the log. The sum of the forces in the vertical direction is

ΣFy : 2 N sin

( α2 ) − 2µ N cos( α2 ) = 0. s

The wedge will not slip out if

µ s ≥ tan

( α2 ) = tan15°

0.3 > 0.268 ⇒ yes. yes.

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Problem 9.68 The weights of the blocks are W A = 100 lb and W B = 25 lb. Between all of the contacting surfaces, µ s = 0.32 and µ k = 0.30. What force F is necessary to move B to the left at a constant rate?

Solution:

We have 7 unknowns, 4 equilibrium equations and 3 fric-

tion equations ΣFAx : N 1 − f 2 cos10 ° − N 2 sin10° = 0 ΣFAy : − f1 + N 2 cos10 ° − f 2 sin10 ° − 100 lb = 0 ΣFBx : f 2 cos10 ° + N 2 sin10 ° + f 3 − F = 0 ΣFBy : f 2 sin10 ° − N 2 cos10 ° + N 3 − 25 lb = 0 f1 = 0.3 N 1 , f 2 = 0.3 N 2 , f 3 = 0.3 N 3 Solving we find F = 102 lb

108

F B

100 lb N2

f2 N2 f2

B 25 lb

f3 N3

Problem 9.69 The masses of the blocks are m A = 30 kg and m B = 70 kg. Between all of the contacting surfaces, µ s = 0.1. What is the largest force F that can be applied without causing the blocks to slip?

The equilibrium equations for block B are ΣFx : P sin 20 ° − µ s P cos 20 ° − µ s N = 0, ΣFy : −P cos 20 ° − µ s P sin 20 ° − m B g + N = 0.

308 F

Substituting the given values and solving yields R = 212 N, P = 456 N, N = 1130 N, and F = 197 N. F = 197 N.

208 B

Solution:

The equilibrium equations for block A are

ΣFx : R − F sin 30 ° + µ S P cos 20 ° − P sin 20 ° = 0, ΣFy : µ s R − F cos30 ° − m A g + P cos 20 ° + µ s P cos 20 ° = 0.

678

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Problem 9.70 Each block weighs 200 lb. Between all of the contacting surfaces, µ s = 0.1. What is the largest force F that can be applied without causing block B to slip upward?

Solution:

Let W = 200 lb. The equilibrium equations for block B are

ΣFx : R cos10 ° + µ s R sin10 ° − P cos10 ° − µ s P sin10 ° = 0, ΣFy : R sin10 ° − µ s R cos10 ° + P sin10 ° − µ s P cos10 ° − W = 0. The equilibrium equations for block C are ΣFx : P cos10 ° + µ s P sin10 ° + µ s N − F = 0, ΣFy : N + µ s P cos10 °

− P sin10 ° − W = 0. A

C 808

808

BandF_6e_ISM_CH09.indd 679

Substituting the given values and solving yields P = R = 1330 lb, N = 300 lb, and F = 1360 lb. F = 1360 lb.

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Problem 9.71 Small wedges called shims can be used to hold an object in place. The coefficient of kinetic friction between the contacting surfaces is 0.4. What force F is needed to push the shim downward until the horizontal force exerted on the object A is 200 N? F

Shims

Solution: fL = µk N L

(1)

fR = µk N R

(2)

N L = 200 N

(3)

ΣFx : N L − N R cos 5 ° + f R sin 5 ° = 0

(4)

ΣFy : −F + f L + f R cos 5 ° + N R sin 5° = 0

(5)

Unknowns: f L , N L , FR , N R , F (5 equations in 5 unknowns)

Solving, F = 181 N A

58 fR

Problem 9.72 The coefficient of static friction between the contacting surfaces is 0.44. If the shims are in place and exert a 200-N horizontal force on the object A, what upward force must be exerted on the left shim to loosen it? F

Shims 58

Solution: FL = µ s N L

(1)

FR = µ s N R

(2)

µ s = 0.44 N L = 200 N ΣFx : N L − N R cos 5 ° − f R sin 5 ° = 0

(3)

ΣFy : F − f L − f R cos 5 ° + N R sin 5 ° = 0

(4)

Unknowns F , f L , f R , N R Solving, F = 156 N 58 y

NR fL

fR x

680

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Problem 9.73 The crate A weighs 600 lb. Between all contacting surfaces, µ s = 0.32 and µ k = 0.30. Neglect the weights of the wedges. What force F is required to move A to the right at a constant rate?

Solution:

The active sliding contact surfaces are between the wall and the left wedge, between the wedges, between the floor and the bottom of the right wedge, and between the crate and the floor. Leftmost wedge: Denote the normal force exerted by the wall by Q, and the normal force between the wedges by N. The equilibrium conditions for the left wedge moving at a constant rate are: ΣFy = −F + µ k N cos α + N sin α + µ kQ = 0. ΣFx = Q − N cos α + µ k N sin α = 0.

For the right wedge: Denote the normal force exerted by the crate by A, and the normal force exerted by the floor by P.

ΣFy = −N sin α − µ k N cos α + P = 0. ΣFx = N cos α − µ k N sin α − µ k P − A = 0. A

For the crate: Denote the weight of the crate by W. Σ Fx = A − µ kW = 0. These five equations are solved for the five unknowns by iteration: Q = 204.4 lb, N = 210.7 lb P = 81.34 lb, A = 180 lb, and F = 142.66 lb F 58

A 58

F mkQ Q

mkN

BandF_6e_ISM_CH09.indd 681

A mkN P

N mkP

mkW

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Problem 9.74 Suppose that between all contacting surfaces, µ s = 0.32 and µ k = 0.30. Neglect the weights of the 5° wedges. If a force F = 800 N is required to move A to the right at a constant rate, what is the mass of A?

Solution:

The free body diagrams of the left wedge and the combined right wedge and crate are as shown. The equilibrium equations are Wedge: ΣFx = N − P cos 5 ° + 0.3P sin 5° = 0, ΣFy = 0.3 N + P sin 5 ° + 0.3P cos 5 ° − F = 0, Wedge and box: ΣFx = P cos 5 ° − 0.3P sin 5 ° − 0.3Q = 0,

ΣFy = Q − P sin 5 ° − 0.3P cos 5 ° − 9.81 m = 0. Solving them, we obtain

P = 1180 N, N = 1150 N, Q = 3820 N,

and m = 343 kg. 58 F 0.3 N N

0.3 P P

m (9.81)

0.3 P 58

Problem 9.75 The box A has a mass of 80 kg, and the wedge B has a mass of 40 kg. Between all contacting surfaces, µ s = 0.15 and µ k = 0.12. What force F is required to raise A at a constant rate?

Solution:

0.3 Q Q

From the free-body diagrams shown, the equilibrium equa-

tions are Box A: ΣFx = 0: Q − N sin10 ° − µ k N cos10 ° = 0, ΣFy = 0 N cos10 ° − µ k N sin10 ° − µ kQ − W = 0. Wedge B: ΣFx = 0 P sin10 ° + µ k P cos10 ° + N sin10 ° + µ k N cos10 ° − F = 0

ΣFy = 0: P cos10 ° − µ k P sin10 ° − N cos10 ° + µ k N sin10° − W w = 0. Solving with

108 B 108

W = (80)(9.81) N, W w = (40)(9.81) N, and µ k = 0.12, we obtain N = 845 N, Q = 247 N, P = 1252 N, and F = 612 N.

N W

mkQ

682

B WW P

mkN

mkN F

mk P

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Problem 9.76 Suppose that A weighs 800 lb and B weighs 400 lb. The coefficients of friction between all of the contacting surfaces are µ s = 0.15 and µ k = 0.12. Will B remain in place if the force F is removed?

Solution: The equilibrium conditions are: For the box A: Denote the normal force exerted by the wall by Q, and the normal force exerted by the wedge by N. The friction forces oppose motion. ΣFy = −W + N cos α + µ s N sin α + µ sQ = 0, ΣFx = +µ s N cos α − N sin α + Q = 0. For the wedge B. Denote the normal force on the lower surface by P. ΣFx = −µ s N cos α − µ s P cos α + P sin α + N sin α = 0. ΣFy = −N cos α + P cos α − µ s N sin α + µ s P sin α − W w = 0.

108 B

108

(A comparison with the equilibrium conditions for Problem 9.75 will show that the friction forces are reversed, since for slippage the box A will move downward, and the wedge B to the right.) The strategy is to solve these equations for the required µ s to keep the wedge B in place when F = 0. The solution Q = 0, N = 787.8 lb, P = 1181.8 lb and µ s = 0.1763. Since the value of µ s required to hold the wedge in place is greater than the value given, the wedge will slip out.

mS Q

mS P N

Problem 9.77 Between A and B, µ s = 0.20, and between B and C, µ s = 0.18. Between C and the wall, µ s = 0.30. The weights W B = 20 lb and WC = 80 lb. What force F is required to start C moving upward?

mS N

Solution:

The active contact surfaces are between the wall and C, between the wedge B and C, and between the wedge B and A. For the weight C: Denote the normal force exerted by the wall by Q, and the normal force between B and C by N. Denote the several coefficients of static friction by subscripts. The equilibrium conditions are: ΣFy = −WC + N − µ CW Q = 0, ΣFx = −Q + µ BC N = 0. For the wedge B: Denote the normal force between A and B by P. ΣFy = −N + P cos α − µ AB P sin α − W B = 0. ΣFx = F − µ BC N − µ AB P cos α − P sin α = 0.

These four equations in four unknowns are solved: Q = 15.2 lb, N = 84.6 lb, P = 114.4 lb,

and F = 66.9 lb 158 msQ

WC N

BandF_6e_ISM_CH09.indd 683

ms N F

WB msP

ms N

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Problem 9.78 The masses of A, B, and C are 8 kg, 12 kg, and 80 kg, respectively. Between all contacting surfaces, µ s = 0.4. What force F is required to start C moving upward?

Solution:

The active contact surfaces are between A and B, between A and the wall, between B and the floor, and between B and C. Assume that the roller supports between C and the wall exert no friction forces. For the wedge A: Denote the normal force exerted by the wall as Q and the normal force between A and B as N. The weight is W A = 8 g = 78.48 N. The equilibrium conditions: ΣFy = −F + µ sQ + µ s N cos α + N sin α − W A = 0 ΣFy = Q − N cos α + µ s N sin α = 0.

For wedge B: Denote the normal force exerted on B by the floor by P, and the normal exerted by the weight C as S. The weight of B is W B = 12 g = 117.72 N. The equilibrium conditions:

ΣFy = −N sin α − S cos β + P − µ s N cos α + µ s S sin β − W B = 0. A

ΣFx = N cos α − µ s N sin α − µ s P − µ s S cos β − S sin β = 0. B

108

128

For the weight C: The weight is WC = 80 g = 784.8 N. The equilibrium conditions: ΣFy = −WC + S cos β − µ s S sin β = 0. These five equations in five unknowns are solved: Q = 1157.6 N, N = 1293.5 N, S = 857.4 N, P = 1677.5, and F = 1160 N

F msQ W msN A

N m

Solution:

Problem 9.79 In Practice Example 9.5, suppose that the pitch of the thread is changed from p = 0.2 in to p = 0.24 in. What is the slope of the thread? What is the magnitude of the couple that must be applied to the collar C to cause it to turn at a constant rate and move the suspended object upward?

R S

msS

WB msP

b S

msS

The slope α is determined from the relation

p 0.24 in tan α = = = 0.0239 2πr 2π (1.6 in ) ⇒ α = 1.37 °. Using this value, the required couple is M = rF tan(θ k + α) = (1.6 in)(200 lb) tan (12.4 ° + 1.37 °) = 78.5 in M = 78.5 in-lb.

684

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Problem 9.80 The pitch of the threaded shaft is p = 2 mm and the mean radius of the thread is r = 20 mm. The coefficients of friction between the thread and the mating groove are µ s = 0.22 and µ k = 0.20. The weight W = 500 N. Neglect the weight of the threaded shaft. What couple must be applied to the threaded shaft to lower the weight at a constant rate?

Solution:

The angle of kinetic friction is

θ k = tan −1 (0.2) = 11.31 °. The angle of pitch is α = tan −1

2   = 0.9118 °. ( 2πpr ) = tan  2π(20) −1

The moment required to lower the weight at a constant rate is M = 0.02(500) tan (11.31 − 0.9118) = 1.835 N-m.

Problem 9.81 The position of the horizontal beam can be adjusted by turning the machine screw A. Neglect the weight of the beam. The pitch of the screw is p = 1 mm, and the mean radius of the thread is r = 4 mm. The coefficients of friction between the thread and the mating groove are µ s = 0.20 and µ k = 0.18. If the system is initially stationary, determine the couple that must be applied to the screw to cause the beam to start moving (a) upward; (b) downward.

Solution:

The sum of the moments about the pin support is

ΣM = −0.4 F + (0.3)400 = 0, from which the force exerted by the screw is F = 300 N. The pitch angle is  1  α = tan −1  = 2.28 °.  2π (4)  The static friction angle is θ s = tan −1 (0.2) = 11.31 °. (a) The moment required to start motion upward is M = 0.004(300) tan (11.31 ° + 2.28 °) = 0.29 N-m (b) The moment required to start motion downward is

400 N

M = 0.004(300) tan (11.31 ° − 2.28 °) = 0.19 N-m A

400 N

100 mm

300 mm

BandF_6e_ISM_CH09.indd 685

100 mm

300 mm

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Problem 9.82 The pitch of the threaded shaft of the C clamp is p = 0.05 in, and the mean radius of the thread is r = 0.15 in. The coefficients of friction between the threaded shaft and the mating collar are µ s = 0.18 and µ k = 0.16. (a) What maximum couple must be applied to the shaft to exert a 30-lb force on the clamped object? (b) If a 30-lb force is exerted on the clamped object, what couple must be applied to the shaft to begin loosening the clamp?

Solution: F = 30 lb, r = 0.15 in, p = 0.05 in, µ s = 0.18, µ k = 0.16 θ s = tan −1 (µ s ) = 10.2 °, θ k = tan −1 (0.16) = 9.09 ° α = tan −1

( 2πpr ) = 3.04°

(a)

M = rF tan (θ k + α) = 0.967 in − lb

(b)

M = rF tan (θ s − α) = 0.566 in − lb

20 lb W F 5W 120 lb

Problem 9.83 The mass of block A is 60 kg. Neglect the weight of the 5° wedge. The coefficient of kinetic friction between the contacting surfaces of the block A, the wedge, the table, and the wall is µ k = 0.4. The pitch of the threaded shaft is 5 mm, the mean radius of the thread is 15 mm, and the coefficient of kinetic friction between the thread and the mating groove is 0.2. What couple must be exerted on the threaded shaft to raise the block A at a constant rate?

Solution:

Denote the wedge angle by β = 5 ° and the normal force on the top by N and on the lower surface by P. The free body diagrams of the wedge and block are as shown. The equilibrium equations for wedge: ΣFx = F − µ k P − N sin 5 ° − µ k N cos 5 ° = 0, ΣFy = P − N cos 5 ° + µ k N sin 5 ° = 0. For the Block: ΣFx = N sin 5 ° + µ k N cos 5 ° − Q = 0, ΣFy = N cos 5 ° − µ k N sin 5 ° − µ kQ − W = 0. Solving them, we obtain F = 668 N. From Eq. (9.9), the couple necessary to rotate the threaded shaft when it is subjected to the axial force F is M = rF tan (θ k + α) r is the radius 15 mm = 0.015 m. θ k is the angle of kinetic friction θ k = arctan(0.2) = 11.31 °.

From Eq. (9.7), the slope is given in terms of the pitch by α = arctan

5   = 3.04 °. ( 2Pπr ) = arctan  2π(15) 

The couple is M = (0.015 m)(668 N) tan (11.31 ° + 3.04°) = 2.56 N-m.

mkQ W F

686

mkN

mkP

mkN

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Problem 9.84 The vise exerts 80-lb forces on A. The threaded shafts are subjected only to axial loads by the jaws of the vise. The pitch of their threads is p = 1/8 in, the mean radius of the threads is r = 1 in, and the coefficient of static friction between the threads and the mating grooves is 0.2. Suppose that you want to loosen the vise by turning one of the shafts. Determine the couple you must apply (a) to shaft B; (b) to shaft C.

Solution:

Isolate the left jaw. The sum of the moments about C:

ΣM C = −4 B + 8(80) = 0, from which B = 160 lb (T ). The sum of the forces: ΣFx = −80 + B − C = 0, from which C = 80 lb (C ). The pitch angle is α = tan −1

( 161π ) = 1.14°.

The static friction angle is θ s = tan −1 (0.2) = 11.31 °. The moments required to loosen the vise are

4 in

MB =

( 121 )(160) tan (11.31° − 1.14°) = 2.39 ft lb,

and M C = rC tan (θ s − α) = 1.2 ft-lb. B 80 lb

4 in

Problem 9.85 Suppose that you want to tighten the vise in Problem 9.84 by turning one of the shafts. Determine the couple you must apply (a) to shaft B; (b) to shaft C. A

4 in

Solution:

Use the solution to Problem 9.84. (a) The moment on shaft B required to tighten the vise is M B = rB tan (θ s + α). Note that 1 r = , B = 160 lb, 12 α = tan −1

4 in

( 161π ) = 1.14°

and θ s = tan −1 (0.2) = 11.31 °, then M B = 2.94 ft lb

B 4 in

(b) For shaft C , M C = rC tan (θ s + α), M C = 1.47 ft-lb.

where

C = 80 lb,

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Problem 9.86 The threaded shaft has a ball-and-socket support at B. The 400-lb load A can be raised or lowered by rotating the threaded shaft, causing the threaded collar at C to move relative to the shaft. Neglect the weights of the members. The pitch of the shaft is p = 14 in, the mean radius of the thread is r = 1 in, and the coefficient of static friction between the thread and the mating groove is 0.24. If the system is stationary in the position shown, what couple is necessary to start the shaft rotating to raise the load?

Solution: Denote the lower right pin support by D. The length of the connecting member CD is L CD = 9 2 + 12 2 = 15 in. The angle between the threaded shaft and member CD is β = 2 tan −1

( 129 ) = 73.74°.

The sum of the moments about D is ΣM D = L CD F cos(90 − β ) − 18W = 0, from which F = 500 lb. The pitch angle is α = tan −1

( 2πpr ) = 2.28°.

The angle of static friction is θ s = tan −1 (0.24) = 13.5 °. The moment needed to start the threaded collar in motion is

9 in

M = rF tan (θ s + α) =

( 121 )(500) tan (13.5° + 2.28°)

= 11.77 ft-lb A

12 in

B F

b LCD 9 in

9 in

Dx 18 in

18 in

Problem 9.87 In Problem 9.86, if the system is stationary in the position shown, what couple is necessary to start the shaft rotating to lower the load?

Solution:

Use the results of the solution to Problem 9.86. The moment is M = rF tan (θ s − α), where r =

( 121 ) ft,

F = 500 lb, θ s = 13.5 °,

9 in

and α = 2.28 °,

from which M = 8.26 ft lb A

12 in B

9 in

688

9 in

18 in

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Problem 9.88 The car jack is operated by turning the horizontal threaded shaft at A. The threaded shaft fits into a mating threading collar at B. As the shaft turns, points A and B move closer together or farther apart, thereby raising or lowering the jack. The pitch of the threaded shaft is p = 0.1 in, the mean radius of the thread is r = 0.2 in, and the coefficient of kinetic friction between the threaded shaft and the mating collar at B is 0.15. What couple must be applied at A to rotate the shaft at a constant rate and raise the jack when it is in the position shown if the load L = 1400 lb?

Solution:

Note that β = tan −1 (3/5) = 31.0 °.

Because of symmetry we know that FBC = FAC = FBD . The equlibrium equations are ΣFy : −L − 2 FBC sin β = 0 ΣFx : 2 FBC cos β + F = 0 Solving yields F = L cot β = (1400 lb) cos(31.0 °) = 2330 lb. The angle of kinetic friction is θ k = tan −1 (µ k ) = tan −1 (0.15) = 8.53 °. The slope of the thread is α = tan −1

in   = 4.55 °. ( 2πpr ) = tan  2π0.1 (0.2 in)  −1

Therefore, the required couple is M = rF tan (θ k + α) = (0.2 in)(2330 lb) tan (8.53 ° + 4.55 °) = 108 in-lb. M = 108 in-lb.

3 in

5 in

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Problem 9.89 The car jack is operated by turning the horizontal threaded shaft at A. The threaded shaft fits into a mating threading collar at B. As the shaft turns, points A and B move closer together or farther apart, thereby raising or lowering the jack. The pitch of the threaded shaft is p = 0.1 in, the mean radius of the thread is r = 0.2 in, and the coefficient of kinetic friction between the threaded shaft and the mating collar at B is 0.15. What couple must be applied at A to rotate the shaft at a constant rate and lower the jack when it is in the position shown if the load L = 1400 lb?

Solution:

Note that β = tan −1 (3/5) = 31.0 °. Because of symmetry we know that FBC = FAC = FBD . The equlibrium equations are ΣFy : −L − 2 FBC sin β = 0 ΣFx : 2 FBC cos β + F = 0 Solving yields F = L cot β = (1400 lb) cot (31.0 °) = 2330 lb. The angle of kinetic friction is θ k = tan −1 (µ k ) = tan −1 (0.15) = 8.53 °. The slope of the threads is α = tan −1

in   = 4.55 °. ( 2πpr ) = tan  2π0.1 (0.2 in)  −1

Therefore, the required couple is M = rF tan (θ k − α) = (0.2 in)(2330 lb) tan (8.53 ° + 4.55 °) = 108 in-lb. M = 32.5 in-lb.

3 in

5 in

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Solution:

Problem 9.90 A turnbuckle, used to adjust the length or tension of a bar or cable, is threaded at both ends. Rotating it draws threaded ends of the bar or cable together or moves them apart. Suppose that the pitch of the threads is p = 0.05 in, their mean radius is r = 0.25 in, and the coefficient of static friction between the threads and the mating grooves is 0.24. If T = 200 lb, what couple must be exerted on the turnbuckle to start tightening it?

α = tan −1

The slope of the threads is

in   = 1.82 °. ( 2πpr ) = tan  2π0.05 (0.25 in)  −1

The angle of static friction is θ k = tan −1 (µ s ) = tan −1 (0.24) = 13.5 °. Using these values, one half of the required couple is M = rF tan (θ k + α) = (0.25 in )(200 lb) tan (13.5 ° + 1.82 °) = 13.7 in-lb. The required couple is 2 M = 2(13.7 in-lb) = 27.4 in-lb.

27.4 in-lb. T

Problem 9.91 Suppose that the pitch of the threads of the turnbuckle is p = 0.05 in, their mean radius is r = 0.25 in, and the coefficient of static friction between the threads and the mating grooves is 0.24. If T = 200 lb, what couple must be exerted on the turnbuckle to start loosening it?

Solution:

The slope of the threads is

in   = 1.82 °. ( 2πpr ) = tan  2π0.05 (0.25 in) 

α = tan −1

−1

The angle of static friction is θ k = tan −1 (µ s ) = tan −1 (0.24) = 13.5 °. Using these values, one half of the required couple is M = rF tan (θ k − α) = (0.25 in)(200 lb) tan (13.5 ° − 1.82 °) = 10.3 in-lb.

T T

BandF_6e_ISM_CH09.indd 691

The required couple is 2 M = 2(10.3 in-lb) = 20.7 in-lb. 20.7 in-lb.

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Problem 9.92 Member BE of the frame has a turnbuckle. (See Problem 9.90.) The threads have pitch p = 1 mm, their mean radius is r = 6 mm, and the coefficient of static friction between the threads and the mating grooves is 0.2. What couple must be exerted on the turnbuckle to start loosening it? 0.4 m A

Unknowns: A x , A y , D x , D y , TBE , FCF (6 equations, 6 unknowns) Solving, TBE = 2017 N Now to analyze the turnbuckle tan θ s = µ s = 0.2 θ s = 11.31 °

600 N

1.0 m B

tan α =

p 1 = 2πr 2π (6)

α = 1.52 ° For one screw, to loosen

0.5 m D

M = rTBE tan (θ s − α)

M = 2.09 N

F 0.8 m

For two screws (turnbuckle)

0.4 m

M TOTAL = 4.18 N-m

Solution:

This problem has two parts. First, we find the tension in the two force member BE. Then we analyze the turnbuckle. tan θ =

0.5 0.4

0.4 m

600

θ = 51.3 ° tan φ =

1.4 m

0.5 0.2

TBE

0.5

FCF 0.5

φ = 68.2 °

0.4

ΣFx : A x + TBE cos θ + FCF cos φ = 0

(1)

ΣFy : A y − TBE sin θ + FCF sin φ − 600 = 0

(2)

Σ M A : −0.4TBE sin θ + 1.4 FCF sin φ − 1.4(600) = 0

(3)

ΣFx : D x − TBE cos θ − FCF cos φ = 0

(4)

Σ Fy : D y + TBE sin θ − FCF sin φ = 0

(5)

ΣM D : 0.8TBE sin θ − 1.2 FCF sin φ = 0

(6)

Problem 9.93 In Problem 9.92, what couple must be exerted on the turnbuckle to start tightening it? 0.4 m A

0.8 m

TBE

FCF

f 0.4 m

Solution:

In Problem 9.92, the tension in the turnbuckle was

TBE = 2017 N

600 N

1.0 m

0.2

r = 0.006 m p = 0.001 mm tan θ s = µ s = 0.2 θ s = 11.31 °

0.5 m D

tan α =

E F 0.8 m

0.4 m

p 2πr

α = 1.52 ° For one screw, to tighten, M = rTBE tan (θ s + α) M = 2.756 N-m For two screws (turnbuckle) M = 5.51 N-m

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Problem 9.94 Members CD and DG of the truss have turnbuckles. (See Problem 9.90.) The pitch of the threads is p = 4 mm, their mean radius is r = 10 mm, and the coefficient of static friction between the threads and the mating grooves is 0.18. What couple must be exerted on the turnbuckle of member CD to start loosening it? C

These equations are solved: The results in detail: GH = −3.54 kN, FH = 2.5 kN, DF = 2.5 kN, GF = 0, EG = −5 kN, DG = 3.54 kN, DE = 0,

CD = 2.12 kN, 2m H

AB = 3.5 kN. 4 kN

Solution:

The complete structure as a free body: The equilibrium

conditions: ΣM A = −2(2) − 4(4) + 8 H = 0, from which H =

BC = 2 kN, BD = 3.5 kN,

2 kN

AC = −4.95 kN,

20 = 2.5 kN. 8

ΣFy = A y + H − 2 − 4 = 0,

The members of interest here are CD = 2.12 kN and DG = 3.54 kN. The pitch angle is   4 α = tan −1  = 3.64 °.  (2π )(10)  The static friction angle is θ s = tan −1 (0.18) = 10.20 °. The moment required to loosen the turnbuckle is M = 2(0.01)(2.12) tan (10.2 ° − 3.64 °) = 0.00488 kN-m m = 4.88 N-m

from which AY = 3.5 kN. Σ Fx = A x = 0. The method of joints: The interior angles GHF , DFG, BDC , and BAC are each β = 45 °. Joint H : (1) 0 = −GH cos β − FH ,

AX AY

2 kN

4 kN

(2) 0 = GH sin β + H . Joint F: (3) 0 = FH − DF , GH

(4) GF = 0, Joint G: (5) 0 = GH cos β − EG − DG cos β ,

b FH

(6) 0 = −GH sin β − DG sin β . Joint E: (7) 0 = EG − CE, (8) 0 = DE, Joint D: (9) 0 = DG cos β − CD cos β + DF − DB,

JointH DE CD

DG DF 4 kN JointD

JointF

CE GF

JointG

BC CD

JointC

EG b

BD 2 kN

JointB

EG DE

JointE AC b AB AX AY JointA

(10) 0 = DG sin β + CD sin β + DE − 4. Joint C: (11) 0 = CD cos β − AC cos β + CE , (12) 0 = −CD sin β − AC sin β − BC . Joint B: (13) 0 = BD − AB, (14) 0 = BC − 2. Joint A: (15) 0 = AB + A x + AC cos β , (16) 0 = A y + AC sin β .

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Problem 9.95 In Problem 9.94, what couple must be exerted on the turnbuckle of member DG to start loosening it? C

Solution:

Use the results of the solution of Problem 9.94. The moment required to loosen the turnbuckle is M = 2rT tan (θ s − α), where r = 0.01 m, T = DG = 3.54 kN, θ s = 10.2 °, and α = 3.64 ° M = 2(0.01)(3.54) tan (10.2 ° − 3.64 °) = 0.00813 kN-m m = 8.13 N-m

2m H

2 kN 4 kN

Problem 9.96* The load W = 800 N can be raised or lowered by rotating the threaded shaft. The distances are b = 75 mm and h = 200 mm. The pinned bars are each 300 mm in length. The pitch of the threaded shaft is p = 5 mm, the mean radius of the thread is r = 15 mm, and the coefficient of kinetic friction between the thread and the mating groove is 0.2. When the system is in the position shown, what couple must be exerted to turn the threaded shaft at a constant rate, raising the load?

Isolate DH : (6) ΣFy = −D y − Fy + H y = 0, (7) ΣFx = −D x − Fx + H x = 0, (8) ΣM H =

( h4 ) F + ( L2 ) F + LD + ( h2 ) D = 0 x

Isolate AE : (9) ΣFy = − A + C y + H y = 0, (10) ΣFx = C x + E x = 0,

( h4 )(C ) − ( L2 )C + LA = 0.

(11) ΣM E = −

Isolate EG : (12) ΣFy = −E y + Fy + G y = 0,

(13) ΣFx = −E x + Fx + G x = 0,

( )

h L h (14) ΣM G = − Fx + F − LE y + E = 0. 4 2 y 2 x These 14 equations in 14 unknowns are to be solved to determine the reaction G x , which is the force that the threaded shaft must overcome to raise the load at a constant rate. An analytic solution is obtained as follows:

B B

D A

Solution:

The vertical distances HE , BE, AD, DG are 100 mm. The included angle ABC is β = sin −1

E E F

50 mm ( 150 ) = 19.47°. mm

The distance L = 300 cos β mm. Isolate the members and write the equilibrium equations, beginning at the top: Isolate the frame AB which supports the load W. The equilibrium conditions are (1) ΣM B = − AL + Wb = 0, (2) ΣFy = A − W + B y = 0, ΣFx = B x = 0 Isolate BD : (3) ΣFy = −B y − C y + D y = 0, (4) ΣFx = −C x + D x = 0, (5) ΣM D =

694

( h4 )C − ( L2 )C − LB = 0 x

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Problem 9.96* (Continued) From (1) A =

Check: This value of G y is expected from the overall equilibrium conditions. check. From (13)

Wb . L

Gx =

(

)

b . L

From (2) B = W 1 − From (8) and (11) C y = W Cx =

( 2Lb − 1),

2WL . h

4WL . h

Substitute: Gx =

1200W cos β N . h

Note that

(

)

b . L

From (9) E y = W 1 −

cos β =

1−

2 From (10) E x = − WL. h

from which

2 From (4) D x = WL h

G x = 2W

From (3) D y =

600 2 − h 2 . h

The pitch angle is

Wb . L

(

)

2b , From (8) and (14) Fy = W 1 − L 6WL . Fx = h From (12) G y =

h ( 600 ),

 5  α = tan −1  = 3.037 °.  2π (15)  The angle of kinetic friction is θ k = tan −1 (0.2) = 11.31 °. The moment required to raise the load at a constant rate:

Wb L

M = rG x tan (θ k + α) = 0.003836G x = 17.36 N m.

Problem 9.97 In Practice Example 9.6, suppose that the placement of the winch at A is changed so that the angle between the rope from A to P and the horizontal increases from 45 ° to 60 °. If the suspended load weighs 1500 lb, what tension must the winch exert on the rope to raise the load at a constant rate?

Solution:

The vector sum of the forces exerted on the pulley by the

rope is F =

(W + T sin 60 °) 2 + (T cos60 °) 2 .

The clockwise couple exerted on the pulley by the rope is M = (6 in)(T − W ). The angle of kinetic friction is θ k = tan −1 (µ k ) = tan −1 (0.2) = 11.3 °. Applying Eq. (9.12), M = rF sin θ k (0.6 in)(T − W ) = (0.5 in) (W + T sin 60 °) 2 + (T cos60 °) 2 sin11.3 °. Setting W = 1500 lb and solving yields T = 1550 lb.

Problem 9.98 The radius of the pulley is 4 in. The pulley is rigidly attached to the horizontal shaft, which is supported by two journal bearings. The radius of the shaft is 1 in, and the combined weight of the pulley and shaft is 20 lb. The coefficients of friction between the shaft and the bearings are µ s = 0.30 and µ k = 0.28. Determine the largest weight W that can be suspended as shown without causing the stationary shaft to slip in the bearings.

Solution: θ s = tan −1 (0.3) = 16.70 °, d = r sin θ s = 0.287 in ΣM P : (W + 20 lb)(4 in − d ) − (20 lb)(4 in) = 0 Solving we find W = 1.548 lb

20 lb W

W F 5W 120 lb

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Problem 9.99 In Problem 9.98, suppose that the weight W = 4 lb. What couple would have to be applied to the horizontal shaft to raise the weight at a constant rate?

Solution: θ k = tan −1 (0.28) = 15.64 °, d = r sin θ k = 0.270 in, W = 4 lb ΣM P : −(20 lb)(4 in) − M + (24 lb)(4 in + d ) = 0 Solving we find M = 22.5 in lb M

W P

20 lb W F 5W 120 lb

Problem 9.100 The pulley is mounted on a horizontal shaft supported by journal bearings. The coefficient of kinetic friction between the shaft and the bearings is µ k = 0.3. The radius of the shaft is 20 mm, and the radius of the pulley is 150 mm. The mass m = 10 kg. Neglect the masses of the pulley and shaft. What force T must be applied to the cable to move the mass upward at a constant rate?

