Engineering Mechanics, Statics, 15th Edition by Russell Hibbeler SOLUTUINS MANUAL by ACADEMIAMILL

© 2022 by R. C. Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1–1. Round off the following numbers to three significant figures: (a) 3.45555 m, (b) 45.556 s, (c) 5555 N, (d) 4525 kg.

SOLUTION Ans.

(a) 3.46 m (b) 45.6 s (c) 5.56 kN (d) 4.52 Mg

Ans: a) 3.46 m b) 45.6 s c) 5.56 kN d) 4.52 Mg 1

1–2. Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) Mg>mm, (b) mN >ms, (c) mm # Mg.

SOLUTION a) kN>ms = 103N> ( 10 - 6 ) s = GN>s

Ans.

b) Mg>mN = 106g>10 - 3 N = Gg>N

Ans.

c) MN>(kg # ms) = 10 N>kg(10 6

-3

s) = GN>(kg # s)

Ans.

Ans: a) GN>s b) Gg>N c) GN>(kg # s) 2

1–3. Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) Mg/mm, (b) mN>ms, (c) mm # Mg.

SOLUTION a) Mg>mm = b) mN>ms =

103 kg -3

10 m 10 - 3 N 10 - 6 s

106 kg = Gg>m m

Ans.

103 N = kN>s s

Ans.

c) mm # Mg = C 10 -6 m D # C 103 kg D = (10)-3 m # kg = mm # kg

Ans.

Ans: a) Gg>m b) kN>s c) mm # kg 3

*1–4. What is the weight in newtons of an object that has a mass of (a) 8 kg, (b) 0.04 kg, and (c) 760 Mg?

SOLUTION Ans.

a) W = 9.81(8) = 78.5 N b) W = 9.81(0.04) ( 10 - 3 ) = 3.92 ( 10 - 4 ) N = 0.392 mN

Ans.

c) W = 9.81(760) ( 10 ) = 7.46 ( 10 ) N = 7.46 MN

Ans.

Ans: a) W = 78.5 N b) W = 0.392 mN c) W = 7.46 MN 4

1–5. Represent each of the following as a number between 0.1 and 1000 using an appropriate prefix: (a) 45 320 kN, (b) 568(105) mm, and (c) 0.00563 mg.

SOLUTION a) 45 320 kN = 45.3 MN

Ans.

b) 568 ( 105 ) mm = 56.8 km

Ans.

c) 0.00563 mg = 5.63 mg

Ans.

Ans: a) 45.3 MN b) 56.8 km c) 5.63 mg 5

1–6. Round off the following numbers to three significant figures: (a) 58 342 m, (b) 68.534 s, (c) 2553 N, and (d) 7555 kg.

SOLUTION a) 58.3 km

b) 68.5 s

c) 2.55 kN

Ans.

d) 7.56 Mg

Ans: a) 58.3 km b) 68.5 s c) 2.55 kN d) 7.56 Mg 6

1–7.

Represent each of the following quantities in the correct SI form using an appropriate prefix: (a) 0.000 431 kg, (b) 35.3(103) N, and (c) 0.005 32 km.

SOLUTION a) 0.000 431 kg = 0.000 431 A 103 B g = 0.431 g

Ans.

b) 35.3 A 103 B N = 35.3 kN

Ans.

c) 0.005 32 km = 0.005 32 A 103 B m = 5.32 m

Ans.

Ans: a) 0.431 g b) 35.3 kN c) 5.32 m 7

*1–8. Represent each of the following combinations of units in the correct SI form: (a) Mg>ms, (b) N>mm, and (c) mN>(kg # ms).

SOLUTION a)

Mg 103 kg = 106 kg>s = Gg>s = ms 10-3 s

Ans.

1N N = 103 N>m = kN>m = mm 10-3 m

Ans.

mN 10-3 N = kN>(kg # s) = # (kg ms) 10-6 kg # s

Ans.

Ans: a) Gg>s b) kN>m c) kN>(kg # s) 8

1–9.

Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) m>ms, (b) mkm, (c) ks>mg, and (d) km # mN.

SOLUTION a) m>ms = ¢

1102 m m = ¢ ≤ = km>s -3 ≤ s 1102 s

Ans.

b) mkm = 1102-611023 m = 1102-3 m = mm 3

c) ks>mg = d) km # mN =

1102 s 1102

-6

10 3 m

Ans.

1102 s

= Gs>kg

Ans.

10 -6 N = 10 -3 m # N = mm # N

Ans.

Ans: a) km>s b) mm c) Gs>kg d) mm # N 9

1–10. Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) GN # mm, (b) kg>mm, (c) N>ks2, (d) kN>ms

SOLUTION

a) GN # mm = 109 ( 10-6 ) N # m = kN # m

Ans.

b) kg>mm = 103 g>10-6 m = Gg>m 2

6 2

c) N>ks = N>10 s = 10

-6

Ans. 2

Ans.

N>s = mN>s

d) kN>ms = 103 N>10-6 s = 109 N>s = GN>s

Ans.

Ans: a) kN # m b) Gg>m c) mN>s2 d) GN>s 10

1–11. Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) mMN, (b) N>mm, (c) MN>ks2, and (d) kN>ms.

SOLUTION a) mMN = 10 - 6 (106) N = N

Ans.

N N = 106 N>m = MN>m = mm 10-6 m

Ans.

106 N MN = = N>s2 2 ks (103)2 s2

Ans.

kN 103 N N = = 106 = MN>s -3 s ms 10 s

Ans.

Ans: a) N b) MN>m c) N>s2 d) MN>s 11

*1–12. Water has a density of 1.94 slug>ft 3. What is the density expressed in SI units? Express the answer to three significant figures.

SOLUTION Using Table 1–2, we have rw = a

1.94 slug ft

14.5938 kg 1 slug

1 ft 3 b 0.30483 m3

= 999.8 kg>m3 = 1.00 Mg>m3

Ans.

Ans: rw = 1.00 Mg>m3 12

1–13. The density (mass> volume) of aluminum is 5.26 slug>ft3. Determine its density in SI units. Use an appropriate prefix.

SOLUTION 5.26 slug>ft3 = a

5.26 slug 3

3 14.59 kg ft b a b 0.3048 m 1 slug

= 2.71 Mg>m3

Ans.

Ans: 2.71 Mg>m3 13

1–14. Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (212 mN)2, (b) (52800 ms)2, and (c) [548(106)]1>2 ms.

SOLUTION a) (212 mN)2 = 3 212(10)-3 N 4 2 = 0.0449 N2 = 44.9(10)-3 N2

b) (52 800 ms)2 = 3 52 800(10)-3 4 2 s2 = 2788 s2 = 2.79 ( 103 ) s2 c) 3 548(10) 4 ms = (23 409)(10) 6

1 2

-3

s = 23.4(10) (10) s = 23.4 s

Ans. Ans. Ans.

Ans: a) 44.9(10)-3 N2 b) 2.79 ( 103 ) s2 c) 23.4 s 14

1–15.

Using the SI system of units, show that Eq. 1–2 is a dimensionally homogeneous equation which gives F in newtons. Determine to three significant figures the gravitational force acting between two spheres that are touching each other. The mass of each sphere is 200 kg and the radius is 300 mm.

SOLUTION Using Eq. 1–2, F = G N = a

m 1 m2 r2

kg # kg kg # m m3 ba b = 2 2 kg # s m s2

F = G

(Q.E.D.)

m 1 m2 r2

= 66.73 A 10 - 12 B c

200(200) 0.62

= 7.41 A 10 - 6 B N = 7.41 mN

Ans.

Ans: 7.41 mN 15

*1–16. Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (200 kN)2, (b) (0.005 mm)2, and (c) (400 m)3.

SOLUTION a) (200 kN)2 = 40 000 A 106 B N2 = 0.04 A 1012 B N2 = 0.04 MN2

Ans.

b) (0.005 mm)2 = 25 A 10 - 12 B m2 = 25mm2

Ans.

c) (400 m)3 = 0.064 A 109 B m3 = 0.064 km3

Ans.

Ans: a) 0.04 MN2 b) 25mm2 c) 0.064 km3 16

1–17. If a car is traveling at 55 mi>h, determine its speed in kilometers per hour and meters per second.

SOLUTION 55 mi>h = a

1 km 55 mi 5280 ft 0.3048 m ba ba ba b 1h 1 mi 1 ft 1000 m Ans.

= 88.5 km>h 88.5 km>h = a

1h 88.5 km 1000 m ba ba b = 24.6 m>s 1h 1 km 3600 s

Ans.

Ans: 88.5 km>h 24.6 m>s 17

1–18. Evaluate (204 mm)(0.00457 kg)> (34.6 N) to three significant figures and express the answer in SI units using an appropriate prefix.

SOLUTION (204 mm)(0.004 57 kg)>(34.6 N) = ¢ = a

C 204 A 10 - 3 B m D C 4.57 A 10 - 3 B kg D 34.6 N 26.9 A 10 - 6 B m # kg 1N

≤

= 26.9 mm # kg>N

Ans.

Ans: 26.9 mm # kg>N 18

1–19.

The specific weight (wt.> vol.) of brass is 520 lb>ft3. Determine its density (mass> vol.) in SI units. Use an appropriate prefix.

SOLUTION 520 lb>ft3 = a

3 1 kg 520 lb 1 ft 4.448 N b a ba b ba 3 0.3048 m 1 lb 9.81 N ft

= 8.33 Mg>m3

Ans.

Ans: 8.33 Mg>m3 19

*1–20. If a man weighs 155 lb on earth, specify (a) his mass in slugs, (b) his mass in kilograms, and (c) his weight in newtons. If the man is on the moon, where the acceleration due to gravity is gm = 5.30 ft>s2, determine (d) his weight in pounds, and (e) his mass in kilograms.

SOLUTION a) m =

155 = 4.81 slug 32.2

b) m = 155 c

Ans.

14.59 kg d = 70.2 kg 32.2

Ans.

c) W = 15514.44822 = 689 N

Ans.

5.30 d = 25.5 lb 32.2

Ans.

14.59 kg d = 70.2 kg 32.2

Ans.

d) W = 155c e) m = 155c Also, m = 25.5

14.59 kg 5.30

Ans.

= 70.2 kg

Ans: a) 4.81 slug b) 70.2 kg c) 689 N d) 25.5 lb e) 70.2 kg 20

1–21. Two particles have a mass of 8 kg and 12 kg, respectively. If they are 800 mm apart, determine the force of gravity acting between them. Compare this result with the weight of each particle.

SOLUTION F = G

m1 m2 r2

Where G = 66.73 A 10-12 B m3>(kg # s2) F = 66.73 A 10 - 12 B B

8(12) (0.8)2

R = 10.0 A 10 - 9 B N = 10.0 nN

Ans.

W1 = 8(9.81) = 78.5 N

Ans.

W2 = 12(9.81) = 118 N

Ans.

Ans: F = 10.0 nN W1 = 78.5 N W2 = 118 N 21

2–1. y

Determine the magnitude of the resultant force FR = F1 + F2 and its orientation u, measured counterclockwise from the positive x axis.

F1 5 260 lb

12 5

SOLUTION

Sine Law:

458

sin (45° + a) sin 67.62° = 320 260

a = 3.728°

u = 180° - a = 176°

Ans. F 5 310 lb 2

12 tan-1 = 67.38° 5 67.38° + 45° = 112.38° 360° - a (112.38°) = 67.62° a Cosine Law: FR = 23102 + 2602 - 2(310)(260) cos 67.62° = 320 lb

Ans.

Ans: FR = 320 lb u = 176° 22

2–2. y

Determine the magnitude of the resultant force FR′ = F1 - F2 and its orientation u, measured counterclockwise from the positive x axis.

F1 5 260 lb

12 5

SOLUTION

tan-1

12 = 67.38° 5

458

67.38° + 45° = 112.38° F2 5 310 lb

Cosine Law:

Sine Law:

FR = 23102 + 2602 - 2(310)(260) cos 112.38° = 474 lb sin (u - 45°) sin 112.38° = 474 260

Ans.

u = 75.4°

Ans: FR = 474 lb u = 75.4° 23

2–3. Two forces are applied at the end of a screw eye in order to remove the post. Determine the angle u 10° … u … 90°2 and the magnitude of force F so that the resultant force acting on the post is directed vertically upward and has a magnitude of 750 N.

y F 500 N

30°

SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. (a). Trigonometry: Using law of sines [Fig. (b)], we have sin f sin 30° = 750 500 sin f = 0.750 f = 131.41° 1By observation, f 7 80°2

Thus,

Ans.

u = 180° - 30° - 131.41° = 18.59° = 18.6° F 500 = sin 18.59° sin 30°

Ans.

F = 319 N

Ans: u = 18.6° F = 319 N 24

*2–4. Determine the magnitudes of the components of F F acts downward The vertical force at two on the two-membered along ABthe and AC. Set F of = the 500two N. components of frame.members Determine magnitudes and . Set F directed along the axes of 500 N.

SOLUTION

Parallelogram Law: Law: The parallelogram law of addition is shown in Fig. a. Parallelogram Trigonometry: Trigonometry: Using the law of sines (Fig. b), we have sin 60°

500 sin 75°

Ans.

448 N sin 45°

500 sin 75° Ans.

366 N

Ans: FAB = 448 N FAC = 366 N 25

2–5. Solve Prob. 2-4 with F = 350 lb.

SOLUTION

Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using the law of sines (Fig. b), we have

FAB 350 = sin 60° sin 75°

30 C

Ans.

FAB = 314 lb FAC 350 = sin 45° sin 75°

Ans.

FAC = 256 lb

Ans: FAB = 314 lb FAC = 256 lb 26

2–6. Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured clockwise from the positive u axis.

v 30 75

F1 4 kN 30 u F2 6 kN

SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a, Trigonometry: Applying Law of cosines by referring to Fig. b, FR = 242 + 62 - 2(4)(6) cos 105° = 8.026 kN = 8.03 kN

Ans.

Using this result to apply Law of sines, Fig. b, sin u sin 105° = ; 6 8.026

u = 46.22°

Thus, the direction f of FR measured clockwise from the positive u axis is Ans.

f = 46.22° - 45° = 1.22°

Ans: FR = 8.03 kN f = 1.22° 27

2–7.

Resolve the force F1 into components acting along the u and v axes and determine the magnitudes of the components.

30 75

F1 4 kN 30 u F2 6 kN

SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a, Trigonometry: Applying the sines law by referring to Fig. b. (F1)v sin 45° (F1)u sin 30°

4 ; sin 105°

(F1)v = 2.928 kN = 2.93 kN

Ans.

4 ; sin 105°

(F1)u = 2.071 kN = 2.07 kN

Ans.

Ans: (F1)v = 2.93 kN (F1)u = 2.07 kN 28

*2–8.

Resolve the force F2 into components along the u and v axes and determine the magnitudes of the components.

30 75

F1 4 kN 30 u F2 6 kN

SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a, Trigonometry: Applying the sines law of referring to Fig. b, (F2)u sin 75° (F2)v sin 30°

6 ; sin 75°

(F2)u = 6.00 kN

Ans.

6 ; sin 75°

(F2)v = 3.106 kN = 3.11 kN

Ans.

Ans: (F2)u = 6.00 kN (F2)v = 3.11 kN 29

2–9. Determine the magnitude of the resultant force FR = F1 + F2 and its orientation u, measured clockwise from the positive x axis.

y F3 5 250 N 308

x 308

SOLUTION

458

Cosine Law:

F2 5 360 N 2

Sine Law:

F1 5 400 N

FR = 2360 + 400 - 2(360)(400) cos 75° = 463.75 = 464 N sin (u - 30°) sin 75° = 463.75 360

Ans.

u = 78.6°

Ans: FR = 464 N u = 78.6° 30

2–10. Determine the magnitude of the resultant force FR = F1 + F3 and its orientation u, measured counterclockwise from the positive x axis.

y F3 5 250 N 308

x 308

SOLUTION

458

Cosine Law:

F2 5 360 N 2

Sine Law:

F1 5 400 N

FR = 2250 + 400 - 2(250)(400) cos 30° = 222.02 = 222 N sin (30° + u) sin 30° = 222.02 250

Ans.

u = 4.26°

Ans: FR = 222 N u = 4.26° 31

2–11. If u =plate 60°, determine the to magnitude the resultant forceBand The is subjected the twoofforces at A and as its direction clockwise from the horizontal. = 60°, determine shown. If u measured the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal.

FA u

8 kN

SOLUTION Parallelogram Law: Law: The parallelogram law of addition is shown in Fig. a. Parallelogram Trigonometry: Trigonometry: Using law of cosines (Fig. b), we have FR = 282 + 62 - 2(8)(6) cos 100°

Ans.

= 10.80 kN = 10.8 kN

The angle u can be determined using law of sines (Fig. b).

40 B FB

6 kN

sin 100° sin u = 6 10.80 sin u = 0.5470 u = 33.16° Thus, the direction f of FR measured from the x axis is Ans.

f = 33.16° - 30° = 3.16°

Ans: FR = 10.8 kN f = 3.16° 32

*2–12. Determine the angle of u for connecting member A to the plate so that the resultant force of FA and FB is directed horizontally to the right. Also, what is the magnitude of the resultant force?

FA u

8 kN

SOLUTION Parallelogram Law: Law: The parallelogram law of addition is shown in Fig. a. Parallelogram Trigonometry: Trigonometry: Using law of sines (Fig .b), we have sin (90° - u) sin 50° = 6 8

40 B

sin (90° - u) = 0.5745 Ans.

u = 54.93° = 54.9°

6 kN

From the triangle, f = 180° - (90° - 54.93°) - 50° = 94.93°. Thus, using law of cosines, the magnitude of FR is FR = 282 + 62 - 2(8)(6) cos 94.93°

Ans.

= 10.4 kN

Ans: u = 54.9° FR = 10.4 kN 33

2–13. The pelvis P is connected to the femur F at A using three different muscles, which exert the forces shown on the femur. Determine the resultant force and specify its orientation u, measured counterclockwise from the positive x axis.

y 120 N

60 N

13 12 5

80 N A

SOLUTION

308

Cosine Law:

Sine Law:

FR1 = 2602 + 802 - 2(60)(80) cos 120° = 121.66 N sin f sin 120° = 121.66 60

f = 25.28°

Cosine Law: FR = 21202 + 121.662 - 2(120)(121.66) cos 167.90°

Ans.

= 240.31 = 240 N

Sine Law: sin a sin 167.90° = 120 240.31

a = 6.01° Ans.

u = 55.28° + a = 61.3°

Ans: FR = 240 N u = 61.3° 34

2–14. If the resultant force acting on the support is to be 1200 lb, directed horizontally to the right, determine the force F in rope A and the corresponding angle u.

F A u B 60

900 lb

SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a, Trigonometry: Applying the law of cosines by referring to Fig. b, F = 29002 + 12002 - 2(900)(1200) cos 30° = 615.94 lb = 616 lb

Ans.

Using this result to apply the sines law, Fig. b, sin u sin 30° = ; 900 615.94

u = 46.94° = 46.9°

Ans.

Ans: F = 616 lb u = 46.9° 35

2–15. The plate is subjected to the forces acting on members A and B as shown. If u = 60°, determine the magnitude of the resultant of these forces and its direction measured clockwise from the positive x axis.

y FA 400 lb

SOLUTION

Cosine Cosine law: Law:

u B

FR = 25002 + 4002 - 2(500)(400) cos 60° = 458 lb

Ans.

Law: Sine law:

FB 500 lb

sin (60°- a) sin 60° = 458 400

Ans.

a = 10.9°

Ans: FR = 458 lb a = 10.9° 36

*2–16. Determine the angle u for connecting member B to the plate so that the resultant angle of FA and FB is directed along the positive x axis. What is the magnitude of the resultant force?

y FA 400 lb

SOLUTION

Sine law: Law:

sin 60° sin u = 400 500 FR 500 = sin (120°- 43.9°) sin 60°

u = 43.9°

Ans.

FR = 561 lb

Ans.

FB 500 lb

Ans: u = 43.9° FR = 561 lb 37

2–17. Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first finding the finding FR = F¿ + F3. resultant F¿ = F1 + F2 and then forming

y F1

30 N 3

50 N

SOLUTION

20 N

F¿ = 2(20)2 + (30)2 - 2(20)(30) cos 73.13° = 30.85 N 30 30.85 = ; sin 73.13° sin (70° - u¿)

u¿ = 1.47°

FR = 2(30.85)2 + (50)2 - 2(30.85)(50) cos 1.47° = 19.18 = 19.2 N 30.85 19.18 = ; sin 1.47° sin u

Ans.

u = 2.37°

Ans: FR = 19.2 N u = 2.37° c 38

2–18. Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first finding the resultant F¿ = F2 + F3 and then forming FR = F¿ + F1.

y F1

30 N 3

50 N

SOLUTION ¿

20 N

F = 2(20) + (50) - 2(20)(50) cos 70° = 47.07 N 20

sin u

47.07 ; sin 70°

u¿ = 23.53°

FR = 2(47.07)2 + (30)2 - 2(47.07)(30) cos 13.34° = 19.18 = 19.2 N 30 19.18 = ; sin 13.34° sin f

Ans.

f = 21.15° Ans.

u = 23.53° - 21.15° = 2.37°

Ans: FR = 19.2 N u = 2.37° c 39

2–19. The subjected to the at A and u =B60°, Theplate plateis is subjected to two the forces two forces at B. A Ifand as determine of thethe resultant force of andthe itsresultant direction = magnitude 60°, determine shown. If uthe magnitude measured fromforces the positive x axis. of these two and its direction measured clockwise from the horizontal.

FA u

8 kN

SOLUTION Law: The parallelogram law of addition is shown in Fig. a. Parallelogram Law: Trigonometry: Trigonometry: Using law of cosines (Fig. b), we have FR = 282 + 62 - 2(8)(6) cos 100°

Ans.

= 10.80 kN = 10.8 kN

The angle u can be determined using law of sines (Fig. b).

40 B FB

6 kN

sin 100° sin u = 6 10.80 sin u = 0.5470 u = 33.16° Thus, the direction f of FR measured from the x axis is Ans.

f = 33.16° - 30° = 3.16°

Ans: FR = 10.8 kN f = 3.16° 40

*2–20. Determine the angle angleofu ufor the Determine the forconnecting connectingmember member A A to to the plate FA and directed the plate so so that thatthe theresultant resultantofforce of FFBAisand FB isalong directed positive x axis. whatAlso, is the magnitude of the resultant horizontally to Also, the right. what is the magnitude of the force? resultant force?

FA u

8 kN

SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of sines (Fig .b), we have sin (90° - u) sin 50° = 6 8

40 B

sin (90° - u) = 0.5745 Ans.

u = 54.93° = 54.9°

6 kN

From the triangle, f = 180° - (90° - 54.93°) - 50° = 94.93°. Thus, using law of cosines, the magnitude of FR is FR = 282 + 62 - 2(8)(6) cos 94.93°

Ans.

= 10.4 kN

Ans: u = 54.9° FR = 10.4 kN 41

2–21. Determine the design angle 90° 2 for for member member angle uu 10° (0° … u … 90°) AB so that the 400-lb 400-lb horizontal horizontal force force has has aa component component of of 500 lb directed from A toward C. What What is is the the component component of of force acting along along member member AB? AB?Take Takeff == 40°. 40°.

400 lb

SOLUTION C

Sine law: Law: sin 40° sin u = 500 400 FAB 400 = sin (180° - 40° - 53.5°) sin 40°

Ans.

u = 53.5°

Ans.

FAB = 621 lb

Ans: u = 53.5° FAB = 621 lb 42

2–22. Determine the 90° 2 between between the design design angle angle ff10° (0° …… f … 90°) members so that 400-lb horizontal force struts ABAB andand ACAC so that thethe 400-lb horizontal force hashas a acomponent component lb which upthe to left, the right, the of of 600600 lb which actsacts up to in the in same direction from B toward A. Also calculate the as from B towards A. Take u = 30° . magnitude of the force component along AC. Take u = 30°.

400 lb A u f f

400 lb

u B

SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a.

Trigonometry: Using law of cosines (Fig. b), we have

FAC = 24002 + 6002 - 2(400)(600) cos 30° = 322.97 lb

The angle f can be determined using law of sines (Fig. b). sin f sin 30° = 400 322.97 sin f = 0.6193

Ans.

f = 38.3°

Ans: FAC = 323 lb f = 38.3° 43

2–23. Determine the magnitude of the two towing forces FB and FC if the resultant force has a magnitude FR = 10 kN and is directed along the positive x axis. Set u = 15°.

C FC 208 x u

A FB

SOLUTION

Sine Law: FC 10 = sin 15° sin 145°

FC = 4.51 kN

Ans.

FB 10 = sin 20° sin 145°

FB = 5.96 kN

Ans.

Ans: FC = 4.51 kN FB = 5.96 kN 44

*2–24. If the resultant FR of the two forces acting on the jet aircraft is to be directed along the positive x axis and have a magnitude of 10 kN, determine the angle u of the cable attached to the truck at B so that FB is a minimum. What is the magnitude of force in each cable when this occurs?

C FC 208 x u

A FB

SOLUTION

u = 90° - 20° = 70°

Ans.

FB = 10 sin 20° = 3.42 kN

Ans.

FC = 10 cos 20° = 9.40 kN

Ans.

Ans: u = 70° FB = 3.42 kN FC = 9.40 kN 45

2–25. The 500-lb force acting on the frame is to be resolved into two components acting along the axis of the struts AB and AC. If the component of force along AC is required to be 300 lb, directed from A to C, determine the magnitude of force acting along AB and the angle u of the 500-lb force.

F = 500 lb A

60°

SOLUTION

Parallelogram Law: The parallelogram law of addition is shown in Fig. (a).

45° C

Trigonometry: Using law of sines [Fig. (b)], we have sin f sin 75° = 300 500 sin f = 0.5796 Ans.

f = 35.42° Thus, 45° + u + 75° + 35.42° = 180° u = 24.58° = 24.6° FAB 500 = sin145° + 24.58°2 sin 75°

Ans.

FAB = 485 lb

Ans: f = 35.42° FAB = 485 lb 46

2–26. y

Determine the magnitude and direction u of FA so that the resultant force is directed along the positive x axis and has a magnitude of 1250 N.

θ O

30° B

SOLUTION

FB = 800 N

+ FR = ΣFx; S x

FRx = FA sin u + 800 cos 30° = 1250

+ c FRy = ΣFy;

FRy = FA cos u - 800 sin 30° = 0 u = 54.3°

Ans.

FA = 686 N

Ans.

Ans: u = 54.3° FA = 686 N 47

2–27. Determine of the force measured acting on Determine the themagnitude magnitude andresultant direction, the ring at O, if F = 750 N and u = 45°. What is its A counterclockwise from the positive x axis, of the resultant direction, force actingmeasured on the ringcounterclockwise at O, if FA = 750from N andthe u = positive 45°. x axis?

θ O

30° B

SOLUTION

FB = 800 N

Scalar Notation: Suming the force components algebraically, we have + FR = ΣFx; S x

FRx = 750 sin 45° + 800 cos 30° = 1223.15 N S

+ c FRy = ΣFy;

FRy = 750 cos 45° - 800 sin 30° = 130.33 N c

The magnitude of the resultant force FR is FR = 3F 2Rx + F 2Ry = 21223.152 + 130.332 = 1230 N = 1.23 kN

Ans.

The directional angle u measured counterclockwise from positive x axis is u = tan-1

FRy FRx

= tan-1 a

130.33 b = 6.08° 1223.15

Ans.

Ans: FR = 1.23 kN u = 6.08° 48

*2–28. If F1 = F2 = 30 lb, determine the angles u and f so that the resultant force is directed along the positive x axis and has a magnitude of FR = 20 lb.

x θ φ

SOLUTION 30 30 = sin f sin u f = u (30)2 = (30)2 + (20)2 - 2(30)(20) cos u Ans.

f = u = 70.5°

Ans: u = f = 70.5° 49

2–29. If the resultant force of the two tugboats is 3 kN, directed along the positive x axis, determine the required magnitude of force FB and its direction u.

2 kN 30

u C

SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.

Applying the law of cosines to Fig. b, FB = 222 + 32 - 2(2)(3)cos 30°

Ans.

= 1.615kN = 1.61 kN

Using this result and applying the law of sines to Fig. b, yields sin 30° sin u = 2 1.615

Ans.

u = 38.3°

Ans: FB = 1.61 kN u = 38.3° 50

2–30. If FB = 3 kN and u = 45°, determine the magnitude of the resultant force of the two tugboats and its direction measured clockwise from the positive x axis.

2 kN 30

u C

SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.

Applying the law of cosines to Fig. b, FR = 222 + 32 - 2(2)(3) cos 105°

Ans.

= 4.013 kN = 4.01 kN

Using this result and applying the law of sines to Fig. b, yields sin 105° sin a = 3 4.013

a = 46.22°

Thus, the direction angle f of FR, measured clockwise from the positive x axis, is Ans.

f = a - 30° = 46.22° - 30° = 16.2°

Ans: FR = 4.01 kN f = 16.2° 51

2–31. If the resultant force of the two tugboats is required to be directed towards the positive x axis, and FB is to be a minimum, determine the magnitude of FR and FB and the angle u.

2 kN 30

SOLUTION

For FB to be minimum, it has to be directed perpendicular to FR. Thus, Ans.

u = 90°

The parallelogram law of addition and triangular rule are shown in Figs. a and b, respectively. By applying simple trigonometry to Fig. b, FB = 2 sin 30° = 1 kN

Ans.

FR = 2 cos 30° = 1.73 kN

Ans.

Ans: u = 90° FB = 1 kN FR = 1.73 kN 52

*2–32. The three cable forces act on the eyebolt. Determine two possible magnitudes for P so that the resultant force has a magnitude of 800 N.

y 1000 lb 5

4 3

SOLUTION

308

3 + FRx = ΣFx; FRx = P - 400 cos 30° - 1000 a b = P - 946.41 S 5

400 lb

4 + c FRy = ΣFy; FRy = 1000 a b - 400 sin 30° = 600 5

FR = 2(P - 946.41)2 + (600)2 = 800

(P - 946.41)2 + (600)2 = (800)2 (P - 946.41)2 = 280 000 P - 946.41 = {529.15

P - 946.41 = 529.15; P = 1475.6 lb = 1.48 kip

Ans.

P - 946.41 = -529.15; P = 417 lb

Ans.

Ans: P = 1.48 kip P = 417 lb 53

2–33. Four concurrent forces act on the plate. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.

F2 5 100 lb

SOLUTION

308

4 + FRx = ΣFx; S FRx = 60 - 38 cos 30° + 5 (50) = 67.09 lb

F3 5 38 lb

3 + c FRy = ΣFy; FRy = 100 - 38 sin 30° - (50) = 51 lb 5 FR = 2(67.09)2 + (51)2 = 84.3 lb

Ans.

u = tan-1 a

Ans.

51 b = 37.2° 67.09

F1 5 60 lb 3

5 4

F4 5 50 lb

Ans: FR = 84.3 lb u = 37.2° 54

2–34. Express F1 and F2 as Cartesian vectors.

y F1 5 40 lb 308

SOLUTION

x 208

Sine Law: F1 = {40 cos 60°i + 40 sin 60°j} lb = {20i + 34.6j} lb

Ans.

F2 = {50 cos 20°i - 50 sin 20°j} lb = {47.0i - 17.1j} lb

Ans.

F2 5 50 lb

Ans: F1 = {20i + 34.6j} lb F2 = {47.0i - 17.1j} lb 55

2–35. y

Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.

F1 5 40 lb 308

If F1 = 300 N and u = 10°, determine the magnitude and direction, measured counterclockwise from the positive x¿ axis, of the resultant force acting on the bracket.

y F2 200 N

SOLUTION

x 208

Scalar Analysis: x¿

+ FRx = ΣFx; FRx = 40 cos 60° + 50 cos 20° S

SOLUTION

F3 180 N

= 66.98 lb S

Magnitude 5 FRy = 300 cos 70° + 200 + (180) = 371.8 N + c FRy = ©Fy ; 13 FR = 2FRx2 + FRy2 = 266.982 + 17.542 = 69.2 lb

Ans.

FR = 2(115.8)2 + (371.8)2 = 389 N

F 17.54 - 1 Ry 1 371.8 u = tanf = -tan 14.7° = tan B - 1 R ==72.71° FRx 66.98 115.8

+ c FRy = ΣFy; FRy = 40 sin 60° - 50 sin 20° 12 + F = ©Fx ; FRx = 300 sin 70° (180) = 115.8 N : Rx =13 17.54 lb c

Direction

F2 5 50 lb

Ans. Ans.

au Ans.

f¿ = 72.71° - 30° = 42.7°

Ans: FR = 69.2 lb u = 14.7° au 56

*2–36. Resolve each force acting on the gusset plate into its x and y components, and express each force as a Cartesian vector.

y 650 N

F3 3

750 N

F1 = {900(+i)} = {900i} N

Ans.

F2 = {750 cos 45°(+i) + 750 sin 45°(+j)} N = {530i + 530j} N

Ans.

F3 = e 650a

3 4 b (+i) + 650 a b (-j) f N 5 5

= {520 i - 390j)} N

x 900 N

Ans.

Ans: F1 = 5900i6 N F2 = 5530i + 530j 6 N F3 = 5520i - 390j 6 N

2–37. Determine the magnitude magnitude of of the the resultant resultant force force acting acting on on Determine the the plate its direction, measured counterclockwise the gusset plate and itsand direction, measured counterclockwise from from the positive the positive x axis.x axis.

y 650 N

F3 3

SOLUTION Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, and F3 can be written as (F1)x = 900 N

750 N

45 F1

x 900 N

(F1)y = 0

(F2)x = 750 cos 45° = 530.33 N 4 (F3)x = 650 a b = 520 N 5

(F2)y = 750 sin 45° = 530.33 N 3 (F3)y = 650a b = 390 N 5

Resultant Resultant Force: Summing the force components algebraically along the x and y axes, we have + : ©(FR)x = ©Fx;

(FR)x = 900 + 530.33 + 520 = 1950.33 N :

+ c ©(FR)y = ©Fy;

(FR)y = 530.33 - 390 = 140.33 N c

The magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 21950.332 + 140.332 = 1955 N = 1.96 kN Ans.

The direction angle u of FR, measured clockwise from the positive x axis, is u = tan-1 c

(FR)y 140.33 b = 4.12° d = tan-1 a (FR)x 1950.33

Ans.

Ans: FR = 1.96 kN u = 4.12° 58

2–38. The three forces are applied to the bracket. Determine the range of values for the magnitude of force P so that the resultant of the three forces does not exceed 2400 N.

800 N 3000 N

90 60 P

SOLUTION : FRx = ©Fx;

FRx = P + 800 cos 60° - 3000 cos 30° = P - 2198.08

+ c FRy = ©Fy;

FRy = 800 sin 60° + 3000 sin 30° = 2192.82 FR = 2(P - 2198.08)2 + (2192.82)2 … 2400 (P - 2198.08)2 + (2192.82)2 … (2400)2 |(P - 2198.08)| … 975.47 - 975.47 … P - 2198.08 … 975.47 1222.6 N … P … 3173.5 N

Ans: 1222.6 N … P … 3173.5 N 59

2–39. y

Determine the x and y components of F1 and F2.

45 F1 200 N

SOLUTION F1x = 200 sin 45° = 141 N

Ans.

F1y = 200 cos 45° = 141 N

Ans.

F2x = - 150 cos 30° = - 130 N

Ans.

F2y = 150 sin 30° = 75 N

Ans.

F2 150 N x

Ans: F1x = 141 N F1y = 141 N F2x = - 130 N F2y = 75 N 60

*2–40. y

Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. 45

F1 200 N

SOLUTION +R FRx = ©Fx;

FRx = - 150 cos 30° + 200 sin 45° = 11.518 N

Q +FRy = ©Fy;

FRy = 150 sin 30° + 200 cos 45° = 216.421 N

F2 150 N x

FR = 2 (11.518)2 + (216.421)2 = 217 N

Ans.

u = tan - 1 ¢

Ans.

216.421 ≤ = 87.0° 11.518

Ans: FR = 217 N u = 87.0° 61

2–41. F3 8 kN

Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

F2 5 kN

60 45

F1 4 kN

SOLUTION Scalar Notation: Summing the force components along x and y axes algebraically by referring to Fig. a, + (F ) = ΣF ; (F ) = 4 + 5 cos 45° - 8 sin 15° = 5.465 kN S S R x

R x

+ c (FR)y = ΣFy; (FR)y = 5 sin 45° + 8 cos 15° = 11.263 kN c By referring to Fig. b, the magnitude of the resultant force FR is FR = 2(FR)2x + (FR)2y = 25.4652 + 11.2632 = 12.52 kN = 12.5 kN

Ans.

And the directional angle u of FR measured counterclockwise from the positive x axis is u = tan - 1 c

(FR)y (FR)x

d = tan - 1 a

11.263 b = 64.12° = 64.1° 5.465

Ans.

Ans: FR = 12.5 kN u = 64.1° 62

2–42. y

Express F1, F2, and F3 as Cartesian vectors. F3 750 N

SOLUTION F1 =

F1 850 N

F2 625 N

3 4 (850) i - (850) j 5 5 Ans.

= {680 i - 510 j} N F2 = - 625 sin 30° i - 625 cos 30° j

Ans.

= { -312 i - 541 j} N F3 = - 750 sin 45° i + 750 cos 45° j { -530 i + 530 j} N

Ans.

Ans: F1 = {680i - 510j} N F2 = { - 312i - 541j} N F3 = { - 530i + 530j} N 63

2–43. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.

y F3 750 N

SOLUTION

30 F2 625 N

+ F = ©F ; : Rx x

FRx =

+ c FRy = ©Fy ;

3 FRy = - (850) - 625 cos 30° + 750 cos 45° = -520.94 N 5

4 (850) - 625 sin 30° - 750 sin 45° = - 162.83 N 5

FR = 2( - 162.83)2 + ( - 520.94)2 = 546 N f = tan-1 a

520.94 b = 72.64° 162.83

F1 850 N

Ans.

u = 180° + 72.64° = 253°

Ans: FR = 546 N u = 253° 64

*2–44. The three threeconcurrent concurrentforces forces acting produce acting on on the the postpost produce a zeroa resultant force FR = 0. If F2 = 12 F1, and F1 is to be 90° from F2 as shown, determine the required magnitude of F3 expressed in terms of F1 and the angle u.

SOLUTION ©FRx¿ = 0;

F3 cos (u - 90°) = F1

©FRy¿ = 0;

F3 sin (u - 90°) = F2 tan (u - 90°) =

F2 1 = F1 2

u - 90° = 26.57° Ans.

u = 116.57° = 117° F3 =

F1 cos(116.57° - 90°) Ans.

F3 = 1.12 F1

Ans: u = 117° F3 = 1.12 F1 65

2–45. Three forces act on the ring. Determine the range of values for the magnitude of P so that the magnitude of the resultant force does not exceed 2500 N. Force P is always directed to the right.

1500 N

600 N 60 45 P

SOLUTION + F = ©F ; : Rx x

FRx = P + 600 cos 45° + 1500 cos 105° = P + 36.0355

+ c FRy = ©Fy ;

FRy = 600 sin 45° + 1500 sin 105° = 1873.153 (2500)2 = (P + 36.0355)2 + (1873.153)2 ; 1655.687 = P + 36.0355

Choose the positive root, thus P = 1.62 kN Ans.

0 … P … 1.62 kN

Ans: 0 … P … 1.62 kN 66

2–46. y

Determine the magnitude and orientation u of FB so that the resultant force is directed along the positive y axis and has a magnitude of 1500 N.

FA 700 N 30

u x

SOLUTION Scalar Notation: Suming the force components algebraically, we have + F = ΣF ; S Rz x

0 = 700 sin 30° - FB cos u (1)

FB cos u = 350 + c FRy = ΣFy;

1500 = 700 cos 30° + FB sin u (2)

FB sin u = 893.8 Solving Eq. (1) and (2) yields u = 68.6°

Ans.

FB = 960 N

Ans: u = 68.6° FB = 960 N 67

2–47. If FB = 600 N and u = 20°, determine the magnitude of the Determine the magnitude and orientation, measured resultant force and its direction measured counterclockwise counterclockwise from the positive y axis, of the resultant from the positive y axis. force acting on the bracket, if FB = 600 N and u = 20°.

y FB

FA 700 N 30

u x

SOLUTION Scalar Notation: Suming the force components algebraically, we have + F = ΣF ; S Rx x

FRx = 700 sin 30° - 600 cos 20° = -213.8 N = 213.8 N d

+ c FRy = ΣFy;

FRy = 700 cos 30° + 600 sin 20° = 811.4 N c

The magnitude of the resultant force FR is FR = 2F 2Rx + F 2Ry = 2213.82 + 811.42 = 839 N

Ans.

The directional angle u measured counterclockwise from positive y axis is u = tan-1

FRx FRy

= tan-1 a

213.8 b = 14.8° 811.4

Ans.

Ans: FR = 839 N u = 14.8° 68

*2–48. Three forces act on the bracket. Determine the magnitude and direction u of F1 so that the resultant force is directed along the positive x¿ axis and has a magnitude of 800 N.

y F2 200 N

x¿ F3 180 N

SOLUTION

+ F = ©F ; : Rx x

12 800 sin 60° = F1 sin(60° + u) (180) 13

+ c FRy = ©Fy ;

800 cos 60° = F1 cos(60° + u) + 200 +

5 (180) 13

60° + u = 81.34° u = 21.3°

Ans.

F1 = 869 N

Ans.

Ans: u = 21.3° F1 = 869 N 69

2–49. the If F and uu == 10°, determinethe themagnitude magnitudeofand F11 = = 300 N and 10°, determine resultant force and itscounterclockwise direction measuredfrom counterclockwise direction, measured the positive from theofpositive x′ axis.force acting on the bracket. axis, the resultant x¿

y F2 200 N

x¿ F3 180 N

SOLUTION

+ F = ©Fx ; : Rx

12 FRx = 300 sin 70° (180) = 115.8 N 13

+ c FRy = ©Fy ;

FRy = 300 cos 70° + 200 +

5 (180) = 371.8 N 13

FR = 2(115.8)2 + (371.8)2 = 389 N f = tan - 1 B

371.8 R = 72.71° 115.8

Ans.

au Ans.

f¿ = 72.71° - 30° = 42.7°

Ans: FR = 389 N f′ = 42.7° 70

2–50. Express F1 and F2 as Cartesian vectors.

y F1 5 60 lb F2 5 50 lb 308

208

SOLUTION F1 = - 60 cos 30° i + 60 sin 30° j = { -52.0 i + 30.0 j} lb

Ans.

F2 = - 50 cos 20° i - 50 sin 20° j = { - 47.0 i - 17.1 j} lb

Ans.

Ans: F1 = { - 52.0 i + 30.0 j} lb F2 = { - 47.0 i - 17.1 j} lb 71

2–51. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.

y F1 5 60 lb F2 5 50 lb 308

208

SOLUTION F1 = - 60 cos 30° i + 60 sin 30° j = { -52.0 i + 30.0 j} lb F2 = - 50 cos 20° i - 50 sin 20° j = { - 47.0 i - 17.1 j} lb FR = F1 + F2 = { - 98.9 i + 12.9 j} lb FR = 2( -98.9)2 + (12.9)2 = 99.8 lb u′ = tan - 1 a

12.9 b = 7.43° 98.9

u = 180° - 7.43° = 173°

Ans.

Ans: FR = 99.8 lb u = 173° 72

*2–52. The four concentric forces act on the post. Determine the resultant force and its direction, measured counterclockwise from the positive x axis.

F2 5 600 N F1 5 300 N

308

SOLUTION

Scalar Notation: Summing the force components along x and y axes algebraically by referring to Fig. a, + (F ) = ΣF ; (F ) = 300 cos 30° + 450 a 5 b - 250 sin 60° S R x x R x F4 5 250 N 13 = 216.38 N S + c (FR)y = ΣFy; (FR)y = 300 sin 30° + 600 - 450 a = 209.62 N c

12 b - 250 cos 60° 13

608

12 5

F3 5 450 N

By referring to Fig. b, the magnitude of the resultant force FR is FR = 2(FR)2x + (FR)2y = 2216.382 + 209.622 = 301.26 N = 301 N

Ans.

And the directional angle u of FR measured counterclockwise from the positive x axis is u = tan - 1 c

(FR)y (FR)x

d = tan - 1 a

209.62 b = 44.09° = 44.1° 216.38

Ans.

Ans: FR = 301 N u = 44.1° 73

2–53. Three forces act on the bracket. Determine the magnitude and direction u of F2 so that the resultant force is directed along the positive u axis and has a magnitude of 50 lb.

y F3 52 lb 13

SOLUTION

12 5

Scalar Notation: Suming the force components algebraically, we have + F = ©F ; : Rx x

F1 80 lb

5 50 cos 25° = 80 + 52a b + F2 cos (25° + u) 13

+ c FRy = ©Fy;

-50 sin 25° = 52 a

[1]

F2 cos (25° + u) = - 54.684 12 b - F2 sin (25° + u) 13

u F2

[2]

F2 sin (25° + u) = 69.131 Solving Eq.[1] and [2] yields 25° + u = 128.35°

Ans.

u = 103°

Ans.

F2 = 88.1 lb

Ans: u = 103° F2 = 88.1 lb 74

2–54. If F the andu u==55°, , determinethe themagnitude magnitudeofand 150lblband F22 == 150 55°determine resultant force and clockwise its direction measured clockwise direction measured from the positive x axis offrom the the positive x axis. resultant force of the three forces acting on the bracket.

yy F3F5 52 lb 3 52 lb 13 13 12 12 5 5

SOLUTION Scalar Notation: Notation: Suming the force components algebraically, we have + F = ©F ; : Rx x

F1F5 80 lb 1 80 lb xx

5 FRx = 80 + 52 a b + 150 cos 80° 13

258 25

= 126.05 lb : FRy = 52 a

+ c FRy = ©Fy;

uu F2F2

12 b - 150 sin 80° 13

= - 99.72 lb = 99.72 lb T The magnitude of the resultant force FR is FR = 2F 2Rx + F 2Ry = 2126.052 + 99.72 2 = 161 lb

Ans.

The directional angle u measured clockwise from positive x axis is u = tan - 1

FRy FRx

= tan - 1 a

99.72 b = 38.3° 126.05

Ans.

Ans: FR = 161 lb u = 38.3° 75

2–55. y

Determine the resultant force acting on the hook, and its direction measured clockwise from the positive x axis.

SOLUTION Scalar Notation: Summing the force components along x and y axes algebraically by referring to Fig. a, + (F ) = ΣF ; (F ) = 500 sin 60° - 800 a 4 b S R x x R x 5 (FR)x = - 206.99 N = 206.99 N d

5 4

F2 5 800 N

608

F1 5 500 N

3 + c (FR)y = ΣFy; (FR)y = - 500 cos 60° - 800 a b 5 (FR)y = - 730 N = 730 N T By referring to Fig. b, the magnitude of the resultant force FR is FR = 2(FR)2x + (FR)2y = 2206.992 + 7302 = 758.78 N = 759 N

Ans.

And the directional angle u of FR measured clockwise from the positive x axis is

Then

u a = tan - 1 c

(FR)y (FR)x

d = tan - 1 a

730 b = 74.17° 206.99 Ans.

u = 180° - a = 180° - 74.17° = 105.83° = 106°

Ans: FR = 759 N u = 106° 76

*2–56. If the magnitude of the resultant force acting on the bracket is to be 450 N directed along the positive u axis, determine the magnitude of F1 and its direction f.

y u

f 30

SOLUTION

Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, Rectangular F3, and FR can be written as (F1)x = F1 sin f

(F1)y = F1 cos f

(F2)x = 200 N

( F2)y = 0

(F3)x = 260 ¢

5 ≤ = 100 N 13

(F3)y = 260 ¢

(FR)x = 450 cos 30° = 389.71 N

x 200 N

12 5

260 N

12 ≤ = 240 N 13

(FR)y = 450 sin 30° = 225 N

Resultant Force: Summing the force components algebraically along the x and y axes, + ©(F ) = ©F ; : R x x

389.71 = F1 sin f + 200 + 100 (1)

F1 sin f = 89.71 + c ©(FR)y = ©Fy;

225 = F1 cos f - 240 (2)

F1 cos f = 465 Solving Eqs. (1) and (2), yields f = 10.9°

Ans.

F1 = 474 N

Ans: f = 10.9° F1 = 474 N 77

2–57. If the resultant force acting on the bracket is required to be a minimum, determine the magnitudes of F1 and the resultant force. Set f = 30°.

y u

f 30

SOLUTION

Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, Rectangular and F3 can be written as (F1)x = F1 sin 30° = 0.5F1

(F1)y = F1 cos 30° = 0.8660F1

(F2)x = 200 N

(F2)y = 0

(F3)x = 260a

5 b = 100 N 13

(F3)y = 260a

x 200 N

12 5

260 N

12 b = 240 N 13

Resultant Force: Force: Summing the force components algebraically along the x and y axes, + ©(F ) = ©F ; : R x x

(FR)x = 0.5F1 + 200 + 100 = 0.5F1 + 300

+ c ©(FR)y = ©Fy;

(FR)y = 0.8660F1 - 240

The magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2

= 2(0.5F1 + 300)2 + (0.8660F1 - 240)2 = 2F 21 - 115.69F1 + 147 600

Thus,

(1)

FR2 = F 21 - 115.69F1 + 147 600

(2)

The first derivative of Eq. (2) is 2FR For FR to be minimum,

dFR = 2F1 - 115.69 dF1

(3)

dFR = 0. Thus, from Eq. (3) dF1

2FR

dFR = 2F1 - 115.69 = 0 dF1 Ans.

F1 = 57.846 N = 57.8 N from Eq. (1), FR = 2(57.846)2 - 115.69(57.846) + 147 600 = 380 N

Ans.

Ans: FR = 380 N F1 = 57.8 N 78

2–58. 4 kN

Three forces act on the bracket. Determine the magnitude and direction u of F so that the resultant force is directed along the positive x′ axis and has a magnitude of 8 kN.

F 15

30 x 6 kN

SOLUTION Scalar Notation: Equating the force components along the x and y axes algebraically by referring to Fig. a, + (F ) = ΣF ; 8 cos 30° = F sin u + 6 - 4 sin 15° S R x

F sin u = 1.9635 (1) + c (FR)y = ΣFy ; 8 sin 30° = F cos u + 4 cos 15° F cos u = 0.1363 (2) Divide Eq (1) by (2) tan u = 14.406

u = 86.03° = 86.0°

Ans.

Substitute this result into Eq (1) F sin 86.03° = 1.9635 Ans.

F = 1.968 kN = 1.97 kN

Ans: u = 86.0° F = 1.97 kN 79

2–59. y

If F = 5 kN and u = 30°, determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

4 kN

F 15

30 x 6 kN

SOLUTION Scalar Notation: Summing the force components along x and y axes algebraically by referring to Fig. a, + (F ) = ΣF ; S (F ) = 5 sin 30° + 6 - 4 sin 15° = 7.465 kN S R x

R x

+ c (FR)y = ΣFy;

(FR)y = 4 cos 15° + 5 cos 30° = 8.194 kN c

By referring to Fig. b, the magnitude of the resultant force is FR = 2(FR)2x + (FR)2y = 27.4652 + 8.1942 = 11.08 kN = 11.1 kN Ans.

And its directional angle u measured counterclockwise from the positive x axis is u = tan - 1 c

(FR)y (FR)x

d = tan - 1 a

8.194 b = 47.67° = 47.7° 7.465

Ans.

Ans: FR = 11.1 kN u = 47.7° 80

*2–60. Determine of the x, y, components of the the The force Fthe hasmagnitudes a magnitude of 80 lb zand acts within force octantF. shown. Determine the magnitudes of the x, y, z components of F.

z Fz F 80 lb

b 45

SOLUTION 2

a 60

1 = cos 60° + cos 45° + cos g

Solving for the positive root, g = 60°

Fx = 80 cos 60° = 40.0 lb

Ans.

Fy = 80 cos 45° = 56.6 lb

Ans.

Fz = 80 cos 60° = 40.0 lb

Ans.

Ans: Fx = 40.0 lb Fy = 56.6 lb Fz = 40.0 lb 81

2–61. The bolt is subjected to the force F, which has components acting along the x, y, z axes as shown. If the magnitude of F is 80 N, and a = 60° and g = 45°, determine the magnitudes of its components.

Fz g

F Fy

SOLUTION

b y

a Fx

cosb = 21 - cos2 a - cos2g

= 21 - cos2 60° - cos2 45°

b = 120°

Fx = |80 cos 60°| = 40 N

Ans.

Fy = |80 cos 120°| = 40 N

Ans.

Fz = |80 cos 45°| = 56.6 N

Ans.

Ans: Fx = 40 N Fy = 40 N Fz = 56.6 N 82

2–62. z

The cable at the end of the beam exerts a force of 450 lb on the beam. Express F as a Cartesian vector.

608 508

SOLUTION x

cos2 a + cos2 b + cos2 g = 1 cos2 50° + cos2 60° + cos2 g = 1 From the figure,

cos g = -0.5804

F 5 450 lb

cos g = {0.5804 g = 125.48°

Force Vector: F = 450(cos 50°i + cos 60°j + cos 125.48°k) Ans.

= {289i + 225j - 261k} lb

Ans: F = {289i + 225j - 261k} lb 83

2–63. The force F acts on the bracket within the octant shown. If F = 400 N, b = 60°, and g = 45°, determine the x, y, z components of F.

z g

SOLUTION

Coordinate Direction Angles: Since b and g are known, the third angle a can be determined from 2

a x y

cos a + cos b + cos g = 1 cos2 g + cos2 60° + cos2 45° = 1 cos a = ; 0.5 Since F is in the octant shown in Fig. a, ux must be greater than 90°. Thus, a = cos - 1(- 0.5) = 120°. Rectangular Components: By referring to Fig. a, the x, y, and z components of F can be written as Fx = F cos a = 400 cos 120° = - 200N

Ans.

Fy = F cos b = 400 cos 60° = 200N

Ans.

Fz = F cos g = 400 cos 45° = 283N

Ans.

The negative sign indicates that Fx is directed towards the negative x axis.

Ans: Fx = - 200 N Fy = 200 N Fz = 283 N 84

*2–64. The force F acts on the bracket within the octant shown. If the magnitudes of the x and z components of F are Fx = 300 N and Fz = 600 N, respectively, and b = 60°, determine the magnitude of F and its y component. Also, find the coordinate direction angles a and g.

z g

SOLUTION

b a

Rectangular Components: The magnitude of F is given by x

F = 2Fx 2 + Fy 2 + Fz 2

F = 23002 + Fy 2 + 6002 F 2 = Fy 2 + 450 000

(1)

The magnitude of Fy is given by (2)

Fy = F cos 60° = 0.5F Solving Eqs. (1) and (2) yields F = 774.60 N = 775 N

Ans.

Fy = 387 N

Ans.

Coordinate Direction Angles: Since F is contained in the octant so that Fx is directed towards the negative x axis, the coordinate direction angle ux is given by a = cos - 1 a

- Fx - 300 b = cos - 1 a b = 113° F 774.60

Ans.

The third coordinate direction angle is g = cos - 1 a

- FZ 600 b = cos - 1 a b = 39.2° F 774.60

Ans.

Ans: F = 775 N Fy = 387 N a = 113° g = 39.2° 85

2–65. Express in Cartesian then The screweach eye isforce subjected to the twovector forces form shown.and Express determine the resultant force. Find magnitude each force in Cartesian vector form andthe then determineand the coordinate direction angles of the resultant force. direction resultant force. Find the magnitude and coordinate angles of the resultant force.

F1 = 300 N

60° 120°

45° y

SOLUTION 45°

F1 = 300( - cos 60° sin 45° i + cos 60° cos 45° j + sin 60°k) = {-106.07 i + 106.07 j + 259.81 k} N

60°

Ans.

= {- 106 i + 106 j + 260 k} N

F2 = 500 N

F2 = 500(cos 60° i + cos 45° j + cos 120° k) = {250.0 i + 353.55 j - 250.0k} N = {250 i + 354 j - 250 k} N

Ans.

FR = F1 + F2 = - 106.07 i + 106.07 j + 259.81 k + 250.0 i + 353.55 j - 250.0 k = 143.93 i + 459.62 j + 9.81k = {144 i + 460 j + 9.81 k} N

Ans.

FR = 2143.932 + 459.622 + 9.812 = 481.73 N = 482 N uFR =

Ans.

143.93i + 459.62j + 9.81k FR = 0.2988i + 0.9541j + 0.02036k = FR 481.73

cos a = 0.2988

a = 72.6°

Ans.

cos b = 0.9541

b = 17.4°

Ans.

cos g

88.8°

Ans.

0.02036

Ans: F1 = { - 106i + 106j + 260k} N F2 = {250i + 354j - 250k} N FR = {144i + 460j + 9.81k} N FR = 482 N a = 72.6° b = 17.4° g = 88.8° 86

2–66. Determine the coordinate direction angles of F1.

F1 = 300 N

60° 120°

45° y

SOLUTION 45°

F1 = 300(- cos 60° sin 45° i + cos 60° cos 45° j + sin 60° k) = { - 106.07 i + 106.07 j + 259.81 k} N

60°

= { -106 i + 106 j + 260 k} N F2 = 500 N

F1 = - 0.3536 i + 0.3536 j + 0.8660 k u1 = 300 a1 = cos-1 (- 0.3536) = 111°

Ans.

b 1 = cos-1 (0.3536) = 69.3°

Ans.

g1 = cos-1 (0.8660) = 30.0°

Ans.

Ans: a1 = 111° b1 = 69.3° g1 = 30.0° 87

2–67. The two forces F1 and F2 have a resultant force of FR = 5 - 100k6 lb. Determine the magnitude and coordinate direction angles of F2. The two forces F1 and F2 acting at A have a resultant force of FR = 5 -100k6 lb. Determine the magnitude and coordinate direction angles of F2.

B z 308

B 508

SOLUTION

Cartesian Vector Vector Notation: Notation: Cartesion

FR = { -100k} lb

F1 = 60{ -cos 50°cos 30°i + cos 50°sin 30°j - sin 50°k} lb

50 F1 5 60 lb

F1 60 lb

= { -33.40i + 19.28j - 45.96k} lb F2 = {F2xi + F2y j + F2z k} lb Resultant Force: Force: Resultant FR = F1 + F2 - 100k = {(F2x - 33.40) i + (F2y + 19.28)j + (F2z - 45.96) k} Equating i, j and k components, we have F2x - 33.40 = 0

F2x = 33.40 lb

F2y + 19.28 = 0

F2y = - 19.28 lb

F2z - 45.96 = - 100

F2z = - 54.04 lb

The magnitude of force F2 is F2 = 2F 22x + F 22y + F 22z = 233.402 + (- 19.28)2 + ( - 54.04)2 Ans.

= 66.39 lb = 66.4 lb The coordinate direction angles for F2 are cos a = cos b = cos g =

F2x F2 F2y F2 F2z F2

33.40 66.39

-19.28 66.39

b = 107°

Ans.

- 54.04 66.39

g = 144°

Ans.

a = 59.8°

Ans: F2 = 66.4 lb a = 59.8° b = 107° g = 144° 88

*2–68. thecoordinate coordinatedirection direction angles of force F1 and Determine the angles of the F1 indicate themthem on the and indicate on figure. the figure.

B 30

SOLUTION A

Unit Vector For Foce FF1: : Unit Vector For Force 1 uF1 = - cos 50°cos 30°i + cos 50°sin 30°j - sin 50°k

= - 0.5567i + 0.3214j - 0.7660k

F1 60 lb

Coordinate Direction Direction Angles: Angles: From the unit vector obtained above, we have cos a = - 0.5567

a = 124°

Ans.

cos b = 0.3214

b = 71.3°

Ans.

cos g = - 0.7660

g = 140°

Ans.

Ans: a = 124° b = 71.3° g = 140° 89

2–69. The bracket is subjected to the two forces shown. Express each force in Cartesian vector form and then determine the resultant force FR. Find the magnitude and coordinate direction angles of the resultant force.

z F2 = 400 N 60°

45° 120°

SOLUTION

y 25°

Cartesian Vector Vector Notation: Notation: Cartesion 35°

F1 = 2505cos 35° sin 25°i + cos 35° cos 25°j - sin 35°k6 N x

= 586.55i + 185.60j - 143.39k6 N

F1 = 250 N

Ans.

= 586.5i + 186j - 143k6 N

F2 = 4005cos 120°i + cos 45°j + cos 60°k6 N = 5 -200.0i + 282.84j + 200.0k6 N

Ans.

= 5-200i + 283j + 200k6 N

Resultant Force: Force: FR = F1 + F2

= 5186.55 - 200.02i + 1185.60 + 282.842j + 1- 143.39 + 200.02 k6 = 5 - 113.45i + 468.44j + 56.61k6 N

Ans.

= 5- 113i + 468j + 56.6k6 N

The magnitude of the resultant force is FR = 2F2Rx + F2Ry + F2Rz

= 21 -113.4522 + 468.442 + 56.612 = 485.30 N = 485 N

Ans.

The coordinate direction angles are cos a = cos b = cos g =

FRx FR F Ry FR FRz FR

-113.45 485.30

a = 104°

Ans.

468.44 485.30

b = 15.1°

Ans.

56.61 485.30

g = 83.3°

Ans.

Ans: F1 = {86.5i + 186j - 143k} N F2 = { - 200i + 283j + 200k} N FR = { - 113i + 468j + 56.6k} N FR = 485 N a = 104° b = 15.1° g = 83.3° 90

2–70. Determine the magnitude and coordinate direction angles of the resultant force, and sketch this vector on the coordinate system.

F2 525 N

120

45 x

y 5

4 3

F1 450 N

SOLUTION Cartesian Vector Notation: For F1 and F2, 4 3 F1 = 450 a j - kb = {270j - 360k} N 5 5

F2 = 525 (cos 45°i + cos 120°j + cos 60°k) = {371.23i - 262.5j + 262.5k} N Resultant Force: FR = F1 + F2 = {270j - 360k} + {371.23i - 262.5j + 262.5k} = {371.23i + 7.50j - 97.5k} N The magnitude of the resultant force is FR = 2(FR)2x + (FR)2y + (FR)2z = 2371.232 + 7.502 + ( - 97.5)2 = 383.89 N = 384 N

Ans.

The coordinate direction angles are cos a = cos b = cos g =

(FR)x FR (FR)y FR (FR)z FR

371.23 ; 383.89

a = 14.76° = 14.8°

Ans.

7.50 ; 383.89

b = 88.88° = 88.9°

Ans.

- 97.5 ; 383.89

g = 104.71° = 105°

Ans.

Ans: FR = 384 N a = 14.8° b = 88.9° g = 105° 91

2–71. Determine magnitude coordinate direction angles Specify the the magnitude andand coordinate direction angles a1, a , b , g of F so that the resultant of the three forces acting 1 1 1 1 that the resultant of the three forces acting b 1, g1 of F1 so on - 350k6 350k6 lb. lb. Note Note that that F F3 lies lies in in the the on the the bracket bracket is is FFRR == 55 3 x–y x–y plane. plane.

F3 = 400 lb

γ1

30° F2 = 200 lb

SOLUTION

β1 α1

F1 = Fx i + Fy j + Fz k F2 = - 200 j

F3 = - 400 sin 30° i + 400 cos 30° j = - 200 i + 346.4 j FR = ©F -350 k = Fx i + Fy j + Fz k - 200 j - 200 i + 346.4 j 0 = Fx - 200 ;

Fx = 200 lb

0 = Fy - 200 + 346.4 ;

Fy = - 146.4 lb

Fz = - 350 lb F1 = 2(200)2 + (- 146.4)2 + (- 350)2

Ans.

F1 = 425.9 lb = 429 lb a1 = cos-1 a b 1 = cos-1 a g1 = cos-1

200 b = 62.2° 428.9

Ans.

- 146.4 b = 110° 428.9

- 350 428.9

Ans. Ans.

= 145°

Ans: F1 = 429 lb a1 = 62.2° b1 = 110° g1 = 145° 92

*2–72. Two forces F1 and F2 act on the screw eye. If the resultant force FR has a magnitude of 150 lb and the coordinate direction angles shown, determine the angle (g) and the magnitude of F2 and its coordinate direction angles.

z F2

FR 150 lb

g 130

120

F1 80 lb

SOLUTION Cartesian Vector Notation: For FR, g can be determined from cos2 a + cos2 b + cos2 g = 1 cos2 120° + cos2 50° + cos2 g = 1 cos g = {0.5804 Here g 6 90°, then Ans.

g = 54.52° Thus FR = 150(cos 120°i + cos 50°j + cos 54.52°k) = { - 75.0i + 96.42j + 87.05k} lb Also F1 = {80j} lb Resultant Force: FR = F1 + F2 { -75.0i + 96.42j + 87.05k} = {80j} + F2 F2 = { - 75.0i + 16.42j + 87.05k} lb Thus, the magnitude of F2 is F2 = 2(F2)x + (F2)y + (F2)z = 2( - 75.0)2 + 16.422 + 87.052 = 116.07 lb = 116 lb

Ans.

And its coordinate direction angles are cos a2 = cos b2 = cos g2 =

(F2)x F2 (F2)y F2 (F2)z F2

- 75.0 ; 116.07

a2 = 130.25° = 130°

Ans.

16.42 ; 116.07

b2 = 81.87° = 81.9°

Ans.

87.05 ; 116.07

g2 = 41.41° = 41.4°

Ans.

Ans: g = 54.52° FR = 116 lb cos a2 = 130° cos b 2 = 81.9° cos g2 = 41.4° 93

2–73. Express each force in Cartesian vector form.

F3 200 N

F1 90 N 3

F2 150 N

SOLUTION x

Cartesian Vector Notation: For F1, F2 and F3, 3 4 F1 = 90 a i + kb = {72.0i + 54.0k} N 5 5

Ans.

F2 = 150 (cos 60° sin 45°i + cos 60° cos 45°j + sin 60°k) = {53.03i + 53.03j + 129.90k} N Ans.

= {53.0i + 53.0j + 130k} N F3 = {200 k}

Ans.

Ans: F1 = {72.0i + 54.0k} N F2 = {53.0i + 53.0j + 130k} N F3 = {200 k} 94

2–74.

Determine the magnitude and coordinate direction angles of the resultant force, and sketch this vector on the coordinate system.

F3 200 N

F1 90 N 3

F2 150 N

SOLUTION x

Cartesian Vector Notation: For F1, F2 and F3, 3 4 F1 = 90 a i + kb = {72.0i + 54.0k} N 5 5

F2 = 150 (cos 60° sin 45°i + cos 60° cos 45°j + sin 60°k) = {53.03i + 53.03j + 129.90k} N F3 = {200 k} N Resultant Force: F = F1 + F2 + F3 = (72.0i + 54.0k) + (53.03i + 53.03j + 129.90k) + (200k) = {125.03i + 53.03j + 383.90} N The magnitude of the resultant force is FR = 2(FR)2x + (FR)2y + (FR)2z = 2125.032 + 53.032 + 383.902 = 407.22 N = 407 N

Ans.

And the coordinate direction angles are cos a = cos b = cos g =

(FR)x FR (FR)y FR (FR)z FR

125.03 ; 407.22

a = 72.12° = 72.1°

Ans.

53.03 ; 407.22

b = 82.52° = 82.5°

Ans.

383.90 ; 407.22

g = 19.48° = 19.5°

Ans.

Ans: FR = 407 N a = 72.1° b = 82.5° g = 19.5° 95

2–75. The forces. Express The spur spur gear gear isissubjected subjectedtotothe thetwo two forces causedeach by force as a Cartesian vector. contact with other gears. Express each force as a Cartesian

vector.

SOLUTION F1 =

180 lb

135

24 7 (50)j (50)k = {14.0j - 48.0k} lb 25 25

Ans.

F2 = 180 cos 60°i + 180 cos 135°j + 180 cos 60°k

y x

25 24 7

Ans.

= {90i - 127j + 90k} lb

50 lb

Ans: F1 = {14.0j - 48.0k} lb F2 = {90i - 127j + 90k} lb 96

*2–76. Determine the resultant of the forces and express The spur gear is subjected to two the two forces caused the by result as a Cartesian vector. contact with other gears. Determine the resultant of the two

forces and express the result as a Cartesian vector.

SOLUTION

180 lb

135

FRx = 180 cos 60° = 90 FRy =

y x

7 (50) + 180 cos 135° = -113 25

25 24 7

24 FRz = - (50) + 180 cos 60° = 42 25

FR = {90i - 113j + 42k} lb

50 lb

Ans.

Ans: FR = {90i - 113j + 42k} lb 97

2–77. z

Specify the magnitude F3 and directions a3, b3, and g3 of F3 so that the resultant force of the three forces is FR = {9 j} kN.

F3 g3

10 kN

y 30

SOLUTION

F1 = 12 cos 30° j - 12 sin 30° k = 10.392 j - 6 k F2 = -

12 kN

12 5 (10) i + (10) k = -9.231 i + 3.846 k 13 13

Require FR = F1 + F2 + F3 9 j = 10.392 j - 6 k - 9.231 i + 3.846 k + F3 F3 = 9.231 i - 1.392 j + 2.154 k Hence, F3 = 2(9.231)2 + ( - 1.392)2 + (2.154)2 Ans.

F3 = 9.581 kN = 9.58 kN a3 = cos-1 a

9.231 b = 15.5° 9.581

b3 = cos-1 a

-1.392 b = 98.4° 9.581

g3 = cos-1 a

2.154 b = 77.0° 9.581

Ans.

Ans: F3 = 9.58 kN a3 = 15.5° b3 = 98.4° g3 = 77.0° 98

2–78. Determine the coordinate angle g for F2 and then express each force acting on the bracket as a Cartesian vector.

F1 450 N

SOLUTION

Rectangular Components: Since then cos2 a2 + cos2 b 2 + cos2g2 = 1, cos g2z = ; 21 - cos2 45° - cos2 60° = ; 0.5. However, it is required that g2 7 90°, thus, g2 = cos - 1( -0.5) = 120°. By resolving F1 and F2 into their x, y, and z components, as shown in Figs, a and b, respectively F1 and F2 can be expressed in Cartesian vector form as

x F2 600 N

F1 = 450 cos 45° sin 30°(- i) + 450 cos 45° cos 30°(+j) + 450 sin 45°(+ k) Ans.

= { -159i + 276j + 318k}N F2 = 600 cos 45°i + 600 cos 60°j + 600 cos 120°k

Ans.

= {424i + 300j - 30k}N

Ans: F1 = { - 159i + 276j + 318k}N F2 = {424i + 300j + 30k}N 99

2–79. Determine the magnitude and coordinate direction angles of the resultant force acting on the bracket.

F1 450 N

SOLUTION

Rectangular Components: Since then cos2 a2 + cos2 b 2 + cos2g2 = 1, cos g2z = ; 21 - cos2 45° - cos2 60° = ; 0.5. However, it is required that a2 7 90°, thus, g2 = cos - 1( -0.5) = 120°. By resolving F1 and F2 into their x, y, and z components, as shown in Figs. a and b, respectively, F1 and F2, can be expressed in Cartesian vector form, as

x F2 600 N

F1 = 450 cos 45° sin 30°(-i) + 450 cos 45° cos 30°(+ j) + 450 sin 45°(+ k) Ans.

= { -159.10i + 275.57j + 318.20k}N F2 = 600 cos 45°i + 600 cos 60°j + 600 cos 120°k

Ans.

= {424i + 300j - 300k} N Resultant Force: By adding F1 and F2 vectorally, we obtain FR. FR = F1 + F2 = (-159.10i + 275.57j + 318.20k) + (424.26i + 300j - 300k) = {265.16i + 575.57j + 18.20k} N The magnitude of FR is FR = 2(FR)x 2 + (FR)y 2 + (FR)z 2 = 2265.162 + 575.572 + 18.202 = 633.97 N = 634N

Ans.

The coordinate direction angles of FR are a = cos - 1 B

(FR)x 265.16 R = cos - 1 ¢ ≤ = 65.3° FR 633.97

b = cos - 1 B

(FR)y

g = cos - 1 B

(FR)z

FR FR

Ans.

R = cos - 1 ¢

575.57 ≤ = 24.8° 633.97

Ans.

R = cos - 1 ¢

18.20 ≤ = 88.4° 633.97

Ans.

Ans: FR = 634 N a = 65.3° b = 24.8° g = 88.4° 100

*2–80. z

Three forces act on the ring. Determine the magnitude and coordinate direction angles of the resultant force.

F3 5 100 N 608

608

F1 5 400 N

F2 5 500 N

458

SOLUTION

y 1208

Force Vector: F1 = 400 (cos 120°i + cos 45°j + cos 60°k) 308

= { - 200i + 282.84j + 200k} N F2 = 500 (sin 60°cos 30° i - sin 60° sin 30°j + cos 60°k)

= {375i - 216.51j + 250k} N F3 = {100k} N Resultant Force Vector: FR = F1 + F2 + F3 = { - 200i + 282.84j + 200k} + {375i - 216.51j + 250k} + {100k} = {175i + 66.33j + 550k} N Magnitude of FR : FR = 21752 + 66.332 + 5502 = 580.97 N = 581 N

Ans.

Coordinate Direction Angles: uFR =

175i + 66.33j + 550k FR = FR 580.97

= 0.3012i + 0.1142j + 0.9467k

cos a = 0.3012

a = 72.5°

Ans.

cos b = 0.1142

b = 83.4°

Ans.

cos g = 0.9467

g = 18.8°

Ans.

Ans: FR = 581 N a = 72.5° b = 83.4° g = 18.8° 101

2–81. z

If the coordinate direction angles for F3 are a3 = 120°, b 3 = 60° and g3 = 45°, determine the magnitude and coordinate direction angles of the resultant force acting on the eyebolt.

F3 800 lb

5 4 3

SOLUTION Force Vectors: Vector: By resolving F1, F2 and F3 into their x, y, and z components, as shown in Figs. a, b, and c, respectively, F1, F2 and F3 can be expressed in Cartesian vector form as

F2 600 lb

30 y

x F1 700 lb

F1 = 700 cos 30°(+ i) + 700 sin 30°(+j) = 5606.22i + 350j6 lb 4 3 F2 = 0i + 600 a b( + j) + 600 a b ( + k) = 5480j + 360k6 lb 5 5

F3 = 800 cos 120°i + 800 cos 60°j + 800 cos 45°k = 3-400i + 400j + 565.69k4 lb

Resultant Force: Force: By adding F1, F2 and F3 vectorally, we obtain FR. Thus, Resultant FR = F1 + F2 + F3

= (606.22i + 350j) + (480j + 360k) + ( - 400i + 400j + 565.69k) = 3206.22i + 1230j + 925.69k4 lb

The magnitude of FR is

FR = 3(FR)x 2 + (FR)y 2 + (FR)z 2

= 3(206.22)2 + (1230)2 + (925.69)2 = 1553.16 lb = 1.55 kip

Ans.

The coordinate direction angles of FR are a = cos-1 c

(FR)x 206.22 b = 82.4° d = cos-1 a FR 1553.16

b = cos-1 c

g = cos-1 c

(FR)y

(FR)z FR

d = cos-1 a d = cos-1 a

Ans.

1230 b = 37.6° 1553.16

Ans.

925.69 b = 53.4° 1553.16

Ans.

Ans: FR = 1.55 kip a = 82.4° b = 37.6° g = 53.4° 102

2–82. z

If the coordinate direction angles for F3 are a3 = 120°, b 3 = 45° and g3 = 60°, determine the magnitude and coordinate direction angles of the resultant force acting on the eyebolt.

F3 800 lb

5 4 3

SOLUTION Force Vectors: By resolving F1, F2 and F3 into their x, y, and z components, as shown in Figs. a, b, and c, respectively, F1, F2, and F3 can be expressed in Cartesian vector form as

F2 600 lb

30 y

x F1 700 lb

F1 = 700 cos 30°( + i) + 700 sin 30°(+j) = 5606.22i + 350j6 lb 3 4 F2 = 0i + 600 a b (+ j) + 600 a b( + k) = 5480j + 360k6 lb 5 5

F3 = 800 cos 120°i + 800 cos 45°j + 800 cos 60°k = 5- 400i + 565.69j + 400k6 lb FR = F1 + F2 + F3

= 606.22i + 350j + 480j + 360k - 400i + 565.69j + 400k = 5206.22i + 1395.69j + 760k6 lb FR = 3(206.22)2 + (1395.69)2 + (760)2

Ans.

= 1602.52 lb = 1.60 kip

a = cos-1 a b = cos-1 a g = cos-1 a

206.22 b = 82.6° 1602.52

Ans.

1395.69 b = 29.4° 1602.52

Ans.

760 b = 61.7° 1602.52

Ans.

Ans: FR = 1.60 kip a = 82.6° b = 29.4° g = 61.7° 103

If the direction of the resultant force acting on the eyebolt 2–83. is defined by the unit vector uFR = cos 30°j + sin 30°k, If of force acting the eyebolt determine the coordinate direction of F the If the the direction direction of the the resultant resultant force angles acting on on the eyebolt 3 and is defined by the unit vector u = cos 30°j + sin F magnitude . unit vector uFRR = cos 30°j + sin 30°k is defined of byFRthe 30°k,, determine determine the the coordinate coordinate direction direction angles angles of of F and the the F33 and magnitude .. magnitude of of F FR R

F3 800 lb F 800 800 lb lb F33

SOLUTION Force Vectors: By resolving F1, F2 and F3 into their x, y, and z components, as shown SOLUTION SOLUTION in Figs. a, b, and c, respectively, F1, F2, and F3 can be expressed in Cartesian vector Force Vectors: Forceas Vectors: By By resolving resolving F F11,, F F22 and and F F33 into into their their x, x, y, y, and and zz components, components, as as shown shown form in Figs. a, b, and c, respectively, , , and can be expressed in Cartesian F F F 1 2 3 in Figs. a, b, and c, respectively, F1, F2, and F3 can be expressed in Cartesian vector vector form as F 700 cos 30°( + i) + 700 sin 30°(+j) = 5606.22i + 350j6 lb 1 = as form

30 x xx

30 30 F1 700 lb

5 4 3

F2 600 lb

5 5 4 33 4

F 600 600 lb lb F22 y yy

F F11 700 700 lb lb

F1 = = 700 700 cos cos 30°( 30°( + i) i) + + 700 700 sin sin 330°( 30°(+j) = 5606.22i 5606.22i + + 350j6 350j6 lb lb 4+ F +j) = 1 F 2 = 0i + 600a b( +j) + 600 a b ( +k) = 5480j + 360k6 lb 5 5 3 4 b (+ k) = = 5480j 5480j + + 360k6 360k6 lb lb b( +j) +j) + + 600 aa 3 b( = 0i 0i + 600a 600a 4 b( F +k) = F22 = F 800+cos a3i55+ 800 cos600 b3 j + 3 55 800 cos g3k

F3 = = 800 800 cos a a3ii + 800 b g FRcos Since the direction is defined by ucos F cos 800 cos b 33 jj + + 800 800 cos g33k kcos 30°j + sin 30°k, it can be written in FR = 3 3 + of Cartesian vector form as Since is defined by uF = FR = cos cos 30°j 30°j + + sin sin 30°k 30°k,, it Since the the direction direction of of F it can can be be written written in in R is defined by uFR R F = FRform (cos 30°j Cartesian as R = FRuFvector R Cartesian vector form as + sin 30°k) = 0.8660FR j + 0.5FR k FR = = F FRu uFR = = F FRBy (cos 30°j 0.8660F j + Rk Resultant adding , F230°k) , and F= obtain F 30°j + +F1sin sin 30°k) =3 vectorally, 0.8660FR + 0.5F 0.5F R R FForce: R(cos R j we R k FR. Thus, R Resultant Force: By adding F ,, F F F2 + F 2, and F3 vectorally, we obtain FR. Thus, R = F1 +Force: 3 adding F1 Resultant By 1 F2, and F3 vectorally, we obtain FR. Thus, F = F F1R+ +j +F F20.5F + FR3 k = (606.22i + 350j) + (480j + 360k) + (800 cos a3i + 800 cos b 3 j + 800 cos g3k) R = F0.8660F R 1 2 + F3 + 800 cos a3(480j )i + (350 + 480 (800 + 800 cos b )j 800 + (360 b+3 j800 cos g3)kg3k) 0.8660FRR j + 0.5FRR k = (606.22 0.8660FR j + 0.5FR k = (606.22i (606.22i + + 350j) 350j) + + (480j + + 360k) 360k) + + (800 cos cos a a33ii 3+ + 800 cos cos b 3 j + + 800 800 cos cos g3k) j + 0.5F k = (606.22 + 800 cos a )i + (350 + 480 + 800 cos b )j + (360 + 800 cos g 0.8660F R kk= Equating j, and components, we have j + i,0.5F (606.22 + 800 cos a3)i + (350 + 480 + 800 cos b 3)j + (360 + 800 cos g3)k )k 0.8660FRthe R

the jj,, and we 0Equating = 606.22 cosk 3 Equating the+ ii,,800 and kacomponents, components, we have have 800 cos a3 = -606.22 = 606.22 606.22 + + 800 800 cos cos a a3 00 = + 480 3+ 800 cos b 3 0.8660F 800 cos a -606.22 R 3 == 350 800 cos a = -606.22 800 cos b33 = 0.8660FR - 830 = 350 350 + + 480 480 + + 800 800 cos cos b b3 0.8660F R = 0.8660FR 3 + 800 cos g3830 0.5F 800 cos = 0.8660F R =b360 3 R 800 cos b 3 = 0.8660FR - 830 800 cos g3 = 0.5FR - 360 = 360 360 + + 800 800 cos cos g g3 0.5F R = 0.5FR 3 800 cos g = 0.5F 360 3 R Squaring and then adding 800 cos g3 = 0.5FR - 360Eqs. (1), (2), and (3), yields

(1) (1) (1) (2) (2) (2) (3) (3) (3)

2 2 2 (3), yields Squaring 800 [cos2and a3 +then cosadding b 3 + Eqs. cos2 (1), g3] (2), = Fand 1797.60F R -(3), Squaring and then adding Eqs. (1), (2), and yields R + 1,186,000

(4)

2 2 2 2 2 2 However, . Thus, fromREq. acos 800 [cos22 cos a3 + b 3 +b cos g3] =g3F=R 2211797.60F + (4) 1,186,000 3 +2 cos 3 +2 cos 8002 [cos a 3 + cos b 3 + cos g3] = FR - 1797.60FR + 1,186,000 2 2 2 2 F 546,000 = 0 2 g3 = 1. Thus, from Eq. (4) However, cos aR3 + R - 1797.60F However, cos2 a + cos cos2 b b3 + + cos cos g = 1. Thus, from Eq. (4)

(4) (4)

FR 22 - 1797.60F 1797.60F + 546,000 = R quadratic Solving the above we have two positive roots F = 00 R R + 546,000equation, Solving the quadratic equation, F N = 387 N R = 387.09 Solving the above above quadratic equation, we we have have two two positive positive roots roots

Ans.

1.41NkN FRR = 1410.51 387.09 N NN= ==387 387 F = 387.09 N = Eq. 1410.51 N = = 1.41 1.41 kN kN FR = From (1), N 1410.51 F R

Ans. Ans. Ans. Ans.

From Eq. a 3 = 139° From Eq. (1), (1),

Ans.

= 139° 139° FR = 387.09 N into Eqs. (2), and (3), yields a Substituting a33 =

Ans. Ans.

b 102° (3), yields Substituting 3 = 128° FR = 387.09 N intogEqs. 3 = (2), Substituting (2), and and (3), yields FR = 387.09 N into Eqs. Substituting into Eqs. (2), and (3), yields F = 1410.51 N b = 128° g = 102° R g33 = 102° b 33 = 128° g = 64.4° b Substituting 3 = 60.7° FR = 1410.51 N into Substituting Eqs. (2), (2), and and (3), (3), yields yields F = 1410.51 N into3Eqs.

Ans.

Ans. Ans. Ans.

b = 60.7° 60.7° b 33 =

g g33 = = 64.4° 64.4°

104

Ans. Ans.

Ans: a3 = 139° b3 = 128°, g3 = 102°, if FR = 387 N b3 = 60.7°, g3 = 64.4°, if FR = 1.41 kN

*2–84. z

The pipe is subjected to the force F, which has components acting along the x, y, z axes. If the magnitude of F is 12 kN, and a = 120° and g = 45°, determine the magnitudes of its three components.

Fz F Fx

SOLUTION

cos2 a + cos2 b + cos2 g = 1 cos2 120° + cos2 b + cos2 45° = 1

cos b = {0.5

From the figure, cos b = + 0.5 b = 60° Fx = - F cos a = |12 cos 120°| = 6 kN

Ans.

Fy = - F cos b = |12 cos 60°| = 6 kN

Ans.

Fz = - F cos g = |12 cos 45°| = 8.49 kN

Ans.

Ans: Fx = 6 kN Fy = 6 kN Fz = 8.49 kN 105

2–85. The pipe is subjected to the force F, which has components Fx = 1.5 kN and Fz = 1.25 kN. If b = 75°, determine the magnitude of F and Fy.

Fz F Fx

SOLUTION Fx = F cos a = 1.5 [1] Fz = F cos g = 1.25 [2]

Fy y

Squaring Eqs.[1] and [2] and rearranging yields: 2.25 [3] F2 1.5625 [4] cos2 g = F2 cos2 a =

cos2 a + cos2 b + cos2 g = 1 2.25 1.5625 + cos275° + = 1 F2 F2 F = 2.021 kN = 2.02 kN

Ans.

Fy = F cos b = 2.021 cos 75° = 0.523 kN

Ans.

Ans: F = 2.02 kN Fy = 0.523 kN 106

2–86. y

Determine the length of the connecting rod AB by first formulating a position vector from A to B and then determining its magnitude.

300 mm

SOLUTION O

Position Vector: The coordinates of points A and B are A( -150 cos 30°, - 150 sin 30°) mm and B(0, 300) mm respectively. Then

rAB = [0 - ( - 150 cos 30°)]i + [300 - ( - 150 sin 30°)]j

A 150 mm

= {129.90i + 375j} mm Thus, the magnitude of rAB is rAB = 2129.902 + 3752 = 396.86 mm = 397 mm

Ans.

Ans: rAB = 397 mm 107

2–87. Determine the distance between the end points A and B on the pipe assembly.

458

A 2 ft

SOLUTION

4 ft

Position Vector: r = (4 - 0)i + (3 - 0)j + ( -2 - 2)k

3 ft

= {4i + 3j - 4k} ft

2 ft B

Magnitude: r = 242 + 32 + ( - 4)2 = 6.40 ft

Ans.

Ans: r = 6.40 ft 108

*2–88. z

Express each force as a Cartesian vector, and then determine the magnitude and coordinate direction angles of the resultant force.

13 12 5

2.5 ft 4 ft

SOLUTION

rAC F1 = 80 lb a b = - 26.20 i - 41.93 j + 62.89 k rAC = { - 26.2 i - 41.9 j + 62.9 k} lb

6 ft

Ans.

rAB = {2 i - 4 j - 6 k} ft rAB b = 13.36 i - 26.73 j - 40.09 k rAB

= {13.4 i - 26.7 j - 40.1 k} lb

F2 50 lb

12 rAC = e - 2.5 i - 4 j + (2.5) k f ft 5

F2 = 50 lb a

F1 80 lb

2 ft

Ans.

FR = F1 + F2 = -12.84 i - 68.65 j + 22.80 k = { - 12.8 i - 68.7 j + 22.8 k } lb FR = 2( - 12.84)2 ( - 68.65)2 + (22.80)2 = 73.47 = 73.5 lb a = cos - 1 a b = cos - 1 a g = cos - 1 a

- 12.84 b = 100° 73.47

Ans. Ans.

- 68.65 b = 159° 73.47

Ans.

22.80 b = 71.9° 73.47

Ans.

Ans: F1 = { - 26.2 i - 41.9 j + 62.9 k} lb F2 = {13.4 i - 26.7 j - 40.1 k} lb FR = 73.5 lb a = 100° b = 159° g = 71.9° 109

2–89. z

The positions of point A on the flag pole and point B on the light have been measured relative to the electronic distance meter at O. Determine the distance between A and B.

A 232 ft 308

138 ft 508

458

208

SOLUTION Position Vector:

rA = 232( - cos 30° sin 45°i + cos 30° cos 45°j + sin 30° k) = { - 142.070i + 142.070j + 116k} ft rB = 138 (cos 50° cos 20°i - cos 50° sin 20°j + sin 50°) k = {83.355i - 30.339j + 105.714k} ft rAB = rA - rB = ( - 142.070i + 142.070j + 116k) - (83.355i - 30.339j + 105.714k) = { - 225.425i + 172.409j + 10.286k} ft Magnitude: rAB = 2( - 225.425)2 + 172.4092 + 10.2862 = 284 ft

Ans.

Ans: rAB = 284 ft 110

2–90. The 8-m-long cable is anchored to the ground at A. If x = 4 m and y = 2 m, determine the coordinate z to the highest point of attachment along the column.

SOLUTION

r = {4i + 2j + zk} m r = 2(4)2 + (2)2 + (z)2 = 8

Ans.

z = 6.63 m

Ans: 6.63 m 111

2–91. z

The 8-m-long cable is anchored to the ground at A. If z = 5 m, determine the location + x, + y of the support at A. Choose a value such that x = y.

SOLUTION

r = {xi + yj + 5k} m r = 2(x)2 + (y)2 + (5)2 = 8

x = y, thus

2x2 = 82 - 52

Ans.

x = y = 4.42 m

Ans: 4.42 m 112

*2–92. z

Determine the magnitude and coordinate direction angles of the resultant force.

0.75 m A FAB 250 N FAC 400 N

y 40

2m 1m

Unit Vectors: The coordinates for points A, B and C are (0, - 0.75, 3) m, B(2 cos 40°, 2 sin 40°, 0) m and C(2, -1, 0) m respectively. uAB =

SOLUTION

(2 cos 40° - 0)i + [2 sin 40° - ( -0.75)]j + (0 - 3)k rAB = rAB 2(2 cos 40° - 0)2 + [2 sin 40° - ( -0.75)]2 + (0 - 3)2 = 0.3893i + 0.5172j - 0.7622k

uAC =

(2 - 0)i + [ - 1 - ( - 0.75)]j + (0 - 3)k rAC = rAC 2(2 - 0)2 + [ - 1 - ( - 0.75)]2 + (0 - 3)2 = 0.5534i - 0.0692j - 0.8301k

Force Vectors: FAB = FAB uAB = 250 (0.3893i + 0.5172j - 0.7622k) = {97.32i + 129.30j - 190.56k} N = {97.3i + 129j - 191k} N

Ans.

FAC = FAC uAC = 400 (0.5534i - 0.06917j - 0.8301k) = {221.35i - 27.67j - 332.02k} N = {221i - 27.7j - 332k} N

Ans.

Resultant Force: FR = FAB + FAC = {97.32i + 129.30j - 190.56k} + {221.35i - 27.67j - 332.02k} = {318.67i + 101.63j - 522.58 k} N The magnitude of FR is FR = 2(FR)2x + (FR)2y + (FR)2z = 2318.672 + 101.632 + ( - 522.58)2 = 620.46 N = 620 N

And its coordinate direction angles are cos a = cos b = cos g =

(FR)x FR (FR)y FR (FR)z FR

318.67 ; 620.46

a = 59.10° = 59.1°

Ans.

101.63 ; 620.46

b = 80.57° = 80.6°

Ans.

- 522.58 ; 620.46

g = 147.38° = 147°

Ans.

113

Ans: FR = 620 N cos a = 59.1° cos b = 80.6° cos g = 147°

2–93. z

The 80-ft-long cable is attached to the boom at A. If x = 60 ft, determine the coordinates y and z to the end of the cable B. Take z = 3y.

z y x

SOLUTION y

r = (60 - 0)i + (y - 0)j + (0 - 3y)k

= 60i + yj - 3yk Magnitude of r: r = 80 = 2602 + y2 + ( - 3y)2 6400 = 3600 + 10y2

Ans.

z = 3y = 3(16.733) = 50.2 ft

Ans.

y = 16.733 ft = 16.7 ft

Ans: y = 16.7 ft z = 50.2 ft 114

2–94. z

The cable is attached to the boom at A and to point B, which has coordinates of x = 80 ft, y = 40 ft, z = 40 ft. Determine the length of the cable and the angle between the positive z axis and the cable.

z y x

SOLUTION y

rAB = (80 - 0)i + (40 - 0)j + (0 - 40)k

= 80i + 40j - 40k Magnitude of rAB: rAB = 2802 + 402 + ( - 40)2 = 97.98 ft = 98.0 ft

Ans.

Unit Vector: uAB =

80i + 40j - 40k rAB = = 0.8165i + 0.4082j - 0.4082k rAB 97.98

cos g = -0.4082

Ans.

g = 114°

Ans: rAB = 98.0 ft g = 114° 115

2–95. z

At a given instant, the position of a plane at A and a train at B are measured relative to a radar antenna at O. Determine the distance d between A and B at this instant. To solve the problem, formulate a position vector, directed from A to B, and then determine its magnitude.

5 km 60° 35° O

SOLUTION

Position Vector: The coordinates of points A and B are

40° 25° 2 km B

A( -5 cos 60° cos 35°, - 5 cos 60° sin 35°, 5 sin 60°) km = A( - 2.048, - 1.434, 4.330) km B(2 cos 25° sin 40°, 2 cos 25° cos 40°, - 2 sin 25°) km = B(1.165, 1.389, - 0.845) km The position vector rAB can be established from the coordinates of points A and B. rAB = {[1.165 - ( -2.048)] i + [1.389 - ( - 1.434)] j + ( - 0.845 - 4.330) k} km = {3.213 i + 2.822 j - 5.175 k} km The distance between points A and B is d = rAB = 23.2132 + 2.8222 + ( - 5.175)2 = 6.71 km

Ans.

Ans: d = 6.71 km 116

*2–96. Represent each cable force as a Cartesian vector.

z C

2m E

2m B

FE 350 N 3m

FC 400 N FB 400 N

SOLUTION

D 2m

rC = (0 - 5)i + ( - 2 - 0)j + (3 - 0)k = { - 5i - 2j + 3k} m

rC = 2( -5)2 + ( -2)2 + 32 = 238 m

rB = (0 - 5)i + (2 - 0)j + (3 - 0)k = { - 5i + 2j + 3k} m rB = 2( -5)2 + 22 + 32 = 238 m

rE = (0 - 2)i + (0 - 0)j + (3 - 0)k = { - 2i + 0j + 3k} m rE = 2( -2)2 + 02 + 32 = 213 m r F = Fu = F a b r FC = 400 a FB = 400 a FE = 350 a

-5i - 2j + 3k 138

b = { - 324i - 130j + 195k} N

Ans.

138

b = { - 324i + 130j + 195k} N

Ans.

113

b = { - 194i + 291k} N

Ans.

-5i + 2j + 3k -2i + 0j + 3k

Ans: FC = { - 324i - 130j + 195k} N FB = { - 324i + 130j + 195k} N FE = { - 194i + 291k} N 117

2–97. z

Determine the magnitude and coordinate direction angles of the resultant of the two forces acting at point A.

2m E

2m B

FE 350 N 3m

FC 400 N FB 400 N

SOLUTION

D 2m

rC = (0 - 5)i + ( - 2 - 0)j + (3 - 0)k = { -5i - 2j + 3k}

rC = 2( - 5)2 + ( - 2)2 + (3)2 = 238 m

( - 5i - 2j + 3k) rC FC = 400 a b = 400 a b rC 138

FC = ( -324.4428i - 129.777j + 194.666k) rB = (0 - 5)i + (2 - 0)j + (3 - 0)k = { - 5i + 2j + 3k} rB = 2( - 5)2 + 22 + 32 = 238 m FB = 400 a

( - 5i + 2j + 3k) rB b = 400 a b rB 138

FB = ( -324.443i + 129.777j + 194.666k) FR = FC + FB = ( -648.89i + 389.33k)

FR = 2( -648.89)2 + (389.33)2 + 02 = 756.7242

Ans.

FR = 757 N a = cos-1 a b = cos-1 a g = cos-1 a

- 648.89 b = 149.03 = 149° 756.7242

Ans.

0 b = 90.0° 756.7242

Ans.

389.33 b = 59.036 = 59.0° 756.7242

Ans.

Ans: FR = 757 N a = 149° b = 90.0° g = 59.0° 118

2–98. The force F has a magnitude of 80 lb and acts at the midpoint C of the thin rod. Express the force as a Cartesian vector.

z B

6 ft C F 80 lb

SOLUTION rAB = ( -3i + 2j + 6k) rCB =

O 3 ft

1 rAB = ( - 1.5i + 1j + 3k) 2

rCO = rBO + rCB

2 ft

= -6k - 1.5i + 1j + 3k

= -1.5i + 1j - 3k rCO = 3.5 F = 80 a

rCO b = { - 34.3i + 22.9j - 68.6k} lb rCO

Ans.

Ans: F = { -34.3i + 22.9j - 68.6k} lb 119

2–99. Position vectors along the robotic arm from O to B and B to A are rOB = 5 100i + 300j + 400k6 mm and rBA = 5350i + 225j - 640k6 mm, respectively. Determine the distance from O to the grip at A.

rBA x

SOLUTION rOA = rOB + rBA

rOA = {450i + 525j - 240k} mm rOA = 2(450)2 + (525)2 + ( - 240)2 = 732 mm

Ans.

Ans: rOA = 732 mm 120

*2–100. rOA = 50.5i + 4j + 0.25k6 m and rOB = 50.3i + If 2j + 2k6 m, express rBA as a Cartesian vector.

rBA x

SOLUTION rOA = rOB + rBA

0.5i + 4j + 0.25k = 0.3i + 2j + 2k + rBA rBA = {0.2i + 2j - 1.75k} m

Ans.

rAD = 2( -1)2 + 12 + ( - 0.5)2 = 1.50 m

Ans.

rBD = 212 + ( - 1)2 + 0.52 = 1.50 m

Ans.

rCD = 212 + 12 + ( - 1)2 = 1.73 m

Ans.

Ans: rBA = {0.2i + 2j - 1.75k} m rAD = 1.50 m rBD = 1.50 m rCD = 1.73 m 121

2–101. Determine the position (x, y, 0) for fixing cable BA so that the resultant force exerted on the pole is directed along its axis, from B toward O. Also, what is the magnitude of the resultant force?

z B F2 5 250 N F1 5 350 N

SOLUTION

F3 5 300 N

Force Vector:

r1 = ( -3 - 0)i + (0 - 0)j + (0 - 4)k = { - 3i + 0j - 4k} m r1 = 2( -3)2 + 02 + ( - 4)2 = 5 m

y x

r1 - 3i - 4k F1 = F1 a b = 350 a b = { - 210i - 280k} N r1 5

r2 = (2 - 0)i + ( - 3 - 0)j + (0 - 4)k = {2i - 3j - 4k} m r2 = 222 + ( - 3)2 + ( - 4)2 = 229 m

2i - 3j - 4k r2 F2 = F2 a b = 250 a b r2 229

= {92.8477i - 139.2715j - 185.6953k} N

r3 = (x - 0)i + (y - 0)j + (0 - 4)k = {xi + yj - 4k} m r3 = 2x2 + y2 + ( - 4)2 = 216 + x2 + y2

300xi + 300yj - 1200k xi + yj - 4k r3 F3 = F3 a b = 300 a b = 2 2 r3 216 + x + y 216 + x2 + y2 FR = -FR k

Resultant Force Vector: FR = F1 + F2 + F3 -FR k = { -210i - 280k} + {92.8477i - 139.2715j - 185.6953k} + -FR k = a - 117.1523 +

300x 216 + x2 + y2

300xi + 300yj - 1200k 216 + x2 + y2

bi + a - 139.2715 +

+ a - 465.6953 -

Equating i, j, k components: -117.1523 + -139.2715 +

300x 216 + x2 + y2 300y

216 + x2 + y2

-FR = - 465.6953 Solving Eqs. [1], [2] and [3] yields: y = 2.34 m

300y

216 + x2 + y2 1200

216 + x2 + y2

= 0 [1] = 0 [2]

1200

216 + x2 + y2

[3]

Ans.

x = 1.96 m

Ans.

FR = 704 N

122

Ans: y = 2.34 m x = 1.96 m FR = 704 N

2–102. Determine the magnitude and coordinate direction angles a, b, g of the resultant force acting on the pole. Set x = 4 m, y = 2 m.

z B F2 5 250 N F1 5 350 N 4m

F3 5 300 N 2m

SOLUTION Force Vector:

3m O

r1 = ( -3 - 0) i + (0 - 0) j + (0 - 4) k = { - 3i + 0j - 4k} m

r1 = 2( -3)2 + 02 + ( - 4)2 = 5 m

r1 - 3i - 4k F1 = F1 a b = 350 a b = { - 210i - 280k} N r1 5 r2 = (2 - 0) i + ( -3 - 0) j + (0 - 4) k = {2i - 3j - 4k} m r2 = 222 + ( -3)2 + ( - 4)2 = 229 m

2i - 3j - 4k r2 F2 = F2 a b = 250 a b r2 229

= {92.8477i - 139.2715j - 185.6953k} N

r3 = (4 - 0) i + (2 - 0) j + (0 - 4) k = {4i + 2j - 4k} m r3 = 242 + 22 + ( - 4)2 = 6 m

- 4i + 2j - 4k r3 b = {200i + 100j - 200k} N F3 = F3 a b = 300 a r3 6 Resultant Force Vector: FR = F1 + F2 + F3 = { - 210i - 280k} + {92.8477i - 139.2715j - 185.6953k} + {200i + 100j - 200k} = {82.8477i - 39.2715j - 665.6953k} N FR = 282.84772 + ( - 39.2715)2 + 665.69532 = 671.98 N = 672 N

Ans.

Coordinate Direction Angles: uR =

FR 82.8477i - 39.2715j - 665.6953k = FR 671.98 = 0.1233i - 0.05844j - 0.9906k

cos a = 0.1233

a = 82.9°

Ans.

cos b = - 0.05844

b = 93.4°

Ans.

cos g = - 0.9906

g = 172°

Ans.

123

Ans: FR = 672 N a = 82.9° b = 93.4° g = 172°

2–103. Given the three vectors A, B, and D, show that A # 1B + D2 = 1A # B 2 + 1A # D2.

0.2 m

0.4 m

Given the three vectors A, B, and D, show that A # (B + D) = (A # B) + (A # D).

Since the component of (B + D) is equal to the sum of the components of B and D, then

0.6 m

A # (B + D) = A # B + A # D

= A x (Bx + Dx) + A y (By + Dy) + A z (Bz + Dz)

# + x(A+ DDx)i+ +A(BDy ++ DAy)jD +) (Bz + Dz)k] A # (B + D)= =(A(ABx i ++ A A yBj ++ A Azk) B )[(B x x

y y

z z

(Bx+ +(AD#xD) ) + A y (By + Dy) + A z (Bz + Dz) = =(AA# xB)

(QED)

= (A xBx + A yBy + A zBz) + (A xDx + A yDy + A zDz) = (A # B) + (A # D)

(QED)

124

2m C 2m

(QED) x Since the component of (B + D) is equal to the sum of the components of B and D, then Also, A # (B + D) = A # B + A # D (QED) A # (B + D) = (A x i + A y j + A zk) # [(Bx + Dx)i + (By + Dy)j + (Bz + Dz)k] Also,

0.8 m B

SOLUTION SOLUTION

D A z

Given the three vectors A, B, and D, show that A # (B + D) = (A # B) + (A # D).

0.5 m

2m 2 m2 m D

B C F 600 N

2m 32mm D

C E

F 600 N

3m E

*2–104. Determine the angles u and f between the wire segments.

0.4 m

0.2 m B 0.5 m

SOLUTION rBA = { -0.4j - 0.5k} m ;

rBA = 0.640 m

rBC = {0.8i + 0.2j - 0.5k} m ;

rBC = 0.964 m

rBA # rB C = 0 + ( -0.4) (0.2) + ( -0.5) ( -0.5) = 0.170 m2 u = cos-1 ¢

0.170 ≤ = 74.0° (0.640)(0.964)

0.6 m

0.8 m

Ans.

rCB = { -0.8i - 0.2j + 0.5k} m ;

rCB = 0.964m

rCD = { -0.8i } m ;

rCD = 0.800 m

rCB # rCD = (- 0.8)( - 0.8) = 0.640 m2 f = cos-1 ¢

0.640 ≤ = 33.9° (0.9464) (0.800)

Ans.

Ans: u = 74.0° f = 33.9° 125

2–105. Determine the magnitude of the projected component of force F along the pole.

z F = {–20i + 50j – 10k} lb A

6 ft

SOLUTION SOLUTION rOA = {{3i ft ft 3 i -- 2j2 j+ +6k} 6 k}

3 ft

+ 6 k) # # (3i(3-2i 2j- 2+ j 6k) Proj 20 i++50j 50 j -10k) 10k) Proj F F = = F ## uuOA OA == ((--20i 2 2 2 2(3) ( -2) + 2(6)2 2(3) + (+- 2) + (6)

2 ft x

Proj F F = = - 31.4 lb |Proj F|

Ans.

31.4 lb

Ans: |Proj F| = 31.4 lb 126

2–106. Determine the components of F that act along rod AC and perpendicular to it. Point B is located at the midpoint of the rod.

SOLUTION rAC = (-3i + 4j - 4k), rAB =

rAC = 2( - 3)2 + 42 + ( -4)2 = 241 m

-3i + 4j + 4k rAC = = - 1.5i + 2j - 2k 2 2

4m C

600 N

rAD = rAB + rBD rBD = rAD - rAB = (4i + 6j - 4k) - ( - 1.5i + 2j - 2k) = {5.5i + 4j - 2k} m rBD = 2(5.5)2 + (4)2 + ( -2)2 = 7.0887 m F = 600a

rBD b = 465.528i + 338.5659j - 169.2829k rBD

Component of F along rAC is F| | F| | =

(465.528i + 338.5659j - 169.2829k) # ( -3i + 4j - 4k) F # rAC = rAC 241

F| | = 99.1408 = 99.1 N

Ans.

Component of F perpendicular to rAC is F F2 + F2|| = F2 = 6002 F2 = 6002 - 99.14082 F

Ans.

= 591.75 = 592 N

Ans: F = 99.1 N F# = 592 N 127

2–107. Determine the components of F that act along rod AC and perpendicular to it. Point B is located 3 m along the rod from end C.

SOLUTION

rCA = 3i - 4j + 4k

4m C

600 N 3m

rCA = 6.403124 rCB =

3 (r ) = 1.40556i - 1.874085j + 1.874085k 6.403124 CA

rOB = rOC + rCB = -3i + 4j + r CB = -1.59444i + 2.1259j + 1.874085k rOD = rOB + rBD rBD = rOD - rOB = (4i + 6j) - rOB = 5.5944i + 3.8741j - 1.874085k rBD = 2(5.5944)2 + (3.8741)2 + ( -1.874085)2 = 7.0582 F = 600(

rBD ) = 475.568i + 329.326j - 159.311k rBD

rAC = (- 3i + 4j - 4k),

rAC = 241

Component of F along rAC is F| | F| | =

(475.568i + 329.326j - 159.311k) # (- 3i + 4j - 4k) F # rAC = rAC 241

F| | = 82.4351 = 82.4 N

Ans.

Component of F perpendicular to rAC is F F2 + F2|| = F2 = 6002 F2 = 6002 - 82.43512 F

Ans.

= 594 N

Ans: F = 82.4 N F# = 594 N 128

*2–108. The force F = 525i - 50j + 10k6 lb acts at the end A of the pipe assembly. Determine the magnitude of the F2 which act along axis ofAB AB and and components FF and F2 which actthealong 1 1and perpendicular to it.

F2 F

A F1

SOLUTION

6 ft

Unit Vectors: The unit unit vector vector along along AB AB axis axis is is Vector: The uAB =

10 - 02i + 15 - 92j + 10 - 62k

210 - 022 + 15 - 922 + 10 - 622

Projected Component Component of F Along AB Axis:

F1 = F # uAB = 125i - 50j + 10k2 # 1- 0.5547j - 0.8321k2

= 1252102 + 1 -5021 -0.55472 + 11021-0.83212 = 19.415 lb = 19.4 lb

Component of F Perpendicular to AB AB Axis: Component of Perpendicular to Axis: F = 2252 + 1 - 5022 + 102 = 56.789 lb. F2 =

F 2 - F 21 =

= - 0.5547j - 0.8321k

5 ft

4 ft

Ans.

The magnitude of force F is

56.7892 - 19.414 2 = 53.4 lb

Ans.

Ans: F1 = 19.4 lb F2 = 53.4 lb 129

2–109. A force of F = {-40k} lb acts at the end of the pipe. Determine the magnitudes of the components F1 and F2 which are directed along the pipe’s axis and perpendicular to it.

y 3 ft

A force of F = 5 - 40k6 lb acts at the end of the pipe. Determine the magnitudes of the components F1 and F2 which are directed along the pipe’s axis and perpendicular to it.

z5 ft

3 ft

x O

y 3 ft

5 ft

SOLUTION uOA =

3 i + 5 j - 3k 2

23 + 5 + (- 3)

F1 = F # uOA = ( - 40 k) # a

3 ft F 5 {240 k} lb

3i + 5j - 3k

243

3i + 5j - 3k 243

Ans.

= 18.3 lb

F1 F { 40 k} lb

F2 = 2F2 - F12 F2 = 2402 - 18.32 = 35.6 lb

Ans.

Ans: F1 = 18.3 lb F2 = 35.6 lb 130

2–110. Determine the magnitudes of the components of F = 600 N acting along and perpendicular to segment DE of the pipe assembly.

the magnitudes of the the components of of the components of Determine magnitudes cting along andFperpendicular to segment DE = 600 N acting along and perpendicular to segment DE ssembly. of the pipe assembly.

2m 2m

A x

SOLUTION

3ym

C (0 - 4)i + (2 - 5)j r+ [0 - (0 (- 2)]k - 4)i += (2 - 5)j +- [00.5571j - (- 2)]k EB 0.7428i + 0.3714k D u = = = - 0.7428i - 0.5571j + 0.3714kF 2 2(0 - 4)2 + (2EB- 5)r2EB+ [0 2(0 - ( -2)]4) 2 + (2 - 5)2 + [0 - (- 2)]2 3m

uED = -j

D 2m

x m : The unit vectors uEBVectors: and uED must bevectors determined first. Fig. Unit The unit u and u From must bea,determined first. From 2Fig. a, ED

A 2m 2m 2m C

F 5 600 N y

E C

600 N D

600 N

3m E

ce vector F is given bythe force vector F is given by Thus, 600 A - 0.7428i = [- 445.66i - 334.25j + 222.83k] N F 0.5571j = Fu +=0.3714k) 600 A - 0.7428i - 0.5571j + 0.3714k) = [- 445.66i - 334.25j + 222.83k] N EB

Product: The magnitude of the component of F parallel to segment DE of F parallel to segment DE Vector Dot Product: Product: The magnitude of the component ssembly is of the pipe assembly is F # uED = A -445.66i B # A - j B - 334.25j + 222.83k B # A -j B = F # u+ 222.83k = A -445.66i (F )- 334.25j ED paral

= ( - 445.66)(0) + ( -334.25)( + (222.83)(0) = -1) ( - 445.66)(0) + ( -334.25)( -1) + (222.83)(0) = 334.25 = 334 N

= 334.25 = 334 N

Ans.

ent of F perpendicular to segment of the pipe assembly is DE of the pipe assembly is The component of DE F perpendicular to segment

er =

2 2 2 2F 2 - (FED)paral - 334.25 Ans. 2 = 498 N 2 (F = )2600 = 2F - (F =) 4982 N= 26002 - 334.25 ED per

ED paral

Ans.

Ans: (FED) = 334 N (FED) # = 498 N 131

2–111. z

Determine the angle u between the two cables.

F2 40 N

u F1 70 N 2m

3m 2m

SOLUTION Unit Vectors: Here, the coordinates of points A, B and C are A(2, -3, 3) m, B(0, 3, 0) and C( - 2, 3, 4) m respectively. Thus, the unit vectors along AB and AC are uAB =

uAC =

3m x

(0 - 2)i + [3 - ( -3)]j + (0 - 3)k

2 6 3 = - i + j - k 7 7 7 2(0 - 2) + [3 - ( - 3)] + (0 - 3) 2

( - 2 - 2)i + [3 - ( - 3)]j + (4 - 3)k

2( - 2 - 2)2 + [3 - ( - 3)]2 + (4 - 3)2

The Angle U Between AB and AC:

= -

253

i +

253

j +

1 253

2 6 3 4 6 1 uAB # uAC = a - i + j - kb # a i + j + kb 7 7 7 253 253 253 = a=

2 4 6 6 3 1 ba b + a b + a - ba b 7 7 7 253 253 253

7253

Then u = cos - 1 ( uAB # uAC ) = cos-1a

41 7253

b = 36.43° = 36.4°

Ans.

Ans: u = 36.4° 132

*2–112. Determine the magnitude of the projection of the force F1 along cable AC.

z C

F2 40 N

u F1 70 N 2m

3m 2m

3m x

SOLUTION Unit Vectors: Here, the coordinates of points A, B and C are A(2, -3, 3)m, B(0, 3, 0) and C( - 2, 3, 4) m respectively. Thus, the unit vectors along AB and AC are uAB =

uAC =

(0 - 2)i + [3 - ( - 3)]j + (0 - 3)k

2 6 3 = - i + j - k 7 7 7 2(0 - 2) + [3 - ( - 3)] + (0- 3) 2

( - 2 - 2)i + [3 - ( -3)]j + (4 - 3)k 2

2( - 2 - 2) + [3 - ( - 3)] + (4 - 3)

= -

253

Force Vector: For F1,

i +

253

j +

1 253

2 6 3 F1 = F1 uAB = 70 a - i + j - kb = { -20i + 60j - 30k} N 7 7 7

Projected Component of F1: Along AC, it is

(F1)AC = F1 # uAC = ( - 20i + 60j - 30k) # a = ( - 20)a -

253

b + 60a

= 56.32 N = 56.3 N

253

4 253

b + ( - 30)a

i + 1

253

6 253

j +

1 253

Ans.

The positive sign indicates that this component points in the same direction as uAC.

Ans: (F1)AC = 56.3 N 133

2–113. If F {16i 10j 14k} N, determine the magnitude of the projection of F along the axis of the pole and perpendicular to it.

SOLUTION

y 2m

rOA = 2 i + 4 j + a 2(4)2 + (2)2 tan 60° b k uOA = a

60°

4m x

rOA b = 0.2236 i + 0.4472 j + 0.8660 k rOA

Proj F = FuOA = (16)(0.2236) + (10)(0.4472) + ( -14)(0.8660) Ans.

|Proj F| = |- 4.0745 N| = 4.07 N F =

(16)2 + (10)2 + ( - 14)2 = 23.49 N

F =

(23.49)2 - (4.0745)2 = 23.1 N

Ans.

Ans: F = 23.49 N F# = 23.1 N 134

2–114. Determine magnitude projection of F1 along the the Two cablestheexert forces ofonthethe pipe. Determine line of action of F . magnitude of the 2projected component of F along the line

of action of F2.

25 lb

SOLUTION

Force Vector: uF1 = cos 30° sin 30°i + cos 30° cos 30°j - sin 30°k

= 0.4330i + 0.75j - 0.5k

30 lb

F1 = FRuF1 = 30(0.4330i + 0.75j - 0.5k) lb = {12.990i + 22.5j - 15.0k} lb Unit Vector: Vector: One can obtain the angle a = 135° for F2 using Eq. 2–8. cos2 a + cos2 b + cos2 g = 1, with b = 60° and g = 60°. The unit vector along the line of action of F2 is uF2 = cos 135°i + cos 60°j + cos 60°k = - 0.7071i + 0.5j + 0.5k F22: Projected Component Component of F11 Along the Line of Action of F (F1)F2 = F1 # uF2 = (12.990i + 22.5j - 15.0k) # (- 0.7071i + 0.5j + 0.5k) = (12.990)(- 0.7071) + (22.5)(0.5) + (- 15.0)(0.5) = - 5.44 lb Negative sign indicates that the projected component of (F1)F2 acts in the opposite sense of direction to that of uF2. Ans.

The magnitude is (F1)F2 = 5.44 lb

Ans: (F1)F2 = 5.44 lb 135

2–115. Determine the angle u between the two cables attached to the pipe.

25 lb

SOLUTION Unit Vectors:

uF1 = cos 30° sin 30°i + cos 30° cos 30°j - sin 30°k = 0.4330i + 0.75j - 0.5k

uF2 = cos 135°i + cos 60°j + cos 60°k F1

= -0.7071i + 0.5j + 0.5k

30 lb

The Angles Angles Between BetweenTwo Two Vectors VectorsuU: : uF1 # uF2 = (0.4330i + 0.75j - 0.5k) # ( -0.7071i + 0.5j + 0.5k) = 0.4330(- 0.7071) + 0.75(0.5) + ( -0.5)(0.5) = -0.1812 Then, u = cos - 1 A uF1 # uF2 B = cos - 1( -0.1812) = 100°

Ans.

Ans: u = 100° 136

*2–116. Determine the magnitudes of the components of the force acting along the axis AB of the wrench handle and perpendicular to it.

z F 5 80 N z

A force of F = 80 N is applied to the handle of the wrench. Determine the angle u between the tail of the force and the handle AB.

308

300 mm

45 A

SOLUTION

80 N 458

500 mm

300 mm

uF = -cos 30° sin 45° i + cos 30° cos 45° j + sin 30° k x

= -0.6124 i + 0.6124 j + 0.5 k

500 mm

uAB = -j cos u = uF # u AB = ( - 0.6124 i + 0.6124 j + 0.5 k) # ( - j) = - 0.6124 Ans.

u = 128°

Ans: u = 128° 137

2–117. z

Determine the magnitude of the projected component of the 100-lb force acting along the axis BC of the pipe.

A B 3 ft u 8 ft x

6 ft 4 ft

SOLUTION

2 ft C

100 lb

^ 6ft rBC = 56i^ + 4j^ - 2k F = 100

5- 6i^ + 8j^ + 2k^ 6

2( - 6)2 + 82 + 22

^ 6 lb = 5 - 58.83i^ + 78.45j^ + 19.61k

- 78.45 r rBC Fp = F # uBC = F ∙ = = - 10.48 |rBC| 7.483 Fp = 10.5 lb

Ans.

Ans: Fp = 10.5 lb 138

2–118. z

Determine the angle u between pipe segments BA and BC.

A B 3 ft u 8 ft x

SOLUTION

6 ft 4 ft D

2 ft C

100 lb

rBC = {6i + 4j - 2k} ft uBA = { -3i} ft u = cos-1 a u = 143°

rBC # rBA - 18 b = cos-1 a b |rBC| # |rBA| 22.45

Ans.

Ans: u = 143° 139

2–119. Determine the angle u between the cables AB and AC.

z 1m B

2m C

D F

SOLUTION A

Unit Vectors: Here, the coordinates of points A, B and C are A(6, 0, 0) m, x B(0, -1, 2) m and C(0, 1, 3) respectively. Thus, the unit vectors along AB and AC are uAB =

uAC =

(0 - 6)i + ( -1 - 0)j + (2 - 0)k 2

2(0 - 6) + ( - 1 - 0) + (2 - 0) (0 - 6)i + (1 - 0)j + (3 - 0)k 2

2(0 - 6) + (1 - 0) + (3 - 0)

= -

The Angle U Between AB and AC: uAB # uAC = a -

6 241

= a= Then

241

i -

1 241

ba -

241

kb # a -

b + a-

j +

246

241

6 241 6

246 6 246 1

246

i -

i +

b +

1 241 1

246 1 246 2

j +

j + a

2 241 3

246 3 246

241 246

21886

u = cos - 1 ( UAB # UAC ) = cos-1a

41 21886

b = 19.24998° = 19.2°

Ans.

Ans: u = 19.2° 140

*2–120. z

Determine the magnitude of the projected component of the force F = {400i - 200j + 500k} N acting along the cable BA.

1m B

2m C

D F

SOLUTION Unit Vector: Here, the coordinates of points A and B are A(6, 0, 0) m and x B(0, -1, 2) m respectively. Thus the unit vector along BA is uBA =

(6 - 0)i + [0 - ( - 1)]j + (0 - 2)k rBA 6 1 2 = = i + j k rBA 2(6 - 0)2 + [0 - ( - 1)]2 + (0- 2)2 241 241 241

Projected Component of F: Along BA, it is

FBA = F # uBA = (400i - 200j + 500k) # a = 400 a

241

b + ( - 200)a

= 187.41 N = 187 N

6 241 1

241

i +

1 241

j -

b + 500a -

2 241

241

Ans.

The positive sign indicates that this component points in the same direction as uBA.

Ans: FBA = 187 N 141

2–121. Determine the magnitude of the projected component of the force F = {400i - 200j + 500k} N acting along the cable CA.

z 1m B

2m C

D F

SOLUTION Unit Vector: Here, the coordinates of points A and C are A(6, 0, 0) m and C(0, 1, 3) m respectively. Thus, the unit vector along CA is uCA =

(6 - 0)i + (0 - 1)j + (0 - 3)k rCA 6 1 3 = = i j k 2 2 2 rCA 2(6 - 0) + (0 - 1) + (0 - 3) 246 246 246

Projected Component of F: Along CA, it is

FCA = F # uCA = (400i - 200j + 500k) # a = 400a

246

6 246

b + ( - 200)a -

= 162.19 N = 162 N

i -

246

1 246

j -

b + 500 a -

3 246 3

246

Ans.

The positive sign indicates that this component points in the same direction as uCA.

Ans: FCA = 162 N 142

2–122. Determine the magnitudes of the components of F acting along and perpendicular to segment BC of the pipe assembly.

3 ft

4 ft 2 ft

4 ft B

SOLUTION Unit Vector: The unit vector uCB must be determined first. From Fig. a uCB =

{30i

45j

50k} lb 4 ft

(3 - 7)i + (4 - 6)j + [0 - ( -4)]k rCB 2 1 2 = = - i- j + k rCB 3 3 3 2 2 2 3(3 - 7) + (4 - 6) + [0 - ( -4)]

Vector Dot Product: Product: The magnitude of the projected component of F parallel to segment BC of the pipe assembly is (FBC)pa = F # uCB = (30i - 45j + 50k) # ¢-

1 2 2 i - j + k≤ 3 3 3

2 1 2 = (30) ¢ - ≤ + ( -45) ¢ - ≤ + 50 ¢ ≤ 3 3 3 Ans.

= 28.33 lb = 28.3 lb

The magnitude of F is F = 330 2 + ( - 45) 2 + 50 2 = 25425 lb. Thus, the magnitude of the component of F perpendicular to segment BC of the pipe assembly can be determined from (FBC)pr = 3F2 - (FBC)pa2 = 25425 - 28.332 = 68.0 lb

Ans.

Ans: ( FBC ) = 28.3 lb ( FBC ) # = 68.0 lb 143

2–123. z

Determine the magnitude of the projected component of F lineExpress AC. Express this component as a Cartesian along AC. this component as a Cartesian vector. vector.

3 ft

4 ft 2 ft

4 ft B

SOLUTION Unit Vector: The unit vector uAC must be determined first. From Fig. a uAC =

(7 - 0)i + (6 - 0)j + ( - 4 - 0)k 3(7 - 0)2 + (6 - 0)2 + ( - 4 - 0)2

{30i

45j

50k} lb 4 ft

= 0.6965 i + 0.5970 j - 0.3980 k

Vector Dot Product: The magnitude of the projected component of F along line AC is FAC = F # uAC = (30i - 45j + 50k) # (0.6965i + 0.5970j - 0.3980k) = (30)(0.6965) + (- 45)(0.5970) + 50( -0.3980) Ans.