Solution:

The angle of kinetic friction is θ k = tan −1 (µ k ) = 16.7 °. The moment required to turn the shaft is M = (mg + T )r sin θ k . The applied moment is M = (T − mg) R where R is the radius of the pulley. Equating and reducing:  1 + r sin θ  k    = (98.1) 1.0383 = 105.92 N R T = mg   1 − r sin θ  0.9617  k   R

(

)

m T

Problem 9.101 In Problem 9.100, what force T must be applied to the cable to lower the mass at a constant rate?

Solution: Form the solution to Problem 9.100, θ k = tan −1 (µ k ) = 16.7 °, and M = (mg + T )r sin θ k . The applied moment is M = (mg − T ) R. Substitute and reduce:  1 − r sin θ  k    = (98.1) 0.9617 = 90.86 N R T = mg   1 + r sin θ  1.0383  k   R

(

)

m T

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Problem 9.102 The pulley of 8-in radius is mounted on a shaft of 1-in radius. The shaft is supported by two journal bearings. The coefficient of static friction between the bearings and the shaft is µ s = 0.15. Neglect the weights of the pulley and shaft. The 50-lb block A rests on the floor. If sand is slowly added to the bucket B, what do the bucket and sand weigh when the shaft slips in the bearings?

Solution:

(See Problem 9.100). The angle of static friction is θ s = tan −1 (µ s ) = 8.53 °. The moment required to start rotation for both bearings is M = r ( B + W )sin θ s . The applied moment is M = ( B − W ) R, where R is the radius of the pulley. Substitute and reduce:  1 + r sin θ  s    = (50) 1.0185 = 51.9 lb R B = W   1 − r sin θ  0.9815  s   R

(

8 in

)

1 in

B A B

Problem 9.103 The pulley of 50-mm radius is mounted on a shaft of 10-mm radius. The shaft is supported by two journal bearings. The mass of the block A is 8 kg. Neglect the weights of the pulley and shaft. If a force T = 84 N is necessary to raise block A at a constant rate, what is the coefficient of kinetic friction between the shaft and the bearings? 50 mm

208 10 mm

Solution: F =

(W + T sin α) 2 + (T cos α) 2 ,

where α = 20 °. F =

107.2 2 + 78.9 2 = 133.13 N.

The moment required to raise the mass at constant rate for both bearings is M = rF sin θ k = 1.33sin θ k . The applied moment is M = (T − W ) R = 0.276 N m. Substitute and reduce: sin θ k =

The weight is W = mg = 78.5 N. The force on the pul-

ley is

(T − W ) R 0.276 = = 0.2073, 1.33 rF

from which θ k = 11.96 ° and µ k = tan (11.96 °) = 0.2119

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Problem 9.104 The mass of the suspended object is 4 kg. The pulley has a 100-mm radius and is rigidly attached to a horizontal shaft supported by journal bearings. The radius of the horizontal shaft is 10 mm and the coefficient of kinetic friction between the shaft and the bearings is 0.26. What tension must the person exert on the rope to raise the load at a constant rate? 258

To Find F, we must find the forces acting on the shaft. ΣFx : O x − T cos 25 ° = 0

(1)

ΣFy : O y − T sin 25 ° − mg = 0

(2)

F =

Ox 2 + Oy 2

(3)

ΣM o : RT − Rmg − M s = 0

(4)

M s = rF sin θ k

(5)

Unknowns: O x , O y , T , M s , F Solving, we get

100 mm

T = 40.9 N Also, F = 67.6 N, M s = 0.170 N-m O x = 37.1 N, O y = 56.5 N

Solution: R = 0.1 m

R 5 0.1 m

µ k = 0.26 258

Shaft radius 0.01 m µ k (shaft ) = 0.26

tan θ k = µ k

mK 5 0.26

Shaft radius 0.01 m mK (shaft) 5 0.26

MS OX OY

θ k = 14.57 ° M s = rF sin θ k

m = 4 kg

Problem 9.105 In Problem 9.104, what tension must the person exert to lower the load at a constant rate? 258

Solution:

This problem is very much like Problem 9.104—only the direction of M s is changed. The analysis is the same except Eq. (4), which becomes RT − Rmg + M s = 0

(4)

We again have 5 equations in 5 unknowns. Solving, 100 mm

T = 37.6 N Also F = 64.8 N, M s = 0.163 N-m O x = 34.1 N, O y = 55.1 N

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Problem 9.106 The radius of the pulley is 200 mm, and it is mounted on a shaft of 20-mm radius. The coefficient of static friction between the pulley and shaft is µ s = 0.18. If FA = 200 N, what is the largest force FB that can be applied without causing the pulley to turn? Neglect the weight of the pulley. y

Solution:

The magnitude of the force on the shaft supporting the

pulley is F =

( FA + FB cos 40 °) 2 + ( FB sin 40 °) 2 .

(1)

The couple exerted on the pulley by the rope is M = (0.2 m)( FB − FA ). (2) From Eq. (9.12), the largest couple which will not cause the shaft to slip is M = rF sin θ s , where r = 0.02 m and θ s = arctan(0.18) = 10.2 °. Substituting Eqs. (1) and (2) into Eq. (3), we obtain (0.2)( FB − 200)

= (0.02)sin (10.2 °) (200 + FB cos 40 °) 2 + ( FB sin 40°) 2 .

408

Solving this equation, we obtain FB = 206.8 N.

FA x

Problem 9.107 The masses of the boxes are m A = 15 kg and m B = 60 kg. The coefficient of static friction between boxes A and B and between box B and the inclined surface is 0.12. The pulley has a radius of 60 mm and is mounted on a shaft of 10-mm radius. The coefficient of static friction between the pulley and shaft is 0.16. What is the largest force F for which the boxes will not slip?

Solution:

Like 9.20, but different tensions. We have 7 unknowns, 4 equilibrium equations, 2 friction equations and one pulley equation. ΣFA : T1 − F − (147.15 N)sin 20 ° + f 2 = 0

ΣFA : N 2 − (147.15 N) cos 20 ° = 0 ΣFB : T2 − (588.6 N)sin 20 ° − f1 − f 2 = 0 ΣFB : N 1 − N 2 − (588.6 N) cos 20 ° = 0 f1 = 0.12 N 1 , f 2 = 0.12 N 2 For the pulley θ s = tan −1 (0.16), d = 10 mm sin θ s

Thus 60 mm (T1 − T2 ) = (T1 + T2 )d ⇒ F = 283 N B 147.15 N 208

A f2

F N2

588.6 N f1 208 N1

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Problem 9.108 The two pulleys have a radius of 4 in and are mounted on shafts of 1-in radius supported by journal bearings. Neglect the weights of the pulleys and shafts. The tension in the spring is 40 lb. The coefficient of kinetic friction between the shafts and the bearings is µ k = 0.3. What couple M is required to turn the left pulley at a constant rate?

Solution:

The angle of kinetic friction is θ k = tan −1 (0.3) = 16.7 °. The load on the bearings is F = 40 lb. The moment required to turn both pulleys at constant rate is M = 2rF sin θ k . This is equal to the applied moment, M applied = 2rF sin θ k = (2)

( 121 )40 sin (16.7°) = 1.92 ft lb

4 in

Problem 9.109 The weights of the boxes are W A = 65 lb and W B = 130 lb. The coefficient of static friction between boxes A and B and between box B and the floor is 0.12. The pulley has a radius of 4 in and is mounted on a shaft of 0.8-in radius. The coefficient of static friction between the pulley and shaft is 0.16. What is the largest force F for which the boxes will not slip?

Solution:

Like 9.22 with different tensions

We have 7 unknowns, 4 equilibrium equations, 2 friction equations and one pulley equation. ΣFAx : −F + T1 cos 20 ° + f 2 = 0 ΣFAy : N 2 − T1 sin 20 ° − 65 lb = 0 ΣFBx : T2 − f1 − f 2 = 0 ΣFBy : N 1 − N 2 − 130 lb = 0

208

f1 = 0.12 N 1 , f 2 = 0.12 N 2 For the pulley we have θ s = tan −1 (0.16), d = 0.8 in sin θ s

The total force on the pulley is Fpulley =

(T1 cos30 ° + T2 ) 2 + (T1 sin 30 °) 2

Thus M = Fpulley d = (T1 − T2 )4 in ⇒ F = 43.4 lb 65 lb

208

A T1 N2

T2 130 lb B f1 N1

700

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Problem 9.110 The coefficient of kinetic friction between the 100-kg box and the inclined surface is 0.35. Each pulley has a radius of 100 mm and is mounted on a shaft of 5-mm radius supported by journal bearings. The coefficient of kinetic friction between the shafts and the journal bearings is 0.18. Determine the tension T necessary to pull the box up the surface at a constant rate.

Solution:

Working through the pulleys

θ s = tan −1 (0.18), d = 5 mm sin θ s (T − T2 )100 mm = d (T + T2 ) (T2 − T1 )100 mm = d (T2 + T1 ) T3 = T2 + T3 Now do equilibrium and friction on the box ΣF : T1 + T − (981 N)sin60 ° − f = 0

ΣF : N − (981 N)cos60 ° = 0 f = 0.35 N Solving we find T = 346 N T4

608 T T2 T1

T3 981 N

N f

Problem 9.111 In Practice Example 9.7, suppose that the diameters D o = 3 12 in and D i = 1 12 in and the angle α = 72 °. What couple is required to turn the shaft at a constant rate?

Solution:

The radii ro = 1.75 in and ri = 0.75 in.

The required couple is given by Eq. (9.13): M = =

2µ k F  ro 3 − ri 3    3cos α  ro 2 − ri 2  2(0.18)(200 lb)  (1.75 in) 3 − (0.75 in) 3    = 153 in-lb.  (1.75 in) 2 − (0.75 in) 2  3cos 72 °

M = 153 in-lb.

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Problem 9.112 The circular flat-ended shaft is pressed into the thrust bearing by an axial load of 600 lb. The weight of the shaft is negligible. The coefficients of friction between the end of the shaft and the bearing are µ s = 0.20 and µ k = 0.15. What is the largest couple M that can be applied to the stationary shaft without causing it to rotate in the bearing?

Solution: by µ s . M =

The required couple is given by Eq. (9.14) with µ k replaced

2 2 µ Fr = (0.2)(600 lb)(2 in) = 160 in-lb. 3 s 3

M = 160 in-lb.

600 lb 2 in

Problem 9.113 The circular flat-ended shaft is pressed into the thrust bearing by an axial load of 600 lb. The weight of the shaft is negligible. The coefficients of friction between the end of the shaft and the bearing are µ s = 0.20 and µ k = 0.15. What couple M is required to rotate the shaft at a constant rate?

Solution: M =

The required couple is given by Eq. (9.14).

2 2 µ Fr = (0.15)(600 lb)(2 in) = 120 in-lb. 3 k 3

M = 120 in-lb.

600 lb 2 in

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Problem 9.114 The disk D is rigidly attached to the vertical shaft. The shaft has flat ends supported by thrust bearings. The disk and the shaft together have a mass of 220 kg and the diameter of the shaft is 50 mm. The vertical force exerted on the end of the shaft by the upper thrust bearing is 440 N. The coefficient of kinetic friction between the ends of the shaft and the bearings is 0.25. What couple M is required to rotate the shaft at a constant rate?

Solution:

There are two thrust bearings, one at the top and one at

the bottom FU = 440 N m = 220 kg ΣFy : FL − FU − mg = 0 FL = 2598.2 N. The couple necessary to turn D at a constant rate is the sum of the couples for the two bearings.

M D

MU =

2 µ F r 3 k U

ML =

2 µ F r 3 k L

r = 0.025 m D

µ k = 0.25 Solving, M U = 1.833 N-m M L = 10.826 M TOTAL = 12.7 N-m

D mg

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Problem 9.115 Suppose that the ends of the shaft in Problem 9.114 are supported by thrust bearings of the type shown in Fig. 9.14, where ro = 25 mm, ri = 6 mm, α = 45 °, and µ k = 0.25. What couple M is required to rotate the shaft at a constant rate?

D mg M

Solution:

There are two thrust bearings, one at the top and one at

the bottom. FU = 440 N

m = 220 kg ΣFy : FL − FU − mg = 0

Thus,

FL = 2598.2 N. The couple necessary to turn D at a constant rate is the sum of the couples for the two bearings. For the bearings used m =

MU =

2µ k FU (ro3 − ri3 ) = 2.7 N-m 3cos α (ro2 − ri2 )

ML =

2µ k FL (ro3 − ri3 ) = 16.0 N-m 3cos α (ro2 − ri2 )

M TOTAL = M U + M L = 18.7 N-m

2µ k F (ro 3 − ri 3 ) 3cos α (ro 2 − ri 2 )

α = 45 °, ro = 0.025 m µ k = 0.25, ri = 0.006 m

Problem 9.116 The shaft is supported by thrust bearings that subject it to an axial load of 800 N. The coefficients of kinetic friction between the shaft and the left and right bearings are 0.20 and 0.26, respectively. What couple is required to rotate the shaft at a constant rate?

Solution:

The left bearing: The parameters are

ro = 38 mm, ri = 0, α = 45 °, µ k = 0.2, and F = 800 N. The moment required to sustain a constant rate of rotation is

15 mm 38 mm

M left =

2µ k F  ro3 − ri3   = 5.73 N m. 3cos α  ro2 − ri2 

The right bearing: This is a flat-end bearing. The parameters are µ k = 0.26, r = 15 mm, and F = 800 N. The moment required to sustain a constant rate of rotation is M right = 38 mm

704

2µ k Fr = 2.08 N m. 3

The sum of the moments: M = 5.73 + 2.08 = 7.81 N m

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Problem 9.117 A motor is used to rotate a paddle for mixing chemicals. The shaft of the motor is coupled to the paddle using a friction clutch of the type shown in Fig. 9.17. The radius of the disks of the clutch is 120 mm, and the coefficient of static friction between the disks is 0.6. If the motor transmits a maximum torque of 15 N-m to the paddle, what minimum normal force between the plates of the clutch is necessary to prevent slipping?

Solution:

The moment necessary to prevent slipping is

2µ s Fr 2(0.6)(0.12) F = = 15 N m. 3 3

M = Solve:

F = 312.5 N

Clutch Paddle

Problem 9.118 The thrust bearing is supported by contact of the collar C with a fixed plate. The area of contact is an annulus with an inside diameter D1 = 40 mm and an outside diameter D 2 = 120 mm. The coefficient of kinetic friction between the collar and the plate is µ k = 0.3. The force F = 400 N. What couple M is required to rotate the shaft at a constant rate? (See Example 9.8.)

This is a thrust bearing with parameters

µ k = 0.3, α = 0, ro = 60 mm, ri = 20 mm, and F = 400 N. The moment required to sustain rotation at a constant rate is 2µ k F  ro3 − ri3   = 5.2 N m 3  ro2 − ri2 

M =

Solution:

M C

M C D1 D2

Problem 9.119 An experimental automobile brake design works by pressing the fixed red annular plate against the rotating wheel. If µ k = 0.6, what force F pressing the plate against the wheel is necessary to exert a couple of 200 N-m on the wheel?

Solution:

This is a thrust bearing with parameters

µ k = 0.6, α = 0, ro = 90 mm, ri = 50 mm, and M = 200 N m. The moment is 2µ k F  ro3 − ri3  . 3  ro2 − ri2 

M = F

Solve: F =

50 mm

90 mm

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Problem 9.120 An experimental automobile brake design works by pressing the fixed red annular plate against the rotating wheel. Suppose that µ k = 0.65 and the force pressing the plate against the wheel is F = 2 kN. (a) What couple is exerted on the wheel? (b) What percentage increase in the couple exerted on the wheel is obtained if the outer radius of the brake is increased from 90 mm to 100 mm?

Solution:

Use the results of the solution to Problem 9.119, with parameters µ k = 0.65, F = 2 kN. (a) The moment is M =

2µ k F  ro3 − ri3   = 0.0935 kN-m = 93.5 N-m 3  ro2 − ri2 

(b) The new moment is M =

2µ k F  ro3 − ri3   = 0.1011 kN-m = 101.1 N-m. 3  ro2 − ri2 

The percentage increase is ∆M% =

− 93.5 ( 101.193.5 )100 = 8.17%

50 mm

90 mm

Problem 9.121 The coefficient of static friction between the plates of the car’s clutch is 0.8. If the plates are pressed together with a force F = 2.60 kN, what is the maximum torque the clutch will support without slipping?

Solution: M =

2µ s F (ro3 − ri3 ) 3cos α (ro2 − ri2 )

where α = 90 °, (cos α ≡ 1) F = 2600 N, ro = 0.15 m

75 mm

ri = 0.075 m

150 mm

Solving for M, M = 243 N-m

Problem 9.122* The “Morse taper” is used to support the workpiece on a machinist’s lathe. The taper is driven into the spindle and is held in place by friction. If the spindle exerts a uniform pressure p = 15 psi on the taper and µ s = 0.2, what couple must be exerted about the axis of the taper to loosen it? Spindle 1.25 in

Taper

Solution:

The outer radius of the taper is ro = 1 in, and the inner radius is ri = 0.6.5 in. The angle of the taper is r − ri 1 − 0.625 α = 90 − tan −1 o = 90 − tan −1 = 87.6 °. L 9 The active area of contact of the taper is the area of a truncated cone:

(

A =

)

(

)

π (ro2 − ri2 ) = 45.99 in 2 . cos α

Check: This expression can be verified using the Pappus-Guldinus Theorem (see Example 7.15) where ro + ri , 2 r − ri and L = o . cos α check. The normal force on the taper is N = pA = 689.8 lb, and the axial force on the taper is F = N cos α = 28.72 lb. The taper is equivalent to a thrust bearing with the parameters µ s = 0.2, α = 87.6 °, ro = 1 in, ri = 0.625 in, and F = 28.72 lb. The moment required to initiate slip is y =

2 in 9 in

M =

706

2µ s F  ro3 − ri3   = 114.1 in lb = 9.51 ft-lb 3cos α  ro2 − ri2 

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Problem 9.123 In Practice Example 9.9, suppose that the left fixed cylinder is replaced by a pulley. Assume that the tensions in the rope on each side of the pulley are approximately equal. What is the smallest force the woman needs to exert on the rope to support the stationary box?

Solution:

Problem 9.124 Suppose that you want to lift a 50-lb crate off the ground by using a rope looped over a tree limb as shown. The coefficient of static friction between the rope and the limb is 0.2, and the rope is wound 135° around the limb. What force must you exert to begin lifting the crate?

Solution:

Problem 9.125 Winches are used on sailboats to help support the forces exerted by the sails on the ropes (sheets) holding them in position. The winch shown is a post that will rotate in the clockwise direction (seen from above), but will not rotate in the counterclockwise direction. The sail exerts a tension TS = 800 N on the sheet, which is wrapped two complete turns around the winch. The coefficient of static friction between the sheet and the winch is µ s = 0.2. What tension TC must the crew member exert on the sheet to prevent it from slipping on the winch?

Solution:

The tension in the rope between the pulley and the right cylinder is now equal to the weight W. Apply Eq. (9.17) to the right cylinder, assuming that slip in the direction of the force W is impending. W = Fe µ s β ⇒ F = We −µ s β = (100 lb)e −(0.4)( π / 2) = 53.3 lb 53.3 lb.

The force is given by Eq. (9.17) with T1 = 50 lb, µ s = 0.2 and β = (135/180)π = 2.36 rad.

T2 = T1e µ s β = (50 lb)e (0.2)(2.36) = 80.1 lb 80.1 lb.

Ts = Tc e µ s β Ts = 800 N µ s = 0.2 β = 4π Solving, Tc = 64.8 N

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Problem 9.126 The coefficient of kinetic friction between the sheet and the winch in Problem 9.125 is µ k = 0.16. If the crew member wants to let the sheet slip at a constant rate, releasing the sail, what initial tension TC must he exert on the sheet as it begins slipping?

Solution: Ts = Tc e µ k β Ts = 800 N, µ k = 0.16 β = 4π Solving Tc = 107.1 N

Problem 9.127 The box A weighs 20 lb. The rope is wrapped one and one-fourth turns around the fixed wooden post. The coefficients of friction between the rope and post are µ s = 0.15 and µ k = 0.12. (a) What minimum force does the man need to exert to support the stationary box? (b) What force would the man have to exert to raise the box at a constant rate?

Solution: (a)

T = (20 lb)e −0.15(5π / 2) = 6.16 lb

(b)

T = (20 lb)e 0.12(5π / 2) = 51.3 lb

708

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Problem 9.128 The weight of block A is W. The disk is supported by a smooth bearing. The coefficient of kinetic friction between the disk and the belt is µ k . What couple M is necessary to turn the disk at a constant rate?

Solution: The angle is β = π radians. The tension in the left belt when the belt is slipping on the disk is Tleft = We µ k β . The tension in the right belt is Tright = W. The moment applied to the disk is M = R(Tleft − Tright ) = R(We µ k β − W ) = RW (e µ k π − 1). This is the moment that is required to rotate the disk at a constant rate.

Problem 9.129 The couple required to turn the wheel of the exercise bicycle is adjusted by changing the weight W. The coefficient of kinetic friction between the wheel and the belt is µ k . Assume the wheel turns clockwise. (a) Show that the couple M required to turn the wheel is M = WR(1 − e −3.4 µ k ). (b) If W = 40 lb and µ k = 0.2, what force will the scale S indicate when the bicycle is in use?

Solution:

Let β be the angle in radians of the belt contact with wheel. The tension in the top belt when the belt slips is Tupper = We −µ k β . The tension in the lower belt is Tlower = W. The moment applied to the wheel is M = R(Tlower − Tupper ) = RW (1 − e −µ k β ). This is the moment required to turn the wheel at a constant rate. The angle β in radians is π β = π + (30 − 15) = 3.40 radians, 180

(

)

from which M = RW (1 − e −3.4 µ k ). (b) The upper belt tension is S

Tupper = 40 e −3.4(0.2) = 20.26 lb. This is also the reading of the scale S.

158 R

308

Problem 9.130 The box B weighs 50 lb. The coefficients of friction between the cable and the fixed round supports are µ s = 0.4 and µ k = 0.3. (a) What is the minimum force F required to support the box? (b) What force F is required to move the box upward at a constant rate?

Solution:

The angle of contact between the cable and each round π support is β = radians. 2 (a) Denote the tension in the horizontal part of the cable by H. The tension in H is H = We −µs β . The force F is F = He −µs β = We −2µs β , from which F = 14.23 lb is the force necessary to hold the box stationary. (b) As the box is being raised, H = We µ k β , and F = He µ k β = We 2µ k β , from which F = 128.32 lb

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Problem 9.131 The coefficient of static friction between the 50-lb box and the inclined surface is 0.10. The coefficient of static friction between the rope and the fixed cylinder is 0.05. Determine the force the woman must exert on the rope to cause the box to start moving up the inclined surface.

Solution:

The contact angle between the rope and the fixed cylinder 180 − 45 − 30 7π is β = π = 180 12 We have 2 equilibrium and 2 friction equations ΣF : T2 cos 25 ° − f − 50 lbsin 20 ° = 0 ΣF : T2 sin 25 ° + N − 50 lb cos 20 ° = 0 f = 0.1 N T = T2 e 0.05(7 π /12)

458

Solving: T = 25.2 lb

A 208

308

50 lb f

Problem 9.132 In Problem 9.131, what is the minimum force the woman must exert on the rope to hold the box in equilibrium on the inclined surface?

Solution:

See 9.131 - Change the friction force

ΣF : T2 cos 25 ° + f − 50 lbsin 20 ° = 0 ΣF : T2 sin 25 ° + N − 50 lbcos20 ° = 0 f = 0.1 N T2 = Te 0.05(7 π /12) Solving T = 13.10 lb

458 A

T2 208

308

50 lb f

710

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Problem 9.133 Blocks B and C each have a mass of 20 kg. The coefficient of static friction at the contacting surfaces is 0.2. Block A is suspended by a rope that passes over a fixed cylinder and is attached to block B. The coefficient of static friction between the rope and the cylinder is 0.3. What is the largest mass block A can have without causing block B to slip to the left?

208 B

Solution:

First we determine what tension in the rope attached to B will cause B to be on the verge of slipping to the left. Assuming slip of blocks B and C to be impending, the equilibrium equations for block B are ΣFx : −T1 + P sin 20 ° + µ s P cos 20 ° + µ s N = 0, ΣFy : −P cos 20 ° + µ s P sin 20 ° + N − mg = 0. The equilibrium equations for block C are ΣFx : R − P sin 20 ° − µ s P cos 20 ° = 0, ΣFy : −µ s R + P cos 20 ° − µ s P sin 20 ° − mg = 0. Substituting the given values and solving yields T1 = 220 N, N = 420 N, P = 256 N, R = 136 N. Now assume that block A is on the verge of slipping downward. From Eq. (9.17), T2 = T1e µ s β = (220 N)e (0.3)( π / 2) = 352 N. Thus mA =

352 N = 35.9 kg 9.81 m/s 2

35.9 kg.

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Problem 9.134 If the force F in Example 9.10 is increased to 400 N, what are the largest values of the couples M A and M B for which the belt will not slip?

Solution:

From Example 9.10, b = 500 mm, µ s = 0.8, R a = 200 mm, R b = 100 mm. The angle of contact for pulley A is β a = π + 2α. The angle of contact for pulley B is β b = π − 2α, where α = sin −1

0.1 ) = 0.2014 radians. ( R −b R ) = sin ( 0.5 a

−1

The belt contact is less for pulley B, so it is most likely to slip first. The couples are in opposition so that the tension in the upper belt is greater than the tension in the lower belt: For belt B: Tupper = Tlower e µs β b = 8.945Tlower . The force is F = (Tupper + Tlower ) cos α, from which Tlower =

F = 41.05 N, (1 + e µs β b ) cos α

and Tupper = 8.945Tlower = 367.19 N. The couples are M b = R b (Tupper − Tlower ) = 32.61 N-m. M a = R a (Tupper − Tlower ) = 65.23 N-m

Problem 9.135 The spring exerts a 320-N force on the left pulley. The coefficient of static friction between the flat belt and the pulleys is µ s = 0.5. The right pulley cannot rotate. What is the largest couple M that can be exerted on the left pulley without causing the belt to slip?

Solution: α = sin −1

The angle of the belt relative to the horizontal is

− 40 ( 100260 ) = 0.2329 radians.

For the right pulley the angle of contact is β right = π − 2α = 2.676 radians. The sum of the horizontal components of the tensions equals the force exerted by the spring: F = (Tupper + Tlower ) cos α = 320 N.

100 mm M

40 mm

Since the angle of contact is less on the right pulley, it should slip there first. At impending slip, the tensions are related by Tupper = Tlower e µs β right = 3.811Tlower . Substitute and solve: Tlower (1 + e 0.5(2.676) ) =

260 mm

320 , cos α

from which Tlower = 68.34 N, and Tupper = 260.48 N. The moment applied to the wheel on the right is M applied = R(Tupper − Tlower ) = 0.1(192.16) = 19.22 N-m

712

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Problem 9.136 The weight of the box is W = 30 lb, and the force F is perpendicular to the inclined surface. The coefficient of static friction between the box and the inclined surface is µ s = 0.2. (a) If F = 30 lb, what is the magnitude of the friction force exerted on the stationary box? (b) If F = 10 lb, show that the box cannot remain at rest on the inclined surface. F

Solution:

The maximum friction force is defined to be f = µ s N, where N is the normal force.

(a) The box is stationary, hence the friction force is equal to the force acting to move the box down the plane: ΣFP = f − W P = 0, from which f = W P = W sin α = 10.26 lb (b) The component of force parallel to the surface is W P = W sin α = 10.26 lb acting to move the box down the plane. The friction force is f = µ s (10 + 30 cos α) = 7.638 lb, acting to hold the box in place. Since W P > f , the box will move.

F 208

W a

208

W f N

Problem 9.137 In Problem 9.136, what is the smallest force F necessary to hold the box stationary on the inclined surface?

Solution:

At impending slip, the sum of the forces parallel to the sur-

face is ΣFP = f − W P = 0, from which f = W P . The friction force is f = µ s ( F + W cos α), and W P = W sin α. Equate and solve:

(

)

 sin α  sin 20 ° F = W  − cos α  = 30 − cos 20 ° = 23.1 lb  µs  0.2

W 208

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Problem 9.138 The homogeneous bar AB weighs 20 lb, and the homogeneous disk weighs 30 lb. The coefficient of static friction between the disk and the floor at C is µ s = 0.24. What is the magnitude of the couple that would have to be exerted on the disk to cause it to start rotating in the counterclockwise direction?

Solution:

The free-body diagrams of the disk and bar are By

Bx x

Ay M WD

WB Ax

f N 4 ft A C

where M is the counterclockwise couple applied to the disk and W D and W B are the weights of the disk and bar. The equilibrium equations for the disk are

2 ft

ΣFx = A x − f = 0, ΣFy = A y − W D + N = 0,

8 ft

ΣM point A = −(2 ft) f + M = 0. For the bar we write the equilibrium equation ΣM point B = −(4 ft) A x + (8 ft) A y + (4 ft)W B = 0. Setting f = µ s N and solving, we obtain A x = 8.57 lb, A y = −5.71 lb, N = 35.7 lb, M = 17.1 ft-lb. 17.1 ft-lb.

Problem 9.139 The homogeneous bar AB weighs 20 lb, and the homogeneous disk weighs 30 lb. The coefficient of static friction between the disk and the floor at C is µ s = 0.24. What is the magnitude of the couple that would have to be exerted on the disk to cause it to start rotating in the clockwise direction?

Solution:

The free-body diagrams of the disk and bar are By

Bx x

Ay M WD

WB Ax

4 ft where M is the clockwise couple applied to the disk and W D and W B are the weights of the disk and bar. The equilibrium equations for the disk are

A C

ΣFx = A x + f = 0,

2 ft

ΣFy = A y − W D + N = 0, 8 ft

ΣM point A = (2 ft) f − M = 0. For the bar we write the equilibrium equation ΣM point B = −(4 ft) A x + (8 ft) A y + (4 ft)W B = 0. Setting f = µ s N and solving, we obtain A x = −10.9 lb, A y = −15.5 lb, N = 45.5 lb, M = 21.8 ft-lb. 21.8 ft-lb.

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Problem 9.140 The masses of crates A and B are 25 kg and 30 kg, respectively. The coefficient of static friction between the contacting surfaces is µ s = 0.34. What is the largest value of α for which the crates will remain in equilibrium?

Solution: Choose a coordinate system with the x-axis parallel to the inclined surface. Denote the tension in the cables by T. Suppose that at impending slip the lower box tends to move up the plane and the upper box tends to move down the plane. Thus, for the lower box: ΣFx = −W B sin α − µ s (W A + W B ) cos α − µ sW A cos α + 2T = 0. For the upper box, ΣFx = +µ sW A cos α − W A sin α + T = 0. Eliminate T from the two equations, and reduce: (W B − 2W A )sin α + µ s (4W A + W B ) cos α = 0,

from which  µ (4W A + W B )  0.34(1275.3) α = tan −1  s = 65.65 °  = tan −1  (2W A − W B )  196.2

(

B a

)

WAcos a

mSWAcosa T

mS(WA+WB)cosa

mSWAcos a

(WA+WB)cos a

WAcos a

Problem 9.141 The side of a soil embankment has a 45° slope (Fig. a). If the coefficient of static friction of soil on soil is µ s = 0.6, will the embankment be stable or will it collapse? If it will collapse, what is the smallest slope that can be stable? Strategy: Draw a free-body diagram by isolating part of the embankment as shown in Fig. b.

Solution:

The strategy is to analyze the free body diagram formed by isolating part of the embankment, as shown. The sum of the force parallel to the slope are: ΣFx = −W sin θ + µ sW cos θ = 0, from which the required value of the coefficient of static friction is: µ s = tan θ = tan 45 ° = 1. Since the coefficient of static friction of soil on soil is less than the required value, the embankment will collapse. The smallest slope that will be stable is α = tan −1 (0.6) = 30.96 = 31 ° This problem is very similar to the problem of the box on an incline. u W

458

mSWcos u

Wcos u

(a)

(b)

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Problem 9.142 The mass of the van is 2250 kg, and the coefficient of static friction between its tires and the road is 0.6. If its front wheels are locked and its rear wheels can turn freely, what is the largest value of α for which it can remain in equilibrium?

Solution: Choose a coordinate system with the x-axis parallel to the incline. The weight of the van is W = mg = 22072.5 N. The moment about the point of contact of the rear wheels is ΣM R = (3 − 1.2)W cos α + 1W sin α − 3 N = 0, from which the normal force at the front wheels is N =

W (1.8cos α + sin α) . 3

The sum of the forces parallel to the inclined surface is ΣFx = +µ s N − W sin α = 0. Combine and reduce: µs 1m

( 1.83 )cos α + ( µ3 − 1)sin α = 0, s

from which  1.8µ s  α = tan −1   = tan −1 (0.45) = 24.2 °  3 − µ s 

1.2 m

Problem 9.143 In Problem 9.142, what is the largest value of α for which the van can remain in equilibrium if it points up the slope?

Solution:

mSN 1.2 m

1.8 m

The sum of the moments about the point of contact of the

rear wheels is ΣM R = −1.8W cos α + 1W sin α + 3 N = 0. The normal force is N =

W (1.8cos α − sin α) . 3

The sum of forces parallel to the incline is ΣFx = +µ s N − W sin α = 0. Combine and reduce: 1m

(

)

1.8µ s µ cos α − s + 1 sin α = 0, 3 3 from which

1.2 m

 1.8µ s  α = tan −1   = 16.7 °  µ s + 3 

1m R

716

ms N

W N 1.2 m 1.8 m

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Problem 9.144 The shelf is designed so that it can be placed at any height on the vertical beam. The shelf is supported by friction between the two horizontal cylinders and the vertical beam. The combined weight of the shelf and camera is W. If the coefficient of static friction between the vertical beam and the horizontal cylinders is µ s , what is the minimum distance b necessary for the shelf to stay in place?