= 25.87 lb = 25.9 lb Thus, FAC expressed in Cartesian vector form is FAC = FAC uAC = - 25.87(0.6965i + 0.5970j - 0.3980k)

Ans.

= { - 18.0i - 15.4j + 10.3k} lb

Ans: FAC = 25.9 lb FAC = { - 18.0i - 15.4j + 10.3k} lb 144

*2–124. Determine the angle u between the pipe segments BA and BC.

3 ft

4 ft 2 ft

4 ft B

SOLUTION

Position Position Vectors: Vectors: The position vectors rBA and rBC must be determined first. From Fig. a,

{30i

45j

50k} lb 4 ft C

rBA = (0 - 3)i + (0 - 4)j + (0 - 0)k = { - 3i - 4j} ft rBC = (7 - 3)i + (6 - 4)j + (- 4 - 0)k = {4i + 2j - 4k} ft The magnitude of rBA and rBC are rBA = 3(- 3)2 + (- 4)2 = 5 ft

rBC = 342 + 22 + (- 4)2 = 6 ft

Product: Vector Dot Product: rBA # rBC = (- 3i - 4j) # (4i + 2j - 4k) = (- 3)(4) + ( -4)(2) + 0(- 4) = - 20 ft2 Thus, u = cos-1 a

rBA # rBC -20 b = cos-1 c d = 132° rBA rBC 5(6)

Ans.

Ans: u = 132° 145

2–125. Determine the magnitudes of the projection of the force F = 5 - 80i + 30j + 20k6 lb in the direction of the cables AB and AC.

z F A

3 ft

8 ft

C 4 ft y

SOLUTION uAC =

uBA =

( - 4 - 0)i + (3 - 0)j + (0 - 8)k

2( - 4 - 0)2 + (3 - 0)2 + (0 - 8)2

(0 - 5)i + [0 - ( - 4)]j + (8 - 0)k 2

2(0 - 5) + [0 - ( - 4)] + (8 - 0)

FAC = F # uAC = ( -80i - 30j + 20k) # a =

289

- 5i + 4j + 8k 2105

( -80)( - 5) + 30(4) + 20(8)

= 66.4 lb

2105

2105 b

289

FBA = F # uBA = ( -80i + 30j + 20k) # a

4 ft

- 5i + 4j + 8k

- 4i + 3j - 8k

5 ft

289

( -80)( - 4) + 30(3) + 20( - 8)

= 26.5 lb

- 4i + 3j - 8k

Ans.

Ans: FAC = 26.5 lb FBA = 66.4 lb 146

2–126. Determine components of the force Determine the the magnitudes magnitudesofofthethe components of force acting parallel and perpendicular to diagonal AB of the F = 90 lb acting parallel and perpendicular to diagonal AB crate. of the crate.

z F

90 lb 60

1 ft

SOLUTION

Force and and Unit Unit Vector: Vector: The force vector F and unit vector uAB must be determined x first. From Fig. a

3 ft

1.5 ft

F = 90(- cos 60° sin 45°i + cos 60° cos 45°j + sin 60°k) = {- 31.82i + 31.82j + 77.94k} lb uAB =

(0 - 1.5)i + (3 - 0)j + (1 - 0)k rAB 6 2 3 = = - i- j + k rAB 7 7 7 3(0 - 1.5)2 + (3 - 0)2 + (1 - 0)2

Product: The magnitude of the projected component of F parallel to the Vector Dot Product: diagonal AB is [(F)AB]pa = F # uAB = (- 31.82i + 31.82j + 77.94k) # ¢-

6 2 3 i + j + k≤ 7 7 7

3 6 2 = (- 31.82) ¢ - ≤ + 31.82 ¢ ≤ + 77.94 ¢ ≤ 7 7 7 Ans.

= 63.18 lb = 63.2 lb The magnitude of the component F perpendicular to the diagonal AB is [(F)AB]pr = 3F2 - [(F)AB]pa2 = 2902 - 63.182 = 64.1 lb

147

Ans.

Ans: 3(F )AB 4 = 63.2 lb 3(F )AB 4 # = 64.1 lb

2–127. Determine the magnitudes magnitudes of ofthe theprojected projections of the force components of acting along x and y axes. the force F =the300 N acting along the x and y axes.

300 N

30 300 mm

SOLUTION

Force Vector: The force vector F must be determined first. From Fig. a, F = -300 sin 30°sin 30°i + 300 cos 30°j + 300 sin 30°cos 30°k

300 mm

300 mm y

= [-75i + 259.81j + 129.90k] N Dot Product: Product: The magnitudes of the projected component of F along the x Vector Dot and y axes are Fx = F # i = A -75i + 259.81j + 129.90k B # i = - 75(1) + 259.81(0) + 129.90(0) = - 75 N Fy = F # j = A - 75i + 259.81j + 129.90k B # j = - 75(0) + 259.81(1) + 129.90(0) = 260 N The negative sign indicates that Fx is directed towards the negative x axis. Thus Fx = 75 N,

Ans.

Fy = 260 N

Ans: Fx = 75 N Fy = 260 N 148

*2–128. Determine the magnitude of the projection of the force acting along line OA.

308

F 5 300 N

308 300 mm

Determine the magnitude of the projected component of the force F = 300 N acting along line OA.

O x

z 300 mm

30030 mm

300 N

30 300 mm

SOLUTION

300 mm

300 mm y

Force and Unit Vector: The force vector F and unit vector uOA must be determined first. From Fig. a F = (- 300 sin 30° sin 30°i + 300 cos 30°j + 300 sin 30° cos 30°k) = { - 75i + 259.81j + 129.90k} N uOA =

(-0.45 - 0)i + (0.3 - 0)j + (0.2598 - 0)k rOA = = -0.75i + 0.5j + 0.4330k rOA 2( - 0.45 - 0)2 + (0.3 - 0)2 + (0.2598 - 0)2

Vector Dot Product: Product: The magnitude of the projected component of F along line OA is FOA = F # uOA = A - 75i + 259.81j + 129.90k B # A -0.75i + 0.5j + 0.4330k B = ( - 75)(-0.75) + 259.81(0.5) + 129.90(0.4330)

Ans.

= 242 N

Ans: FOA = 242 N 149

2–129. z

Determine the angle u between the two cables. 8 ft C

10 ft

B 4 ft

10 ft

SOLUTION u = cos-1 a = cos-1 c

= cos-1 a

u = 82.9°

rAC # rAB b rAC rAB

(2 i - 8 j + 10 k) # ( -6 i + 2 j + 4 k)

122 + ( - 8)2 + 102 1( - 6)2 + 22 + 42

12 b 96.99

FAB 12 lb 6 ft

8 ft

Ans.

Ans: u = 82.9° 150

2–130. Determine the projection of the force acting in the direction of cable AC. Express the result as a Cartesian vector.

z 8 ft C 10 ft

B 4 ft

10 ft FAB 12 lb 6 ft

SOLUTION rAC = {2 i - 8 j + 10 k} ft

8 ft

rAB = { - 6 i + 2 j + 4 k} ft FAB = 12 a

rAB 6 2 4 b = 12 a i + j + kb rAB 7.483 7.483 7.483

FAB = { - 9.621 i + 3.207 j + 6.414 k} lb uAC =

2 8 10 i j + k 12.961 12.961 12.961

Proj FAB = FAB # uAC = - 9.621 a = 1.4846

2 8 10 b + 3.207 a b + 6.414 a b 12.961 12.961 12.961

Proj FAB = FAB uAC Proj FAB = (1.4846) c

2 8 10 i j + kd 12.962 12.962 12.962

Proj FAB = {0.229 i - 0.916 j + 1.15 k} lb

Ans.

Ans: Proj FAB = {0.229 i - 0.916 j + 1.15 k} lb 151

3–1. The members of a truss are pin connected at joint O. Determine the magnitudes of F1 and F2 for equilibrium. Set u = 60°.

5 kN

70 30

x O

SOLUTION + ©F = 0; : x

5 4

4 F2 sin 70° + F1 cos 60° - 5 cos 30° - (7) = 0 5

7 kN

0.9397F2 + 0.5F1 = 9.930 + c ©Fy = 0;

F2 cos 70° + 5 sin 30° - F1 sin 60° -

3 (7) = 0 5

0.3420F2 - 0.8660F1 = 1.7 Solving: F2 = 9.60 kN

Ans.

F1 = 1.83 kN

Ans.

Ans: F2 = 9.60 kN F1 = 1.83 kN 152

3–2. The members of a truss are pin connected at joint O. Determine the magnitude of F1 and its angle u for equilibrium. Set F2 = 6 kN.

5 kN

70 30

x O

SOLUTION + ©F = 0; : x

5 4

4 6 sin 70° + F1 cos u - 5 cos 30° - (7) = 0 5

7 kN

F1 cos u = 4.2920 + c ©Fy = 0;

6 cos 70° + 5 sin 30° - F1 sin u -

3 (7) = 0 5

F1 sin u = 0.3521 Solving: u = 4.69°

Ans.

F1 = 4.31 kN

Ans.

Ans: u = 4.69° F1 = 4.31 kN 153

3–3. Knowing the forces in members A and C, determine the forces FB and FD acting on members B and D that are required for equilibrium. The force system is concurrent at point O.

4 kN FD A

60˚ 45˚ 2 kN C

FB B

SOLUTION +cΣFy = 0; - FD sin 45° + 4 sin 60° = 0 FD = 4.90 kN

Ans.

+ ΣFx = 0; 4.90 cos 45° + 4 cos 60° - 2 - FB = 0 FB = 3.46 kN Ans. S

Ans: FD = 4.90 kN FB = 3.46 kN 154

*3–4. Determine the maximum weight W that can be supported in the position shown if each chain AC and AB can support a maximum tension of 600 lb before it fails.

308

SOLUTION If AC = 600 lb

+ ΣFx = 0; - FAB a 5 b + 600 sin 30° = 0 S 13

FAB = 780 lb 7 600 lb (N. G!)

If AB = 600 lb + ΣFx = 0; - 600 a 5 b + FAC sin 30° = 0 S 13

+c ΣFy = 0;

FAC = 461.54 lb

12 (600) + 461.54 cos 30° - W = 0 13 Ans.

W = 953.55 = 954 lb

Ans: W = 954 lb 155

3–5. If the forces of area truss concurrent at point The members are connected to O, the determine gusset plate.the If magnitudes T for equilibrium.Take u = 90°. the the forces of areF and concurrent at point O, determine magnitudes of F and T for equilibrium. Take u = 90°.

9 kN F A 5 3 B 4

SOLUTION 3 f = 90° - tan - 1 a b = 53.13° 4

+ ©F = 0; : x

4 T cos 53.13° - Fa b = 0 5

+ c ©Fy = 0;

3 9 - T sin 53.13° - Fa b = 0 5

C T

Solving, T = 7.20 kN

Ans.

F = 5.40 kN

Ans.

Ans: T = 7.20 kN F = 5.40 kN 156

3–6. The gusset plate is subjected the forcesCof three members. Determine the tension force to in member and its angle u for Determine tension force member C at andpoint its angle u for equilibrium.the The forces areinconcurrent O. Take equilibrium. F = 8 kN. The forces are concurrent at point O. Take F = 8 kN.

9 kN F A 5 3 B 4

SOLUTION + ©F = 0; : x

4 T cos f - 8a b = 0 5

(1)

+ c ©Fy = 0;

3 9 - 8 a b - T sin f = 0 5

(2)

Rearrange then divide Eq. (1) into Eq. (2):

C T

tan f = 0.656, f = 33.27° T = 7.66 kN

Ans.

3 u = f + tan - 1 a b = 70.1° 4

Ans.

Ans: T = 7.66 kN u = 70.1° 157

3–7. C

The man attempts to pull down the tree using the cable and small pulley arrangement shown. If the tension in AB is 60 lb, determine the tension in cable CAD and the angle u which the cable makes at the pulley.

θ 20°

A B 30°

SOLUTION + R©Fx¿ = 0;

60 cos 10° - T - T cos u = 0

+Q©Fy¿ = 0;

T sin u - 60 sin 10° = 0

Thus, T(1 + cos u) = 60 cos 10° u T(2cos2 ) = 60 cos 10° 2 2T sin

(1)

u u cos = 60 sin 10° 2 2

(2)

Divide Eq.(2) by Eq.(1) tan

u = tan 10° 2 Ans.

u = 20° T

Ans.

30.5 lb

Ans: u = 20° T = 30.5 lb 158

*3–8. Determine the force in cables AB and AC necessary to support the 12-kg traffic light.

C 7

25 24

12°

SOLUTION Equilibrium : Equations of Equilibrium: + ©F = 0; : x

FAB cos 12° - FAC a

24 b = 0 25 [1]

FAB = 0.9814FAC + c ©Fy = 0;

FAB sin 12° + FAC a

7 b - 117.72 = 0 25 [2]

0.2079FAB + 0.28FAC = 117.72 Solving Eqs.[1] and [2] yields FAB = 239 N

Ans.

FAC = 243 N

Ans: FAB = 239 N FAC = 243 N 159

3–9. The 30-kg block is supported by two springs having the stiffness shown. Determine the unstretched length of each spring after the block is removed.

0.6 m

0.4 m

B 0.5 m

kAC 1.5 kN/m

kAB 1.2 kN/m A

SOLUTION + ©F = 0; : x + c ©Fy = 0;

0.4 2(0.4)2 + (0.5)2 0.5 2(0.4)2 + (0.5)2

FAB FAB +

0.6 2(0.6)2 + (0.5)2 0.5 2(0.6)2 + (0.5)2

FAC = 0 FAC - 30(9.81) = 0

Solving, FAC = 183.9 N FAB = 226.1 N Then FAB = kABxAB xAB =

226.1 = 0.188 m 1200

FAC = kACxAC xAC =

183.9 = 0.123 m 1500

lAB = 2(0.4)2 + (0.5)2 = 0.640 m lAC = 2(0.6)2 + (0.5)2 = 0.781 m œ lAB = lAB - xAB = 0.640 - 0.188 = 0.452 m

Ans.

œ lAC = lAC - xAC = 0.781 - 0.123 = 0.658 m

Ans.

Ans: l′AB = 0.452 m l′AC = 0.658 m 160

3–10. If the spring DB has an unstretched length of 2 m, determine the stiffness of the spring to hold the 40-kg crate in the position shown.

SOLUTION

Equations of Equilibrium: Referring to the FBD shown in Fig. a, + ΣFx = 0; TBD a 3 b - TCDa 1 b = 0 S 113 12

+cΣFy = 0; TBD a

Solving Eqs (1) and (2)

2 1 b + TCDa b - 40(9.81) = 0 113 12

(2)

TBD = 282.96 N TCD = 332.96 N The stretched length of the spring is l = 232 + 22 = 213 m

Then, x = l - l0 = ( 113 - 2 ) m. Thus, Fsp = kx;

282.96 = k ( 113 - 2 )

Ans.

k = 176.24 N>m = 176 N>m

Ans: k = 176 N>m 161

3–11. Determine the unstretched length of DB to hold the 40-kg crate in the position shown. Take k = 180 N>m.

SOLUTION

Equations of Equilibrium: Referring to the FBD shown in Fig. a, + ΣFx = 0; TBD a 3 b - TCDa 1 b = 0 S 113 12

+cΣFy = 0; TBD a

Solving Eqs (1) and (2)

(1)

2 1 b + TCDa b - 40(9.81) = 0 113 12

(2)

TBD = 282.96 N TCD = 332.96 N The stretched length of the spring is l = 232 + 22 = 213 m

Then, x = l - l0 = 113 - l0. Thus Fsp = kx;

282.96 = 180 ( 113 - l0 )

Ans.

l0 = 2.034 m = 2.03 m

Ans: l0 = 2.03 m 162

*3–12. Determine the force in each cord for equilibrium of the 60-kg bucket.

4m C

SOLUTION

Equilibrium of bucket F:

+cΣFy = 0; FAF - 60(9.81) = 0 FAF = 589 N

Ans.

Equilibrium of joint A: F

4 F - 588.6 = 0 FAC = 736 N 5 AC

Ans.

+ ΣFx = 0; FAB - 3 (735.75) = 0 FAB = 441 N S 5

Ans.

+cΣFy = 0;

Equilibrium of joint C: 4 (735.75) = 0 FCE = 589 N 5

Ans.

+ ΣFx = 0; - FCD + 3 (735.75) = 0 FCD = 441 N S 5

Ans.

+cΣFy = 0; FCE -

Ans: FAF = 589 N FAC = 736 N FAB = 441 N FCE = 589 N FCD = 441 N 163

3–13. The lift sling is used to hoist a container having a mass of 500 kg. Determine the force in each of the cables AB and AC as a function of u. If the maximum tension allowed in each cable is 5 kN, determine the shortest lengths of cables AB and AC that can be used for the lift. The center of gravity of the container is located at G.

F A

SOLUTION

Free-Body Diagram: Free-Body Diagram: By observation, the force F1 has to support the entire weight of the container. Thus, F1 = 50019.812 = 4905 N.

1.5 m

Equations of Equilibrium: + ©F = 0; : x

FAC cos u - FAB cos u = 0

+ c ©Fy = 0;

4905 - 2F sin u = 0

FAC = FAB = F G

(2452.5 csc F = 52452.5 cos u) u6 N

Thus, FAC = FAB = F = 52.45 csc cos u6 kN

Ans.

If the maximum allowable tension in the cable is 5 kN, then 2452.5 csc cos u = 5000 u = 29.37° From the geometry, l =

1.5 and u = 29.37°. Therefore cos u l =

1.5 = 1.72 m cos 29.37°

Ans.

Ans: FAC = {2.45 cos u} kN l = 1.72 m 164

3–14. Determine the stretch in each spring for equlibrium of the 2-kg block. The springs are shown in the equilibrium position.

B kAC 20 N/m kAB 30 N/m

SOLUTION FAD = 2(9.81) = xAD(40) + ©F = 0; : x + c ©Fy = 0;

xAD = 0.4905 m

Ans.

4 1 FAB a b - FAC a b = 0 5 22 FAC a

kAD 40 N/m

3 b + FAB a b - 2(9.81) = 0 5 22

FAC = 15.86 N xAC =

15.86 = 0.793 m 20

Ans.

FAB = 14.01 N xAB =

14.01 = 0.467 m 30

Ans.

Ans: xAD = 0.4905 m xAC = 0.793 m xAB = 0.467 m 165

3–15. The unstretched length of spring AB is 3 m. If the block is held in the equilibrium position shown, determine the mass of the block at D.

B 20 N/m

kAC

kAB

SOLUTION

F = kx = 30(5 - 3) = 60 N + ©F = 0; : x

4 Tcos 45° - 60 a b = 0 5 T = 67.88 N

+ c ©Fy = 0;

30 N/m

3 -W + 67.88 sin 45° + 60 a b = 0 5 W = 84 N m =

84 = 8.56 kg 9.81

Ans.

Ans: m = 8.56 kg 166

*3–16. Determine the stretch of each spring for equilibrium of the 20-kg cylinder.

4m A

kAB = 300 N/m kAC = 200 N/m 3m

kAD = 400 N/m

SOLUTION D

Equilibrium:

C 3m

Spring AD +cΣFy = 0; FAD - 20(9.81) = 0 FAD = 196.2 N Spring AB and AC +cΣFy = 0;

4 F - FAC sin 45° = 0 5 AB

(1)

+ ΣFx = 0; 3 FAB - FAC cos 45° - 196.2 = 0 S 5

(2)

Solving Eq. (1) and (2) yields: FAC = 158.55 N FAB = 140.14 N Spring Elongation: x =

F k

xAB =

140.14 = 0.467 m 300

Ans.

xAC =

158.55 = 0.793 m 200

Ans.

xAD =

196.2 = 0.490 m 400

Ans.

Ans: xAB = 0.467 m xAC = 0.793 m xAD = 0.490 m 167

3–17. ■The spring spring has has aa stiffness stiffness of of k = 800 N>m and an The an unstretched length of 200 mm. Determine the force in cables BC and BD when the spring is held in the position shown.

400 mm A

k = 800 N m

300 mm

D 500 mm

400 mm

SOLUTION The Force in The the Spring: The spring stretches s = l - l0 = 0.5 - 0.2 = 0.3 m. Applying Eq.3–2, we have Fsp = ks = 800(0.3) = 240 N Equations of Equilibrium: + ΣFx = 0; S

4 FBC cos 45° + FBD a b - 240 = 0 5

(1)

0.7071FBC + 0.8FBD = 240

+ c ΣFy = 0;

3 FBC sin 45° - FBD a b = 0 5

(2)

FBC = 0.8485FBD

Solving Eqs. (1) and (2) yields, FBD = 171 N

Ans.

FBC = 145 N

Ans: FBD = 171 N FBC = 145 N

168

3–18. The lamp has a weight of 15 lb and is supported by the six cords connected together as shown. Determine the tension in each cord and the angle u for equilibrium. Cord BC is horizontal.

E D u

30 B

45 A

SOLUTION Equations of Equilibrium: Considering the equilibrium of Joint A by referring to its FBD shown in Fig. a, + ΣFx = 0; TAC cos 45° - TAB cos 60° = 0 S

(1)

+cΣFy = 0; TAC sin 45° + TAB sin 60° - 15 = 0

(2)

Solving Eqs (1) and (2) yield Ans.

TAB = 10.98 = 11.0 lb TAC = 7.764 lb = 7.76 lb Then, joint B by referring to its FBD shown in Fig. b +cΣFy = 0;

TBE sin 30° - 10.98 sin 60° = 0

+ ΣFx = 0; S

TBC + 10.98 cos 60° - 19.02 cos 30° = 0

TBE = 19.02 lb = 19.0 lb

TBC = 10.98 lb = 11.0 lb Finally joint C by referring to its FBD shown in Fig. c

Ans. Ans.

+ ΣFx = 0; TCD cos u - 10.98 - 7.764 cos 45° = 0 S (3)

TCD cos u = 16.4711 +cΣFy = 0; TCD sin u - 7.764 sin 45° = 0

(4)

TCD sin u = 5.4904 Divided Eq (4) by (3)

Ans.

tan u = 0.3333 u = 18.43° = 18.4° Substitute this result into Eq (3) TCD cos 18.43° = 16.4711 TCD = 17.36 lb = 17.4 lb

Ans.

Ans: TAB = 11.0 lb TAC = 7.76 lb TBC = 11.0 lb TBE = 19.0 lb TCD = 17.4 lb u = 18.4° 169

3–19. Each cord can sustain a maximum tension of 20 lb. Determine the largest weight of the lamp that can be supported. Also, determine u of cord DC for equilibrium.

E D u

30 B

45 A

SOLUTION Equations of Equilibrium: Considering the equilibrium of Joint A by referring to its FBD shown in Fig. a, + ΣFx = 0; TAC cos 45° - TAB cos 60° = 0 S

(1)

+cΣFy = 0; TAC sin 45° - TAB sin 60° - W = 0

(2)

Solving Eqs (1) and (2) yield TAB = 0.7321 W TAC = 0.5176 W Then, joint B by referring to its FBD shown in Fig. b, +cΣFy = 0; TBE sin 30° - 0.7321W sin 60° = 0 + ΣFx = 0; S

TBE = 1.2679 W

TBC + 0.7321 W cos 60° - 1.2679 W cos 30° = 0

Finally, joint C by referring to its FBD shown in Fig. c,

TBC = 0.7321 W

Ans.

+ ΣFx = 0; TCD cos u - 0.7321 W - 0.5176 W cos 45° = 0 S (3)

TCD cos u = 1.0981 W +cΣFy = 0; TCD sin u - 0.5176 W sin 45° = 0

(4)

TCD sin u = 0.3660 W Divided Eq (4) by (3)

Ans.

tan u = 0.3333 u = 18.43° = 18.4° Substitute this result into Eq (3), TCD cos 18.43° = 1.0981 W TCD = 1.1575 W

Here cord BE is subjected to the largest tension. Therefore, its tension will reach the limit first, that is TBE = 20 lb. Then 20 = 1.2679 W;

W = 15.77 lb = 15.8 lb

Ans.

Ans: u = 18.4° W = 15.8 lb 170

*3–20. The 10-lb weight is supported by the cord AC and roller and by the spring that has a stiffness of k = 10 lb>in. and an unstretched length of 12 in. Determine the distance d to where the weight is located when it is in equilibrium.

12 in. θ

SOLUTION

+ ΣFx = 0; S

-TAC + F1 cos u = 0

+ c ΣFy = 0;

Fx sin u - 10 = 0

Fx = kx;

Fx = 10 a

12 - 12b cos u

= 120 (sec u - 1) Thus, 120 (sec u - 1) sin u = 10 (tan u - sin u) =

1 12

Solving, u = 30.71° Ans.

d = 12 tan 30.71° = 7. 13 in.

Ans: d = 7.13 in. 171

3–21. The 10-lb weight is supported by the cord AC and roller and by a spring. If the spring has an unstretched length of 8 in., and the weight is in equilibrium when d = 4 in., determine the stiffness k of the spring.

12 in. θ

SOLUTION

+ c ΣFy = 0;

Fx sin u - 10 = 0

Fx = kx;

Fx = k a

tan u =

4 ; 12

12 - 8b cos u

u = 18.435°

Thus, ka

12 - 8b sin 18.435° = 10 cos 18.435° Ans.

k = 6.80 lb>in.

Ans: k = 6.80 lb/in. 172

3–22. The springs BA and BC each have a stiffness of 500 N>m and an unstretched length of 3 m. Determine the horizontal force F applied to the cord which is attached to the small ring B so that the displacement of the ring from the wall is d = 1.5 m.

A k

500 N/m B

SOLUTION + ©F = 0; : x

1.5 211.25

500 N/m

(T)(2) - F = 0

C d

T = ks = 500(232 + (1.5)2 - 3) = 177.05 N Ans.

F = 158 N

Ans: F = 158 N 173

3–23. The springs BA and BC each have a stiffness of 500 N>m and an unstretched length of 3 m. Determine the displacement d of the cord from the wall when a force F = 175 N is applied to the cord.

A k

500 N/m B

SOLUTION + ©F = 0; : x

500 N/m

175 = 2T sin u

T sin u = 87.5 TC

d 2

23 + d 2

S = 87.5

T = ks = 500( 232 + d 2 - 3) d a1 -

3 29 + d2

b = 0.175

By trial and error: Ans.

d = 1.56 m

Ans: d = 1.56 m 174

*3–24. Determine the distances x and y for equilibrium if F1 = 800 N and F2 = 1000 N.

C y

SOLUTION Equations of Equilibrium: The tension throughout rope ABCD is constant, that is F1 = 800 N. Referring to the FBD shown in Fig. a,

A x

+cΣFy = 0; 800 sin f - 800 sin u = 0 f = 0 + ΣFx = 0; 1000 - 2[800 cos u] = 0 u = 51.32° S Referring to the geometry shown in Fig. b, y = 2 m

Ans.

2 = tan 51.32°; x = 1.601 m = 1.60 m x

Ans.

and

Ans: y = 2m x = 1.60 m 175

3–25. Determine the magnitude of F1 and the distance y if x = 1.5 m and F2 = 1000 N.

C y

SOLUTION Equations of Equilibrium: The tension throughout rope ABCD is constant, that is F1. Referring to the FBD shown in Fig. a, +c ΣFy = 0; F1a

y 2

2y + 1.5 y

b - F1a 2

2y + 1.5

A x

2 b = 0 2.5

2 2.5

Ans.

y = 2m + ΣFx = 0; 1000 - 2c F1a 1.5 b d = 0 S 2.5

Ans.

F1 = 833.33 N = 833 N

Ans: y = 2m F1 = 833 N 176

3–26. The 30-kg pipe is supported at A by a system of five cords. Determine the force in each cord for equilibrium.

C B 60°

SOLUTION At H: + c ΣFy = 0;

THA - 30(9.81) = 0 Ans.

THA = 294 N At A: + c ΣFy = 0;

TAB sin 60° - 30(9.81) = 0 Ans.

TAB = 339.83 = 340 N + ΣFx = 0; S

TAE - 339.83 cos 60° = 0 Ans.

TAE = 170 N At B: + c ΣFy = 0;

3 TBD a b - 339.83 sin 60° = 0 5

Ans.

TBD = 490.50 = 490 N + ΣFx = 0; S

4 490.50 a b + 339.83 cos 60° - TBC = 0 5

Ans.

TBC = 562 N

Ans: THA = 294 N TAB = 340 N TAE = 170 N TBD = 490 N TBC = 562 N 177

3–27. Each cord can sustain a maximum tension of 500 N. Determine the largest mass of pipe that can be supported.

C B 60°

SOLUTION

At H: + c ©Fy = 0;

FHA = W

At A: + c ©Fy = 0;

FAB sin 60° - W = 0 FAB = 1.1547 W

+ ©F = 0; : x

FAE - (1.1547 W) cos 60° = 0 FAE = 0.5774 W

At B: + c ©Fy = 0;

3 FBD a b - (1.1547 cos 30°)W = 0 5 FBD = 1.667 W

+ ©F = 0; : x

4 - FBC + 1.667 W a b + 1.1547 sin 30° = 0 5 FBC = 1.9107 W

By comparison, cord BC carries the largest load. Thus 500 = 1.9107 W W = 261.69 N m =

261.69 = 26.7 kg 9.81

Ans.

Ans: m = 26.7 kg 178

*3–28. If the cords suspend the two buckets in the equilibrium position, determine the weight of bucket B. Bucket A has a weight of 60 lb.

658

C 208 408

SOLUTION

208

At Point F: + ΣFx = 0; TFC cos 20° - TFE cos 40° = 0 S +c ΣFy = 0;

TFC sin 20° + TFE sin 40° - 60 = 0 TFC = 53.07 lb

At Point C: + ΣFx = 0; TCD cos 65° - 53.07 cos 20° = 0 S

TCD = 118.01 lb

+c ΣFy = 0; 118.01 sin 65° - 53.07 sin 20° - WB = 0

Ans.

WB = 88.8 lb

Ans: WB = 88.8 lb 179

3–29. If the bucket weighs 50 lb, determine the tension developed in each of the wires.

C B

A 5

SOLUTION

30 E

Equations of Equilibrium: First, we will apply the equation of equilibrium along the x and y axes to the free-body diagram of joint E shown in Fig. (a). + ©F = 0; : x

3 FED cos 30° - FEB a b = 0 5

(1)

+ c ©Fy = 0;

4 FED sin 30° + FEB a b - 50 = 0 5

(2)

Solving Eqs. (1) and (2), yields FED = 30.2 lb

Ans.

FEB = 43.61 lb = 43.6 lb

Using the result FEB = 43.61 lb and applying the equation of equilibrium to the free-body diagram of joint B shown in Fig. (b), + c ©Fy = 0;

4 FBC sin 30° - 43.61 a b = 0 5 Ans.

FBC = 69.78 lb = 69.8 lb + ©F = 0; : x

3 69.78 cos 30° + 43.61 a b - FBA = 0 5 Ans.

FBA = 86.6 lb

Ans: FED = 30.2 lb FEB = 43.6 lb FBC = 69.8 lb FBA = 86.6 lb 180

3–30. Determine the maximum weight of the bucket that the wire system can support so that no single wire develops a tension exceeding 100 lb.

C B A D

SOLUTION

Equations of Equilibrium: First, we will apply the equations of equilibrium along the x and y axes to the free-body diagram of joint E shown in Fig. (a). + ©F = 0; : x

3 FED cos 30° - FEB a b = 0 5

(1)

+ c ©Fy = 0;

4 FED sin 30° + FEB a b - W = 0 5

(2)

Solving, FEB = 0.8723W

FED = 0.6043W

Using the result FEB = 0.8723W and applying the equations of equilibrium to the free-body diagram of joint B shown in Fig. (b), + c ©Fy = 0;

4 FBC sin 30° - 0.8723W a b = 0 5 FBC = 1.3957W

+ ©F = 0; : x

3 1.3957W cos 30° + 0.8723W a b - FBA = 0 5 FBA = 1.7320W

From these results, notice that wire BA is subjected to the greatest tensile force. Thus, it will achieve the maximum allowable tensile force first. FBA = 100 = 1.7320W Ans.

W = 57.7 lb

Ans: W = 57.7 lb 181

3–31.

Determine the forces in cables AC and AB needed to hold the 20-kg cylinder D in equilibrium. Set F = 300 N and d = 1 m.

1.5 m C d A 2m

SOLUTION f = tan-1 a

2 b = 38.660° 2.5

1 u = tan-1 a b = 26.565° 2 + ©F = 0; : x

- FAB sin 38.660° - FAC cos 26.565° + 300 = 0 0.6247FAB + 0.8944FAC = 300

+ c ©Fy = 0;

FAC sin 26.565° + FAB cos 38.660° - 20(9.81) = 0 0.7809FAB + 0.4472FAC = 196.2

Solving: FAC = 267 N

Ans.

FAB = 98.6 N

Ans.

Ans: FAC = 267 N FAB = 98.6 N 182

*3–32.

The cylinder D has a mass of 20 kg. If a force of F = 100 N is applied horizontally to the ring at A, determine the largest dimension d so that the force in cable AC is zero.

1.5 m C d

A 2m

SOLUTION + ©F = 0; : x

- FAB cos u + 100 = 0

+ c ©Fy = 0;

FAB sin u - 20(9.81) = 0

Solving: u = 62.99°

FAB = 220.2 N

From geometry: tan u =

1.5 + d 2

tan 62.99° =

1.5 + d 2

Ans.

d = 2.42 m

Ans: d = 2.42 m 183

3–33. Blocks D and F weigh 5 lb each and block E weighs 8 lb. Determine the sag s for equilibrium. Neglect the size of the pulleys.

4 ft

s A

SOLUTION + c ©Fy = 0;

2(5) sin u - 8 = 0

u = sin - 1(0.8) = 53.13° tan u =

s 4 Ans.

s = 4 tan 53.13° = 5.33 ft

Ans: s = 5.33 ft 184

3–34. If blocks D and F weigh 5 lb each, determine the weight of block E if the sag s = 3 ft. Neglect the size of the pulleys.

4 ft

s A

SOLUTION + c ©Fy = 0;

3 2(5)a b - W = 0 5

Ans.

W = 6 lb

Ans: W = 6 lb 185

3–35. The ring of negligible size is subjected to a vertical force of 200 lb. Determine the required length l of cord AC such that the tension acting in AC is 160 lb. Also, what is the force in cord AB? Hint: Use the equilibrium condition to determine the required angle u for attachment, then determine l using trigonometry applied to triangle ABC.

40°

θ l

2 ft A

SOLUTION

200 lb

+ ©F = 0; : x

FAB cos 40° - 160 cos u = 0

+ c ©Fy = 0;

160 sin u + FAB sin 40° - 200 = 0

Thus, sin u + 0.8391 cos u = 1.25 Solving by trial and error, u = 33.25° Ans.

FAB = 175 lb l 2 = sin 33.25° sin 40°

Ans.

l = 2.34 ft Also, u = 66.75°

Ans.

FAB = 82.4 lb 2 l = sin 66.75° sin 40° l

Ans.

1.40 ft

Ans: FAB = 175 lb l = 2.34 ft FAB = 82.4 lb l = 1.40 ft 186

*3–36. If the bucket and its contents have a mass of 10 kg and are suspended from the cable by means of a small pulley at C, determine the location x of the pulley for equilibrium. The cable is 6 m long.

4m x A

1m C

SOLUTION Equilibrium: + ΣFx = 0; T cos u - T cos f = 0 cos u = cos f u = f S Geometry: Since u = f, the two triangles are similar. y 2x2 + y2 x = = 4 - x y - 1 2(4 - x)2 + (y - 1)2

(1)

Also,

2x2 + y2 + 2(4 - x)2 + (y - 1)2 = 6 2x2 + y2 + 2(4 - x)2 + (y - 1)2 ¢

From Eq. (1)

≤ = 6

2x2 + y2

2x2 + y2 + ¢

Squaring, we get

2x2 + y2

2(4 - x)2 + (y - 1)2

However, from Eq. (1)

Divide both side by

2x2 + y2

4 - x x

4 - x ≤ 2x2 + y2 = 6 x

4 6 = 2 x 2x + y2

16y2 = 20x2 y = 21.25

(2)

y x = 4 - x y - 1 (3)

4y + x - 2xy = 0 Substituting Eq. (2) into (3) yields: 4¢ 21.25≤ + x - 2x¢ 21.25≤ = 0 xc ¢421.25 + 1≤ - 221.25 d = 0 since x ≠ 0 ¢421.25 + 1≤ - 221.25 = 0

Ans.

x = 2.45 m

187

Ans: x = 2.45 m

3–37. A 4-kg sphere rests on the smooth parabolic surface. Determine the normal force it exerts on the surface and the mass mB of block B needed to hold it in the equilibrium position shown.

SOLUTION Geometry: Geometry: The angle u which the surface make with the horizontal is to be determined first. dy = tan u ` = 5.0x ` = 2.00 ` dx x = 0.4m x = 0.4 m x = 0.4 m

A y 2.5x2 0.4 m x 0.4 m

u = 63.43° Free-Body Free Body Diagram: The tension in the cord is the same throughout the cord and is equal to the weight of block B, WB = mB (9.81). Equations of Equilibrium: + ©F = 0; : x

mB (9.81) cos 60° - Nsin 63.43° = 0 [1]

N = 5.4840mB + c ©Fy = 0;

mB (9.81) sin 60° + Ncos 63.43° - 39.24 = 0 [2]

8.4957mB + 0.4472N = 39.24 Solving Eqs. [1] and [2] yields mB = 3.58 kg

Ans.

N = 19.7 N

Ans: mB = 3.58 kg N = 19.7 N 188

3–38. The block has a weight of 20 lb and is being hoisted at uniform velocity. Determine the angle u for equilibrium and the required force in each cord.

B 30 A

SOLUTION

u T

Point A: + ©F = 0; : x

TAB cos 60° - 20 sin u = 0

TAB cos 60° = 20 sin u + c ©Fy = 0;

TAB sin 60° - 20 - 20 cos u = 0

TAB sin 60° = 20(1 + cos u) tan 60° =

1 + cos u sin u

tan 60° sin u = 1 + cos u Ans.

u = 60° TAB =

20 sin 60° = 34.6 lb cos 60°

Ans.

Also: ©Fy = 0;

Requires

u = 30° 2 Ans.

u = 60° Q + ©Fx = 0;

TAB - 2[20 cos 30°] = 0 Ans.

TAB = 34.6 lb

Ans: u = 60° TAB = 34.6 lb u = 60° TAB = 34.6 lb 189

3–39. Determine the maximum weight W of the block that can be suspended in the position shown if each cord can support a maximum tension of 80 lb. Also, what is the angle u for equilibrium?

B 30 A

SOLUTION

u T

1) Assume TAB = 80 lb + c ©Fy = 0;

80 sin 60° - W - W sin u = 0 (1)

80 sin 60° = W(1 + cos u) + ©F = 0; : x

80 cos 60° - W sin u = 0 (2)

80 cos 60° = W sin u tan 60° =

1 + cos u sin u

tan 60° sin u = 1 + cos u Ans.

u = 60° W =

80 cos 60° = 46.188 lb 6 80 lb sin 60°

(O.K!)

2) Assume W = 80 lb + c ©Fy = 0;

Tsin 60° - 80 - 80 cos u = 0 (3)

T sin 60° = 80(1 + cos u) + ©F = 0; : x

Tcos 60° - 80 sin u = 0 (4)

T cos 60° = 80 sin u tan 60° =

1 + cos u sin u

tan 60° sin u = 1 + cos u u = 60° T =

80 sin 60° = 138.6 lb 7 80 lb cos 60°

(N.G!) Ans.

Thus, W = 46.2 lb

Ans: u = 60° W = 46.2 lb 190

*3–40. F

The 200-lb uniform tank is suspended by means of a 6-ftlong cable, which is attached to the sides of the tank and passes over the small pulley located at O. If the cable can be attached at either points A and B, or C and D, determine which attachment produces the least amount of tension in the cable. What is this tension?

O B

1 ft

C D A 2 ft

2 ft

SOLUTION Free-Body Diagram: By observation, the force F has to support the entire weight of the tank. Thus, F = 200 lb. The tension in cable AOB or COD is the same throughout the cable. Equations of Equilibrium: + ©F = 0; : x

T cos u - T cos u = 0

+ c ©Fy = 0;

200 - 2T sin u = 0

( Satisfied!) T =

100 sin u

(1)

From the function obtained above, one realizes that in order to produce the least amount of tension in the cable, sin u hence u must be as great as possible. Since the attachment of the cable to point C and D produces a greater u A u = cos - 113 = 70.53° B

as compared to the attachment of the cable to points A and B A u = cos - 1 23 = 48.19° B , the attachment of the cable to point C and D will produce the least amount of tension in the cable.

Ans.

Thus, T =

100 = 106 lb sin 70.53°

Ans.

Ans: T = 106 lb 191

3–41. If the mass of the block at A is 20 kg, determine the mass of the block at B and at C for equilibrium. Also, determine the angles u and f.

F 3m

SOLUTION

Equations of Equilibrium: Since the pulleys are smooth, the tension throughout the cord that wrap around the pulley remain constant. Referring to the FBD of pulley D, Fig. a,

C A

+ ΣFx = 0; 20(9.81) cos 45° - TD sin u = 0 S (1)

TD sin u = 138.73 +cΣFy = 0; TD cos u - 20(9.81) - 20(9.81)sin 45° = 0

(2)

TD cos u = 334.93 Divide Eq. 1 by Eq. 2,

TD sin u 138.73 = TD cos u 334.93

tan u = 0.4142 Ans.

u = 22.5° Referring to the FBD of ring E, Fig. b, + ΣFx = 0; mC (9.81)a 4 b - 20(9.81) cos 45° = 0 S 5

Ans.

mC = 17.68 kg = 17.7 kg

3 +cΣFy = 0; 20(9.81) sin 45° + 17.68(9.81)a b - mB(9.81) = 0 5

mB = 24.74 kg = 24.7 kg

Ans.

Referring to the FBD of pulley F, Fig. c, + ΣFx = 0; TF sinf - 17.68(9.81)a 4 b = 0 S 5

(3)

TF sin f = 138.73

3 +cΣFy = 0; TF cos f - 17.68(9.81) - 17.68(9.81)a b = 0 5 TF cos f = 277.47

(4)

Divide Eq. 3 by Eq. 4,

TF sin f 138.73 = TF cos f 277.47

tan f = 0.5 Ans.

f = 26.57° = 26.6°

Ans: u = 22.5° mC = 17.7 kg mB = 24.7 kg f = 26.6° 192

3–42. A “scale” is constructed with a 4-ft-long cord and the 10-lb block D. The cord is fixed to a pin at A and passes over two small pulleys at B and C. Determine the weight of the suspended block at B if the system is in equilibrium when s = 1.5 ft.

1 ft

1.5 ft

SOLUTION Free-Body Diagram: The tension force in the cord is the same throughout the cord, that is, 10 lb. From the geometry, u = sin-1 a

D B

0.5 b = 23.58° 1.25

Equations of Equilibrium: + ©F = 0; : x

10 sin 23.58° - 10 sin 23.58° = 0

+ c ©Fy = 0;

2(10) cos 23.58° - WB = 0

(Satisfied!)

Ans.

WB = 18.3 lb

Ans: WB = 18.3 lb 193

3–43. Determine the magnitude of the force P and the orientation of the 200-lb force required to keep the particle in equilibrium.

z ( 1 ft, 7 ft, 4 ft) F1 = 360 lb P F2 = 120 lb

20˚

F4 = 300 lb F3 = 200 lb

SOLUTION

Force Vector: F1 = 360¢

-i - 7j + 4k

≤ = {-44.313i - 310.191j + 177.252k} lb 2( - 1)2 + (- 7)2 + 42

F2 = 5 - 120j 6 lb

F3 = 200(cos ai + cos bj + cos gk) lb F4 = 5 - 300k6 lb

P = Pcos 20°j + P sin 20°k = 0.93969Pj + 0.34202Pk Equilibrium: ΣF = 0; F1 + F2 + F3 + F4 + P = 0 {- 44.313i - 310.191j + 177.252k} + { - 120j} + 200(cos ai + cos bj + cos gk) + { - 300k} + (0.93969Pj + 0.34202Pk) = 0 1 -44.313 + 200 cos a 2i + 1 - 310.191 - 120 + 200 cos b + 0.93969P 2j + (177.252 + 200 cos g - 300 + 0.34202P)k = 0

Equating the respective i, j, k components to zero yields: ΣFx = 0; - 44.313 + 200 cos a = 0

(1)

ΣFy = 0; - 310.191 - 120 + 200 cos b + 0.93969P = 0

(2)

ΣFz = 0; 177.252 + 200 cos g - 300 + 0.34202P = 0

(3)

Also, cos2a + cos2b + cos2g = 1

(4)

Solving Eqs. (1) to (4) yields: Ans.

a = 77.2° b = 148° g = 119° P = 639 lb

Ans: a = 77.2° b = 148° g = 119° P = 639 lb 194

*3–44. z

Determine the magnitudes of F1, F2, and F3 for equilibrium of the particle. 150 lb

225 lb

135° 45°

SOLUTION ©Fx = 0;

F1 cos 60° + 150 cos 45° cos 30° - 225 = 0 Ans.

-F3 - 150 cos 45° sin 30° + 266.29 cos 60° = 0 Ans.

F3 = 80.1 lb ©Fz = 0;

60°

60° F1

F1 = 266.2 = 266 lb ©Fy = 0;

30°

F2 + 150 sin 45° + 226.29 cos 135° = 0 Ans.

F2 = 82.2 lb

Ans: F1 = 266 lb F3 = 80.1 lb F2 = 82.2 lb 195

3–45. z

If a vertical force of 2.5 kN is applied to the hook at A, determine the tension in each of the three cables for equilibrium. Set d = 1 m.

F = 2.5 kN

A 3m

D 3m

C 1.5 m

SOLUTION

0.75 m y

Force Vector: FAD = FAD ¢

FAC = FAC ¢

FAB = FAB ¢

- 0.75i + 1j - 3k 2( - 0.75)2 + 12 + ( - 3)2 1i + 1.5j - 3k

212 + 1.52 + ( - 3)2

≤ = - 0.2308FADi + 0.3077FADj - 0.9231FADk

≤ = 0.2857FACi + 0.4286FAC j - 0.8571FACk

1i - 3j - 3k

21 + ( - 3)2 + ( - 3)2

≤ = 0.2294FABi - 0.6882FAB j - 0.6882FABk

F = 5 2.5k6 kN Equilibrium:

ΣF = 0; FAD + FAB + FAC + F = 0 1 -0.2308FADi + 0.3077FAD j - 0.9231FADk2

+ 10.2294FABi - 0.6882FAB j - 0.6882FABk2

+ 10.2857FACi + 0.4286FAC j - 0.8571FACk2 + 12.5k2 = 0 1 -0.2308FAD + 0.2294FAB + 0.2857FAC 2i

+ 10.3077FAD - 0.6882FAB + 0.4286FAC 2j

+ 1 -0.9231FAD - 0.6882FAB - 0.8571FAC + 2.52k = 0 Equating the respective i, j, k components to zero yields:

ΣFx = 0; - 0.2308FAD + 0.2294FAB + 0.2857FAC = 0

(1)

ΣFy = 0; 0.3077FAD - 0.6882FAB + 0.4286FAC = 0

(2)

ΣFz = 0; - 0.9231FAD - 0.6882FAB - 0.8571FAC + 2.5 = 0

(3)

Solving Eqs. (1), (2) and (3) yields: Ans.

FAB = 0.980 kN FAC = 0.463 kN FAD = 1.55 kN

Ans: FAB = 0.980 kN FAC = 0.463 kN FAD = 1.55 kN 196

3–46. z

Determine the force in each cable needed to support the load. 500-lb cylinder.

B C

6 ft

3 ft

SOLUTION

6 ft

6 6 3 FAB = FAB a - i - j + k b 9 9 9

6 ft

2 ft

A 3 ft

5 ft

= - 0.3333 FAB i - 0.6667 FAB j + 0.6667 FAB k

3 6 2 FAC = FAC a - i + j + k b 7 7 7 = - 0.2857 FAC i + 0.4286 FAC j + 0.8571 FAC k FAD = FAD i © Fx = 0;

- 0.3333 FAB - 0.2857 FAC + FAD = 0

- 0.6667 FAB + 0.4286 FAC = 0

0.6667 FAB + 0.8571 FAC - 500 = 0 FAB = 250 lb

Ans.

FAC = 389 lb

Ans.

FBD = 194 lb

Ans.

Ans: FAB = 250 lb FAC = 389 lb FBD = 194 lb 197

3–47. z

The shear leg derrick is used to haul the 200-kg net of fish onto the dock. Determine the compressive force along each of the legs AB and CB and the tension in the winch cable DB. Assume the force in each leg acts along its axis.

5.6 m 4m B

SOLUTION FAB = FAB a-

4 4 2 i + j + kb 6 6 6

A x

= - 0.3333 FAB i + 0.6667 FAB j + 0.6667 FAB k

2m y

4 4 2 FCB = FCB a i + j + k b 6 6 6 = 0.3333 FCB i + 0.6667 FCB j + 0.6667 FCB k FBD = FBD a-

9.6 4 j kb 10.4 10.4

= - 0.9231 FBD j - 0.3846 FBD k W = -1962 k ©Fx = 0;

- 0.3333FAB + 0.3333 FCB = 0

0.6667 FAB + 0.6667 FCB - 0.9231 FBD = 0

0.6667 FAB + 0.6667 FCB - 0.3846 FBD - 1962 = 0 FAB = 2.52 kN

Ans.

FCB = 2.52 kN

Ans.

FBD = 3.64 kN

Ans.

Ans: FAB = 2.52 kN FCB = 2.52 kN FBD = 3.64 kN 198

*3–48. Determine the tension in the cables in order to support the 100-kg crate.in the equilibrium position shown.

D 1m

A 2.5 m

SOLUTION

Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. (a) in Cartesian vector form as

2m y

FAB = FAB i FAC = - FAC j FAD = FAD B

(- 2 - 0)i+ (2 - 0)j+ (1 - 0)k 2

2(- 2 - 0) + (2 - 0) + (1 - 0)

R = - FAD i + 2 3

2 1 F j + FAD k 3 AD 3

W = [ -100(9.81)k]N = [ - 981 k]N Equations of Equilibrium: Equilibrium requires ©F = 0;

FAB + FAC + FAD + W = 0

2 2 1 FAB i + ( -FAC j) + a - FAD i + FAD j + FAD kb + ( -981k) = 0 3 3 3 aFAB -

2 2 1 F b i + a - FAC + FAD b j + a FAD - 981b k = 0 3 AD 3 3

Equating the i, j, and k components yields FAB -

2 F = 0 3 AD

(1)

- FAC +

2 F = 0 3 AD

(2)

1 F - 981 = 0 3 AD

(3)

Solving Eqs. (1) through (3) yields FAD = 2943 N = 2.94 kN

Ans.

FAB = FAC = 1962 N = 1.96 kN

Ans.

Ans: FAD = 2.94 kN FAB = 1.96 kN 199

3–49. Determine the maximum mass of the crate so that the tension developed in any cable does not exceeded 3 kN.

D 1m

A 2.5 m

SOLUTION

Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. (a) in Cartesian vector form as

2m y

FAB = FAB i FAC = - FAC j FAD = FAD B

(- 2 - 0)i+ (2 - 0)j+ (1 - 0)k 2(-2 - 0)2 + (2 - 0)2 + (1 - 0)2

R = - FAD i + 2 3

2 1 F j + FAD 3 AD 3

W = [ -m(9.81)k] Equations of Equilibrium: Equilibrium requires ©F = 0;

FAB + FAC + FAD + W = 0

2 2 1 FAB i + ( -FAC j) + a - FAD i + FAD j + FAD kb + [- m(9.81)k] = 0 3 3 3 a FAB -

2 2 1 F b i + a -FAC + FAD b j + a FAD - 9.81m bk = 0 3 AD 3 3

Equating the i, j, and k components yields FAB -

2 F = 0 3 AD

(1)

- FAC +

2 F = 0 3 AD

(2)

1 F - 9.81m = 0 3 AD

(3)

When cable AD is subjected to maximum tension, FAD = 3000 N. Thus, by substituting this value into Eqs. (1) through (3), we have FAB = FAC = 2000 N Ans.

m = 102 kg

Ans: m = 102 kg 200

3–50. z

Determine the force in each cable if F = 500 lb.

F A

1 ft 6 ft C

SOLUTION

2 ft

3 ft

Equations of Equilibrium: Referring to the FBD shown in Fig. a,

1 ft

3 ft x

3 3 2 ΣFx = 0; FABa b - FAC a b - FADa b = 0 7 7 146

(1)

2 1 3 ΣFy = 0; - FABa b - FAC a b + FADa b = 0 7 7 146 6 6 6 ΣFz = 0; - FABa b - FAC a b - FADa b + 500 = 0 7 7 146

(2) (3)

Solving Eqs (1), (2) and (3)

FAC = 113.04 lb = 113 lb

Ans.

FAB = 256.67 lb = 257 lb

Ans.

FAD = 210 lb

Ans.

Ans: FAC = 113 lb FAB = 257 lb FAD = 210 lb 201

3–51. z

Determine the greatest force F that can be applied to the ring if each cable can support a maximum force of 800 lb.

F A

1 ft 6 ft C

SOLUTION

2 1 3 b + FADa b = 0 ΣFy = 0; - FABa b - FAC a 7 7 146

2 ft

(1)

2 ft

3 ft

Equations of Equilibrium: Referring to the FBD shown in Fig. a, 3 3 2 ΣFx = 0; FABa b - FAC a b - FADa b = 0 7 7 146

1 ft

3 ft x

(2)

6 6 6 b - FADa b + F = 0 ΣFz = 0; - FABa b - FAC a 7 7 146

(3)

Solving Eqs (1), (2) and (3)

FAC = 0.2261 F FAB = 0.5133 F FAD = 0.42 F Since cable AB is subjected to the greatest tension, its tension will reach the limit first that is FAB = 800 lb. Then 800 = 0.5133 F Ans.

F = 1558.44 lb = 1558 lb

Ans: F = 1558 lb 202

*3–52. z

Determine the tension developed in cables AB and AC and the force developed along strut AD for equilibrium of the 400-lb crate.

2 ft 2 ft B

4 ft

5.5 ft

SOLUTION Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. a in Cartesian vector form as FAB = FAB C FAC = FAC C FAD = FAD C

(- 2 - 0)i + ( -6 - 0)j + (1.5 - 0)k 2( -2 - 0)2 + ( - 6 - 0)2 + (1.5 - 0)2 (2 - 0)i + ( -6 - 0)j + (3 - 0)k 2

2(2 - 0) + ( - 6 - 0) + (3 - 0)

S = -

S =

2.5 ft 6 ft y

4 12 3 F i F j + F k 13 AB 13 AB 13 AB

2 6 3 F i - FAC j + FAC k 7 AC 7 7

(0 - 0)i + [0 - ( -6)]j + [0 - (- 2.5)]k 2

2(0 - 0) + [0 - (- 6)] + (0 - (- 2.5)]

S =

12 5 F j + F k 13 AD 13 AD

W = {- 400k} lb Equations of Equilibrium: Equilibrium requires gF = 0;

¢¢-

FAB + FAC + FAD + W = 0

2 12 4 12 3 6 3 5 F i F j + F k ≤ + ¢ FAC i - FAC j + FAC k ≤ + ¢ FAD j + F k ≤ + ( -400 k) = 0 13 AB 13 AB 13 AB 7 7 7 13 13 AD 4 2 12 6 12 3 5 3 F F ≤ j + ¢ FAB + FAC + F - 400 ≤ k = 0 + FAC ≤ i + ¢ - FAB - FAC + 13 AB 7 13 7 13 AD 13 7 13 AD

Equating the i, j, and k components yields 2 4 F + FAC = 0 13 AB 7 6 12 12 F = 0 - FAB - FAC + 13 7 13 AD 3 3 5 F + FAC + F - 400 = 0 13 AB 7 13 AD

(1)

(2) (3)

Solving Eqs. (1) through (3) yields Ans. Ans. Ans.

FAB = 274 lb FAC = 295 lb FAD = 547 lb

Ans: FAB = 274 lb FAC = 295 lb FAD = 547 lb 203

3–53. z

If the tension developed in each of the cables cannot exceed 300 lb, determine the largest weight of the crate that can be supported. Also, what is the force developed along strut AD?

2 ft 2 ft B

4 ft

5.5 ft

SOLUTION Vectors: We can express each of the forces on the free-body diagram Force Vectors: shown in Fig. a in Cartesian vector form as FAB = FAB C

(-2 - 0)i + (- 6 - 0)j + (1.5 - 0)k 2

2(-2 - 0) + ( -6 - 0) + (1.5 - 0)

FAC = FAC C

(2 - 0)i + ( -6 - 0)j + (3 - 0)k 2(2 - 0)2 + ( -6 - 0)2 + (3 - 0)2

FAD = FAD C

S = -

S =

D x

2.5 ft 6 ft y

4 12 3 F i F j + F k 13 AB 13 AB 13 AB

2 6 3 FAC i - FAC j + FAC k 7 7 7

(0 - 0)i + [0 - ( -6)]j + [0 - (- 2.5)]k 2(0 - 0)2 + [0 - ( - 6)]2 + [0 - ( - 2.5)]2

S =

12 5 FAD j + FAD k 13 13

W = -Wk Equations of Equilibrium: Equilibrium requires gF = 0;

¢¢-

FAB + FAC + FAD + W = 0

12 3 6 3 5 4 2 12 FAB i FAB j + FAB k ≤ + ¢ FAC i - FAC j + FAC k ≤ + ¢ FAD j + FAD k ≤ + ( -Wk) = 0 13 13 13 7 7 7 13 13 2 12 6 12 3 5 4 3 FAB + FAC ≤ i + ¢- FAB - FAC + FAD ≤ j + ¢ FAB + FAC + FAD - W ≤ k = 0 13 7 13 7 13 13 7 13

Equating the i, j, and k components yields -

2 4 F + FAC = 0 13 AB 7

(1)

6 12 12 F - FAC + F = 0 13 AB 7 13 AD

(2)

3 3 5 FAB + FAC + FAD - W = 0 13 7 13

(3)

Let us assume that cable AC achieves maximum tension first. Substituting FAC = 300 lb into Eqs. (1) through (3) and solving, yields FAB = 278.57 lb FAD = 557 lb

Ans.

W = 407 lb

Since FAB = 278.57 lb 6 300 lb, our assumption is correct. Ans: FAD = 557 lb W = 407 lb 204

3–54. z

Determine the tension developed in each cable for equilibrium of the 300-lb crate. 2 ft

3 ft

2 ft

C 3 ft

D 6 ft

4 ft

A 3 ft

SOLUTION Force Vectors: We can express each of the forces shown in Fig. a in Cartesian vector form as FAB = FAB C FAC = FAC C

3 6 2 S = - FAB i - FAB j + FAB k 7 7 7 2( - 3 - 0) + (-6 - 0) + (2 - 0) ( -3 - 0)i + (-6 - 0)j + (2 - 0)k 2

(2 - 0)i + ( -6 - 0)j + (3 - 0)k 2

2(2 - 0) + ( -6 - 0) + (3 - 0)

FAD = FAD C

(0 - 0)i + (3 - 0)j + (4 - 0)k 2

2(0 - 0) + (3 - 0) + (4 - 0)

S =

2 6 3 F i - FAC j + FAC k 7 AC 7 7

3 4 F j + FAD k 5 AD 5

W = {-300k} lb Equations of Equilibrium: Equilibrium requires g F = 0;

¢ - FAB i 3 7

FAB + FAC + FAD + W = 0

2 3 6 2 6 3 4 F j + FAB k ≤ + ¢ FAC i - FAC j + FAC k ≤ + ¢ FAD j + FAD k ≤ + ( -300k) = 0 7 AB 7 7 7 7 5 5

Equating the i, j, and k components yields 2 3 - FAB + FAC = 0 7 7 6 3 6 - FAB - FAC + FAD = 0 7 7 5 2 3 4 F + FAC + FAD - 300 = 0 7 AB 7 5

(1) (2) (3)

Solving Eqs. (1) through (3) yields FAB = 79.2 lb

FAC = 119 lb

Ans.

FAD = 283 lb

Ans: FAB = 79.2 lb FAC = 119 lb FAD = 283 lb 205

3–55. z

Determine the maximum weight of the crate that can be suspended from cables AB, AC, and AD so that the tension developed in any one of the cables does not exceed 250 lb. 2 ft

3 ft

2 ft

C 3 ft

D 6 ft

4 ft

A 3 ft

SOLUTION Force Vectors: We can express each of the forces shown in Fig. a in Cartesian vector form as FAB = FAB C

3 6 2 S = - FAB i - FAB j + FAB k 7 7 7 2(- 3 - 0) + ( -6 - 0) + (2 - 0) (- 3 - 0)i + ( -6 - 0)j + (2 - 0)k 2

FAC = FAC C

(2 - 0)i + ( -6 - 0)j + (3 - 0)k 2

2(2 - 0) + (- 6 - 0) + (3 - 0)

FAD = FAD C

(0 - 0)i + (3 - 0)j + (4 - 0)k 2

2(0 - 0) + (3 - 0) + (4 - 0)

S =

2 6 3 FAC i - FAC j + FAC k 7 7 7

3 4 F j + FAD k 5 AD 5

W = -WC k Equations of Equilibrium: Equilibrium requires g F = 0;

¢ - FAB i 3 7

¢ - FAB + 3 7

FAB + FAC + FAD + W = 0

6 2 6 3 4 2 3 FAB j + FAB k ≤ + ¢ FAC i - FAC j + FAC k ≤ + ¢ FAD j + FAD k ≤ + ( -WC k) = 0 7 7 7 7 7 5 5

2 6 6 3 3 4 2 FAC ≤ i + ¢ - FAB - FAC + FAD ≤ j + ¢ FAB + FAC + FAD - WC ≤ k = 0 7 7 7 5 7 7 5

Equating the i, j, and k components yields 2 3 - FAB + FAC = 0 7 7 6 3 6 - FAB - FAC + FAD = 0 7 7 5 2 3 4 FAB + FAC + FAD - WC = 0 7 7 5

(1) (2) (3)

Assuming that cable AD achieves maximum tension first, substituting FAD = 250 lb into Eqs. (2) and (3), and solving Eqs. (1) through (3) yields FAB = 70 lb WC = 265 lb

FAC = 105 lb

Ans.

Since FAB = 70 lb 6 250 lb and FAC = 105 lb, the above assumption is correct.

Ans: WC = 265 lb 206

*3–56. The 25-kg flowerpot is supported at A by the three cords. Determine the force acting in each cord.for equilibrium.

z C

B 60

30 45

SOLUTION

FAD = FAD (sin 30°i - cos 30° sin 60°j + cos 30° cos 60°k) = 0.5FADi - 0.75FADj + 0.4330FAD k FAC = FAC ( -sin 30°i - cos 30° sin 60°j + cos 30° cos 60°k)

A x

= -0.5FAC i - 0.75FACj + 0.4330FACk FAB = FAB(sin 45°j + cos 45°k) = 0.7071FAB j + 0.7071FAB k F = -25(9.81)k = {- 245.25k} N ©F = 0 ;

FAD + FAB + FAC + F = 0

(0.5FAD i - 0.75FAD j) + 0.4330FAD k + (0.7071FAB j + 0.7071FAB k) + ( -0.5FACi - 0.75FACj + 0.4330FACk) + ( -245.25k) = 0 (0.5FAD - 0.5FAC)i + ( -0.75FAD + 0.7071FAB - 0.75FAC) j + (0.4330FAD + 0.7071FAB + 0.4330FAC - 245.25) k = 0 Thus, ©Fx = 0;

0.5FAD - 0.5FAC = 0

[1]

©Fy = 0;

-0.75FAD + 0.7071FAB - 0.75FAC = 0

[2]

©Fz = 0;

0.4330FAD + 0.7071FAB + 0.4330FAC - 245.25 = 0

[3]

Solving Eqs. [1], [2], and [3] yields: FAD = FAC = 104 N

Ans.

FAB = 220 N

Ans: FAD = FAC = 104 N FAB = 220 N 207

3–57. If each cord can sustain a maximum tension of 50 N before it fails, determine the greatest weight of the flowerpot the cords can support.

z C

B 60

30 45

SOLUTION

FAD = FAD (sin 30°i - cos 30° sin 60°j + cos 30° cos 60°k)

= 0.5FAD i - 0.75FAD j + 0.4330FAD k

FAC = FAC (- sin 30°i - cos 30° sin 60° j + cos 30° cos 60° k) = -0.5FAC i - 0.75FAC j + 0.4330FAC k FAB = FAB (sin 45° j + cos 45° k) = 0.7071FAB j + 0.7071FAB k W = -Wk ©Fx = 0;

0.5FAD - 0.5FAC = 0 (1)

FAD = FAC ©Fy = 0;

-0.75FAD + 0.7071FAB - 0.75FAC = 0 (2)

0.7071FAB = 1.5FAC ©Fz = 0;

0.4330FAD + 0.7071FAB + 0.4330FAC - W = 0 0.8660FAC + 1.5FAC - W = 0 2.366FAC = W

Assume FAC = 50 N then FAB =

1.5(50) = 106.07 N 7 50 N (N . G!) 0.7071

Assume FAB = 50 N. Then FAC =

0.7071(50) = 23.57 N 6 50 N (O. K!) 1.5

Thus, Ans.

W = 2.366(23.57) = 55.767 = 55.8 N

Ans: W = 55.8 N 208

3–58. The triangular frame ABC can be adjusted vertically between the three equal-length cords. If it remains in a horizontal plane, determine the required distance s so that the tension in each of the cords, OA, OB, and OC, equals 20 N.The lamp has a mass of 5 kg.

B 0.5 m A

60 60 C

SOLUTION © Fz = 0 ;

3 (20) cos g - 5(9.81) = 0 O

g = 35.16° d =

0.25 = 0.28868 cos 30°

s =

0.28868 = 0.4098 m = 410 mm tan 35.16°

Ans.

Ans: s = 410 mm 209

3–59. z

The 500-lb crate is suspended from cable shown. Determine the force in cables AB,the AC, CD,system CE, and CF Determine the forcethe in 500-lb each segment of the cable, i.e., AB, needed to support crate. Hint: First analyze the AC, CD, CE, Hint:using First the analyze equilibrium equilibrium ofand pointCF. A, then resultthe for AC, analyze of A, then usingC.the result for AC, analyze the the point equilibrium of point equilibrium of point C.

24 ft

10 ft

y x

C A

SOLUTION

At A:

24 ft

+ ©F = 0; : x

FAB cos 20° - FAC cos 35° = 0

+ c ©Fy = 0;

FAB sin 20° + FAC sin 35° - 500 = 0

E 7 ft 7 ft

FAC = 573.58 = 574 lb

Ans.

FAB = 500 lb

Ans.

At C: ©Fx = 0;

10 F + 573.58 cos 35° = 0 26 CD Ans.

FCD = 1221.6 lb = 1.22 kip ©Fy = 0;

FCE = FCF

©Fz = 0;

24 24 (1221.6) (2)FCE - 573.58 sin 35° = 0 26 25 Ans.

FCE = FCF = 416 lb

Ans: FAC = 574 lb FAB = 500 lb FCD = 1.22 kip FCE = 416 lb 210

*3–60. The 9500-lb pipe is supported by three cables. Determine the force in each cable for equilibrium of the hook at A.

9500 lb

A 2.4 ft 1358 1 ft B

SOLUTION FAD = FAD c C

908 1358

j - 2.4k 212 + ( - 2.4)2

= {0.3846 FAD j - 0.9231 FAD k} lb

FAB = FAB c C

- cos 45°i - sin 45°j - 2.4k 2( -cos 45°)2 + ( - sin 45°)2 + ( - 2.4)2

= { - 0.2720 FAB i - 0.2720 FAB j - 0.9231 FAB k} lb

FAC = FAC c C

cos 45°i - sin 45°j - 2.4k 2(cos 45°)2 + ( - sin 45°)2 + ( - 2.4)2

= {0.2720 FAC i - 0.2720 FAC j - 0.9231 FAC k} lb ΣFx = 0; - 0.2720FAB + 0.2720FAC = 0 ΣFy = 0; 0.3846FAD - 0.2720FAB - 0.2720FAC = 0 ΣFz = 0; - 0.9231FAD - 0.9231FAB - 0.9231FAC + 9500 = 0 Solving, FAB = 3.01 kip

Ans.

FAC = 3.01 kip

Ans.

FAD = 4.26 kip

Ans.

Ans: FAB = 3.01 kip FAC = 3.01 kip FAD = 4.26 kip 211

3–61. z

Determine the force in each of the three cables needed to lift the machine that has a weight of 10 kN.

10 kN A 3m

SOLUTION FAB = FAB ¢

1.25 i ++ 2j 2 j--3k 3k 1.25i 2(1.25)2 + 22 + (- 3)2

≤

= 0.3276 FAB i + 0.5241 FAB j - 0.7861 FAB k FAC = FAC ¢

1.25 i ++ 2j 2 j--3k 3k - 1.25i 2

2(-1.25) + 2 + ( - 3)

≤

= - 0.3276 FAC i + 0.5241 FAC j - 0.7861 FAC k FAD = FAD ¢

- 1j - 3k 2(- 1)2 + ( - 3)2

1.25 m

B 1.25 m

≤

= - 0.3162 FAD j - 0.9487 FAD k ©Fx = 0;

0.3276 FAB - 0.3276 FAC = 0

©Fy = 0;

0.5241 FAB + 0.5241 FAC - 0.3162 FAD = 0

©Fz = 0;

- 0.7861 FAB - 0.7861 FAC - 0.9487 FAD + 10 = 0 FAB = 2.12 kN

Ans.

FAC = 2.12 kN

Ans.

FAD = 7.03 kN

Ans.

Ans: FAB = 2.12 kN FAC = 2.12 kN FAD = 7.03 kN 212

3–62. Determine the tension developed in cables OD and OB and the strut OC, required to support the 50-kg crate. The spring OA has an unstretched length of 0.8 m and a stiffness kOA = 1.2 kN/m. The force in the strut acts along the axis of the strut.

z B C 4m 3m 2m

SOLUTION Free Body Diagram Free-Body Diagram:: The spring stretches s = l - l0 = 1 - 0.8 = 0.2 m. Hence, the spring force is Fsp = ks = 1.210.22 = 0.24 kN = 240 N.

kOA = 1.2 kN/m 4m

Cartesian Vector VectorNotation Notation:: FOB = FOB ¢ FOC = FOC ¢ FOD = FOD ¢

- 2i - 4j + 4k

21 - 222 + 1 - 422 + 42 - 4i + 3k 2

21 - 42 + 3

≤ = - FOCi + 4 5

2i + 4j + 4k 2

22 + 4 + 4

Fsp = 5- 240j6 N

≤ = - FOBi -

≤ =

1 3

2 2 F j + FOBk 3 OB 3

1m O

x 4m

3 F k 5 OC

1 2 2 F i + FODj + FODk 3 OD 3 3

F = 5 - 490.5k6 N

Equilibrium : Equations of Equilibrium: ©F = 0;

FOB + FOC + FOD + Fsp + F = 0

4 1 2 2 1 a - FOB - FOC + FOD b i + a - FOB + FOD - 240b j 3 5 3 3 3 2 3 2 + a FOB + FOC + FOD - 490.5b k = 0 3 5 3 Equating i, j and k components, we have 4 1 1 - FOB - FOC + FOD = 0 3 5 3

[1]

2 2 - FOB + FOD - 240 = 0 3 3

[2]

2 3 2 F + FOC + FOD - 490.5 = 0 3 OB 5 3

[3]

Solving Eqs.[1], [2] and [3] yields FOB = 120 N FOC = 150 N

Ans.

FOD = 480 N

Ans: FOB = 120 N FOC = 150 N FOD = 480 N 213

3–63. Determine the force in each cable used to lift the 9.50-Mg surge arrester at constant velocity.

z F

SOLUTION F = 9.50(10)3 (9.81) = {93.195 k} kN

0.5 i - 2 k b = 0.2425 FB i - 0.9701FB k FB = FB a 2.062

B 0.5 m

D y

- 0.5 cos 45° i - 0.5 sin 45° j - 2 k b FC = FC a 2.062 = -0.1715 FC i - 0.1715 FC j - 0.9701 FC k FD = FD a

0.5 j - 2 k b = 0.2425 FD j - 0.9701FD k 2.062

0.2425 FB - 0.1715 FC = 0

- 0.1715 FC + 0.2425 FD = 0

93.195 - 0.9701 FB - 0.9701 FC - 0.9701 FD = 0

Solving FB = 28.1 kN

Ans.

FC = 39.8 kN

Ans.

FD = 28.1 kN

Ans.

Ans: FB = 28.1 kN FC = 39.8 kN FD = 28.1 kN 214

*3–64. If cable AD is tightened by a turnbuckle and develops aa tension of 1300 lb, determine the tension developed in in cables AB thethe force developed alongalong the z axis ABand andAC, ACand and force developed the of the antenna tower antenna tower AE at AE. point A.

z A 30 ft

C 10 ft B

E 15 ft

SOLUTION

12.5 ft

10 ft

15 ft y

Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. a in Cartesian vector form as FAB = FAB C

(10 - 0)i + (- 15 - 0)j + (- 30 - 0)k 2

2(10 - 0) + ( -15 - 0) + ( - 30 - 0)

S =

2 3 6 F i - FAB j - FAB k 7 AB 7 7

( -15 - 0)i + (- 10 - 0)j + (- 30 - 0)k

FAC = FAC C

3 2 6 S = - FAC i - FAC j - FAC k 7 7 7 2( - 15 - 0) + ( -10 - 0) + ( -30 - 0)

FAD = FAD C

(0 - 0)i + (12.5 - 0)j + ( -30 - 0)k

2(0 - 0)2 + (12.5 - 0)2 + ( -30 - 0)2

S = {500j - 1200k} lb

FAE = FAE k Equations of Equilibrium: Equilibrium requires g Σ F = 0;

FAB + FAC + FAD + FAE = 0

¢ FAB i 2 7

3 6 3 2 6 FAB j - FAB k ≤ + ¢- FAC i - FAC j - FAC k ≤ + (500j - 1200k) + FAE k = 0 7 7 7 7 7

¢ FAB -

3 3 2 6 6 F ≤ i + ¢- FAB - FAC + 500 ≤ j + ¢ - FAB - FAC + FAE - 1200 ≤ k = 0 7 AC 7 7 7 7

2 7

Equating the i, j, and k components yields 2 3 FAB - FAC = 0 7 7 2 3 - FAB - FAC + 500 = 0 7 7 6 6 - FAB - FAC + FAE - 1200 = 0 7 7

(1) (2) (3)

Solving Eqs. (1) through (3) yields Ans. Ans. Ans.

FAB = 808 lb FAC = 538 lb FAE = 2354 lb = 2.35 kip

Ans: FAB = 808 lb FAC = 538 lb FAE = 2.35 kip 215

© 2022 by R. C. Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws they currently exist. No portion of this material may be cannot reproduced, in any form or by any means, without permission zin writing from the publisher. Ifasthe tension developed in either cable AB or AC

exceed 1000 lb, determine the maximum tension that can be developed in cable AD when it is tightened by the 3–65. turnbuckle. Also, what is the force developed along the antenna tower at point A? If the tension developed in either cable AB or AC cannot exceed 1000 lb, determine the maximum tension that can be developed in cable AD when it is tightened by the turnbuckle. Also, what is the force developed along the antenna tower at point A?

A z

30 ft

A C

SOLUTION

E 15 ft

Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. a in Cartesian vector form as x 10 ft

SOLUTION

10 ft

30 ft

10 ft

12.5 ft

(10 - 0)i + (- 15 - 0)j + (- 30 - 0)k 2 3 6 15 ft FAB =Vectors: FAB C We can express each of the forces on the free-body S = FAB i - F AB j -in FAB k diagram shown Force 2 2 2 7 7 7 2(10 - 0) + ( -15 - 0) + (- 30 - 0) Fig. a in Cartesian vector form as x

D C

E 12.5 ft

15 ft 10 ft

15 ft y

( - 15 - 0)i + ( -10 - 0)j + (-30 - 0)k 3 2 6 (10 - 0)i + ( - 15 - 0)j + (- 30 - 0)k S = -2 FAC i - 3 FAC j - 6 FAC k FAC = FAC C 2 2 2 FAB = FAB2(C 15 - 0) + (-10 - 0) + (- 30 - 0) S = 7FAB i - 7FAB j - 7FAB k 7 7 7 2(10 - 0)2 + ( -15 - 0)2 + (- 30 - 0)2 (0 - 0)i + (12.5 - 0)j + ( -30 - 0)k 12 5 ( - 15 - 0)i + ( -10 - 0)j + (-30S-=0)k Fj - 3 F k FAD = FC 2 6 13 S = -13 FAC i - FAC j - FAC k FAC = F2(0 AC C - 0)2 + (12.5 - 0)2 + (-30 - 0)2 2 2 2 7 7 7 2(- 15 - 0) + (-10 - 0) + ( -30 - 0)

FAE = FAE k

(0 - 0)i + (12.5 - 0)j + ( -30 - 0)k 12 5 Fj Fk = Equilibrium: FC S = FAD of Equations Equilibrium requires 2 2 2 13 13 2(0 - 0) + (12.5 - 0) + (-30 - 0) F + FAC + FAD + FAE = 0 g F = 0; k FAE = FAE AB

12 3 6 3 2 6 2 5 requires Equations F k ≤ + FAE k = 0 FAB j - FABEquilibrium FAC j - FAC k ≤ + ¢ Fj k≤ + ¢ - F ¢ FAB i - of Equilibrium: AC i 13 7 7 7 7 7 7 13

¢ FAB -

FAB + FAC 3 + FAD 2+ FAE = 50 6 6 12 2 g F = 0;3 F ≤ i + ¢ - FAB - FAC + F ≤ j + ¢ - FAB - FAC F + FAE ≤ k = 0 7 AC 7 7 13 7 7 13 7 12 3 6 3 2 6 2 5 ¢ FAB i - FAB j - FAB k ≤ + ¢ - FAC i - FAC j - FAC k ≤ + ¢ Fj - F k ≤ + FAE k = 0 7 7 7 7 7 7 13 13 Equating the i, j, and k components yields 3 3 2 5 6 6 12 2 ¢ F - F ≤ i + ¢ - FAB - FAC + F ≤ j + ¢ - FAB - FAC - F + FAE ≤ k = 0 2 7 AB3 7 AC 7 7 13 7 7 13 F - FAC = 0 (1) 7 7 AB 2 5 3 + k components F = 0 - FAB -the F (2) Equating i,AC j, and yields 7 13 7 6 12 6 - 3FAC - 0 F + FAE = 0 - 2FF (3)(1) AB - 7 FAC = 13 77 AB 7 2 5 AB achieves maximum tension first. Substituting Let us- 3assume + FAB - that FAC cable F = 0 (2) 7 7 13 FAB = 1000 lb into Eqs. (1) through (3) and solving yields 6 12 6 F + FAE = 0 - FAB - FAC (3) FAC 7= 666.67 7lb 13 F = 1610 lb = 1.61 kip Ans. FAE = 2914 lb = 2.91 kip Let us assume that cable AB achieves maximum tension first. Substituting (1) through (3) and solving yields FABF = 1000 lb into Since is correct. = 666.67 lb Eqs. 6 1000 lb, our assumption AC

FAC = 666.67 lb FAE = 2914 lb = 2.91 kip

Ans.

F = 1610 lb = 1.61 kip

Since FAC = 666.67 lb 6 1000 lb, our assumption is correct. Ans: FAE = 2.91 kip F = 1.61 kip 216

3–66. Determine the force in each of the three cables needed to support the tractor tread assembly, which has a mass of 8 Mg.

A 3m x D B 1.25 m

SOLUTION

1.25 m

C 2m

Force Vector: FAD = FAD ¢

FAC = FAC ¢

FAB = FAB ¢

- 1j - 3k 2( - 1)2 + ( - 3)2

≤ = - 0.3162FADj - 0.9487FADk

- 1.25i + 2j - 3k

2( - 1.25)2 + 22 + ( - 3)2 1.25i + 2j - 3k

21.252 + 22 + ( - 3)2

≤ = - 0.3276FACi + 0.5241FAC j - 0.7861FACk

≤ = 0.3276FABi + 0.5241FAB j - 0.7861FABk

F = 8(10)3(9.81)k = 5 78.48k6 kN

Equilibrium:

ΣF = 0; FAD + FAB + FAC + F = 0 1- 0.3162FADj + 0.9487FADk2 + 10.3276FABi + 0.5241FABj + 0.7861FABk2

+ 1- 0.3276FACi + 0.5241FACj - 0.7861FACk2 + 178.48k2 = 0

10.3276FAB - 0.3276FAC 2i + 1- 0.3162FAD + 0.5241FAB + 0.5241FAC 2j

+ 1- 0.9487FAD - 0.7861FAB - 0.7861FAC + 78.482k = 0 Equating the respective i, j, k components to zero yields: ΣFx = 0; 0.3276FAB - 0.3276FAC = 0

(1)

ΣFy = 0; - 0.3162FAD + 0.5241FAB + 0.5241FAC = 0

(2)

ΣFz = 0; - 0.9487FAD - 0.7861FAB - 0.7861FAC + 78.48 = 0

(3)

Solving Eqs. (1), (2) and (3) yields: Ans.

FAB = FAC = 16.6 kN FAD = 55.2 kN

Ans: FAB = 16.6 kN FAD = 55.2 kN 217

3–67. Three 10-lb balls and one 15-lb ball are suspended from the pulley-and-cable system. If the pulleys are frictionless and the centers of all of them lie in the same horizontal plane, determine the sag s for equilibrium of the system.

20 in. 20 in. 20 in. s

10 lb 10 lb 10 lb

SOLUTION

15 lb

Force Vector: F1 = 10¢

F2 = 10¢

11.547j + sk 2

211.547 + s

≤ =

115.47j + 10sk 2133.33 + s2

10i - 5.773j + sk 2

210 + ( - 5.773) + s

≤ =

-10i - 5.773j + sk 2

2( - 10) + ( - 5.773) + s

100i - 57.73j + 10sk 2133.33 + s2

≤ =

- 100i - 57.73j + 10sk 2133.33 + s2

F4 = 5 - 15k6 lb

Equilibrium: Check if ΣFx = ΣFy = 0

ΣFx = ΣFy =

100 2133.33 + s

2133.33 + s

115.47

ΣFz = 0;

10s

-100 2133.33 + s2 -57.73

2133.33 + s

Solving yields:

= 0 (O.K!) +

10s

-57.73 2133.33 + s2

2133.33 + s

= 0 (O.K!)

10s

2133.33 + s2

- 15 = 0

Ans.

s = 6.67 in.

Ans: s = 6.67 in. 218

4–1. Determine the moment about point A of each of the three forces acting on the beam.

F2 = 500 lb

F1 = 375 lb 5

4 3

8 ft

6 ft

0.5 ft

5 ft

SOLUTION

30˚ F3 = 160 lb

a + 1MF12A = - 375182

= - 3000 lb # ft = 3.00 kip # ft (Clockwise)

Ans.

4 a + 1MF22A = - 500 a b 1142 5

= -5600 lb # ft = 5.60 kip # ft (Clockwise)

Ans.

a + 1MF32A = - 1601cos 30°21192 + 160 sin 30°10.52

= - 2593 lb # ft = 2.59 kip # ft (Clockwise)

Ans.

Ans:

( MF1 ) A = 3.00 kip # ft b ( MF2 ) A = 5.60 kip # ft b ( MF3 ) A = 2.59 kip # ft b

219

4–2. Determine the moment about point B of each of the three forces acting on the beam.

F2 = 500 lb

F1 = 375 lb 5

8 ft

SOLUTION

4 3

6 ft

0.5 ft

5 ft 30˚ F3 = 160 lb

a + 1MF12B = 3751112

= 4125 lb # ft = 4.125 kip # ft (Counterclockwise)

Ans.

4 a + 1MF22B = 500 a b 152 5

= 2000 lb # ft = 2.00 kip # ft (Counterclockwise)

Ans.

a + 1MF32B = 160 sin 30°10.52 - 160 cos 30°102 = 40.0 lb # ft (Counterclockwise)

Ans.

Ans: ( MF1 ) B = 4.125 kip # ftd ( MF2 ) B = 2.00 kip # ftd ( MF3 ) B = 40.0 lb # ftd 220

4–3. The man exerts the two forces on the handle of the shovel. Determine the resultant moment of these forces about the blade at A.

658

F1 5 12 N

F2 5 30 N 308

800 mm

208

SOLUTION a+ MRA = ΣMA;

450 mm

MRA = 12 sin 65°(1.25) - 30 cos 30°(0.45) = 1.90 N # m a

Ans.

Ans: MRA = 1.90 N # m a 221

*4–4. Determine the moment of each of the three forces about point A.

250 N 30

300 N

A 2m

SOLUTION The moment arm measured perpendicular to each force from point A is d1 = 2 sin 60° = 1.732 m

5 3

d2 = 5 sin 60° = 4.330 m

500 N

d3 = 2 sin 53.13° = 1.60 m Using each force where MA = Fd, we have a + 1MF12A = - 25011.7322

= - 433 N # m = 433 N # m (Clockwise)

Ans.

a + 1MF22A = - 30014.3302

= - 1299 N # m = 1.30 kN # m (Clockwise)

Ans.

a + 1MF32A = - 50011.602

= - 800 N # m = 800 N # m (Clockwise)

Ans.

Ans: ( MF1 ) A = 433 N # mb ( MF2 ) A = 1.30 kN # mb ( MF3 ) A = 800 N # mb 222

4–5. Determine the moment of each of the three forces about point B.

250 N 30

300 N

A 2m

SOLUTION

The forces are resolved into horizontal and vertical component as shown in Fig. a. For F11 , a + MB = 250 cos 30°(3) - 250 sin 30°(4) = 149.51 N # m = 150 N # m d

Ans.

5 3

For F22 ,

500 N

a + MB = 300 sin 60°(0) + 300 cos 60°(4) = 600 N # m d

Ans.

Since the line of action of F33 passes through B, its moment arm about point B is zero. Thus Ans.

MB = 0

Ans: MB = 150 N # md MB = 600 N # md MB = 0 223

4–6. Determine the moment of each force about the bolt located at A. Take FB = 40 lb, FC = 50 lb.

0.75 ft B

2.5 ft

30 FC

SOLUTION a +MB = 40 cos 25°(2.5) = 90.6 lb # ft d

Ans.

a +MC = 50 cos 30°(3.25) = 141 lb # ftd

Ans.

Ans: MB = 90.6 lb # ftb MC = 141 lb # ftd 224

4–7. If FB = 30 lb and FC = 45 lb, determine the resultant moment about the bolt located at A.

0.75 ft B

2.5 ft

30 FC

20 FB

SOLUTION a + MA = 30 cos 25°(2.5) + 45 cos 30°(3.25) = 195 lb # ft d

Ans: MA = 195 lb # ftd 225

*4–8. The cable exerts a force of P = 6 kN at the end of the 8-m-long crane boom. If u = 30°, determine the placement x of the hook at B so that this force creates a maximum moment about point O. What is this moment?

A P 6 kN 8m u

B x

SOLUTION In order to produce the maximum moment about point O, P must act perpendicular to the boom’s axis OA as shown in Fig. a. Thus

a+ (MO)max = 6 (8) = 48.0 kN # m (Counterclockwise) Referring to the geometry of Fig. a,

Ans.

8 + tan 30° = 9.814 m = 9.81 m cos 30°

Ans.

x = x' + x" =

Ans: (MO)max = 48.0 kN # m d x = 9.81 m 226

4–9. A

The cable exerts a force of P = 6 kN at the end of the 8-m-long crane boom. If x = 10 m, determine the position u of the boom so that this force creates a maximum moment about point O. What is this moment?

P 6 kN 8m u

B x

SOLUTION In order to produce the maximum moment about point O, P must act perpendicular to the boom’s axis OA as shown in Fig. a. Thus, a+ (MO)max = 6 (8) = 48.0 kN # m (Counterclockwise) Referring to the geometry of Fig. a, x = x' + x";

10 =

8 + tan u cos u

10 =

8 sin u + cos u cos u

10 cos u - sin u = 8 10 1 8 cos u sin u = 1101 1101 1101

Ans.

(1)

From the geometry shown in Fig. b, a = tan-1 a sin a = Then Eq (1) becomes

1 b = 5.711° 10

1 1101

cos a =

10 1101

cos u cos 5.711° - sin u sin 5.711° =

8 1101

Referring that cos (u + 5.711°) = cos u cos 5.711° - sin u sin 5.711° cos (u + 5.711°) =

8 1101

u + 5.711° = 37.247° u = 31.54° = 31.5°

Ans.

Ans: (MO)max = 48.0 kN # m d u = 31.5° 227

4–10. Determine the angle u at which the 500-N force must act at A so that the moment of this force about point B is equal to zero.

B 0.3 m A 1m

θ 500 N

SOLUTION This problem requires that the resultant moment about point B be equal to zero. a +MRs = ©Fd;

MRs = 0 = 500 cos u10.32 - 500 sin u122 Ans.

u = 8.53°

Also note that if the line of action of the 500 N force passes through point B, it produces zero moment about point B. Hence, from the geometry u = tan-1

0.3 2

= 8.53°

Ans: u = 8.53° 228

4–11. Two forces act on the skew caster. Determine the resultant moment of these forces about point A and about point B.

B 2.5 in.

2 in.

608 10 lb

1.25 in.

SOLUTION

a+ MA = 80 sin 30°(1.25) + 10 cos 30°(1.25) = 60.8 lb # in

Ans.

80 lb

a+ MB = 10(2 + 2.5 sin 30° + 1.25 cos 30°) - 80(2.5 cos 30° - 1.25 sin 30°)

= - 79.8798 lb # in. MB = -79.9 lb # in. b

Ans.

Ans: MA = 60.8 lb # in. d MB = 79.9 lb # in. b 229

*4–12. Two men exert forces of F = 80 lb and P = 50 lb on the ropes. Determine the moment of each force about A. Which way will the pole rotate, clockwise or counterclockwise?

6 ft

45 B

12 ft

5 4

SOLUTION

4 c + (MA)C = 80 a b (12) = 768 lb # ftb 5

Ans.

a + (MA)B = 50 (cos 45°)(18) = 636 lb # ftd

Ans.

Since (MA)C 7 (MA)B Ans.

(Clockwise) Clockwise

Ans: (MA)C = 768 lb # ftb (MA)B = 636 lb # ftd 230

4–13. If the man at B exerts a force of P = 30 lb on his rope, determine the magnitude of the force F the man at C must exert to prevent the pole from rotating, i.e., so the resultant moment about A of both forces is zero.

6 ft

45 B

SOLUTION a+

12 ft

5 4

4 30 (cos 45°)(18) = Fa b(12) = 0 5

Ans.

F = 39.8 lb

Ans: F = 39.8 lb 231

4–14. A force of 80 N acts on the handle of the paper cutter at A. Determine the moment created by this force about the hinge at O, if u = 60°. At what angle u should the force be applied so that the moment it creates about point O is a maximum (clockwise)? What is this maximum moment?

F = 80 N A

400 mm

30°

SOLUTION a + Mo = ©Fd;

10 mm

Mo = - 80 cos u10.012 - 80 sin u10.42 = - 10.800 cos u + 32.0 sin u2 N # m = 10.800 cos u + 32.0 sin u2 N # m (Clockwise)

At u = 60°,

Mo = 0.800 cos 60° + 32.0 sin 60° = 28.1 N # m (Clockwise)

Ans.

In order to produce the maximum and minimum moment about point A,

dMo = 0 du

dMo = 0 = - 0.800 sin u + 32.0 cos u du Ans.

u = 88.568° = 88.6° d 2MA du2 Since

= - 0.800 cos u - 32.0 sin u

d 2MA du2

= - 0.800 cos 88.568° - 32.0 sin 88.568° = - 32.00 is a u = 88.568°

negative value, indeed at u = 88.568°, the 80 N produces a maximum clockwise moment at O. This maximum clockwise moment is 1MO2max = 0.800 cos 88.568° + 32.0 sin 88.568° = 32.0 N # m (Clockwise)

Ans.

Ans: Mo = 28.1 N # m b u = 88.6° (MO)max = 32.0 N # m b 232

4–15. Determine the orientation u 10° … u … 180° 2 of the 40-lb force F so that it produces (a) the maximum moment about point A and (b) no moment about point A. Find the moment in case (a).

2 ft

F 5 40 lb 8 ft

SOLUTION (a) MA = 40 cos u(2) + 40 sin u(8) = 80 cos u + 320 sin u dMA = - 80 sin u + 320 cos u du Target the maximum (or minimum) moment, set - 80 sin u + 320 cos u = 0 d 2MA du 2

tan u =

dMA = 0. du

320 80

u = 76.0°

Ans.

= - 80 cos u - 320 cos u = - 80 cos 76.0° - 320 sin 76.0° = -330 (negative value)

Since

d 2MA

has a negative value at u = 76.0°, the force will produce the du 2 maximum moment at point A at this angle. Note also that this requires the force line of action be perpendicular to the distance from A to the force, 2 Ex, f = tan-1 = 14.0°, u = 90° - 14° = 76°. Ans. 8

(b) In order to have zero moment at A, the line of action of the 40 lb force must pass through point A. tan f =

2 8

f = 14.0°

u = 180° - f = 166°

Ans.

Ans: u = 76.0° u = 76° f = 14.0°, u = 166° 233

*4–16. Determine the moment of the force F about point A as a function of u. Plot the result of M (ordinate) versus u (abscissa) for 0° … u … 180°.

2 ft

u F 5 40 lb

8 ft

SOLUTION Ans.

cMA = 40 cos u(2) + 40 sin u(8) = 80 cos u + 320 sin u

Ans: cMA = 40 cos u(2) + 40 sin u(8) = 80 cos u + 320 sin u 234

4–17. The handle hammerof isthesubjected to subjected the forceto oftheF force = 20 lb. hammer is of Determine the momentthe of this force of about point A. the F = 20 lb. Determine moment this the force about point A.

F 30

5 in. 18 in.

SOLUTION Resolving the 20-lb force into components parallel and perpendicular to the hammer, Fig. a, and applying the principle of moments,

A B

a +MA = - 20 cos 30°(18) - 20 sin 30°(5) = -361.77 lb # in = 362 lb # in (Clockwise)

Ans.

Ans: MA = 362 lb # inb 235

4–18. F F

In order to pull out the nail at B, the force F exerted on the handle of the hammer must produce a clockwise moment of 500 lb # in. about point A. Determine the required magnitude of force F.

30 308

5 in. 5 in.

18 18 in. in.

SOLUTION Resolving force F into components parallel and perpendicular to the hammer, Fig. a, and applying the principle of moments,

AA B B

a + MA = - 500 = -F cos 30°(18) - F sin 30°(5) Ans.

F = 27.6 lb

Ans: F = 27.6 lb 236

4–19. If the tension in the belt is 52 lb, determine the moment of each of these forces about the pin at A.

F1 5 52 lb

308

6 in. 8 in.

SOLUTION

208

F2 5 52 lb

c +(MA)1 = 52 cos 30°(8 + 6 cos 30°) - 52 sin 30°(5 - 6 sin 30°) = 542 lb # in.

5 in.

Ans.

c +(MA)2 = 52 cos 20°(8 - 6 cos 20°) - 52 sin 20°(5 + 6 sin 20°) = 10.0 lb # in.

Ans.

Ans: (MA)1 = 542 lb # in. (MA)2 = 10.0 lb # in. 237

*4–20. The to hoist hoist the the 2-Mg 2-Mg load load upward upward at at The tower tower crane crane is is used used to constant The 1.5-Mg 1.5-Mg jib 0.5-Mg jib jib BC, BC, and and constant velocity. velocity. The jib BD, BD, 0.5-Mg 6-Mg counterweight C C have have centers centers of of mass mass at at G G1,, G G2,, and and 6-Mg counterweight 1 2 G moment produced produced G33,, respectively. respectively. Determine Determine the the resultant resultant moment by the load load and and the the weights weight of jibs about about by the of the the tower tower crane crane jibs point point A A and and about about point point B. B.

4m G2

C G3

9.5m

7.5 m

D 12.5 m

23 m

SOLUTION Since the moment arms of the weights and the load measured to points A and B are the same, the resultant moments produced by the load and the weight about points A and B are the same. a + (MR)A = (MR)B = ©Fd;

(MR)A = (MR)B = 6000(9.81)(7.5) + 500(9.81)(4) - 1500(9.81)(9.5)

- 2000(9.81)(12.5) = 76 027.5 N # m = 76.0 kN # m (Counterclockwise)

Ans.

Ans: (MR)A = (MR)B = 76.0 kN # md 238

4–21. The tower crane is used to hoist a 2-Mg load upward at constant velocity. The 1.5-Mg jib BD and 0.5-Mg jib BC have centers of mass at G1 and G2 , respectively. Determine the required mass of the counterweight C so that the resultant moment produced by the load and the weight of the tower crane jibs about point A is zero. The center of mass for the counterweight is located at G3 .

4m G2 C G3

9.5m B

7.5 m

D 12.5 m

23 m

SOLUTION a + (MR)A = ©Fd;

mC(9.81)(7.5) + 500(9.81)(4) - 1500(9.81)(9.5) - 2000(9.81)(12.5) 0 = M Ans.

m C = 4966.67 kg = 4.97 Mg M

Ans: mC = 4.97 Mg 239

4–22. If A, B, and D are given vectors, prove the distributive law for the vector cross product, i.e., A : (B + D) = (A : B) + (A : D).

SOLUTION Consider the three vectors; with A vertical. Note obd is perpendicular to A. od = ƒ A * (B + D) ƒ = ƒ A ƒ ƒ B + D ƒ sin u3 ob = ƒ A * B ƒ = ƒ A ƒ ƒ B ƒ sin u1 bd = ƒ A * D ƒ = ƒ A ƒ ƒ D ƒ sin u2 Also, these three cross products all lie in the plane obd since they are all perpendicular to A. As noted the magnitude of each cross product is proportional to the length of each side of the triangle. The three vector cross products also form a closed triangle o¿b¿d¿ which is similar to triangle obd. Thus from the figure, A * (B + D) = (A * B) + (A * D)

(QED)

Note also, A = Ax i + Ay j + Az k B = Bx i + By j + Bz k D = Dx i + Dy j + Dz k A * (B + D) = 3

i Ax Bx + Dx

j Ay By + Dy

k Az 3 Bz + Dz

= [A y (Bz + Dz) - A z(By + Dy)]i - [A x(Bz + Dz) - A z(Bx + Dx)]j + [A x(By + Dy) - A y(Bx + Dx)]k = [(A y Bz - A zBy)i - (A x Bz - A z Bx)]j + (A x By - A y Bx)k + [(A y Dz - A z Dy)i - (A x Dz - A z Dx)j + (A x Dy - A y Dx)k i = 3 Ax Bx

j Ay By

k i Az 3 + 3 Ax Bz Dx

j Ay Dy

k Az 3 Dz

= (A * B) + (A * D)

(QED)

240

4–23. Prove the triple scalar A # (B : C) = (A : B) # C.

product

identity

SOLUTION As shown in the figure Area = B(C sin u) = |B * C| Thus, Volume of parallelepiped is |B * C||h| But, |h| = |A # u(B * C)| = ` A # a

B * C b` |B * C|

Thus, Volume = |A # (B * C)| Since |(A * B) # C| represents this same volume then A # (B : C) = (A : B) # C

(QED)

Also, LHS = A # (B : C) i = (A x i + A y j + A z k) # 3 Bx Cx

j By Cy

k Bz 3 Cz

= A x (ByCz - BzCy) - A y (BxCz - BzCx) + A z (BxCy - ByCx) = A xByCz - A xBzCy - A yBxCz + A yBzCx + A zBxCy - A zByCx RHS = (A : B) # C i = 3 Ax Bx

j Ay By

k A z 3 # (Cx i + Cy j + Cz k) Bz

= Cx(A y Bz - A zBy) - Cy(A xBz - A zBx) + Cz(A xBy - A yBx) = A xByCz - A xBzCy - A yBxCz + A yBzCx + A zBxCy - A zByCx Thus, LHS = RHS A # (B : C) = (A : B) # C

(QED)

241

*4–24. Given the three nonzero vectors A, B, and C, show that if A # (B : C) = 0, the three vectors must lie in the same plane.

SOLUTION Consider, |A # (B * C)| = |A| |B * C | cos u = (|A| cos u)|B * C| = |h| |B * C| = BC |h| sin f = volume of parallelepiped. If A # (B * C) = 0, then the volume equals zero, so that A, B, and C are coplanar.

242

4–25. z

The force F = {400i - 100j - 700k} lb acts at the end of the beam. Determine the moment of this force about point O.

A F O

8 ft

1.5 ft

SOLUTION Position Vector: The coordinates of point B are B(8, 0.25, 1.5) ft.

0.25 ft

Thus, rOB = {8i + 0.25j + 1.5k} ft Moments of F About Point O: MO = rOB * F i j = † 8 0.25 400 - 100

k 1.5 † - 700

= { - 25i + 6200j - 900k} lb # ft

Ans.

Ans: MO = { - 25i + 6200j - 900k} lb # ft 243

4–26. z

The force F = {400i - 100j - 700k} lb acts at the end of the beam. Determine the moment of this force about point A.

A F O

8 ft

1.5 ft

SOLUTION Position Vector: The coordinates of points A and B are A (0, 0, 1.5) ft and B (8, 0.25, 1.5) ft, respectively. Thus,

0.25 ft

rAB = (8 - 0)i + (0.25 - 0)j + (1.5 - 1.5)k = {8i + 0.25j} ft Moment of F About Point A: MA = rAB * F i = † 8 400

j 0.25 - 100

k 0 † - 700

= { - 175i + 5600j - 900k} lb # ft

Ans.

Ans: MA = { - 175i + 5600j - 900k} lb # ft 244

4–27. z

Determine the moment of the force F about point O. Express the result as a Cartesian vector.

F {–6i + 4 j 8k} kN

4m P 3m

6m O 1m

SOLUTION x

Position Vector: The coordinates of point A are (1, -2, 6) m. Thus, rOA = {i - 2j + 6k} m The Moment of F About Point O: MO = rOA * F i = † 1 -6

j -2 4

k 6† 8

= { - 40i - 44j - 8k} kN # m

Ans.

Ans: MO = { - 40i - 44j - 8k} kN # m 245

*4–28. z

Determine the moment of the force F about point P. Express the result as a Cartesian vector.

F {–6i + 4 j 8k} kN

4m P 3m

6m O 1m

SOLUTION Position Vector: The coordinates of points A and P are A (1, - 2, 6) m and P (0, 4, 3) m, respectively. Thus

rPA = (1 - 0)i + ( - 2 - 4)j + (6 - 3)k = {i - 6j + 3k} m The Moment of F About Point P: MP = rPA * F i j = † 1 -6 -6 4

k 3† 8

= { - 60i - 26j - 32k} kN # m

Ans.

Ans: MP = { - 60i - 26j - 32k} kN # m 246

4–29.

The man pulls on the rope with a force of F = 20 N. Determine the moment of this force about the base of the pole at O. Solve the problem two ways, i.e., by using a position vector from O to A, then O to B.

A F

10.5 m

B 1.5 m

SOLUTION

O 4m

F = 20 £

4i - 3j - 9k 2(4)2 + ( - 3)2 + ( -9)2

≥

= {7.77i - 5.83j - 17.5k} N i MO = † 0 7.77

j 0 - 5.83

k 10.5 † - 17.5

MO = {61.2i + 81.6j} N # m i MO = † 4 7.77

j -3 - 5.83

Ans.

k 1.5 † - 17.5

MO = {61.2i + 81.6j} N # m

Ans.

Ans: MO = {61.2i + 81.6j} N # m MO = {61.2i + 81.6j} N # m

247

4–30. Determine the smallest force F that must be applied to the rope in order to create a moment of M = 900 N # m at point O.

z A F

10.5 m

SOLUTION

1.5 m

F = F£

4i - 3j - 9k 2(4)2 + ( -3)2 + ( - 9)2

≥ x

F = F(0.3885i - 0.29139j - 0.87416k) MO = F †

i 0 0.3885

j 0 - 0.29139

O 4m y

k 10.5 † -0.87416

MO = (3.05955i + 4.07940j)F F =

MO 2

2(3.05955) + (4.07940)

900 = 176 N # m 5.09925

Ans.

Ans: F = 176 N # m 248

4–31. Determine the moment of the force F about point P. Express the result as a Cartesian vector.

z P 2m

2m 3m

SOLUTION Position Vector: The coordinates of points A and P are A (3, 3, - 1) m and P ( -2, - 3, 2) m, respectively. Thus

1m A

F {2i 4j 6k} kN

rPA = [3 - ( - 2)]i + [3 - ( - 3)]j + ( - 1 - 2)k = {5i + 6j - 3k} m The Moment of F About Point P: MP = rPA * F i = †5 2

j 6 4

k -3 † -6

= { - 24i + 24j + 8k} kN # m

Ans.

Ans:

MP = { - 24i + 24j + 8k} kN # m

249

*4–32. tude If F = 5 50i + 60j + 30k6 lb, determine the magni and coordinate direction angles of the moment of F about point A.

F B

A 1 ft

3 ft y 458

2 ft x

SOLUTION Position Vector: rA = (1 - 0)i + 3(2 + 3 cos 45°) - 04j + (3 sin 45°)k = {1i + 4.1213j + 2.1213k} ft

Moment about Point A: M A = rA * F i = † 1 50

j 4.1213 60

k 2.1213 † 30

= 34.1213(30) - 60(2.1213) 4 i - 31(30) - 50(2.1213) 4j

+ 31(60) - 50(4.1213) 4k

= { - 3.640i + 76.066j - 146.066k} lb # ft

Magnitude: MA = 2( -3.640)2 + 76.0662 + ( - 146.066)2 = 165 lb # ft

Ans.

Coordinate direction angles:

cos a =

- 3.639 a = 91.3° 165

Ans.

cos b =

76.065 b = 62.5° 165

Ans.

cos g =

- 146.065 g = 152° 165

Ans.

Ans: a = 91.3° b = 62.5° g = 152° 250

4–33. Determine the coordinate direction angles of the force F applied at the end of the pipe such that the moment of F about point A is zero.

F B

A 1 ft

3 ft y 458

2 ft x

SOLUTION In order to create no moment about point A, the line of action of force F has to pass through point A. In this problem, the line of action of force F directed along rA creates no moment about point A. rA = (1 - 0)i + 3(2 + 3 cos 45°) - 04j + (3 sin 45°)k = {1i + 4.1213j + 2.1213k} ft

uA = uF =

1i + 4.1213j + 2.1213k

212 + 4.12132 + 2.12132

= 0.2109i + 0.8691j + 0.4474k Coordinate direction angles:

Ans.

cos a = 0.2109 a = 77.8° cos b = 0.8691 b = 29.6°

Ans.

cos g = 0.4474 g = 63.4°

Ans.

Ans: a = 77.8° b = 29.6° g = 63.4° 251

4–34. z

Determine the moment of the force of F = 600 N about point A.

x 6m

F C

SOLUTION

Position Vectors and Force Vector: The coordinates of points A, B and C are A (0, 0, 4) m, B (4 sin 45°, 0, 4 cos 45°) m and C (6, 6, 0) m, respectively. Thus

rAB = (4 sin 45° - 0)i + (0 - 0)j + (4 cos 45° - 4)k = {2.8284i - 1.1716k} m rAC = (6 - 0)i + (6 - 0)j + (0 - 4)k = {6i + 6j - 4k} m rBC = (6 - 4 sin 45°)i + (6 - 0)j + (0 - 4 cos 45°)k = {3.1716i + 6j - 2.8284k} m F = Fa

3.1716i + 6j - 2.8284k rBC b = 600£ ≥ rBC 23.17162 + 62 + ( -2.8284)2

= {258.82i + 489.63j - 230.81k} N The Moment of Force F About Point A: MA = rAB * F i = † 2.8284 258.82

j 0 489.63

k - 1.1716 † - 230.81

= {573.64i + 349.62j + 1384.89k} N # m = {574i + 350j + 1385k} N # m

Ans.

OR MA = rAC * F = †

i 6 258.82

j 6 489.63

k -4 † - 230.81

= {573.64i + 349.62j + 1384.89k} N # m Ans.

= {574i + 350j + 1385k}

Ans: MA = {574i + 350j + 1385k} N # m 252

4–35. z

Determine the smallest force F that must be applied along the rope in order to develop a moment of M = 1500 N # m at A.

x 6m

F C

SOLUTION

Position Vectors and Force Vector: The coordinates of points A, B and C are A (0, 0, 4) m, B (4 sin 45°, 0, 4 cos 45°) m and C (6, 6, 0) m, respectively.

Thus, rAB = (4 sin 45° - 0)i + (0 - 0)j + (4 cos 45° - 4)k = {2.8284i - 1.1716k} m rAC = (6 - 0)i + (6 - 0)j + (0 - 4)k = {6i + 6j - 4k} m rBC = (6 - 4 sin 45°)i + (6 - 0)j + (0 - 4 cos 45°)k = {3.1716i + 6j - 2.8284k} m F = Fa

3.1716i + 6j - 2.8284k rBC b = F£ ≥ rBC 23.17162 + 62 + ( -2.8284)2

= 0.4314F i + 0.8161Fj - 0.3847F k The Moment of Force F About Point A: MA = rAB * F i = † 2.8284 0.4314F

j 0 0.8161F

k - 1.1716 † -0.3847F

= 0.9561F i + 0.5827Fj + 2.3081F k OR MA = rAC * F = †

i 6 0.4314F

j 6 0.8161F

k -4 † -0.3847F

= 0.9561F i + 0.5827F j + 2.3081F k The magnitude of MA is MA = 2(MA)2x + (MA)2y + (MA)2z = 2(0.9561F)2 + (0.5827F)2 + (2.3081F)2

= 2.5654F It is required that MA = 1500 N # m, then 1500 = 2.5654F

Ans.

F = 584.71 N = 585 N

Ans: F = 585 N 253

*4–36. z

Determine the coordinate direction angles a, b, g of force F, so that the moment of F about O is zero. O

y 0.4 m A 0.5 m

0.3 m

SOLUTION

Position and Force Vectors: Let’s denote the point of application of force F as B. The coordinates of point B are B (0.4, 0.5, - 0.3) m. Thus, rOB = {0.4i + 0.5j - 0.3k} m u OB =

0.4i + 0.5j - 0.3k rOA 4 5 3 = = i + j k 2 2 2 rOA 150 150 150 20.4 + 0.5 + ( - 0.3)

rBO = { - 0.4i - 0.5j + 0.3k} m u BO =

- 0.4i - 0.5j + 0.3k rBO 4 5 3 = = i j + k 2 2 2 rBO 150 150 150 2( - 0.4) + ( - 0.5) + 0.3

Moment of F About Point O: To produce zero moment about point O, the line of action of F must pass through point O. Thus, F must directed from O to B (direction defined by uOB). Thus, 4 a ; = 55.56° = 55.6° 150 5 cos b = ; b = 45° 150 -3 cos g = ; g = 115.10° = 115° 150

Ans.

cos a = -

Ans. Ans.

OR F must directed from B to O (direction defined by uBO). Thus 4 ; a = 124.44° = 124° 150 5 cos b = ; b = 135° 150

Ans.

cos a = -

cos g =

Ans.

3 ; g = 64.90° = 64.9° 150

Ans.

Ans: a = 55.6° b = 45° g = 115° OR a = 124° b = 135° g = 64.9° 254

4–37. z

Determine the moment of force F about point O. The force has a magnitude of 800 N and coordinate direction angles of a = 60°, b = 120°, g = 45°. Express the result as a Cartesian vector.

y 0.4 m A 0.5 m

0.3 m

SOLUTION

Position and Force Vectors: Let’s denote the point of application of force F as B. The coordinates of point B are B (0.4, 0.5, - 0.3) m. Thus rOB = {0.4i + 0.5j - 0.3k} m F = FuF = 800 (cos 60°i + cos 120°j + cos 45°k) = {400i - 400j + 565.69k} N Moment of F About Point O: MO = rOB * F i = † 0.4 400

j 0.5 - 400

k -0.3 † 565.69

= {162.84 i - 346.27j - 360 k} N # m = {163i - 346j - 360k} N # m

Ans.

Ans: MO = {163i - 346j - 360k} N # m 255

4–38. z

Determine the moment of the force F about the door hinge at A. Express the result as a Cartesian vector.

4 ft

7 ft

F 80 lb

1.5 ft

SOLUTION Position Vectors and Force Vector: The coordinates of points A, C and D are A ( -6.5, - 3, 0) ft, C [0, - (3 + 4 cos 45°), 4 sin 45°] ft and D ( - 5, 0, 0) ft, respectively. Thus, rAC = [0 - ( -6.5)]i + [ - (3 + 4 cos 45°) - ( -3)] j + (4 sin 45° - 0)k

1.5 ft 3 ft x

5 ft

= {6.5i - 2.8284j + 2.8284k} ft rAD = [ -5 - ( - 6.5)]i + [0 - ( - 3)]j + (0 - 0)k = {1.5i + 3j} ft rCD = ( -5 - 0)i + {0 - [ - (3 + 4 cos 45°)]} j + (0 - 4 sin 45°)k = { - 5i + 5.8284j - 2.8284k} ft F = Fa

- 5i + 5.8284j - 2.8284k rCD b = 80 £ ≥ rCD 2( -5)2 + 5.82842 + ( -2.8284)2

= { - 48.88 i + 56.98j - 27.65k} lb Moment of F About Point A: MA = rAC * F = †

i 6.5 - 48.88

j -2.8284 56.98

k 2.8284 † - 27.65

= { - 82.9496i + 41.47j + 232.10k} lb # ft = { - 82.9i + 41.5j + 232k} lb # ft

Ans.

OR MA = rAD * F = †

i 1.5 - 48.88

j 3 56.98

k 0 † -27.65

= { - 82.9496i + 41.47j + 232.10 k} lb # ft = { - 82.9i + 41.5j + 232k} lb # ft

Ans.

Ans: MA = { - 82.9i + 41.5j + 232k} lb # ft 256

4–39. z

Determine the moment of the force F about the door hinge at B. Express the result as a Cartesian vector.

4 ft

7 ft

F 80 lb

1.5 ft

SOLUTION Position Vectors and Force Vector: The coordinates of points B, C and D are B ( - 1.5, -3, 0) ft, C [0, - (3 + 4 cos 45°), 4 sin 45°] ft and D ( - 5, 0, 0) ft, respectively. Thus, rBC = [0 - ( - 1.5)]i + [ - (3 + 4 cos 45°) - ( - 3)] j + (4 sin 45° - 0)k

1.5 ft 3 ft x

5 ft

= {1.5i - 2.8284j + 2.8284k} ft rBD = [ - 5 - ( - 1.5)]i + [0 - ( - 3)]j + (0 - 0)k = { -3.5i + 3j} ft rCD = ( - 5- 0)i + {0 - [ -(3 + 4 cos 45°)]} j + (0 - 4 sin 45°)k = { - 5i + 5.8284j - 2.8284k} ft F = Fa

- 5i + 5.8284j - 2.8284k rCD b = 80 £ ≥ rCD 2( -5)2 + 5.82842 + ( -2.8284)2

= { - 48.88 i + 56.98j - 27.65k} lb Moment of F About Point B: MB = rBC * F = †

i 1.5 - 48.88

j -2.8284 56.98

k 2.8284 † - 27.65

= { - 82.9496i - 96.77j - 52.78k} lb # ft = { - 82.9i - 96.8j - 52.8k} lb # ft

Ans.

or MB = rBD * F i = † - 3.5 - 48.88

j 3 56.98

k 0 † -27.65

= { - 82.9496i - 96.77j - 52.78 k} lb # ft = { - 82.9i - 96.8j - 52.8k} lb # ft

Ans.

Ans: MB = { - 82.9i - 96.8j - 52.8k} lb # ft 257

*4–40. The curved rod has a radius of 5 ft. If a force of 60 lb acts at its end as shown, determine the moment of this force about point C.

z C

5 ft

60°

5 ft

SOLUTION

y 60 lb

Position and Force Force Vector: Vector: Position Vectors Vector and

6 ft

rCA = 515 sin 60° - 02j + 15 cos 60° - 52k6 m

7 ft

= 54.330j - 2.50k6 m FAB = 60 ¢

16 - 02i + 17 - 5 sin 60°2j + 10 - 5 cos 60°2k 216 - 022 + 17 - 5 sin 60°22 + 10 - 5 cos 60°22

≤ lb

= 551.231i + 22.797j - 21.346k6 lb Point C: C: Applying Eq. 4–7, we have Moment of Force FAB About Point M C = rCA * FAB =

i 0 51.231

j 4.330 22.797

k -2.50 - 21.346

-35.4i - 128j - 222k lb # ft

Ans.

Ans: MC = { - 35.4i - 128j - 222k} lb # ft 258

4–41. z

Determine the smallest force F that must be applied along the rope in order to cause the curved rod, which has a radius of 5 ft, to fail at the support C. This requires a moment of M = 80 lb # ft to be developed at C.

5 ft

y 60 lb

SOLUTION

6 ft

Position and Force Force Vector: Vector: Position Vectors Vector and

7 ft

rCA = {(5 sin 60° - 0)j + (5 cos 60° - 5)k} m = {4.330j - 2.50 k} m FAB = F a

(6 - 0)i + (7 - 5 sin 60°)j + (0 - 5 cos 60°)k 2(6 - 0)2 + (7 - 5 sin 60°)2 + (0 - 5 cos 60°)2

b lb

= 0.8539Fi + 0.3799Fj - 0.3558Fk Moment of Force FAB About Point C: C: M C = rCA * FAB = 3

i 0 0.8539F

j 4.330 0.3799F

k - 2.50 3 - 0.3558F

= - 0.5909Fi - 2.135j - 3.697k Require 80 = 2(0.5909)2 + (- 2.135)2 + ( -3.697)2 F F = 18.6 lb.

Ans.

Ans: F = 18.6 lb 259

4–42. The curved rod has a radius of 4 ft. If the cable AB exerts a force of 80 lb on the rod, determine the moment of this force about point C. Solve the problem by using two different position vectors.

z B

4 ft C y

F 5 80 lb 4 ft

SOLUTION

Position Vector:

r1 = (0 - 0)i + (0 - 4) j + (4 - 0)k = { - 4j + 4k} ft r2 = (4 - 0)i + (0 - 4) j + (0 - 0)k = {4i - 4j} ft Force Vector: F = 80 £

(0 - 4)i + (0 - 0)j + (4 - 0)k 2(0 - 4)2 + (0 - 0)2 + (4 - 0)2

≥

= { - 56.569 i + 56.569k} lb Moment About Point C: M C = r1 * F = †

i 0 - 56.569

j -4 0

k 4 † 56.569

= [( - 4)(56.569) - (0)(4)]i - [(0)(56.569) - ( -56.569)(4)]j + [(0)(0) - ( -56.569)( -4)]k

= { - 226i - 226j - 226k} lb # ft

Ans.

Or M C = r2 * F = †

i 4 - 56.569

j -4 0

k 0 † 56.569

= [( - 4)(56.569) - (0)(0)]i - [(4)(56.569) - ( -56.569)(0)]j + [(4)(0) - ( -56.569)( -4)]k

= { - 226i - 226j - 226k} lb # ft

Ans.

Ans:

MC = { - 226i - 226j - 226k} lb # ft

260

4–43. The pipe assembly is subjected to the 80-N force. Determine the moment of this force about point A.

A 400 mm B x

SOLUTION

300 mm

Position Position Vector Vector and AndForce ForceVector: Vector:

200 mm

rAC = {(0.55 - 0)i + (0.4 - 0)j + ( -0.2 - 0)k} m

200 mm

250 mm

= {0.55i + 0.4j - 0.2k} m 40

F = 80(cos 30° sin 40°i + cos 30° cos 40°j - sin 30°k) N

= (44.53i + 53.07j - 40.0k} N

80 N

Moment of Force F About Point A: A: Applying Eq. 4–7, we have MA = rAC * F i = 3 0.55 44.53

j 0.4 53.07

k - 0.2 3 - 40.0

= {- 5.39i + 13.1j + 11.4k} N # m

Ans.

Ans: MA = {-5.39i + 13.1j + 11.4k} N # m 261

*4–44. The pipe assembly is subjected to the 80-N force. Determine the moment of this force about point B.

A 400 mm B x

SOLUTION

300 mm

Position Position Vector Vector and AndForce ForceVector: Vector:

200 mm

rBC = {(0.55 - 0) i + (0.4 - 0.4)j + ( -0.2 - 0)k} m

200 mm

250 mm

= {0.55i - 0.2k} m 40

F = 80 (cos 30° sin 40°i + cos 30° cos 40°j - sin 30°k) N

= (44.53i + 53.07j - 40.0k} N

80 N

B: Applying Eq. 4–7, we have Moment of Force F About Point B: MB = rBC * F i = 3 0.55 44.53

j 0 53.07

k - 0.2 3 - 40.0

= {10.6i + 13.1j + 29.2k} N # m

Ans.

Ans: MB = {10.6i + 13.1j + 29.2k} N # m 262

4–45. A force of F = 5 6i - 2j + 1k 6 kN produces a moment of M O = 54i + 5j - 14k6 kN # m about the origin of coordinates, point O. If the force acts at a point having an x coordinate of x = 1 m, determine the y and z coordinates. . Note: The figure shows F and MO in an arbitrary position.

z F

P MO z

O 1m

SOLUTION y

MO = r * F i 3 4i + 5j - 14k = 1 6

j y -2

k z3 1

4 = y + 2z 5 = - 1 + 6z -14 = - 2 - 6y y = 2m

Ans.

z = 1m

Ans.

Ans: y = 2m z = 1m 263

4–46. The force F = 56i + 8j + 10k6 N creates a moment about point O of M O = 5 -14i + 8j + 2k6 N # m. If the force passes through a point having an x coordinate of 1 m, determine the y and z coordinates of the point.Also, realizing that MO = Fd, determine the perpendicular distance d from point O to the line of action of F. Note: The figure shows F and MO in an arbitrary position.

z F

P MO z

O 1m

SOLUTION i -14i + 8j + 2k = 3 1 6

j y 8

k z 3 10

- 14 = 10y - 8z 8 = -10 + 6z 2 = 8 - 6y y = 1m

Ans.

z = 3m

Ans.

MO = 2( -14)2 + (8)2 + (2)2 = 16.25 N # m F = 2(6)2 + (8)2 + (10)2 = 14.14 N d =

16.25 = 1.15 m 14.14

Ans.

Ans: y = 1m z = 3m d = 1.15 m 264

4–47. The cable exerts a 140@N force on the telephone pole. Determine the moment of this force about point A. Solve the problem using two different position vectors.

B F 5 140 N

SOLUTION Position Vector:

rAB = {6k} m rAC = {2i - 3j} m A

Force Vector:

F = 140 £

(2 - 0)i + ( - 3 - 0)j + (0 - 6)k 2(2 - 0)2 + ( - 3 - 0)2 + (0 - 6)2

≥

= {40 i - 60j - 120k} N Moment about point A: MA = r * F Use r = rAB i MA = † 0 40

j 0 - 60

k 6 † -120

= [0( - 120) - ( -60)(6)]i - [0( - 120) - 40(6)]j + [0( - 60) - 40(0)]k = {360i + 240j} N # m

Ans.

Use r = rAC i MA = † 2 40

j -3 - 60

k 0 † - 120

= [( -3)( - 120) - ( - 60)(0)]i - [2( -120) - 40(0)]j + [2( - 60) - 40( -3)]k = {360i + 240j} N # m

Ans.