Solution:

Take the sum of the moments about the lower cylinder

ΣM LC = +bW − hFNU + tµ s FNU = 0. The sum of the forces ΣFy = FNU − FNL = 0, from which the normal forces at the two cylinders are equal, and FN =

bW . (h − µ s t )

The force causing slippage is the weight, which is balanced by the friction force: 2µ s bW = 0, (h − tµ s )

0 = −W + 2µ s FN = −W + from which W

b =

 h − tµ s 1  h = − t    2µ s 2  µ s

( )

msFNU W

FNU FNL

msFNL b

Problem 9.145 The 20-lb homogeneous object is supported at A and B. The distance h = 4 in, friction can be neglected at B, and the coefficient of static friction at A is 0.4. Determine the largest force F that can be exerted without causing the object to slip.

The center of mass is located at x =

2(24) + (4/3)4 = 1.90 in. 28

The moment about B is ΣM B = −hF + W (4 − x ) − 4 A = 0, from which the normal force at A is

2 in F

A =

−hF + (4 − x )W . 4

The sum of the forces parallel to x is ΣFx = −µ s A + F = 0, 6 in

from which

(

F 1+

and F = A

)

µ sh µ (4 − x )W − s = 0, 4 4 µ s (4 − x )W = 2.993 = 2.99 lb (4 + µ s h)

4 in

Solution:

Choose a coordinate system with origin at A and the x-axis parallel to the floor. Divide the object into a “rectangular” volume and a “triangular” (wedge) volume. The volume of the lower rectangular portion is V1 = 4(6)(1) = 24 in 3 . The centroidal coordinates are x 1 = 2 in, y 1 = 3 in. The wedge has a volume V2 =

( 12 )(4)(2)(1) = 4 in ,

mSAA

A mSBB B x 4 in

and the centroid is at x 2 = (4/3) in, 2 y 2 = 6 + in. 3

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Problem 9.146 In Problem 9.145, suppose that the coefficient of static friction at B is 0.36. What is the largest value of h for which the object will slip before it tips over?

Solution:

Use the solution to Problem 9.145, as applicable. Tipping is imminent when the normal force at A becomes zero. From the solution to Problem 9.145, A =

−hF + (4 − x )W , 4

from which 2 in

h tip =

(4 − x )W . F

The sum of the forces parallel to the y-axis is

ΣFy = A + B − W = 0, from which the normal force at B = W − A. The sum of the forces parallel to the x-axis

6 in

ΣFx = −µ sA A − µ sB B + F = 0, from which, for A = 0, F = µ sBW . Combine and reduce to obtain A

h tip =

(4 − x ) = 5.82 in µ sB

4 in

Problem 9.147 The 180-lb climber is supported in the “chimney” by the normal and friction forces exerted on his shoes and back. The static coefficients of friction between his shoes and the wall and between his back and the wall are 0.8 and 0.6, respectively. What is the minimum normal force his shoes must exert?

Solution:

Choose a coordinate system with the x-axis horizontal and y-axis vertical. Let N s be the normal force exerted by shoes, N b the normal force exerted by his back. The sum of the forces: ΣFx = N s cos 4 ° − N b cos3 ° − µ ss N s sin 4 ° + µ sB N b sin 3 ° = 0, ΣFy = N s sin 4 ° + N b sin 3 ° + µ ss N s cos 4 ° + µ sB N b cos ° − W = 0.

Reduce to two simultaneous equations in two unknowns: a11 N s + a12 N b = 0, and a 21 N s + a 22 N b = W, where a11 = cos 4 ° − µ ss sin 4° = 0.9418, a12 = − cos3 ° + µ sB sin 3 ° = −0.9672, a 21 = sin 4° + µ ss cos 4° = 0.8678, and a 22 = sin 3 ° + µ sB cos3 ° = 0.6515. The equations have the solutions

a W N s = − 12 , det

and N b =

a11W , det

where det = a11a 22 − a12 a 21 = 1.4529 is the determinant of the coefficients. The result: N s = 119.83 lb

718

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Problem 9.148 The sides of the 200-lb door fit loosely into grooves in the walls. Cables at A and B raise the door at a constant rate. The coefficient of kinetic friction between the door and the grooves is µ k = 0.3. What force must the cable at A exert to continue raising the door at a constant rate if the cable at B breaks?

Solution:

Since the door fits loosely in the grooves, assume that the moment due to the unbalance when cable B breaks causes the door to contact the upper right and lower left corners. Thus the normal force on the sides occur at these corners. The friction forces at the corners oppose movement. The sum of the moments about the lower left corner is ΣM L = 2 A − 5W + 6 N R − 10µ k N R = 0. The sum of the forces:

3 ft

ΣFx = N L − N R = 0,

3 ft

from which N L = N R , and

ΣFy = A − W − µ k N R − µ k N L = 0. 3 ft

Combine to obtain the two simultaneous equations: 2 A + (6 − 10µ k ) N R = 5W ,

3 ft

and A − 2µ k N R = W . These have the solution: A = −

5 ft

(2µ k )(5W ) (6 − 10µ k )W − , det det

where det = 6(µ k − 1) is the determinant of the coefficients. Reducing: A =

(−10µ k − (6 − 10µ k ))W W = = 285.71 lb −6(1 − µ k ) 1 − µk

mNR

NR 69

NL mNL

Problem 9.149 The coefficients of static friction between the tires of the 1000-kg tractor and the ground and between the 450-kg crate and the ground are 0.8 and 0.3, respectively. Starting from rest, what torque must the tractor’s engine exert on the rear wheels to cause the crate to move? (The front wheels can turn freely.)

Solution:

The weight of the crate is W = mg = 4414.5 N. The force required to produce imminent slip of the crate on level ground is Fc = µ sW = 0.3(4414.5) = 1324.5 N. This is the friction force exerted by the ground on the tires, ΣFx = −Fc + f tires = 0. The friction force is related to the torque about the axle (both wheels) by ΣM axle = T − 0.8 f tires = 0, from which T = 0.8Fc = 1059.5 N m

0.8 m

0.4 m

1.4 m

0.8 m

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Problem 9.150 In Problem 9.149, what is the most massive crate the tractor can cause to move from rest if its engine can exert sufficient torque? What torque is necessary?

Solution:

The weight of the tractor is Wt = m t g = 9810 N. The sum of the moments about the front wheels is

ΣM F = +0.8Wt + 0.4µ scWc − 2.2 N = 0, where N is the normal force on the rear wheels. The sum of the forces at imminent tire slip is ΣFx = −µ scWc + µ st N = 0, from which

0.8 m

µ  N =  sc Wc .  µ st  Substitute into the first equation and reduce:

0.4 m

0.8µ st Wt = 11131.9 N, (2.2µ sc − 0.4µ scµ st )

Wc = 1.4 m

0.8 m

from which the mass of the crate is mc =

Wc = 1134.75 = 1134.8 kg . g

The friction force on the tires is f tires = µ scWc = 3339.6 N, from which the torque on the axle (both wheels) is T = 0.8 f tires = 2671.7 N m

Problem 9.151 The mass of the vehicle is 900 kg, it has rear-wheel drive, and the coefficient of static friction between its tires and the surface is 0.65. The coefficient of static friction between the crate and the surface is 0.4. If the vehicle attempts to pull the crate up the incline, what is the largest value of the mass of the crate for which it will slip up the incline before the vehicle’s tires slip?

Solution:

The normal force between the crate and the incline is N c = Wc cos θ, where θ = 20 °. The drawbar force parallel to the incline is Fd = −Wc sin θ − µ sc N c . For brevity write ψ = sin θ + µ sc cos θ. The horizontal component of the drawbar force at the tractor is Fdh = Fd cos θ = −Wc ψ cos θ. The vertical component of the drawbar force at the tractor is Fdv = Fd sin θ = −Wc ψ sin θ. The weight of the tractor is Wt = 900 g = 8829 N. The sum of the moments about the front wheels is ΣM F = 1Wt − 0.8Fdh − 3.7 Fdv − 2.5 N = 0, from which 0 = Wt + (0.8cos θ + 3.7sin θ)Wc ψ − 2.5 N = 0. The sum of the force parallel to the ground is ΣFx = Fdh + µ st N = 0,

0.8 m 208

from which N =

Wc ψ cos θ . µ st

Substitute and reduce: 208

1.2 m

1.5 m 2.5 m

µ stWt ((2.5 − 0.8µ st ) cos θ − 3.7µ st sin θ)(sin θ + µ sc cos θ) = 7701.1 N. The mass of the crate is Wc =

mc =

720

Wc = 785 kg g

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Problem 9.152 Each 12-lb bar has length L = 4 ft. Assume that each bar’s weight acts at its center. If the angle α = 30 °, what is the magnitude of the friction force exerted on the right bar by the floor at C?

Solution:

Consider the free-body diagrams of the bars:

B a

W a

Ax C

C N

We write the equilibrium equations 1 ΣM point A = −( L cos α) B x + ( L sin α) B y − ( L sin α)W = 0, 2 1 ΣM point C = ( L cos α) B x + ( L sin α) B y + ( L sin α)W = 0. 2 Adding the first equation to the second one, we see that B y = 0. From 1 the second equation, we obtain B x = − W tan α. Then we can see from 2 1 the right free-body diagram that N = W and f = −B x = W tan α. 2 The friction force is f = =

1 W tan α 2 1 (12 lb) tan30 ° 2

= 3.46 lb. 3.46 lb.

Problem 9.153 Each 12-lb bar has length L = 4 ft. Assume that each bar’s weight acts at its center. If the coefficient of static friction between the right bar and the floor at C is µ s = 0.4, what is the largest value of the angle α at which the system can be in equilibrium? B

Solution:

Consider the free-body diagrams of the bars:

We write the equilibrium equations ΣM point A = −( L cos α) B x + ( L sin α) B y − ΣM point C = ( L cos α) B x + ( L sin α) B y +

( 12 L sin α )W = 0,

( 12 L sin α )W = 0.

Adding the first equation to the second one, we see that B y = 0. From 1 the second equation, we obtain B x = − W tan α. Then we can see from 2 1 the right free-body diagram that N = W and f = −B x = W tan α. 2 a

The maximum friction force that can be supported is f = µ s N. Setting C

1 W tan α f = 2 = µ s = 0.4, N W The largest angle is α = 38.7 °. 38.7 °.

Bx a

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y x

W f

C N

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Problem 9.154 The collars A and B each have a mass of 2 kg. If friction between collar B and the bar can be neglected, what minimum coefficient of static friction between collar A and the bar is necessary for the collars to remain in equilibrium in the position shown?

Solution:

The weight of each collar is W = mg = 19.62 N. Denote θ = 45 °, α = 20 °. The unit vector parallel to the bar holding B is e B = −i sin θ + j cos θ.

The weight of B is W = − j W . The components of the weight of B parallel to the bar is W PB = (e B ⋅ W)e B = − W cos θe B . This force must be balanced by a component of the force in the connecting wire for B to remain stationary. The components of the tension in the wire are

T = T ( i cos(180 ° + α) + j sin (180 ° + α)) = T (−i cos α − j sin α), from which the component of T parallel to the bar supporting B is

458 208 A

TPB = (e B ⋅ T)e B = T sin (θ − α)e B . The sum of forces along e B : T sin (θ − α)sin θ − W cos θ sin θ = 0, from which: sin(θ − α) W . = T cosθ The unit vector parallel to the bar supporting A is e A = j. The component of T parallel to the bar supporting A is TPA = − j ⋅ T = T sin α, and the force exerted by T on the slider A perpendicular to the bar is N A = −i ⋅ T = T cos α, where the negative sign is used because the tension at A is in opposition to the tension at B (tension is reversed). The sum of forces parallel to the bar is ΣFA = +µ s N A − W + TPA = 0, from which µ s T cos α − W + T sin α = 0, and µ s =

(

)

W 1 − tan α. T cos α

Substitute and reduce: µs =

( θ − α) ( sin ) − tan α = 0.272 cos α cos θ mSNA T NA

208

mg NB 458 458 mSNB

208 T

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Problem 9.155 If the coefficient of static friction has the same value µ s between the 2-kg collars A and B and the bars, what minimum value of µ s is necessary for the collars to remain in equilibrium in the position shown? (Assume that slip impends at A and B.)

Solution:

The weight of each collar is W = mg = 19.62 N. Denote θ = 45 °, α = 20 °. Isolate Collar A. The sum of forces:

ΣFx = T cos α − N A = 0, ΣFy = T sin α + µ s N A − mg = 0. Isolate Collar B: The sum of forces: ΣFx = N B cos θ + µ s N B cos θ − T cos α = 0. ΣFy = N B sin θ − µ s N B sin θ − T sin α = 0. These are four equations in four unknowns. Solve: T = 45.4 N, N A = 42.6 N, N B = 55 N, and µ s = 0.0963 . This is the minimum coefficient of friction required to maintain equilibrium.

mS NA 458

208 A

Problem 9.156 The clamp presses two pieces of wood together. The pitch of the threads is p = 2 mm, the mean radius of the thread is r = 8 mm, and the coefficient of kinetic friction between the thread and the mating groove is 0.24. What couple must be exerted on the threaded shaft to press the pieces of wood together with a force of 200 N?

Solution:

125 mm

458 458 mS NB mg

208 T

The free-body diagram of the upper arm of the clamp is

shown. From the equilibrium equation ΣM (ptc ) = −(0.25)(200) − 0.1BE = 0, we find that BE = −500 N. The compressive load in BE is 500 N. The slope of the thread is α = arctan

50 mm

NB T 208

( 2Pπr )

 0.002  = arctan    2π(0.008)  = 2.279 °. The angle of friction is

50 mm

θ k = arctan(0.24) = 13.496 °. C

50 mm D

From Eq. (9.9) with θ s = θ k , the required couple is M = rF tan (θ k + α) = (0.008)(500) tan (13.496 ° + 2.279 °) = 1.13 N-m.

BE Cy 200 N

0.1 m Cx

0.25 m

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Problem 9.157 In Problem 9.156, the coefficient of static friction between the thread and the mating groove is 0.28. After the threaded shaft is rotated sufficiently to press the pieces of wood together with a force of 200 N, what couple must be exerted on the shaft to loosen it? 125 mm

125 mm

B 50 mm

50 mm D

First, find the forces in the parts of the clamp. Then analyze the threaded shaft. BE is a two force member ΣFx : BE + C x = 0

(1)

ΣFy : 200 + C y = 0

(2)

ΣM A :

(3)

−0.05BE + 0.05C x + 0.25C y = 0

Solving, we get BE = −500 N (compression) C x = 500 N

50 mm

125 mm

Solution:

C y = −200 N We don’t have to solve for additional forces because we used the fact that member BE was a two force member. From Problem 9.170, P = 2 mm, r = 8 mm. We have µ s = 0.28. We want to loosen the clamp (Turn the clamp such that the motion is in the direction of the axial force. To do this, M = rF tan (θ s − α) where tan θ s = µ s = 0.28 θ s = 15.64 ° tan α = P /2πr =

2 2π (8)

α = 2.28 ° F = BE = 500 N Solving M = 0.950 N-m

0.125 m B

0.05 m A

0.125 m

200 N

BE C

0.05 m

x CX

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Problem 9.158 The axles of the tram are supported by journal bearings. The radius of the wheels is 75 mm, the radius of the axles is 15 mm, and the coefficient of kinetic friction between the axles and the bearings is µ k = 0.14. The mass of the tram and its load is 160 kg. If the weight of the tram and its load is evenly divided between the axles, what force P is necessary to push the tram at a constant speed?

Solution:

Assume that there are two bearings per axle. The weight of the tram is W = mg = 1569.6 N. This load is divided between four bearings: F =

W = 392.4 N. 4

The angle of kinetic friction is θ k = tan −1 (µ k ) = 7.97 °. The moment required to turn each bearing at a constant rate is M = Fr sin θ k = 0.8161 N m, and the force per wheel is Pw =

M 0.8161 = = 10.88 N. R 0.075

The total force required to push the tram is P = 4 Pw = 43.5 N

Problem 9.159 The two pulleys have a radius of 6 in and are mounted on shafts of 1-in radius supported by journal bearings. Neglect the weights of the pulleys and shafts. The coefficient of kinetic friction between the shafts and the bearings is µ k = 0.2. If a force T = 200 lb is required to raise the man at a constant rate, what is his weight?

Solution:

Denote the tension in the horizontal portion of the cable by H. The angle of kinetic friction is θ k = tan −1 (0.2) = 11.31 °

Consider the right pulley: The force on the right pulley is F =

T 2 + H 2.

The magnitude of the moment required to turn the shaft in the bearing is M right = r T 2 + H 2 sin θ k . The applied moment is M applied = (T − H ) R, from which (T − H ) R = r T 2 + H 2 sin θ k . Square both sides and reduce to obtain the quadratic:     1  + T 2 = 0,  2  1 − r sin 2θ   k  R

H 2 − 2TH 

( )

or H 2 − 2(200.214) H + 40000 = 0. This has the solutions: H = 200.214 ± 200.214 2 − 40000 = 209.46, 190.95. The lesser root corresponds to the horizontal tension, H = 190.97 = 191 lb.

Consider the left pulley: The force on the pulley is Fleft = W 2 + H 2 . The applied moment is M applied = −( H − W ) R, from which ( H − W ) R = r W 2 + H 2 sin θ k . Square both sides and reduce to the quadratic:     H W + H 2 = 0, W 2 − 2 2  r  1 − 2θ   sin k    R

( )

or W 2 − 2(191.1)W + 36431.9 = 0. This has the solutions: W1,2 = 191.166 ± 191.166 2 − 36431.9 = 199.9 lb, 182.25 lb. By an analogous argument to that used in Problem 10.92, the lesser root corresponds to the weight of the man, Wraised = 182.3 lb

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Problem 9.160 If the man in Problem 9.159 weighs 160 lb, what force T is necessary to lower him at a constant rate?

Solution:

Use the solution to Problem 9.160, with W = 160 lb. Begin with the left pulley: the quadratic relation between the weight and the horizontal tension is     W  H + W 2 = 0, H 2 − 2 2  r  1 − 2 sin θ k   R

( )

or H 2 − 2(160.171) H + 25600 = 0. This has the solutions: H 1,2 = 160.171 ±

160.171 2 − 25600 = 167.57 lb, 152.77 lb.

The lesser root corresponds to the horizontal tension: H = 152.77 lb Consider the right pulley: The quadratic relation between the tension and the horizontal tension is

H   T + H 2 = 0, T 2 − 2 r 2 2θ   1 − sin  k   R

( )

or T 2 − 2(152.93)T + 23338.4 = 0. This has the solutions: T1,2 = 152.93 ±

152.93 2 − 23338.4 = 160 lb, 145.87 lb.

By previous arguments, the lesser root corresponds to the tension when the man is being lowered at a constant rate, Tlower = 145.9 lb.

Problem 9.161 If the two cylinders are held fixed, what is the range of W for which the two weights will remain stationary? ms 5 0.30 mk 5 0.28

ms 5 0.34 mk 5 0.32

100 lb

Problem 9.162 If the system is initially stationary and the left cylinder is slowly rotated, determine the largest weight W that can be (a) raised; (b) lowered. ms 5 0.30 mk 5 0.28

ms 5 0.34 mk 5 0.32

Solution:

Denote the tension in the horizontal part of the rope by H. The angle of contact with each cylinder is π β = . 2

Begin on the left with the known weight. Suppose that the known weight is on the verge (imminent slip) of being lowered. The tension is T L = 100 lb. The tension in the horizontal portion of the rope is H = T L e −0.34 β = 58.62 lb. The tension in the right part of the rope is T R = He −0.30 β = 36.59 lb. This is the minimum weight for which the system will remain stationary. Suppose that the weight on the left is on the verge of being raised. The horizontal tension is H = T L e 0.34 β = 170.59 lb. The tension on the right is T R = He 0.30 β = 273.3 lb. This is the maximum weight for which the system will remain stationary. Thus 36.59 ≤ W ≤ 273.3 (lb)

Solution: Assume that the rope does not slip on the slowly rotating cylinder, but is always at the point of imminent slip. The tension in the horizontal part as the cylinder is rotated is H = T L e 0.34 β = 170.6 lb, where the static coefficient of friction is used, since the rope does not slip. The tension in the right portion of the rope is T R = He −0.28β = 109.88 = 110 lb, where the kinetic coefficient of friction is used, since the rope slips at a steady rate. Thus (a) W = 110 lb . (b) If the weight W is slowly being lowered, the tension in the horizontal portion is also H = T L e 0.34 β = 170.6 lb, and the tension in the right portion of the rope is T R = He 0.28β = 264.8 lb, from which W = 264.8 lb

100 lb

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Chapter 10 Problem 10.1 The beam has a built-in support at A and is subjected to an 80-lb force at C. The free-body diagram of the entire beam is also shown. Determine the internal forces and moment at B.

Let us draw the free-body diagram of the part of the beam to the left of B: y 640 ft-lb

A 80 lb

y 80 lb B

Ay 3 ft

ΣFx = PB = 0, ΣM point B = −(3 ft)(80 lb) + M B + 640 ft-lb = 0,

ΣFy = −V B + 80 lb = 0,

80 lb

5 ft

we obtain PB = 0, VB = 80 lb, M B = −400 ft-lb. Alternatively, we could have drawn the free-body diagram of the part of the beam to the right of B: y VB

80 lb

MB PB

Solution:

From the equilibrium equations x

The equilibrium equations for the entire beam are

ΣFx = A x = 0, ΣFy = A y − 80 lb = 0, ΣM point A = −(8 ft)(80 lb) + M A = 0, yielding A x = 0, A y = 80 lb, M A = 640 ft-lb.

The equilibrium equations are ΣFx = −PB = 0, ΣFy = VB − 80 lb = 0, ΣM point B = −(5 ft)(80 lb) − M B = 0, giving PB = 0, VB = 80 lb, M B = −400 ft-lb. PB = 0, VB = 80 lb, M B = −400 ft-lb.

Problem 10.2 The magnitude of the triangular distributed load is w 0 = 2 kN/m. Determine the internal forces and moment at A.

Solution:

The free-body diagram of the beam is shown in Fig. a. From the equilibrium equations

0.4 m 0.6 m

0.6 m

ΣFx : B x = 0, ΣFy : B y + C − 600 N = 0, ΣM B : C (1.2 m) − (600 N)(1 m) = 0, We obtain B x = 0, B y = 100 N, C = 500 N. Passin a plane through the beam at A and isolating the part of the beam to the left of A (Fig. b), we obtain ΣFx : PA = 0, ΣFy : 100 N − V A = 0, ΣM A : M A − (100 N)(0.4 m) = 0. Solving yields PA = 0, V A = 100 N, M A = 40 N-m.

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Problem 10.3 The C clamp exerts 30-lb forces on the clamped object. Determine the internal forces and moment in the clamp at A.

Solution: ΣFx : −30 lb + PA = 0 ΣFy : −V A = 0 ΣM A : (30 lb)(2 in) + M A = 0 Solving: PA = 30 lb, V A = 0, M A = −60 in-lb 30 lb

PA VA

2 in

Problem 10.4 moment at A. y

Determine the internal forces and

ΣFx : −PA = 0,

400 lb

100 lb

ΣFy : V A − 400 lb = 0,

900 ft-lb

ΣM A : −M A + 900 ft-lb − (400 lb)(7 ft) = 0. x

3 ft

728

4 ft

3 ft

Solution: Passing a plane through the beam at A and writing the equilibrium equations for the part of the beam to the right of A, we obtain

4 ft

Solving yields PA = 0, V A = 400 lb, M A = −1900 ft-lb.

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Problem 10.5 The pipe has a fixed support at the left end. Determine the internal forces and moment at A.

Solution:

Use the right section

ΣFx : −PA + 2 kN cos 20 ° = 0 ΣFy : V A + 2 kN sin 20 ° + 2 kN = 0

2 kN

208

ΣM A : −M A − (2 kN cos 20 °)(0.2 m)

0.2 m

Solving:

+ (2 kN sin 20 °)(0.2 m) + (2 kN)(0.4 m) = 0

y 2 kN

2 kN

0.2 m

PA = 1.88 kN, V A = −2.68 kN, M A = 0.561 kN-m

208 2 kN 0.2 m VA PA 0.2 m

Problem 10.6 Determine the internal forces and moment at A for each loading.

0.2 m

Make a cut at A: Isolate the left hand part. The sum of moments: ΣM = M A − 4(1) = 0, from which M A = 4 kN-m V A = 4 kN PA = 0

(b) Determine the reaction at B: The sum of the moments about R:

8 kN

ΣM R = −∫ 2 x dx + 4 B = 0, 0

from which

B =

2 4

( 14 )  2 x2  = 164 = 4 kN. 0

The reaction at R:

(a)

ΣFy = R y − ∫ 2 dx + B = 0,

from which R y = 8 − 4 = 4 kN,

2 kN/m

ΣFx = R x = 0. x A

Make a cut at A: Isolate the left hand part. The sum of moments: 1

ΣM = M A − (1) R y + ∫ 2 x dx = 0,

4m ( b)

from which M A = R y − 1 = 3 kN m. 1

V A = R y − ∫ 2 dx = 4 − 2 = 2 kN 0

Solution: (a) Denote the reaction at the pinned left end by R, and the reaction at the roller support by B. The reaction at B:

PA = 0.

Ry 5 4 kN

ΣM = −2(8) + B(4) = 0, from which B = 4 kN. The reaction at R: ΣFy = R y − 8 + B = 0, from which R y = 4 kN.

(a) 1m (b)

2 kN/m 4 kN

VA MA

ΣFx = R x = 0.

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Problem 10.7 Model the ladder rung as a simply supported (pin supported) beam and assume that the 750-N load exerted by the person’s shoe is uniformly distributed. Determine the internal forces and moment at A.

Solution: ΣFy = B + C − 750 = 0, ΣM (pt. B ) = 0.375C − (0.25)(750) = 0. Solving, B = 250 N, C = 500 N.

A 250 N 250 mm 200 mm

0.2 m

500 N

0.1 m

x MA

100 mm 375 mm

250 N

0.2 m

0.05 m

0.05 MA 250 N

0.025 m VA 0.25 m

The distributed load is w = (750 N)/(0.1 m) = 7500 N/m. From the equilibrium equations ΣFx = PA = 0, ΣFy = 250 − V A − 0.05(7500) = 0, ΣM (rightend) = M A − (250)(0.25) + (0.05)(7500)(0.025) = 0, we obtain PA = 0, V A = −125 N, M A = 53.1 N-m. 750 N

0.25 m 0.375 m

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Problem 10.8 In Example 10.2, suppose that the distance from point A to point B is increased from 3 m to 4 m. Draw a sketch of the beam with B in its new position. Determine the internal forces and moment at B.

Solution:

Let us pass a plane through the beam at B. Using similar triangles, the magnitude of the distributed load at B is 4 (60 N/m) = 40 N/m. 6

If we represent the distributed load to the left of B by an equivalent force, its magnitude is 1 (40 N/m)(4 m) = 80 N, 2 and it acts at the centroid of the distributed load to the left of B. The distance from A to the centroid is 2 (4 m) = 2.67 m. 3 The equilibrium equations for the part of the beam to the left of B are ΣFx : PB = 0, ΣFy : 120 N − 80 N − VB = 0, ΣM B : M B − (120 N)(4 m) + (80 N)(1.33 m) = 0. Solving yields PB = 0, VB = 40 N, M B = 373 N-m.

Problem 10.9 If x = 3 ft, what are the internal forces and moment at A? y

ΣFx : −PA = 0, ΣFy : V A + 900 lb = 0,

600 lb/ft

ΣM A : −M A + (900 lb)(1 ft) = 0. x

A 600 lb/ft 3 ft

Solving yields PA = 0, V A = −900 lb, M A = 900 ft-lb.

3 ft

BandF_6e_ISM_CH10.indd 731

Solution: Isolating the part of the beam to the right of A, we represent the distributed load by an equivalent force. From this free-body diagram, we write the equilibrium equations:

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Problem 10.10 If x = 4 ft, what are the internal forces and moment at A?

Solution:

Isolating the part of the beam to the right of A, we represent the distributed load by an equivalent force. We can obtain the magnitude of the distributed load by similar triangles:

2 (600 lb/ft) = 400 lb/ft. 3

600 lb/ft x x

A 600 lb/ft 3 ft

3 ft

If we represent the distributed load to the right of point A by a single equivalent force, its magnitude is 1 (400 lb/ft)(2 ft) = 400 lb, 2 and it acts at the centroid of the distributed load to the right of point A. The distance from A to the centroid is 1 (2 ft) = 0.667 ft. 3 From this free-body diagram, we write the equilibrium equations: ΣFx : −PA = 0, ΣFy : V A + 400 lb = 0, ΣM A : −M A + (400 lb)(0.667 ft) = 0. Solving yields PA = 0, V A = −400 lb, M A = 267 ft-lb.

732

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Problem 10.11 Determine the internal forces and moment at A for the loadings (a) and (b).

(a)

Use the left section with the distributed loading ΣFx : PA = 0 ΣFy : C − 180 lb − V A = 0 ΣM A : −C (3 ft) + (180 lb)(1.5 ft) + M A = 0

60 lb/ft

Solving PA = 0, V A = 84 lb, M A = 522 ft-lb

(a)

3 ft 4 ft

5 ft 6 ft

240 lb

180 lb

(b) 3 ft 2 ft

180 lb

5 ft 4 ft

4 ft

3 ft VA

Solution:

The external reactions are the same for either loading

condition ΣM A : −(240 lb)(2 ft) − (180 lb)(6 ft) + D(10 ft) = 0 ⇒ D = 156 lb ΣFy : C + D − 240 lb − 180 lb = 0 ⇒ C = 264 lb

(b) Use the left section with the discrete load ΣFx : PA = 0 ΣFy : C − 240 lb − V A = 0 ΣM A : −C (3 ft) + (240 lb)(1.0 ft) + M A = 0 Solving PA = 0, V A = 24 lb, M A = 552 ft-lb

240 lb

180 lb 240 lb

MA PA C

3 ft

D C

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Problem 10.12 Determine the internal forces and moment at B for the loadings (a) and (b).

(a)

Use the right section with the distributed loading ΣFx : −PB = 0 ΣFy : VB − 125 lb + D = 0 ΣM B : −M B − 125 lb

60 lb/ft (a)

5 ft 6 ft

240 lb

3 ft 2 ft

125 lb MB

180 lb

(b)

Solving PB = 0, VB = −31 lb, M B = 572 ft-lb

3 ft 4 ft

( 13 5 ft ) + D(5 ft) = 0

B PB

5 ft 4 ft

4 ft

5 ft D

Solution:

The external reactions are the same for either loading

(b) Use the right section with the discrete loads

condition

ΣFx : − PB = 0

ΣM A : −(240 lb)(2 ft) − (180 lb)(6 ft) + D(10 ft) = 0 ΣFy : C + D − 240 lb − 180 lb = 0 ⇒ C = 264 lb 240 lb

ΣFy : VB − 180 lb + D = 0 ΣM B : −M B − (180 lb)(1 ft) + D(5 ft) = 0

⇒ D = 156 lb

Solving PB = 0, VB = 24 lb, M B = 600 ft-lb 180 lb

180 lb

MB 4 ft PB VB C

734

5 ft

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Problem 10.13 moment at A.

Determine the internal forces and

Solution:

Use the whole body to find the reactions

ΣM C : −B(8 ft) + (1600 lb)(4 ft) + (400 lb)(2.67 ft) − (600 lb)(1.33 ft) = 0 300 lb/ft

200 lb/ft

⇒ B = 833 lb 400 lb

600 lb

1600 lb 6 ft 8 ft

4 ft

Now examine the section to the left of the cut ΣFx : PA = 0 ΣFy : B − 1200 lb − 225 lb − V A = 0 ΣM A : −B(6 ft) + (1200 lb)(3 ft) + (225 lb)(2 ft) + M A = 0 Solving PA = 0, V A = −592 lb, M A = 950 ft-lb

225 lb 1200 lb

275 lb/ft 200 lb/ft MA

6 ft

BandF_6e_ISM_CH10.indd 735

PA VA

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Problem 10.14 moment at A.

Determine the internal forces and

from which M A = −10 + 16 = 6 kN-m V A = 8 − 10 = −2 kN, PA = 0

y 10 kN

10 kN CX

B 1m

Solution:

The complete structure as a free body: The sum of the moments about the right end:

10 kN

ΣM = 3(10) − 5 R = 0, 30 from which R = = 6 kN. The sum of forces in the y-Direction: 5

CX A

ΣFy = R y + C y − 10 = 0, from which C y = 4 kN. The element CA as a free body: The sum of the moments about C: F1

The sum of the forces:

10 kN

ΣM C = −4 F1 + 10(3) − F2 = 0.

ΣFy = C y + F2 − 10 + F1 = 0. Solve the simultaneous equations: F1 = 8 kN, F2 = −2 kN. Make a cut at A: Isolate the left end of CA. The sum of the moments about A: ΣM = M A − 2 F1 + 10 = 0,

Problem 10.15 moment at B.

Determine the internal forces and

Solution: Use the solutions to Problem 10.14. Make a cut at point B: Isolate the left part. The sum of the moments about B: ΣM B = M B + 2 F1 − 3 R y = 0,

y 10 kN

from which A

M B = −16 + 18 = 2 kN-m VB = R y − F1 = 6 − 8 = −2 kN PB = 0

B 1m

F1 5 8

MB PB

VB 1m

736

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Problem 10.16 moment at A.