Ans:

MA = {360i + 240j} N # m

265

*4–48. Determine the moment about point O of each force acting on the pipe assembly. Add these moments and calculate the magnitude and coordinate direction angles of the resultant moment.

z F1 5 {30i 1 20j 2 30k} N

F2 5 {60i 2 10j 2 35k} N

0.8 m

O 1.20 m

0.4 m y

SOLUTION Position Vector:

r1 = (0 - 0)i + (1.2 - 0) j + (0.8 - 0)k = {1.2j + 0.8k} m r2 = ( -0.4 - 0) i + (1.2 - 0) j + (0.8 - 0)k = { -0.4i + 1.2j + 0.8k} m Moment about point O: M1 = r1 * F1 i = † 0 30

j 1.2 20

k 0.8 † - 30

= [1.2( -30) - 20(0.8)]i - [0( -30) - 30(0.8)]j + [0(20) - 30(1.2)]k = { -52i + 24j - 36k} N # m

Ans.

M2 = r2 * F2 i = † - 0.4 60

j 1.2 -10

k 0.8 † - 35

= [1.2( -35) - ( -10)(0.8)]i - [( -0.4)( - 35) - 60(0.8)]j + [( - 0.4)( - 10) - 60(1.2)]k

= { - 34i + 34j - 68k} N # m

Ans.

Resultant Moment About Point O: (MR)O = ΣMO; (MR)O = M1 + M2 = ( - 52i + 24j - 36k) + ( - 34i + 34j - 68k)

Magnitude:

= { - 86i + 58j - 104k} N # m

(MR)O = 2( - 86)2 + 582 + ( - 104)2 = 147 N # m

Ans.

Coordinate direction angles:

cos a =

- 86 a = 126° 147

Ans.

cos b =

58 b = 66.7° 147

Ans.

cos g =

- 104 g = 135° 147

Ans.

Ans:

M1 = { - 52i + 24j - 36k} N # m M2 = { - 34i + 34j - 68k} N # m (MR)O = 147 N # m a = 126° b = 66.7° g = 135°

266

4–49. Determine the moment of each force about point A. Add these moments and calculate the magnitude and coordinate direction angles of the resultant moment.

z F1 5 {30i 1 20j 2 30k} N

F2 5 {60i 2 10j 2 35k} N

0.8 m

O 1.20 m

0.4 m y

SOLUTION Position Vector:

r1 = (0 - 0)i + (1.2 - 0) j + (0.8 - 0.8)k = {1.2j} m r2 = ( -0.4 - 0) i + (1.2 - 0) j + (0.8 - 0.8)k = { - 0.4i + 1.2j} m Moment about point A: M1 = r1 * F1 i = † 0 30

j 1.2 20

k 0 † - 30

= [1.2( -30) - 20(0)]i - [0( - 30) - 30(0)]j + [0(20) - 30(1.2)]k = { - 36i - 36k} N # m

Ans.

M2 = r2 * F2 i = † - 0.4 60

j 1.2 -10

k 0 † - 35

= [1.2( -35) - ( -10(0)]i - [( - 0.4)( - 35) - 60(0)]j + [( - 0.4)( - 10) - 60(1.2)]k

= { - 42i - 14j - 68k} N # m

Ans.

Resultant Moment About Point A: (MR)A = ΣMA; (MR)A = M1 + M2 = ( - 36i - 36k) + ( -42i - 14j - 68k) Magnitude:

= { - 78i - 14j - 104k} N # m

(MR)A = 2( - 78)2 + ( - 14)2 + ( -104)2 = 131 N # m

Ans.

Coordinate direction angles:

- 78 a = 127° 131 - 14 cos b = b = 96.1° 131 - 104 cos g = g = 143° 131

Ans.

cos a =

Ans. Ans.

Ans:

M1 = { - 36i - 36k} N # m

M2 = { - 42i - 14j - 68k} N # m (MR)A = 131 N # m a = 127° b = 96.1° g = 143° 267

4–50. Using a ring collar the 75-N force can act in the vertical Using a ring collar the 75-N force can act in the vertical plane at various angles u. Determine the magnitude of the plane at various angles u. Determine the magnitude of moment it produces about point A, plt the result of the moment it produces about point A, plot the result of M (ordinate) verses u (abscissa) for 0° … u … 180°, and M (ordinate) versus u (abscissa) for 0° … u … 180°, and specify the angles that give the maximum and minimum specify the angles that give the maximum and minimum moment. moment.

A 2m

SOLUTION i MA = 3 2 0

j 1.5 75 cos u

1.5 m

x 75 N

k 3 0 75 sin u

= 112.5 sin u i - 150 sin u j + 150 cos u k MA = 2(112.5 sin u)2 + ( - 150 sin u)2 + (150 cos u)2 = 212 656.25 sin2 u + 22 500 1

dMA 1 2 = ( 12 656.25 sin2 u + 22 500 ) (12 656.25)(2 sin u cos u) = 0 du 2

sin u cos u = 0;

Ans.

u = 0°, 90°, 180°

Mmax = 187.5 N # m at u = 90° Mmin = 150 N # m at u = 0°, 180°

Ans: u = 0°, 90°, 180° 268

4–51. The curved pipe has a radius of 5 ft. If a force of 80 lb acts at its end as shown, determine the moment of this force about point C. Solve the problem by using two different position vectors.

z 2 ft

B 6 ft F 5 80 lb y

758

SOLUTION

5 ft A

Position Vector:

r1 = (0 - 5)i + (2 - 0) j + (6 - 0)k = { -5i + 2j + 6k} ft r2 = (5 cos 75° - 5) i + (5 sin 75° - 0) j + (8 - 0)k = { -3.706i + 4.8302j} ft

Force Vector: F = 80 °

(0 - 5 cos 75°)i + (2 - 5 sin 75°)j + (6 - 0)k 2(0 - 5 cos 75°)2 + (2 - 5 sin 75°)2 + (6 - 0)2

= { - 15.317i - 33.493j + 71.018k} lb Moment about point C: M C = r1 * F = †

i -5 - 15.317

j 2 - 33.493

k 6 † 71.018

= [2(71.018) - ( - 33.493)(6)]i - [( - 5)(71.018) - ( -15.317)(6)]j + [( - 5)( - 33.493) - ( -15.317)(2)]k

= {343i + 263j + 198k} lb # ft

Ans.

Or M C = r2 * F i = † - 3.706 - 15.317

j 4.830 - 33.493

k 0 † 71.018

= [4.830(71.018) - ( - 33.493)(0)]i - [( - 3.706)(71.018) - ( -15.317)(0)]j + [( - 3.706)( - 33.493) - ( -15.317)(4.830)]k

= {343i + 263j + 198k} lb # ft

Ans.

Ans: MC = {343i + 263j + 198k} lb # ft 269

*4–52. Determine the moment of force F about the x, y, and z axes. Solve the problem (a) using a Cartesian vector approach and (b) using a scalar approach.

F 5 {80i 2 40j 2 120k} N z

0.6 m

SOLUTION O

Vector Analysis Position Vector: x

r = ( - 0.3 - 0)i + (0.8 - 0)j + (0.6 - 0)k = ( - 0.3i + 0.8j + 0.6k) m

0.8 m

0.3 m y

Moment about x, y and z axes: 1 Mx = i # (r * F) = † - 0.3 80

0 0.8 -40

0 0.6 † - 120

= 1[0.8( - 120) - ( - 40)(0.6)] = -72 N # m

0 My = j # (r * F) = † - 0.3 80

1 0.8 -40

0 0.6 † - 120

= - 1[( - 0.3)( -120) - (80)(0.6)] = 12 N # m

0 Mx = k # (r * F) = † - 0.3 80

0 0.8 -40

Ans.

1 0.6 † - 120

= 1[( - 0.3)( - 40) - (80)(0.8)] = -52 N # m

Ans.

Negative signs indicate that the sense of Mx and Mz is opposite to their respective (positive) axes. Scalar Analysis Mx = ΣMx;

Mx = 40(0.6) - 120(0.8) = - 72 N # m

Ans.

My = ΣMy;

My = 80(0.6) - 120(0.3) = 12 N # m

Ans.

Mz = ΣMz;

Mz = 40(0.3) - 80(0.8) = -52 N # m

Ans.

Ans: Mx = - 72 N # m My = 12 N # m Mx = - 52 N # m 270

4–53. Determine the moment of force F about an axis extending between O and A. Express the result as a Cartesian vector.

F 5 {80i 2 40j 2 120k} N z

0.6 m

SOLUTION Position Vector:

r = { - 0.3i} m Unit vector along OA axis: uOA =

0.3 m

0.8 m

(0 - 0)i + (0.8 - 0)j + (0.6 - 0)k

2(0 - 0)2 + (0.8 - 0)2 + (0.6 - 0)2 = 0.8j + 0.6k Moment of The Force About OA Axis: MOA = uOA # (r * F) 0 MOA = † - 0.3 80

0.8 0 -40

0.6 0 † - 120

= 03(0)( -120) - ( -40)(0) 4 - (0.8) 3( - 0.3)( - 120) - 80(0) 4 = - 21.6 N # m

+ 0.63( - 0.3)( - 40) - 80(0) 4

Cartesian vector form: MOA = MOA uOA = - 21.6(0.8j + 0.6k)

= { - 17.3j - 13.0k} N # m

Ans.

Ans:

MAO = { - 17.3j - 13.0k} N # m

271

4–54. Determine the forcebyF about diagonal Determine the the moment moment ofproduced force Ftheabout the AF of the block.block. Express the the result as asa diagonal AF rectangular of the rectangular Express result Cartesian vector. a Cartesian vector.

SOLUTION

D O 3m

10k} N

A C

{ 6i

1.5 m G

3m F

Moment About Diagonal AF: AF: Either position vector rAB or rFB, Fig. a, can be used to find the moment of F about diagonal AF. rAB = (0 - 0)i + (3 - 0)j + (1.5 - 1.5)k = [3j] m rFB = (0 - 3)i + (3 - 3)j + (1.5 - 0)k = [- 3i + 1.5k] m The unit vector uAF , Fig. a, that specifies the direction of diagonal AF is given by uAF =

(3 - 0)i + (3 - 0)j + (0 - 1.5)k 2(3 - 0)2 + (3 - 0)2 + (0 - 1.5)2

2 1 2 i + j - k 3 3 3

The magnitude of the moment of F about diagonal AF axis is 2 3 MAF = uAF # rAB * F = 5 0 -6

2 3 3 3

1 3 0 5 10

2 1 2 [3(10) - (3)(0)]- [0(10) - ( - 6)(0)] + a - b[0(3) - (- 6)(3)] 3 3 3

= 14 N # m or 2 3 MAF = uAF # rAB * F = 5 - 3 -6

2 3 0 3

1 3 1.5 5 10 -

2 1 2 [0(10) - (3)(1.5)]- [( -3)(10) - ( - 6)(1.5)] + a - b( -3)(3) - ( - 6)(0)] 3 3 3

= 14 N # m Thus, MAF can be expressed in Cartesian vector form as 2 2 1 MAF = MAF uAF = 14 a i + j - k b = [9.33i + 9.33j - 4.67k] N # m 3 3 3

Ans.

Ans:

MAF = [9.33i + 9.33j - 4.67k] N # m

272

4–55. Determine the forcebyF about diagonal Determine the the moment moment ofproduced force Ftheabout the OD of the block.block. Express the the result as asa diagonal OD rectangular of the rectangular Express result Cartesian vector. a Cartesian vector.

SOLUTION

D O 3m

10k} N

A C

{ 6i

1.5 m G

3m F

Moment About Diagonal OD: Either position vector rOB or rDB, Fig. a, can be used to find the moment of F about diagonal OD. rOB = (0 - 0)i + (3 - 0)j + (1.5 - 0)k = [3j + 1.5j] m rDB = (0 - 3)i + (3 - 3)j + (1.5 - 1.5)k = [- 3i] m The unit vector uOD, Fig. a, that specifies the direction of diagonal OD is given by u OD =

(3 - 0)i + (3 - 0)j + (1.5 - 0)k 2(3 - 0)2 + (3 - 0)2 + (0 - 1.5)2

2 2 1 i + j - k 3 3 3

The magnitude of the moment of F about diagonal OD is 2 3 MOD = uOD # rOB * F = 5 0 -6 =

2 3 3 3

1 3 1.5 5 10 -

2 1 2 [3(10) - (3)(1.5)]- [0(10) - ( - 6)(1.5)] + [0(3) - ( -6)(3)] 3 3 3

= 17 N # m or

2 3 MOD = uOD # rDB * F = 5 - 3 -6 =

2 3 3 3

1 3 0 5 10

1 2 2 [0(10) - (3)(0)]- [- 3(10) - ( -6)(0)] + [- 3(3) - ( - 6)(0)] 3 3 3

= 17 N # m

Thus, MOD can be expressed in Cartesian vector form as 2 2 1 MOD = MOD uOD = 17 a i + j + k b = [11.3i + 11.3j + 5.67k] N # m 3 3 3

Ans.

Ans:

MOD = [11.3i + 11.3j + 5.67k] N # m

273

*4–56. The force of F = 80 lb acts along the edge DB of the tetrahedron. Determine the magnitude of the moment of this force about the edge CA.

z D

5 ft

15 ft

8 ft

SOLUTION

y 6 ft

Force Vector: F = 80¢

B 10 ft

(0 - 5)i + (20 - 10)j + (0 - 15)k 2(0 - 5)2 + (20 - 10)2 + (0 - 15)2

C 4 ft

≤ = { -21.381i + 42.762j - 64.143k} lb

Position Vector:

r = (5 - 0) i + (10 - 0) j + (15 - 0)k = (5i + 10j + 15k) ft Unit Vector along edge AC: uAC =

(13 - 0)i + (16 - 0)j + (0 - 0)k 2(13 - 0)2 + (16 - 0)2 + (0 - 0)2

= 0.6306i + 0.7761j

Magnitude of moment of the force about edge AC: MAC = uAC # (r * F) = †

0.6306 5 - 21.381

0.7761 10 42.762

0 15 † - 64.143

= 0.6306310( - 64.143) - (42.762)(15) 4

- 0.7761 [5( -64.143) - ( - 21.381)(15)] + 0

= -809 lb # ft

Ans.

Negative sign indicates that the sense of MAC is opposite to that of uAC.

Ans:

MAC = -809 lb # ft

274

4–57. If the moment of the force F about the edge CA of the tetrahedron has a magnitude of M = 200 lb # ft and is directed from C toward A, determine the magnitude of F.

z D

5 ft

15 ft

8 ft

SOLUTION

B 10 ft

y 6 ft

Force Vector: F = F¢

(0 - 5)i + (20 - 10)j + (0 - 15)k 2(0 - 5)2 + (20 - 10)2 + (0 - 15)2

C 4 ft

≤

= { - 0.2673F i + 0.5345F j - 0.8018F k} lb Position Vector:

r = (5 - 0) i + (10 - 0) j + (15 - 0)k = (5i + 10j + 15k) ft Unit vector along edge CA: uCA =

(0 - 13)i + (0 - 16)j + (0 - 0)k 2(0 - 13)2 + (0 - 16)2 + (0 - 0)2

= 0.6306i + 0.7761j

Magnitude of moment of the force about edge AC: MAC = uAC # (r * F) 200 = †

- 0.6306 5 - 0.2673F

-0.7661 10 0.5345F

0 † 15 - 0.8018F

200 = - 0.6306310( -0.8018F) - (0.5345F)(15) 4

- ( - 0.7761) 35( - 0.8018F) - ( -0.2673F)(15) 4 + 0 F = 19.8 lb

Ans.

Ans: F = 19.8 lb 275

4–58. The cross luglug wrench in the The board board is is used used to tohold holdthe theend endof ofthe a four-way wrench in position shown when the man applies a force of F = 100 the position shown when the man applies a force of F N. = Determine the magnitude of theofmoment produced by this N. Determine the magnitude the moment produced by 100 force about the xthe axis. Force F liesF in plane. this force about x axis. Force liesa vertical in a vertical plane.

F 60

SOLUTION

250 mm y

250 mm

Vector Analysis Vector Analysis Moment About the x Axis: The position vector rAB, Fig. a, will be used to determine the moment of F about the x axis. rAB = (0.25 - 0.25)i + (0.25 - 0)j + (0 - 0)k = {0.25j} m The force vector F, Fig. a, can be written as F = 100(cos 60°j - sin 60°k) = {50j - 86.60k} N Knowing that the unit vector of the x axis is i, the magnitude of the moment of F about the x axis is given by 1 Mx = i # rAB * F = 3 0 0

0 0.25 50

0 0 3 - 86.60

= 1[0.25(- 86.60) - 50(0)] + 0 + 0 = - 21.7 N # m

Ans.

The negative sign indicates that Mx is directed towards the negative x axis. Scalar Scalar Analysis Analysis This problem can be solved by summing the moment about the x axis Mx = ©Mx;

Mx = -100 sin 60°(0.25) + 100 cos 60°(0) = - 21.7 N # m Ans.

Ans: Mx = 21.7 Ν # m 276

4–59. The board is used to hold hold the the end end of of athe cross lug four-way lugwrench wrenchin # m about # mN about the position shown. If a torque x axisisis in position. If a torque of 30ofN30 thethe x axis required to required magnitude to tighten tightenthe thenut, nut,determine determinethe the required magniof theofforce F needed turn the wrench. Force F lies in a tude the force F thattothe man’s foot must apply on the vertical plane. end of the wrench in order to turn it. Force F lies in a vertical plane.

F 60

250 mm y

250 mm

SOLUTION Vector Vector Analysis Analysis Moment Axis: The position vector rAB, Fig. a, will be used to deterMoment About About the x Axis: mine the moment of F about the x axis. rAB = (0.25 - 0.25)i + (0.25 - 0)j + (0 - 0)k = {0.25j} m The force vector F, Fig. a, can be written as F = F(cos 60°j - sin 60°k) = 0.5Fj - 0.8660Fk Knowing that the unit vector of the x axis is i, the magnitude of the moment of F about the x axis is given by 1 Mx = i # rAB * F = 0 0

0 0.25 0.5F

0 0 - 0.8660F

= 1[0.25( -0.8660F) - 0.5F(0)] + 0 + 0 Ans.

= - 0.2165F

The negative sign indicates that Mx is directed towards the negative x axis. The magnitude of F required to produce Mx = 30 N # m can be determined from 30 = 0.2165F Ans.

F = 139 N Scalar Scalar Analysis Analysis This problem can be solved by summing the moment about the x axis Mx = ©Mx;

- 30 = - F sin 60°(0.25) + F cos 60°(0) Ans.

F = 139 N

Ans: F = 139 Ν 277

*4–60. The bevel gear is subjected to the force F which is caused from contact with another gear. Determine the moment of this force about the y axis of the gear shaft.

z F {20i 8j 15k} N

40 mm 30 mm

SOLUTION u = j F = 20 i + 8 j - 15 k

r = - 0.04 i + 0.03 k 0 My = 3 - 0.04 20

1 0 8

0 0.03 3 = 0 - 15

Ans.

Ans: My = 0 278

4–61. Determine the magnitude of the moment of the force F = {50i - 20j - 80k} N about the base line AB of the tripod.

z F D 4m

2m 1.5 m

SOLUTION uAB =

{3.5i + 0.5j}

2(3.5)2 + (0.5)2

0.9899 MAB = uAB # ( rAD * F ) = † 2.5 50 MAB = 136 N # m

0.1414 0 -20

2.5 m

uAB = {0.9899i + 0.1414j} 0 4 † -80

0.5 m

1m B

Ans.

Ans: MAB = 136 N # m 279

4–62. Determine the magnitude of the moment of the force F = {50i - 20j - 80k} N about the base line BC of the tripod.

z F D 4m

2m 1.5 m

SOLUTION uBC =

{ -1.5i - 2.5j}

2( - 1.5)2 + ( - 2.5)2

uBC = { - 0.5145i - 0.8575j}

- 0.5145 MBC = uBC # ( rCD * F ) = † 0.5 50 MBC = 165 N # m

- 0.8575 2 - 20

2.5 m

0 4 † - 80

0.5 m

1m B

Ans.

Ans: MBC = 165 N # m 280

4–63. Determine the magnitude of the moment of the force F = {50i - 20j - 80k} N about the base line CA of the tripod.

z F D 4m

2m 1.5 m

SOLUTION uCA =

{ -2i + 2j}

2( - 2)2 + (2)2

uCA = { -0.707i + 0.707j} MCA = uCA # ( rAD * F ) = † MCA = 226 N # m

- 0.707 2.5 50

0.707 0 -20

2.5 m

0 4 † - 80

0.5 m

1m B

Ans.

Ans: MCA = 226 N # m 281

*4–64. Determine the moment of each force acting on the handle of the wrench about the a axis.Take F1 = 5- 2i + 4j - 8k6 lb, F2 = 53i + 2j - 6k6 lb.

z a

SOLUTION ua = sin 45° i + cos 45°k

= 0.7071 i + 0.7071 k

4 in.

6 in. y

rCA = 4 cos 45° i - 4 sin 45° k

3.5 in.

= 2.828 i - 2.828 k rCB = 7.5 cos 45°i - 7.5 sin 45°k = 5.303 i - 5.303 k 0.7071 (Ma)1 = 3 5.303 -2

0 0 4

0.7071 - 5.303 3 = 30 lb # in. -8

Ans.

0.7071 (Ma)2 = 3 2.828 3

0 0 2

0.7071 - 2.828 3 = 8 lb # in. -6

Ans.

Ans: (Ma)1 = 30 lb # in. (Ma)2 = 8 lb # in. 282

4–65. Determine the moment of each force acting on the handle of the wrench about the z axis.Take F1 = 5 - 2i + 4j - 8k6 lb, F2 = 53i + 2j - 6k6 lb.

z a

SOLUTION u = k

rA = (6 sin 45° + 4 cos 45°) i + (6 cos 45° - 4 sin 45°) k

6 in. y

= 7.0711 i + 1.4142 j

4 in.

3.5 in.

rB = (6 sin 45° + 7.5 cos 45°)i + (6 cos 45° - 7.5 sin 45°)k = 9.546 i - 1.0607 k 0 (Ma)1 = 3 9.546 -2

0 0 4

1 - 1.0607 3 = 38.2 lb # in. -8

Ans.

0 (Ma)2 = 3 7.0711 3

0 0 2

1 1.4142 3 = 14.1 lb # in. -6

Ans.

Ans: (Ma)1 = 38.2 lb # in. (Ma)2 = 14.1 lb # in. 283

4–66. The Rollerball rollerball skate is an in-line tandem skate that uses two large spherical wheels on each skate, rather than traditional wafer-shape wheels. During skating the two forces acting on the wheel of one skate consist of a 78-lb normal force and a 13-lb friction force. Determine the moment of both of these forces about the axle AB of the wheel.

SOLUTION

30 A

uAB = cos 30° i + cos 120° j

1.25 in.

= 0.8660 i - 0.5 j

r = {- 1.25 j} in.

13 lb 78 lb

F = {13 i + 78 j} lb 0.866 MAB = 3 0.625 13

- 0.5 - 1.25 78

0 03 = 0 0

Ans.

Also, by inspection, both of these forces pass through the AB axis since they lie in the plane of the axis. Ans.

MAB = 0

Ans: MAB = 0 284

4–67. z

Determine the resultant moment of the two forces about the Oa axis. Express the result as a Cartesian vector.

F1 = 80 lb

45˚

60˚

120˚

F2 = 50 lb

6 ft O 5 ft

4 ft

y 30˚

30˚ a x

SOLUTION F1 = 80(cos 120° i + cos 60° j + cos 45° k) = 5 - 40 i + 40 j + 56.569 k6 lb

F2 = 550 k6 lb

r1 = (4 sin 30° - 0) i + (4 cos 30° - 0) j + (6 - 0) k = 5 2 i + 3.464 j + 6 k 6 ft

r2 = ( -5 sin 30°) j = 5 - 2.5 j 6 ft

MR = r1 * F1 + r2 * F2 i = † 2 - 40

j 3.464 40

k 6 † + ( - 2.5 j) * (50 k) 56.569

= [3.464(56.569) - 40(6)] i - [2(56.569) - ( - 40)(6)] j + [2(40) - ( -40)(3.464)] k - 125 i = 5 -169.044 i - 353.138 j + 218.560 k6 lb # ft

uOa = cos 30° i - sin 30° j = 0.8660 i - 0.5 j

(MR)Oa = uOa # MR = (0.8660 i - 0.5 j) # ( - 169.044 i - 353.138 j + 218.560 k) = (0.8660)( - 169.044) + ( -0.5)( -353.138) + 0(218.560) = 30.173 lb # ft (MR)Oa = (MR)OauOa = 30.173(0.8660 i - 0.5 j) = 5 26.1 i - 15.1 j 6 lb # ft

Ans.

Ans: (MR)Oa = {26.1i - 15.1j} lb # ft 285

*4–68.

The pipe assembly is secured on the wall by the two brackets. If the flower pot has a weight of 50 lb, determine the magnitude of the moment produced by the weight about the OA axis.

4 ft A O 60

3 ft

4 ft 3 ft

SOLUTION

Moment About About the the OA OA Axis: axis: The coordinates of point B are [(4 + 3 cos 30°) cos 60°, (4 + 3 cos 30°) sin 60°, 3 sin 30°]ft = (3.299, 5.714, 1.5) ft. Either position vector rOB or rAB can be used to determine the moment of W about the OA axis. rOB = (3.299 - 0)i + (5.714 - 0)j + (1.5 - 0)k = [3.299i + 5.714j + 1.5k] ft rAB = (3.299 - 0)i + (5.714 - 4)j + (1.5 - 3)k = [3.299i + 1.714j - 1.5k] ft Since W is directed towards the negative z axis, we can write W = [- 50k] lb The unit vector uOA, Fig. a, that specifies the direction of the OA axis is given by uOA =

(0 - 0)i + (4 - 0)j + (3 - 0)k 2(0 - 0)2 + (4 - 0)2 + (3 - 0)2

3 4 j + k 5 5

The magnitude of the moment of W about the OA axis is given by 0 MOA = uOA # rOB * W = 5 3.299 0

= 0 -

4 5 5.714 0

3 5 1.5 5 - 50

3 4 [3.299( -50) - 0(1.5)] + [3.299(0) - 0(5.714)] 5 5

= 132 lb # ft

Ans.

or 0 MOA = uOA # rAB * W = 5 3.299 0

= 0 -

4 5 1.714 0

3 5 - 1.5 5 - 50

4 3 [3.299( - 50) - 0( - 1.5)] + [3.299(0) - 0(1.714)] 5 5

= 132 lb # ft

Ans.

Ans: MOA = 132 lb # ft 286

4–69. The pipe assembly is secured on the wall by the two brackets. If the frictional force of both brackets can resist a maximum moment of 150 lb # ft, determine the largest weight of the flower pot that can be supported by the assembly.without causing it to rotate about the OA axis.

z 4 ft A O 60

SOLUTION

3 ft

4 ft 3 ft

Moment About About the theOA OAAxis: axis: The coordinates of point B are

[(4 + 3 cos 30°) cos 60°, (4 + 3 cos 30°) sin 60°, 3 sin 30°]ft = (3.299, 5.174, 1.5) ft. Either position vector rOB or rOC can be used to determine the moment of W about the OA axis. rOA = (3.299 - 0)i + (5.714 - 0)j + (1.5 - 0)k = [3.299i + 5.714j + 1.5k] ft rAB = (3.299 - 0)i + (5.714 - 4)j + (1.5 - 3)k = [3.299i + 1.714j - 1.5k] ft Since W is directed towards the negative z axis, we can write W = -Wk The unit vector uOA, Fig. a, that specifies the direction of the OA axis is given by (0 - 0)i + (4 - 0)j + (3 - 0)k

uOA =

2(0 - 0) + (4 - 0) + (3 - 0)

4 3 j + k 5 5

Since it is required that the magnitude of the moment of W about the OA axis not exceed 150 ft # lb, we can write MOA = uOA # rOB * W 0

150 = 5 3.299 0

150 = 0 -

4 5 5.714 0

3 5 1.5 5 -W

3 4 [3.299( -W) - 0(1.5)] + [3.299(0) - 0(5.174)] 5 5 Ans.

W = 56.8 lb or MOA = uOA # rOB * W 0

150 = 5 3.299 0

150 = 0 -

4 5 5.714 0

3 5 0 5 -W

4 3 [3.299( -W) - 0(0)] + [3.299(0) - 0(5.714)] 5 5 Ans.

W = 56.8 lb

Ans: W = 56.8 lb 287

4–70. The tool is used to shut off gas valves that are difficult to access. If the force F is applied to the handle, determine the component of the moment created about the z axis of the valve.

0.25 m

{ 60i

20j

15k} N

SOLUTION u = k

0.4 m

r = 0.25 sin 30°i + 0.25 cos 30°j

= 0.125i + 0.2165j 0 3 Mz = 0.125 - 60

0 0.2165 20

1 0 3 = 15.5 N # m 15

Ans.

Ans: Mz = 15.5 N # m 288

4–71. z

The A-frame is being hoisted into an upright position by the vertical force of F = 80 lb. Determine the moment of this force about the y′ axis passing through points A and B when the frame is in the position shown.

F C A x¿

6 ft x

SOLUTION

6 ft

y B y¿

Scalar Analysis:

My′ = 80 (6 cos 15°) = 464 lb # ft

Vector analysis : uAB = cos 60° i + cos 30° j Coordinates of point C: x = 3 sin 30° - 6 cos 15° cos 30° = - 3.52 ft y = 3 cos 30° + 6 cos 15° sin 30° = 5.50 ft z = 6 sin 15° = 1.55 ft rAC = - 3.52 i + 5.50 j + 1.55 k F = 80 k sin 30° My′ = † - 3.52 0

My′ = 464 lb # ft

cos 30° 5.50 0

0 1.55 † 80 Ans.

Ans: My′ = 464 lb # ft 289

*4–72. z

The A-frame is being hoisted into an upright position by the vertical force of F = 80 lb. Determine the moment of this force about the x axis when the frame is in the position shown.

F C A x¿

6 ft x

SOLUTION

6 ft

y B y¿

Using x′, y′, z : ux = cos 30° i′ + sin 30° j′ rAC = - 6 cos 15° i′ + 3 j′ + 6 sin 15° k F = 80 k cos 30° Mx = † - 6 cos 15° 0 Mx = 440 lb # ft

sin 30° 3 0

0 6 sin 15° † = 207.85 + 231.82 + 0 80

Also, using x, y, z, Coordinates of point C: x = 3 sin 30° - 6 cos 15° cos 30° = - 3.52 ft y = 3 cos 30° + 6 cos 15° sin 30° = 5.50 ft z = 6 sin 15° = 1.55 ft rAC = - 3.52 i + 5.50 j + 1.55 k F = 80 k 1 Mx = † - 3.52 0

0 5.50 0

0 1.55 † = 440 lb # ft 80

Ans.

Ans: Mx = 440 lb # ft 290

4–73. The crossbar wrench is used to remove a lug nut from the automobile wheel. The mechanic applies a couple to the wrench such that his hands are a constant distance apart. Is it necessary that a b in order to produce the most effective turning of the nut? Explain. Also, what is the effect of changing the shaft dimension c in this regard? The forces act in the vertical plane.

–F

SOLUTION

Couple moment, MC = F (a + b) a

The couple moment depends on the total distance between grips, not a = b. No.

Ans.

Changing dimension c has no effect on turning the nut.

Ans.

Ans: No. Changing dimension c has no effect on turning the nut. 291

4–74. The hanger H is used to support the end of the floor beam B. If the loading on the hanger consists of two couples, determine the magnitude of the horizontal forces F and -F so that the resultant couple moment on the hanger is zero.

H B

1 in. 1.5 in. 1 in. 2F

SOLUTION a+ ΣM = 0; F(5) - 150(2.25) = 0 F = 67.5 lb

Ans.

6 in.

150 lb 150 lb F 0.75 in.

Ans: F = 67.5 lb 292

4–75. Determine the magnitude of the forces F and - F, so that the resultant couple moment is 400 N # m clockwise.

600 N 250 N 408

600 N

408

SOLUTION a+ MR = ΣM; - 400 = 600 a

250 N

0.5 0.5 b - Fa b - 250(1) cos 40° cos 40° F = 830 N

Ans.

Ans: F = 830 N 293

*4–76. Two couples on the beam. If F = 125 lb , determine 125 lb,act determine the resultant couple moment. the resultIf F = ant couple moment.

200 lb

F 30 1.25 ft

1.5 ft

SOLUTION

200 lb

125 lb couple is resolved in to their horizontal and vertical components as shown in Fig. a.

2 ft

a + (MR)C = 200(1.5) + 125 cos 30° (1.25) = 435.32 lb # ft = 435 lb # ftd

Ans.

Ans: (MR)C = 435 lb # ft d 294

4–77. Determine of FDetermine so that thethe resultant couple Two couplesthe actmagnitude on the beam. magnitude of # ft, counterclockwise. # ft, moment is 450 Whereison 450 the lb beam F so that thelb resultant couple moment does the resultant couple act? does the resultant counterclockwise. Where moment on the beam couple moment act?

200 lb

F 30 1.25 ft

1.5 ft

200 lb 2 ft

SOLUTION a +MR = ©M ;

450 = 200(1.5) + Fcos 30°(1.25) Ans.

F = 139 lb

The resultant couple moment is a free vector. It can act at any point on the beam.

Ans: F = 139 lb 295

4–78. The road exerts a torque of MA = 400 N # m and MB = 200 N # m on the brushes of the road sweeper. Determine the magnitude of the couple forces that are developed by the road on the rear wheels of the sweeper, so that the resultant couple moment on the sweeper is zero. What is the magnitude of these forces if the brush at B is turned off?

1.5 m

B 2F

SOLUTION Both brushes are on: Ans.

a+ 400 - 200 - F(1.5) = 0 F = 133 N When the brush B is off:

Ans.

a+ 400 - F(1.5) = 0 F = 267 N

Ans: F = 133 N F = 267 N 296

4–79. The cord passing over the two small pegs A and B on the board is subjected to a tension of 10 lb. Determine the minimum tension P and the orientation u of the cord passing over pegs C and D, so that the resultant couple produced by both cords is 20 lb · in. clockwise.

B 15 in.

308

2P 10 lb 10 lb 458

308 A

P 15 in.

u D

SOLUTION In order to yield a maximum counterclockwise couple moment a minimum force P which acts perpendicular to CD is needed. Thus Ans.

u = 45°

a+ MR = ΣM; - 20 = P(30) - 10 cos 15°(30) Ans.

P = 8.99 lb

Ans: u = 45° P = 8.99 lb 297

*4–80. The man tries to open the valve by applying the couple forces of F = 75 N to the wheel. Determine the couple moment produced.

150 mm

SOLUTION a + Mc = ©M;

Mc = - 75(0.15 + 0.15) = - 22.5 N # m = 22.5 N # m b

Ans. F

Ans: MC = 22.5 N # mb 298

4–81. If the valve can be opened with a couple moment of 25 N # m, determine the required magnitude of each couple force which must be applied to the wheel.

150 mm

SOLUTION a + Mc = ©M;

- 25 = -F(0.15 + 0.15) F = 83.3 N

Ans. F

Ans: F = 83.3 N 299

4–82. Two couples act on the beam. If the resultant couple is to be zero, determine the magnitudes of P and F, and the distance d between A and B.

308 0.2 m

300 N

500 N

608

SOLUTION

In order to have two couples, F = 300 N

Ans.

P = 500 N

Ans.

c+ MR = 500 (2) - 300 cos 30° (d) + 300 sin 30° (0.2) = 0 Ans.

d = 3.96 m

Ans: F = 300 N P = 500 N d = 3.96 m 300

4–83. Three couple moments act on the pipe assembly. Determine the magnitude of the resultant couple moment if M2 = 50 N # m and M3 = 35 N # m.

1208

M3 1208

SOLUTION + M = 50 sin 30° + 80 sin 30° - 35 S Rx′

M15 80 N ? m

= 30 N # m

+ c MRy′ = 50 cos 30° - 80 cos 30° = - 25.98 N # m

MR = 2(30)2 + ( - 25.98)2 = 39.7 N # m

Ans.

Ans: MR = 39.7 N # m 301

*4–84. Three couple moments act on the pipe assembly. Determine the magnitudes of M2 and M3 so that the resultant couple moment is zero.

1208

SOLUTION + c MRy′ = 0; + M = 0; S Rx′

1208

M2 cos 30° - 80 cos 30° = 0 M2 = 80 N # m

Ans.

80 sin 30° + 80 sin 30° - M3 = 0

M15 80 N ? m

M3 = 80 N # m

Ans.

Ans: M3 = 80 N # m 302

4–85. Two couples act on the frame. If the resultant couple moment is to be zero, determine the distance d between the 100-lb couple forces.

100 lb 150 lb

308 3 ft

3 ft

a+ MR = 0 = ΣM; 0 = 100 cos 30°(d) -

4 (150)(4) 5

4 ft 308

SOLUTION

5 3 4

100 lb

150 lb

5 3 4

Ans.

d = 5.54 ft

Ans: d = 5.54 ft 303

4–86. Two couples act on the frame. If d = 4 ft, determine the resultant couple moment by (a) summing the moments of the two couples and (b) resolving each force into x and y components and then summing the moments about A of all the force components.

100 lb 150 lb

308 3 ft

3 ft

(a)

M2 =

100 lb

150 lb

M1 = 100 cos 30°(4) = 346.4 lb # ft d

4 ft 308

SOLUTION

5 3 4

4 (150)(4) = 480 lb # ft c 5

MR = - M2 + M1 = - 480 + 346.4 = 134 lb # ft c

Ans.

(b)

a+ MR = ΣMA; MR = 100 cos 30°(3) +

3 4 (150)(10) + (150)(4) 5 5 - 100 cos 30°(7) -

MR = 134 lb # ft c

3 (150)(10) 5 Ans.

Ans: MR = 134 lb # ft c 304

4–87. Two couples act on the frame. If d = 6 ft, determine the resultant couple moment by (a) summing the moments of the two couples and (b) resolving each force into x and y components and then summing the moments about B of all the force components.

100 lb 150 lb

308 3 ft

3 ft

(a)

M2 =

100 lb

150 lb

M1 = 100 cos 30°(6) = 519.6 lb # ft d

4 ft 308

SOLUTION

5 3 4

4 (150)(4) = 480 lb # ft c 5

MR = M1 - M2 = 519.6 - 480 = 39.6 lb # ft d

Ans.

(b) c+ MR = ΣMB; MR = 100 cos 30°(3) +

4 (150)(4) - 100 cos 30°(9) 5

MR = - 39.6 lb # ft = 39.6 lb # ft d

Ans.

Ans: MR = 39.6 lb # ft d 305

*4–88. z

Express the moment of the couple acting on the pipe assembly in Cartesian vector form. What is the magnitude of the couple moment?

20 lb x

3 ft B 1 ft 1.5 ft 20 lb 2 ft 1 ft

SOLUTION

rCB = { -3i - 2.5j} ft MC = rCB * F i = † -3 0

j -2.5 0

k 0 † 20

MC = { -50i + 60j} lb # ft

Ans.

MC = 2( -50) + (60) = 78.1 lb # ft

Ans.

Ans: MC = 78.1 lb # ft 306

4–89. If M1 = 180 lb # ft, M2 = 90 lb # ft, and M3 = 120 lb # ft, determine the magnitude and coordinate direction angles of the resultant couple moment.

z 150 lb ft M3 1 ft 2 ft

SOLUTION Since the couple moment is a free vector, it can act at any point without altering its effect. Thus, the couple moments M1, M2, M3, and M4 acting on the gear deducer can be simplified, as shown in Fig. a. Expressing each couple moment in Cartesian vector form,

2 ft

45 45

x 2 ft

y 3 ft

M2 M1

M 1 = [180j]lb # ft M 2 = [-90i]lb # ft M 3 = M3u = 120 C

(2 - 0)i + ( - 2 - 0)j + (1 + 0)k 2(2 - 0)2 + ( -2 - 0)2 + (1 - 0)2

S = [80i - 80j + 40k]lb # ft

M 4 = 150[cos 45° sin 45°i - cos 45° cos 45°j - sin 45°k] = [75i - 75j - 106.07k]lb # ft The resultant couple moment is given by (M c)R = M 1 + M 2 + M 3 + M 4

(M c)R = ©M;

= 180j - 90i + (80i - 80j + 40k) + (75i - 75j - 106.07k) = [65i + 25j - 66.07k]lb # ft The magnitude of (M c)R is (Mc)R = 2[(Mc)R]x 2 + [(Mc)R]y 2 + [(Mc)R]z 2 = 2(65)2 + (25)2 + (- 66.07)2 = 95.99 lb # ft = 96.0 lb # ft

Ans.

The coordinate angles of (M c)R are a = cos - 1 ¢ b = cos - 1 ¢ g = cos - 1 ¢

[(Mc)R]x 65 b = 47.4° ≤ = cos a (Mc)R 95.99 [(Mc)R]y (Mc)R

[(Mc)R]z (Mc)R

≤ = cos a

Ans.

25 b = 74.9° 95.99

Ans.

- 66.07 b = 133° 95.99

Ans.

Ans: MR = 96.0 lb # ft, a = 47.4°, b = 74.9°, g = 133° 307

4–90. Determine the magnitudes of couple moments M1, M2, and M3 so that the resultant couple moment is zero.

z 150 lb ft M3 1 ft 2 ft

2 ft

45 45

SOLUTION

2 ft

Since the couple moment is a free vector, it can act at any point without altering its effect. Thus, the couple moments M1, M2, M3, and M4 acting on the gear deducer can be simplified, as shown in Fig. a. Expressing each couple moment in Cartesian vector form,

y 3 ft

M2 M1

M 1 = M1j M 2 = -M2i M 3 = M3u = M3 C

(2 - 0)i + ( -2 - 0)j + (1 + 0)k 2(2 - 0)2 + ( -2 - 0)2 + (1 - 0)2

S =

2 2 1 M i - M3j + M3k 3 3 3 3

M 4 = 150[cos 45° sin 45°i - cos 45° cos 45°j - sin 45°k] = [75i - 75j - 106.07k]lb # ft The resultant couple moment is required to be zero. Thus, 0 = M1 + M2 + M3 + M4

(M c)R = ©M;

2 2 1 0 = M1j + (-M2i) + a M3i - M3j + M3kb + (75i - 75j - 106.07k) 3 3 3 0 = a -M2 +

Equating the i, j, and k components,

0 = - M2 + 0 = M1 0 =

2 2 1 M + 75b i + aM1 - M3 - 75b j + a M3 - 106.07b k 3 3 3 3

2 M + 75 3 3

(1)

2 M - 75 3 3

(2)

1 M - 106.07 3 3

(3)

Solving Eqs. (1), (2), and (3) yields

M3 = 318 lb # ft

Ans.

M1 = M2 = 287 lb # ft

Ans.

Ans: M3 = 318 lb # ft, M1 = M2 = 287 lb # ft 308

4–91. Determine the magnitude and coordinate direction angles of the resultant couple moment.

40 lb ft

z 20

30 15

30 lb ft

SOLUTION M1 = 40 cos 20° sin 15° i + 40cos 20° cos 15° j - 40 sin 20° k

= 9.728 i + 36.307 j - 13.681 k M 2 = -30 sin 30° i + 30 cos 30° j = -15 i + 25.981 j MR = M1 + M2 = - 5.272 i + 62.288 j - 13.681 k MR = 2(- 5.272)2 + (62.288)2 + ( -13.681)2 = 63.990 = 64.0 lb # ft

Ans.

a = cos-1 a

-5.272 b = 94.7° 63.990

Ans.

b = cos-1 a

62.288 b = 13.2° 63.990

Ans.

g = cos-1 a

-13.681 b = 102° 63.990

Ans.

Ans: MR = 64.0 lb # ft a = 94.7° b = 13.2° g = 102° 309

*4–92. If F = 80 N, determine the magnitude and coordinate direction angles of the couple moment. The pipe assembly lies in the x–y plane.

300 mm 300 mm F

x 200 mm

SOLUTION It is easiest to find the couple moment of F by taking the moment of F or –F about point A or B, respectively, Fig. a. Here the position vectors rAB and rBA must be determined first.

200 mm

300 mm

rAB = (0.3 - 0.2)i + (0.8 - 0.3)j + (0 - 0)k = [0.1i + 0.5j] m rBA = (0.2 - 0.3)i + (0.3 - 0.8)j + (0 - 0)k = [ -0.1i - 0.5j] m The force vectors F and –F can be written as F = {80 k} N and - F = [-80 k] N Thus, the couple moment of F can be determined from i M c = rAB * F = 3 0.1 0

j 0.5 0

i Mc = rBA * -F = 3 -0.1 0

j - 0.5 0

k 0 3 = [40i - 8j] N # m 80

or k 0 3 = [40i - 8j] N # m - 80

The magnitude of Mc is given by Mc = 2Mx 2 + My 2 + Mz 2 = 2402 + ( -8)2 + 02 = 40.79 N # m = 40.8 N # m

Ans.

The coordinate angles of Mc are a = cos - 1 ¢ b = cos - 1 ¢ g = cos - 1 ¢

Mx 40 ≤ = cos ¢ ≤ = 11.3° M 40.79 My M Mz M

-8 ≤ = 101° 40.79

Ans.

0 ≤ = 90° 40.79

Ans.

≤ = cos ¢

Ans.

Ans: Mc = 40.8 N # m a = 11.3° b = 101° g = 90° 310

4–93. If the magnitude of the couple moment acting on the pipe assembly is 50 N # m, determine the magnitude of the couple forces applied to each wrench. The pipe assembly lies in the x–y plane.

300 mm 300 mm F

x 200 mm

SOLUTION It is easiest to find the couple moment of F by taking the moment of either F or –F about point A or B, respectively, Fig. a. Here the position vectors rAB and rBA must be determined first.

200 mm

300 mm

rAB = (0.3 - 0.2)i + (0.8 - 0.3)j + (0 - 0)k = [0.1i + 0.5j] m rBA = (0.2 - 0.3)i + (0.3 - 0.8)j + (0 - 0)k = [- 0.1i - 0.5j] m The force vectors F and –F can be written as F = {Fk} N and - F = [-Fk]N Thus, the couple moment of F can be determined from i M c = rAB * F = 3 0.1 0

j 0.5 0

k 0 3 = 0.5Fi - 0.1Fj F

The magnitude of Mc is given by Mc = 2Mx 2 + My 2 + Mz 2 = 2(0.5F)2 + (0.1F)2 + 02 = 0.5099F Since Mc is required to equal 50 N # m, 50 = 0.5099F Ans.

F = 98.1 N

Ans: F = 98.1 N 311

4–94. Express the themoment moment of couple the couple acting on the of the acting on the frame in frame as vector a Cartesian Theapplied forcesperpendicular are applied Cartesian form. Thevector. forces are perpen dicularWhat to the frame. What isofthe of the to the frame. is the magnitude themagnitude couple moment? couple Take Fmoment? = 50 N. Take F = 50 N

O F

y 3m

1.5 m

SOLUTION

MC = 80(1.5) = 75 N # m

Ans.

MC = - 75(cos 30° i + cos 60° k) = {- 65.0i - 37.5k} N # m

Ans.

Ans: MC = { - 65.0i - 37.5k} N # m 312

4–95. If is If the the component component of of the thiscouple couplemoment momentalong alongthe thex axis x axis M N #Nm,# m, determine thethemagnitude the is xM=x 5=-20i6 5-20i6 determine magnitudeFF of of the couple couple forces. forces.

O F

y 3m

SOLUTION MC = F (1.5)

1.5 m

Thus

20 = F (1.5) cos 30° Ans.

F = 15.4 N

Ans: F = 15.4 N 313

*4–96. Express Express the the moment moment of of the thecouple coupleacting actingon onthe thepipe pipeasina Cartesian vector vector.form. What is the magnitude couple What is the magnitudeofofthis the couple moment? Take F F= = 125 N.

z O –F

150 mm 600 mm A

SOLUTION

MC = rAB * (125 k)

x 150 mm

MC = (0.2i + 0.3j) * (125 k)

200 mm

MC = {37.5i - 25j} N # m

Ans.

MC = 2(37.5)2 + (- 25)2 = 45.1 N # m

Ans.

Ans: MC = {37.5i - 25j} N # m MC = 45.1 N # m 314

4–97. If the couple moment acting on the pipe has a magnitude of 300 N # m, determine the magnitude F of the forces applied to the wrenches.

z O –F

150 mm 600 mm A

SOLUTION

MC = rAB * (F k)

200 mm

x 150 mm

= (0.2i + 0.3j) * (F k)

= {0.2Fi - 0.3Fj} N # m MC = F 2(0.2F)2 + ( - 0.3F) = 0.3606 F 300 = 0.3606 F Ans.

F = 832 N

Ans: F = 832 N 315

4–98. If M1 = 500 N # m, M2 = 600 N # m, and M3 = 450 N # m, determine the magnitude and coordinate direction angles of the resultant couple moment.

z M3

M2 30

SOLUTION

Since the couple moment is a free vector, it can act at any point without altering its effect. Thus, the couple moments M1, M2 and M3 acting on the gear deducer can be simplified, as shown in Fig. a. Expressing each couple moment in Cartesian vector form,

M1 = [500j] N # m M 2 = 600(- cos 30°i - sin 30°k) = {- 519.62i - 300k} N # m M 3 = [ -450k] N # m The resultant couple moment is given by (M c)R = M 1 + M 2 + M 3

(M c)R = ©M;

= 500j + ( - 519.62i - 300k) + ( -450k) = [ - 519.62i + 500j - 750k]N # m The magnitude of (Mc)R is (Mc)R = 2[(Mc)R]x 2 + [(Mc)R]y 2 + [(Mc)R]z 2 = 2( -519.62)2 + 5002 + ( - 750)2 = 1040.43N # m = 1.04 kN # m

Ans.

The coordinate angles of (M c)R are a = cos - 1 ¢

[(MC)R]x - 519.62 b = 120° ≤ = cos a (MC)R 1040.43

b = cos - 1 ¢

[(MC)R]y

g = cos - 1 ¢

[(MC)R]z

(MC)R

Ans.

≤ = cos a

500 b = 61.3° 1040.43

Ans.

≤ = cos a

- 750 b = 136° 1040.43

Ans.

Ans: (Mc)R = 1.04 kN # m a = 120° b = 61.3° g = 136° 316

4–99. Determine the required magnitude of couple moments M1, M2, and M3 so that the resultant couple moment is MR = 5- 300i + 450j - 600k6 N # m.

z M3

M2 30

SOLUTION

Since the couple moment is a free vector, it can act at any point without altering its effect. Thus, the couple moments M 1, M 2, and M 3 acting on the gear deducer can be simplified, as shown in Fig. a. Expressing each couple moment in Cartesian vector form,

M 1 = M1j M 2 = M2(- cos 30°i - sin 30°k) = - 0.8660M2i - 0.5M2k M 3 = -M3k The resultant couple moment is given by (M c)R = ©M;

(M c)R = M 1 + M 2 + M 3 ( - 300i + 450j - 600k) = M1j + ( -0.8660M2i - 0.5M2k) + (- M3k) - 300i + 450j - 600k = -0.8660M2i + M1 j - (0.5M2 + M3)k

Equating the i, j, and k components yields -300 = -0.8660M2

M2 = 346.41 N # m = 346 N # m M1 = 450N # m

600 = - 0.5(346.41) + M3

Ans. Ans.

M3 = 427 N # m

Ans.

Ans:

M2 = 346 N # m M1 = 450N # m

M3 = 427 N # m 317

*4–100. If the resultant couple moment acting on the triangular block is to be zero, determine the magnitudes F and P.

z 2F

3 in.

10 lb P 10 lb

3 in. 4 in.

308

6 in.

SOLUTION Couple moment of the 10 lb force: The sense of direction of this couple moment is outward perpendicular to the inclined plane. u =

3 4 j + k 5 5

3 4 M = Mu = 10(3) a j + kb = {18j + 24k} lb # in 5 5

Couple moment of P and F:

MF = - 6F j MP = -6Pk Resultant moment: MR = 0 = ΣM; 0 = M + MF + MP 0 = (18j + 24k) + ( -6F j) + ( - 6Pk) 0 = (18 - 6F) j + (24 - 6P)k Equating j, k components: (18 - 6F ) = 0 F = 3 lb

Ans.

(24 - 6P) = 0 P = 4 lb

Ans.

Ans: F = 3 lb P = 4 lb 318

4–101. If F = 80 N, determine the magnitude and coordinate direction angles of the couple moment. The pipe assembly lies in the x–y plane.

300 mm 300 mm F

x 200 mm

SOLUTION It is easiest to find the couple moment of F by taking the moment of F or –F about point A or B, respectively, Fig. a. Here the position vectors rAB and rBA must be determined first.

200 mm

300 mm

j 0.5 0

i Mc = rBA * -F = 3 - 0.1 0

j - 0.5 0

k 0 3 = [40i - 8j] N # m 80

or k 0 3 = [40i - 8j] N # m - 80

The magnitude of Mc is given by Mc = 2Mx 2 + My 2 + Mz 2 = 2402 + ( -8)2 + 02 = 40.79 N # m = 40.8 N # m

Ans.

The coordinate angles of Mc are a = cos - 1 ¢ b = cos - 1 ¢ g = cos - 1 ¢

Mx 40 ≤ = cos ¢ ≤ = 11.3° M 40.79 My M Mz M

-8 ≤ = 101° 40.79

Ans.

0 ≤ = 90° 40.79

Ans.

≤ = cos ¢

Ans.

Ans: Mc = 40.8 N # m a = 11.3° b = 101° g = 90° 319

4–102. If the magnitude of the couple moment acting on the pipe assembly is 50 N # m, determine the magnitude of the couple forces applied to each wrench. The pipe assembly lies in the x–y plane.

300 mm 300 mm F

x 200 mm

200 mm

300 mm

rAB = (0.3 - 0.2)i + (0.8 - 0.3)j + (0 - 0)k = [0.1i + 0.5j] m rBA = (0.2 - 0.3)i + (0.3 - 0.8)j + (0 - 0)k = [ - 0.1i - 0.5j] m The force vectors F and –F can be written as F = {Fk} N and - F = [-Fk]N Thus, the couple moment of F can be determined from i M c = rAB * F = 3 0.1 0

j 0.5 0

k 0 3 = 0.5Fi - 0.1Fj F

The magnitude of Mc is given by Mc = 2Mx 2 + My 2 + Mz 2 = 2(0.5F)2 + (0.1F)2 + 02 = 0.5099F Since Mc is required to equal 50 N # m, 50 = 0.5099F Ans.

F = 98.1 N

Ans: F = 98.1 N 320

4–103. y

Replace the force system by an equivalent resultant force and couple moment at point O. 455 N

12 13

2.5 m O

0.75 m

0.75 m 60

SOLUTION Equivalent Resultant Force and Couple Moment At O: 12 + (FR)x = ΣFx; (FR)x = 600 cos 60° - 455 a b = -120 N = 120 N d S 13 5 + c (FR)y = ΣFy; (FR)y = 455 a b - 600 sin 60° = -344.62 N = 344.62 NT 13

P 1m

600 N

As indicated in Fig. a

And

Also,

FR = 2 (FR)2x + (FR)2y = 21202 + 344.622 = 364.91 N = 365 N u = tan-1 c

(FR)y (FR)x

d = tan-1 a

a+(MR)O = ΣMO; (MR)O = 455 a

344.62 b = 70.80° = 70.8° d 120

Ans.

12 b(2) + 600 cos 60° (0.75) + 600 sin 60° (2.5) 13

= 2364.04 N # m

= 2364 N # m (Counterclockwise)

Ans.

Ans: FR = 365 N u = 70.8° d (MR)O = 2364 N # m d 321

*4–104. y

Replace the force system by an equivalent resultant force and couple moment at point P. 455 N

12 13

2.5 m O

0.75 m

0.75 m 60

SOLUTION Equivalent Resultant Force and Couple Moment At P: + (FR)x = ΣFx; S + c (FR)y = ΣFy;

12 (FR)x = 600 cos 60° - 455 a b = -120 N = 120 N d 13 5 (FR)y = 455 a b - 600 sin 60° = - 344.62 N = 344.62 N T 13

P 1m

600 N

As indicated in Fig. a,

FR = 2 (FR)2x + (FR)2y = 21202 + 344.622 = 364.91 N = 365 N

Ans.

And

u = tan-1 c

(FR)y (FR)x

Also,

a+ (MR)P = ΣMP;

d = tan-1 a

344.62 b = 70.80° = 70.8° d 120

(MR)P = 455 a

Ans.

12 5 b(2.75) - 455 a b(1) + 600 sin 60° (3.5) 13 13

= 2798.65 N # m

= 2799 N # m (Counterclockwise)

Ans.

Ans: FR = 365 N u = 70.8° d (MR)P = 2799 N # md 322

4–105. Replace the force system acting on the beam by an equivalent resultant force and couple moment at point B.

260 lb 13

30 in.

4 in. 12 in.

458

16 in.

300 lb 3 in. 10 in. 308

SOLUTION Force Summation:

250 lb

12 (260) = 327.13 lb 13

+ FRx = ΣFx; S

FRx = 300 cos 45° - 250 sin 30° +

+ c FRy = ΣFy;

FRy = - 300 sin 45° - 250 cos 30° +

5 (260) = - 328.64 lb 13

2 2 FR = 2 FRx + FRy = 2(327.13)2 + ( - 328.64)2 = 463.70 lb

Ans.

u = tan-1

Ans.

FRy FRx

= tan-1

328.64 = 45.1° cu 327.13

Moment Summation: a+ MRB = ΣMB;

MRB = -

12 5 (260)(32) + (260)(30) - 300 sin 45°(30) 13 13 - 300 cos 45°(16) - 250 cos 30°(20) - 250 sin 30°(3)

= - 19143 lb # in. = 1595 lb # ft c

Ans.

Ans: FR = 463.70 lb u = 45.1° c

MRB = 1595 lb # ft c

323

4–106. Replace the force system acting on the beam by an equivalent resultant force and couple moment at point A.

260 lb 13

30 in.

4 in. 12 in.

16 in.

SOLUTION

458

300 lb 3 in.

Force Summation:

10 in.

+ FRx = ΣFx; S

12 FRx = 300 cos 45° - 250 sin 30° + (260) = 327.13 lb 13

+ c FRy = ΣFy;

FRy = - 300 sin 45° - 250 cos 30° +

308 250 lb

5 (260) = - 328.64 lb 13

2 2 FR = 2 FRx + FRy = 2(327.13)2 + ( - 328.64)2 = 463.70 lb

Ans.

u = tan-1

Ans.

FRy FRx

= tan-1

328.64 = 45.1° cu 327.13

Moment Summation: a+ MRA = ΣMA;

MRA = -

12 5 (260)(4) + (260)(30) - 300 sin 45°(30) 13 13 + 300 cos 45°(12) - 250 cos 30°(20) - 250 sin 30°(31)

= - 9983.5 lb # in. = -832 lb # ft c

Ans.

Ans: F = 464 lb u = 45.1° c MRA = 832 lb # ft c 324

4–107. Replace the forces acting on the gear by an equivalent resultant force and couple moment acting at point O.

608 3 kN

175 mm 208

175 mm

Force Summation: + FRx = ΣFx; S + c FRy = ΣFy;

40 (2.25) - 3 sin 60° = - 0.40295 kN 41 9 FRy = (2.25) - 3 cos 60° = -1.0061 kN 41

FRx =

2.25 kN

FR = 2 FR2 x + FR2 y = 2( - 0.40295)2 + ( - 1.0061)2 = 1.08 kN

Ans.

u = tan-1

Ans.

FRy FRx

SOLUTION

= tan-1

1.0061 = 68.2° d 0.40295

Moment Summation: a+ MRO = ΣMO;

MRO = 3 cos 10°(0.175) + = 0.901 kN # m a

40 (2.25)(0.175) 41 Ans.

Ans: FR = 1.08 kN u = 68.2° d

MRO = 0.901 kN # m a 325

*4–108. Replace the force and couple moment system acting on the beam by an equivalent resultant force and couple moment at point A.

y 308 A

500 N

0.2 m

200 N · m x

O 1.5 m

1.5 m 200 N

SOLUTION Force Summation: + FRx = ΣFx; d

FRx = 500 sin 30° = 250 N

+ c FRy = ΣFy;

FRy = 200 - 500 cos 30° = -233.0 N

FR = 2 FR2 x + FR2 y = 22502 + ( - 233.0)2 = 342 N

Ans.

u = tan-1

Ans.

FRy FRx

= tan-1

233 = 43.0° d 250

Moment Summation: a+ MRA = ΣMA;

MRA = 200(3.5) - 500 cos 30°(1.5) - 200 = - 150 N # m = 150 N # m b

Ans.

Ans: FR = 342 N u = 43.0° d

MRA = 150 N # m b 326

4–109. Replace the force and couple moment system acting on the beam by an equivalent resultant force and couple moment at point O.

y 308 A

500 N

0.2 m

200 N · m x

O 1.5 m

1.5 m 200 N

SOLUTION Force Summation: + FRx = ΣFx; d

FRx = 500 sin 30° = 250 N

+ c FRy = ΣFy;

FRy = 200 - 500 cos 30° = -233.0 N

FR = 2 FR2 x + FR2 y = 2( - 250)2 + ( - 233.0)2 = 342 N

Ans.

u = tan-1

Ans.

FRy FRx

= tan-1

233 = 43.0° d 250

Moment Summation: a+ MRO = ΣMO;

MRO = 200(3.5) - 500 cos 30°(1.5) + 500 sin 30°(0.2) - 200 = - 99.5 N # m = 99.5 N # m b

Ans.

Ans: FR = 342 N u = 43.0° d

MRO = 99.5 N # m b 327

4–110. Replace the force system acting on the beam by an equivalent resultant force, and specify its point of application on the beam measured from point O.

y 3 ft

2 ft

2 ft x

O 308 10 lb

5 4

20 lb

60 lb

SOLUTION Force Summation: + FRx = ΣFx; S + c FRy = ΣFy;

4 (60) - 20 sin 30° = 38 lb 5 3 FRy = - 10 - (60) - 20 cos 30° = - 63.32 lb 5

FRx =

FR = 2 FR2 x + FR2 y = 2382 + ( - 63.32)2 = 73.8 lb

Ans.

u = tan-1

Ans.

FRy FRx

= tan-1

63.32 = 59.0° c 38

Moment Summation: a+ MRO = ΣMO;

- 63.32d = - 10(2) - 20 cos 30°(5) -

3 (60)(7) 5 Ans.

d = 5.66 ft

Ans: FR = 73.8 lb u = 59.0° c d = 5.66 ft 328

4–111. Determine the magnitude and orientation of u of force F and its placement d on the beam so the loading system is equivalent to a resultant force of 15 kN acting vertically downward at point O and a clockwise couple moment of 60 kN # m.

y 5 kN O

F 10 kN 3

5 4

x 2 kN · m

3m d

SOLUTION Force Summation: 4 (10) [1] 5

+ FRx = ΣFx; S

0 = - F cos u +

+ c FRy = ΣFy;

- 15 = - 5 - F sin u -

3 (10) [2] 5

Solving Eqs. [1] and [2] yields: u = 26.6°

Ans.

F = 8.94 kN

Moment Summation: a+ MRO = ΣMO;

- 60 = - 5(2) - 8.94 sin 26.6°(d) -

3 (10)(5) - 2 5 Ans.

d = 4.5 m

Ans: F = 8.94 kN d = 4.5 m 329

*4–112. Determine the magnitude and orientation of u of force F and its placement d on the beam so the loading system is equivalent to a resultant force of 20 kN acting vertically downward at point O and a clockwise couple moment of 80 kN # m.

y 5 kN O

F 10 kN 3

5 4

x 2 kN · m

3m d

SOLUTION Force Summation: 4 (10) [1] 5

+ FRx = ΣFx; S

0 = - F cos u +

+ c FRy = ΣFy;

- 20 = -5 - F sin u -

3 (10) [2] 5

Solving Eqs. [1] and [2] yields: u = 48.4°

Ans.

F = 12.0 kN

Moment Summation: a+ MRO = ΣMO;

- 80 = - 5(2) - 12.0 sin 48.4°(d) -

3 (10)(5) - 2 5 Ans.

d = 4.22 m

Ans: F = 12.0 kN d = 4.22 m 330

4–113. Replace the force system by an equivalent resultant force and couple moment at point O. Set F = 20 lb.

20 lb 308 F 5 4

6 in. O

SOLUTION + FRx = ΣFx; S

4 FRx = (20) - 20 sin 30° = 6 lb 5

+ c FRy = ΣFy;

FRy = 20 cos 30° +

408

2 in.

3 (20) = 29.32 lb 5

FR = 2 FR2 x + FR2 y = 262 + 29.32)2 = 29.9 lb

Ans.

u = tan-1

Ans.

FRx

1.5 in.

Force Summation:

FRy

= tan-1

29.32 = 78.4° a 6

Moment Summation: a+ MRO = ΣMO;

MRO = 20 sin 30°(6 sin 40°) + 20 cos 30°(3.5 + 6 cos 40°) 4 3 - (20)(6 sin 40°) + (20)(3.5 + 6 cos 40°) 5 5 = 214 lb # in. a

Ans.

Ans: FR = 29.9 lb u = 78.4° a

MRO = 214 lb # in. a 331

4–114. Replace the force system by an equivalent resultant force and couple moment at point O. Set F = 15 lb.

20 lb 308 F 5 4

6 in. O

SOLUTION

408

2 in.

4 + FRx = ΣFx; FRx = (15) - 20 sin 30° = 2 lb S 5 3 + c FRy = ΣFy; FRy = 20 cos 30° + (15) = 26.32 lb 5 FR = 2 FR2 x + FR2 y = 222 + 26.322 = 26.4 lb

Ans.

u = tan-1

Ans.

FRx

1.5 in.

Force Summation:

FRy

= tan-1

26.32 = 85.7° a 2

Moment Summation: a+ MRO = ΣMO;

MRO = 20 sin 30°(6 sin 40°) + 20 cos 30°(3.5 + 6 cos 40°) 4 3 - (15)(6 sin 40°) + (15)(3.5 + 6 cos 40°) 5 5 = 205 lb # in. a

Ans.

Ans: FR = 26.4 lb u = 85.7° a

MRO = 205 lb # in. a 332

4–115. Replace the loading acting on the frame by an equivalent resultant force and couple moment acting at point A.

80 lb 3 ft

60 lb 3 ft

3 ft

3 ft A

B 2 ft

4 ft 40 lb 75 lb

SOLUTION Force Summation: + FRx = ΣFx; S

FRx = - 40 lb = 40 lb d

+ c FRy = ΣFy;

FRy = - 80 - 60 - 75 = -215 lb = 215 lb

FR = 2 FR2 x + FR2 y = 2402 + 2152 = 219 lb

Ans.

u = tan-1

Ans.

FRy FRx

= tan-1

215 = 79.5° d 40

Moment Summation: a+ MRA = ΣMA;

MRA = 80(9) + 75(6) + 60(3) - 40(4) = 1190 lb # ft = 1.19 kip # ft a

Ans.

Ans: FR = 219 lb u = 79.5° d

MRA = 1.19 kip # ft a 333

*4–116. Replace the loading acting on the frame by an equivalent resultant force and couple moment acting at point B.

80 lb 3 ft

60 lb 3 ft

3 ft

3 ft A

B 2 ft

4 ft 40 lb 75 lb

SOLUTION Force Summation: + FRx = ΣFx; S

FRx = - 40 lb = 40 lb d

+ c FRy = ΣFy;

FRy = - 80 - 60 - 75 = -215 lb = 215 lb

FR = 2 FR2 x + FR2 y = 2402 + 2152 = 219 lb

Ans.

u = tan-1

Ans.

FRy FRx

= tan-1

215 = 79.5° d 40

Moment Summation: a+ MRB = ΣMB;

MRB = - 80(3) - 75(6) - 60(9) - 40(4) = - 1390 lb # ft = 1.39 kip # ft b

Ans.

Ans: FR = 219 lb u = 79.5° d

MRB = 1.39 kip # ft b 334

4–117. Replace the force system acting on the post by a resultant force and couple moment at point O.

300 lb 30 150 lb 3

2 ft

5 4

SOLUTION Equivalent Resultant ResultantForce: Force: Forces F1 and F2 are resolved into their x and y comEquivalent ponents, Fig. a. Summing these force components algebraically along the x and y axes, we have + ©(F ) = ©F ; : R x x

4 (FR)x = 300 cos 30° - 150 a b + 200 = 339.81 lb : 5

+ c (FR)y = ©Fy;

3 (FR)y = 300 sin 30° + 150a b = 240 lb c 5

2 ft 200 lb 2 ft O

The magnitude of the resultant force FR is given by FR = 2(FR)x2 + (FR)y2 = 2339.812 + 2402 = 416.02 lb = 416 lb

Ans.

The angle u of FR is u = tan-1 c

(FR)y (FR)x

d = tan-1 c

240 d = 35.23° = 35.2° a 339.81

Ans.

Equivalent Resultant Resultant Couple Couple Moment: Applying the principle of moments, Equivalent Figs. a and b, and summing the moments of the force components algebraically about point A, we can write a + (MR)A = ©MA;

4 (MR)A = 150 a b (4) - 200(2) - 300 cos 30°(6) 5

= - 1478.85 lb # ft = 1.48 kip # ft (Clockwise) Ans.

Ans: FR = 416 lb u = 35.2° a ( MR ) A = 1.48 kip # ft b 335

4–118. Determine the magnitudes of F1 and F2 and the angle u so that the loading creates a zero-resultant force and couple moment on the handle.

80 lb · ft 308

308

SOLUTION

0.75 ft

+ ΣFx = 0; S

F2 - 30 cos 60° - F1 cos u = 0 [1]

+ c ΣFy = 0;

F1 sin u - 30 sin 60° = 0 [2]

a+ ΣMO = 0;

80 - F2(0.75) - 30(0.75) - F1 sin (u + 30°)(0.75) = 0 [3]

30 lb

Solving Eqs. [1], [2], and [3] yields: F2 = 41.1 lb

u = 44.9°

Ans.

F1 = 36.8 lb

Ans: F2 = 41.1 lb u = 44.9° F1 = 36.8 lb 336

4–119. Replace the force system acting on the truss by a resultant force and couple moment at point C.

200 lb

2 ft

150 lb

2 ft

100 lb

2 ft

2 ft B 5

3 4

500 lb

6 ft

SOLUTION Equivalent Resultant Forces: The 500-lb force is resolved into its x and y components, Fig. a. Summing these force components algebraically along the x and y axes, 4 (FR)x = 500 a b = 400 lb : 5

+ ©(F ) = ©F ; : R x x + c (FR)y = ©Fy;

3 (FR)y = - 200 - 150 - 100 - 500a b = - 750 lb = 750 lb T 5

The magnitude of the resultant force FR is given by FR = 2(FR)x 2 + (FR)y 2 = 24002 + 7502 = 850 lb

Ans.

The angle u of FR is u = tan - 1 B

(FR)y (FR)x

R = tan - 1 c

750 d = 61.93° = 61.9° 400

Ans.

Equivalent Couple Moment: Summing the moment of the forces and force components, Fig. a, algebraically about point C, a + (MR)C = ©MC ;

4 3 (MR)C = - 200(2) - 150(4) - 100(6) - 500a b(8) - 500 a b(6) 5 5 = - 6400 lb # ft = 6.40 kip # ft (Clockwise)

Ans.

Ans: FR = 850 lb u = 61.9° (MR)C = 6.40 kip # ft b 337

*4–120. A biomechanical model of the lumbar region of the human trunk is shown. The forces acting in the four muscle groups consist of FR = 35 N for the rectus, FO = 45 N for the oblique, FL = 23 N for the lumbar latissimus dorsi, and FE = 32 N for the erector spinae. These loadings are symmetric with respect to the y–z plane. Replace this system of parallel forces by an equivalent force and couple moment acting at the spine, point O. Express the results in Cartesian vector form.

FO FE

FE FO

FL O

75 mm

15 mm 45 mm

SOLUTION

50 mm

30 mm 40 mm

FR = ©Fz ;

FR = {2(35 + 45 + 23 + 32)k } = {270k} N

MROx = ©MOx ;

MR O = [ - 2(35)(0.075) + 2(32)(0.015) + 2(23)(0.045)]i MR O = { - 2.22i} N # m

Ans.

338

Ans: FR = 5270k6 N MRO = 5 -2.22i6 N # m

4–121. The resultant force of the wind and the weights of the various components act on the sign. Determine the equivalent force and moment acting at its base, A.

z 12 ft

700 lb 12 ft

1800 lb

SOLUTION FR = ΣFR; FR = - 400k - 1800k - 700j

16 ft

FR = { - 700j - 2200k} lb MRA = ΣMA;

400 lb

Ans. A

MRA = ( - 12i + 28k) * ( -700j - 1800k) = {19.6i - 21.6j + 8.40k} kip # ft

Ans.

Ans: FR = { - 700j - 2200k} lb

MRA = {19.6i - 21.6j + 8.40k} kip # ft

339

4–122. Replace the force system acting on the pipe assembly by a resultant force and couple moment at point O. Express the results in Cartesian vector form.

z F2 F1

{ 20i

10j

{ 10i

25j

20k} lb

25k}lb 2 ft

y 1.5 ft

2 ft

SOLUTION

2 ft

Resultant Force: Force: The resultant force FR can be determined from Equivalent Resultant FR = ©F; FR = F1 + F2 = (- 20i - 10j + 25k) + (- 10i + 25j + 20k) = [ - 30i + 15 j + 45 k]lb

Ans.

Equivalent Resultant Couple Moment: The position vectors rOA and rOB, are rOA = (1.5 - 0)i + (2 - 0)j + (0 - 0)k = [1.5i + 2 j] ft rOB = (1.5 - 0)i + (4 - 0)j + (2 - 0)k = [1.5i + 4 j + 2k] ft Thus, the resultant couple moment about point O is M W = ©M O ;

(M R)O = rOA * F1 + rOB * F2 =

i 1.5 -20

j 2 - 10

k 0 + 25

i 1.5 -10

= [80i - 87.5 j + 102.5k] lb # ft

j 4 25

k 2 20 Ans.

Ans: FR = [ - 30i + 15j + 45k]lb

(MR)O = [80i - 87.5j + 102.5k]lb # ft 340

4–123. The crate is on the ground and is to be hoisted using the three slings shown. Replace the system of forces acting on the slings by an equivalent resultant force and couple moment at point O. The force F1 is vertical.

F2 5 500 N

F3 5 400 N

F1 5 600 N

458 458 O 608

1208

308 1m 0.25 m

SOLUTION

0.25 m

Force Vector: F1 = {600k} N F2 = 500(cos 45° cos 30°i + cos 45° sin 30°j + sin 45°k) = {306.2i + 176.8j + 353.6k} N F3 = 400 (cos 60°i + cos 120°j + cos 45°k) = {200i - 200j + 282.8k} N Force Summation: FR = ΣF; FR = F1 + F2 + F3 = (600k) + (306.2i + 176.8j + 353.6k) + (200i - 200j + 282.8k) FR = {506i - 23.2j + 1236k} N

Ans.

Moment Summation: MRO = ΣM;

MRO = (1i) * (600k) + ( -0.25j) * (306.2i + 176.8j + 353.6k) + (0.25j) * (200i - 200j + 282.8k)

MRO = { - 17.7i - 600j + 26.5k} N # m

Ans.

Ans: FR = {506i - 23.2j + 1236k} N

MRO = { - 17.7i - 600j + 26.5k} N # m 341

*4–124.

The forces and couple moments exerted on the toe and heel plates of a snow ski are Ft = { - 50i + 80j - 158k} N, Mt = { -6i + 4j + 2k} N # m, and Fh = { - 20i + 60j - 250k} N, Mh = { - 20i + 8j + 3k} N # m, respectively. Replace this system by an equivalent force and couple moment acting at point O. Express the results in Cartesian vector form.

P Fh Ft

120 mm

Mh 800 mm

SOLUTION FR = Ft + Fh FR = { - 70i + 140j - 408k} N

Ans.

MRO = (rOFt * Ft) + Mt + Mh i = † 0.12 - 50

j 0 80

k 0 † + ( - 6i + 4j + 2k) + ( - 20i + 8j + 3k) -158

MRO = { -26i + 31.0j + 14.6k} N # m

Ans.

Ans: FR = { - 70i + 140j - 408k} N

MRO = { - 26i + 31.0j + 14.6k} N # m

342

4–125.

The forces and couple moments exerted on the toe and heel plates of a snow ski are Ft = { - 50i + 80j - 158k} N, Mt = { - 6i + 4j + 2k} N # m, and Fh = { - 20i + 60j - 250k} N, Mh = { - 20i + 8j + 3k} N # m, respectively. Replace this system by an equivalent force and couple moment acting at point P. Express the results in Cartesian vector form.

P Fh Ft

120 mm

Mh 800 mm

SOLUTION FR = Ft + Fh = { - 70i + 140j - 408k} N i MRP = † 0.8 - 20

j 0 60

k i 0 † + † 0.92 - 250 - 50

j 0 80

Ans. k 0 † + ( -6i + 4j + 2k) + ( -20i + 8j + 3k) - 158

MRP = (200j + 48k) + (145.36j + 73.6k) + ( - 6i + 4j + 2k) + ( -20i + 8j + 3k) MRP = { - 26i + 357.36j + 126.6k} N # m MRP = { - 26i + 357j + 127k} N # m

Ans.

Ans: FR = { - 70i + 140j - 408k} N MRP = { - 26i + 357j + 127k} N # m 343

4–126. The various components of theoftruck shown. The weights weightsofofthethe various components the are truck are Replace this system of forces equivalent resultant force shown. Replace this systembyofanforces by an equivalent and specify its location, measured from B. resultant force and specify its location measured from B.

+ c FR = ©Fy;

FR = -1750 - 5500 - 3500

= - 10 750 lb = 10.75 kip T a +MRA = ©MA ;

3500 lb

SOLUTION

5500 lb 14 ft

3 ft

1750 lb

6 ft 2 ft

Ans.

-10 750d = -3500(3) - 5500(17) - 1750(25) d = 13.7 ft

Ans.

Ans: FR = 10.75 kip T d = 13.7 ft 344

4–127. The various components of the are shown. The weights weightsofofthethe various components oftruck the truck are Replace this system of system forces byofanforces equivalent resultant force shown. Replace this by an equivalent and specifyforce its location, measured point measured A. resultant and specify itsfrom location from point A.

5500 lb 14 ft

Equivalent Force: Force: Equivalent + c FR = ©Fy ;

3500 lb

SOLUTION

3 ft

1750 lb

6 ft 2 ft

FR = - 1750 - 5500 - 3500 = - 10 750 lb = 10.75 kipT

Ans.

Location of of Resultant ResultantForce ForceFrom FromPoint PointA: A: Location a +MRA = ©MA ;

10 750(d) = 3500(20) + 5500(6) - 1750(2) d = 9.26 ft

Ans.

Ans: FR = 10.75 kip T d = 9.26 ft 345

*4–128. Replace the force and couple-moment system by an equivalent resultant force and specify its coordinate point of application (x, 0) on the x axis.

M2 5 100 lb ? ft

F1 5 20 lb

3 ft

F3 5 50 lb

1 ft O

4 ft

5 ft

SOLUTION

M1 5 170 lb ? ft

F2 5 10 lb

+ F = ΣF ; F = - 50 a 3 b - 20 = - 50 lb S Rx x Rx 5

4 + c FRy = ΣFy; FRy = - 50 a b + 10 = - 30 lb 5

FR = 2( -50)2 + ( - 30)2 = 58.3 lb

Ans.

u = tan-1 a

Ans.

30 b = 31.0° d 50

4 3 a+MRO = ΣMO; -30(x) = -50 a b(3) + 50 a b(1) + 20(3) + 100 - 170 - 10(5) 5 5

Ans.

x = 5.00 ft

Ans: FR = 58.3 lb u = 31.0° d x = 5.00 ft 346

4–129. Replace the force and couple-moment system by an equivalent resultant force and specify its coordinate point of application (0, y) on the y axis.

M2 5 100 lb ? ft

F1 5 20 lb

3 ft

F3 5 50 lb

1 ft O

4 ft

5 ft

SOLUTION

M1 5 170 lb ? ft

F2 5 10 lb

+ F = ΣF ; F = - 50 a 3 b - 20 = -50 lb S Rx x Rx 5

4 + c FRy = ΣFy; FRy = - 50 a b + 10 = -30 lb 5

FR = 2( - 50)2 + ( - 30)2 = 58.3 lb

Ans.

u = tan-1 a

Ans.

30 b = 31.0° d 50

4 3 a+ MRO = ΣMO; 50(y) = - 50 a b(3) + 50 a b(1) + 20(3) + 100 - 170 - 10(5) 5 5

Ans.

y = - 3.00 ft

Ans: FR = 58.3 lb u = 31.0° d y = -3.00 ft 347

4–130. The system of four forces acts on the roof truss. Determine the equivalent resultant force and specify its location along AB, measured from point A.

200 lb 275 lb 4 ft 300 lb 4 ft

30 B

150 lb 4 ft

SOLUTION b+FRx = ©Fx ;

FRx = 200 sin 30° = 100 lb

R+FRy = ©Fy ;

FRy = 150 + 300 + 275 + 200 cos 30° = 898.2 lb

FR = 2(100)2 + (898.2)2 = 904 lb u = tan - 1 a

Ans.

100 b = 6.35° 898.2 Ans.

f = 30° - 6.35° = 23.6° c +MRA = ©MA ;

898.2 (d) = 4 (300) + 8 (275) + 12 cos 30° (200) Ans.

d = 6.10 ft

Ans: FR = 904 lb

f = 23.6° d = 6.10 ft 348

4–131. Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from end A.

700 N 450 N

300 N

A 2m

1500 N m

SOLUTION ;

+ F = ©F ; : Rx x

FRx = 450 cos 60° - 700 sin 30° = - 125 N = 125 N

+ c FRy = ©Fy ;

FRy = - 450 sin 60° - 700 cos 30° - 300 = - 1296 N = 1296 N

F = 2(-125)2 + ( -1296)2 = 1302 N

Ans.

1296 b = 84.5° 125

Ans.

u = tan-1 a

c + MRA = ©MA ;

1296(x) = 450 sin 60°(2) + 300(6) + 700 cos 30°(9) + 1500 Ans.

x = 7.36 m

Ans: F = 1302 N u = 84.5° d x = 7.36 m 349

*4–132. Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from B.

700 N 450 N

300 N

A 2m

1500 N m

SOLUTION ;

+ F = ©F ; : Rx x

FRx = 450 cos 60° - 700 sin 30° = - 125 N = 125 N

+ c FRy = ©Fy ;

FRy = - 450 sin 60° - 700 cos 30° - 300 = - 1296 N = 1296 N

F = 2(- 125)2 + ( - 1296)2 = 1302 N u = tan-1 a

Ans.

1296 b = 84.5° d 125

c + MRB = ©MB ;

Ans.

1296(x) = - 450 sin 60°(4) + 700 cos 30°(3) + 1500 Ans.

x = 1.36 m (to the right)

Ans: F = 1302 N u = 84.5° d x = 1.36 m (to the right) 350

4–133. Replace the loading system acting on the post by an equivalent resultant force and couple moment at point O.

30 lb P 1 ft

SOLUTION 3 FRx = 40 - 60 a b = 4 lb 5

+ T FRy = ©Fy ;

4 60aa b + 30 = 78 lb FRy = 60 5

2 ft O

2(4) + (78) = 78.1 lb

Ans.

FRy

78 b = 87.1° c c 4

Ans.

FRx

40 lb

2F2Rx + F2Ry =

u = tan-1 a

b = tan-1 a

a + MRo = ©MO ;

3 ft

60 lb

+ F = ©F ; : Rx x

FR =

3 (5) = 100 lb # ftd MRo = - 40(2) + 60 a bb(5) 5

Ans.

Ans: FR = 78.1 lb u = 87.1° c

MRO = 100 lb # ft d 351

4–134. Replace the loading system acting on the post by an equivalent resultant force and couple moment at point P.

30 lb P 1 ft

SOLUTION 3 FRx = 40- 60 a b = 4 lb 5

+ T FRy = ©Fy ;

4 FRy = 60 a b + 30 = 78 lb 5

2 ft O

2(4) + (78) = 78.1 lb

Ans.

FRy

78 b = 87.1° c 4

Ans.

FRx

40 lb

2F2Rx + F2Ry =

u = tan-1 a

b = tan-1 a

a + MRp = ©Mp ;

3 ft

60 lb

+ F = ©F ; : Rx x

FR =

4 3

3 MRp = 40(4) -60 a b (1) = 124 lb # ftd 5

Ans.

Ans: FR = 78.1 lb u = 87.1° c

MRp = 124 lb # ft d 352

4–135. Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A.

300 N 1m C

250 N

400 N m 60

SOLUTION + ©F = F ; : x Rx

4 FRx = - 250a b - 500(cos 60°) = -450 N = 450 N ; 5

+ c ©Fy = ©Fy ;

3 FRy = - 300 - 250 a b - 500 sin 60° = -883.0127 N = 883.0127 N T 5

FR = 2(- 450)2 + (-883.0127)2 = 991 N u = tan-1 a

500 N

Ans.

883.0127 b = 63.0° d 450

a + MRA = ©MA ;

3 4 450y = 400 + (500 cos 60°)(3) + 250a b(5) - 300(2) - 250 a b(5) 5 5 y =

800 = 1.78 m 450

Ans.

Ans: FR = 991 N u = 63.0° d y = 1.78 m d 353

*4–136. Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member CD, measured from end C.

300 N 1m C

400 N m 60

SOLUTION 4 FRx = - 250a b - 500(cos 60°) = - 450 N = 450 N ; 5

+ c ©Fy = ©Fy ;

3 FRy = - 300 - 250 a b - 500 sin 60° = -883.0127 N = 883.0127 N T 5

FR = 2( -450)2 + (- 883.0127)2 = 991 N

500 N

+ ©F = F ; : x Rx

u = tan-1 a

250 N

Ans.

883.0127 b = 63.0° d 450

c + MRA = ©MC ;

3 883.0127x = -400 + 300(3) + 250 a b(6) + 500 cos 60°(2) + (500 sin 60°)(1) 5 x =

2333 = 2.64 m 883.0127

Ans.

Ans: FR = 991 N u = 63.0° d x = 2.64 m 354

4–137. Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant’s line of action intersects member AB, measured from point A.

35 lb

20 lb

30 4 ft

B 2 ft 3 ft 25 lb

SOLUTION + F = ©F ; : Rx x

FRx = 35 sin 30° + 25 = 42.5 lb

+ TFRy = ©Fy ;

2 ft

FRy = 35 cos 30° + 20 = 50.31 lb 2

FR = 2(42.5) + (50.31) = 65.9 lb

Ans.

50.31 b = 49.8° c 42.5

Ans.

u = tan - 1 a c + MRA = ©MA ;

50.31 (d) = 35 cos 30° (2) + 20 (6) - 25 (3) Ans.

d = 2.10 ft

Ans: FR = 65.9 lb u = 49.8° c d = 2.10 ft 355

4–138. Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant’s line of action intersects member BC, measured from point B.

35 lb

20 lb

30 4 ft

B 2 ft 3 ft 25 lb

SOLUTION + F = ©F ; : Rx x

FRx = 35 sin 30° + 25 = 42.5 lb

+ TFRy = ©Fy ;

FRy = 35 cos 30° + 20 = 50.31 lb 2

FR = 2(42.5) + (50.31) = 65.9 lb

Ans.

50.31 b = 49.8° c 42.5

Ans.

u = tan - 1 a c +MRA = ©MA ;

2 ft

50.31 (6) - 42.5 (d) = 35 cos 30° (2) + 20 (6) - 25 (3) Ans.

d = 4.62 ft

Ans: FR = 65.9 lb u = 49.8° c d = 4.62 ft 356

4–139. Replace the force system acting on the frame by an equivalent resultant force, and specify where the resultant’s line of action intersects member AB, measured from point A.

50 lb

5 4 3

5 ft

40 lb

2 ft

20 lb

5 ft

SOLUTION

Force Summation:

+ F = ΣF ; F = 40 + 20 + 3 (50) = 90 lb S S Rx x Rx 5 + c FRy = ΣFy; FRy =

4 (50) = 40 lb c 5

2 2 FR = 2FRx + FRy = 2902 + 402 = 98.5 lb

Ans.

u = tan-1

Ans.

FRy

FRx

= tan - 1

40 = 24.0° au 90

Moment Summation: a+ MRA = ΣMA; - 90d = - 20(5) - 40(7) +

4 3 (50)(5) - (50)(7) 5 5 Ans.

d = 4.33 ft

Ans: FR = 98.5 lb u = 24.0° a d = 4.33 ft 357

*4–140. Replace the force system acting on the frame by an equivalent resultant force, and specify where the resultant’s line of action intersects member BC, measured from point B.

50 lb

5 4 3

5 ft

40 lb

2 ft

20 lb

SOLUTION

5 ft

Force Summation: + F = ΣF ; F = 40 + 20 + 3 (50) = 90 lb S S Rx x Rx 5 + c FRy = ΣFy; FRy =

4 (50) = 40 lb c 5

2 2 FR = 2FRx + FRy = 2902 + 402 = 98.5 lb

Ans.

u = tan-1

Ans.

FRy

FRx

= tan - 1

40 = 24.0° au 90

Moment Summation: 4 (50)(5) 5 d = 6.0 ft

a+ MRB = ΣMB; 40d = 20(2) +

Ans.

Ans: FR = 98.5 lb u = 24.0° a d = 6.0 ft 358

4–141. Replace the the force loading the frame Replace andacting coupleon system actingby onan theequivalent frame by resultant force and specify where the resultant’s linethe of an equivalent resultant force and specify where action intersects AB, measured fromAB, A. measured resultant’s line ofmember action intersects member from A.

A 2 ft

3 4

150 lb

4 ft

SOLUTION + F = ©F ; : Rx x + c FRy = ©Fy ;

4 FRx = 150 a b + 50 sin 30° = 145 lb 5

3 FRy = 50 cos 30° + 150 a b = 133.3 lb 5

FR = 2(145)2 + (133.3)2 = 197 lb u = tan - 1 a

a + MRA = ©MA ;

133.3 b = 42.6° 145

500 lb ft B

C 3 ft

Ans.

30 50 lb

Ans.

4 145 d = 150 a b (2) - 50 cos 30° (3) + 50 sin 30° (6) + 500 5 Ans.

d = 5.24 ft

Ans: FR = 197 lb u = 42.6° a d = 5.24 ft 359

4–142. the force loading the frame Replace the andacting coupleon system actingby onan theequivalent frame by resultant force and specifyforce where resultant’s linethe of an equivalent resultant andthespecify where action intersects BC, measured fromBC, B. measured resultant’s line ofmember action intersects member from B.

A 2 ft

3 4

150 lb

SOLUTION + F = ©F ; : Rx x

4 FRx = 150 a b + 50 sin 30° = 145 lb 5

+ c FRy = ©Fy ;

3 FRy = 50 cos 30° + 150 a b = 133.3 lb 5

500 lb ft

3 ft

Ans.

133.3 b = 42.6° 145

a +MRA = ©MA ;

FR = 2(145)2 + (133.3)2 = 197 lb u = tan - 1 a

4 ft

30 50 lb

Ans.

4 145 (6) - 133.3 (d) = 150 a b (2) - 50 cos 30° (3) + 50 sin 30° (6) + 500 5 Ans.

d = 0.824 ft

Ans: FR = 197 lb u = 42.6°a d = 0.824 ft 360

4–143. IIff FFAA == 77kN kNand andFFBB== 55kN, kN,represent representhe t thforce e forcsystem e systeby m a resultant force, and specify its location on the plane.its acting on the corbels by a resultant force, andx–y specify location on the x–y plane.

z 150 mm

6 kN

100 mm 650 mm

SOLUTION

FB 750 mm O

8kN 700 mm 100 mm

600 mm

150 mm

Equivalent Resultant Force: By equating the sum of the forces in Fig. a along the z axis to the resultant force FR, Fig. b, + c FR = ©Fz;

- FR = -6 - 5- 7- 8 Ans.

FR = 26 kN

Point of Application: By equating the moment of the forces shown in Fig. a and FR, Fig. b, about the x and y axes, (MR)x = ©Mx;

-26(y) = 6(650) + 5(750) - 7(600) - 8(700) Ans.

y = 82.7 mm (MR)y = ©My;

26(x) = 6(100) + 7(150) - 5(150) - 8(100) Ans.

x = 3.85 mm

Ans: FR = 26 kN y = 82.7 mm x = 3.85 mm 361

*4–144. of FofA and the that resultant Determine the themagnitudes magnitudes FA Fand FB so the B so that force passes through O. point O of the column. resultant force passespoint through

z 150 mm

6 kN

100 mm 650 mm

SOLUTION

FB 750 mm O

8kN 700 mm 100 mm

600 mm

150 mm

Equivalent Resultant Force: By equating the sum of the forces in Fig. a along the z axis to the resultant force FR, Fig. b, + c FR = ©Fz;

- FR = - FA - FB - 8 - 6 (1)

FR = FA + FB + 14

Point of Application: Since FR is required to pass through point O, the moment of FR about the x and y axes are equal to zero. Thus, (MR)x = ©Mx;

0 = FB (750) + 6(650) - FA (600) - 8(700) (2)

750FB - 600FA - 1700 = 0 (MR)y = ©My;

0 = FA (150) + 6(100) - FB (150) - 8(100) (3)

159FA - 150FB + 200 = 0 Solving Eqs. (1) through (3) yields FA = 18.0 kN

FB = 16.7 kN

FR = 48.7 kN

Ans.

Ans: FA = 18.0 kN FB = 16.7 kN FR = 48.7 kN 362

4–145. The tube supports the four parallel forces. Determine the magnitudes of forces FC and FD acting at C and D so that the equivalent resultant force of the force system acts through the midpoint O of the tube.

600 N D FC

A 400 mm

SOLUTION Since the resultant force passes through point O, the resultant moment components about x and y axes are both zero. ©Mx = 0;

500 N C

400 mm z B

200 mm 200 mm y

FD(0.4) + 600(0.4) - FC(0.4) - 500(0.4) = 0 (1)

FC - FD = 100 ©My = 0;

500(0.2) + 600(0.2) - FC(0.2) - FD(0.2) = 0 (2)

FC + FD = 1100 Solving Eqs. (1) and (2) yields: FC = 600 N

Ans.

FD = 500 N

Ans: FC = 600 N FD = 500 N 363

4–146. Three parallel bolting forces act on the circular plate. Determine the resultant force, and specify its location (x, z) on the plate. FA = 200 lb, FB = 100 lb, and FC = 400 lb.

1.5 ft

SOLUTION

Equivalent Equivalent Force: Force: FR = ©Fy;

30 B

-FR = - 400 - 200 - 100 Ans.

FR = 700 lb Location Location of of Resultant ResultantForce: Force: MRx = ©Mx;

700(z) = 400(1.5) - 200(1.5 sin 45°) - 100(1.5 sin 30°) Ans.

z = 0.447 ft MRz = ©Mz;

- 700(x) = 200(1.5 cos 45°) - 100(1.5 cos 30°) Ans.

x = - 0.117 ft

Ans: FR = 700 lb z = 0.447 ft x = -0.117 ft 364

4–147. The three parallel bolting forces act on the circular plate. If the force at A has a magnitude of FA = 200 lb, determine the magnitudes of FB and FC so that the resultant force FR of the system has a line of action that coincides with the y axis. Hint: This requires ©Mx = 0 and ©Mz = 0.

1.5 ft

SOLUTION

Since FR coincides with y axis, MRx = MRy = 0. MRz = ©Mz;

30 B

0 = 200(1.5 cos 45°) - FB (1.5 cos 30°) Ans.

FB = 163.30 lb = 163 lb Using the result FB = 163.30 lb, MRx = ©Mx;

0 = FC (1.5) - 200(1.5 sin 45°) - 163.30(1.5 sin 30°) Ans.

FC = 223 lb

Ans: FB = 163 lb FC = 223 lb 365

*4–148. z

Replace the three forces acting on the plate by a wrench. Specify the force and couple moment for the wrench and the point P(x, y) where its line of action intersects the plate.

12 ft

12 ft y A

SOLUTION

Resultant Force Vector:

107.7

B FB 5 {260j} lb

FR = 2( - 80)2 + ( - 60)2 + 402 = 107.7 lb = 180 lb - 80 i - 60 j + 40 k

FC 5 {40k} lb

FA 5 {280i} lb

FR = { - 80 i - 60 j + 40 k} lb

uFR =

x P

Ans.

= - 0.7428 i - 0.5571 j + 0.3714 k

The line of action of MR of the wrench is parallel to the line of action of FR; also assume the both MR and FR have the same sense. Therefore uMR = -0.7428 i - 0.5571 j + 0.3714 k (MR)x′ = ΣMx′;

- 0.7428MR = 40(12 - y) (1)

(MR)y′ = ΣMy′;

0.5571MR = 40x (2)

(MR)z′ = ΣMz′;

0.3714MR = - 60(12 - x) - 80y (3)

Solving Eqs. (1), (2), and (3) yields: MR = - 624 lb # ft

x = 8.69 ft

Ans.

y = 0.414 ft

Negative sign indicates that MR has a sense which is opposite to that of FR.

Ans: FR = 180 lb MR = -624 lb # ft x = 8.69 ft y = 0.414 ft 366

4–149. Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(y, z) where its line of action intersects the plate.

4m C y

FC 5 {150j} N

P z

SOLUTION x

Resultant Force Vector:

FA 5 {250k} N

FR = { -400 i + 150 j + 250 k} N FR = 24002 + 1502 + 2502 = 495 N uFR =

- 400 i + 150 j + 250 k 495

B FB 5 {2400i} N

Ans.

= - 0.8081 i + 0.3030 j + 0.5051 k

The line of action of MR of the wrench is parallel to the line of action of FR; also assume the both MR and FR have the same sense. Therefore uMR = -0.8081 i - 0.3030 j + 0.5051 k (MR)x′ = ΣMx′;

- 0.8081MR = - 150(6 - z) - 250y (1)

(MR)y′ = ΣMy′;

0.3030MR = 400z (2)

(MR)z′ = ΣMz′;

0.5051MR = 400(4 - y) (3)

Solving Eqs. (1), (2), and (3) yields: MR = 1535.4 N # m = 1.54 kN # m

z = 1.16 m

y = 2.06 m

Ans.

Ans: FR = 495 N

MR = 1.54 kN # m z = 1.16 m y = 2.06 m

367

4–150. The pipe assembly is subjected to the action of a wrench at B and a couple at A. Determine the magnitude F of the couple forces so that the system can be simplified to a wrench acting at point C.

2Fi A

C 0.6 m

0.8 m {60k} N 0.25 m

B 0.25 m

SOLUTION

{240i} N

0.3 m 0.3 m

0.5 m y

0.7 m

{260k} N

An equivalent loading at point C : FR = { - 40 i} N MRC = - 60(0.5)i + 40(0.8)k - F(0.6)k MRC = { - 30i + (32 - 0.6F)k} N # m Require 32 - 0.6F = 0 Ans.

F = 53.3 N The wrench at C is therefore FRW = { - 40 i} N

Ans.

MRW = { -30i} N # m

Ans.

Ans: F = 53.3 N FRW = { - 40 i} N MRW = { -30i} N # m 368

4–151. The pipe assembly is subjected to the action of a wrench at B and a couple at A. Simplify this system to a resultant wrench and specify the location of the wrench along the axis of pipe CD, measured from point C. Set F = 40 N.

2Fi A

C 0.6 m

0.8 m {60k} N 0.25 m

SOLUTION

B 0.25 m {240i} N

0.3 m 0.3 m

0.5 m y

0.7 m

{260k} N

An equivalent loading at point C: FR = { -40 i} N MRC = - 60(0.5)i - 40(0.8)k - 40(0.6)k MRC = { - 30i + 8k} N # m

The k component of MRC can be eliminated since it is perpendicular to FR. Thus, y =

8 = 0.20 m 40

Ans.

(Measured from C towards D)

The resultant wrench consists of FRW = { -40 i} N

Ans.

MRW = { - 30i} N # m

Ans.

Ans: y = 0.20 m FRW = { -40 i} N MRW = { -30i} N # m 369

*4–152. Replace the loading by an equivalent resultant force and couple moment acting at point O.

50 lb/ft 9 ft O 9 ft 50 lb/ft

SOLUTION + c FR = ΣF ;

Ans.

FR = 0

a+ MRO = ΣMO ; MRO = 225 (6) = 1350 lb # ft = 1.35 kip # ft

Ans.

Ans: FR = 0 MRO = 1.35 kip # ft 370

4–153. Replace loading with with an an equivalent equivalentresultant resultant Replace the the distributed distributed loading force, location on on thethe beam, measured fromfrom O. force,and andspecify specifyitsits location beam measured point O.

3 kN/m

SOLUTION

1.5 m

Loading: The distributed loading can be divided into two parts as shown in Fig. a. Loading: Equations of of Equilibrium: Equilibrium: Equating the forces along the y axis of Figs. a and b, Equations we have + T FR = ©F;

FR =

1 1 (3)(3) + (3)(1.5) = 6.75 kN T 2 2

Ans.

If we equate the moment of FR, Fig. b, to the sum of the moment of the forces in Fig. a about point O, we have a + (MR)O = ©MO;

1 1 (3)(3)(2) - (3)(1.5)(3.5) 2 2 x = 2.5 m

- 6.75(x) = -

Ans.

Ans: FR = 6.75 kN x = 2.5 m 371

4–154. w

Replace the loading by an equivalent resultant force and specify its location on the beam, measured from A.

5 kN/m 2 kN/m A

B 4m

SOLUTION Equivalent Resultant Force: Summing the forces along the y axis by referring to Fig. a 1 +c (FR)y = ΣFy; - FR = - 2(6) - (3)(6) Ans. 2 Ans.

FR = 21.0 kNT Location of the Resultant Force: Summing the moments about point A, 1 a + (MR)A = ΣMA; - 21.0(d) = - 2(6)(3) - (3)(6)(4) 2 d = 3.429 m = 3.43 m

Ans.

Ans: FR = 21.0 kN d = 3.43 m 372

4–155. Currently eighty-five percent of all neck injuries are caused by rear-end car collisions. To alleviate this problem, an automobile seat restraint has been developed that provides additional pressure contact with the cranium. During dynamic tests the distribution of load on the cranium has been plotted and shown to be parabolic. Determine the equivalent resultant force and its location, measured from point A.

A 12 lb>ft

0.5 ft A 0.5 ft

18 lb>ft

0.5

w(x) dx =

12 A 1 + 2 x2 B dx = 12 c x +

0.5

x =

x w(x) dx

= w(x)dx

w 12(1 2x2) lb/ft

B 18 lb/ft

SOLUTION L

FR =

w 12 lb/ft w 5 12(1 1 2x2) lb>ft

x(12) A 1 + 2 x2 B dx 7

12c =

2 3 0.5 x d = 7 lb 3 0

Ans.

x4 0.5 x2 + (2) d 2 4 0 7 Ans.

x = 0.268 ft

Ans: FR = 7 lb x = 0.268 ft 373

*4–156. Determine the length b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is 8 kN # m clockwise.

b 6 kN/m

2 kN/m

SOLUTION Equivalent Resultant Force and Couple Moment At Point A: Summing the forces along the y axis by referring to Fig. a, with the requirement that FR = 0, + c (FR)y = ΣFy; 0 = 2(a + b) -

1 (6)(b) 2

2a - b = 0 (1)

Summing the moments about point A, with the requirement that (MR)A = 8 kN # m, a+ (MR)A = ΣMA; - 8 = 2(a + b) c4 -

1 1 1 (a + b)d - (6)(b) a4 - bb 2 2 3

- 8 = 8a - 4b - 2ab -a2 (2)

Solving Eqs (1) and (2), a = 1.264 m = 1.26 m

Ans.

b = 2.530 m = 2.53 m

Ans.

Ans: a = 1.26 m b = 2.53 m 374

4–157. Replace the loading by an equivalent resultant force and specify its location measured from A.

50 lb

20 lb>ft 5 lb>ft A B 9 ft

5 ft

SOLUTION + T FR = ΣFy; a+ MRA = ΣMA;

1 (15)(9) + 5(9) + 50 = 162 lb T 2 1 - 162.5(d) = - (15)(9)(3) - 5(9)(4.5) - 50(14) 2

FR =

Ans.

d = 6.80 ft

Ans: FR = 162 lb T d = 6.80 ft 375

4–158. The distribution of soil loading on the bottom of a building slab is shown. Replace this loading by an equivalent resultant force and specify its location, measured from point O.

O 50 lb/ft

12 ft

SOLUTION + c FR = ©Fy;

300 lb/ft 9 ft

FR = 50(12) + 12 (250)(12) + 12 (200)(9) + 100(9) = 3900 lb = 3.90 kip c

a + MRo = ©MO;

100 lb/ft

Ans.

3900(d) = 50(12)(6) + 12 (250)(12)(8) + 12 (200)(9)(15) + 100(9)(16.5) Ans.

d = 11.3 ft

Ans: FR = 3.90 kip d = 11.3 ft 376

4–159.

2 kip>ft 1.5 kip>ft

Determine the magnitude of the equivalent resultant force of the distributed loading and specify its location on the beam measured from point A.

4.5 ft

6 ft

SOLUTION + T FR = ΣFy; a+ MRA = ΣMA;

1 (2)(4.5) + 1.5(6) = 13.5 kip 2 1 - 13.5(d) = - (2)(4.5)(3) - 1.5(6)(7.5) 2

FR =

Ans.

d = 6.0 ft

Ans: FR = 13.5 kip d = 6.0 ft 377

*4–160. 6 kN/m

Replace the distributed loading by an equivalent resultant force and couple moment acting at point A.

6 kN/m 3 kN/m

B 3m

SOLUTION Equivalent Resultant Force and Couple Moment About Point A: Summing the forces along the y axis by referring to Fig. a, + c (FR)y = ΣFy;

1 1 FR = - (3)(3) - 3(6) - (3)(3) 2 2 Ans.

= - 27.0 kN = 27.0 kN T Summing the moments about point A, a+ (MR)A = ΣMA;

1 1 (MR)A = - (3)(3)(1) - 3(6)(3) - (3)(3)(5) 2 2

= - 81.0 kN # m = 81.0 kN # m (Clockwise)

Ans.

Ans: FR = 27.0 kN (MR)A = 81.0 kN # m b 378

4–161.

Determine the resultant moment of both the 100-lb force and the triangular distributed load about point O.

20 lb>ft

100 lb

608 A 1 ft

12 ft

SOLUTION 1 2 a+ MRO = ΣMO; MRO = - (20)(12)a b(12) - 100 sin 60 °(12) 2 3

+ 100 cos 60°(1)

= - 1949 lb # ft = 1.95 kip # ft b

Ans.

Ans: MRO = 1.95 kip # ftb 379

4–162. If the soil exerts a trapezoidal distribution of load on the bottom of the footing, determine the intensities w1 and w2 of this distribution needed to support the column loadings.

80 kN

60 kN

50 kN 2.5 m

3.5 m

w2 w1

SOLUTION + T FR = ΣFy; FR =

1 1 (5)(3) + 5(3) + (3)(3) = 27 kN 2 2

a+ MRB = ΣMB; 27d =

Ans.

1 1 (3)(3)(1) + 5(3)(1.5) + (5)(3)(4) 2 2 Ans.

d = 2.11 m

Ans: FR = 27 kN d = 2.11 m 380

4–163. The form is used to cast a concrete wall having a width of 5 m. Determine the equivalent resultant force the wet concrete exerts on the form AB if the pressure distribution due to the concrete can be approximated as shown. Specify the location of the resultant force, measured from point B.

p (4z 2 ) kPa 4m

SOLUTION L

dA =

1 2

8 kPa

4z dz

4 3 2 = c (4)z2 d 3 0

= 21.33 kN>m Ans.

FR = 21.33(5) = 107 kN L

zdA =

4z2 dz

4 5 2 = c (4)z2 d 5 0

= 51.2 kN z =

51.2 = 2.40 m 21.33

Ans.

Also, from the back of the book, A =

2 2 ab = (8)(4) = 21.33 3 3

FR = 21.33 (5) = 107 kN

Ans.

z = 4 - 1.6 = 2.40 m

Ans.

Ans: FR = 107 kN z = 2.40 m 381

*4–164. Replace the loading by an equivalent resultant force and couple moment acting at point O.

200 lb>ft w 5 2 x2 lb>ft

O x

10 ft

SOLUTION + T FR = ΣFy; FR = 1 dA = 10 2x dx = 10

x- =

1A xdA = 1A dA

10 3 10 2x dx

667

3 10

2x ` 3 0

= 667 lb T

Ans.

x4 10 ` 2 0 = = 7.50 ft 667

a+ MRO = FR(10 - x) = 667(10 - 7.5) = 1667 lb # ft = 1.67 kip # ft

382

Ans.

4–165. Replace the loading on the beam by an equivalent resultant force and specify its location, measured from point A.

2.5 kN/m

1.5 kN 0.5 kN/m

SOLUTION + T FR = ©F ; c +MRA = ©MA ;

Ans.

FR = 3 + 1.5 + 1.5 = 6 kN T 6(d) = 3 (1) + 1.5 (1.5) + 1.5 (4)

Ans.

d = 1.875 = 1.88 m

Ans: FR = 6 kN d = 1.88 m 383

4–166. Replace the loading by an equivalent resultant force and couple moment acting at point A.

400 N/m

A 3m

SOLUTION Equivalent Resultant Force and Couple Moment At Point A: Summing the forces along the y axis by referring to Fig. a, + c (FR)y = ΣFy;

FR = - 400(3) -

1 (400)(3) 2 Ans.

= - 1800 N = 1.80 kN T Summing the moment about point A, a+ (MR)A = ΣMA;

(MR)A = - 400(3)(1.5) -

1 (400)(3)(4) 2

= - 4200 N # m = 4.20 kN # m (Clockwise) Ans. a+ MRO = ΣMO;

1 2 MRO = - (20)(12)a b(12) - 100 sin 60°(12) + 100 cos 60°(1) 2 3 = - 1949 lb # ft = 1.95 kip # ft b

Ans.

Ans: FR = 1.80 kN

(MR)A = 4.20 kN # m b

384

4–167. 400 N/m

Replace the loading by a single resultant force, and specify its location on the beam measured from point A.

A 3m

SOLUTION Equivalent Resultant Force: Summing the forces along the y axis by referring to Fig. a, + c (FR)y = ΣFy;

- FR = - 400(3) -

1 (400)(3) 2 Ans.

FR = 1800 N = 1.80 kN T

Location of Resultant Force: Summing the moment about point A by referring to Fig. a, 1 a+ (MR)A = ΣMA; - 1800 d = - 400(3)(1.5) - (400)(3)(4) 2 Ans.

d = 2.333 m = 2.33 m

Ans: FR = 1.80 kN d = 2.33 m 385

*4–168. 3 kN/m

Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects a horizontal line along member AB, measured from A.

A 3m

2 kN/m

SOLUTION Equivalent Resultant Force: Summing the forces along the x and y axes by referring to Fig. a, + (FR)x = ΣFx; (FR)x = - 2(4) = - 8 kN = 8 kN d S + c (FR)y = ΣFy; (FR)y = - 3(3) = - 9 kN = 9 kN T Then

And

FR = 2(FR)2x + (FR)2y = 282 + 92 = 12.04 kN = 12.0 kN u = tan-1 c

(FR)y (FR)x

Ans.

9 d = tan-1 a b = 48.37° = 48.4° d Ans. 8

Location of the Resultant Force: Summing the moments about point A, by referring to Fig. a, a+ (MR)A = ΣMA ; - 8x - 9y = -3(3)(1.5) - 2(4)(2) 8x + 9y = 29.5

(1)

386

*4–168. Continued

Along AB, x = 0. Then Eq (1) becomes 8(0) + 9y = 29.5 y = 3.278 m Thus, the inter section point of the line of action of FR on AB measured to the right from point A is d = y = 3.28 m Ans.

Ans: FR = 12.0 kN u = 48.4° d y = 3.28 m 387

4–169. Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects a vertical line along member BC, measured from C.

3 kN/m B

A 3m

2 kN/m

And

FR = 2(FR)2x + (FR)2y = 282 + 92 = 12.04 kN = 12.0 kN u = tan-1 c

(FR)y (FR)x

Ans.

9 d = tan-1 a b = 48.37° = 48.4° d Ans. 8

Location of the Resultant Force: Summing the moments about point A, by referring to Fig. a, a+ (MR)A = ΣMA; - 8x - 9y = - 3(3)(1.5) - 2(4)(2) 8x + 9y = 29.5

(1)

388

4–169. Continued

Along BC, y = 3 m. Then Eq (1) becomes 8x + 9(3) = 29.5 x = 0.3125 m Thus, the intersection point of line of action of FR on BC measured upward from point C is d = 4 - x = 4 - 0.3125 = 3.6875 m = 3.69 m

Ans.

Ans: FR = 12.0 kN u = 48.4° d d = 3.69 m 389

4–170. Determine the the triangular triangular load and and its its Determine the length length b of the thatthat the equivalent resultant force position aa on onthe thebeam beamsosuch the equivalent resultant is zero and and the the resultant couple kN # m m force is zero resultant couplemoment moment isis 88 kN clockwise. clockwise.

b 4 kN/m

A 2.5 kN/m

SOLUTION + c FR = 0 = ©Fy ; a + MRA = ©MA ;

0 =

1 1 (2.5)(9) - (4)(b) 2 2

b = 5.625 m

Ans.

1 2 1 - 8 = - (2.5)(9)(6) + (4)(5.625) a a + (5.625) b 2 2 3 Ans.

a = 1.54 m

Ans: b = 5.625 m a = 1.54 m 390

4–171. Determine the magnitude of the equivalent resultant force of the distributed load and specify its location on the beam measured from point A.

w 420 lb/ft

w = (5 (x – 8)2 +100) lb/ft 100 lb/ft

120 lb/ft

SOLUTION 8 ft

Equivalent Equivalent Resultant ResultantForce: Force:

2 ft

+ c FR = ©Fy ;

-FR = -

dA = -

wdx

10ft

FR =

351x - 822 + 1004dx

Ans.

= 1866.67 lb = 1.87 kip T Location Location of of Equivalent EquivalentResultant ResultantForce: Force: x

' LA x =

xdA = dA

L0 L0

xwdx x

wdx

10ft

x351x - 822 + 1004dx

10ft

351x - 822 + 1004dx

L0 10ft

15x3 - 80x2 + 420x2dx 10ft

351x - 822 + 1004dx Ans.

= 3.66 ft

Ans: FR = 1.87 kip x = 3.66 ft 391

*4–172. If the distribution of the ground reaction on the pipe per foot of length can be approximated as shown, determine the magnitude of the resultant force due to this loading.

25 lb/ft 2.5 ft u w 25 (1 cos u) lb/ft

SOLUTION

50 lb/ft

Resultant Components: The magnitude of the differential force dFR is equal to the area of the element shown shaded in Fig. a. dFR = wr du = 25(1 + cos u)(2.5 du) = 62.5(1 + cos u)du The horizontal and vertical components of dFR are given by + ;

(dFR)x = dFR sin u = 62.5(1 + cos u)sin u du = 62.5 a sin u +

sin 2u bdu 2

(dFR)y = dFR cos u = 62.5(1 + cos u)cos u du - 62.5 a cos u +

cos 2u + 1 bdu 2

-p p rad to u = rad gives the horizontal 2 2 and vertical components of the resultant for FR. Integrating 1dFR2x and 1dFR2y from u = + (F ) = ; R x

p>2

62.5 a sin u +

sin 2u + 1 cos 2u 2 p>2 b du = 62.5a cos u + b = 0 2 4 -p>2

62.5a cos u +

p>2 cos 2u + 1 sin 2u 1 p b du = 62.5 asin u + + ub 2 = 62.5a 2 + b = 223.17 lb c 2 4 2 2 -p>2

L-p>2 p>2

+ c (FR)y =

L-p>2

Thus, FR = (FR)y = 223.17 lb = 223 lb c

Ans.

Ans: FR = 223 lb 392

4–173. Replace the loading by an equivalent resultant force and specify its location on the beam, measured from point O.

5 kN/m

4 kN/m 2 kN/m

SOLUTION + T FR = ©Fy;

FR =

1 1 (3)(6) + 2(9) + (2)(3) 2 2 Ans.

= 30 kN T c +MRO = ©MO;

1 1 - 30d = - (3)(6)(2) - 2(9)(4.5) - (2)(3)(8) 2 2 Ans.

d = 4.10 m

Ans: FR = 30 kN d = 4.10 m 393

4–174. Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point A.

w (x2 3x 100) lb/ft

370 lb/ft

100 lb/ft A B

SOLUTION Resultant: The magnitude of the differential force dFR is equal to the area of the element shown shaded in Fig. a. Thus,

15 ft

dFR = w dx = a x2 + 3x + 100b dx Integrating dFR over the entire length of the beam gives the resultant force FR. L

FR =

dFR =

a x2 + 3x + 100 bdx = ¢

15 ft 3x2 x3 + + 100x ≤ ` 3 2 0

Ans.

= 2962.5 lb = 2.96 kip

Location: Location: The location of dFR on the beam is xc = x measured from point A. Thus, the location x of FR measured from point A is given by 15 ft

x =

xcdFR

= dFR

xa x2 + 3x + 100b dx 2962.5

15 ft x4 + x3 + 50x2 ≤ ` 4 0

2962.5

= 9.21 ft Ans.

Ans: FR = 2.96 kip x = 9.21 ft 394

4–175. w

Replace the loading by an equivalent resultant force and couple moment acting at point O.

p x w w0 cos ( 2L (

O L

SOLUTION Equivalent Resultant Force and Couple Moment About Point O: The differential p xbdx. Thus, summing the force indicated in Fig. a is dFR = w dx = aw0 cos 2L forces along the y axis, L p + c (FR)y = ΣFy; FR = - dFR = ¢w0 cos x≤dx 2L L LO L 2Lw0 p asin xb ` p 2L O 2Lw0 2Lw0 = = T p p Summing the moments about point O,

= -

a+ (MR)O = ΣMO;

(MR)O = = -

xdFR

x aw0 cos

= - w0 a = -a = a

Ans.

p xb dx 2L

L 2L p 4L2 p x sin x + 2 cos xb ` p 2L 2L p O

2p - 4 bw0L2 p2

2p - 4 bw0L2 (Clockwise) p2

Ans.

Ans: FR =

2Lw0 p

(MR)O = a 395

2p - 4 bw0L2 b p2

*4–176. pp

Wet concrete exerts a pressure distribution along the wall of the form. Determine the resultant force of this distribution and specify the height h where the bracing strut should be placed so that it lies through the line of action of the resultant force. The wall has a width of 5 m.

1>2

p 5 (40 z 1/)2 kPa p (4 z ) kPa

4m 4m

SOLUTION Equivalent Resultant ResultantForce: Force: Equivalent

h h

+ F = ©F ; : R x

-FR = - LdA = 4m

FR =

80 kPa 8 kPa

wdz

3 3 (200)(10 ) dz a20z2 b A 10 B dz

z z

= A 103)B N = 107 = 106.67 1066.7(10 1.07kN mN;d

Ans.

Location of of Equivalent EquivalentResultant ResultantForce: Force: Location z

z =

zdA = dA

zwdz

wdz

zc(200z (1033))ddz A 20z2 2B)(10

L0 4m

1 1

33 A 20z2 B2(10 (200z )(10)dz )dz 33

c(200z (10 33))ddz A 20z22B)(10

1 1

3 (200z )(103)dz )dz A 20z2 B2(10

= 2.40 m Thus,

Ans.

h = 4 - z = 4 - 2.40 = 1.60 m

Ans: FR = 1.07 kN h = 1.60 m 396

*5–1.

75 N/m

Draw the free-body diagram for the following problems. a) The beam in Prob. 5–10.

800 N · m

b) The beam in Prob. 5–11. B c) 8The m beam in Prob. 5–12. d) The beam in Prob. 5–14.

SOLUTION a)

4 kN

30 2m

e beam, Fig. a, NB can bout point A. c) Ans.

ium along the x and

Ans.

Ans. d) 800 N/m A

B 1m

397

*5–2. Draw the free-body diagram for the following problems. a) The beam in Prob. 5–15. b) The linkage in Prob. 5–16. c) The frame in Prob. 5–17. d) The beam in Prob. 5–18.

SOLUTION a)

b) 4m

450 N 600 N · m

ctly by

A 2m

Ans. Ans.

Ans.

398

*5–3. Draw the free-body diagram for the following problems. a) The lamp in Prob. 5–19. b) The rod in Prob. 5–20. c) The assembly in Prob. 5–21.

u B

d) The beam in Prob. 5–22. r A

SOLUTION a)

Ans.

399

100 lb

*5–4.

3 ft for the following problems. Draw the3 ft free-body diagram 200 lb ft

a) The rod in Prob. 5–25. b) The bar in Prob. 5–27.

2 ft

c) The4disk in Prob. 5–28. 5

13 12

5 b (2) = 0 SOLUTION 3 a)

Ans. Ans. Ans. b)

400

*5–5. Draw the free-body diagram for the following problems. a) The clamp in Prob. 5–32. b) The jib crane in Prob. 5–33. c) The crane in Prob. 5–35. d) The beam in Prob. 5–36.

SOLUTION a)

A 0.75 m

G F B

Ans.

401

*5–6. Draw the free-body diagram for the following problems. 8 ft

B jib crane in Prob. 5–37. a) The T b) The bar in Prob. 5–39.

c) The 4 ft davit in Prob. 5–41. x

d) The boom in Prob. 5–42. 780 lb

g. a, we tion of y axis,

SOLUTION a)

Ans. Ans.

, Ans.

5 4

1.5 m

D F2

Ans.

30 A

Ans.

d) Ans.

402

*5–7. Draw the free-body diagram for the following problems. a) The rod in Prob. 5–44. b) The hand truck and load when it is lifted in Prob. 5–45. c) The beam in Prob. 5–47. d) The boom frame in Prob. 5–52.

SOLUTION a)

403

*5–8.

B Draw the free-body diagram for the following problems. A

a) The beam in Prob. 5–53. 100 mm b) The brake pedal Prob. 5–54.

c) TheBexercise machine Prob.A5–55. k

d) The hatch door in Prob. 5–58. 50 mm d directly by D

SOLUTION Ans.

40 mm

30 C

10 mm

b) Ans.

Ans. Ans. c)

404

*5–9. Draw the free-body diagram for the following problems. a) The crane in Prob. 5–59. b) The smooth pipe in Prob. 5–61. c) The bar in Prob. 5–62.

SOLUTION a)

c) d

405

5–10. Determine the reactions at the supports.

75 N/m 800 N · m A

B 8m

SOLUTION c + ©MA = 0;

600 (4) + 800 - (8) By = 0 Ans.

By = 400 N + ©F = 0; : x

Ax = 0

+ c ©Fy = 0;

Ay - 600 + 400 = 0

Ans.

Ay = 200 N

Ans: By = 400 N Ax = 0 Ay = 200 N 406

5–11. Determine the reactions at the supports.

400 N/m 5

3 4

B A 3m

SOLUTION Equations of Equilibrium: NA and By can be determined directly by writing the moment equations of equilibrium about points B and A, respectively, by referring to the beam’s FBD shown in Fig. a. 1 4 a+ ΣMB = 0; (400)(6)(3) - NA a b(6) = 0 2 5 NA = 750 N

a+ ΣMA = 0; By(6) -

Ans.

1 (400)(6)(3) = 0 2 Ans.

By = 600 N

Using the result of NA to write the force equation of equilibrium along the x axis, + ΣFx = 0; 750 a 3 b - Bx = 0 S 5

Ans.