Determine the internal forces and

Next examine the vertical bar to find the tension in the cable 1 ΣM E : −(600 N)(0.4 m) + T (0.8 m) = 0 ⇒ T = 671 N 5 1 2

0.4 m

600 N T

600 N Ex

0.4 m A

x Ey

0.2 m

0.6 m Finally cut at A and look at the left section 1 T + PA = 0 5 2 ΣFy : C y + T − VA = 0 5 ΣFx : C x +

0.4 m

Solution:

0.4 m

ΣM A : C x (0.6 m) − C y (0.2 m) −

2 T (0.2 m) + M A = 0 5

Solving we have Use the entire structure to find the reactions

ΣFx : C x + 600 N = 0 ⇒ C x = −600

PA = 300 N, V A = −150 N, M A = 330 N-m

ΣM C : −(600 N)(1.0 m) + D(0.8 m) = 0 ⇒ D = 750 N ΣFy : C y + D = 0 ⇒ C y = −750 N

T 2 600 N

1 PA VA D

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Problem 10.17 moment at B.

Determine the internal forces and

Solution:

Use the section to the right of the cut at B

ΣFx : −PB = 0 ΣFy : VB + D = 0 ΣM B : −M B + D(0.2 m) = 0 0.4 m

Solving PB = 0, VB = −750 N, M B = 150 N-m

600 N

0.4 m A

0.2 m

0.4 m

0.6 m

0.4 m

Problem 10.18 The tension in the rope is 10 kN. Determine the internal forces and moment at point A.

Now examine the bent bar (T = 10 kN) 4 (T )(0.6 m) = 0 5

ΣM D : −B x (1.2 m) − B y (1.6 m) + ⇒ B y = 6 kN

y A

0.8 m

0.6 m 0.6 m

3 T

3 kN 0.8 m

0.8 m 3 kN

Solution:

Use the whole structure first

Finally cut the bar at A and examine the left section

ΣM C : −B x (1.2 m) − (3 kN)(1.6 m) = 0 ⇒ B x = −4 kN Bx

MA Bx PA

738

ΣFx : B x + PA = 0

PA = 4 kN

ΣFy : B y − V A = 0

⇒ V A = 6 kN

ΣM A : −B y (0.8 m) + M A = 0

M A = 4.8 kN-m

3 kN

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Problem 10.19 Determine the internal forces and moment at point A of the frame.

Next examine the slanted bar and take advantage of the 2-force member ΣM D : C (0.8 m) − T (0.4 m) = 0 ⇒ T = 3 kN

0.2 m 3 kN A 0.2 m

Dy x

0.2 m

C 0.8 m Now cut the bar at A and look at the lower right section

Solution:

3 4 T + C = 0 5 5 4 3 ΣF : PA + T + C = 0 5 5 2 ΣM A : −T (0.2 m) + C 0.8 m − M A = 0 3 ΣF : V A −

Use the whole structure first

ΣM B : −(3 kN)(0.4 m) + C (0.8 m) = 0 ⇒ C = 1.5 kN

(

3 kN

)

Solving PA = −3.3 kN, V A = 0.6 kN, M A = 0.2 kN-m PA

VA 4

Bx By

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Problem 10.20 moment at A.

Determine the internal forces and

Solution:

The free-body diagrams of the horizontal members are CY

(3 m)(4 kN/m) 5 12 kN

1.5 m

(a) CX 4 kN/m

B DY

u R

(b) DX

T T

The angle θ = arctan(2/1) = 63.4 °.

From free-body diagram (a), ΣFx = C x − R cos θ = 0, x

A 1m

ΣFy = C y − 12 − R sin θ − T = 0, ΣM (pt. C ) = −(1.5)(12) − 2 R sin θ − 3T = 0,

and from free-body diagram (b), ΣFx = D x + R cos θ = 0, ΣFy = D y + R sin θ + T = 0, ΣM (pt. D ) = (1) R sin θ + 3T = 0. Solving, we obtain C x = −9 kN, C y = 0, and T = 6 kN. Cutting member (b) at A,

6 kN

1m we see that PA = 0, V A = −6 kN, M A = (1)(6) = 6 kN-m.

Problem 10.21 moment at B.

Determine the internal forces and

Solution:

See the solution of Problem 10.20. Cutting member (a) at B and including the distributed load acting on the part of the member to the left of B,

4 kN/m MB 4 kN/m

9 kN

4 kN MB

9 kN

A 1m

0.5 m

we see that PB = 9 kN, VB = −4 kN, M B = −(0.5)(4) = −2 kN-m.

740

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Problem 10.22 Determine the shear force and bending moment as functions of x. Strategy: Cut the beam at an arbitrary position x and draw the free-body diagram of the part of the beam to the left of the plane. y

Solution:

Cut the beam at arbitrary position x and look at section to the left of the cut ΣFy : 400 lb − V = 0 ΣM cut : −(400 lb) x + M = 0 Solving we have V = 400 lb M = (400 lb) x

400 lb

3 ft

x V

Problem 10.23 (a) Determine the shear force and bending moment as functions of x. (b) Draw the shear force and bending moment diagrams.

Now that this part has been isolated, we can represent the distributed load by an equivalent force:

y (48 kN/m)x

48 kN/m

144 kN

1x 2 Writing the equilibrium equations ΣFx = P = 0,

Solution: (a)

ΣFy = −V + 144 kN − (48 kN/m) x = 0,

We draw the free-body diagram of the entire beam, representing the distributed load by its equivalent force:

ΣM right end = −x (144 kN) +

we obtain

(6 m)(48 kN/m)

P = 0, V = 144 − 48 x kN, M = 144 x − 24 x 2 kN-m.

(b) The shear and bending-moment diagrams are shown:

( 12 x )(48 kN/m)x + M = 0,

200

1 (6 m) 2

V, kN

100

From the equilibrium equations ΣFx = A x = 0,

0 2100

ΣFy = A y + B − (6 m)(48 kN/m) = 0,

2200

1 ΣM point A = (6 m) B −  (6 m)  [(6 m)(48 kN/m)] = 0, 2 

3 x, m

250

Now we “cut” the beam at an arbitrary position x and draw the free-body diagram of the part of the beam to the left of the cut (including the part of the distributed load that acts on that part), showing the internal forces and moment.

M, kN-m

we obtain the reactions A x = 0, A y = B = 144 kN.

200 150 100 50

0 48 kN/m U

M P

A 144 kN

(a) P = 0, V = 144 − 48 x kN, M = 144 x − 24 x 2 kN-m.

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Problem 10.24 (a) Determine the shear force and bending moment as functions of x. (b) Show that the equations for V and M satisfy the equation V = dM /dx.

Now we “cut” the beam at an arbitrary position x and draw the freebody diagram of the part of the beam to the left of the cut (including the part of the distributed load that acts on that part), showing the internal forces and moment. y

48 kN/m

M 48 kN/m

864 kN-m

288 kN

Now that this part has been isolated, we can represent the distributed load by an equivalent force:

Solution: (a)

(48 kN/m)x

We draw the free-body diagram of the entire beam, representing the distributed load by its equivalent force: y

M 864 kN-m

(6 m)(48 kN/m)

1 x 2

288 kN

Writing the equilibrium equations 1 (6 m) 2

ΣFx = P = 0, ΣFy = −V + 288 kN − (48 kN/m) x = 0,

From the equilibrium equations

ΣM right end = −x (288 kN) +

ΣFx = A x = 0,

+ 864 kN-m + M = 0,

ΣFy = A y − (6 m)(48 kN/m) = 0, 1 ΣM point A = M A −  (6 m)  [(6 m)(48 kN/m)] = 0, 2  we obtain the reactions A x = 0, A y = 288 kN, M A = 864 kN-m.

( 12 x )(48 kN/m)x

we obtain P = 0, V = 288 − 48 x kN, M = −864 + 288 x − 24 x 2 kN-m. (b) Observe that dM = 288 − 48 x = V . dx (a) P = 0, V = 288 − 48 x kN, M = −864 + 288 x − 24 x 2 kN-m.

742

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Problem 10.25 Draw the shear force and bending moment diagrams for the beam.

Now that this part has been isolated, we can represent the distributed load by an equivalent force: y

(48 kN/m)x M

48 kN/m

864 kN-m

Solution:

We draw the free-body diagram of the entire beam, representing the distributed load by its equivalent force:

ΣFx = P = 0, ΣFy = −V + 288 kN − (48 kN/m) x = 0, ΣM right end = −x (288 kN) +

(6 m)(48 kN/m)

1x 2 Writing the equilibrium equations

288 kN

( 12 x )(48 kN/m)x

+ 864 kN-m + M = 0, we obtain P = 0, V = 288 − 48 x kN, M = −864 + 288 x − 24 x 2 kN-m. The shear force and bending-moment diagrams are shown:

1 (6 m) 2

300 V, kN

From the equilibrium equations ΣFx = A x = 0, ΣFy = A y − (6 m)(48 kN/m) = 0,

100

1 ΣM point A = M A −  (6 m)  [(6 m)(48 kN/m)] = 0, 2 

we obtain the reactions A x = 0, A y = 288 kN, M A = 864 kN-m.

48 kN/m

3 x, m

0 M, kN-m

200

2200 2400 2600 2800

M 864 kN-m

288 kN

21000

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Solution:

Problem 10.26 Determine the shear force and bending moment as functions of x for 0 < x < 2 m.

Examine the whole beam first

ΣM B : − A(6 m) − 3600 N-m = 0 ⇒ A = −600 N ΣFy : A + B = 0 ⇒ B = 600 N

Now cut at arbitrary x < 2 m and examine the left cut ΣFy : A − V = 0

3600 N-m

ΣM cut : − Ax + M = 0

Solving V = −600 N, M = −(600 N) x for 0 < x < 2 m 2m

3600 N-m

A M

x V

Problem 10.27 Determine the shear force and bending moment as functions of x for 0 < x < 6 ft.

Now we draw the free-body diagram of the portion of the beam to the left of a position x in the range 0 < x < 6 ft and then represent the distributed load by an equivalent force: x (200 lb/ft) 12 ft

y 200 lb/ft 400 lb

V x

6 ft

Solution:

M The free-body diagram of the entire beam is

1 x x (200 lb/ft) 2 12 ft P 400 lb

1 (12 ft)(200 lb/ft) 2

2x 3

x Ay 8 ft

From the equilibrium equations ΣFx = A x = 0, 1 ΣFy = A y + B − (12 ft)(200 lb/ft) = 0, 2 1 ΣM point A = (6 ft) B − (8 ft)  (12 ft)(200 lb/ft)  = 0, 2 

From the equilibrium equations

B 6 ft

ΣFx = P = 0, 1 x2 (200 lb/ft) = 0, 2 (12 ft)  1  1 x2 ΣM right end = M + x (400 lb) + x  (200 lb/ft)  = 0,  3  2 (12 ft) ΣFy = −V − 400 lb −

( )

we obtain P = 0, V = −400 −

25 2 25 3 x lb, M = −400 x − x ft-lb. 3 9

P = 0, V = −400 − (25/3) x 2 lb, M = −400 x − (25/9) x 3 ft-lb.

we obtain the reactions A x = 0, A y = −400 lb, B = 1600 lb.

744

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Problem 10.28 Determine the shear force and bending moment as functions of x for 6 < x < 12 ft.

Now we draw the free-body diagram of the portion of the beam to the left of a position x in the range 6 < x < 12 ft and then represent the distributed load by an equivalent force:

y 200 lb/ft

x (200 lb/ft) 12 ft

x 6 ft

6 ft

Solution:

6 ft

The free-body diagram of the entire beam is

1600 lb

400 lb

1 x x (200 lb/ft) 2 12 ft

2x 3

x B

1 (12 ft)(200 lb/ft) 2

400 lb

1600 lb

6 ft

6 ft 8 ft From the equilibrium equations ΣFx = A x = 0, 1 ΣFy = A y + B − (12 ft)(200 lb/ft) = 0, 2 1 ΣM point A = (6 ft) B − (8 ft)  (12 ft)(200 lb/ft)  = 0, 2  we obtain the reactions A x = 0, A y = −400 lb, B = 1600 lb.

From the equilibrium equations ΣFx = P = 0, ΣFy = −V − 400 lb + 1600 lb −

1 x2 (200 lb/ft) = 0, 2 (12 ft)

ΣM right end = M + x (400 lb) − ( x − 6 ft)(1600 lb) +

( 13 x )  12 (12x ft) (200 lb/ft)  = 0, 2

we obtain P = 0, V = 1200 −

25 2 25 3 x lb, M = −9600 + 1200 x − x ft-lb. 3 9

P = 0, V = 1200 − (25/3) x 2 lb, M = −9600 + 1200 x − (25/9) x 3 ft-lb.

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Problem 10.29 The loads F = 200 N and C = 800 N-m. (a) Determine the internal forces and moment as functions of x. (b) Draw the shear force and bending moment diagrams. F

ΣM x = M 3 ( x ) + C − R y x − B( x − 8) = 0, from which M 3 ( x ) = −800 − 100 x + 300 x − 2400 = 200 x − 3200 N m. The axial forces are zero, P( x ) = 0 in all intervals.

from which, M 2 ( x ) = −100 x − 800 Nm. Interval 3: The shear is V3 ( x ) = RY + B = −100 + 300 = 200 N. The sum of the moments is

(b) The diagrams are shown.

Shear & Moment Diagram 500

Solution:

250

The reactions at the supports: The sum of the moments about the left end:

Shear Force

ΣM = C + 8B − 16 F = 0,

2250

from which

Bending Moment

2500

B = ( 18 ) (−C + 16 F ) = ( 18 ) (−800 + 16 (200)) = 300 N.

2750

The sum of the forces: ΣFy = R y + B − F = 0,

21000

from which R y = −100 N. The intervals as free bodies: Divide the interval into three parts: 0 ≤ x ≤ 4, 4 < x ≤ 8, and 8 < x ≤ 16. Interval I: The shear is V1 ( x ) = R y = −100 N. The moment is M 1 ( x ) = + R y x = −100 x N m. Interval 2: The shear is V2 ( x ) = R y = −100 N. The sum of the moments is

21250 21500 21750 22000

ΣM x = M 2 ( x ) + C − R y x = 0,

Problem 10.30 The beam will safely support shear forces and bending moments of magnitudes 2 kN and 6.5 kN-m, respectively. On the basis of this criterion, can it safely be subjected to the loads F = 1 kN, C = 1.6 kN-m?

8 x, m

Solution:

From the solution to Problem 10.29, the shear and the moments in the intervals are Interval 1 : V1 = R y , M 1 ( x ) = R y x, Interval 2 : V2 ( x ) = R y , M 2 ( x ) = R y x + C , Interval 3 : V3 ( x ) = R y + B, M 3 ( x ) = ( R y + B) x + C − 8B. The reactions are B = ( 18 ) (16 F − C ),

The maximum shears in each interval have the magnitude rank: x

and R y = F − B.

V1 ( x ) = V2 ( x ) ≤ V3 ( x ) . so that the largest shear for a force F = 1 kN is V3 ( x ) = R y + B = F − B + B = F = 1 kN, which can be safely supported. The maximum moment magnitudes in each interval have the rank: M 1 ( x ) ≤ M 2 ( x ) ≤ M 3 ( x ) . The maximum moment magnitude occurs in the third interval:

B 4m

M 3 ( x ) = ( R y + B) x + C − 8B = Fx + C − (16 F − C ) = Fx − 16 F + 2C . The maximum magnitude occurs at x = 8, M 3 (8) = 8F = 8 kN m and it exceeds the safe limit by 2.5 kN m. NO

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Problem 10.31 Model the ladder rung as a simply supported (pin-supported) beam and assume that the 750-N load exerted by the person’s shoe is uniformly distributed. Draw the shear force and bending moment diagrams.

7500 N/m M P V

0.2 m

250 N

7500 (x20.2) N M P x 200 mm

1– (x20.2) 2

250 N

100 mm 375 mm

V = 250 − 7500( x − 0.2) N, M = 250 x − 12 7500( x − 0.2) 2 N-m. 0.3 < x < 0.375 m

Solution:

See the solution of Problem 10.7. The free-body diagram of the rung is 7500 N/m

P M

500 N

x 250 N

0.1 m

0.2 m

500 N

0.375 m

(0.3752x) V = −500 N, M = 500(0.375 − x ) N-m. 500

0 < x < 0.2 m

P x 250 N

V, N

M 0

V 2500

V = 250 N, M = 250 x N-m. 0.2 < x < 0.3 m M, N–m BandF_6e_ISM_CH10.indd 747

0.1

0.2 x, m

0.3

0.4

0.1

0.2 x, m

0.3

0.4

60 40 20 0

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Problem 10.32 What is the maximum bending moment in the ladder rung in Problem 10.31 and where does it occur? y

Solution:

See the solution of Problem 10.31. The maximum moment occurs in the interval 0.2 < x < 0.3 m, in which M = 250 x − 3750( x − 0.2) 2 N-m. Setting

dM = 250 − 7500( x − 0.2) = 0, dx

we find that the maximum moment occurs at x = 0.233 m. Substituting this value into the expression for M gives M = 54.2 N-m.

x 200 mm

100 mm 375 mm

Problem 10.33 Assume that the surface the beam rests on exerts a uniformly distributed load. Draw the shear force and bending moment diagrams.

The shear and moment diagrams are shown. 4 kN

2 kN

y 4 kN

1 kN/m

2 kN

x 2m

M1(x) P1(x)

V1(x)

x 6 = 1 kN/m. 6 The intervals as free bodies: Divide the beam into three intervals:

Solution:

4 kN

The load density is w =

0 ≤ x < 2 (m), 2 ≤ x < 5 (m),

V2(x)

and 5 ≤ x ≤ 6 (m).

M2(x) P2(x)

4 kN

2 kN

Interval 1: The shear force is x

V1 ( x ) = ∫ w dx = x kN.

M3(x) P3(x) V3(x)

The force to the left is x

F1 ( x ) = ∫ w dx = x kN. 0

The centroid distance from x is d 1 =

x . 2

x2 M 1 ( x ) = F1 ( x )d 1 = kN m. 2 Interval 2: The shear force is V2 ( x ) = V1 ( x ) − 4 = x − 4 kN. The moment is M 2 ( x ) = M 1 ( x ) − 4( x − 2) =

Shear & Moment Diagrams Moment

kN, kN-m

The moment is

2.5 2 1.5 1 .5 0 2.5 21 21.5 22 22.5

x2 − 4 x + 8 kN m. 2

Interval 3: The shear force is

Shear 0

3 4 X, m

V3 ( x ) = x − 4 − 2 = x − 6 kN. The bending moment is M 3 ( x ) = M 2 ( x ) − 2( x − 5) =

748

x2 − 6 x + 18 kN. 2

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Problem 10.34 Determine the shear force and bending moment as functions of x for 0 < x < 3 ft.

From the equilibrium equations ΣFx = A x = 0, 1 ΣFy = A y − (3 ft)(600 lb/ft) = 0, 2 1 ΣM point A = −(4 ft)  (3 ft)(600 lb/ft)  + M A = 0, 2 

600 lb/ft

we obtain the reactions A x = 0, A y = 900 lb, M A = 3600 ft-lb.

Now we draw the free-body diagram of the portion of the beam to the left of a position x in the range 0 < x < 3 ft: 3 ft

3 ft

y 3600 ft-lb

Solution:

We draw the free-body diagram of the entire beam, representing the distributed load by its equivalent force:

M 900 lb

V x

1 (3 ft)(600 lb/ft) 2

From the equilibrium equations

x Ay

ΣFx = P = 0, ΣFy = −V + 900 lb = 0, ΣM right end = M − x (900 lb) + 3600 ft-lb = 0,

4 ft

we obtain P = 0, V = 900 lb, M = −3600 + 900 x ft-lb. P = 0, V = 900 lb, M = −3600 + 900 x ft-lb.

Problem 10.35 Determine the shear force and bending moment as functions of x for 3 < x < 6 ft. y

From the equilibrium equations ΣFx = −P = 0, ΣFy = V −

ΣM left end = −M −

600 lb/ft x

(

)

1 6 ft − x (6 ft − x ) 600 lb/ft = 0, 2 3 ft

(

)

1 1 6 ft − x (6 ft − x )  (6 ft − x) 600 lb/ft  = 0, 2  3 3 ft

we obtain P = 0, V = 100(6 − x ) 2 lb, M = −

100 (6 − x ) 3 ft-lb. 3

P = 0, V = 100(6 − x ) 2 lb, M = −33.3(6 − x ) 3 ft-lb. 3 ft

3 ft

Solution:

Let us draw the free-body diagram of the portion of the beam to the right of a position x in the range 3 < x < 6 ft, then represent the distributed load on that part by its equivalent force: y

x 600 lb/ft 16 ft3 2 ft 2 M P

x V

6 ft 2 x 1 (6 ft 2 x) 6 ft 2 x 600 lb/ft 2 3 ft

M P

x V

1 (6 ft 2 x) 3

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Problem 10.36 Determine the shear force V and bending moment M for the beam as functions of x for 0 < x < 3 ft.

Solution: From the free-body diagram of the entire beam (Fig. a), we obtain the equilibrium equations ΣFx : A x = 0, ΣFy : A y − 900 lb + 900 lb = 0,

ΣM A : M A − (900 lb)(2 ft) + (900 lb)(4 ft) = 0. We see that

600 lb/ft x 600 lb/ft 3 ft

3 ft

A x = A y = 0, M A = −1800 ft-lb. Cutting the beam at an arbitrary position x in the range 0 < x < 3 ft, we obtain a free-body diagram of the part of the beam to the left of x (Fig. b). The distributed load is replaced by the equivalent force R =

(

)

1  600 lb/ft  x  x = (100 lb/ft 2 ) x 2 .  2  3 ft

From the equilibrium equations we have ΣFy : −(100 lb/ft 2 ) x 2 − V ( x ) = 0, ΣM right end : −(1800 lb-ft) + (100 lb/ft 2 ) x 2 ( x /3) + M ( x ) = 0. We obtain V ( x ) = −(100 lb/ft 2 ) x 2 , M ( x ) = −(33.3 lb/ft 2 ) x 3 + (1800 lb-ft).

750

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Problem 10.37 Draw the shear force and bending moment diagrams for the beam.

The resulting diagrams are shown.

y 600 lb/ft x 600 lb/ft 3 ft

3 ft

Solution:

In the solution to Problem 10.36 we found that in the range 0 < x < 3 ft, we have

V ( x ) = −(100 lb/ft 2 ) x 2

600 lb/ft

M ( x ) = −(33.3 lb/ft 2 ) x 3

+ (1800 lb-ft). Cutting the beam at an arbitrary position x in the range 3 ft < x < 6 ft, and isolating the right part (Fig. a), we have the free-body diagram shown. The distributed load is replaced by the equivalent force R =

(

1 lb 600 2  ft

)( 6 ft3 ft− x )  (6 ft − x) = (100 lb/ft )(6 ft − x) 2

600 lb/ft 3 ft

3 ft

From the equilibrium equations we learn ΣFy : V ( x ) + (100 lb/ft 2 )(6 ft − x ) 2 = 0, ΣM left end : −M ( x ) + [(100 lb/ft 2 )(6 ft − x ) 2 ][ 13 (6 ft − x )] = 0. Thus V ( x ) = −(100 lb/ft 2 )(6 ft − x ) 2 M ( x ) = (33.3 lb/ft 2 )(6 ft − x ) 3 .

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Problem 10.38 In preliminary design studies, the vertical forces on an airplane’s wing are modeled as shown. The distributed load models aerodynamic forces and the force exerted by the wing’s weight. The 80-kN force at x = 4.4 m models the force exerted by the weight of the engine. Draw the shear force and bending moment diagrams for the wing for 0 < x < 4.4 m.

The resulting diagrams are shown.

50 kN/m x 80 kN 4.4 m

13.0 m

Solution: From the free-body diagram of the entire wing (Fig. a), we obtain the equilibrium equations

V 4.4 m

ΣFx : A x = 0,

ΣFy : A y + (220 + 325 − 80) kN = 0, ΣM A : M A + (220 kN)(2.2 m) − (80 kN)(4.4 m)

245 kN

+ (325 kN)(17.4 − 8.67) m = 0. Solving yields A x = 0, A y = −465 kN, M A = −2970 kN-m. Cutting the wing at an arbitrary position x in the range 0 < x < 4 m, and representing the distributed load by an equivalent force (Fig. b), the equilibrium equations are ΣFy : −465 kN + (50 kN/m) x − V ( x ) = 0,

465 kN M 2970 kN-m

1408 kN-m

ΣM right end : −2970 kN-m + (465 kN) x − [(50 kN/m) x ]( x /2) + M ( x ) = 0. Therefore

4.4 m

V ( x ) = (50 kN/m) x − 465 kN, M ( x ) = (25 kN/m) x 2 − (465 kN) x + 2970 kN-m.

752

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Problem 10.39 Draw the shear force and bending moment diagrams for the entire wing.

Solution:

The shear force and bending moment diagrams for 0 < x < 4.4 m were obtained in the solution to Problem 10.38. Cut the wing at an arbitrary position x in the range 4.4 m < x < 17.4 m and isolate the right part of the beam (Fig. a). The distributed loading is represented by the equivalent force R = =

(

1  17.4 m − x 2  13.0 m

)( 50 kN )  (17.4 m − x) m 

25 kN ( 13 )(17.4 m − x) m 2

Using the equilibrium equations we find 25 kN ( 13 )(17.4 m − x) , m

V ( x) = −

(

)

1 25 kN M ( x ) = R  ( 17.4 m − x )  = (17.4 m − x ) 3 3  39 m 2

50 kN/m x

The diagrams for the entire beam are shown.

80 kN 4.4 m

13.0 m

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Problem 10.40 Draw the shear force and bending moment diagrams.

Solution:

− (24 kN)(3 m) − (12 kN)(8 m) + B(6 m) = 0

y 20 kN-m

ΣFy : A + B − 6 kN − 24 kN − 12 kN = 0

4 kN/m

6 kN

Start with the reactions

ΣM A : (6 kN)(6 m) + 20 kN-m

Thus A = 23.3 kN, B = 18.67 kN 24 kN

12 kN

20 kN-m

6 kN

The complete diagrams: y 4kN/m

20 kN-m

x 23.3 kN

6kN

20 kN

18.7 kN

10 kN x

0 210kN

M 0

220 kN-m 240 kN-m 260 kN-m

754

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Problem 10.41 Draw the shear force and bending moment diagrams.

Now we write Eq. (10.4) as dV = −w dx = 0 and integrate, obtaining

∫80 dV = 0

80 lb

[ V ]V80 = 0

V = 80 lb. Writing Eq. (10.6) as

dM = V dx = 80 dx and integrating, we obtain

Solution:

The free-body diagram of the entire beam is y

[ M ]−M640 = 80 x

80 lb

∫−640 dM = ∫0 80 dx M = −640 + 80 x ft-lb.

We find that V = 80 lb and M = −640 + 80 x ft-lb. The shear and bending moment diagrams are shown:

Ay 8 ft

100

From the equilibrium equations V, lb

ΣFx = A x = 0, ΣFy = A y − 80 lb = 0,

ΣM point A = −(8 ft)(80 lb) + M A = 0, 0

we obtain the reactions A x = 0, A y = 80 lb, M A = 640 ft-lb.

y 640 ft-lb

4 x, ft

0 2200 2400 2600

P 80 lb

200 M, ft-lb

Before we begin integrating Eqs. (10.4) and (10.6), what are the values of the shear force and bending moment at the left end of the beam? To determine this, we can draw a free-body diagram of an infinitesimal sliver of the left end:

2800

We see that at the left end of the beam, V = 80 lb and M = −640 ft-lb.

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Problem 10.42 Draw the shear force and bending moment diagrams.

Solution:

We must first find the reactions

ΣM B : − A(6 m) − 3600 N-m = 0 ΣFy : A + B = 0

A = −600 N, B = 600 N In the first region 0 < x < 2 m

3600 N-m x

w1 = 0 V1 = A = −600 N M 1 = −(600 N) x

In the second region 2 m < x < 6 m w2 = 0 V2 = −B = −600 N M 2 = −(600 N) x + 3600 N-m 3600 Nm

4m B

A y 3600 N-m x 600 N

600 N

2600 N

M 2400 N-m

21200 N-m

756

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Problem 10.43 This arrangement is used to subject a segment of a beam to a uniform bending moment. Draw the shear force and bending moment diagrams.

Solution:

We first find the reactions

ΣM B : − A(24 in) + (50 lb)(6 in) + (50 lb)(18 in) = 0 ΣFy : A + B − 50 lb − 50 lb = 0 A = B = 50 lb

In the first regions 0 < x < 6 in w1 = 0 V1 = A = 50 lb x

M 1 = (50 lb) x In the second region 6 in < x < 18 in

50 lb

w2 = 0 V2 = 0 M 2 = 300 in lb

6 in

12 in

6 in

In the last region 18 in < x < 24 in w3 = 0 V3 = −B = −50 lb M 3 = −(50 lb) x + 1200 in lb

50 lb

x 50 lb

50 lb

V 50 lb 0

250 lb M 300 in-lb 0

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Problem 10.44 Use the procedure described in Example 10.5 to draw the shear force and bending moment diagrams for the beam.

Solution: We can use the boundary conditions at the right end instead of calculating the reactions at the left end w = 4

kN m

( kN ) x + 24 kN m

V = − 4 4 kN/m

( kN ) x + (24 kN)x − 72 kN-m m

M = − 2

x y

4 kN/m x

72 kN-m 24 kN V

24 kN

M 0

272 kN-m

758

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Problem 10.45 Draw the shear force and bending moment diagrams. y

Solution:

Writing Eq. (10.4) as

(

dV = −w dx = −3 1 −

)

x2 dx 25

and integrating, we obtain

∫0 dV = ∫0 −3(1 − 25 ) dx V

w 5 3(12 x 2/25) kN/m

x 5m

x3  x [ V ]V0 =  −3 x −  75  0 x3 V = −3 x − . 75

( (

) )

Writing Eq. (10.6) as

(

dM = V dx = −3 x −

)

x3 dx 75

and integrating, we obtain

∫0 dM = ∫0 −3( x − 75 ) dx M

x2 x4 x [ M ]0M =  −3 −  2 300  0 x2 x4 − M = −3 . 2 300

( (

) )

The shear force and bending moment distributions are

(

V = −3 x −

)

(

)

x3 x2 x4 , M = −3 − . 75 2 300

The shear and bending moment diagrams are shown: 5 V, KN

0 25 210 215

0.5

1.5

2.5 x, m

3.5

4.5

0.5

1.5

2.5 x, m

3.5

4.5

M, KN-m

20 0 220 240

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Problem 10.46 Draw the shear force and bending moment diagrams.

300 lb

y 100 lb/ft x

6 ft

Solution:

6 ft

The plots y

Find the reactions first

ΣM A : −300 lb(10 ft) + B(12 ft) = 0

100 lb/ft

ΣFy : A + B − 300 lb = 0

A = 50 lb, B = 250 lb

50 lb

For 0 < x < 6 ft w = 0

250 lb

V 50 lb 0

dV = −w = 0 ⇒ V = 50 lb dx

dM = V = 50 lb ⇒ M = (50 lb) x dx For 6 ft < x < 12 ft w =

lb ( 100 ) ( x − 6 ft) = ( 503 ftlb ) ( x − 6 ft) 6 ft

(

)

50 lb ( x − 6 ft) 2 V = − + 50 lb 3 ft 2

( 253 ftlb )(x − 6 ft) + 50 lb

= −

2250 lb M 400 ft-lb 200 ft-lb

( 253 ftlb ) (x − 36 ft) + (50 lb)x

M = −

( 259 ftlb )(x − 6 ft) + (50 lb)x

= −

760

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Problem 10.47 Draw the shear force and bending moment diagrams.

We see that at x = 3 ft, V = 900 lb and M = −900 ft-lb. For 3 < x < 6 ft, we can express the distributed load as a linear equation w = cx + d. Using the conditions that w = 600 lb/ft at x = 3 ft and w = 0 at x = 6 ft to evaluate the constants c and d, we obtain the equation w = −200 x + 1200 lb/ft. Writing Eq. (10.4) as

dV = −w dx = (200 x − 1200) dx

600 lb/ft

and integrating, we obtain

∫900 dV = ∫3 (200 x − 1200) dx 3 ft

3 ft

[ V ]V900 = [ 100 x 2 − 1200 x ]3x V − 900 = 100 x 2 − 1200 x − 100(3) 2 + 1200(3) V = 100 x 2 − 1200 x + 3600 lb.

Solution:

Writing Eq. (10.6) as

We draw the free-body diagram of the entire beam, representing the distributed load by its equivalent force:

dM = V dx = (100 x 2 − 1200 x + 3600) dx

and integrating, we obtain 1 (3 ft)(600 lb/ft) 2

MA Ax

x Ay 4 ft

∫−900 dM = ∫3 (100 x 2 − 1200 x + 3600) dx x

100 x 3 [ M ]−M900 =  − 600 x 2 + 3600 x   3 3 M + 900 =

+ 600(3) 2 − 3600(3)

From the equilibrium equations M =

ΣFx = A x = 0, ΣFy = A y −

1 (3 ft)(600 lb/ft) = 0, 2

1000

V, lb

For 0 < x < 3 ft, we write Eq. (10.4) as

100 x 3 − 600 x 2 + 3600 x − 7200 ft-lb. 3

The shear and bending moment diagrams are shown:

1 ΣM point A = −(4 ft)  (3 ft)(600 lb/ft)  + M A = 0, 2  we obtain the reactions A x = 0, A y = 900 lb, M A = 3600 ft-lb.

100(3) 3 100 x 3 − 600 x 2 + 3600 x − 3 3

500

dV = −w dx = 0 0

and integrate, obtaining

3 x, ft

∫900 dV = 0

= 0

21000

V = 900 lb. Writing Eq. (10.6) as dM = V dx = 900 dx and integrating, we obtain M

M, ft-lb

[ V ]V900

22000 23000 24000

∫−3600 dM = ∫0 900 dx [ M ]−M3600 = 900 x M = −3600 + 900 x ft-lb.

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Problem 10.48 Draw the shear force and bending moment diagrams.

At x = 6 ft, these expressions give V = −700 lb and M = −3000 ft-lb. The reaction due to the support at x = 6 ft will cause a jump in the value of the shear force to V = −700 + 1600 = 900 lb. For 3 < x < 6 ft, we write Eq. (10.4) as

y 200 lb/ft

( 200 ) x dx 12

dV = −w dx = −

and integrate, obtaining

∫900 dV = ∫6 −( 12 ) x dx V

6 ft

200

(

)

100 2  [ V ]V900 =  − x   6 12

Solution: y

( 100 ) x + ( 100 )(6) 12 12

The free-body diagram of the entire beam is

V − 900 = −

1 (12 ft)(200 lb/ft) 2

V = 1200 −

x B

Ay 6 ft

( 253 ) x  dx 2

∫−3000 dM = ∫6  1200 − ( 3 ) x 2  dx M





25 3  x x  [ M ]−M3000 =  1200 x −  6 9

( )

1 (12 ft)(200 lb/ft) = 0, 2 1 ΣM point A = (6 ft) B − (8 ft)  (12 ft)(200 lb/ft)  = 0, 2  ΣFy = A y + B −

we obtain the reactions A x = 0, A y = −400 lb, B = 1600 lb. From these results we see that at x = 0, the shear force V = −400 lb and the bending moment M = 0.

M + 3000 = 1200 x −

( 259 ) x ft-lb. 3

The shear and bending moment diagrams are shown: 1000 500 V, lb

( 200 ) x dx 12

dV = −w dx = −

( 259 ) x − 1200(6) + ( 259 )(6)

M = −9600 + 1200 x −

The distributed load on the beam is described by the linear function w = (200/12) x lb/ft. For 0 < x < 6 ft, we write Eq. (10.4) as

and integrate, obtaining

2500

∫−400 dV = ∫0 −( 12 ) x dx

21000

200

100 2  x [ V ]V−400 =  − x   0 12

)

6 x, ft

0 M, ft-lb

V = −400 −

and integrating, we obtain

ΣFx = A x = 0,

(

( 253 ) x lb.

dM = V dx =  1200 − 

From the equilibrium equations

Writing Eq. (10.6) as

8 ft

( 253 ) x . 2

Writing Eq. (10.6) as

21000 22000

( )

25 2  dM = V dx = −  400 + x  dx   3 and integrating, we obtain

23000

∫0 dM = ∫0 −  400 + ( 3 ) x 2  dx M





25 3  x x  [ M ]0M = −  400 x +  0 9

( )

M = −400 x −

762

( 259 ) x ft-lb. 3

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Problem 10.49 Draw the shear force and bending moment diagrams for the beam AB.

400 N/m B

Solution:

w1 = 400 N/m V1 = −(400 N/m) x

Using free-body diagram, ACD

M 1 = −(200 N/m) x 2

1 TC = 0 5

In the second region 2 m < x < 1 m

ΣM A : −(1200 N)(1.5 m) − ΣFy : A y − 1200 N −

In the first region 0 < x < 2 m

First we must find the reactions

ΣM B : − A x (2 m) − (1200 N)(1.5 m)

ΣFx : A x −

( 25 T )(2 m) − T (3 m) = 0 C

2 TC − T D = 0 5

w 2 = 400 N/m V2 = −(400 N/m)( x − 2 m) + 1800 N M 2 = −(200 N/m)( x − 2 m) 2 + (1800 N)( x − 2 m) − 1600 N-m The plots

We find A x = −900 N, A y = 0

TC = −2012 N, T D = 600 N Now we are ready to construct the diagrams.

400 N/m

1200 N

900 N A 1800 N

900 N

B 600 N

V Ay

1000 N 600 N

2 1

800 N

1200 N

800 N-m C

Ax Ay TC

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Problem 10.50 The cable supports a distributed load w = 12, 000 lb/ft. Using the approach described in Practice Example 10.6, determine the maximum tension in the cable.

Solution:

Equation (10.10) must be satisfied for both attachment

points: y L = 40 ft =

1 1 ax 2 , y R = 90 ft = ax R 2 . 2 L 2

Dividing the second equation by the first yields

xR2 = 2.25. xL 2

The horizontal span of the bridge is x R − x L = 100 ft. Solving these two equations yields x L = −40 ft and x R = 60 ft. 90 ft

Substituting the coordinates of the right attachment point into Eq. (10.10), yR =

40 ft

1 1 ax 2 ⇒ 90 ft = a(60 ft) 2 ⇒ a = 0.05 ft −1 . 2 R 2

Therefore the tension at the lowest point is w

T0 =

w 12, 000 lb/ft = = 240, 000 lb. a 0.05 ft −1

The maximum tension in the cable occurs at its right end. From Eq. (10.11),

100 ft

T = T0 1 + a 2 x 2 = (240 kip) 1 + (0.05 ft −1 ) 2 (60 ft) 2 = 759 kip. 759 kip.

Problem 10.51 The cable is subjected to a load w = 80 lb/ft that is uniformly distributed horizontally. What are the tensions in the cable at the left and right attachment points?

Dividing the first equation by the second one gives yR x 2 = R2 : yL xL x 2 40 ft = R2 . xL 20 ft We know the cable’s span, so

40 ft

x R − x L = 100 ft. (Notice that in writing this equation we have accounted for the fact that x L is negative.)

20 ft

Solving these two equations yields x L = −41.4 ft, x R = 58.6 ft. From Eq. (10.10) for the right attachment point, w 100 ft

1 a(58.6 ft) 2 , 2 the constant a = 0.0233 ft −1 .

We introduce a coordinate system with its origin at the cable’s lowest point:

The tension at the cable’s lowest point is T0 =

xR , yR

= xL , yL

1 ax 2 : 2 R

40 ft =

Solution:

yR =

w a 80 lb/ft 0.0233 ft −1

= 3430 lb.

40 ft

From Eq. (10.11), the tension at the left attachment point is

20 ft x

TL = T0 1 + a 2 x L 2 = 4770 lb,

100 ft

and the tension at the right attachment point is The coordinates of the left and right attachment points are shown. We write Eq. (10.10) for these two points: yR =

764

1 ax 2 , 2 R

yL =

1 ax 2 . 2 L

TR = T0 1 + a 2 x R 2 = 5810 lb. TL = 4770 lb, TR = 5810 lb.

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Problem 10.52 A cable is used to suspend a pipeline above a river. The towers supporting the cable are 36 m apart. The lowest point of the cable is 1.4 m below the tops of the towers. The mass of the suspended pipe is 2700 kg. (a) What is the maximum tension in the cable? (b) What is the suspending cable’s length?

Solution:

The distributed load is

(2700 kg)(9 : 81 m/s 2 ) = 736 N/m. 36 m

w =

y 1.4 m x 18 m Setting x = 18 m, y = 1.4 m in Eq. (10.10),

(a)

1.4 = 12 a(18) 2 , we obtain a =

w = 0.00864 m −1 . T0

Therefore the tension at x = 0 is T0 =

w 736 = = 85,100 N. a 0.00864

From Eq. (10.11), the maximum tension is T = T0 1 + a 2 (18) 2 = 86, 200 N. (b) Setting x = 18 m in Eq. (10.12), the length of the cable is 2s = 18 1 + a 2 (18) 2 +

1 ln(18a + a

1 + a 2 (18) 2 )

= 36.14 m.

Problem 10.53 A cable suspension supports a pipeline above a river. The towers supporting the cable are 200 ft apart. The suspended pipe exerts a uniformly distributed load of 130 lb/ft on the cable. Suppose that the lowest point of the cable is a distance h below the tops of the towers. If the cable will safely support a tension of 100,000 lb, what is the minimum safe value of h?

Solution:

In terms of a coordinate system with its origin at the cable’s lowest point, the coordinates of the right attachment point are x R = 100 ft, y R = h. Solving Eq. (10.10), yR = h =

1 ax 2 , 2 R

for the constant a yields 2h . xR 2

a =

From the definition of a, the tension at the lowest point of the cable is wx R 2 w = . a 2h

T0 =

We substitute these expressions for a and T0 into Eq. (10.11) to obtain the tension at the right attachment point in terms of h: TR = T0 1 + a 2 x R 2 =

 2h  2 wx R 2 1 +  2  x R 2  xR  2h

wx R 2 4h 2 1+ . 2h xR 2

By squaring both sides of this equation, we can solve for h, obtaining h =

xR . TR 2 2 1 − w 2xR 2

Substituting x R = 100 ft, w = 130 lb/ft, and TR = 100, 000 lb, we obtain h = 6.56 ft. h = 6.56 ft.

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Problem 10.54 The cable supports a uniformly distributed load w = 750 N/m. The lowest point of the cable is 0.18 m below the attachment points C and D. Determine the axial loads in the truss members AC and BC.

The free-body diagram of joint C is shown.

T PAC

0.4 m

0.4 m F

From the equations ΣFx = T cos θ − PAC cos 45 ° = 0, ΣFy = −T sin θ − PBC − PAC sin 45 ° = 0, we obtain

1.2 m

0.4 m

PBC

458

0.4 m

PAC = 1061 N, PBC = −1200 N.

Solution:

y = 12 ax 2:

0.18 = 12 a(0.6) 2 .

0.18 m x

From this equation we obtain a = 1 m −1 . Therefore w T0 = = 750 N a

0.6 m

and T = T0 1 + a 2 (0.6) 2 = 875 N. From the equation tan θ = ax = (1)(0.6), we obtain θ = 30.96 °.

Problem 10.55 The cable supports a railway bridge between two tunnels. The distributed load is w = 1 MN/m, and h = 40 m. (a) What is the maximum tension in the cable? (b) What is the length of the cable?

Solution: a = 2

y 40 = 2 2 = 0.06173. x2 36

The tension at the lowest point: T0 =

36 m

The parameter

w 1 × 10 6 = = 16200 kN. a a

The maximum tension: TMAX = T0 1 + a 2 x 2 , x = 36 m, TMAX = 39477 kN. The cable length is

36 m

(

s( x) = x 1 + a 2 x 2 +

1 ln ( ax + a

which, for

)

1 + a 2x 2 ) ,

which, for x = 36 m, L = 112.66 m.

766

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Problem 10.56 The cable in Problem 10.55 will safely support a tension of 40 MN. What is the shortest cable that can be used, and what is the corresponding value of h?

Solution: T0 =

The tension at the lowest point is

w . a

The maximum tension is TMAX = T0 1 + a 2 x 2 .

36 m

Square both sides, substitute and reduce algebraically: To 2 = TMAX 2 − w 2 x 2 . The terms on the right are known: TMAX 2 = 40 2 (10 6 ), and w 2 x 2 = 36 2 (10 6 ). Solve for the parameter a,

36 m

a2 =

10 6 = 3.29 × 10 −3 , (40 2 − 36 2 )(10 6 )

from which a = 0.0574. The height is 1 1 h = ax 2 = (0.0574)(36 2 ) = 37.165 m. 2 2 The length is

( )

(

( )

s( x) = x 1 + a 2 x 2 +

1 ln ( ax + a

)

1 + a 2x 2 ) ,

which, for x = 36 m, L = 108.26 m

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Problem 10.57* An oceanographic research ship tows an instrument package. Assume that the hydrodynamic drag force exerted on the cable is uniformly distributed in the vertical direction and equal to 30 N/m. The tensions in the cable at 1 and 2 are 3800 N and 6000 N, respectively. Determine the distance h.

From the definition of the parameter a in the equation for the cable’s shape, the cable tension at the origin would be T0 = w /a. Let us substitute this expression into the equations for the tensions at positions 1 and 2: T1 =

w 1 + a 2 x1 2 , a

(3)

T2 =

w 1 + a 2x2 2 . a

(4)

Solving Eq. (2) for x 2 and substituting the result into Eq. (4), it becomes

T2 =

2 100 m

w 1 + a 2 ( x 1 + b) 2 , a

(5)

where b = x 2 − x1 = 100 m. Now we solve Eq. (3) for x1 and substitute the result into Eq. (5), obtaining 2

T2 = 1

  T2 w 1 1 + a 2  1 2 − 2 + b  .   w a a

The only unknown quantity in this equation is a. We square both sides and write the resulting equation as

Solution:

Because of the orientation of the distributed load, we need to describe the cable in terms of a coordinate system with its x axis vertical: x

(6)

f (a) =

  T2 T2 2 a 2 1 − a 2  1 2 − 2 + b  − 1 = 0. 2   w w a

(7)

We can draw a graph of this equation as a function of a and use it to estimate the solution:

x2 , y2

0.15 0.1

x1 , y1

f (a)

0.05 0 20.05 20.1 y

20.15 20.2

The equation for the cable’s shape is y =

1 2 ax . 2

(1)

The origin of the coordinate system is the “bottom” of the parabola, so we don’t initially know the coordinates of the positions 1 and 2, although we do know that x 2 − x1 = 100 m.

768

(2)

9.2

9.4

9.6 a

9.8

10.2 31023

From the graph we estimate that a = 0.00945 m −1 . Using software designed to obtain roots of nonlinear algebraic equations, we obtain a = 0.00945946 m −1 . With a known, we can solve Eqs. (3) and (4) for x1 and x 2 , obtaining x1 = 69.8 m and x 2 = 170 m. Then we can solve Eq. (1) for y1 and y 2 , obtaining y1 = 23.0 m and y 2 = 136 m. The distance h = y 2 − y1 = 113 m. h = 113 m.

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Problem 10.58* An oceanographic research ship tows an instrument package. Assume that the hydrodynamic drag force exerted on the cable is uniformly distributed in the vertical direction and equal to 30 N/m. The tensions in the cable at 1 and 2 are 3800 N and 6000 N, respectively. Draw a graph of the curve described by the cable. h

Solving Eq. (2) for x 2 and substituting the result into Eq. (4), it becomes w 1 + a 2 ( x 1 + b) 2 , a

T2 =

(5)

where b = x 2 − x1 = 100 m. Now we solve Eq. (3) for x1 and substitute the result into Eq. (5), obtaining 2

  T2 w 1 1 + a 2  1 2 − 2 + b  .   w a a

T2 =

(6)

The only unknown quantity in this equation is a. We square both sides and write the resulting equation as

f (a) = 100 m

  T2 T22 a 2 1 − a 2  1 2 − 2 + b  − 1 = 0.   w w2 a

(7)

We can draw a graph of this equation as a function of a and use it to estimate the solution: 1

0.15

0.1

Solution:

Because of the orientation of the distributed load, we need to describe the cable in terms of a coordinate system with its x axis vertical:

0.05

x f (a)

x2 , y2

20.05 20.1 20.15 20.2

x1 , y1

9.2

9.4

9.6 a

9.8

10.2 31023

From the graph we estimate that a = 0.00945 m −1 . Using software designed to obtain roots of nonlinear algebraic equations, we obtain a = 0.00945946 m −1 .

With a known, we can solve Eqs. (3) and (4) for x1 and x 2 , obtaining x1 = 69.8 m and x 2 = 170 m. Then we can solve Eq. (1) for y1 and y 2 , obtaining y1 = 23.0 m and y 2 = 136 m. Now we can use Eq. (1) to plot the cable: 150 x2, y2 The equation for the cable’s shape is 1 2 ax . 2

(1)

The origin of the coordinate system is the “bottom” of the parabola, so we don’t initially know the coordinates of the positions 1 and 2, although we do know that x 2 − x1 = 100 m.

(2)

(3) (4)

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100 y, m

y =

50 x1, y1

80 100 x, m

120

140

160

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Problem 10.59 The footbridge is supported by two cables. Its span is 80 ft. Each cable is subjected to an average force per unit of its length of 8 lb/ft. The tension at the center of each cable is 650 lb. Determine (a) the maximum tension in each cable and (b) its length.

Solution: a = =

The parameter

w T0 8 lb/ft 650 lb

= 0.0123 ft −1 . (a)

The maximum tension occurs at the end of the cable, x = 40 ft. From Eq. (10.21), T = T0 cosh ax = (650 lb) cosh[(0.0123 ft −1 )(40 ft)] = 730 lb.

(b) From Eq. (10.22), one-half of the length of the cable is sinh ax a sinh[(0.0123 ft −1 )(40 ft)] = (0.0123 ft −1 ) = 41.6 ft.

s =

The total length is 2s = 82.3 ft.

Source: Courtesy of Esteban De Armas/Alamy Stock Photo.

Problem 10.60 The stationary balloon’s tether is horizontal at point O where it is attached to the truck. The mass per unit length of the tether is 0.45 kg/m. The tether exerts a 50-N horizontal force on the truck. The horizontal distance from point O to point A where the tether is attached to the balloon is 20 m. What is the height of point A relative to point O?

(a) 730 lb. (b) 83.3 ft.

Solution: a =

(9.81)(0.45) w = = 0.0883 m −1 . T0 50

From equation (10.20), 1 [cosh(ax ) − 1] a 1 {cosh[(0.0883)(20)] − 1} = 22.8 m. h = 0.0883 y =

A h

50 N

x 20 m

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Problem 10.61 In Problem 10.60, determine the magnitudes of the horizontal and vertical components of the force exerted on the balloon at A by the tether.

Solution:

From the solution to Problems 10.60, a = 0.0883 m −1 . The value of the tension at x = 20 m is T = T0 cosh(ax ) = 50 cosh[(0.0883)(20)] = 150 N. The slope at x = 20 m (Equation 10.19) is σ = tan θ = sinh(ax ) = sinh[(0.0883)(20)] = 2.84, so θ = arctan 2.84 = 70.6 °. The horizontal and vertical components are T x = (150) cos θ = 50 N T y = (150)sin θ = 142 N. 150 N

50 N 20 m

Problem 10.62 The mass per unit length of lines AB and BC is 2 kg/m. The tension at the lowest point of cable AB is 1.8 kN. The two lines exert equal horizontal forces at B. (a) Determine the sags h1 and h 2 . (b) Determine the maximum tensions in the two lines. A

Solution: (a)

The lines meet the condition for a catenary.

The line AB. The weight density is w = 2(9.81) = 19.62 N/m. The parameter a1 =

w 19.62 = = 0.0109. T AB 1800

The sag is 1 h1 =   (cosh(30 a1 ) − 1) = 4.949 m.  a1  60 m

40 m

The line BC. The horizontal component of the tension at B is T AB = 1.8 kN. Thus the tension at the lowest point in BC is 1.8 kN, and the parameter a for line BC is equal to a1 . The sag is 1 h 2 =   (cosh(20 a1 ) − 1) = 2.189 m.  a1  (b) The line AB. The maximum tension is AB TMAX = T AB cosh(30 a1 ) = 1897.1 N.

The line BC. The maximum tension is BC TMAX = T AB cosh(20 a1 ) = 1842.9 N.

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Problem 10.63* The segment of light string is subjected to an average weight per unit of its length of 0.0006 N/mm. Determine the tension in the segment at its lowest point.

Solution: In terms of a coordinate system with its origin at the string’s lowest point, y 440 mm

880 mm

140 mm

we write Eq. (10.20) for the right endpoint: y = =

1 {cosh ax − 1}: a 1 {cosh[a(440 mm)] − 1}. a

We can draw a graph of this equation to estimate the value of a for which y = 140 mm: 160 150

y, mm

140 130 120 110 100 1.2

1.25

1.3

1.35

1.4 1.45 a, 1/mm

1.5

1.55 1.6 31023

From the graph we estimate that y = 140 mm when a = 0.00140 mm −1 . Using software designed to solve nonlinear algebraic equations, we obtain a = 0.0014013 mm −1 . Using the definition of a, the tension at the lowest point is T0 = =

w a 0.0006 N/mm 0.00140 mm −1

= 0.428 N. T0 = 0.428 N.

772

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Problem 10.64 The boxes suspended at points B and C of the rope each weigh 8 lb. (a) What is the dimension h? (b) What is the tension in the segment AB?

Next, we cut the rope just to the left of point C:

20 in

10 in

12 in

20 in

10 in W

12 in

The sum of the moments about C is

TBC

ΣM point C = hTh − (30 in)Tv + (10 in)W = 0.

(2)

Finally, we cut the rope just to the left of point D:

Solution: (a)

Let us draw a free-body diagram by cutting the rope just to the left of point B:

TCD

Th A

12 in B 20 in

Th A

10 in W

C 10 in W

a The sum of the moments about D is

12 in B 20 in

ΣM point D = −(40 in)Tv + (20 in)W + (10 in)W = 0.

Solving Eqs. (1)-(3) yields Th = 10 lb, Tv = 6 lb, h = 10 in.

TAB

Here Th and Tv are the horizontal and vertical components of the tension at A. The sum of the moments about B is ΣM point B = (12 in)Th − (20 in)Tv = 0.

(3)

(1)

(b) Going back to the first free-body diagram, the angle α = atand(12 in/20 in) = 31.0 . Summing the horizontal forces, T AB cos α − Th = 0, Gives T AB = 11.7 lb. (a) h = 10 in. (b) T AB = 11.7 lb.

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Problem 10.65 Each lamp weighs 12 lb. (a) What is the length of the wire ABCD needed to suspend the lamps as shown? (b) What is the maximum tension in the wire?

Solution: ΣM B : −TV (12 in) + T H (12 in) = 0 ⇒ TV = T H Using the second free-body diagram shown, ΣM C : T H h − TV (30 in) + (12 lb)(18 in) = 0

12 in

18 in

Make a cut at the right attachment point and take moments ΣM D : T H (30 in) − TV (48 in) + (12 lb)(36 in) + (12 lb)(18 in) = 0

Solving together we find

12 in

T H = TV = 36 lb, h = 24 in

30 in

(a)

The length of the cable is then L =

12 2 + 12 2 +

18 2 + (24 − 12) 2 +

18 2 + (30 − 24) 2

= 57.6 in

C D

(b) The maximum tension occurs where the angle is the greatest (AB) Tmax = T AB =

36 2 + 36 2 = 50.9 lb TV

12 in TH 12 in

TBC

12 lb TV

12 in

18 in

TH 12 in

TBC

12 lb

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Problem 10.66 Two weights, W1 = W2 = 50 lb, are suspended from a cable. The vertical distance h1 = 4 ft. (a) Determine the vertical distance h 2 . (b) What is the maximum tension in the cable? 6 ft

10 ft

Finally, we cut the rope just to the left of the right attachment point:

6 ft

3 ft

2 ft

10 ft

3 ft

2 ft

The sum of the moments about A3 is

Solution: (a)

ΣM point A3 = (2 ft)Th − (19 ft)Tv + (13 ft)W1 + (3 ft)W2 = 0. (3)

We use the notation in Fig. 10.18. Let us draw a free-body diagram by cutting the cable just to the left of the weight W1: Tv

6 ft

h1 A1

Here Th and Tv are the horizontal and vertical components of the tension at the left attachment point. The sum of the moments about A1 is ΣM point A1 = h1Th − (6 ft)Tv = 0. (1)

We solve Eqs. (1)-(3) with W1 = W2 = 50 lb and h1 = 4 ft, obtaining Th = 75 lb, Tv = 50 lb, h 2 = 4 ft. (b) The tension in segment 1 is T1 = Tv 2 + Th 2 = 90.1 lb. We can see from the free-body diagrams that the horizontal component of the tension in each segment of the cable equals Th . That implies that the maximum tension occurs in the segment with the greatest slope. The angle between segment 1 and the horizontal is α1 = arctan(h1 /6 ft) = 33.7 . The angle between segment 3 and the horizontal is α 3 = arctan[(h 2 − 2 ft)/3 ft] = 33.7 , so it has the same slope, and tension, as segment 1. The maximum tension is 90.1 lb. (a) h 2 = 4 ft, (b) 90.1 lb.

Next, we cut the rope just to the left of weight W2 : Tv

6 ft

10 ft

Th h1

h2 T2

The sum of the moments about A2 is ΣM point A2 = h 2Th − (16 ft)Tv + (10 ft)W1 = 0. (2)

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Problem 10.67 The weights are W1 = 50 lb and W2 = 100 lb, and the vertical distance h1 = 4 ft. (a) Determine the vertical distance h 2 . (b) What is the maximum tension in the cable? 6 ft

10 ft

The sum of the moments about A2 is ΣM point A2 = h 2Th − (16 ft)Tv + (10 ft)W1 = 0. (2) Finally, we cut the rope just to the left of the right attachment point:

3 ft

Tv 2 ft

6 ft

10 ft

h1 W1

3 ft

Solution:

The sum of the moments about A3 is

(a)

ΣM point A3 = (2 ft)Th − (19 ft)Tv + (13 ft)W1

We use the notation in Fig . 10.18. Let us draw a free-body diagram by cutting the cable just to the left of the weight W1: Tv

6 ft

(3)

α1 = atand(h1 /6 ft) = 33.7 , α 2 = atand[(h 2 − h1 )/10 ft] = 6.01 ,

Here Th and Tv are the horizontal and vertical components of the tension at the left attachment point. The sum of the moments about A1 is

α 3 = atand[(h 2 − 2 ft)/3 ft] = 45.5 . By summing the horizontal forces on each free-body diagram, we obtain the results

ΣM point A1 = h1Th − (6 ft)Tv = 0. (1) Next, we cut the rope just to the left of weight W2 :

6 ft

(b) Going back to the three free-body diagrams, observe that the three angles

h1 A1

We solve Eqs. (1)-(3) with W1 = 50 lb, W2 = 100 lb, and h1 = 4 ft, obtaining Th = 89.1 lb, Tv = 59.4 lb, h 2 = 5.05 ft.

+ (3 ft)W2 = 0.

2 ft

T1 =

Th = 107 lb, cos α1

T2 =

Th = 89.6 lb, cos α 2

T3 =

Th = 127 lb. cos α 3

10 ft

Th h1

(a) h 2 = 5.05 ft, (b) 127 lb. a2 W1

776

T2 A2

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Problem 10.68* Three identical masses m = 10 kg are suspended from the cable. Determine the vertical distances h1 and h3 and draw a sketch of the configuration of the cable. 2m

2m TH

h1 A1

T2 1

4 98.1 N

h1 2m

Solution:

We make 3 cuts and then draw one diagram of the

whole system

ΣM A1 : T H h1 − TV (2 m) = 0 98.1 N

ΣM A2 : T H (2 m) − TV (3 m) + (98.1 N)(1 m) = 0

98.1 N

ΣM A3 : T H h3 − TV (6 m) + (98.1 N)(4 m)

+ (98.1 N)(3 m) = 0

+ (98.1 N)(1 m) = 0

ΣM A4 : −TV (7 m) + (98.1 N)(5 m) + (98.1 N)(4 m) TH

Solving we find

3m h3

h1 = 1.739 m

A3 A2

h3 = 0.957 m T H = 161.2 N

98.1 N

TV = 140.1 N Sketch the configuration

98.1 N

TV 1m

2m TH

T5 A4

2m A1

A3 A2

98.1 N

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Problem 10.69* In Problem 10.68, what are the tensions in cable segments 1 and 2? 2m

Solution: ΣFx : − ΣFx :

Use the solution to 10.68

2m T1 + (2 m) 2 + h12 h1 T1 − (2 m) 2 + h12

1m T2 = 0 (1 m) 2 + (2 m − h1 ) 2 2 m − h1 T2 − 98.1 N = 0 (1 m) 2 + (2 m − h1 ) 2

Solving

T1 = 214 N, T2 = 167 N T1 h1 1m

2 m2h1 T2

98.1 N

778

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Problem 10.70* Three masses are suspended from the cable, where m = 30 kg, and the vertical distance h1 = 400 mm. Determine the vertical distances h 2 and h3 . 500 mm

700 mm

0.7 m

0.3 m T4

m 2 mg

ΣM (ptA3 ) = h3Th − (1.5)Tn + (1)2 mg + (0.3) mg = 0.

Cutting to the right of the left mass, TV

0.5 m

4 3

(2)

Cutting to the right of the right mass, we obtain TV

Solution:

Σm (ptA2 ) = h 2Th − (1.2)Tn + (0.7)2 mg = 0.

300 mm 300 mm 200 mm

we obtain

(3)

Finally, cutting at the right attachment point,

0.5 m

TV Th

0.5 m

0.3 m 0.3 m

0.7 m

0.2 m

A1 T2 2 mg

2 mg we obtain ΣM (ptA1 ) = h1Th − (0.5)Tn = 0.

we obtain

(1)

ΣM (ptA4 ) = (0.2)Th − (1.8)Tn + (1.3) 2 mg + (0.6) mg

Cutting to the right of the middle mass, TV

+ (0.3) mg = 0.

(4)

Solving Equations (1)–(4), we obtain Th = 831 N, Tn = 665 N, h 2 = 464 mm, h3 = 385 mm.

0.7 m

0.5 m

mg mg

Th h2 T3

2 mg

A2 mg

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Problem 10.71* In Problem 10.70, what is the maximum tension in the cable, and where does it occur? 500 mm

700 mm

3 m

h1 400 = = 0.8. 500 500 Slope of segment 4:

2 2m

Slope of segment 1 (see the solution of Problem 10.71):

300 mm 300 mm 200 mm

Solution: The tension is greatest in the segment with the greatest slope, which is either segment 1 or segment 4.

h3 − 200 385 − 200 = = 0.62. 300 300 Cutting segment 1, we obtain TV

Th b

T1 The angle β is β = arctan

h ) = 38.7°. ( 500 ) = arctan( 400 500 1

From the equation ΣFx = −Th + T1 cos β = 0, we obtain T1 =

780

Th 831 = = 1060 N. cos β cos 38.7 °

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Problem 10.72 Each suspended object has the same weight. Determine the vertical distances h 2 and h3 .

Solution: Tv

4 ft P1

(a)

h2 h3

(b)

12 ft

W We obtain free-body diagram (a) by cutting the rope at the left attachment point and just to the right of the left weight. Notice that the tension at the left attachment point is resolved into horizontal and vertical components. The sum of the moments about the point P1 where the weight is attached is

2 ft

3 ft

4 ft

5 ft

ΣM P1 = (4 ft)Th − (2 ft)Tv = 0.

(1)

We next obtain free-body diagram (b) by cutting the rope just to the right of the next weight, obtaining the equation ΣM P2 = h 2Th − (5 ft)Tv + (3 ft)W = 0.

(2)

We progress in this way:

(c)

(d)

From free-body diagram (c) we obtain ΣM P3 = h3Th − (9 ft)Tv + (7 ft)W + (4 ft)W = 0.

(3)

Free-body diagram (d) results from cutting the rope at the right attachment point. The resulting moment equation is ΣM P4 = (14 ft)Th − (14 ft)Tv + (12 ft)W + (9 ft)W + (5 ft)W = 0.

(4)

Solving Eqs. (1)–(4) with W = 1, we obtain Th = 1.63, Tv = 3.25, h 2 = 8.15 ft, h3 = 11.2 ft. h 2 = 8.15 ft, h3 = 11.2 ft.

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Problem 10.73 A cube is suspended below the surface of a pool of stationary water. The distance d = 3 ft and the dimension b = 6 ft. The weight density of the water is γ = 62.4 lb/ft 3 , and atmospheric pressure at the surface is 2120 lb/ft 2 . (a) What is the magnitude of the force exerted by the pressure on the top face of the cube? (b) What is the magnitude of the force exerted on the bottom face?

Solution: (a)

The pressure of the water at the top of the cube is p = p atm + γ d = 2120 lb/ft 2 + (62.4 lb/ft 3 )(3 ft) = 2310 lb/ft 2 . The pressure on the top is uniform, so the force exerted on it is pb 2 = (2310 lb/ft 2 )(6 ft) 2 = 83,100 lb.

(b) The pressure of the water at the bottom of the cube is p = p atm + γ (d + b) = 2120 lb/ft 2 + (62.4 lb/ft 3 )(9 ft) = 2680 lb/ft 2 .

The force exerted on the top is pb 2 = (2680 lb/ft 2 )(6 ft) 2

= 96, 500 lb. (a) 83,100 lb. (b) 96, 500 lb.

Problem 10.74 A cube is suspended below the surface of a pool of stationary water. The distance d = 3 ft and the dimension b = 6 ft. The weight density of the water is γ = 62.4 lb/ft 3 , and atmospheric pressure at the surface is 2120 lb/ft 2 . Use integration to determine the magnitude of the force exerted by pressure on one of the side faces of the cube.

Solution:

Let us define an element of area dA of the face that is at a depth x and has height dx: patm d

dA x

The area dA = b dx. The pressure at the element is p = p atm + γ x. b

The force exerted by pressure on the face is F = ∫ p dA A

d +b

∫d

( p atm + γ x )b dx

d +b 1 =  p atm bx + γ bx 2   d 2

= p atm b(d + b) +

(

1 1 γ b(d + b) 2 − p atm bd − γ bd 2 2 2

= p atm b 2 + γ db 2 +

1 3 b 2

)

= 89,800 lb. Because the pressure on the face varies linearly with depth, we can check this result by determining the product of the area of the face and the value of the pressure at its center:

(

)

1 F = b 2  p atm + γ d + b   2  = 89,800 lb. 89,800 lb.

782

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Problem 10.75 The area shown is subjected to a uniform pressure p atm = 1 × 10 5 Pa. (a) What is the total force exerted on the area by the pressure? (b) What is the moment about the y-axis due to the pressure on the area?