Bx = 450 N

Ans: NA = 750 N By = 600 N Bx = 450 N 407

*5–12. 4 kN

Determine the horizontal and vertical components of reaction at the pin A and the reaction of the rocker B. on the beam.

SOLUTION Equations of Equilibrium: From the free-body diagram of the beam, Fig. a, NB can be obtained by writing the moment equation of equilibrium about point A. a + ©MA = 0;

NB cos 30°(8) - 4(6) = 0 Ans.

NB = 3.464 kN = 3.46 kN

Using this result and writing the force equations of equilibrium along the x and y axes, we have + ©F = 0; : x

A x - 3.464 sin 30° = 0 Ans.

A x = 1.73 kN + c ©Fy = 0;

A y + 3.464 cos 30° - 4 = 0 Ans.

A y = 1.00 kN

Ans: NB = 3.46 kN Ax = 1.73 kN Ay = 1.00 kN 408

5–13. Determine the reactions at the supports.

900 N/m 600 N/m

A 3m

SOLUTION

Equations of Equilibrium: NA and By can be determined directly by writing the moment equations of equilibrium about points B and A, respectively, by referring to the FBD of the beam shown in Fig. a. a+ ΣMB = 0; 600(6)(3) +

1 (300)(3)(5) - NA(6) = 0 2 Ans.

NA = 2175 N = 2.175 kN a+ ΣMA = 0; By(6) -

1 (300)(3)(1) - 600(6)(3) = 0 2 Ans.

By = 1875 N = 1.875 kN

Also, Bx can be determined directly by writing the force equation of equilibrium along the x axis. + ΣFx = 0; Bx = 0 S

Ans.

Ans: NA = 2.175 kN By = 1.875 kN Bx = 0 409

5–14. Determine the reactions at the supports. 800 N/m A

B 1m

NA = 3333.33 N = 3.33 kN

Using this result to write the force equations of equilibrium along the x and y axes, + ΣFx = 0; Bx - 800(5) a 3 b = 0 S 5

Ans.

Bx = 2400 N = 2.40 kN

4 + c ΣFy = 0; 3333.33 - 800 (5)a b - By = 0 5

Ans.

By = 133.33 N = 133 N

Ans: NA = 3.33 kN Bx = 2.40 kN By = 133 N 410

5–15. Determine the reaction at fixed support A.

3 kN 0.5 kN>m 5 kN · m A 6m

SOLUTION + ΣFx = 0; S

Ans.

Ax = 0

1 + c ΣFy = 0; Ay – 0.5(6) – (0.5)(3) – 3 = 0 2 a+ ΣMA = 0; MA - 0.5(6)(3) -

Ay = 6.75 kN

Ans.

1 (0.5)(3)(7) - 3(9) - 5 = 0 2

MA = 46.25 kN # m

Ans.

Ans: Ax = 0 Ay = 6.75 kN MA = 46.25 kN # m 411

*5–16. The linkage supports a force of 500 lb and rides along the top and bottom flanges of the crane rail. Determine the force of each roller on the flange.

8 in.

3 in.

SOLUTION a+ ΣMA = 0;

FB(8) - 500(11) = 0

+ c ΣFy = 0;

500 lb

Ans.

FB = 687.5 = 688 lb 687.5 - 500 - FA = 0

Ans.

FA = 187.5 = 188 lb

Ans: FB = 688 lb FA = 188 lb 412

5–17. Determine the reactions at the roller A and pin B.

450 N 600 N · m

SOLUTION Equilibrium: The vertical reaction Ay can be obtained directly by Equations of Equilibrium: summing moments about point B.

A 2m

a + ©MB = 0;

Ans.

Ay = 390 N + ©F = 0; : x + c ©Fy = 0;

450(3) + 600 - Ay (5) = 0

Ans.

Bx = 0 By + 390 - 450 = 0

Ans.

By = 60.0 N

Ans: Ay = 390 N Bx = 0 By = 60.0 N 413

5–18. Determine the reactions at the roller B, the rocker C, and where the beam contacts the smooth plane at A. Neglect the thickness of the beam.

800 N 3

500 N

608

B 4m

C 2m

SOLUTION + ΣFx = 0; S

3 R - 800 cos 60° = 0 5 A Ans.

RA = 666.7 = 667 N a+ ΣMC = 0;

4 - (666.7)(12) - RB (8) + 500(8) + 800 sin 60°(6) = 0 5 Ans.

RB = 219.6 = 220 N + c ΣFy = 0;

4 (666.7) - 500 + 219.6 - 800 sin 60° + RC = 0 5 Ans.

RC = 440 N

Ans: RA = 667 N RB = 220 N RC = 440 N 414

5–19. The 10-kg lamp has a center of mass at G. Determine the horizontal and vertical components of reaction at A, and the force in the cable BC.

0.15 m

0.5 m

C 0.3 m B

0.4 m

SOLUTION

Equations of Equilibrium: Here link BC is a two force member. FBC can be determined directly by writing the moment equation of equilibrium about point A. Referring to the FBD of the lamp, Fig. a. a+ ΣMA = 0; FBC a

b(0.15) + FBC a

b(0.4) - 10(9.81)(0.65) = 0 25 25 FBC = 203.69 N = 204 N Ans.

Using this results of FBC to write the force equation of equilibrium along x and y axes, + ΣFx = 0; Ax - 203.69 a 1 b = 0 S 25 Ax = 91.09 N = 91.1 N + c ΣFy = 0; 203.69 a

Ans.

b - 10(9.81) - Ay = 0 25 Ay = 84.09 N = 84.1 N

Ans.

Ans: FBC = 204 N Ax = 91.1 N Ay = 84.1 N 415

*5–20. A uniform glass rod having a length L is placed in the smooth hemispherical bowl having a radius r. Determine the angle of inclination u for equilibrium.

u B r A

SOLUTION By observation f = u. Equilibrium: Equations of Equilibrium: L cos ub = 0 2

a + ©MA = 0;

NB (2r cos u) - W a

+Q ©Fx = 0;

NA cos u - W sin u = 0

+a©Fy = 0;

(W tan u) sin u +

WL 4r

NA = W tan u

WL - W cos u = 0 4r

sin2 u - cos2 u +

L cos u = 0 4r

(1 - cos2 u) - cos2 u + 2 cos2 u cos u =

NB =

L cos u = 0 4r

L cos u - 1 = 0 4r

L ; 2L2 + 128r2 16r

Take the positive root cos u =

L + 2L2 + 128r2 16r

u = cos - 1 ¢

L + 2L2 + 128r2 ≤ 16r

Ans.

Ans: u = cos - 1a 416

L + 2L2 + 12r 2 b 16r

5–21. The 2.5-Mg assembly has a center of mass at G and is hoisted with a constant velocity. Determine the force in each cable and the distance d for equilibrium.

P 4m

4m A 3m

C 1m

B G

1.25 m

SOLUTION Equation of Equilibrium: First consider the equilibrium of the entire hoisting system. Referring to the FBD of the system, Fig. a.

+ c ΣFy = 0; P - 2500(9.81) = 0 P = 24.525(103) N a+ ΣMG = 0;

324.525(103) 4d = 0 d = 0

Ans.

Subsequently, using the result of P to consider the equilibrium of ring A, Fig. b, + ΣFx = 0; FAB cos 45° - FAC a 4 b = 0 [1] S 5

3 + c ΣFy = 0; 24.525(103) - FAC a b - FAB sin 45° = 0 [2] 5

Solving Eqs. 1 and 2

FAC = 17.52(103) N = 17.5 kN

Ans.

FAB = 19.82(103) N = 19.8 kN

Ans.

Ans: d = 0 FAC = 17.5 kN FAB = 19.8 kN 417

5–22. If the intensity of the distributed load acting on the beam is w = 3 kN>m, determine the reactions at the roller A and pin B.

A w 30

3m 4m

NA = 3.713 kN = 3.71 kN

Using this result to write the force equation of equilibrium along the x and y axes, + ΣFx = 0; 3.713 sin 30° - Bx = 0 S Ans.

Bx = 1.856 kN = 1.86 kN + c ΣFy = 0; By + 3.713 cos 30° - 3(4) = 0

Ans.

By = 8.7846 kN = 8.78 kN

Ans: NA = 3.71 kN Bx = 1.86 kN By = 8.78 kN 418

5–23. If the roller at A and the pin at B can support a load up to 4 kN and 8 kN, respectively, determine the maximum intensity of the distributed load w, measured in kN>m, so that failure of the supports does not occur.

A w 30

3m 4m

SOLUTION Equations of Equilibrium: NA can be determined directly by writing the moment equation of equilibrium about point B by referring to the FBD of the beam shown in Fig. a. a+ ΣMB = 0; w(4)(2) - NA sin 30° (3 sin 30°) - NA cos 30° (3 cos 30° + 4) = 0 NA = 1.2376 w Using this result to write the force equation of equilibrium along x and y axes, + ΣFx = 0; 1.2376 w sin 30° - Bx = 0 S

Bx = 0.6188 w

+ c ΣFy = 0; By + 1.2376 w cos 30° - w(4) = 0

By = 2.9282 w

Thus, FB = 2Bx2 + By2 = 2(0.6188 w)2 + (2.9282 w)2 = 2.9929 w

It is required that FB 6 8 kN;

2.9929 w 6 8

w 6 2.673 kN>m

1.2376 w 6 4

w 6 3.232 kN>m

And NA 6 4 kN;

Thus, the maximum intensity of the distributed load is Ans.

w = 2.673 kN>m = 2.67 kN>m

Ans: w = 2.67 kN>m 419

*5–24. The relay regulates voltage and current. Determine the force in the spring CD, which has a stiffness of k 120 N m, so that it will allow the armature to make contact at A in figure (a) with a vertical force of 0.4 N. Also, determine the force in the spring when the coil is energized and attracts the armature to E, figure (b), thereby breaking contact at A.

50 mm 50 mm 30 mm 10°

C k

SOLUTION From Fig. (a): a + ©MB = 0;

0.4(100 cos 10°) - Fs (30 cos 10°) = 0

(a)

Ans.

Fs = 1.333 N = 1.33 N Fs = kx;

(b)

1.333 = 120 x x = 0.01111 m = 11.11 mm

From Fig (b), energizing the coil requires the spring to be stretched an additional amount ¢x = 30 sin 10° = 5.209 mm. Thus x¿ = 11.11 + 5.209 = 16.32 mm Ans.

Fs = 120 (0.01632) = 1.96 N

Ans: Fs = 1.33 N Fs = 1.96 N 420

5–25. Determine the reactions on the bent rod which is supported by a smooth surface at B and by a collar at A, which is fixed to the rod and is free to slide over the fixed inclined rod.

100 lb 3 ft

3 ft

200 lb ft

2 ft

3 5

13 12

SOLUTION 12 5 b (6) - NB a b (2) = 0 13 13

a + ©MA = 0;

MA - 100 (3) - 200 + NB a

+ ©F = 0; : x

4 5 NA a b - NB a b = 0 5 13

+ c ©Fy = 0;

3 12 NA a b + NB a b - 100 = 0 5 13

Solving, NA = 39.7 lb

Ans.

NB = 82.5 lb

Ans.

MA = 106 lb # ft

Ans.

Ans: NA = 39.7 lb NB = 82.5 lb MA = 106 lb # ft 421

5–26. The file cabinet contains four uniform drawers, each 2.5 ft long and weighing 30 lb/ft. If the empty cabinet (without drawers) has a weight of 40 lb and a center of mass at G, determine how many of the drawers can be fully pulled out and the extension x of the last drawer that will cause the assembly to be on the verge of tipping over.

2.5 ft

G x

SOLUTION Assume top two drawers are pulled out and third is partially opened as shown. Thus,

1.5 ft

a+ ΣMA = 0; 75(1.25) + 40(1.5) + 75(x′) - 75(1.25) - 75(1.25) = 0 x′ = 0.450 ft Ans.

x = 1.25 - x′ = 1.25 - 0.450 = 0.8 ft

Ans: x = 0.8 ft 422

5–27. Determine the reactions acting on the smooth uniform bar, which has a mass of 20 kg.

30º

60º

SOLUTION Equations of Equilibrium: NB can be determined directly by writing the moment equation of equilibrium about point A by referring to the FBD of the bar shown in Fig. a. a+ ΣMA = 0; NB cos 30°(4) - 20(9.81) cos 30°(2) = 0 Ans.

NB = 98.1 N

Using this result to write the force equation of equilibrium along the x and y axes, + ΣFx = 0; Ax - 98.1 sin 60° = 0 Ax = 84.96 N = 85.0 N Ans. S + c ΣFy = 0; Ay + 98.1 cos 60° - 20(9.81) = 0 Ans.

Ay = 147.15 N = 147 N

Ans: NB = 98.1 N Ax = 85.0 N Ay = 147 N 423

*5–28. The disk has a mass of 10 kg. If it is suspended from a spring having an unstretched length of 400 mm, determine the angle u for equilibrium.

k 5 600 N>m

SOLUTION Geometry:

200 mm

l =

Equilibrium:

0.2 sin u

Fsp = k(l - l0) = 600 a

+ c ΣFy = 0; 600 a

0.2 - 0.4b sin u

0.2 - 0.4b cos u - 10(9.81) = 0 sin u

120 cot u - 240 cos u = 98.1

Solving by trial and error: Ans.

u = 20.376° = 20.4°

Ans: u = 20.4° 424

5–29. Determine the force P needed to pull the 50-kg roller over the smooth step. Take u = 30°.

u A

50 mm

300 mm

SOLUTION Equations of Equilibrium: P can be determined directly by writing the moment equation of Equilibrium about point B, by referring to the FBD of the roller shown in Fig. a. a+ ΣMB = 0; P cos 30°(0.25) + P sin 30° ( 20.32 - 0.252 2 - 50(9.81) 20.32 - 0.252 = 0 Ans.

P = 271.66 N = 272 N

Ans: P = 272 N 425

5–30. Determine the magnitude and direction u of the minimum force P needed to pull the 50-kg roller over the smooth step.

u A

50 mm

300 mm

SOLUTION Equations of Equilibrium: P will be minimum if its orientation produces the greatest moment about point B. This happens when it acts perpendicular to AB as shown in Fig. a. Thus u = f = cos-1 a

0.25 b = 33.56° = 33.6° 0.3

Ans.

Pmin can be determined by writing the moment equation of equilibrium about point B by referring to the FBD of the roller shown in Fig. b. a+ ΣMB = 0; Pmin (0.3) - 50(9.81)(0.3 sin 33.56°) = 0 Ans.

Pmin = 271.13 N = 271 N

Ans: u = 33.6° Pmin = 271 N 426

5–31. The mechanism shown was thought by its inventor to be a perpetual-motion machine. It consists of the stand A, two smooth idler wheels B and C, and in between a uniform hollow cylindrical ring D suspended in the manner shown. The ring has a weight W and it was expected to revolve in the direction indicated by the arrow. Draw a free-body diagram of the ring and, using an appropriate equation of equilibrium, show that it will not rotate.

SOLUTION a+ ΣMO = 0; Therefore, the ring will not turn.

W FB D NC

Ans: The ring will not turn. 427

*5–32. The hold-down clamp exerts a compressive force of 400 N on the wood block at C. If the clamp is loosely bolted to the bench using two symmetrically placed bolts B, one of which is shown, determine the force along the axis of each bolt and the vertical reaction of the bench on the clamp at A.

C A

SOLUTION a+ ΣMA = 0;

400(0.4) - 2FB(0.15) = 0

FB = 533 N

Ans.

+ c ΣFy = 0;

NA + 400 - 2(533.33) = 0

NA = 667 N

Ans.

B 150 mm

250 mm

Ans: FB = 533 N NA = 667 N 428

5–33. If the crane has a mass of 800 kg and a center of mass at G, and the maximum rated force at its end is F = 15 kN, determine the reactions at its bearings. The bearing at A is a journal bearing and supports only a horizontal force, whereas the bearing at B is a thrust bearing that supports both horizontal and vertical components.

3m A 0.75 m 2m

G F

SOLUTION a + ©MB = 0;

Ans.

Ax = 25.4 kN + c ©Fy = 0;

By - 800 (9.81) - 15 000 = 0 Ans.

By = 22.8 kN + ©F = 0; : x

Ax (2) - 800 (9.81) (0.75) - 15 000(3) = 0

Bx - 25.4 = 0 Ans.

Bx = 25.4 kN

Ans: Ax = 25.4 kN By = 22.8 kN Bx = 25.4 kN 429

5–34. The crane has a mass of 800 kg and a center of mass at G. The bearing at A is a journal bearing and can support a horizontal force, whereas the bearing at B is a thrust bearing that supports both horizontal and vertical components. Determine the maximum load F that can be suspended from its end if the selected bearings at A and B can sustain a maximum resultant load of 24 kN and 34 kN, respectively.

3m A 0.75 m 2m

G F

SOLUTION

a + ©MB = 0;

Ax (2) - 800 (9.81) (0.75) - F (3) = 0

+ c ©Fy = 0;

By - 800 (9.81) - F = 0

+ ©F = 0; : x

Bx - Ax = 0

Assume Ax = 24 000 N. Solving, Bx = 24 kN By = 21.9 kN Ans.

F = 14.0 kN FB =

(24)2 + (21.9)2 = 32.5 kN 6 34 kN

Ans: F = 14.0 kN 430

5–35. The crane lifts the 400-kg load L. The primary boom AB has a mass of 1.20 Mg and a center of mass at G1, whereas the secondary boom BC has a mass of 0.6 Mg and a center of mass at G2. Determine the tension in the cable BD and the horizontal and vertical components of reaction at the pin A.

3m C

3m G2

308

358

8m G1 D

SOLUTION a+ ΣMA = 0;

TBD cos 35°(16 sin 45°) - TBD sin 35°(16 cos 45°)

458

- 1200(9.81)(8 cos 45°) - 600(9.81)(16 cos 45° + 3 cos 30°)

- 400(9.81)(16 cos 45° + 6 cos 30°) = 0 Ans.

TBD = 76757.9 N = 76.8 kN + c ΣFy = 0;

Ay - 76757.9 sin 35° - 1200(9.81) - 600(9.81) - 400(9.81) = 0 Ans.

Ay = 65608.5 N = 65.6 kN + ΣFx = 0; Ax - 76757.9 cos 35° = 0 S

Ans.

Ax = 62876.4 N = 62.9 kN

Ans: TBD = 76.8 kN Ay = 65.6 kN Ax = 62.9 kN 431

*5–36. If the beam is horizontal and the springs are unstretched when the load is removed, determine the angle of tilt of the beam when the load is applied.

A 600 N/m

kA = 1 kN/m

kB = 1.5 kN/m D

C 3m

SOLUTION Equations of Equilibrium: FA and FB can be determined directly by writing the moment equations of equilibrium about points B and A, respectively, by referring to the FBD of the beam shown in Fig. a. Assuming that the angle of tilt is small, a+ ΣMA = 0; FB(6) -

1 (600)(3)(2) = 0 FB = 300 N 2

1 a+ ΣMB = 0; (600)(3)(4) - FA(6) = 0 FA = 600 N 2 Thus, the stretches of springs A and B can be determined from FA = kAxA;

600 = 1000 xA

xA = 0.6 m

FB = kB xB;

300 = 1500 xB

xB = 0.2 m

From the geometry shown in Fig. b u = sin-1 a

0.4 b = 3.82° 6

Ans.

The assumption of small u is confirmed.

Ans: u = 3.82° 432

5–37. The cantilevered jib crane is used to support the load of 780 lb. If x = 5 ft, determine the reactions at the supports. Note that the supports are collars that allow the crane to rotate freely about the vertical axis. The collar at B supports a force in the vertical direction, whereas the one at A does not.

8 ft B T 4 ft

SOLUTION

780 lb

Equations of Equilibrium: Referring to the FBD of the jib crane shown in Fig. a, we notice that NA and By can be obtained directly by writing the moment equation of equilibrium about point B and force equation of equilibrium along the y axis, respectively. a+ ©MB = 0;

NA(4) - 780(5) = 0

NA = 975 lb

Ans.

+ c ©Fy = 0;

By - 780 = 0

By = 780

Ans.

Using the result of NA to write the force equation of equilibrium along x axis, + : ©Fx = 0;

975 - Bx = 0

Ans.

Bx = 975 lb

Ans: NA = 975 lb Bx = 975 lb By = 780 lb 433

5–38. The cantilevered jib crane is used to support the load of 780 lb. If the trolley T can be placed anywhere between 1.5 ft … x … 7.5 ft, determine the maximum magnitude of reaction at the supports A and B. Note that the supports are collars that allow the crane to rotate freely about the vertical axis. The collar at B supports a force in the vertical direction, whereas the one at A does not.

8 ft B T 4 ft

x 780 lb

SOLUTION

Require x = 7.5 ft a + ©MA = 0;

- 780(7.5) + Bx (4) = 0 Bx = 1462.5 lb

+ ©F = 0; : x

Ax - 1462.5 = 0 Ans.

Ax = 1462.5 = 1462 lb + c ©Fy = 0;

By - 780 = 0 By = 780 lb FB = 2(1462.5)2 + (780)2 Ans.

= 1657.5 lb = 1.66 kip

Ans: Ax = 1.46 kip FB = 1.66 kip 434

5–39. The bar of negligible weight is supported by two springs, each having a stiffness k = 100 N>m. If the springs are originally unstretched, determine the angle u the bar makes with the horizontal, when the 30-N force is applied to the bar.

k 2m

1m C 30 N

B k

SOLUTION Equations of Equilibrium: FA and FB can be determined directly by writing the moment equation of equilibrium about points B and A respectively by referring to the FBD of the bar shown in Fig. a. a+ ΣMB = 0; 30(1) - FA(2) = 0 FA = 15 N

a+ ΣMA = 0; 30(3) - FB(2) = 0 FB = 45 N Thus, the stretches of springs A and B can be determined from FA = kxA;

15 = 100 xA

xA = 0.15 m

FB = k xB;

45 = 100 xB

xB = 0.45 m

From the geometry shown in Fig. b, d 2 - d = ; 0.45 0.15

d = 1.5 m

Thus u = sin-1 a

0.45 b = 17.46° = 17.5° 1.5

Ans.

Note: The moment equations are set up assuming small u, but even with non-small u the reactions come out with the same FA, FB, and then the rest of the solution goes through as before.

Ans: u = 17.5° 435

*5–40. Determine the stiffness k of each spring so that the 30-N force causes the bar to tip u = 15° when the force is applied. Originally the bar is horizontal and the springs are unstretched. Neglect the weight of the bar.

k 2m

1m C 30 N

B k

a+ ΣMA = 0; 30(3) - FB(2) = 0 FB = 45 N Thus, the stretches of springs A and B can be determined from FA = kxA;

15 = kxA

xA =

15 k

FB = k xB;

45 = kxB

xB =

45 k

From the geometry shown in Fig. b d 2 - d = ; 45>k 15>k

d = 1.5 m

Thus, sin 15° =

45>k 1.5

Ans.

k = 115.91 N>m = 116 N>m

Note: The moment equations are set up assuming small u, but even with non-small u the reactions come out with the same FA, FB, and then the rest of the solution goes through as before.

Ans: k = 116 N>m 436

5–41. The davit is used to suspend the lifeboat over the side of a ship. If the boat exerts a force of 800 lb on the chain at D, determine the force acting on the hydraulic cylinder BC, and the horizontal and vertical components of reaction at the pin A.

8 ft E

B D

12 ft

SOLUTION a+ ΣMA = 0;

FBC cos 10°(14) - FBC sin 10°(12) - 800(6) = 0

+ ΣFx = 0; S

410.13 sin 10° - Ax = 0 Ax = 71.2 lb

+ c ΣFy = 0;

Ay + 410.13 cos 10° - 800 = 0

808

Ans.

FBC = 410.13 lb = 410 lb

Ans.

14 ft

Ans.

Ay = 396 lb

Ans: FBC = 410 lb Ax = 71.2 lb Ay = 396 lb 437

5–42. The boom supports the two vertical loads. Neglect the size of the collars at D and B and the thickness of the boom, and compute the horizontal and vertical components of force at the pin A and the force in cable CB. Set F1 = 800 N and F2 = 350 N.

5 4

SOLUTION a + ©MA = 0;

- 800(1.5 cos 30°) - 350(2.5 cos 30°) +

3 4 F (2.5 sin 30°) + FCB(2.5 cos 30°) = 0 5 CB 5

4 Ax - (781.6) = 0 5

Ay - 800 - 350 +

30 A

Ans.

Ax = 625 N + c ©Fy = 0;

D F2

Ans.

FCB = 781.6 = 782 N + ©F = 0; : x

1.5 m

3 (781.6) = 0 5 Ans.

Ay = 681 N

Ans: FCB = 782 N Ax = 625 N Ay = 681 N 438

5–43. The boom is intended to support two vertical loads, F1 and F2. If the cable CB can sustain a maximum load of 1500 N before it fails, determine the critical loads if F1 = 2F2. Also, what is the magnitude of the maximum reaction at pin A?

5 4

SOLUTION a + ©MA = 0;

- 2F2(1.5 cos 30°) - F2(2.5 cos 30°) +

4 3 (1500)(2.5 sin 30°) + (1500)(2.5 cos 30°) = 0 5 5

1.5 m

F1 = 2F2 = 1448 N

Ax -

30 A

Ans.

F1 = 1.45 kN + ©F = 0; : x

D F2

Ans.

F2 = 724 N

4 (1500) = 0 5

Ax = 1200 N + c ©Fy = 0;

Ay - 724 - 1448 +

3 (1500) = 0 5

Ay = 1272 N FA = 2(1200)2 + (1272)2 = 1749 N = 1.75 kN

Ans.

Ans: F2 = 724 N F1 = 1.45 kN FA = 1.75 kN 439

*5–44. The 10-kg uniform rod is pinned at end A. If it is subjected to a couple moment of 50 N # m, determine the smallest angle u for equilibrium. The spring is unstretched when u = 0, and has a stiffness of k = 60 N>m.

B k 60 N/m

0.5 m

SOLUTION

50 N m

Equations of Equilibrium: Here the spring stretches x = 2 sin u. The force in the spring is Fsp = kx = 60 (2 sin u) = 120 sin u. Write the moment equation of equilibrium about point A by referring to the FBD of the rod shown in Fig. a, a+ ΣMA = 0; 120 sin u cos u (2) - 10(9.81) sin u (1) - 50 = 0 240 sin u cos u - 98.1 sin u - 50 = 0 Solve numerically Ans.

u = 24.598° = 24.6°

Ans: u = 24.6° 440

5–45. If the truck and its contents have a mass of 50 kg with center of gravity at G, determine the normal reaction on both wheels and the magnitude and direction of the minimum force required at the grip B needed to lift the load.

0.4 m B 0.5 m 0.2 m

0.4 m

608 0.4 m A

0.1 m

SOLUTION Equations of Equilibriums: Py can be determined directly by writing the force equation of equilibrium along y axis by referring to the FBD of the hand truck shown in Fig. a. + c ΣFy = 0; Py - 50(9.81) = 0

Py = 490.5 N

Using this result to write the moment equation of equilibrium about point A, a+ ΣMA = 0; Px sin 60°(1.3) - Px cos 60°(0.1) - 490.5 cos 30°(0.1) - 490.5 sin 30°(1.3) - 50(9.81) sin 60°(0.5) + 50(9.81) cos 60°(0.4) = 0 Px = 442.07 N Thus, the magnitude of minimum force P, Fig. b, is P = 2Px2 + Py2 = 2442.072 + 490.52 = 660.32 N = 660 N

Ans.

and the angle is u = tan-1 a

490.5 b = 47.97° = 48.0° b 442.07

Ans.

Write the force equation of equilibrium along x axis, + ΣFx = 0; NA - 442.07 = 0 NA = 442.07 N = 442 N S

Ans.

Ans: P = 660 N NA = 442 N u = 48.0° b 441

5–46. Three uniform books, each having a weight W and length a, are stacked as shown. Determine the maximum distance d that the top book can extend out from the bottom one so the stack does not topple over.

SOLUTION Equilibrium: For top two books, the upper book will topple when the center of gravity of this book is to the right of point A. Therefore, the maximum distance from the right edge of this book to point A is a/2. Equation of Equations of Equilibrium: Equilibrium: For the entire three books, the top two books will topple about point B. a + ©MB = 0;

a W(a-d) -W ad- b = 0 2 d =

3a 4

Ans.

Ans: d = 442

3a 4

5–47. Determine the reactions at the pin A and the tension in cord BC. Set F = 40 kN. Neglect the thickness of the beam.

26 kN

C 13

SOLUTION a+ ΣMA = 0; - 26 a

12 3 b(2) - 40(6) + FBC(6) = 0 13 5

Ans.

FBC = 80 kN

+ ΣFx = 0; 80 a 4 b - Ax - 26 a 5 b = 0 S 5 13

Ans.

Ax = 54 kN

+ c ΣFy = 0; Ay - 26 a

12 3 b - 40 + 80 a b = 0 13 5

Ans.

Ay = 16 kN

Ans: FBC = 80 kN Ax = 54 kN Ay = 16 kN 443

*5–48. If rope BC will fail when the tension becomes 50 kN, determine the greatest vertical load F that can be applied to the beam at B. What is the magnitude of the reaction at A for this loading? Neglect the thickness of the beam.

26 kN

C 13

SOLUTION a+ ΣMA = 0; - 26 a

3 12 b(2) - F(6) + (50)(6) = 0 13 5

Ans.

F = 22 kN

+ ΣFx = 0; 50 a 4 b - Ax - 26 a 5 b = 0 S 5 13

Ans.

Ax = 30 kN

+ c ΣFy = 0; Ay - 26 a

12 3 b - 22 + 50 a b = 0 13 5

Ans.

Ay = 16 kN

Ans: F = 22 kN Ax = 30 kN Ay = 16 kN 444

5–49. The assembly is used to support the 120-kg container having a center of mass at G. If the spring has an unstretched length of 250 mm and stiffness of k = 300 kN>m, determine its height h and the reaction at the rollers A and B.

2m A

SOLUTION

Equations of Equilibrium: Write the force equations of equilibrium along x and y axes by referring to the FBD of the assembly, Fig a, + ΣFx = 0; S

NB - NA = 0

NB = NA = N

+ c ΣFy = 0;

Fsp - 120 (9.81) = 0

Fsp = 1177.2 N

Using the above results to write the moment equation of equilibrium about point C, a+ ΣMC = 0; [120(9.81)](3) - N(2) = 0 Ans.

NA = NB = N = 1765.8 N = 1.77 kN The compression of the spring is x=

Fsp k

1177.2 = 0.003924 m = 3.924 mm 300(103)

Then Ans.

h = 250 - 3.924 = 246.076 mm = 246 mm

Ans: NA = NB = 1.77 kN h = 246 mm 445

5–50. The check valve is used to regulate pressure in the pipe. If the stiffness of the spring is k = 80 kN/m and its uncompressed length is 120 mm, determine the maximum pressure in the tank if the lever is to remain in the horizontal position as shown. The plug at A is circular.

150 mm

300 mm

90 mm

50 mm

SOLUTION Equations of Equilibrium: The pressure force acting on the piston is F = PA = P[p(0.0252)] = 0.625(10-3) pP and the spring develop a force of Fsp = kx = [80(103)] (0.12 - 0.09) = 2400 N Write the moment equation of equilibrium about point B by referring to the FBD of the lever, Fig. a a+ ΣMB = 0; 2400(0.15) - [0.625(10-3)pP](0.3) = 0 P = 611.15(103) N/m2 = 611 kPa

Ans.

Ans: P = 611 kPa 446

5–51. The block has a mass of 80 kg and center of mass at G. Determine the smooth reactions on the supporting platform at A, B, and C.

A 608

B 608

0.3 m

0.8 m

SOLUTION Equations of Equilibrium: NC Can be determind directly by writing the moment equation of equilibrium about point D. Referring to the FBD of the fitting, Fig. a, a+ ΣMD = 0;

NC(1.1) - 80(9.81)(0.3) = 0 Ans.

NC = 214.04 N = 214 N

Using the result of NC to write the force equation of equilibrium along x and y axes, + ΣFx = 0; S

NA sin 60° - NB sin 60° = 0

+ c ΣFy = 0;

2(N cos 60°) + 214.04 - 80(9.81) = 0

NA = NB = N

Ans.

NA = NB = N = 570.76 N = 571 N

Ans: NC = 214 N NA = 571 N 447

*5–52. Calculate the tension in the cable at B required to support the load of 200 kg. The cable passes over a frictionless pulley located at C. Neglect the size of this pulley and weight of the boom.

4.5 m

608 B

C 308

1m A

200 kg

SOLUTION a+ ΣMA = 0;

T cos 30°(1) - 200(9.81)(4.5 cos 30°) = 0 Ans.

T = 8829 N = 8.83 kN + ΣFx = 0; S

Ax - 8829 sin 30° = 0 Ax = 4415 N = 4.41 kN

+ c ΣFy = 0;

Ay - 8.829 cos 30° - 200(9.81) = 0 Ans.

Ay = 9608 N = 9.61 kN

Ans: T = 8.83 kN 448

5–53. The uniform beam has a weight W and length l and is supported by a pin at A and a cable BC. Determine the horizontal and vertical components of reaction at A and the tension in the cable necessary to hold the beam in the position shown.

C l B A

SOLUTION Equations of Equilibrium: The tension in the cable can be obtained directly by Equations of Equilibrium: summing moments about point A. a + ©MA = 0;

l T sin 1f - u2l - W cos u a b = 0 2 T =

Using the result T = + ©F = 0; : x

Ans.

W cos u 2 sin 1f - u2 W cos u b cos f - Ax = 0 2 sin 1f - u2 Ax =

+ c ©Fy = 0;

W cos u 2 sin 1f - u2

Ay + a Ay =

W cos f cos u 2 sin 1f - u2

Ans.

W cos u b sin f - W = 0 2 sin 1f - u2

W1sin f cos u - 2 cos f sin u2

Ans.

2 sin f - u

Ans: W cos u 2 sin(f - u) Wcos f cos u Ax = 2 sin(f - u) W(sin f cos u - 2 cos f sin u) Ay = 2 sin (f - u) T =

449

5–54. When no force is applied to the brake pedal of the lightweight truck, the retainer spring AB keeps the pedal in contact with the smooth brake light switch at C. If the force on the switch is 3 N, determine the unstretched length of the spring if the stiffness of the spring is k = 80 N>m.

100 mm B

50 mm D 40 mm

SOLUTION a + ©MD = 0;

30 C 10 mm

Fs (50) - 3 cos 30° (40) - 3 sin 30° (10) = 0 Fs = 2.378 N

Fs = kx,

2.378 = 80 (x) x = 0.02973 m = 29.73 mm

l = lo + x 100 = lo + 29.73 Ans.

lo = 70.3 mm

Ans: lO = 70.3 mm 450

5–55. The exercise machine consists of a rigid bar pin connected to the frame at A. As the woman pushes up on the bar it stretches the rubber band BC. If the stiffness of the rubber band is k = 500 lb>ft, determine the applied vertical force F needed to hold the bar in the position u = 30°. The rubber band is unstretched when u = 0°.

5 ft

F B u

1 ft A 0.5 ft C

SOLUTION From the Geometry: l = 212 + 0.52 - 2(1)(0.5)cos 120° = 1.3229 ft l0 = 212 + 0.52 = 1.1180 ft

sin f sin 120° = f = 19.106° 0.5 1.3229

Equilibrium: The force in the rubber band T = k(l – l0) = 500(1.3229 – 1.1180) = 102.42 lb. a+ ΣMA = 0;

102.42 sin 19.106°(1) - F(5) = 0 Ans.

F = 6.70 lb

Ans: F = 6.70 lb 451

*5–56. The uniform rod has a length l and weight W. It is supported at one end A by a smooth wall and the other end by a cord of length s which is attached to the wall as shown. Determine the placement h for equilibrium.

h s A

SOLUTION Equilibrium: The tension in the cable can be obtained directly by Equations of Equilibrium: summing moments about point A.

l B

a + ©MA = 0;

l T sin f(l) - W sin u a b = 0 2 T =

Using the result T = + c ©Fy = 0;

W sin u 2 sin f

W sin u , 2 sin f

W sin u cos (u - f) - W = 0 2 sin f (1)

sin u cos (u - f) - 2 sin f = 0 Geometry: Applying the sine law with sin (180° - u) = sin u, we have sin f sin u = s h

sin f =

h sin u s

(2)

Substituting Eq. (2) into (1) yields cos (u - f) =

2h s

(3)

Using the cosine law, l2 = h2 + s2 - 2hs cos (u - f) cos (u - f) =

h2 + s2 - l2 2hs

(4)

Equating Eqs. (3) and (4) yields h2 + s 2 - l2 2h = s 2hs h =

s2 - l2 A 3

Ans.

Ans: h = 452

s2 - l 2 A 3

5–57. A 10-kN load is suspended from the boom at D. Determine the force in the hydraulic cylinder BC and the pin reaction at A.

1.5 m

30° B

10 kN/m

45° C

SOLUTION Equations of Equilibrium: Hydraulic cylinder BC is a two force member. FBC can be determined directly by writing the moment equation of equilibrium about point A. Refering to the FBD of the boom, Fig. a, a+ ΣMA = 0;

FBC sin 15°(1.5) - 10 cos 30°(4.5) = 0 Ans.

FBC = 100.38 kN = 100 kN

Using the result of FBC to write the force equations of equilibrium along x and y axes, + ΣFx = 0; S

100.38 cos 45° - Ax = 0 Ans.

Ax = 70.98 kN = 71.0 kN + c ΣFy = 0;

100.38 sin 45° - 10 - Ay = 0 Ans.

Ay = 60.98 kN = 61.0 kN

Ans: FBC = 100 kN Ax = 71.0 kN Ay = 61.0 kN 453

5–58.

Determine the force in the cable as a function of u, needed to support the 100-kg hatch door. B

2m A

SOLUTION Equations of Equilibriums: P can be determined directly by writing the moment equation of equilibrium about point A. Refering to the FBD of the hatch door, Fig. a, a+ ΣMA = 0;

P sin f (2) - 3100(9.81) cos u 4 (1) = 0 P =

490.5 cos u sin f

(1)

From the geometry shown in Fig. b, 2f + (90° - u) = 180° f = u/2 + 45° Substitute this result into Fig. 1, P =

490.5 cos u sin (u/2 + 45°)

Ans.

454

5–59.

d B

The crane supports the load of 800 kg. Determine the reactions at the supports A and B as a function of position d of the trolley. Plot the reactions vs. d for 0.5 m … d … 3 m.

SOLUTION Equations of Equilibrium: NB and Ay can be determind by writing the moment equation of equilibrium about point A and force equation of equilibrium along y axis respectively. Referring to the FBD of the frame, Fig. a, a+ ΣMA = 0;

NB(5) - 3800(9.81) 4d = 0

Ans.

NB = 1569.6d N = {1.57d} kN + c ΣFy = 0;

Ay - 800(9.81) = 0 Ay = 7848 N = 7.848 kN

Using the result of NB to write the force equation of equilibrium along x axis + ΣFx = 0; S

Ax - 1569.6d = 0 Ax = 1569.6d N = {1.5696d} kN

Then the reaction on pair A is FA = 2Ax2 + Ay2 = 2(1.5696d)2 + 7.8482 = 22.4636d 2 + 61.59

= e 22.46d 2 + 61.6 f kN

Ans.

The values of d and the corresponding values of NB and FA are tabulated below. d(m)

0.5

1.5

2.5

NB(kN)

0.785

1.57

2.35

3.14

3.92

4.71

d(m)

0.5

1.5

2.5

FA(kN)

7.89

8.00

8.19

8.45

8.77

9.15

The plots of NB and FA vs d are shown in Fig. b and c respectively.

Ans: NB = 1569.6d N = {1.57d} kN

455

FA = e 22.46d 2 + 61.6 f kN

*5–60. The smooth uniform rod has a mass m and is placed on the semicircular arch and against the wall. Show that for equilibrium the angle u must satisfy 1 sin u = 1 21 + 3 cos 2u 2 1d - l sin u 2 . r

A l u B r

SOLUTION AB is a three-force member. So the three forces are concurrent at point O. Since d = l sin u + r sin f d - l sin u = r sin f (1) But, OA = 2(l>2 sin u)2 + (l cos u)2 = l =

1 2 sin u + cos2 u A4

l 21 + 3 cos2 u 2

Thus, sin f =

l>2 sin u l>2 21 + 3 cos2 u

From Eq. (1):

d - l sin u = r So that sin u =

sin u 21 + 3 cos2 u

1 ( 21 + 3 cos2 u ) (d - l sin u) r

(QED)

456

5–61. The smooth pipe rests against the wall at the points of contact A, B, and C. Determine the reactions at these points needed to support the vertical force of 45 lb. Neglect the pipe’s thickness in the calculation.

8 in.

908 B

20 in. C 16 in.

SOLUTION a+ ΣMA = 0;

45 cos 30°(36) - 45 sin 30°(8) - RC(20) + RB (8 tan 30°) = 0

+ c ΣFy = 0;

RC cos 30° - RB cos 30° - 45 = 0

+ ΣFx = 0; S

308 45 lb

RC = 63.91 = 63.9 lb

Ans.

RB = 11.95 = 11.9 lb

Ans.

RA + 11.95 sin 30° - 63.91 sin 30° = 0 Ans.

RA = 26.0 lb Also; + ΣFx′ = 0

45 sin 30° - RA cos 30° = 0 Ans.

RA = 26.0 lb + ΣFy′ = 0;

- 45 cos 30° + RC - RB - 25.98 sin 30° = 0

a+ ΣMC = 0;

45 cos 30°(16) - RB (20 - 8 tan 30°) - 25.98(8 cos 30° + 20 sin 30°) = 0 RB = 11.9 lb

Ans.

RC = 63.9 lb

Ans.

Ans: RC = 63.9 lb RB = 11.9 lb RA = 26.0 lb 457

5–62. Determine the distance d for placement of the load P for equilibrium of the smooth bar in the position u as shown. Neglect the weight of the bar.

d P

u a

SOLUTION + c ©Fy = 0;

R cos u - P = 0

a + ©MA = 0;

- P(d cos u) + R a Rd cos2 u = R a d =

a b = 0 cos u

a b cos u

a cos3 u

Ans.

Also; Require forces to be concurrent at point O. AO = d cos u =

a>cos u cos u

Thus, d =

a cos3 u

Ans.

Ans: d = 458

a cos3 u

5–63. A

The rod BC is supported by two cords, each of length a, which are attached to the pin at A. If weights W and 2W are suspended from the ends of the rod, determine the angle u for equilibrium, measured from the horizontal. Express the answer in terms of a and l. Neglect the weight of the rod.

a a

C l

u W

SOLUTION a+ ΣMB = 0;

-W cos u(l) +

p+ΣFx′ = 0;

- 3W sin u +

+rΣFy = 0;

- 3W cos u +

2a2 - ( 2l ) 2 a

l>2

FAB -

2a2 - ( 2l ) 2 a

- 3 cos u + cos u + cos u =

l>2

FAC(l) = 0

FAC = 0

FAB +

2a2 - ( 2l ) 2

6 sin u 2a2 - ( 2l ) 2 l

FAC = 0

+ cos u = 0

6 sin u 2a2 - ( 2l ) 2

u = tan - 1 a

l 62a2 - l4

b Ans.

Ans: u = tan - 1 a 459

l 62a2 - l4

*5–64. The uniform load has a mass of 600 kg and is lifted using a uniform 30-kg strongback beam BAC and four wire ropes as shown. Determine the tension in each segment of rope and the force that must be applied to the sling at A.

F 1.25 m B

SOLUTION

A C

of Equilibrium: Equilibrium: Due to symmetry, all wires are subjected to the same Equations of tension. This condition statisfies moment equilibrium about the x and y axes and force equilibrium along y axis. ©Fz = 0;

1.25 m

1.5 m 1.5 m

4 4Ta b - 5886 = 0 5 Ans.

T = 1839.375 N = 1.84 kN

The force F applied to the sling A must support the weight of the load and strongback beam. Hence ©Fz = 0;

F - 60019.812 - 3019.812 = 0 Ans.

F = 6180.3 N = 6.18 kN

Ans: T = 1.84 kN F = 6.18 kN 460

5–65. The cable of the tower crane is subjected to a force of 840 N. Determine the x, y, z components of reaction at the fixed base A.

B F = 840 N 24 m y

SOLUTION A

15 m

rAC = {12i + 8j - 24k} m F = 840 c

12 i + 8 j - 24 k 2

2(12) + (8) + ( - 24)

= {360 i + 240 j - 720 k} N

10 m x

FR = Ax i + Ay j + Az k Thus ΣF = 0; F + FA = 0; Ax = -360 N

Ans.

Ay = -240 N

Ans.

Az = 720 N

Ans.

ΣM = 0; M A + rAB * F = 0 i MA + † 15 360

j 10 240

k 0 † = 0 - 720

MA - 7200 i + 10 800 j = 0 Thus,

MAx = 7.20 kN # m

Ans.

MAy = - 10.8 kN # m

Ans.

MAz = 0

Ans.

Ans: Ax = - 360 N Ay = - 240 N Az = 720 N MAx = 7.20 kN # m MAy = - 10.8 kN # m MAz = 0 461

5–66. Determine the reaction at the ball supports B and C and the componentsthe of x, they,reaction at the ball-and-socket A (not Determine z components of reaction at the ball shown) for the uniformly loaded plate. A (not shown) for supports B and C and the ball-and-socket the uniformly loaded plate.

z z

2 lb/ft2 2 lb/ft2

4 ft 4 ft

SOLUTION W = (4 ft)(2 ft) ( 2 lb>ft 2 ) = 16 lb

1 ft 1 ft 2 ft x

ΣFx = 0;

Ax = 0

Ans.

ΣFy = 0;

Ay = 0

Ans.

ΣFz = 0;

Az + Bz + Cz - 16 = 0

(1)

ΣM Mxz = 0;

2 Bz - 16(1) + Cz (1) = 0

(2)

ΣMy = 0;

-Bz (2) + 16(2) - Cz (4) = 0

(3)

A B

B 2 ft 2 ft C

2 ft

Solving Eqs. (1) – (3): Ans.

Az = Bz = Cz = 5.33 lb

Ans: Ax = 0 Ay = 0 Az = Bz = Cz = 5.33 lb 462

5–67.

Determine the normal floor reaction on each wheel of the engine stand. The engine weights 750 lb and has a center of gravity at G.

2 ft

4 ft

2 ft

C 2 ft

y 4 ft

SOLUTION ΣMy = 0;

750(2) - NC(4) = 0 NC = 375 lb

ΣMx = 0;

NA(2) - NB(2) = 0 NA = NB [1]

ΣFz = 0;

NA + NB + 375 - 750 = 0 [2]

Ans.

Substituting Eq. [1] into [2] yields: Ans.

NA = NB = 187.5 lb

Ans: NC = 375 lb NA = NB = 187.5 lb 463

*5–68. Determine the components of reaction at the ball-and-socket B and the reaction at the rollers A and C due to the loading shown.

z 800 N 600 N 0.4 m 400 N A

0.3 m 0.3 m

0.4 m C x

SOLUTION ΣMx = 0;

Cz(0.6) - 800(0.3) - 400(0.3) = 0

Cz = 600 N

Ans.

ΣMy = 0;

- Bz(0.8) + 600(0.4) + 400(0.4) = 0

Bz = 500 N

Ans.

ΣFz = 0;

Az + 600 + 500 - 800 - 400 - 600 = 0

Ay = 700 N Ans.

ΣFx = 0;

Bx = 0

Ans.

ΣFy = 0;

By = 0

Ans.

Ans: Cz = 600 N Bz = 500 N Ay = 700 N Bx = 0 By = 0 464

5–69. Determine the x, y, z components of reaction acting on the ball-and-socket at A, the reaction at the roller B, and the tension in the cord CD required for equilibrium of the plate. Determine the x, y, z components of reaction acting on the ball-and-socket at A, the reaction at the roller B, and the tension in the cord CD required for equilibrium of the plate.

D C D 40 lb/ft

40 lb/ft

2 ft B

C 5 ft

100 lb ? ft

100 lb · ft

2 ft

A 5 ft

SOLUTION

©Mx = 0;

100 + Bz (5) - 200 (2.5) = 0;

Bz = 80 lb

Ans.

©My = 0;

200 (2) - Az (2) - 80 (2) = 0;

Az = 120 lb

Ans.

©Fz = 0;

80 + 120 - 200 + TCD = 0;

©Fx = 0;

Ax = 0

Ans.

©Fy = 0;

Ay = 0

Ans.

TCD = 0

Ans: Bz = 80 lb Az = 120 lb TCD = 0 Ax = 0 Ay = 0 465

5–70. z

The 100-lb door has its center of gravity at G. Determine the components of reaction at hinges A and B if hinge B resists only forces in the x and y directions and A resists forces in the x, y, z directions.

18 in.

B 24 in.

SOLUTION Equations of Equilibrium: Equilibrium: From the free-body diagram of the door, Fig. a, By, Bx, and Az can be obtained by writing the moment equation of equilibrium about the x¿ and y¿ axes and the force equation of equilibrium along the z axis.

A 18 in.

-By(48) - 100(18) = 0

©Mx¿ = 0;

24 in.

By = - 37.5 lb

Ans.

©My¿ = 0;

Bx = 0

Ans.

©Fz = 0;

- 100 + A z = 0;

A z = 100 lb

30 x

Ans.

Using the above result and writing the force equations of equilibrium along the x and y axes, we have Ans.

©Fx = 0;

Ax = 0

©Fy = 0;

A y + ( - 37.5) = 0

A y = 37.5 lb

Ans.

The negative sign indicates that By acts in the opposite sense to that shown on the free-body diagram. If we write the moment equation of equilibrium ©Mz = 0, it shows that equilibrium is satisfied.

Ans: By = - 37.5 lb Bx = 0 Az = 100 lb Ax = 0 Ay = 37.5 lb 466

5–71.

z B

Determine the tension in each cable and the components of reaction at D needed to support the load.

C 2m D

A y 30

SOLUTION Force and Position Vectors: The coordinates of points A, B, and C are A(6, 0, 0) m, B(0, -3, 2) m and C(0, 0, 2) m respectively. FAB = FAB a FAC = FAC a

(0 - 6)i + ( - 3 - 0)j + (2 - 0)k rAB 6 3 2 b = FAB c d = - FABi - FAB j + FABk rAB 7 7 7 1(0 - 6)2 + ( - 3 - 0)2 + (2 - 0)2 (0 - 6)i + (2 - 0)k rAC 6 2 b = FAC c d = FAC i + FAC k 2 2 rAC 140 140 1(0 - 6) + (2 - 0)

F = 400 (sin 30°j - cos 30°k) = {200j - 346.41k}N FD = Dxi + Dy j + Dzk rDA = {6i} m

Referring to the FBD of the rod shown in Fig. a, the force equation of equilibrium gives ΣF = 0; FAB + FAC + F + FD = 0 6 6 3 a- FAB FAC + Dx bi + a- FAB + Dy + 200bj 7 7 140

2 2 + a FAB + FAC + Dz - 346.41bk = 0 7 140

467

400 N

5–71. Continued

Equating i, j and k components, 6 6 - FAB FAC + Dx = 0 7 140

(1)

3 - FAB + Dy + 200 = 0 7

(2)

2 2 F + FAC + Dz - 346.41 = 0 7 AB 140

(3)

Moment equation of equilibrium gives

ΣMD = 0; rDA * (FAB + FAC + F) = 0

i 6

j 0

k 0

6 6 a- FAB F b 140 AC 7

3 a- FAB + 200 b 7

2 2 FAC - 346.41 b a FAB + 140 7

5 = 0

2 2 3 -6 a FAB + FAC - 346.41bj + 6 a- FAB + 200b k = 0 7 7 140

Equating j and k Components,

2 2 -6 a FAB + FAC - 346.41b = 0 7 140

(4)

3 6 a - FAB + 200b = 0 7

(5)

Solving Eqs. (1) to (5)

FAB = 466.67 N = 467 N

Ans.

FAC = 673.81 N = 674 N

Ans.

Dx = 1039.23 N = 1.04 kN

Ans.

Dy = 0

Ans.

Dz = 0

Ans.

Ans: FAB = 467 N FAC = 674 N Dx = 1.04 kN Dy = 0 Dz = 0 468

*5–72. Determine the components of reaction acting at the balland-socket A, roller B, and cord CD.

Determine the components of reaction acting at the balland-socket A, roller B, and cord CD.

z 400 N 2m 2m

400 N 2m

SOLUTION

©Fx = 0;

Ax = 0

Ans. x

©Fy = 0;

Ay = 0

Ans.

©Mx = 0;

- 300 (2) + Az (3) = 0;

©My = 0;

400 (2) - Bz (4) = 0;

©Fz = 0;

200 + 200 + TCD - 700 = 0;

300 N 2m 1m CD

300 N 2m 1m

A y

Ans. x

Az = 200 N

Ans.

Bz = 200 N

Ans.

TCD = 300 N

Ans: Ax = 0 Ay = 0 Ay = 200 N Bz = 200 N TCD = 300 N 469

5–73. The triangular plate, having a 90° angle at A, supports the 150-lb load. Determine the tension in cables BD and CD and the x, y, z components of reaction at the ball-and-socket joint at A.

6 ft 150 lb A

B 4 ft

SOLUTION

Force Vector: - 2i - 4j + 4k

FCD = FCD a

1( - 2)2 + ( - 4)2 + 62

FBD = FBD a

1( - 2)2 + (4)2 + 62

Equilibrium: ΣFx = 0; ΣFy = 0;

-2i + 4j + 6k

1 ft 4 ft

2 ft

b = - 0.2673FCDi - 0.5345FCD j + 0.8018FCDk

b = - 0.2673FBDi - 0.5345FBD j + 0.8018FBDk (1)

Ax - 0.2673FCD - 0.2673FBD = 0

(2)

0.5345FBD - 0.5345FCD + Ay = 0

ΣFz = 0;

0.8018FCD + 0.8018FBD + Az - 150 = 0

(3)

ΣMx′ = 0;

0.2673FCD(4) - 0.2673FBD(4) = 0

(4)

ΣMy′ = 0;

150(3) - 0.8018FCD(4) - 0.8018FBD(4) = 0

(5)

Solving Eqs. (1) to (5) yields: Ans.

FBD = FCD = 70.2 1b Az = Ax = 37.5 1b Ay = 0 Notice that ΣMz′ = 0;. Therefore, equilibrium condition is satisfied.

Ans: FBD = 70.2 lb Az = 37.5 lb Ay = 0 470

5–74. z

The bent rod is supported at A, B, and C by smooth journal bearings. Determine the components of reaction at the bearings if the rod is subjected to the force F = 800 N. The bearings are in proper alignment and exert only force reactions on the rod.

C A

SOLUTION

B 0.75 m

30 60

Equations of Equilibrium: The x, y and z components of force F are

Fx = 800 cos 60° cos 30° = 346.41 N

Fy = 800 cos 60° sin 30° = 200 N Fz = 800 sin 60° = 692.82 N Referring to the FBD of the bent rod shown in Fig. a, ΣMx = 0;

- Cy(2) + Bz(2) - 692.82 (2) = 0

(1)

ΣMy = 0;

Bz(1) + Cx(2) = 0

(2)

ΣMz = 0;

- Cy(1.75) - Cx(2) - By(1) - 346.41(2) = 0

(3)

ΣFx = 0;

Ax + Cx + 346.41 = 0

(4)

ΣFy = 0;

200 + By + Cy = 0

(5)

ΣFz = 0;

Az + Bz - 692.82 = 0

(6)

Solving Eqs. (1) to (6) Cy = 800 N Bz = - 107.18 N = 107 N By = 600 N

Ans.

Cx = 53.59 N = 53.6 N Ax = - 400 N Az = 800 N

Ans.

The negative signs indicate that Cy, Bz and Ax are directed in the senses opposite to those shown in FBD.

Ans: Cy = 800 N Bz = 107 N By = 600 N Cx = 53.6 N Ax = - 400 N Az = 800 N 471

5–75. The bent rod is supported at A, B, and C by smooth journal bearings. Determine the magnitude of F which will cause the positive x component of reaction at the bearing C to be Cx = 50 N. The bearings are in proper alignment and exert only force reactions on the rod.

C A

SOLUTION

B 0.75 m

30 60

Equations of Equilibrium: The x, y and z components of force F are

Fx = F cos 60° cos 30° = 0.4330 F

Fy = F cos 60° sin 30° = 0.25 F Fz = F sin 60° = 0.8660 F Here, it is required that Cx = 50. Thus, by referring to the FBD of the beat rod shown in Fig. a, ΣMx = 0;

(1)

-Cy(2) + Bz(2) - 0.8660 F(2) = 0

ΣMy = 0;

Bz(1) + 50(2) = 0

(2)

ΣMz = 0;

-Cy(1.75) - 50(2) - By(1) - 0.4330 F(2) = 0

(3)

ΣFy = 0;

0.25 F + By + Cy = 0

(4)

Solving Eqs. (1) to (4) Ans.

F = 746.41 N = 746 N Cy = - 746.41 N Bz = - 100 N By = 559.81 N

Ans: F = 746 N 472

*5–76. The 50-kg glass tabletop rests on the centrally located equilateral triangular frame which is supported by three legs. Determine the smallest vertical force P that, when applied to the glass, would cause it to lift or topple off the frame. Specify the force location r and the smallest angle u, and also determine the vertical reactions of the legs on the floor when tipping is about to occur.

P m

0.3

0.3 m 0.5 m

SOLUTION

Rotation should occur about the AB axis since this axis is closest to the center of the table (W).

d = 0.15 tan 30° = 0.0866 m Hence, put P at r = 0.5 m

Ans.

u = 60°

Ans.

©MAB = 0;

W (0.0866) - P (0.5 - 0.15 tan 30°) = 0 P = 0.20948 W Ans.

= 0.20948(50)(9.81) = 103 N This would require

Ans.

Cz = 0 ©Ma - a = 0;

Az (0.15) - Bz (0.15) = 0 Az = Bz

©Fz = 0;

2 Az - W - 0.20948 W = 0 Ans.

Az = Bz = 0.6047 W

Ans.

Az = Bz = 0.6047(50)(9.81) = 287 N

Ans: r = 0.5 m u = 60° P = 103 N Cz = 0 Az = Bz = 287 N 473

5–77.

Member AB is supported by a cable BC and at A by a square rod which fits loosely through the square hole in the collar fixed to the member as shown. Determine the components of reaction at A and the tension in the cable needed to hold the rod in equilibrium.

1.5 m 400 N

A y C

200 N

SOLUTION

Force and Position Vectors: The coordinates of points B and C are B(3, 0, -1) m C(0, 1.5, 0) m, respectively. TBC = TBC a

(0 - 3)i + (1.5 - 0)j + [0 - ( -1)]k rBC b = TBC • ¶ rBC 2(0 - 3)2 + (1.5 - 0)2 + [0 - ( -1)]2

6 3 2 = - TBC i + TBC j + T k 7 7 7 BC

F = {200j - 400k} N FA = Ax i + Ay j MA = (MA)x i + (MA)y j + (MA)z k r1{3 i} m r2 = {1.5 j} m Equations of Equilibrium: Referring to the FBD of member AB shown in Fig. a, the force equation of equilibrium gives ΣF = 0; TBC + F + FA = 0 3 2 6 a - TBC + Ax bi + a TBC + 200 + Ayb j + a TBC - 400bk = 0 7 7 7

Equating i, j and k components 6 T + Ax = 0 7 BC

(1)

3 T + 200 + Ay = 0 7 BC

(2)

2 T - 400 = 0 7 BC

(3)

474

5–77. Continued

The moment equation of equilibrium gives ΣMA = O; MA + r1 * F + r2 * TBC = 0 i

( MA ) x i + ( MA ) y j + ( MA ) zk + † 3 0

j 0 200

k 0 † + 5 - 400

i 0 6 - TBC 7

j 1.5 3 TBC 7

k 0 2 TBC 7

3 9 T d i + 3 ( MA ) y + 1200 4 j + c ( MA ) z + TBC + 600 d k = 0 7 BC 7 Equating i, j, and k components, 3 ( MA ) x + TBC = 0 7

5 = 0

c ( MA ) x +

( MA ) y + 1200 = 0

(4) (5)

9 TBC + 600 = 0 7 Solving Eqs. (1) to (6),

( MA ) z +

(6)

TBC = 1400 N = 1.40 kN

Ans.

Ay = 800 N

Ans.

Ax = 1200 N = 1.20 kN

Ans.

( MA ) x = 600 N # m

Ans.

( MA ) y = - 1200 N # m = 1.20 kN # m

Ans.

( MA ) z = - 2400 N # m = 2.40 kN # m

Ans.

The negative signs indicate that Ay, ( MA ) x, ( MA ) y and ( MA ) z are directed in sense opposite to those shown in FBD.

Ans: TBC = 1.40 kN Ay = 800 N Ax = 1.20 kN ( MA ) x = 600 N # m ( MA ) y = 1.20 kN # m ( MA ) z = 2.40 kN # m 475

5–78. z

The member is supported by a pin at A and cable BC. Determine the components of reaction at these supports if the cylinder has a mass of 40 kg.