Solution: 1

N dydx = 66, 700 N = 66.7 kN m2

(a)

F = ∫ ∫ 2 10 5

(b)

M y = ∫ ∫ 2 x 10 5

(

)

N dydx = 25, 000 Nm = 25 kN-m m2

y 5 x2 x

Problem 10.76 The area shown is subjected to a uniform pressure. Determine the coordinates of the center of pressure. y

Solution: 1

R = ∫ ∫ 2 P dy dx =

2P 3

P M y = ∫ ∫ 2 xP dy dx = 0 x 4 1

M x = ∫ ∫ 2 yP dy dx =

My = 0.375 m R ⇒ Mx y = = 0.6 m R x =

2P 5

y 5 x2 x

BandF_6e_ISM_CH10.indd 783

783

01/04/23 12:29 PM

Problem 10.77 The coordinates a = 1 m and b = 3 m. The area is subjected to a uniform pressure p atm = 1E5 Pa. (a) What is the magnitude of the force exerted on the area by the pressure? (b) Use integration to determine the magnitude of the moment about the y-axis due to the pressure.

Solution: (a)

Because the pressure is uniform, the force is the product of the pressure and the total area. Let dA be a vertical strip of width dx: y y 5 x2

y dA b

x dx

y 5 2x

The area dA = (2/x )dx. The total area is

A = ∫ dA A

= ∫

= 2[ ln x ]ba = 2 ln

( ba ).

The force is F = p atm A = 2 p atm ln

( ba )

= 2(1E5 Pa) ln

( 13 mm )

= 220 kN. (b) The magnitude of the moment about the y-axis due to the pressure acting on the element dA is F = x p atm dA. The total moment is M = ∫ x p atm dA A

= p atm ∫ x a

( 2x dx )

= 2 p atm (b − a) = 2(1E5 N/m 2 )(3 m − 1 m) = 400 kN-m. (Note: The units in the last two steps of the preceding equation don’t look consistent. The reason is that in the equation y = 2/x, the terms x and y have units of meters, so the “2” in that equation, and in the equation above, should be considered to have units of m 2 .) (a) 220 kN. (b) 400 kN-m.

784

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Problem 10.78 The coordinates a = 1 m and b = 3 m. The area is subjected to a uniform pressure p atm = 1E5 Pa. What are the coordinates of the center of pressure?

The y coordinate of the midpoint (centroid) of the vertical strip is y strip = (1/2)(2/x ) = 1/x: y y 5 x2

1 x x

y 5 2x

dx The y coordinate of the centroid is

y =

∫ A y dA ∫ A dA

)( x dx ) ∫a ( x b

Solution:

Because the pressure is uniform, the center of pressure is located at the centroid of the area. Let dA be a vertical strip of width dx:

y strip

∫a x dx b

1 2  −   x a = 2[ ln x ]ba

y y 5 x2

= 0.607 m. x = 1.82 m, y = 0.607 m.

dA b

x dx The area dA = (2/x )dx. The x coordinate of the centroid is x =

∫ A x dA ∫ A dA ∫a x ( x ) dx 2

∫a x dx [ 2 x ]ba

2[ ln x ]ba

= 1.82 m.

BandF_6e_ISM_CH10.indd 785

785

01/04/23 12:29 PM

Problem 10.79 The diagram shows a mercury barometer used to measure atmospheric pressure. Assume that there’s a vacuum above the mercury in the inverted tube. The mass density of mercury is ρ Hg = 13,500 kg/m 3 . If the height of the column of mercury within the tube is 760 mm, what is atmospheric pressure in pascals? Strategy: Draw a free-body diagram of the column of mercury.

Solution:

We draw the free-body diagram of the column of mercury:

patm A

Mercury

h patm

The pressures on the sides of the column, which exert horizontal forces and cancel out, are not shown. The pressure on the top of the column is nominally zero. We denote the cross-sectional area by A. The weight per unit volume of the mercury is ρ Hg g, so the weight of the column is W = ρ Hg g( Ah). From the equilibrium equation p atm A = W = ρ Hg gAh, we obtain p atm = ρ Hg gh = (13, 500 kg/m 3 )(9.81 m/s 2 )(0.76 m) = 1.01E5 N/m 2 (Pa). p atm = 1.01E5 Pa.

786

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Problem 10.80 The top of the rectangular plate is 2 m below the surface of a lake. The mass density of the water is ρ = 1000 kg/m 3 . What is the magnitude of the force exerted on a face of the plate by the gage pressure of the water?

Solution:

In terms of a coordinate system with the origin at the surface and the x-axis directed downward, the gauge pressure is p g = γ x, where γ is the weight density γ = ρg = (1000 kg/m 3 )(9.81 m/s 2 ) = 9.81 kN/m 3 . Let dA be a horizontal element of the plate’s area at a distance x below the surface:

x dA

The area dA = (2 m)dx. The magnitude of the force due to the gauge pressure on the plate’s surface is F = ∫ p g dA A 5

= ∫ γ x (2 m) dx. 2

x2 = (2 m)γ    2 2 = (1 m)γ [(5 m) 2 − (2 m) 2 ] = (1 m)(9.81 kN/m 3 )(21 m 2 ) = 206 kN (46,300 lb). 206 kN.

BandF_6e_ISM_CH10.indd 787

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Problem 10.81 The top of the rectangular plate is 2 m below the surface of a lake. The mass density of the water is ρ = 1000 kg/m 3 . What distance below the top of the plate is the center of pressure due to the gage pressure of the water?

The area dA = (2 m)dx. The magnitude of the force due to the gauge pressure on the plate’s surface is F = ∫ p g dA A 5

= ∫ γ x (2 m) dx. 2

x2 = (2 m)γ    2 2 = (1 m)γ[(5 m) 2 − (2 m) 2 ]

= (1 m)(9.81 kN/m 3 )(21 m 2 ) = 206 kN (46,300 lb). The magnitude of the moment about the y-axis due to the gauge pressure is 3m

M = ∫ x p g dA A 5

= ∫ γ x 2 (2 m) dx. 2

x3 = (2 m)γ    3 2 2 m γ[(5 m) 3 − (2 m) 3 ] = 3 2 m (9.81 kN/m 3 )(117 m 3 ) = 3 = 765 kN-m.

( (

Solution:

In terms of a coordinate system with the origin at the surface and the x-axis directed downward, the gauge pressure is p g = γ x, where γ is the weight density γ = ρg

) )

The distance from the surface of the water to the center of pressure is M F 765 kN-m = 206 kN = 3.71 m.

x =

= (1000 kg/m 3 )(9.81 m/s 2 ) = 9.81 kN/m 3 . Let dA be a horizontal element of the plate’s area at a distance x below the surface:

The distance from the top of the plate to the center of pressure is x p = 3.71 m − 2 m = 1.71 m. 1.71 m.

2m x dA

788

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Problem 10.82 The width of the gate (the dimension into the page) is 4 ft. The weight density of the water is γ = 62.4 lb/ft 3 . Atmospheric pressure is p atm = 2120 lb/ft 2 . Determine the force and couple exerted on the gate by its built-in support A.

Solution:

Because atmospheric pressure acts on the left side of the gate, we only need to consider the gauge pressure on the gate to determine the reactions at A. Let dA be a horizontal element of the gate’s area at a distance x below the surface:

y dA

6 ft

4 ft x A

The area dA = (4 ft)dx. The magnitude of the force due to the gauge pressure on the gate’s surface is F = ∫ γ x dA A 6

= ∫ γ x (4 ft) dx 0

x2 = γ (4 ft)    2 0 = 4490 lb. The counterclockwise moment about A due to the gauge pressure is M = ∫ (6 ft − x )γ x dA A 6

= ∫ (6 ft − x )γ x (4 ft) dx 0

= 4 ∫ (6γ x − γ x 2 ) dx 0

x2 x3 6 = 4  6γ − γ   2 3 0 = 8990 ft-lb. 4490 lb to the right, 8990 ft-lb clockwise.

BandF_6e_ISM_CH10.indd 789

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01/04/23 12:29 PM

Problem 10.83 The width of the gate (the dimension into the page) is 4 ft. The weight density of the water is γ = 62.4 lb/ft 3 . Atmospheric pressure is p atm = 2120 lb/ft 2 . Consider the sum of the atmospheric and gage pressures on the right face of the gate. What distance below the top of the gate is the center of pressure?

Solution: Let dA be a horizontal element of the gate’s area at a distance x below the surface: y dA

6 ft

4 ft

6 ft

x The area dA = (4 ft)dx. The magnitude of the force due to the gauge pressure on the gate’s surface is A

F = ∫ ( p atm + γ x ) dA A 6

= ∫ ( p atm + γ x )(4 ft) dx 0

x2 6 = (4 ft)  p atm x + γ   2 0 = 55, 400 lb. The magnitude of the moment about the y-axis is M = ∫ x ( p atm + γ x ) dA A 6

= ∫ x ( p atm + γ x )(4 ft) dx 0

= 4 ∫ ( p atm x + γ x 2 ) dx 0

x2 x3 = 4  p atm + γ   2 3 0 = 171, 000 ft-lb. The distance from the surface to the center of pressure is M xp = = 3.08 ft. F 3.08 ft.

790

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Problem 10.84 The homogeneous gate weighs 100 lb, and its width (the dimension into the page) is 3 ft. The weight density of the water is γ = 62.4 lb/ft 3 , and atmospheric pressure is p atm = 2120 lb/ft 2 . Determine the reactions at A and B. B

along the inner face of the gate from A. The sum of the moments about A is ΣM A = dF + W (1.5)sin 30 ° − 3B = 0, from which B = 135.933 lb, normal to the inner face of the gate. The sum of the forces normal to the gate surface is ΣFN = + A N − F + B − W sin 30 ° = 0, from which A N = 346.4 lb at an angle α = 30 ° relative to the positive x-axis. The sum of the forces acting parallel to the gate surface is ΣFP = AP − W cos(30 °) = 0,

3 ft

308

2 ft

from which AP = 86.6 lb at an angle of 120 °. Thus the components of the reaction at A are A x = −346.3cos(210 °) + 86.6 cos(120 °) = 256.7 lb, to the right, and A y = −346.3sin(210 °) + 86.6sin(120 °) = 248.2 lb upward.

A y

Solution:

The atmospheric pressure acts on both sides of the gate, so it is ignored. The strategy is use the “volume” of the pressure distribution to compute the force acting on the face of the gate. The pressure distribution is a triangle with base 2γ . The pressure acts over an area

308 3ft

2ft A

AN AP

 6  2  cos(30)  = 6.92 ft . Thus the “volume” is F = ( 12 ) (62.4)(2)(6.92) = 432.32 lb. This force acts normally to the surface of the gate, or at an angle of θ = 210 ° relative to the positive x-axis. The centroid of the pressure is d =

1 ( 32 )( cos30 ) = 0.7698 ft °

BandF_6e_ISM_CH10.indd 791

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01/04/23 12:30 PM

Problem 10.85 The width of the gate (the dimension into the page) is 2 m, and there is water of depth d = 1 m on the right side. Atmospheric pressure is p atm = 1 × 10 5 Pa, and the mass density of the water is ρ = 1000 kg/m 3 . Determine the horizontal forces exerted on the gate at A and B.

Solution:

The “volume” of the gage pressure distribution is

1 ρ gd (2) 2 = δ pd 2

F =

= (1000)(9.81)(1) 2 = 9810 N.

Ax A

0.5 m

1 d 5 1 (1) m 3 3

500 mm B

Bx Applying the equilibrium equations, we find that 2 F = 6540 N, 3 1 B x = F = 3270 N. 3

Ax =

Problem 10.86 The gate in Problem 10.85 is designed to rotate and release the water when the depth d exceeds a certain value. What is that depth?

Solution:

See the solution of Problem 10.85. The gate rotates when

1 1 d > m, 3 2 3 d > m. 2

500 mm B

792

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01/04/23 12:30 PM

Problem 10.87* The dam has water of depth 4 ft on one side. The width of the dam (the dimension into the page) is 8 ft. The weight density of the water is γ = 62.4 lb/ft 3 , and atmospheric pressure p atm = 2120 lb/ft 2 . If you neglect the weight of the dam, what are the reactions at A and B?

2 ft

Solution:

To simplify the analysis of the pressure forces, we will draw a free body diagram of the dam and the volume of water shown: The left side and top of the free body diagram are subjected to atmospheric pressure, and the right side is subjected to the sum of atmospheric pressure and the gage pressure of the water, so we only need to consider the gage pressure. Let us represent the pressure force on the right side by an equivalent force: We can determine F by calculating the “volume” of the pressure distribution: 1 1 F = p(4 ft)(8 ft) = γ (4)(4)(8) = 3990 lb 2 2 the complete free body diagram is: The two weights are

A 2 ft

w1 = γ (2 ft)(2 ft)(8 ft) = 2000 lb 1 w 2 = γ (2 ft)(2 ft)(8 ft) = 988 lb. 2 From the equilibrium equations ΣFx = A x + B − F = 0 ΣFy = A y − w1 − w 2 = 0 ΣM (pt.A) = −4 B +

( 43 ) F + (1)W + 13 W = 0, 1

1 (4) ft 3

we obtain A x = 2000 lb, A y = 3000 lb, B = 2000 lb. P 5 r (4 ft) B W1

4 ft 3 Ax

BandF_6e_ISM_CH10.indd 793

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01/04/23 12:30 PM

Problem 10.88* The dam has water of depth 4 ft on one side. The width of the dam (the dimension into the page) is 8 ft. The weight density of the water is γ = 62.4 lb/ft 3 , and atmospheric pressure is p atm = 2120 lb/ft 2 . If you neglect the weight of the dam, what are the reactions at A and B?

Solution: The atmospheric pressure acts on both faces of the dam, so it is ignored. The strategy is to use the “volume” of the pressure distribution to the determine the reactions. The pressure distribution is a triangle of base 4 γ and altitude 4 ft. The force on the vertical faces of the dam is 1 F1 = (4)(4)(γ )8 = 3993.6 lb. 2 The moment about A due to the force on the vertical faces is 4 M1 = F = 5324.8 ft lb. 3 1 The force on the horizontal face of the dam is

( )

F2 = (2)(γ )(2)(8) = 1996.8 lb. 2 ft

The moment about A due to the force on the horizontal face is M 2 = 1F2 = 1996.8 ft lb. The sum of the moments about A : ΣM A = M 1 + M 2 − 4 B = 0, from which B = 1830.4 lb. The sum of the forces: ΣFx = A x − F1 + B = 0, from which A x = 2163.3 lb to the right. ΣFy = A y − F2 = 0, from which A y = 1996.8 lb upward.

2 ft B

1 ft

2 ft

F1 AX

4 ft 3

Problem 10.89 Consider a plane, vertical area A below the surface of a liquid. Let p 0 be the pressure at the surface. (a) Show that the force exerted by pressure on the area is F = pA, where p = p 0 + γ x is the pressure of the liquid at the centroid of the area. (b) Show that the x coordinate of the center of pressure is xp = x +

Solution: (a)

The definition of the centroid of the area is x A = ∫ x dA. A

The force on the plate is F = ∫ pdA = p 0 ∫ dA + γ ∫ x dA = p 0 A + γ x A.

γ I y′ , pA

where I y′ is the moment of inertia of the area about the y ′-axis through its centroid.

But by definition, the pressure at the centroid of the area is p = p 0 + γ x, hence F = pA. (b) The moment about the y-axis is

M y -axis = ∫ px dA = p 0 ∫ x dA + γ ∫ x 2 dA. A

The moment of inertia is defined: –x

I y A = ∫ x 2 dA, A

y9 A

from which the moment is M y -axis = p 0 x A + γ I y . But the moment is also given by M y -axis = Fx P , where F is the force on the plate and x P is the distance to the center of pressure. Thus pAx P = p 0 x A + γ I y , p  γ Iy from which x P =  0  x + .  p  pA From the parallel axis theorem, I y = x 2 A + I y ′ . Substitute and reduce: xP =

γI y′ γI y′ ( p 0 + γ x )x + = x + , p pA pA

which demonstrates the required result.

794

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01/04/23 12:31 PM

Problem 10.90 A circular plate of 1-m radius is below the surface of a stationary pool of water. Atmospheric pressure is p atm = 1.00E5 Pa, and the mass density of the water is ρ = 1000 kg/m 3 . Determine (a) the force exerted on a face of the plate by the pressure of the water; (b) the x coordinate of the center of pressure. (See Problem 10.89.) y 1m

Solution: (a)

From Problem 10.89, the force F exerted on the plate’s area A by the pressure of the water is F = pA, where p is the pressure at the plate’s centroid. For the circular plate, the pressure at the centroid is p = p0 + γ x = 1.00E5 Pa + (9.81 m/s 2 )(1000 kg/m 3 )(2 m) = 1.20E5 Pa. The force on the plate is F = pA = (1.20E5 Pa)[π (1 m) 2 ] = 376 kN (84,500 lb).

(b) From Problem 10.89, the x coordinate of the center of pressure is xp = x +

γ I y′ , pA

where I y′ is the moment of inertia of the area about the horizontal axis through the area’s centroid. For the circular area, 1m I y′ =

1 π (1 m) 4 = 0.785 m 4 , 4

γ I y′ (9.81 m/s 2 )(1000 kg/m 3 )(0.785 m 4 ) = 2m + pA (1.20E5 Pa)π (1 m) 2 = 2.02 m.

xp = x +

(a) 376 kN (84, 500 lb). (b) x p = 2.02 m.

Problem 10.91* The tank consists of a cylinder with hemispherical ends. It is filled with water (ρ = 1000 kg/m 3 ). The pressure of the water at the top of the tank is 140 kPa. Determine the magnitude of the force exerted by the pressure of the water on each hemispherical end of the tank. (See Example 10.12.)

Solution:

The free-body diagram is

18 m

mg 6m

The force exerted by the pressure distribution on the curved surface (equal and opposite to the force exerted on the tank’s hemispherical end) can be determined from the fact that the free-body diagram is in equilibrium. The magnitude of the vertical component is mg = (1000)

(

)

1 4 π R 3 (9.81) = 4.44 × 10 6 N. 2 3

To determine the horizontal component, see Problem 10.89. The pressure at the centroid is p = 140, 000 + (1000)(9.81)(6) = 199, 000 Pa, so the horizontal component is pA = (199, 000)π (6) 2 = 22.5 × 10 6 N. The magnitude of the force is (mg) 2 + ( pA) 2 = 22.9 MN.

BandF_6e_ISM_CH10.indd 795

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01/04/23 12:31 PM

Problem 10.92 An object of volume V and weight W is suspended below the surface of a stationary liquid of weight density γ (Fig. a). Show that the tension in the cord is W − V γ . In other words, show that the pressure distribution on the surface of the object exerts an upward force equal to the product of the object’s volume and the weight density of the water. This result is due to Archimedes (287–212 BCE). Strategy: Draw the free-body diagram of a volume of liquid that has the same shape and position as the object (Fig. b).

Solution:

The result follows from the free body diagram of the space occupied by the object and the development of the force exerted by the pressure in terms of the volume of the object. Let A be the area bounding the volume V. The force exerted on a surface A is F = −∫ np dA, A

where n is a unit vector normal to the elemental surface dA, positive outward from the surface, and the negative sign comes from the equilibrium condition that the reaction force acts oppositely to the unit vector. Choose an x, y, z coordinate system such that the elemental forces are np dA = ip dy dz + jp dx dz + kp dx dy. The force becomes F = −∫ ip dy dz − ∫ jp dx dz − ∫ kp dx dy. A

But these integrals can also be written as

F = −i ∫

∂p

V ∂x

dx dy dz − j∫

∂p

V ∂y

dx dy dz − k ∫

∂p

V ∂z

dx dy dz.

This set of integrals can be collapsed into (a)

(b)

F = −∫ np dA = −∫ ∇p dV , A

where the volume V is bounded by the surface A, ∇p is a shorthand notation for ∇p = i

∂p ∂p ∂p + j + k , ∂x ∂y ∂z

and dV = dx dy dz. The pressure p = p 0 + γ x, from which ∇p = iγ , and the integral becomes F = −∫ np dA = −iγ ∫ dV = −iγV . A

The weight of the object acts in the positive x direction, so the resultant force is FR = i(W − γV )

796

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01/04/23 12:32 PM

Problem 10.93 moment at B.

Determine the internal forces and

y 10 kN

4 kN-m

0.4 m

Solution:

0.4 m

10 kN 0.4 m

0.4 m

The free-body diagram of the entire beam is

y 10 kN

4 kN-m

B Ax

x Ay

10 kN

From the equilibrium equations ΣFx = A x = 0, ΣFy = A y + C − 10 kN + 10 kN = 0, ΣM point A = (2 m)C − (0.4 m)(10 kN) + (0.8 m)(10 kN) − 4 kN-m = 0, we obtain A x = 0, A y = 0, C = 0. The free-body diagram of the part of the beam to the left of B is

y 10 kN

MB PB 10 kN

From the equilibrium equations ΣFx = PB = 0, ΣFy = −V B + 10 kN − 10 kN = 0, ΣM point B = M B + (0.8 m)(10 kN) − (0.4 m)(10 kN) = 0, we obtain PB = 0, VB = 0, M B = −4 kN-m. PB = 0, VB = 0, M B = −4 kN-m.

BandF_6e_ISM_CH10.indd 797

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01/04/23 12:32 PM

Problem 10.94 moment at B. y

Determine the internal forces and

Writing the equilibrium equations ΣFx = PB = 0, ΣFy = −V B + 25 kN − 30 kN = 0,

30 kN

ΣM point B = −(0.6 m)(25 kN) + (0.4 m)(30 kN) + 2 kN-m

2 kN-m

0.2 m

+ MB = 0

0.4 m

gives PB = 0, VB = −5 kN, M B = 1 kN-m. Alternatively, we can cut the beam at B and draw the free-body diagram of the part of the beam to the right of B:

0.2 m

Solution:

The free-body diagram of the entire beam is

30 kN

2 kN-m

5 kN

From the equilibrium equations

x VB

ΣFx = −PB = 0, ΣFy = VB + 5 kN = 0,

From the equilibrium equations

ΣM point B = (0.2 m)(5 kN) − M B = 0

ΣFx = A x = 0,

we again obtain PB = 0, VB = −5 kN, M B = 1 kN-m.

ΣFy = A y + C − 30 kN = 0, ΣM point A = (0.8 m)C − (0.2 m)(30 kN) + 2 kN-m = 0

PB = 0, VB = −5 kN, M B = 1 kN-m.

we obtain the reactions A x = 0, A y = 25 kN, C = 5 kN. Now we “cut” the beam at B and draw the free-body diagram of the part of the beam to the left of B, showing the internal forces and moment:

30 kN

2 kN-m

25 kN

Problem 10.95 If x = 2 ft, what are the internal forces and moment at A?

Solution:

With x = 2 ft, we draw the free-body diagram of the part of the beam to the right of A, representing the distributed load by its equivalent force:

1 (3 ft)(600 1b/ft) 2

600 lb/ft PA

x A

2 ft

3 ft

2 ft

3 ft From the equilibrium equations ΣFx = −PA = 0, ΣFy = V A −

1 (3 ft)(600 lb/ft) = 0, 2

1 ΣM point A = −(2 ft)  (3 ft)(600 lb/ft)  − M A = 0, 2  we obtain PA = 0, V A = 900 lb, M A = −1800 ft-lb. PA = 0, V A = 900 lb, M A = −1800 ft-lb.

798

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Solution:

Problem 10.96 If x = 4 ft, what are the internal forces and moment at A?

With x = 4 ft, we draw the free-body diagram of the part of the beam to the right of A, being careful to “cut” the distributed load and then represent it by its equivalent force:

400 lb/ft MA

600 lb/ft

PA x

2 ft

3 ft

3 ft y

1 (2 ft)(400 1b/ft) 2

PA VA

2 ft 3

From the equilibrium equations ΣFx = −PA = 0, 1 ΣFy = V A − (2 ft)(400 lb/ft) = 0, 2 2 1 ΣM point A = − ft  (2 ft)(400 lb/ft)  − M A = 0, 2  3

(

)

we obtain PA = 0, V A = 400 lb, M A = −267 ft-lb. PA = 0, V A = 400 lb, M A = −267 ft-lb.

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01/04/23 12:33 PM

Problem 10.97 Determine the shear force and bending moment diagrams for the beam. y

The moment is

(

M = ∫ V ( x )dx + C = −10 2 x 3 −

)

x4 + 1440 x + C . 12

The moment is zero at x = 0, hence the constant C = 0, and the moment is

w 5 10(12x 2 x2) lb/ft

(

M = −10 2 x 3 −

)

x4 + 1440 x 12

12 ft

Ax B

12 ft

Solution:

Denote the reactions at the left and right ends by A and B, respectively. The total force due to the load is 12

12 x3  F = ∫ 10(12 x − x 2 ) dx = 10  6 x 2 − = 2880 lb. 0  3  0

Shear Force & Moment Diagram

6000 5000

The moment about the left end due to the load is

4000

12 x4  M LOAD = ∫ 10(12 x − x 2 ) x dx = 10  4 x 3 − . 0  4  0

Bending Moment

3000 2000

M LOAD = 17280 ft lb.

1000 0

The sum of the moments about the left end:

Shear Force

21000

ΣM A = −M + 12 B = 0,

22000

from which B = 1440 lb. The sum of the forces:

X, ft

ΣFy = A y + B − F = 0, from which A y = 1440 lb. Beginning from the left end, the shear is X

V ( x ) = −∫ 10(12 x − x 2 ) + 1440 0

(

V ( x ) = −10 6 x 2 −

)

x3 + 1440. 3

Problem 10.98 Determine V and M as functions of x for the beam ABC.

Solution:

First find the reactions

ΣM c : − A(6 m) + (4 kN)(2 m) = 0 ΣFy : A + C − 4 kN = 0

A = 1.33 kN, C = 2.67 kN In the first region 0 < x < 2 m

w1 = 0

V1 = 1.33 kN M 1 = (1.33 kN) x

4 kN 2m

In the second region 2 m < x < 6 m w2 = 0 V2 = −2.67 kN M 2 = −(2.67 kN) x + 16 kN-m

4 kN

800

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Problem 10.99 moment at B.

Determine the internal forces and

Now we “cut” the beam at B and draw the free-body diagram of the part of the beam to the left of B (including the part of the distributed load that acts on that part!), showing the internal forces and moment: y

y 48 kN/m A

24 kN/m MB A

B 48 kN

VB 3m

Solution:

We draw the free-body diagram of the entire beam, representing the distributed load by its equivalent force: y

Now we can represent the distributed load acting on this part by an equivalent force: y

1 (6 m)(48 kN/m) 2

1 (3 m)(24 kN/m) 2 MB

C 2 (6 m) 3

From the equilibrium equations

48 kN

2 (3 m) 3 Writing the equilibrium equations

ΣFx = A x = 0, ΣFy = A y + C −

1 (6 m)(48 kN/m) = 0, 2

2 1 ΣM point A = (6 m)C −  (6 m)   (6 m)(48 kN/m)  = 0 3 2  we obtain the reactions A x = 0, A y = 48 kN, C = 96 kN.

ΣFx = PB = 0, 1 (3 m)(24 kN/m) = 0, 2 1 1 ΣM point B = −(3 m)(48 kN) +  (3 m)   (3 m)(24 kN/m)  3 2  + M B = 0, ΣFy = −V B + 48 kN −

we obtain PB = 0, VB = 12 kN, M B = 108 kN-m. PB = 0, VB = 12 kN, M B = 108 kN-m.

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Problem 10.100 moment at B.

Determine the internal forces and

Solution:

Consider the free-body diagram of the entire structure:

y F E

8 in E 8 in

400 lb

G From the equilibrium equations 6 in

6 in

8 in

ΣFx = A x + G = 0, ΣFy = A y − 400 lb = 0, ΣM point A = (8 in)G − (28 in)(400 lb) = 0, the reactions are A x = −1400 lb, A y = 400 lb, G = 1400 lb. Now we draw the free-body diagram of the part of the member to the left of point B: y 6 in

1400 lb 400 lb

MB PB

Writing the equilibrium equations ΣFx = PB − 1400 lb = 0, ΣFy = −V B + 400 lb = 0, ΣM point B = −(6 in)(400 lb) + M B = 0 yields PB = 1400 lb, VB = 400 lb, M B = 2400 in-lb. PB = 1400 lb, VB = 400 lb, M B = 2400 in-lb.

802

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Problem 10.101 moment at D.

Determine the internal forces and

Now we consider the free-body diagram of the horizontal member: y

y F

x Cx

400 lb

8 in E A

8 in

1400 lb

400 lb

Here T is the tensile axial force in the two-force member EF. From the equilibrium equation

400 lb

ΣM point C = (16 in)T sin 45 − (16 in)(400 lb) − (12 in)(400 lb) = 0,

the axial force T = 990 lb. 6 in

Solution:

6 in

8 in

To determine the internal forces and moment at D, we can draw the freebody diagram of the part of the horizontal member to the right of point D:

Consider the free-body diagram of the entire structure:

990 lb MD

8 in VD

E Ax

400 lb

Writing the equilibrium equations ΣFx = −PD − (990 lb) cos 45 = 0, ΣFy = V D + (990 lb)sin 45 − 400 lb = 0,

G From the equilibrium equations ΣFx = A x + G = 0,

ΣM point D = (8 in)[(990 lb)sin 45 ] − (8 in)(400 lb) − M D = 0 yields PD = −700 lb, V D = −300 lb, M D = 2400 in-lb. PD = −700 lb, V D = −300 lb, M D = 2400 in-lb.

ΣFy = A y − 400 lb = 0, ΣM point A = (8 in)G − (28 in)(400 lb) = 0, the reactions are A x = −1400 lb, A y = 400 lb, G = 1400 lb.

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Problem 10.102 moment at C.

Determine the internal forces and

where B x and B y are the forces exerted by member AB. The equilibrium equations are ΣFx = B x + D x + 360 lb = 0,

ΣFy = B y − 180 lb + 30 lb = 0,

ΣM point B = −(6 ft) D x + (8 ft)(30 lb) − (3 ft)(360 lb) = 0, 3 ft

720 ft-lb

360 lb B

yielding B x = −220 lb, B y = 150 lb, D x = −140 lb. Now we draw the free-body diagram of the part of member BCD to the left of point C:

3 ft

180 lb 4 ft

Solution:

4 ft

Consider the free-body diagram of the entire structure: y

220 lb

We see that PC = 220 lb, VC = − 30 lb, and M C = − (4 ft)(30 lb) = −120 ft-lb. PC = 220 lb, VC = −30 lb, M C = −120 ft-lb.

360 lb

30 lb

720 ft-lb

4 ft

180 lb

We write the equilibrium equation ΣM point A = (12 ft) D y + (3 ft)(360 lb) − (4 ft)(180 lb) − 720 ft-lb = 0, Obtaining D y = 30 lb. Next, we draw the free-body diagram of the right member, y

30 lb

Dx 360 lb

By Bx

804

180 lb

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Problem 10.103 Draw the shear force and bending moment diagrams.

We cut the beam at a point x in the interval 0 < x < 0.5 m: y

y 20 N-m A

40 N

x This indicates that in this interval, V = −40 N, M = 20 − 40 x N-m. Then we cut the beam in the interval 0.5 m < x < 1 m:

1000 mm Consider the free-body diagram of the entire beam.

20 N-m

40 N

500 mm

Solution:

20 N-m

40 N

1–x

40 N

In this interval, V = −40 N, M = (40 N)(1 − x ) N-m. The shear and bending moment diagrams are shown.

From the equilibrium equations ΣFx = A x = 0, ΣFy = A y + 40 N = 0, ΣM point A = M A + (40 N)(1 m) − 20 N-m = 0, we see that A x = 0, A y = −40 N, M A = −20 N-m.

V, N

220

240

260

0.1

0.2

1.3

0.4

0.5 x, m

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5 x, m

0.6

0.7

0.8

0.9

M, N-m

15 10 5 0

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Problem 10.104 The homogeneous beam weighs 1000 lb. What are the internal forces and bending moment at its midpoint?

Solve the two simultaneous equations to obtain: A = 400 lb, B = 600 lb. ΣFx = 0. Divide the left half of the beam into two parts: 0 ≤ x < 2 (ft) and 2 ≤ x < 5 (ft). Part 1: The shear is V1 ( x ) = −100 x. The moment is

x 2 ft

At x = 0, M 1 (0) = 0, thus C1 = 0, and the moment is M 1 ( x ) = −50 x 2 .

3 ft 10 ft

Part 2: The shear is V2 ( x ) = V1 ( x ) + 400 lb. The moment is

Solution:

Denote the support reactions by A and B. The load distribution is w = 1000/10 = 100 lb/ft. The moment about the left end due to the load of the beam is ML = ∫

10 0

M 1 ( x ) = ∫ V1 ( x ) dx + C1 = −50 x 2 + C1 .

x2 wx dx = 100   = 5000 ft lb.  2 0

The sum of the moments about the left end is

M 2 ( x ) = ∫ V2 ( x ) dx + C 2 = −50 x 2 + 400 x + C 2 . The moment is continuous at x = 2, M 1 (2) = M 2 (2), from which C 2 = −800, and the moment is M 2 ( x ) = −50 x 2 + 400 x − 800. At the midpoint, x = 5 ft, the shear is V2 (5) = −100(5) + 400 = −100 lb and the moment is M 2 (5) = −50(25) + 400(5) − 800 = −50 ft lb. The axial force is zero, P(5) = 0.

Σ M = −M L + 2 A + 7 B = 0 The sum of the forces: ΣFy = A + B − 1000 = 0.

Problem 10.105 The homogeneous beam weighs 1000 lb. Draw the shear force and bending moment diagrams.

Solution:

From the solution for Problem 10.104, the shear force and bending moment are: (1) V1 ( x ) = −100 x lb, M 1 ( x ) = −50 x 2 ft lb, (0 ≤ x < 2 ft). (2) V2 ( x ) = −100 x + 400 lb, M 2 ( x ) = −50 x 2 + 400 x − 800 ft lb, (2 ≤ x < 7 ft). (3) V3 ( x ) − 100 x + 1000 lb, M 3 ( x ) = −50 x 2 + 1000 x − 5000 ft lb,

(7 ≤ x < 10 ft). x 2 ft

The shear force and bending moment diagrams are shown.

3 ft 10 ft

Shear Force & Moment Diagram 300

Shear Force

200 100 0 2100 2200 2300

Bending Moment

2400 2500

806

5 6 X, ft

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Problem 10.106 At A the main cable of the suspension bridge is horizontal and its tension is 1 × 10 8 lb. (a) Determine the distributed load acting on the cable. (b) What is the tension at B? y

Solution: (a)

The parameter a = 2

y 300 = 2 = 7.4074 × 10 −4. x2 900 2

The distributed load is

w = T0 a = (1 × 10 8 )(7.4 × 10 −4 ) = 7.4074 × 10 4 lb/ft (b) The tension at B is

300 ft

TB = T0 1 + a 2 (900) 2 = 1.2 × 10 8 lb

900 ft

Problem 10.107 The power line has a mass of 1.4 kg/m. If the line will safely support a tension of 5 kN, determine whether it will safely support an ice accumulation of 0.4 kg/m. 128

Solution:

The power line meets the conditions for a catenary. The weight density with an ice load is γ = (1.4 + 4)(9.81) = 52.974 N/m. The angle at the attachment point is related to the length and the parameter a by sa = tan θ. But sa = sinh(ax ), and x is known. Thus the parameter can be found from a =

sinh −1 (tan θ) sinh −1 (0.2126) = = 0.01055. x 20

The tension at the lowest point is T0 =

w = 5021.53 = 5.02 kN. a

The maximum tension is

40 m

T = T0 cosh(ax ) = 5133.7 = 5.133 kN. Thus the line will not sustain the load.

Problem 10.108 The water depth at the center of the elliptical window is 20 ft. Determine the magnitude of the net force exerted on the window by the pressure of the seawater (γ = 64 lb/ft 3 ) and the atmospheric pressure of the air on the opposite side. (See Problem 10.89.)

Solution:

The force on the plate is

F = ∫ p dA. A

The pressure is p = p 0 + γ x − p 0 = γ x, where the atmospheric pressure cancels, since it appears on both sides. The force: F = γ ∫ x dA = γ x A, A

hence the force on the window is F = 1280 A lb. The area of the ellipse is A = πab = 21π, and the force is F = 84445 lb

3 ft 6 in L

6 ft

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Problem 10.109 The water depth at the center of the elliptical window is 20 ft. Determine the magnitude of the net moment exerted on the window about the horizontal axis L by the pressure of the seawater (γ = 64 lb/ft 3 ) and the atmospheric pressure of the air on the opposite side. (See Problem 10.89.)

Solution:

The moment is

M = ∫ px dA = γ ∫ x 2 dA = γ I y . The area moment of inertia for an ellipse is Iy =

πab 3 = 202 ft 4 , 4

and the moment is M = γ I y = 12930.8 ft lb.

3 ft 6 in L

6 ft

Problem 10.110* The gate has water of 2-m depth on one side. The width of the gate (the dimension into the page) is 4 m, and its mass is 160 kg. The mass density of the water is ρ = 1000 kg/m 3 , and atmospheric pressure is p atm = 1 × 10 5 Pa. Determine the reactions on the gate at A and B. (The support at B exerts only a horizontal reaction on the gate.) B

The area above the gate is Aa = R 2 − Ab = 0.858 m 2 . The centroid of Aa ∪ Ab is at x = R /2, so x A + x b Ab R = 1 a , Aa + Ab 2 from which we obtain x 1 = 0.447 m. The weight of the water is Q = γ Aa (4) = 33, 700 N. From the equilibrium equations ΣFx = F + A x − B = 0, ΣFy = A y − W − Q = 0,

ΣM (ptA) = RB − y F − x 1Q − x 2W = 0, we obtain

A x = −44.2 kN, A y = 35.3 kN,

Solution:

Consider the free-body diagram shown. The weight of

B = 34.3 kN.

the gate is y

W = (160)(9.81) = 1570 N, and x 2 = R −

2R = 0.727 m. π

The pressure force 1 F = (γ R) R(4) 2 1 = (1000)(9.81)(2) 2 (4) 2 = 78, 480 N, 1 R = 0.667 m. 3 The area below the gate is 1 Ab = π R 2 = 3.142 m 2 , 4 and y =

B Q

W Y AX

and the centroid of Ab is at xb = R −

808

4R = 1.151 m. 3π

R5 2 m

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Problem 10.111 A spherical tank of 400-mm inner radius is full of water (ρ = 1000 kg/m 3 ). The pressure of the water at the top of the tank is 4 × 10 5 Pa. (a) What is the pressure of the water at the bottom of the tank? (b) What is the total force exerted on the inner surface of the tank by the pressure of the water?

Solution:

Strategy: For (b), draw a free-body diagram of the sphere of water in the tank.

W =

The weight density is γ = ρ g = 9810 N/m 3 . ((a) The pressure distribution is P( x ) = p 0 + γ x. ) The pressure at the bottom of the tank is P(0.8) = p 0 + γ (0.5) = 4 × 10 5 + (9810)(0.8) = 4.0785 × 10 5 Pa. (b) From the free body diagram of the sphere of water, the unbalanced force is the weight of water, acting downward:

( 43 ) π R γ = 2629.9 N 3

400 mm

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Chapter 11 Problem 11.1 Determine the reactions at A. Strategy: Subject the beam to three virtual motions: 1. a horizontal displacement δ x; 2. a vertical displacement δ y; and 3. a rotation δθ about A. y

δU = A x δ x = 0 ⇒ A x = 0 δU = (300 N + A y )δ y = 0 ⇒ A y = −300 N δU = ( M A − 800 N-m)δθ + (300 N)(4 m δθ) = 0 ⇒ M A = 400 N-m

300 N

800 N-m

Solution:

x 2m

Problem 11.2 The forces F1 = 1200 lb and F2 = 600 lb. Use virtual work to determine the reactions at the built-in support A.

Solution:

We draw the free-body diagram of the beam, showing the

reactions at A: y MA

400 ft-lb

Ax F2

308

x A

4 ft

If the beam is subjected to a virtual displacement δx, the resulting virtual work is δU = A x δ x + ( F2 sin 30 °) δ x = 0.

4 ft

6 ft

308

400 ft-lb

Subjecting it to a virtual displacement δy, the virtual work is δU = A y δ y + (−F1 ) δ y + ( F2 cos30 °) δ y = 0. Finally, subjecting it to a counterclockwise virtual rotation δα about A, the virtual work is δU = M A + (−400 ft-lb) δα + (−F1 )[(6 ft) δα] + ( F2 cos30 °)[(10 ft) δα] = 0. Solving these three equations yields A x = −300 lb, A y = 680 lb, M A = 2400 ft-lb. A x = −300 lb, A y = 680 lb, M A = 2400 ft-lb.

Problem 11.3

Determine the tension in the cable.

Solution:

When the beam rotates through a counterclockwise angle δθ, the virtual work is δU = 200(1.6δθ) − T sin 60 ° (0.8δθ) = 0,

so 608

200 N

T =

(200)(1.6) = 462 N. 0.8sin 60 ° T

0.8 m

1.6 m

200 N

608

0.8 m

810

1.6 m

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Problem 11.4 The bar is in equilibrium. Determine the total virtual work done by the forces and couple acting on the bar if it is subjected to (a) a virtual displacement δ y; (b) a counterclockwise virtual rotation δα about A.

The virtual work done by the 70-lb force is 70 δy ft-lb and the work done by the 80-lb force is 80 δy ft-lb. The 150-lb force is opposite to the direction of the virtual displacement, so the virtual work is −150 δ y ft-lb. The bar does not rotate, so the couple does no virtual work. The total virtual work is δU = 70 δ y + 80 δ y − 150 δ y ft-lb = 0.

y 100 ft-lb

(b) Next, we subject the beam to a virtual rotation δα about A:

150 lb C B

70 lb 6 ft

150 lb

80 lb

4 ft

(a) When we subject the beam to a virtual displacement δy,

100 ft-lb

70 lb

150 lb C B

da x

70 lb

80 lb

6 ft

Solution:

100 ft-lb

80 lb

4 ft

The rotation causes point B to undergo an upward displacement (6 ft)δα. (Keep in mind that the rotation is infinitesimal; the explanatory diagram necessarily shows a finite rotation.) As a result, the 150-lb force does virtual work (6)(−150)δα ft-lb. Point C undergoes an upward displacement (10 ft)δα, and the 80-lb force does virtual work (10)(80)δα ft-lb. The counterclockwise couple does virtual work 100 δα ft-lb. The total virtual work is δU = (6)(−150)δα + (10)(80)δα + 100 δα ft-lb = 0. (a) 70 δ y + 80 δ y − 150 δ y ft-lb = 0. (b) (6)(−150)δα + (10)(80)δα + 100 δα ft-lb = 0.

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Problem 11.5* The length L = 12 ft and w 0 = 200 lb/ft. Suppose that the beam is given a counterclockwise virtual rotation δα about A. (a) Use integration to determine the resulting virtual work done by the distributed load. (b) Represent the distributed load by an equivalent force and then determine the resulting virtual work done by the equivalent force. (Your answers will be in terms of δα. )

The downward virtual displacement of the beam at the position x is x δα, so the virtual work done on the element of width dx is

( wL x dx )(x dα). 0

We can integrate this expression from x = 0 to x = L to determine the virtual work done by the distributed load: δU = ∫

Lw 0

d α x 2 dx

w0 x3 L d α    3 0 L 1 = w 0 L2 d α 3 = 9600 d α ft-lb. =

w0 A

(b) We represent the distributed load by an equivalent force and then subject the beam to a virtual rotation about A: y

L 1 wL 2 0

Solution: (a) The distributed load is described by the linear function w = ( w 0 /L ) x. We subject the beam to a virtual rotation about A, and consider a element of the beam of width dx at a position x:

da 2 3L

y w

w 5 L0 x

x da

x L

The downward force exerted on the element of width dx is

The downward virtual displacement of the beam where the force is applied is (2/3) L δα, so the virtual work done is δU =

( 12 w L )( 23 L dα ) 0

1 = w 0 L2 d α 3 = 9600 d α ft-lb. (a),(b) δU = 9600 d α ft-lb.

w w dx = 0 x dx. L

812

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Problem 11.6 The length L = 12 ft and w 0 = 200 lb/ft. Use virtual work to determine the reactions at A and B.

Solution: Let us draw the free-body diagram of the beam, showing the reactions at A and B and representing the distributed load by an equivalent force: y

1w L 2 0 w0

If the beam is subjected to a virtual displacement δx, the resulting virtual work is δU = A x δ x = 0. Subjecting it to a virtual displacement δy, the virtual work is δU = A y δ y + B δ y −

( 12 w L ) δ y = 0. 0

Finally, we subject it to a counterclockwise virtual rotation δα about A: y 1 2 w0L Ax

B 2L 3

The downward virtual displacement of the beam where the equivalent force is applied is (2/3) L δα, so the virtual work done is

( 12 w L )( 23 L dα ) = 13 w L dα. 0

The downward virtual displacement of the beam at B is L δα, and the virtual due to the reaction at B is −BL d α. The total virtual work is δU =

1 w L2 d α − BL d α = 0. 3 0

Solving the three virtual work equations yields A x = 0, 1 A y = w 0 L = 400 lb, 6 1 B = w 0 L = 800 lb. 3 A x = 0, A y = 400 lb, B = 800 lb.

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Problem 11.7 The bar is 6 ft long, the distance b = 4 ft, and the angle α = 30 °. The bar’s 50-lb weight acts at its midpoint. What is the tension in the spring?

Let L denote the length of the bar. The angle β is β = arctan

( b +L cosL sinα α )

  (6 ft) cos30 ° = arctan    4 ft + (6 ft)sin 30 °  = 36.6 °. As a result of the virtual rotation, the top undergoes a virtual motion L δα in the direction perpendicular to the bar. The component of that motion in the direction parallel to the spring is a

δ S = ( L δα) cos(α + β ). If T denotes the tension in the spring, the virtual work due to the motion is (the force on the bar due to the tension is opposite to the direction of the virtual motion)

A b

Solution:

δU spring = −T δ S = −T ( L δα) cos(α + β ).

Consider the geometry if the bar is subjected to a virtual

rotation δα:

We can represent the bar’s weight by an equivalent downward force W acting at the midpoint of the bar. The virtual rotation causes the midpoint to undergo a virtual motion (1/2) L δα in the direction perpendicular to the bar. The component of that motion in the direction parallel to the weight is [(1/2) L δα]sin α, so the virtual work done by the weight is

b a

δU weight = W [(1/2) L δα]sin α.

Lda

The total virtual work equals zero, δU = δU spring + δU weight = −T ( L δα) cos(α + β ) + W [(1/2) L δα]sin α = 0,

1 Lda 2 a b

From which we obtain 1 W sin α T = 2 cos(α + β )

1 (50 lb)sin 30 ° 2 cos(30 ° + 36.6 °)

= 31.5 lb. 31.5 lb.

814

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Problem 11.8 support.

Determine the reaction at the roller

Solution:

Assume that B and E remain fixed, and give bar ABC a clockwise virtual rotation δθ:

Notice that (1) δθ = (1.5)δβ . 200 N

δU = 200(1.5)(δθ) − F (1)(δβ ) = 0,

1m 1.5 m

F =

200(1.5)(δθ) δβ

= 200(1.5)(1.5)

= 450 N. B 1.5 m

200 N

1m C

F A

1.5 m

du E

1.5 m B

1.5 m

1m C (1) du

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Problem 11.9 (a) If end A of the bar is subjected to a virtual displacement δ x, show that the resulting virtual displacement of end B of the bar is δy =

Solution: (a) Consider the geometry when the bar is subjected to the virtual displacement δx: y

b δ x. h

(b) The distances b = 3 ft and h = 6 ft. If the force F1 = 200 lb, what is the magnitude of the force F2 needed to keep the bar in equilibrium? (The bar’s weight is negligible.)

x dx

Let L be the length of the bar. We have (neglecting second-order terms in infinitesimals),

L2 = (b − δ x ) 2 + (h + δ y ) 2 = b 2 + h 2 − 2bδ x + 2hδ y = L2 − 2bδ x + 2hδ y. h

we see that δy =

A x

b δ x. h

(b) Assuming the bar to be in equilibrium, the total virtual work resulting from the virtual displacement is δU = F1 δ x − F2 δ y b = F1 δ x − F2 δ x = 0, h

b so

F2 = =

h F b 1 6 ft (200 lb) 3 ft

= 400 lb. (b) F2 = 400 lb.

816

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Problem 11.10 The distances b = 2 m and h = 4 m. The ends of the bar slide in smooth tracks. Suppose that end A is given a virtual displacement δ x. Determine the resulting virtual angular displacement δα of the bar. (Your answer will be in terms of δ x. )

Solution:

Consider the geometry when the bar is subjected to the virtual displacement δx: y dy B

y h a a1da

dx b Observe that

b − δ x = L cos(α + δα). Let us express the cosine in terms of its Taylor series (see Appendix A), neglecting terms of higher order in δα:

b − δ x = L cos α − L (sin α)δα = b − h δα. From this equation we obtain

x A

δα =

1 1 δx = δ x = 0.25 δ x. h 4m

δα = 0.25 δ x.

Problem 11.11 The distances b = 2 m and h = 4 m. The ends of the bar slide in smooth tracks. The bar is held in equilibrium in the position shown by the force F = 400 N. Neglect the bar’s weight. Determine the magnitude of the compressive force exerted on the bar by the spring.

Solution:

Consider the geometry when the bar is subjected to the virtual displacement δx: y B

h a a1da

A dx

Let L be the length of the bar. We have (neglecting second-order terms in infinitesimals), B

L2 = (b − δ x ) 2 + (h + δ y ) 2 = b 2 + h 2 − 2bδ x + 2hδ y = L2 − 2bδ x + 2hδ y. we see that

δy = a

x A

b 2m δx = δ x = 0.5δ x. h 4m

Let B denote the compressive force exerted on the bar by the spring. The total virtual work resulting from the virtual displacement is δU = F δ x − B δ y = (400 N) δ x − B(0.5 δ x ) = (400 N − 0.5B)δ x. The magnitude of the compressive force is B = 800 N. 800 N.

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Problem 11.12* Show that δ x is related to δα by δ x = ( L1 tan β ) δα.

Solution:

When the horizontal bar is subjected to the infinitesimal

rotation, L1

(See Practice Example 11.1.) da

L1da

b h

L1 b

its right end undergoes a downward infinitesimal displacement L1 δα. Observe that (h − L1 δα) 2 + (b + δ x ) 2 = L 2 2 . da

Expanding the two quantities on the left and neglecting second-order infinitesimals, we obtain h 2 − 2hL1 δα + b 2 + 2b δ x = L 2 2 . Because h 2 + b 2 = L22 , we see that δx = dx

h L δα b 1 L 2 sin β L δα L 2 cos β 1

= ( L1 tan β )δα.

Problem 11.13 The horizontal surface is smooth. Determine the horizontal force F necessary for the system to be in equilibrium. (See Practice Example 11.1.)

Solution:

Use the results of 11.12

δU = (400 in-lb)δα − F δ x From 11.12 δ x = ( L1 tan β )δα

9 in

= (400 in-lb)δα − F ((9 in) tan 50°)δα = (400 in-lb − F [(9 in) tan 50 °])δα = 0

6 in 400 in-lb

508

Thus F = 37.3 lb F

818

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Show that δ x is related to δα by

Problem 11.14*

L1 x sin α δα. δx = x − L1 cos α

Solution:

Applying the law of cosines,

L 2 2 = L1 2 + x 2 − 2 L1 x cos α. The derivative of this equation with respect to α is

Strategy: Write the law of cosines (Appendix A) in terms of α and evaluate the derivative of the resulting equation with respect to α. (See Practice Example 11.1.)

0 = 0 + 2x

dx dx − 2 L1 cos α + 2 L1 x sin α. dα dα

We write this equation as 0 = ( x − L1 cos α)dx + L1 x sin α d α, from which we conclude that

δx = −

L1 x sin α δα. x − L1 cos α

(The minus sign indicates that an increase in α results in a decrease in x, which is correct. The δα in the figure is defined to be a decrease in α. )

Problem 11.15 The angles α = 60 ° and β = 45 °. Determine the force F. (See Problem 11.14.)

When the left bar is subjected to a clockwise virtual rotation δα,

L1 da

b 2m

a dx

600 N-m

the virtual work done by the couple is δU couple = (600 N-m)δα.

Solution:

The resulting virtual displacement δx of the bottom of the right bar (see Problem 11.14) is

First, some geometry:

h a

x We can see that h = (2 m)sin α = L 2 sin β , so L 2 = (2 m)sin α / sin β = 2.45 m. Then the distance x is x = (2 m) cos α + (2.45 m) cos β = 2.73 m.

 L1 x sin α  δ x =   δα.  x − L1 cos α  The resulting virtual work done by the force F is δU force = −F δ x  L1 x sin α  = −F   δα.  x − L1 cos α  The total virtual work is δU = δU couple + δU force  L1 x sin α  = (600 N-m)δα − F  δα  x − L1 cos α  = 0. Solving for F yields (600 N-m)( x − L1 cos α) L1 x sin α = 220 N.

F =

F = 220 N.

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Problem 11.16 The linkage is in equilibrium. What is the force F ? (See Practice Example 11.1.) 400 lb

The length L R = (8 ft) 2 + (3 ft) 2 = 8.54 ft and the angles α R = arctan(3/8) = 20.6 °, β R = arctan(3/4) = 36.9 °. The force P is the horizontal force exerted by the other pair of bars. As a result of the virtual displacement δx undergone by the bottom of the left bar of this pair, the right bar will undergo a clockwise virtual rotation δα R . They are related by (See Problem 11.14)

3 ft

 L R x R sin α R  δ x =  δα .  x R − L R cos α R  R 6 ft

4 ft

(2)

The top of the right bar undergoes a virtual displacement L R δα R in the direction perpendicular to the bar, and the force F does virtual work

8 ft

δU R = −F cos α R ( L R δα R ) = −8F δα R ft-lb.

Solution:

We consider the left pair of bars, and subject the left bar to a clockwise virtual rotation δα L :

The total virtual work done on the linkage is δU = δU L + δU R = 1600 δα L − 8F δα R

400 lb

= 0.

(Notice that when we consider the linkage as a whole, the forces P cancel out. Another way of looking at it is that the virtual works done on each pair by the forces P cancel out.) Solving Eqs. (1) and (2) for δα L and δα R and substituting the results into this equation, we obtain

LL daL P

aL daL

The length L L = 5 ft and the angles α L = arctan(3/4) = 36.9 °, β L = arctan(3/6) = 26.6 °. The force P is the horizontal force exerted by the other pair of bars. The top of the left bar undergoes a virtual displacement L L δα L in the direction perpendicular to the bar. As a result, the 400-lb force does virtual work

 L R x R sin α R     x R − L R cos α R  δα = 360 lb. F = 200 L = 200  L L x L sin α L  δα R    x L − L L cos α L  F = 360 lb.

δU L = (400 lb) cos α L ( L L δα L ) = 1600 δα L ft-lb. The virtual rotation causes the bottom of the right bar of this pair to undergo a virtual displacement (See Problem 11.14)  L L x L sin α L  δ x =  δα .  x L − L1 cos α L  L

(1)

Next, we consider the right pair of bars:

F bR

LR daR LR

daR aR

P dx

820

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Problem 11.17 Bar AC is connected to bar BD by a pin that fits in the smooth vertical slot. The masses of the bars are negligible. If M A = 30 N-m, what couple M B is necessary for the system to be in equilibrium?

Solution: x 2 + y 2 = L2 = (0.7) 2 + (0.4) 2 . 2 x dx + 2 y dy = 0. The δy shown is a decrease in y, so

δ x = dx, δ y = −dy C

0.4 m

and δ y =

x 0.7 δx = δ x. y 0.4

δβ =

δx 1 = δ x. y 0.4

MB A

MA 0.7 m

(1)

δθ =

δy 2 δx 2 + δy 2 1 = 1+ δx L L δx

1 0.7 2 1+ δ x. 2 2 0.4 (0.7) + (0.4)

( )

(2)

The virtual work is δU = M Aδθ − M Bδβ = 0. Substituting Eqs. (1) and (2), we obtain ( M A − M B )δ x = 0, so M A = M B = 30 N-m.

MA db x

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Problem 11.18 The force F1 = 800 lb. Determine the magnitude of the force F2 required to keep the structure in equilibrium. (See Problem 11.12.)

Solution:

When the horizontal bar is subjected to a clockwise virtual rotation δα,

608

(2 ft)da (4 ft)da

608 4 ft

dx the midpoint of the bar undergoes a downward infinitesimal displacement (2 ft)δα. The resulting virtual work done by the force F1 is F2

2 ft

F1 (2 ft)δα. The virtual rotation causes the bottom of the slanted bar to undergo a virtual displacement (see Problem 11.12) δ x = (4 ft) tan60 ° δα. As a result, the force F2 does virtual work −F2 (4 ft) tan60 ° δα. The total virtual work is δU = F1 (2 ft)δα − F2 (4 ft) tan 60° δα = 0. We see that the force F2 is F2 =

2 ft F (4 ft) tan60 ° 1

= 231 lb. F2 = 231 lb.

822

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Problem 11.19 The structure is subjected to an 800-lb load and is held in equilibrium by a horizontal cable. Determine the tension in the cable. 800 lb

3 ft 608

608

3 ft

The point A, where the cable is attached, undergoes a virtual displacement (1/2) L δα perpendicular to the bar. The horizontal component of this displacement is (1/2) L δα cos30 °. Denoting the tension in the cable by T, the resulting virtual work is 1 δU cable = −T L δα cos30 °. 2 Point B, the top of the bar, undergoes a virtual displacement L δα. The vertical component of this displacement is L δα sin 30 °. The resulting virtual work done by the 800-lb force is δU force = (800 lb) L δα sin 30 °. The total virtual work on the structure is δU = δU force + δU cable

Solution:

Consider the right slanted bar. Let L be its length. Let us look at what happens when it is subjected to a counterclockwise virtual rotation δα:

1 = (800 lb) L δα sin 30 ° − T L δα cos30 ° = 0. 2 From this equation we obtain T = 2(800 lb) tan 30 ° = 924 lb.

924 lb. 308 L da da

L A

1 L da 2

608

Problem 11.20 If the load on the car jack is L = 6.5 kN, what is the tension in the threaded shaft between A and B?

Solution:

We have the constraint 2

( 2x ) + ( 2y ) = b ⇒ xδ x + yδ y = 0 2

x δ y = −  δ x y

From virtual work we know see 65 mm

 x δU = −Pδ x − Lδ y =  −P + L  δ y = 0  y P = L

65 mm

120 mm

(

)

x 240 mm = (6.5 kN) = 12.0 kN y 130 mm

b y

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Problem 11.21 Determine the reactions at A and B. (Use the equilibrium equations to determine the horizontal components of the reactions, and use the procedure described in Example 11.2 to determine the vertical components.)

Solution: Denote the angle α = 60° and the distance AC by L. The sum of the moments about A is ΣM A =

12 ( 2 tan ) (300) + ( tan12α ) B = 0, α x

from which B x = −150 lb. The sum of the forces: ΣFx = A x + B x + 300 = 0, from which A x = −B x − 300 = −150 lb. The sum of the forces in the vertical direction is

A 608

ΣFy = A y + B y = 0, from which A y = −B y . Denote the distance between A and B by y and the distance from the wall to the point of application of the force as x. Then 300 lb

x =

( 32 ) L − y . 2

Take the variation of both sides, δx =

608

( 32 )

−yδ y 1 = (−yδ y ), 8 L2 − y 2

( )

y = 12 tan α, from which

B 12 in

6 in

( )

δx 1 = − (12 tan α). δy 8 Perform a virtual elongation of the mechanism in the x-direction. The virtual work is δU = 300δ x − A y

δy δy + By = 0, 2 2

from which

( 12 ) ( A − B ) = 300 δδ xy = A . y

The vertical reaction is Ay =

( 32 ) 300 tan α = 779.4 lb,

and B y = − A y = −779.4 lb

824

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Problem 11.22 This device raises a load W by extending the hydraulic actuator DE. The bars AD and BC are each 2 m long, and the distances b = 1.4 m and h = 0.8 m. If W = 4 kN, what force must the actuator exert to hold the load in equilibrium? b

Solution: Perform a virtual vertical displacement of the load. Denote the distance CD by x. The virtual work is δU = −W δ h + Dδ x = 0, from which D = W

δh . δx

The distances are related: h = BC, from which

L2 − x 2 , where L is the length of bar

x δh x = − = − . δx h L2 − x 2

Thus A

D = −W h

x = −W h

L2 − h 2 . h

Substitute numerical values:

4 − 0.8 2 = −9.165 = −9.17 kN, 0.8

D = −4

where the negative sign implies that the force is directed parallel to the negative x axis.

Problem 11.23 Determine the force P necessary for the mechanism to be in equilibrium.

The horizontal position of pt E is x E = ( L AB + 2 L BD ) cos θ. From

dx E = −( L AB + 2 L BD )sin θ, dθ

we see that δ x E = −( L AB + 2 L BD )sin θδθ.

600 mm

The virtual work resulting from a virtual rotation δθ is F

δU = −Pδ y c + F δ y D − F δ x E = −P(2 L AB cos θδθ) + F ( L AD cos θδθ)

600 mm

+ F ( L AB + 2 L BD )sin θδθ = 0. Solving, 800 mm

400 mm

P =

[ L AD cos θ + ( L AB + 2 L BD )sin θ]F . 2 L AB cos θ

Substituting the lengths and θ = arctan(600/800), we obtain

Solution:

The height of pt. C is

P = 1.5 F.

y c = 2 L AB sin θ. From

dy c = 2 L AB cos θ, dθ

we obtain δ y c = 2 L AB cos θδθ. B

The height of pt. D is y D = L AD sin θ.

From dy D = L AD cos θ, dθ we obtain

u x A

δ y D = L AD cos θδθ.

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Problem 11.24 The collar A slides on the smooth vertical bar. The masses are m A = 20 kg and m B = 10 kg. (a) If the collar A is given an upward virtual displacement δ y, what is the resulting downward displacement of the mass B? (b) Use virtual work to determine the tension in the spring.

Solution:

The motion is constrained by the constant length L of the

string. L =

y A2 + d 2 + yB

0 =

y Aδ y A y Aδ y A + δyB ⇒ δyB = − yA2 + d 2 yA2 + d 2

(a) If the motion of a is δ y = −δ y A then we have 0.2 m (−δ y ) = 0.625δ y (0.2 m) 2 + (0.25 m) 2

δyB = −

0.25 m

Where positive means down δU = −(20 kg)(9.81 m/s 2 )δ y + [(10 kg)(9.81 m/s 2 ) − Fspring ](0.625δ y )

0.2 m

= [−196.2 N + (98.1 N)(0.625) − Fspring (0.625)]δ y = 0

Fspring = 216 N

B d5 0.25 m

yA A

Problem 11.25 The potential energy of a conservative system is given by V = 2 x 3 + 3 x 2 − 12 x. (a) For what values of x is the system in equilibrium? (b) Determine whether the equilibrium positions you found in (a) are stable or unstable.

Solution: (a) The system is in equilibrium when dV = 6 x 2 + 6 x − 12 = 0. dx This is a quadratic equation in x: it is put into canonical form by dividing by 6, x 2 + 2bx + c = 0, from which b = 12 and c = −2. The solutions are x = −b ±

b2 − c

x = 1 or x = −2. (b) The stable and unstable positions are determined by the sign of the second derivative: d 2V = 12 x + 6. d 2x d 2V For x = 1,  2  = 18 > 0,  d x  x =1 and the equilibrium is stable at x = 1. For d 2V x = −2,  2  = −18 < 0,  d x  x =−2 and the equilibrium is unstable at x = −2.

826

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Problem 11.26 The potential energy of a conservative system is given by V = 2q 3 − 21q 2 + 72q. (a) For what values of q is the system in equilibrium? (b) Determine whether the equilibrium positions you found in (a) are stable or unstable.

Solution: (a) The equilibrium positions are given by dV = 6q 2 − 42q + 72 = 0. dx In canonical form: q 2 + 2bq + c = 0, where 7 b = − , 2 and c = 12. The solutions: q = −b ±

b2 − c

q = 4 or q = 3. (b) The second derivative is d 2V = 12q − 42. dq 2 For q = 4,  d 2V  = 6 > 0,  2  dq  q = 4 and the equilibrium is stable. For q = 3,  d 2V  = −6 < 0,  2  dq  q = 3 and the equilibrium is unstable.

Problem 11.27 The mass m = 2 kg and the spring constant k = 100 N/m. The spring is unstretched when x = 0. (a) Determine the value of x for which the mass is in equilibrium. (b) Is the equilibrium position stable or unstable? (See Practice Example 11.3.)

Solution: (a) The potential energy is 1 V = kx 2 − mgx. 2 Setting dV = kx − mg = 0, dx we obtain mg (2)(9.81) x = = = 0.196 m. k 100 d 2V (b) = k > 0, so the equilibrium position is stable. dx 2

m x

Problem 11.28 The nonlinear spring exerts a force −kx + ε x 3 on the mass, where k and ε are constants. Determine the potential energy V associated with the force exerted on the mass by the spring. (See Practice Example 11.3.) x

BandF_6e_ISM_CH11.indd 827

Solution:

The potential energy of the spring is

V ( x ) = −∫ (−kx + ε x 2 )dx + C =

1 2 εx 4 kx − + C. 2 4

Choose x = 0 as the datum point, V (0) = 0, hence C = 0, and the potential energy associated with the force is V ( x) =

1 2 εx 4 kx − 2 4

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Problem 11.29 The 1-kg mass is suspended from the nonlinear spring described in Problem 11.28. The constants k = 10 and ε = 1, where x is in meters. (a) Show that the mass is in equilibrium when x = 1.12 m and when x = 2.45 m. (b) Determine whether the equilibrium positions are stable or unstable. (See Practice Example 11.3.)

Solution: In Problem 11.28 the potential energy of the spring is shown to be 1 εx 4 V ( x ) = kx 2 − . 2 4 The potential energy of the mass is V ( x ) = −∫ W dx = −Wx, where the datum is x = 0. The total potential energy is 1 εx 4 Vtotal ( x ) = −Wx + kx 2 − . 2 4 The equilibrium points are determined from dV = −W + kx − ε x 3 = 0. dx This cubic may be solved by iteration or by graphing the function f ( x ) = −W + kx − ε x 3

to find the zero crossings. Both methods were used here: the graph was used to get approximate values, and these values were then refined by iteration (using TK Solver Plus). The results: x = 1.12 m, and x = 2.45 m. (b) The second derivative of the potential energy at x = 1.12 m is  d 2V  = [10 − 3 x 2 ] x = 1.12 = 6.2 > 0,  dx 2  x = 1.12 and the system is stable. For x = 2.45 m,  d 2V  = [10 − 3 x 2 ] x = 2.45 = −7.98 < 0,  dx 2  x = 2.45 and the system is unstable.

Zero crossings

3 2 1 f(x) 0 21

1.122

2.448

1.2 1.4 1.6 1.8 2 X, meters

2.2 2.4 2.6

22 23

828

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Problem 11.30 The two straight segments of the bar are each of weight W and length L. Determine whether the equilibrium position shown is stable if (a) 0 < α 0 < 90 °; (b) 90 ° < α 0 < 180 °.

Solution:

From a heuristic argument, if the bars hang straight down (α 0 = 0) they are equivalent to one bar suspended at one end, and they should be stable in this position. If the bars are straight out (α 0 = 90 °) they are equivalent to a bar suspended at its midpoint, and as a balanced bar they should be neutrally stable. If the bars are positioned upward (90 ° < α 0 < 180 °), the bars will be unbalanced everywhere, and the system will be unstable in every position. This heuristic reasoning suggests the strategy to treat the composite bar as if it were a mass point suspended (supported) at the pin support. The mass point will be located at the center of weight of the composite. It will behave like a pendulum under the action of gravity. Choose a coordinate system with origin at the pin support with the x axis positive downward. Denote the right and left bars by the subscripts R and L. The center of weight of each bar has the coordinates: L cos α 0 , 2 L y R = sin α 0 ; 2 L x L = cos α 0 , 2 L y L = − sin α 0 . 2 xR =

The coordinates of the center of weight of the composite is x =

Wx R + Wx L L = cos α 0 , 2W 2

WyR + WyL = 0. 2W Suppose that a small angular displacement θ occurs. The potential energy of the equivalent system under this displacement is and y =

V = −(2W )

( L2 cos α )cos θ = −WL cos α cos θ. 0

The equilibrium points are determined by dV = WL cos α 0 sin θ = 0, dθ from which θ = 0, and θ = π. The stability of the equilibrium points is determined from  d 2V  = [WL cos α 0 cos θ] θ = 0 = WL cos α 0 .  d 2  θ θ=0 The stability depends upon the value of α 0 : (a) in the interval 0 < α 0 < 90 °, WL cos α 0 > 0, and the system is stable. (b) At the point α 0 = 0, WL > 0, and the system is stable. (c) In the interval 90 ° < α 0 < 180 °, WL cos α 0 < 0, and the system is unstable. If the mechanical constraints permit the system to reach the equilibrium point at θ = π, the results above are reversed: (a) is unstable, (b) is unstable, and (c) is stable.

L sin a 0 2 2W

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Problem 11.31 The homogeneous composite object consists of a hemisphere and a cylinder. It is at rest on the plane surface. Show that this equilibrium position is stable only if L < R/ 2. (See Example 11.4.)

Suppose that the system is subjected to a small angular rotation about the point of contact with the floor. The length of the equivalent pendulum is R − y. The potential energy due to this rotation is V = −( R − y) W cos θ. The point of equilibrium is dV = ( R − y) W sin θ = 0, dθ from which θ = 0 is a point of equilibrium. The stability is determined by

 d 2V   d θ 2 

θ=0

= [( R − y) W cos θ] θ = 0 = ( R − y) W .

Note that R and y are both positive numbers, from which, if R > y, the system is stable; if R < y, the system is unstable. Reduce algebraically:

( L ( L2 + R ) + 125 R ) 2

Solution:

An angular disturbance will cause the composite system to rock about the radial center of the hemisphere. The change in potential energy, if it occurs, must be a change in the height of the composite mass caused by motion about the radial center. This suggests the pendulum analogy: a point mass at the composite mass center, suspended (supported) from the radial center.

Choose a coordinate system with the y axis along the axis of the cylinder, positive upward, the x axis parallel to the floor, and the origin at the point of contact with the floor. The center of mass of the cylinder is located on the axis of the cylinder at y =

L + R. 2

RL +

L +

2 R 3

3 2 L2 L2 2 2 5 2 − RL − R − R R − 2 . 3 2 12 = 12 2 2 L + R L + R 3 3

Retain only the numerator, since the denominator must be positive always: The condition for stability is 3 2 L2 R > . 12 2 Take the positive square root: R > or L <

The center of mass of the hemisphere is located at y =

R−y = R−

2L , R . 2

5R . 8

The mass of the cylinder is m cyl = ρπ R 2 L, where ρ is the mass density in kg/m 3 . The mass of the hemisphere is 2 πρ R 3 . 3 The location of the mass centroid of the composite is y =

830

( L2 + R ) + 23 πρ R ( 58R ) = L ( L2 + R ) + 125 R . 2 2 πρ R L + ( ) πρ R L + R 3 3

πρ R 2 L

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Problem 11.32 The homogeneous composite object consists of a half-cylinder and a triangular prism. It is at rest on the plane surface. Show that this equilibrium position is stable only if h < 2 R. (See Example 11.4.)

Solution: y =

The center of mas is located

( 12 ρπ R 2 )

( 43πR ) − (ρ h[2 R]) (h /3) = 4 R − 2h 1 2

1 2 2 ρπ R

+ ρ 12 h[2 R]

6h + 3π R

When the object rotates through an angle θ, the potential energy is V = mgR − mgy cos θ dV = mgy sin θ > 0 ⇒ y > 0 dθ Imposing this condition we find that stability requires To be stable we must have

4 R 2 − 2h 2 > 0 ⇒ h <

R y R

Problem 11.33 The homogeneous bar has weight W, and the spring is unstretched when the bar is vertical (α = 0). (a) Use potential energy to show that the bar is in equilibrium when α = 0. (b) Show that the equilibrium position α = 0 is stable only if 2kL > W .

Solution: (a) The potential energy of the spring is Vspring = ∫ ks ds =

where the datum is α = 0. Noting s = Lα, then Vspring =

1 2 ks , 2

k 2 2 Lα . 2

The height of the center of the bar relative to the pinned end is L cos α; 2

using the pinned end as the datum: L

Vbar =

WL cos α, 2

from which Vtot =

k 2 2 WL Lα + cos α. 2 2

The equilibrium point is dVtot WL = kL2α − sin α = 0, dα 2 from which 2kL α − sin α = 0. W This is a transcendental equation in α, with at least one solution α = 0. (b) Stability is determined by  d 2Vtot  2kL 2kL =  − cos α  = − 1,    W  α= 0 W  dα 2  α = 0 from which if 2kL − 1 > 0, W the system is stable. Thus the condition for stability at α = 0 is 2kL > W

BandF_6e_ISM_CH11.indd 831

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Problem 11.34 Suppose that the bar in Problem 11.33 is in equilibrium when α = 20 °. (a) Show that the spring constant k = 0.490 W / L. (b) Determine whether the equilibrium position is stable.

Solution:

Use the results of the solution to Problem 11.33. The equilibrium condition is 2kL α − sin α = 0, W

from which, for α ≠ 0, divide by: k

2kL sin α = , α W or k = a

( WL ) sin2αα = 0.490 ( WL )

(where α is in radians.) (b) The condition for stability is L

 2kL − cos α  > 0,  W  α = 20 ° from which 2(0.4899) − 0.9397 > 0, which is satisfied. The system is stable. Check: From a heuristic argument, two sets of equilibrium conditions apply: If the spring force is weaker than the gravity force in the neighborhood of α = 0, the bar should rotate under the action of gravity until this tendency is balanced by the increased spring force, at which point by analogy with the mass-spring system (see Problem 11.29) the system should be in stable equilibrium. If, however, the spring force is greater that the gravity force in the neighborhood of α = 0, the system should remain close to α = 0, which will be a position of stable equilibrium. This heuristic argument is supported as follows: (1) The equilibrium condition 2kL α − sin α = 0, W for α ≠ 0, is satisfied if 2kL sin α = , α W (0 < α < 90 °), which is the first set of equilibrium conditions. The condition for stability is sin α − cos α > 0, α which is satisfied for all 0 < α < 90 °, and the system is stable for α > 0. Taking the limit as α goes to zero, the equilibrium position at α = 0 is neutrally stable if 2kL sin α = . α W (2) If 2kL > W , a second set of conditions apply: the α = 0 position is stable (see Problem 11.33), and this is the only equilibrium position since 2kL α − sin α = 0 W has no solution except α = 0 for 2kL > 1. W check.

832

BandF_6e_ISM_CH11.indd 832

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Problem 11.35 The bar AB has mass m and length L. The spring is unstretched when the bar is vertical (α = 0). The light collar C slides on the smooth vertical bar so that the spring remains horizontal. Show that the equilibrium position α = 0 is stable only if 2kL > mg.

Solution:

The potential energy is 1 L V = k ( L sin α) 2 + mg cos α, 2 2

so and

(1)

d 2V L = kL2 (cos 2 α − sin 2 α) − mg cos α dα 2 2 L = kL2 (2 cos 2 α − 1) − mg cos α. 2

(2)

Notice that dV /d α = 0 at α = 0. Substituting α = 0 into Eq. (2) yields

L dV = kL2 sin α cos α − mg sin α dα 2

L d 2V = kL2 − mg . dα 2 2 Therefore d 2V > 0 dα 2

only if 2kL > mg.

A L sin a

Problem 11.36 The bar AB in Problem 11.35 has mass m = 4 kg, length 2 m, and the spring constant is k = 12 N/m. (a) Determine the value of α in the range 0 < α < 90 ° for which the bar is in equilibrium. (b) Is the equilibrium position determined in part (a) stable?

L cos a 2

Solution: (a) From Eq. (1) at the solution of Problem 11.35,

(

)

dV L = kL2 cos α − mg sin α. dα 2 Setting kL2 cos α − mg

L = 0, 2

we obtain α = arccos

( 2mgkL )

 (4)(9.81)  = arccos    2(12)(2)  = 35.2 °. (b) From Eq. (2) of the solution of Problem 11.35,

d 2V L = kL2 (2 cos 2 α − 1) − mg cos α. dα 2 2 Substituting α = 35.2 °,

d 2V = (12)(2) 2 (2cos 2 35.2 ° − 1) dα 2 − (4)(9.81)(1)cos35.2 ° = −15.9 N-m. This equilibrium position is unstable.

BandF_6e_ISM_CH11.indd 833

833

01/04/23 11:50 AM

Problem 11.37 The bar AB has weight W and length L. The spring is unstretched when the bar is vertical (α = 0). The light collar C slides on the smooth horizontal bar so that the spring remains vertical. Show that the equilibrium position α = 0 is unstable. C

Solution:

The potential energy of the spring is

Vspring = ∫ ks ds =

1 2 ks , 2

where the datum is α = 0. Noting s = L(1 − cos α), then Vspring =

k 2 L (1 − cos α) 2 . 2

The height of the center of the bar above the pin joint is k

L cos α. 2 With the pin joint as the datum, the potential energy of the bar is

WL cos α, 2 k WL from which Vtot = L2 (1 − cos α) 2 + cos α. 2 2 The equilibrium point is Vbar =

dVtot WL = kL2 (1 − cos α)sin α − sin α = 0, dα 2

1m A

from which

( 2WkL (1 − cos α) − 1)sin α = 0, which has at least one solution: α = 0. Stability is determined by

(

) (

)

2kL 2kL  d 2V  sin 2 α + (1 − cos α) − 1 cos α  =  = −1.  d 2   W  α= 0 W α α= 0 d 2V Since  2  = −1 < 0,  dα  α = 0 the system is unstable at the equilibrium position α = 0.

Problem 11.38 The bar AB described in Problem 11.37 has a mass of 2 kg, and the spring constant is k = 80 N/m. (a) Determine the value of α in the range 0 < α < 90 ° for which the bar is in equilibrium. (b) Is the equilibrium position determined in (a) stable? C

Substitute numerical values to obtain 2kL = 8.1549. W The zero crossing of a graph of f (α) =

( 2WkL ) (1 − cos α) − 1

was determined approximately over the interval 0 < α < 90 °, and this crossing value was then refined by iteration (using TK Solver Plus). k

The equilibrium point occurs at: α = 28.67 = 28.7 °. The condition for

(

) (

)

2kL 2kL  d 2V  sin 2 α + (1 − cos α) − 1 cos  . =   d 2   W  α = αi W α α = αi For α = 28.7 °,  d 2V  = 1.88 > 0,  d α 2  α = 28.7

so the system is stable at this equilibrium point 1m

f(a) vs a

1.5 1

Solution:

.5 Use the solution to Problem 11.37. The condition for

equilibrium is 2kL (1 − cos α) − 1 sin α = 0, W

2.5

from which the non-zero position of equilibrium is determined by

21.5 0

(

)

( 2WkL (1 − cos α) − 1) = 0,

834

f(a) 0 28.678

21 20

40 60 a , deg

100

BandF_6e_ISM_CH11.indd 834

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Problem 11.39 The homogeneous bar is 4 ft long and weighs 6 lb. Its ends slide in smooth tracks. The spring is unstretched when the angle α = 90 ° and the spring constant k = 5 lb/ft. Determine the angle α > 0 at which the bar is in equilibrium. y

Solution: The stretch of the spring is L − L sin α, so the potential energy of the spring is 1 k ( L − L sin α) 2 . 2 The height of the center of mass of the bar above the origin is (1/2) L sin α, so the potential energy associated with the weight W of the bar is 1 WL sin α. 2 The total potential energy is V =

1 1 k ( L − L sin α) 2 + WL sin α. 2 2

The bar is in equilibrium when

1 dV = k ( L − L sin α)(−L cos α) + WL cos α = 0. dα 2 This equation can be solved for sin α: W sin α = 1 − . 2kL 6 lb = 1− 2(5 lb/ft)(4 ft) = 0.85.

We obtain α = 58.2 °. α = 58.2 °. a x A

BandF_6e_ISM_CH11.indd 835

835

01/04/23 11:50 AM

Problem 11.40 The homogeneous bar is 4 ft long and weighs 6 lb. Its ends slide in smooth tracks. The spring is unstretched when the angle α = 90 ° and the spring constant k = 5 lb/ft. Show that the angle α > 0 at which the bar is in equilibrium is a stable position.

This equation can be solved for sin α: sin α = 1 −

W . 2kL

6 lb 2(5 lb/ft)(4 ft) = 0.85. = 1−

The equilibrium position is α = 58.2 °.

The second derivative of the potential energy is d 2V 1 = k (−L cos α) 2 + k ( L − L sin α) L sin α − WL sin α dα 2 2 1 2 2 2 = kL (cos α + sin α − sin α) − WL sin α. 2 At the equilibrium position α = 58.2 °, d 2V /d α 2 = 22.2 ft-lb. The equilibrium position is stable. Graphs of the potential energy and its derivatives are shown. Notice from the graphs that the equilibrium position α = 90 ° is unstable.

It is stable.

V, ft-lb

20 15 10 30

a x

60 a, degrees

60 70 a, degrees

60 a, degrees

Solution: The stretch of the spring is L − L sin α, so the potential energy of the spring is 1 k ( L − L sin α) 2 . 2

dv , ft-lb da

220 230

The height of the center of mass of the bar above the origin is (1/2) L sin α, so the potential energy associated with the weight W of the bar is

The total potential energy is V =

1 1 k ( L − L sin α) 2 + WL sin α. 2 2

The bar is in equilibrium when dV 1 = k ( L − L sin α)(−L cos α) + WL cos α = 0. dα 2

836

100 d2v, ft-lb da2

1 WL sin α. 2

0 210

50 0 250

BandF_6e_ISM_CH11.indd 836

01/04/23 11:50 AM

Problem 11.41 The pinned bars are held in place by the linear spring. Each bar has weight W and length L. The spring is unstretched when α = 90 °. Determine the value of α in the range 0 < α < 90 ° for which the system is in equilibrium. (See Example 11.5.)

Solution:

The potential energy is 1 L V = 2 W sin α + k (2 L cos α) 2 2 2 dV For equilibrium we must have = WL cos α − 4 kL2 cos α sin α = 0 dα

(

)

Solving we find α = sin −1

( 4WkL )

L k

a a

Problem 11.42 Determine whether the equilibrium position found in Problem 11.41 is stable or unstable. (See Example 11.5.)

Solution:

See 11.41

d 2V = −WL sin α − 4 kL2 cos 2 α + 4 kL2 sin 2 α dα 2 = −WL sin α + 8kL2 sin 2 − 4 kL2 We need to evaluate this expression at the equilibrium position (given in problem 11.41)

L k

a a

(

)

(

)

W W 2 W2 d 2V = −WL + 8kL2 − 4 kL2 = − 4 kL2 > 0 dα 2 4 kL 4 kL 4k For stability we therefore need W > 4 kL Assuming that an equilibrium position exists (Problem 11.41) then this condition cannot be met. We conclude that the equilibrium position is unstable

Problem 11.43 The bar weighs 15 lb. The spring is unstretched when α = 0. The bar is in equilibrium when α = 30 °. Determine the spring constant k.

Solution:

Let L = 2 ft be the length of the bar. (Notice that the unstretched length of the spring equals the length of the bar.) From the geometry, the stretch of the spring at an angle α is S = =

(2 L − L cos α) 2 + ( L sin α) 2 − L 5 L2 − 4 L2 cos α − L.

The potential energy of the spring is 1 2 kS 2 2 1 = k ( 5 L2 − 4 L2 cos α − L ) . 2 The height of the center of mass of the bar above the lower support is (1/2) L cos α, so the potential energy associated with the weight W of the bar is

Vspring =

4 ft a

Vweight =

1 WL cos α. 2

The total potential energy is V = Vspring + V weight 2 1 1 k ( 5 L2 − 4 L2 cos α − L ) + WL cos α. 2 2 The bar is in equilibrium when

2 ft

dV 1 = 2kL2 [(5 − 4 cos α) 1/ 2 − 1](5 − 4 cos α) −1/ 2 sin α − WL sin α dα 2 = 0. Setting α = 30 ° and solving for k, we obtain k = 9.71 lb/ft.

BandF_6e_ISM_CH11.indd 837

837

01/04/23 11:50 AM

Problem 11.44* The bar is 6 ft long and its 50-lb weight acts at its midpoint. The distance b = 4 ft. The spring is unstretched when the bar is vertical and the spring constant k = 20 lb/ft. Determine the angle α > 0 at which the bar is in equilibrium.

k a A

S =

(b + L sin α) 2 + ( L cos) 2 −

b 2 + L2 + 2bL sin α −

b 2 + L2

b 2 + L2 .

The potential energy of the spring is 1 2 kS 2 1 = k  b 2 + L2 + 2bL sin α − 2

Vspring =

b 2 + L2  .

The height of the center of mass of the bar above the supports is (1/2) L cos α, so the potential energy associated with the weight W of the bar is Vweight =

1 WL cos α. 2

The total potential energy is

Solution:

the stretch of the spring at an angle α is

V = Vspring + V weight

Let L be the length of the bar. From the geometry,

1  2 k b + L2 + 2bL sin α − 2 

b 2 + L2  +

1 WL cos α. 2

The bar is in equilibrium when dV = k[(b 2 + L2 + 2bL sin α) 1/ 2 − (b 2 + L2 ) 1/ 2 ] dα 1 (b 2 + L2 + 2bL sin α) −1/ 2 bL cos α − WL sin α 2 = 0. Consider graphs of the potential energy and its derivative: 153

152

V, ft–lb

151

150

20 25 a, degrees

dV, ft–lb da

10 0

210 220

From the graph we estimate that the bar is in equilibrium at α = 27.5 °. By using software designed to find roots of nonlinear algebraic equations we obtain α = 27.4913 °. (Notice from the shape of the potential energy curve that the equilibrium position is unstable.) α = 27.5 °.

838

BandF_6e_ISM_CH11.indd 838

01/04/23 11:50 AM

Problem 11.45 (a) Determine the couple exerted on the beam at A. (b) Determine the vertical force exerted on the beam at A.

(a)

Perform a virtual rotation about A: δU = M Aδθ − 200δθ + 2(100)sin 30 °δθ = 0, from which

100 N

200 N-m

Solution:

( M A − 200 + 2(100)sin 30 °)δθ = 0,

308

from which M A = 200 − 2(50) = 100 N m. 2m

(b)

Perform a virtual translation of the bar in the y-direction: δU = A y δ y + 100(sin 30 °)δ y = 0, from which ( A y + 50)δ y = 0, or A y = −50 N

Problem 11.46 The structure is subjected to a 20 kN-m couple. Determine the horizontal reaction at C.

Solution:

The interior angle at B is 100 °. Denote the axial force in BC by R BC . Do a virtual rotation of member AB about A:

δU = (−20 + 2 R BC sin100 °)δθ = 0,

from which

20 kN-m

R BC =

20 = 10.15 kN. 2 cos10 °

The horizontal component of the axial force is 408

408

R x = R BC cos 40 ° = 7.778 = −7.8 k Ν (directed parallel to the negative x axis.)

Problem 11.47 The “rack and pinion” mechanism is used to exert a vertical force on a sample at A for a stamping operation. If a force F = 30 lb is exerted on the handle, use the principle of virtual work to determine the force exerted on the sample.

Solution: Perform a virtual rotation of the handle. The virtual work is δU = 8F δθ + Aδ x = 0, from which A = − 8F

δθ . δx

The angular rotation is related to the vertical translation by 2δθ = −δ x, from which δθ 1 = − , δx 2 and A = 4 F = 120 lb

2 in

8 in A

BandF_6e_ISM_CH11.indd 839

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01/04/23 11:50 AM

Problem 11.48 If you were assigned to calculate the force exerted on the bolt by the pliers when the grips are subjected to forces F as shown in Fig. a, you could carefully measure the dimensions, draw free-body diagrams, and use the equilibrium equations. But another approach would be to measure the change in the distance between the jaws when the distance between the handles is changed by a small amount. If your measurements indicate that the distance d in Fig. b decreases by 1 mm when D is decreased 8 mm, what is the approximate value of the force exerted on the bolt by each jaw when the forces F are applied?

Solution: Let L be the distance between the points of application of the forces F and the point of application of the gripping force at the jaw. The ratio of the motions indicates that the “effective” axis of rotation of a jaw is located 89 L from the point of application of F. Perform a virtual rotation about this axis: δU = ( 89 ) LF δθ − ( 19 ) LF δθ = 0, from which (8F − f )δθ = 0, or f = 8F . This result is approximate because some work is done by the mechanism as the handle is closed. In addition, the closure of the handles produces a translation of one handle relative to the other in the direction required to close the jaws, and this translation does work since it is associated with moment about the effective axis; hence not all of the virtual work due to a virtual rotation is included in the above expression.

F (a)

d D

(b)

Problem 11.49 The system is in equilibrium. The total weight of the suspended load and assembly A is 300 lb. (a) By using equilibrium, determine the force F. (b) Using the result of (a) and the principle of virtual work, determine the distance the suspended load rises if the cable is pulled downward 1 ft at B.

Solution: (a)

Isolate the assembly A. The sum of the forces: ΣFy = −W − 3F = 0,

where F is the tension in the cable, from which W F = = 100 lb. 3 (b) Perform a virtual translation of the assembly A in the vertical direction. The virtual work: δU = −W δ y + F δ x = 0, from which δx W = = 3. F δy The ratio of translations of the assembly A and the point B is 1 = 3, from which y A = 13 ft yA

840

BandF_6e_ISM_CH11.indd 840

01/04/23 11:51 AM

Problem 11.50 The system is in equilibrium. (a) By drawing free-body diagrams and using equilibrium equations, determine the couple M. (b) Using the result of (a) and the principle of virtual work, determine the angle through which pulley B rotates if pulley A rotates through an angle α. 200 mm

200 N-m 100 mm

(a) For pulley A: (1) (T3 − T4 )(0.1) = 200 N m, For pulley B (2) M = (T1 − T2 )(0.2). For the center pulley, (3) (T1 − T2 )(0.1) = (T3 − T4 )(0.2). Combine and solve: M = (4)(200) = 800 N m (b) Perform a virtual rotation of the pulley A. The virtual work of the system is δU = M 1δα − M δθ = 0, from which M δθ 200 1 = 1 = = , δα M 800 4 α from which θ = 4

100 mm A

Solution: The pulleys are frictionless and the belts do not slip. Denote the left pulley by A and the right pulley by B. Denote the upper and lower tensions in the belts at pulley A by T3 , T4 , at B by T1 , T2 .

200 mm

Problem 11.51 The mechanism is in equilibrium. Neglect friction between the horizontal bar and the collar. Determine M in terms of F, α, and L.

200 mm 200 T3 Nm 100 A mm T4 100 mm

T1 M B

200 mm

Solution:

Perform a virtual rotation about the left pin support. δU = M δα − F δ x = 0, from which M = F

δx . δα

Using the results of the solution to Problem 11.14, 2L

L M

M = F

a F

( x −xL Lsincosα α ).

From the dimensions given and the cosine law, 4 L2 = x 2 + L2 − 2 Lx cos α, from which x 2 − 2 xL cos α − 3 L2 = 0, which has the solution x = L cos α ±

L2 cos 2 α + 3 L2 = L (cos α ±

cos 2 α + 3).

Since a negative value of x has no meaning here, x = L(cos α +

cos 2 α + 3).

Substitute into the expression for the moment, and reduce:

(

M = FL sin α 1 +

)

cos α . cos 2 α + 3

From the identity, cos 2 α = 1 − sin 2 α, an alternate form of the solution is cos α M = FL sin α 1 + 4 − sin 2 α

(

BandF_6e_ISM_CH11.indd 841

)

841

01/04/23 11:51 AM

Problem 11.52 In an injection casting machine, a couple M applied to arm AB exerts a force on the injection piston at C. Given that the horizontal component of the force exerted at C is 4 kN, use the principle of virtual work to determine M.

350 mm

300 mm

Solution:

Perform a virtual rotation of the crank. The virtual work is

δU = M δθ + F δ x = 0, from which δx M = −F . δθ Denote the interior angle ACB by β, and the interior angle ABC by α. From the sine law, 0.35 0.3 = , sin 45 ° sin β from which β = sin −1

458

A M

0.3 sin 45 ° ) = 37.3 °, ( 0.35

and α = 180 ° − 45 ° − 37.3 ° = 97.69 °.

The distance AC is x 0.35 = , sin α sin 45 ° from which x = 0.490 m. From the solution to Problem 11.14, 0.3(0.49)sin 45 ° δx = = 0.374. δθ 0.49 − 0.3cos 45 ° The moment is M = −0.374 F = −0.374(4) = −1.5 kN m

Problem 11.53 Show that if bar AB is subjected to a clockwise virtual rotation δα, bar CD undergoes a counterclockwise virtual rotation (b /a) δα.

Solution:

The coordinates of pts B and C are

B x = 400sin α, B y = 400 cos α, C x = (a + b) − b cos β ,

C y = −b sin β .

400 mm 6 kN-m

We know that

(C x − B x ) 2 + (C y − B y ) 2 = constant. The derivative of this equation with respect to α is

600 mm

2(C x − B x )

( dCdα − dBdα ) + 2(C − B ) dCdα − dBdα  x

(

= 2(a + b − b cos β − 400sin α) b sin β

(

dβ − 400 cos α dα

)

dβ + 400sin α = 0. + 2(−b sin β − 400 cos α) −b cos β dα At α = 0, β = 0, this equation is

(

2a(−400) + 2(−400) −b

)

dβ = 0, dα

from which we obtain a δβ = δα. b y B a

x b C

842

BandF_6e_ISM_CH11.indd 842

01/04/23 11:51 AM

Problem 11.54 The system is in equilibrium, a = 800 mm, and b = 400 mm. Use the principle of virtual work to determine the force F.

Solution:

See the solution of Problem 11.53. When bar AB undergoes a clockwise virtual rotation δα, the virtual work is

δU = 6δα − F (0.6δβ ) a = 6δα − F 0.6 δα = 0, b so

(

B F

400 mm 6 kN-m

A a

F =

)

6b = 5 kN. 0.6a

600 mm

Problem 11.55 Show that if bar AB is subjected to a clockwise virtual rotation δα, bar CD undergoes a clockwise virtual rotation [ad /(ac + bc − bd )] δα. C

Solution: Denote the interior acute angle formed by BC with the horizontal by β, the obtuse interior angle at C by γ, and the interior acute angle at D by ψ. Perform a virtual translation δX parallel to the bar BC. (Note: This is often a useful step where cranks- and connecting-rod-like mechanisms are involved.) Then δ X cos β = d δα, and δ X sin γ = CDδθ, from which

d cos β δθ = . δα CD sin γ

24 N-m

d A

Noting that D

sin γ = sin (90 ° − β + 90 ° − ψ) = sin (β + ψ) = sin β cos ψ + sin ψ cos β ,

and

a , BC c sin ψ = , CD c−d sin β = , BC b cos ψ = . CD cos β =

Substitute: δθ = δα CD

Problem 11.56 The system is in equilibrium, a = 300 mm, b = 350 mm, c = 350 mm, and d = 200 mm. Use the principle of virtual work to determine the couple M.

(

Solution:

a ( BC )

(c − d ) a c b × + × BC CD CD BC

)

ad ac + bc − bd

Perform a virtual rotation of the crank at A. The virtual

work is δU = −M 1δα + M δθ = 0, from which

M δθ = 1. δα M

From the solution Problem 11.55, δθ ad = , δα ac + bc − bd

from which

24 N-m

d A

M 1 (ac + bc − bd ) = (24)2.625 = 63 N m ad

BandF_6e_ISM_CH11.indd 843

M =

843

01/04/23 11:51 AM

Problem 11.57 The mass of the bar is 10 kg, and it is 1 m in length. Neglect the masses of the two collars. The spring is unstretched when the bar is vertical (α = 0), and the spring constant is k = 100 N/m. Determine the values of α at which the bar is in equilibrium.

Solution: V =

1 2

The potential energy of the spring is

ks 2 .

Noting that s = L(1 − cos α), then V = 12 kL2 (1 − cos α) 2 . The potential energy of the bar is WL cos α, 2

Vbar =

where the datum point is the lower pin joint. From which kL2 WL (1 − cos α) 2 + cos α. 2 2

Vtot =

The condition for equilibrium is

(

)

dV 2kL = (1 − cos α) − 1 sin α = 0. dα W

The equilibrium points are α = 0, and the value of α determined by 2kL (1 − cos α) − 1 = 0, W from which cos α = 1 −

W . 2kL

Substitute numerical values: cos α = 0.5095, or α = cos −1 (0.5095) = 59.369 = 59.4 °

Problem 11.58 Determine whether the equilibrium positions of the bar in Problem 11.57 are stable or unstable.

Solution:

Use the solution to Problem 11.57. The equilibrium con-

dition is

(

)

dV 2kL = (1 − cos α) − 1 sin α = 0. dα W The stability condition is determined by

(

)

(

)

2kL 2kL  d 2V  =  (sin 2 α − cos 2 α) + − 1 cos α  .  2   W  α = αi d α α = αi W For α = 0,  d 2V  = −1 < 0,  d α 2  α= 0

so the equilibrium point is unstable. For α = 59.4 °,  d 2V  = 1.51 > 0,  d α 2  α = 59.4 ° so the equilibrium point is stable.

844

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Problem 11.59 The spring is unstretched when α = 90 °. Determine the value of α in the range 0 < α < 90 ° for which the system is in equilibrium.

Solution:

Choose a coordinate system such that the equilibrium position of the spring occurs at x = 0 and at y = L. The potential energy is

V = mgy +

kx 2 . 2

Noting that

1 L 2

y = L sin α and x = L cos α,

then kL2 cos 2 α. 2 The equilibrium condition is

1 L 2

1 L 2 a

V = mgL sin α +

dV = 0 = (mg − kL sin α) cos α = 0. dα This has two solutions: cos α = 0, mg . kL In the interval 0 < α < 90 °, only one solution exists, mg α = sin −1 kL

and sin α =

( )

Problem 11.60 Determine whether the equilibrium position found in Problem 11.59 is stable or unstable.

Solution:

Use the solution to Problem 11.59. The stability condition

is d 2V = −mg sin α + kL (sin 2 α − cos 2 α) = kL (sin 2 α − 1) < 0. dα 2

1 L 2

The system is unstable.

m 1 L 2

1 L 2 a

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Problem 11.61 The hydraulic cylinder C exerts a horizontal force at A, raising the weight W. Determine the magnitude of the force the hydraulic cylinder must exert to support the weight in terms of W and α.

Solution:

The distance x = 2.5b, so δ x = 2.5δ b.

Notice that y 2 + b 2 = constant. Taking the derivative of this equation with respect to y, 2 y + 2b

db = 0, dy

we obtain b δ y = − δ b. y

The virtual work is δU = −W δ y − F δ x  b  = −W  − δ b  − F (2.5δ b) = 0,  y 

so F =

b W . W = 2.5 y 2.5 tan α y

1 b 2

F a

x x

Problem 11.62 The homogeneous composite object consists of a hemisphere and a cone. It is at rest on the plane surface. Show that this equilibrium position is stable only if h < 3R.

For a small angular rotation θ the length of the equivalent pendulum is R − y. The potential energy due to this rotation is V = −( R − y) W cos θ. The point of equilibrium is dV = ( R − y) W sin θ = 0, dθ from which θ = 0 is a point of equilibrium. The stability is determined by

 d 2V  = [( R − y) W cos θ] θ = 0 = ( R − y) W .  d θ 2  θ=0

from which, if R > y, the system is stable; if R < y, the system is unstable. Reduce:

Solution:

Use the same strategy used to solve Problem 11.31: a point mass at the composite mass center, suspended (supported) at the radial center. Choose a coordinate system with the y axis along the axis of the cone, positive upward, the x axis parallel to the floor, and the origin at the point of contact with the floor. The mass of the cone is 1 m cyl = ρπ R 2 h, 3 where ρ is the mass density. The center of mass of the cone is h y = + R. 4 The mass of the hemisphere is 2 πρ R 3 . 3 5R The center of mass of the hemisphere is located at y = . 8 The location of the mass centroid of the composite is h h h h 2 5R 5 2 + R + πρ R 3 + R + πρ R 2 R 3 4 3 8 = 3 4 12 . y = h 2 h 2 + R πρ R 2 + πρ R 3 3 3 3 3

(

)

( )

(

R2 h2 Rh 2 h2 Rh 5 2 + R2 − − − − R 3 3 12 3 12 4 12 R−y = = , h 2 2R h + R + 3 3 3 3 from which, (since the denominator is always positive) if R2 h2 − > 0, 4 12 the system is stable. Thus h2 R2 > , 3 or, taking the positive square root:

3R > h

)

( )

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