0.5 m B

1m A D

SOLUTION

FCB = FCB a

(0 - 3)i + ( - 0.5 - 1)j + (1 - 0)k rCB b = FCB c d rCB 2(0 - 3)2 + ( - 0.5 - 1)2 + (1 - 0)2 6 3 2 = - FCBi - FCBj + FCBk 7 7 7

W = { - 40(9.81)k} N = { - 392.4k} N. FA = Ax i + Ay j + Az k MA = ( MA ) x i + ( MA ) z k rAC = {3i + j} m

rAD = {3i - j} m

Equations of Equilibrium: Referring to the FBD of the assembly shown in Fig. a. the force equation of equilibrium gives ΣF = 0; FCB + W + FA = 0; 6 3 2 a - FCB + Ax bi + a - FCB + Ay bj + a FCB + Az - 392.4bk = 0 7 7 7

Equating i, j and k components

6 - FCB + Ax = 0 7 3 - FCB + Ay = 0 7 2 F + Az - 392.4 = 0 7 CB The moment equation of equilibrium gives

(1) (2) (3)

ΣMA = 0; rAC * FCB + rAD * W + MA = 0 i 3 6 - FCB 7

y 1m

Force and Position Vectors: The coordinates of points B, C and D are B(0, -0.5, 1) m, x C(3, 1, 0) m and D(3, - 1, 0) m, respectively.

j 1 3 - FCB 7

k 0 2 FCB 7

i 5 + †3 0

j -1 0

k 0 † + ( MA ) x i + ( MA ) Z k = 0 - 392.4

2 6 9 6 c FCB + 392.4 + ( MA ) x d i + a - FCB + 1177.2bj + c - FCB + FCB + ( MA ) z d k = 0 7 7 7 7

476

5–78. Continued

Equating i, j and k components, 2 F + 392.4 + ( MA ) x = 0 7 CB 6 - FCB + 1177.2 = 0 7 9 6 - FCB + FCB + ( MA ) z = 0 7 7 Solving Eqs (1) to (6),

(4) (5) (6)

Ans.

FCB = 1373.4 N = 1.37 kN

( MA ) x = - 784.8 N # m = 785 N # m

Ans.

( MA ) z = 588.6 N # m = 589 N # m

Ans.

Ax = 1177.2 N = 1.18 kN

Ans.

Ay = 588.6 N = 589 N

Ans.

Az = 0

Ans.

Ans: FCB = 1.37 kN ( MA ) x = 785 N # m ( MA ) z = 589 N # m Ax = 1.18 kN Ay = 589 N Az = 0 477

5–79. The forked rod is supported by a collar at A, a thrust bearing at B, and a cable CD. Determine the tension within cable CD and the x, y, z components of reaction at supports A and B due to the loading shown. The supports at A and B are in proper alignment and exert only force reactions on the rod.

z D

SOLUTION 0.5 m

Vector and Position Vectors: Force Vector Vectors: FA = Ay j + Az k FCD = FCD B

2(0 - 0.5)2 + (0 - 2)2 + (2 - 0)2

F = {50i + 40j - 80k} N

1m 1m

0.1741 CDi - 0.6963FCD j + 0.6963FCDk = - 0.1714F r1 = {0.5i + 2j} m

FB = Bx i + By j + Bz k

(0 - 0.5) i + (0 - 2)j + (2 - 0) k

0.5 m

C F = {50i + 40j – 80k} N

r2 = {1i} m

Equations of Equilibrium: Equilibrium: Force equilibrium requires ©F = 0;

FA + FB + F + FCD = 0

(Bx + 50 - 0.1741FCD) i + A Ay + By + 40 - 0.6963FCD B j + (Az + Bz - 80 + 0.6963FCD) k = 0 Equating i, j and k components, we have ©Fx = 0;

Bx + 50 - 0.1741 0.1714FCD = 0

[1]

©Fy = 0;

Ay + By + 40 - 0.6963FCD = 0

[2]

©Fz = 0;

Az + Bz - 80 + 0.6963FCD = 0

[3]

Moment equilibrium requires ©MB = 0;

r1 * (F + FCD) + r2 * FA = 0

(0.5i + 2j) * [(50 - 0.1741 0.1714FCD)i + (40 - 0.6963FCD)j + (0.6963FCD - 80) k] + 1i * A Ay j + Az k B = 0 Equating i, j and k components, we have ©Mx = 0;

1.3926FCD - 160 = 0

[4]

©My = 0;

-(0.3482FCD - 40) - Az = 0

[5]

©Mz = 0;

Ay - 80.0 = 0

[6]

Solving Eqs.[1], [2], [3], [4], [5] and [6] yields FCD = 115 N

Ay = 80.0 N

Az = 0

Ans.

Bx = -30.0 N

By = - 40.0 N

Bz = 0

Ans.

Negative signs indicate that the reaction components act in the opposite sense to those shown on FBD.

478

Ans: Az = 0 Bz = 0 FCD = 115 N, Ay = 80.0 N Bx = -30.0 N, By = - 40.0 N

*5–80. The bent rod is supported at A, B, and C by journal bearings. Determine the components of reaction at bearings if the rod is subjected to the 200-lb vertical force and the 30 lb-ft couple moment. The bearings are in proper alignment and exert only force reactions on the rod.

1 ft

30 lb · ft 1 ft

458

2 ft

SOLUTION

ΣMx = 0;

- By(1) + Cz(1) = 0

(1)

ΣMy = 0;

- Az(2) + Bx(1) + Cx(3) + 30 cos 45° = 0

(2)

ΣMz = 0;

Ay(2) - Cx(1) + 30 sin 45° = 0

(3)

ΣFx = 0;

Bx + Cx = 0

(4)

ΣFx = 0;

Ay + By = 0

(5)

ΣFz = 0;

Az + Cz - 200 = 0

(6)

1 ft A x

200 lb 2 ft

Solving Eqs.[1] to [6] yields: Cx = 358 lb Ay = 168 lb By = -168 lb

Ans.

Bx = - 358 lb Cz = - 168 lb Az = 368 lb

Ans.

Negative signs indicate that the reaction components act in the opposite sense of those shown on FBD.

Ans: Cx = 358 lb Ay = 168 lb By = - 168 lb Bx = - 358 lb Cz = - 168 lb Az = 368 lb 479

5–81. The bent rod is supported at A, B, and C by smooth journal bearings. Compute the x, y, z components of reaction at the bearings if the rod is subjected to forces F1 = 300 lb and F2 = 250 lb. F1 lies in the y–z plane. The bearings are in proper alignment and exert only force reactions on the rod.

1 ft A

4 ft

SOLUTION

5 ft

F1 = (- 300 cos 45°j - 300 sin 45°k)

2 ft

3 ft

= {-212.1j - 212.1k} lb

F2 = (250 cos 45° sin 30°i + 250 cos 45° cos 30°j - 250 sin 45°k)

= {88.39i + 153.1j - 176.8k} lb ©Fx = 0;

Ax + Bx + 88.39 = 0

©Fy = 0;

Ay + Cy - 212.1 + 153.1 = 0

©Fz = 0;

Bz + Cz - 212.1 - 176.8 = 0

©Mx = 0;

-Bz (3) - Ay (4) + 212.1(5) + 212.1(5) = 0

©My = 0;

Cz (5) + Ax (4) = 0

©Mz = 0;

Ax (5) + Bx (3) - Cy (5) = 0 Ax = 633 lb

Ans.

Ay = - 141 lb

Ans.

Bx = - 721 lb

Ans.

Bz = 895 lb

Ans.

Cy = 200 lb

Ans.

Cz = - 506 lb

Ans.

Ans: Ax = 633 lb Ay = - 141 lb Bx = - 721 lb Bz = 895 lb Cy = 200 lb Cz = - 506 lb 480

5–82. The bent rod is supported at A, B, and C by smooth journal bearings. Determine the magnitude of F2 which will cause the reaction C y at the bearing C to be equal to zero. The bearings are in proper alignment and exert only force reactions on the rod. Set F1 = 300 lb.

1 ft A

4 ft

SOLUTION

5 ft

F1 = ( - 300 cos 45°j - 300 sin 45°k)

2 ft

= { -212.1j - 212.1k} lb

F2 = (F2 cos 45° sin 30°i + F2 cos 45° cos 30°j - F2 sin 45°k)

= {0.3536F2 i + 0.6124F2 j - 0.7071F2 k} lb ©Fx = 0;

Ax + Bx + 0.3536F2 = 0

©Fy = 0;

Ay + 0.6124F2 - 212.1 = 0

©Fz = 0;

Bz + Cz - 0.7071F2 - 212.1 = 0

©Mx = 0;

- Bz (3) - Ay (4) + 212.1(5) + 212.1(5) = 0

©My = 0;

Cz (5) + Ax (4) = 0

©Mz = 0;

Ax (5) + Bx (3) = 0

3 ft

Ax = 357 lb Ay = -200 lb Bx = -595 lb Bz = 974 lb Cz = - 286 lb Ans.

F2 = 674 lb

Ans: Ax = 357 lb Ay = - 200 lb Bx = - 595 lb Bz = 974 lb Cz = - 286 lb F2 = 674 lb 481

5–83. The bar AB is supported by two smooth collars. At A the connection is with a ball-and-socket joint and at B it is a rigid attachment. If a 50-lb load is applied to the bar, determine the x, y, z components of reaction at A and B.

D B z C

SOLUTION

6 ft

4 ft

6 ft

ΣFx = 0; Ax + Bx = 0 (1) ΣFy = 0; By + 50 = 0

50 lb

5 ft x

By = - 50 lb

Ans.

ΣFz = 0; Az + Bz = 0

(2)

ΣMz = 0; MBz = 0

Ans.

3 ft

A F

ΣMx = 0; MBx + 50(6) = 0

MBx = - 300 lb # ft

Ans.

rCD = - 5i + 3j uCD = - 0.8575i + 0.5145j Require FB # uCD = 0

(Bxi - 50j + Bzk) # ( -0.8575i + 0.5145j) = 0 -0.8575Bx - 50(0.5145) = 0 Ans.

Bx = - 30.0 lb From Eq. (1);

Ans.

Ax = 30.0 lb Require MB # uCD = 0

( -300i + MByj) # ( - 0.8575i + 0.5145j) = 0 300(0.8575) + MBy(0.5145) = 0 MBy = - 500 lb # ft ΣMy = 0;

Ans.

- 500 + Bz(5) - 30.0(6) = 0

Bz = 136 lb From Eg. (z): Az = - 136 lb Ans: Bx = -30.0 lb By = -50 lb Bz = 136 lb MBz = 0 MBx = -300 lb # ft MBy = -500 lb # ft Bx = -30.0 lb Ax = 30.0 lb Az = -136 lb 482

*5–84. The rod has a weight of 6 lb>ft. If it is supported by a balland-socket joint at C and a journal bearing at D, determine the x, y, z components of reaction at these supports and the moment M that must be applied along the axis of the rod to hold it in the position shown.

z D

60 A 0.5 ft 45

SOLUTION

ΣFx = 0; Cx + Dx - 15 sin 45° = 0

(1)

ΣFy = 0; Cy + Dy = 0

(2)

1 ft

ΣFz = 0; Cz - 15 cos 45° = 0 Ans.

Cz = 10.6 lb ΣMx = 0; - 3 cos 45°(0.25 sin 60°) - Dy(2) = 0 Dy = - 0.230 lb

Ans.

Cy = 0.230 lb

Ans.

From Eq. (2);

ΣMy = 0;

- (12 sin 45°)(1) - (3 sin 45°)(1) + (3 cos 45°)(0.25 cos 60°) + Dx(2) = 0 Dx = 5.17 lb

Ans.

Cx = 5.44 lb

Ans.

From Eq. (1);

ΣMz = 0;

- M + (3 sin 45°)(0.25 sin 60°) = 0

M = 0.459 lb # ft

Ans.

Ans: Cz = 10.6 lb Dy = - 0.230 lb Cy = 0.230 lb Dx = 5.17 lb Cx = 5.44 lb M = 0.459 lb # ft 483

5–85. The sign has a mass of 100 kg with center of mass at G. Determine the x, y, z components of reaction at the ball-andsocket joint A and the tension in wires BC and BD.

z 1m

SOLUTION

1m 2m

Equations of Equilibrium: Expressing the forces indicated on the free-body diagram, Fig. a, in Cartesian vector form, we have

FA = A xi + A yj + A zk

W = {- 100(9.81)k} N = {- 981k} k N FBD = FBDuBD = FBD ≥

FBC = FBCuBC = FBC ≥

(- 2 - 0)i + (0 - 2)j + (1 - 0)k 2

2(- 2 - 0) + (0 - 2) + (1 - 0)

¥ = a-

2 2 1 FBDi - FBDj + FBDkb 3 3 3

(1 - 0)i + (0 - 2)j + (2 - 0)k

1 2 2 ¥ = a FBCi - FBCj + FBCkb 3 3 3 2(1 - 0) + (0 - 2) + (2 - 0) 2

Applying the forces equation of equilibrium, we have ©F = 0;

FA + FBD + FBC + W = 0

2 2 1 1 2 2 (A xi + A yj + A zk) + a - FBDi - FBDj + FBDk b + a FBCi - FBCj + FBCkb + ( - 981 k k) = 0 3 3 3 3 3 3 a Ax -

2 1 2 2 1 2 F + FBC bi + a A y - FBD - FBC b j + aA z + FBD + FBC - 981bk = 0 3 BD 3 3 3 3 3

Equating i, j, and k components, we have Ax -

2 1 F + FBC = 0 3 BD 3

(1)

Ay -

2 2 F - FBC = 0 3 BD 3

(2)

Az +

1 2 FBD + FBC - 981 = 0 3 3

(3)

In order to write the moment equation of equilibrium about point A, the position vectors rAG and rAB must be determined first. rAG = {1j} m (the k component is not needed for the moment of w.) rAB = {2j} m

484

5–85. Continued Thus, ©M A = 0; rAB * (FBC + FBD) + (rAG * W) = 0 1 2 2 2 2 1 (2j) * c a FBC - FBD b i - a FBC + FBD b j + a FBC + FBD bk d + (1j) * ( - 981k) = 0 3 3 3 3 3 3 2 4 2 4 a FBC + FBD - 981bi + a FBD - FBC b k = 0 3 3 3 3 j, and k components have Equating i, i and k components wewe have 2 4 FBD - 981 = 0 F + F 3 BC 3 BC

(4)

2 4 F FBD - FBC = 0 3 BC 3

(5)

Solving Ans.

FBD = 294 N FBC = 589 N Substituting into Eqs. 1, 2, 3, Ax = 0 Ay = 589 N

Ans.

Az = 490.5 N

Ans.

Ans: FBD = 294 N FBC = 589 N Ax = 0 Ay = 589 N Az = 490.5 N FBD = 589 N Ax = 0 Ay = 589 N Az = 490.5 N 485

5–86. z

Both pulleys are fixed to the shaft and as the shaft turns with constant angular velocity, the power of pulley A is transmitted to pulley B. Determine the horizontal tension T in the belt on pulley B and the x, y, z components of reaction at the journal bearing C and thrust bearing D if u = 0°. The bearings are in proper alignment and exert only force reactions on the shaft.

200 mm 250 mm

300 mm C

SOLUTION

80 mm A

©Fx = 0; ©Fy = 0;

165 + 80210.452 - Cz 10.752 = 0 Ans.

150 + 58.0210.22 - Cy 10.752 = 0 Cy = 28.8 N

Ans.

Dx = 0

Ans.

Dy + 28.8 - 50 - 58.0 = 0 Ans.

Dy = 79.2 N ©Fz = 0;

80 N

Ans.

Cz = 87.0 N ©Mz = 0;

65 N

6510.082 - 8010.082 + T10.152 - 5010.152 = 0 T = 58.0 N

©My = 0;

150 mm B T

Equations of of Equilibrium: Equilibrium: Equations ©Mx = 0;

50 N D

Dz + 87.0 - 80 - 65 = 0 Ans.

Dz = 58.0 N

Ans: T = 58.0 N Cz = 87.0 N Cy = 28.8 N Dx = 0 Dy = 79.2 N Dz = 58.0 N 486

5–87. z

Both pulleys are fixed to the shaft and as the shaft turns with constant angular velocity, the power of pulley A is transmitted to pulley B. Determine the horizontal tension T in the belt on pulley B and the x, y, z components of reaction at the journal bearing C and thrust bearing D if u = 45°. The bearings are in proper alignment and exert only force reactions on the shaft.

200 mm 250 mm

300 mm C

SOLUTION

80 mm A

©Fx = 0; ©Fy = 0;

165 + 80210.452 - 50 sin 45°10.22 - Cz 10.752 = 0 Ans.

58.010.22 + 50 cos 45°10.22 - Cy 10.752 = 0 Cy = 24.89 N = 24.9 N

Ans.

Dx = 0

Ans.

Dy + 24.89 - 50 cos 45° - 58.0 = 0 Ans.

Dy = 68.5 N ©Fz = 0;

80 N

Ans.

Cz = 77.57 N = 77.6 N ©Mz = 0;

65 N

6510.082 - 8010.082 + T10.152 - 5010.152 = 0 T = 58.0 N

©My = 0;

150 mm B T

Equations Equations of ofEquilibrium: Equilibrium: ©Mx = 0;

50 N D

Dz + 77.57 + 50 sin 45° - 80 - 65 = 0 Ans.

Dz = 32.1 N

Ans: T = 58.0 N Cz = 77.6 N Cy = 24.9 N Dy = 68.5 N Dz = 32.1 N 487

*5–88. Member AB is supported by a cable BC and at A by a square rod which fits loosely through the square hole at the end joint of the member as shown. Determine the components of reaction at A and the tension in the cable needed to hold the 800-lb cylinder in equilibrium.

z C

2 ft

x 3 ft

3 6 2 FBC = FBC a i - j + k b 7 7 7 ©Fx = 0;

6 ft

SOLUTION

3 FBC a b = 0 7 FBC = 0

Ans.

©Fy = 0;

Ay = 0

Ans.

©Fz = 0;

Az = 800 lb

Ans.

©Mx = 0;

(MA)x - 800(6) = 0 (MA)x = 4.80 kip # ft

Ans.

©My = 0;

(MA)y = 0

Ans.

©Mz = 0;

(MA)z = 0

Ans.

Ans: FBC = 0 Ay = 0 Az = 800 lb (MA)x = 4.80 kip # ft (MA)y = 0 (MA)z = 0 488

6–1. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 2 kN and P2 = 1.5 kN.

SOLUTION

30°

Method of Joints: In this case, the support reactions are not required for determining the member forces.

Joint C: + c ΣFy = 0;

FCB sin 30° - 1.5 = 0

C 3m P2

Ans.

FCB = 3.00 kN (T) + ΣFx = 0; S

30°

FCD - 3.00 cos 30° = 0 Ans.

FCD = 2.598 kN (C) = 2.60 kN (C) Joint D: + ΣFx = 0; S

FDE - 2.598 = 0

FDE = 2.60 kN (C)

Ans.

+ c ΣFy = 0;

FDB - 2 = 0

FDB = 2.00 kN (T)

Ans.

Joint B: + ΣFy′ = 0;

FBE cos 30° - 2.00 cos 30° = 0 Ans.

FBE = 2.00 kN (C) + ΣFx′ = 0;

(2.00 + 2.00) sin 30° + 3.00 - FBA = 0 Ans.

FBA = 5.00 kN (T)

Note: The support reactions at support A and E can be determined by analyzing Joints A and E respectively using the results obtained above.

Ans: FCB = 3.00 kN (T) FCD = 2.60 kN (C) FDE = 2.60 kN (C) FDB = 2.00 kN (T) FBE = 2.00 kN (C) FBA = 5.00 kN (T) 487

6–2. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = P2 = 4 kN.

SOLUTION

30°

Method of Joints: In this case, the support reactions are not required for determining the member forces.

Joint C: + c ΣFy = 0;

FCB sin 30° - 4 = 0

C 3m P2

Ans.

FCB = 8.00 kN (T) + ΣFx = 0; S

30°

FCD - 8.00 cos 30° = 0 Ans.

FCD = 6.928 kN (C) = 6.93 kN (C) Joint D: + ΣFx = 0; S

FDE - 6.928 = 0

FDE = 6.93 kN (C)

Ans.

+ c ΣFy = 0;

FDB - 4 = 0

FDB = 4.00 kN (T)

Ans.

Joint B: Q + ΣFy′ = 0;

FBE cos 30° - 4.00 cos 30° = 0 Ans.

FBE = 4.00 kN (C) R + ΣFx′ = 0;

(4.00 + 4.00) sin 30° + 8.00 - FBA = 0 Ans.

FBA = 12.0 kN (T)

Note: The support reactions at support A and E can be determined by analyzing Joints A and E respectively using the results obtained above.

Ans: FCB = 8.00 kN (T) FCD = 6.93 kN (C) FDE = 6.93 kN (C) FDB = 4.00 kN (T) FBE = 4.00 kN (C) FBA = 12.0 kN (T) 488

6–3. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 200 lb, P2 = 500 lb.

4 ft

SOLUTION

3 ft

Joint B: + ΣFx = 0; S

C 4 ft

3 - FBA a b - FBC cos 45° + 500 = 0 5

4 + c ΣFy = 0; - FBA a b + FBC sin 45° - 200 = 0 5 FBA = 214 lb (T)

Ans.

FBC = 525.28 lb = 525 lb (C)

Ans.

Joint C: + ΣFx = 0; 525.28 cos 45° - FCA = 0 S Ans.

FCA = 371 lb (T)

Ans: FBA = 214 lb (T) FBC = 525 lb (C) FCA = 371 lb (T) 489

*6–4.

Determine the force in each member of the truss and state if the members are in tension or compression.

2 kip

1.5 kip A

SOLUTION Ay (40) + 1.5(4) + 2(12) - 3(10) - 3(20) = 0

: ©Fx = 0;

1.5 + 2 - Ex = 0

+ c ©Fy = 0;

Ey + 1.5 - 3 - 3 = 0

Ay = 1.5 kip

Ex = 3.5 kip Ey = 4.5 kip

Joint A: + c ©Fy = 0;

1.5 - FAl sin 21.80° = 0 FAl = 4.039 kip = 4.04 kip (C)

+ ©F = 0; : x

FAB - 4.039 cos 21.80° = 0

FAB = 3.75 kip (T)

Ans. Ans.

Joint E: + c ©Fy = 0;

4.5 - FEF sin 21.80° = 0 FEF = 12.12 kip = 12.1 kip (C)

+ ©F = 0; : x

Ans.

- FED - 3.5 + 12.12 cos 21.80° = 0 Ans.

FED = 7.75 kip (T) Joint B: + c ©Fy = 0;

FBI = 0

+ ©F = 0; : x

FBC - 3.75 = 0

Ans. Ans.

FBC = 3.75 kip (T)

Joint D: + c ©Fy = 0;

FDF = 0

+ ©F = 0; : x

- FDC + 7.75 = 0

Ans. Ans.

FDC = 7.75 kip (T)

Joint F: +Q©Fy¿ = 0;

FFC cos 46.40° - 3 cos 21.80° = 0 Ans.

FFC = 4.039 kip = 4.04 kip (C) +R©Fx¿ = 0;

FFG + 3 sin 21.80° + 4.039 sin 46.40° - 12.12 = 0 Ans.

FFG = 8.078 kip = 8.08 kip (C) Joint H: + ©F = 0; : x

+ c ©Fy = 0;

2 - FHG cos 21.80° = 0 FHG = 2.154 kip = 2.15 kip (C)

Ans.

2.154 sin 21.80° - FHI = 0

Ans.

FHI = 0.8 kip (T)

490

F B

10 ft

c+ ©ME = 0;

3 kip G

8 ft 4 ft

3 kip

C 10 ft

D 10 ft

10 ft

*6–4. Continued

Joint C: + ©F = 0; : x

- FCI cos 21.80° - 4.039 cos 21.80° - 3.75 + 7.75 = 0 FCI = 0.2692 kip = 0.269 kip (T)

+ c ©Fy = 0;

Ans.

FCG + 0.2692 sin 21.80° - 4.039 sin 21.80° = 0 FCG = 1.40 kip (T)

Ans.

Joint G: +Q©Fy¿ = 0;

FGI cos 46.40° - 3 cos 21.80° - 1.40 cos 21.80° = 0 Ans.

FGI = 5.924 kip = 5.92 kip (C) +R©Fx¿ = 0;

2.154 + 3 sin 21.80° + 5.924 sin 46.40° + 1.40 sin 21.80° - 8.081 = 0 (Check)

Ans: FAl = 4.04 kip (C) FAB = 3.75 kip (T) FEF = 12.1 kip (C) FED = 7.75 kip (T) FBI = 0 FBC = 3.75 kip (T) FDF = 0 FDC = 7.75 kip (T) FFC = 4.04 kip (C) FFG = 8.08 kip (C) FHG = 2.15 kip (C) FHI = 0.8 kip (T) FCI = 0.269 kip (T) FCG = 1.40 kip (T) FGI = 5.92 kip (C) 491

6–5. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 7 kN, P2 = 7 kN.

2m A

P1 2m

SOLUTION

Joint A: + c ΣFy = 0;

FAD sin 45° - 7 = 0 Ans.

FAD = 9.899 = 9.90 kN (C) + ΣFx = 0; S

FAB - 9.899 cos 45° = 0 Ans.

FAB = 7 kN (T) Joint D: +p ΣFy = 0;

FDB - 7 sin 45° = 0 Ans.

FDB = 4.95 kN (T) + ΣFx = 0; S

9.899 - FDC + 7 cos 45° = 0 Ans.

FDC = 14.8 kN (C) Joint C: + c ΣFy = 0;

FCB - 14.8 cos 45° = 0 Ans.

FCB = 10.5 kN (T)

Ans: FAD = 9.90 kN (C) FAB = 7 kN (T) FDB = 4.95 kN (T) FDC = 14.8 kN (C) FCB = 10.5 kN (T) 492

6–6. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 8 kN, P2 = 10 kN.

2m A

P1 2m

SOLUTION

Joint A: + c ΣFy = 0;

FAD sin 45° - 8 = 0 Ans.

FAD = 11.314 = 11.3 kN (C) + ΣFx = 0; S

FAB - 11.314 cos 45° = 0 Ans.

FAB = 8 kN (T) Joint D: +p ΣFy′ = 0;

FDB - 10 sin 45° = 0 Ans.

FDB = 7.07 kN (T) +q ΣFx′ = 0;

11.314 - FDC + 10 cos 45° = 0 Ans.

FDC = 18.385 = 18.4 kN (C) Joint C: + c ΣFy = 0;

FCB - 18.385 cos 45° = 0 Ans.

FCB = 13.0 kN (T)

Ans: FAD = 11.3 kN (C) FAB = 8 kN (T) FDB = 7.07 kN (T) FDC = 18.4 kN (C) FCB = 13.0 kN (T) 493

6–7. 4 kN

Determine the force in each member of the truss and state if the members are in tension or compression.

3m A

SOLUTION 5 kN

a + ©MD = 0;

4162 + 5192 - Ey 132 = 0

+ c ©Fy = 0;

23.0 - 4 - 5 - D y = 0

+ ©F = 0 : x

Ey = 23.0 kN Dy = 14.0 kN

Dx = 0

Method of Joints: Joint D: FDE ¢

5 234

≤ - 14.0 = 0

FDE = 16.33 kN 1C2 = 16.3 kN 1C2 + ©F = 0; : x

16.33 ¢

3 234

Ans.

≤ - FDC = 0

FDC = 8.40 kN 1T2

Ans.

Joint E: + ©F = 0; : x

FEA ¢

3 210

≤ - 16.33 ¢

3 234

≤ = 0

FEA = 8.854 kN 1C2 = 8.85 kN 1C2 + c ©Fy = 0;

23.0 - 16.33 ¢

5 234

≤ - 8.854 ¢

1 210

Ans.

≤ - FEC = 0 Ans.

FEC = 6.20 kN 1C2 Joint C: + c ©Fy = 0; + ©F = 0; : x

F E

Support Reactions:

+ c ©Fy = 0;

6.20 - FCF sin 45° = 0 FCF = 8.768 kN 1T2 = 8.77 kN 1T2

Ans.

8.40 - 8.768 cos 45° - FCB = 0 Ans.

F CB = 2.20 kN T

494

6–7. Continued

Joint B: + ©F = 0; : x

2.20 - FBA cos 45° = 0 FBA = 3.111 kN 1T2 = 3.11 kN 1T2

+ c ©Fy = 0;

Ans.

FBF - 4 - 3.111 sin 45° = 0 FBF = 6.20 kN 1C2

Ans.

+ c ©Fy = 0;

8.768 sin 45° - 6.20 = 0

(Check!)

+ ©F = 0; : x

8.768 cos 45° - FFA = 0

Joint F:

FFA = 6.20 kN 1T2

Ans.

Ans: FDE = 16.3 kN (C) FDC = 8.40 kN (T) FEA = 8.85 kN (C) FEC = 6.20 kN (C) FCF = 8.77 kN (T) FCB = 2.20 kN (T) FBA = 3.11 kN (T) FBF = 6.20 kN (C) FFA = 6.20 kN (T) 495

*6–8. Determine the force in each member of the truss and state if the members are in tension or compression.

A 13

3 ft

3 ft F

4 ft

130 lb

4 ft

SOLUTION Joint A: + c ΣFy = 0;

4 12 (FAC) (130) = 0 5 13 Ans.

FAC = 150 lb (C) + ΣFx = 0; FAB - 3 (150) - 5 (130) = 0 S 5 13

Ans.

FAB = 140 lb (T) Joint B: + ΣFx = 0; FBD - 140 = 0 S

Ans.

FBD = 140 lb (T) + c ΣFy = 0; FBC = 0

Ans.

Joint C: 4 4 + c ΣFy = 0; a bFCD - a b 150 = 0 5 5

Ans.

FCD = 150 lb (T)

+ ΣFx = 0; - FCE + 3 (150) + 3 (150) = 0 S 5 5 Ans.

FCE = 180 lb (C) Joint D: + c ΣFy = 0; FDE -

4 (150) = 0 5 Ans.

FDE = 120 lb (C) + ΣFx = 0; FDE + 140 - 3 (150) = 0 S 5

Ans.

FDF = 230 lb (T) Joint E: + ΣFx = 0; 180 - 3 (FEF) = 0 S 5

Ans.

FEF = 300 lb (C)

496

Ans: FAC = 150 lb (C) FAB = 140 lb (T) FBD = 140 lb (T) FBC = 0 FCD = 150 lb (T) FCE = 180 lb (C) FDE = 120 lb (C) FDF = 230 lb (T) FEF = 300 lb (C)

6–9. Determine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression.

B P 3 —a 4

SOLUTION a + ©MA = 0;

-P(a) + Cy (2a) - P(a) = 0

1 —a 4 C

Cy = P Joint C: + ©F = 0; : x + c ©Fy = 0;

1 22 P +

FBC 1 217

FCD = 0

217

FCD -

1 22

FBC = 0

FBC = 1.886P = 1.89P

(C)

Ans.

FCD = 1.374P = 1.37P

(T)

Ans.

Joint B: + ©F = 0; : x

P -

1 22

(1.886P) +

FAB = 0 Ans.

FAB = 0.471P (C) + c ©Fy = 0;

1 22

(0.471P) +

1 22

(1.886P) - FBD = 0

FBD = 1.67P (T)

Ans.

+ ©F = 0; : x

FAD = 1.374P = 1.37P (T)

Ans.

+ c ©Fy = 0;

1.667P - P -

Joint D:

1 217

(1.374P)(2) = 0

Check!

Ans: FBC = 1.89P (C) FCD = 1.37P (T) FAB = 0.471P (C) FBD = 1.67P (T) FAD = 1.37P (T)

497

The 6–10.maximum allowable tensile force in the members of the truss is 1Ft2max = 1500 lb, and the maximum allowable The maximumforce allowable tensile force in the members of the compressive is 1F c2max = 800 lb. Determine the truss is and maximum allowable 2 = 1500 lb, 1F t max maximum magnitude P of the twothe loads that can be applied compressive force to the truss. Take a = is8 ft.1Fc2max = 800 lb. Determine the maximum magnitude P of the two loads that can be applied to the truss. Take a = 8 ft.

SOLUTION

SOLUTION a + ©MA = 0;

Cy (2a) - P(a) - P(a) = 0

a + ©MA = 0; + c ©Fy = 0;

Cy (2a) = P - P(a) - P(a) = 0 C y A Cyy += C Py - P = 0

+ c ©Fy = 0; + ©F = 0; :

Ay += C 0 - P = 0 A y y P Ax = P A -=A0x = 0;

+ ©F = 0; : Joint C:x + ©F Joint C:x = 0; : + ©F = 0; : x

+ c ©Fy = 0; + c ©Fy = 0;

Joint B: Joint B: + ©F : x = 0;

Ax = P

FCB a

b - FCD a

+ ©F Joint D:x = 0; :

FDA =

+ c ©Fy = 0;

D a a

D P P

1 —a 4 1 C a — 4 C

b = 0 22 217 1 4 FCB a b - FCD a b = 0 4 22 22 217 FCB = FCD 4217 22 FCB = 1FCD 1 P - FCB a b + FCD a b = 0 217 22 217 1 1 P - FCB a b + FCD a b = 0 217 22 217 FCD = P = 1.37P (T) 3 217 FCD = 4 22 P = 1.37P (T) FCB = 3 P = 1.89P (C) 3 4 22 FCB = P = 1.89P (C) 3

Joint D:

+ c ©Fy = 0;

3 —a 4 3 —a 4

P - Ax = 0;

4 22 1 1 Pb a b + FAB a b = 0 3 22 22 4 22 1 1 P- a Pb a b + FAB a b = 0 322 22 22 FAB = P = 0.471 P (C) 3 22 FAB = 1 22 P = 0.471 1 P4(C) 22 a 3 Pb + a P b - FBD = 0 3 3 22 22 1 22 1 422 a Pb + a P b - FBD = 0 3 22 53 22 FBD = P = 1.67P (T) 3 5 FBD = P = 1.67P (T) 3

+ ©F = 0; : x

P- a

217 P = 1.37 P(T) 3 217 +Assume : ©Fx = F0;BD =lb lb P = 1.37 P(T) 1) Assume FBD= F= 1500 1) 1500 DA 3

1) Assume FBDP == 1500 1500;lb P = 900 lb FBD = 1.667 1.667PP==1.89(900) 1500; = 900 lb7 800 lb (N.G!) =P1697.05 FBD CB = 1.89 1.89(900) = 1697.05 7 800 lb (N.G!) FCB = 1.89 2) Assume lb FCBP= =800 2) Assume FCB ==800 800;lb FCB = 1.89P

P = 424.264 lb

= 800; = P = 424.264 lb lb (O.K!) 1.667(424.264) 707.11 lb 6 1500 FCB BD = 1.89P = 1.667(424.264) = 707.11 lb 6 1500 lb (O.K!) FBD P = 424 lb P = 424 lb

Ans. 498

Ans.

Ans: P = 424 lb

6–11. Determine the zero-force members in the Pratt roof truss. Explain your answers using appropriate joint free-body diagrams.

400 N 300 N

A L

SOLUTION

F K

12 m, 6 @ 2 m

Method of Joints: Joint L: + c ©Fy = 0;

FLB = 0

+ c ©Fy¿ = 0;

FLB = 0

Joint B:

Joint K: + c ©Fy = 0;

FKC = 0

+ c ©Fy = 0;

FHF = 0

+ ©Fy¿ = 0;

FFI cos u = 0

+ c ©Fy = 0;

FIE = 0

+ ©Fy¿ = 0;

FEJ cos u = 0

Joint H:

Joint F: FFI = 0

Joint I:

Joint E: FEJ = 0

Hence, members LB, BK, KC, HF, FI, IE and EJ are zero force member.

Ans: LB, BK, KC, HF, FI, IE and EJ 499

*6–12. Determine the force in each member of the truss and state if the members are in tension or compression. Assume all member are pin connected.

SOLUTION

Method of Joints: The support reactions are not required. The joint equilibrium analysis will be carried out in the sequence of C, B and D. D

Joint C: Fig. a, + ΣFx = 0; S + c ΣFy = 0;

FCB a FCB a

Solving Eqs. 1 and 2

b - FCD a

25 1

b + FCD a

2 25 1

b = 0 (1) b - 20 = 0 (2)

FCB = 1025 kN (C) = 22.4 kN (C)

Ans.

FCD = 1025 kN (T) = 22.4 kN (T)

Ans.

1m C

A 2m

2m 20 kN

Joint B: Fig. b, + ΣFx = 0; S

FBAa

2 25

b - a10 25ba

2 25

b =0

Ans.

FBA = 10 25 kN (C) = 22.4 kN (C)

+ c ΣFy = 0;

FBD - 2 a10 25ba FBD = 20 kN (T)

1 25

b =0

Ans.

Joint D: Fig. c, + ΣFx = 0; S + c ΣFy = 0;

FDAa

2 25 1

Solving Eqs. 1 and 2,

b - FDE a

2 213

b + FDE a

b + (10 25)a

213

2 25

b - (10 25)a

b = 0 (1)

b - 20 = 0 (2)

FDE = 10213 kN (T) = 36.1 kN (T)

Ans.

FDA = 0

Ans.

Ans: FCB = 22.4 kN (C) FCD = 22.4 kN (T) FBA = 22.4 kN (C) FBD = 20 kN (T) FDE = 36.1 kN (T) FDA = 0 500

6–13. Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression.

3 — a 4

1 — a 4

D A

SOLUTION Joint A: + ©F = 0; : x + c ©Fy = 0;

C a

4 214

(FAD) -

a P

1 22

FAB = 0

1 P 1 (FAB) + (FAD) = 0 2 22 217 FCD = FAD = 0.687 P (T)

Ans.

FCB = FAB = 0.943 P (C)

Ans.

Joint D: + c ©Fy = 0;

FDB - 0.687 P ¢

1 217

≤ -

1 217

(0.687 P) - P = 0 Ans.

FDB = 1.33 P (T)

Ans: FCD = FAD = 0.687P (T) FCB = FAB = 0.943P (C) FDB = 1.33P (T) 501

6–14. Members AB and BC can each support a maximum compressive force of 800 lb, and members AD, DC, and BD can support a maximum tensile force of 1500 lb. If a = 10 ft , determine the greatest load P the truss can support.

3 — a 4

1 — a 4

D A

SOLUTION

C a

Assume FAB = 800 lb (C)

a P

Joint A: + ©F = 0; : x

- 800 ¢

1 22

≤ + FAD ¢

4 217

≤ = 0 OK

FAD = 583.0952 lb 6 1500 lb + c ©Fy = 0;

1 P 1 (800) + (583.0952) = 0 2 22 217 P = 848.5297 lb

Joint D: + c ©Fy = 0;

- 848.5297 - 583.0952(2) ¢

1 217

≤ + FDB = 0

FBD = 1131.3724 lb 6 1500 lb

Pmax = 849 lb

Ans.

Thus,

Ans: Pmax = 849 lb 502

6–15. Members AB and BC can each support a maximum compressive force of 800 lb, and members AD, DC, and BD can support a maximum tensile force of 2000 lb. If a = 6 ft, determine the greatest load P the truss can support.

3 — a 4

1 —a 4

D A

C a

SOLUTION

a 3

1) Assume FAB = 800 lb (C) Joint A: + ΣFx = 0; S

- 800 a

1 22

b + FAD a

4 217

b = 0

FAD = 583.0952 lb 6 2000 lb + c ΣFy = 0;

P 1 1 (800) + (583.0952) = 0 2 22 217

P = 848.5297 lb Joint D: + c ΣFy = 0;

Thus,

-848.5297 - 583.0952(2) a

1 217

b + FDB = 0

FBD = 1131.3724 lb 6 2000 lb

Pmax = 849 lb

Ans.

Ans: Pmax = 849 lb 503

*6–16. Determine the of of thethe truss andState state the force forceinineach eachmember member truss. if the members are in tension compression. Set P = 8 kN. whether the members are inortension or compression. Set P = 8 kN.

4m B

SOLUTION

60°

+ ©F = 0; : x

FDC sin 60° - 8 = 0 Ans.

FDE - 9.238 cos 60° = 0 FDE = 4.619 kN 1C2 = 4.62 kN 1C2

Ans.

Joint C: + c ©Fy = 0; + ©F = 0; : x

FCE sin 60° - 9.238 sin 60° = 0 FCE = 9.238 kN 1C2 = 9.24 kN 1C2

Ans.

219.238 cos 60°2 - FCB = 0 FCB = 9.238 kN 1T2 = 9.24 kN 1T2

Ans.

Joint B: + c ©F y = 0;

FBE sin 60° - FBA sin 60° = 0 FBE = FBA = F

+ ©F = 0; : x

9.238 - 2F cos 60° = 0 F = 9.238 kN

Thus, FBE = 9.24 kN 1C2

FBA = 9.24 kN 1T2

Ans.

Joint E: + c ©F y = 0;

Ey - 219.238 sin 60°2 = 0

+ ©F = 0; : x

FEA + 9.238 cos 60° - 9.238 cos 60° - 4.619 = 0

Ey = 16.0 kN

FEA = 4.62 kN 1C2

Ans.

Note: The support reactions Ax and Ay can be determinedd by analyzing Joint A using the results obtained above.

504

4m P

Joint D:

FDC = 9.238 kN 1T2 = 9.24 kN 1T2

60°

Method of Joints: In this case, the support reactions are not required for determining the member forces.

+ c ©Fy = 0;

Ans: FDC = 9.24 kN (T) FDE = 4.62 kN (C) FCE = 9.24 kN (C) FCB = 9.24 kN (T) FBA = 9.24 kN (T) FEA = 4.62 kN (C)

6–17.

If the maximum force that any member can support is 8 kN in tension and 6 kN in compression, determine the maximum force P that can be supported at joint D.

4m C

SOLUTION

60°

Method of Joints: In this case, the support reactions are not required for determining the member forces.

60°

4m P

Joint D: + c ©F y = 0;

FDC sin 60° - P = 0

+ ©F = 0; : x

FDE - 1.1547P cos 60° = 0

FDC = 1.1547P 1T2 FDE = 0.57735P 1C2

Joint C: + c ©Fy = 0; + ©F = 0; : x

FCE sin 60° - 1.1547P sin 60° = 0 FCE = 1.1547P 1C2 211.1547P cos 60°2 - FCB = 0

FCB = 1.1547P 1T2

+ c ©F y = 0;

FBE sin 60° - FBA sin 60° = 0

FBE = FBA = F

+ ©F = 0; : x

1.1547P - 2F cos 60° = 0

Joint B:

F = 1.1547P

Thus, FBE = 1.1547P 1C2

FBA = 1.1547P 1T2

Joint E: + ©F = 0; : x

FEA + 1.1547P cos 60° - 1.1547P cos 60° - 0.57735P = 0 FEA = 0.57735P 1C2

From the above analysis, the maximum compression and tension in the truss member is 1.1547P. For this case, compression controls which requires 1.1547P = 6 Ans.

P = 5.20 kN

Ans: P = 5.20 kN 505

6–18. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 10 kN, P2 = 8 kN.

1m P1

SOLUTION Support Reactions: Not required. Method of Joints: We will perform the joint equilibrium according to the sequence of joints D, C, E, B and F. Joint D: Fig. a + c ΣFy = 0;

FDE a

b - 8 = 0 FDE = 4 25 kN (T) = 8.94 kN (T) Ans.

+ ΣFx = 0; FDC - a4 25ba 1 b = 0 FDC = 4.00 kN (C) S 25

Ans.

Joint C: Fig. b

+ ΣFx = 0; FCB - 4.00 = 0 FCB = 4.00 kN(C) S

Ans.

+ c ΣFy = 0;

Ans.

FCE = 0

Joint E: Fig. c + c ΣFy = 0; FEB a

1 22

b - a425ba

2 25

b =0

FEB = 8 22 kN (C) = 11.3 kN (C)

Ans.

FEF = 12.0 kN (T)

Ans.

+ ΣFx = 0; a822ba 1 b + a425ba 1 b - FEF = 0 S 22 25

506

D 1m P2

6–18. Continued

Joint B: Fig. d + ΣFx = 0; FBA - 4.00 - a8 22ba 1 b = 0 FBA = 12.0 kN (C) Ans. S 22

+ c ΣFy = 0; FBF - 10 - a8 22ba

Joint F: Fig. e + c ΣFy = 0;

FFA a

2 25

b - 18.0 = 0

b = 0 FBF = 18.0 kN (T)

Ans.

FFA = 9 25 kN (C) = 20.1 kN (C) Ans.

+ ΣFx = 0; 12.0 + a9 25ba 1 b - FFG = 0 FFG = 21.0 kN (T) S 25

Ans.

Ans: FDE = 8.94 kN (T) FDC = 4.00 kN (C) FCB = 4.00 kN (C) FCE = 0 FEB = 11.3 kN (C) FEF = 12.0 kN (T) FBA = 12.0 kN (C) FBF = 18.0 kN (T) FFA = 20.1 kN (C) FFG = 21.0 kN (T) 507

6–19. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 8 kN, P2 = 12 kN.

1m P1

SOLUTION Support Reactions: Not required. Method of Joints: We will perform the joint equilibrium according to the sequence of joints D, C, E, B and F. Joint D: Fig. a + c ΣFy = 0; + ΣFx = 0; S Joint C: Fig. b

FDE a

2 25

b - 12 = 0

FDC - a6 25ba

FDE = 6 25 kN (T) = 13.4 kN (T)

b = 0

FDC = 6.00 kN (C)

Ans. Ans.

+ ΣFx = 0; FCB - 6.00 = 0 FCB = 6.00 kN (C) S

Ans.

+ c ΣFy = 0; FCE = 0

Ans.

Joint E: Fig. c + c ΣFy = 0;

FEBa

1 22

b - a6 25ba

2 25

b =0

Ans.

FEB = 12 22 kN (C) = 17.0 kN (C)

+ ΣFx = 0; S

a12 22ba

1 22

b + a6 25ba

FEF = 18.0 kN (T)

1 25

b - FEF = 0

Ans.

508

D 1m P2

6–19. Continued

Joint B: Fig. d + ΣFx = 0; FBA - 6.00 - a12 22ba 1 b = 0 FBA = 18.0 kN (C) Ans. S 22

+ c ΣFy = 0; FBF - 8 - a12 22ba

Joint F: Fig. e + c ΣFy = 0;

FFA a

2 25

b - 20.0 = 0

b = 0 FBF = 20.0 kN (T)

Ans.

FFA = 10 25 kN (C) = 22.4 kN (C) Ans.

+ ΣFx = 0; a10 25ba 1 b + 18.0 - FFG = 0 FFG = 28.0 kN(T) S 25

Ans.

Ans: FDE = 13.4 kN (T) FDC = 6.00 kN (C) FCB = 6.00 kN (C) FCE = 0 FEB = 17.0 kN (C) FEF = 18.0 kN (T) FBA = 18.0 kN (C) FBF = 20.0 kN (T) FFA = 22.4 kN (C) FFG = 28.0 kN (T) 509

*6–20. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 20 kN, P2 = 10 kN.

2m E

A G 1.5 m

1.5 m P1

SOLUTION Entire Truss: a + ΣMA = 0;

- 20(1.5) - 10(4.5) + Ey(6) = 0 Ey = 12.5 kN

+ c ΣFy = 0; Ay - 20 - 10 + 12.5 = 0 Ay = 17.5 kN + ΣFx = 0; Ax = 0 S Joint A: + c ΣFy = 0;

17.5 -

4 F = 0 5 AB Ans.

FAB = 21.875 = 21.9 kN (C) + ΣFx = 0; FAG - 3 (21.875) = 0 S 5

Ans.

FAG = 13.125 = 13.1 kN (T) Joint B: + ΣFx = 0; 3 (21.875) - FBC = 0 S 5

Ans.

FBC = 13.125 = 13.1 kN (C) 4 + c ΣFy = 0; (21.875) - FBG = 0 5 FBG = 17.5 kN (T)

Ans.

Joint G: + c ΣFy = 0; 17.5 - 20 +

4 F = 0 5 CG Ans.

FCG = 3.125 = 3.12 kN (T) + ΣFx = 0; 3 (3.125) + FFG - 13.125 = 0 S 5

Ans.

FFG = 11.25 = 11.2 kN (T) Joint C: 4 4 + c ΣFy = 0; FCF - (3.125) = 0 5 5

Ans.

FCF = 3.125 = 3.12 kN (C) + ΣFx = 0; 13.125 - 3 (3.125) - 3 (3.125) - FCD = 0 S 5 5

Ans.

FCD = 9.375 = 9.38 kN (C) 510

F 1.5 m

1.5 m P2

*6–20. Continued

Joint D: + ΣFx = 0; 9.375 - 3 FDE = 0 S 5 Ans.

FDE = 15.625 = 15.6 kN (C) 4 + c ΣFy = 0; (15.625) - FDF = 0 5 FDF = 12.5 kN (T)

Ans.

Joint F: + ΣFx = 0; 3 (3.125) - 11.25 + FEF = 0 S 5 Ans.

FEF = 9.38 kN (T) + c ΣFy = 0; 12.5 - 10 -

4 (3.125) = 0 Check! 5

Ans: FAB = 21.9 kN (C) FAG = 13.1 kN (T) FBC = 13.1 kN (C) FBG = 17.5 kN (T) FCG = 3.12 kN (T) FFG = 11.2 kN (T) FCF = 3.12 kN (C) FCD = 9.38 kN (C) FDE = 15.6 kN (C) FDF = 12.5 kN (T) FEF = 9.38 kN (T) 511

6–21. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 40 kN, P2 = 20 kN.

2m E

A G 1.5 m

F 1.5 m

1.5 m

1.5 m P2

SOLUTION a + ΣMA = 0;

- 40(1.5) - 20(4.5) + Ey (6) = 0 Ey = 25 kN

+ c ΣFy = 0;

Ay - 40 - 20 + 25 = 0 Ay = 35 kN

+ ΣFx = 0; Ax = 0 S Joint A: + c ΣFy = 0;

35 -

4 F = 0 5 AB Ans.

FAB = 43.75 = 43.8 kN (C) + ΣFx = 0; S

FAG -

3 (43.75) = 0 5 Ans.

FAG = 26.25 = 26.2 kN (T) Joint B: + ΣFx = 0; S

+ c ΣFy = 0;

3 (43.75) - FBC = 0 5 FBC = 26.25 = 26.2 kN (C)

Ans.

4 (43.75) - FBG = 0 5 FBG = 35.0 kN (T)

Ans.

Ans: FAB = 43.8 kN (C) FAG = 26.2 kN (T) FBC = 26.2 kN (C) FBG = 35.0 kN (T) 512

6–22. Determine the force in each member of the truss and indicate whether the members are in tension or compression. Assume that all members are pin connected.

SOLUTION

+ c ΣFy = 0; Ey + 7.5 - 5 - 5 - 5 = 0 + ΣFx = 0; S Joint A: + c ΣFy = 0;

Ay = 7.5 kN

Ey = 7.5 kN

Ex = 0

7.5 - FAB sin 45° = 0 FAB = 10.61 kN = 10.6 kN (C)

+ ΣFx = 0; S

FAH - 10.61 cos 45° = 0

Joint E: + c ΣFy = 0;

7.5 - FEF sin 45° = 0

Ans.

FAH = 7.5 kN (T)

Ans.

FEF = 10.61 kN = 10.6 kN (T)

Ans.

+ ΣFx = 0; S

FED - 10.61 cos 45° = 0

Joint B: + ΣFx = 0; S

10.61 cos 45° - FBC = 0

+ c ΣFy = 0;

10.61 sin 45° - FBH = 0

Joint F: + ΣFx = 0; S

10.61 cos 45° - FFG = 0

+ c ΣFy = 0;

10.61 sin 45° - 5 - FFD = 0

3m E

3m A

Support Reactions: a + ΣME = 0; 5(3) + 5(6) + 5(9) - Ay(12) = 0

FED = 7.5 kN (C)

FBC = 7.50 kN (C) FBH = 7.50 kN (T)

FFG = 7.50 kN (T) FFD = 2.50 kN (C)

513

Ans.

Ans. Ans.

5 kN

6–22. Continued

Joint H: + c ΣFy = 0; 7.5 - 5 - FHC sin 45° = 0 FHC = 3.536 kN = 3.54 kN (C) + ΣFx = 0; S

FHG - 3.536 cos 45° - 7.5 = 0

Joint D: + c ΣFy = 0;

- FDG sin 45° + 2.5 = 0

FHG = 10.0 kN (T)

Ans. Ans.

FDG = 3.536 kN = 3.54 kN (T) Ans. + ΣFx = 0; S

FDC - 3.536 cos 45° - 7.5 = 0

Joint C: + c ΣFy = 0;

3.536 sin 45° - FCG = 0

+ ΣFx = 0; S

7.5 + 3.536 cos 45° - 10 = 0

FDC = 10.0 kN (C)

Ans.

FCG = 2.50 kN (T) (Check !)

Ans: FAB = 10.6 kN (C) FAH = 7.5 kN (T) FEF = 10.6 kN (T) FED = 7.5 kN (C) FBC = 7.50 kN (C) FBH = 7.50 kN (T) FFG = 7.50 kN (T) FFD = 2.50 kN (C) FHC = 3.54 kN (C) FHG = 10.0 kN (T) FDG = 3.54 kN (T) FCG = 2.50 kN (T) 514

6–23. The Fink truss supports the loads shown. Determine the force in each member and state if the members are in tension or compression. Approximate each joint as a pin.

500 lb 1000 lb 1000 lb 1000 lb

SOLUTION

Entire Truss: + ©F = 0; : x

Ex = 2000 lb a + ©ME = 0;

- Ay (10) cos 30° + 500 (10) + 1000 (7.5) + 1000 (5) + 1000 (2.5) = 0 Ay = 2309.4 lb

+ c ©Fy = 0;

2309.4 - (500 + 1000 + 1000 + 1000 + 500) cos 30° + Ey = 0 Ey = 1154.7 lb

Joint A: + c ©Fy = 0;

- 500 cos 30° + 2309.4 - FAB sin 30° = 0 Ans.

FAB = 3752.78 lb = 3.75 kip (C) + ©F = 0; : x

500 sin 30° + FAH - 3752.78 cos 30° = 0 FAH = 3000 lb = 3 kip (T)

Ans.

+ Q©Fx = 0;

FBC = 3752.78 lb = 3.75 kip (C)

Ans.

+ a©Fy = 0;

FBH = 1000 lb = 1 kip (C)

Ans.

Joint B:

Joint H: + c ©Fy = 0;

FHC sin 60° - 1000 sin 60° = 0 Ans.

FHC = 1000 lb = 1 kip (T) + ©F = 0; : x

FGH + 1000 cos 60° + 1000 cos 60° - 3000 = 0 Ans.

FGH = 2000 lb = 2 kip (T) Joint E: +a©Fy = 0;

- 500 + 2000 sin 30° + 1154.7 cos 30° - FEF sin 30° = 0 Ans.

FEF = 3000 lb = 3 kip (T) + Q©Fx = 0;

1154.7 sin 30° - 2000 cos 30° - 3000 cos 30° + FED = 0 Ans.

FED = 3752.78 lb = 3.75 kip (C)

515

2.5 ft

30 F

B 30 H

(500 + 1000 + 1000 + 1000 + 500) sin 30° - Ex = 0

2.5 ft

500 lb 2.5 ft

2.5 ft

6–23. Continued Joint D: + Q©Fx = 0;

FDC = 3752.78 lb = 3.75 kip (C)

Ans.

+a©Fy = 0;

FDF = 1000 lb = 1 kip (C)

Ans.

+Q©Fx = 0;

FCF = 1000 lb = 1 kip (T)

Ans.

+a©Fy = 0;

- 1000 - 1000 cos 60° (2) + FCG = 0

Joint C:

Ans.

FCG = 2000 lb = 2 kip (C) Joint F: +Q©Fx = 0;

3000 - 1000 cos 60° (2) - FFG = 0 Ans.

FFG = 2000 lb = 2 kip (T) Joint G: + ©F = 0; : x

2000 cos 60° - 2000 + 2000 cos 60° = 0

Check!

+ c Fy = 0;

2000 sin 60° - 2000 sin 60° = 0

Check!

Ans: FAB = 3.75 kip (C) FAH = 3 kip (T) FBC = 3.75 kip (C) FBH = 1 kip (C) FHC = 1 kip (T) FGH = 2 kip (T) FEF = 3 kip (T) FED = 3.75 kip (C) FDC = 3.75 kip (C) FDF = 1 kip (C) FCF = 1 kip (T) FCG = 2 kip (C) FFG = 2 kip (T)

516

*6–24. Determine the force in each member of the truss and state if the member is in tension or compression.

2m 3 kN

SOLUTION Support Reactions: Write the moment equation of equilibrium about point A by referring to the FBD of the entire truss, Fig. a, a + ΣMA = 0; NC (6) - 6(3) - 3(4) = 0 NC = 5 kN

6 kN

Method of Joint: The joint equilibrium analysis will be carried out in the sequence of C, D and B. Joint C: Fig. b + ΣFx = 0; S + c ΣFy = 0;

FCD a FCD a

Solving Eqs. 1 and 2

3 b - FCB a b = 0 (1) 5 213 2

4 b - FCB a b + 5 = 0 (2) 5 213 Ans.

FCB = 12.5 kN (T) FCD =

5 213 kN (C) = 9.01 kN (C) 2

Ans.

Joint D: Fig. c + ΣFx = 0; S

+ c ΣFy = 0;

FDA a

3 213

5 3 b - a 213ba b = 0 2 213

FDA =

5 213 kN = 9.01 kN (C) 2

Ans.

5 2 FDB - 2c a 213ba bd = 0 2 213

Ans.

3 3 12.5 a b - 3 - FBAa b = 0 5 5

Ans.

FDB = 10 kN (C)

Joint B: Fig. d + ΣFx = 0; S

FBA = 7.50 kN (T)

+ ΣFy = 0; S

4 4 12.5 a b + 7.5 a b - 10 - 6 = 0 (O.K!!) 5 5

Ans: FDA = 9.01 kN (C) FDB = 10 kN (C) FBA = 7.50 kN (T) FCB = 12.5 kN (T) FCD = 9.01 kN (C) 517

6–25. Determine the force in members BF and FD of the truss and state if the members are in tension or compression. Set P1 = 200 lb, P2 = 150 lb.

P1 C P2

60°

10 ft

SOLUTION

D F

30°

Joint B: +a©Fy = 0;

FBF sin 75° - 150 = 0

10 ft

FBF = 155 lb (C)

Ans.

FDF = 0

Ans.

10 ft

Joint D: + b©Fy = 0;

Ans: FBF = 155 lb (C) FDF = 0

518

6–26. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 800 lb , P2 = 600 lb.

P1 C P2

60°

10 ft

SOLUTION AF =

D F

30°

10 = 11.547 ft cos 30° 10 ft

AB 11.547 = sin 60° sin 105°

10 ft

AB = 10.3528 ft + ©F = 0; : x

600 sin 45° - Ax = 0 Ax = 424.264 lb

a + ©MA = 0;

E (20) - 800(10) - 600(10.3528) = 0 Ey = 710.583 lb

+ c ©Fy = 0;

Ay + 710.583 - 800 - 600 cos 45° = 0 Ay = 513.681 lb

Joint E: + ©F = 0; : x

- FEF cos 30° + FED cos 45° = 0

+ c ©Fy = 0;

710.583 - FED sin 45° + FEF sin 30° = 0 FED = 2377.66 lb = 2.38 kip (C)

Ans.

FEF = 1941.35 lb = 1.94 kip (T)

Ans.

Joint D: +b©Fy = 0;

FDF sin u = 0 Ans.

FDF = 0 +a©Fx = 0;

2.38 - FDC = 0 Ans.

FDC = 2.38 kip (C)

Ans: FED = 2.38 kip (C) FEF = 1.94 kip (T) FDF = 0 FDC = 2.38 kip (C) 519

6–27. Determine the force in members DC, HC, and HI of the truss and state if the members are in tension or compression.

50 kN

40 kN 2m

2m D

F G

SOLUTION

Support Reactions: Applying the moment equation of equilibrium about point A to the free - body diagram of the truss, Fig. a, a + ©MA = 0;

H B

1.5 m 30 kN 1.5 m 40 kN 1.5 m

40(1.5) + 30(3) + 40(2) - Fy(4) = 0 Fy = 57.5 kN

+ ©F = 0; : x

Ax - 30 - 40 = 0;

Ax = 70 kN

+ c ©Fy = 0;

57.5 - 40 - 50 + Ay = 0;

Ay = 32.5 kN

Method of Sections: Using the bottom portion of the free - body diagram, Fig. b. a + ©MC = 0;

70(3) - 32.5(2) - 40(1.5) - FHI(2) = 0 Ans.

FHI = 42.5 kN (T) a + ©MD = 0;

70(4.5) - 40(3) - 30(1.5) - FHC(1.5) = 0 Ans.

FHC = 100 kN (T) + c ©Fy = 0;

3 32.5 + 42.5 - FDC ( ) = 0 5 Ans.

FDC = 125 kN (C)

Ans: FHI = 42.5 kN (T) FHC = 100 kN (T) FDC = 125 kN (C) 520

© 2022 by R. C.the Hibbeler. by Pearson Education, Inc., NJ. All rights reserved. This material is protected 50 kNunder all copyright laws Determine force Published in members ED, EH, and GH of Hoboken, the as they currently exist. No portion of this material may be reproduced, in any form or by any means, without 40 permission in writing from the publisher. kN truss and state if the members are in tension or compression. 2m 2m 2m

Determine the force in members ED, EH, and GH of the truss and state if the members are in tension or compression.

SOLUTION

40 kN 2m G 2 mH

A H

2m D

40(1.5) + 30(3) + 40(2) - Fy(4) = 0

Fy = 57.5 kN Support Reactions: Applying the moment equation of equilibrium about point A to + ©F = 0; : - 40Fig. = a, 0; Ax = 70 kN the freex - body diagramAof the30truss, x c ©F ++ ©MyA==0;0; a

50 kN

Support Reactions: Applying the moment equation of equilibrium about point A to the free - body diagram of the truss, Fig. a, a + ©MA = 0;

*6–28.

1.5 m 30 kN

1.5 m B 40 kN

1.5mm 1.5 C

30 kN

1.5 m 40 kN 1.5 m

57.5 - 40 - 50 ++ A y = 0; 40(1.5) + 30(3) 40(2) - Fy(4) = 0Ay = 32.5 kN

57.5 Fy =the Method of Sections: Using leftkN portion of the free - body diagram, Fig. b. ++ ©F : a ©MxE==0;0;

A 0; = 0 - 57.5(2) + 40 FGH=(1.5) x - 30 -

Ax = 70 kN

+ c ©Fy = 0;

57.5 50 (T) + Ay = 0; = 40 76.7-kN F GH -

Ay = 32.5 kN

Ans.

a + ©MHof=Sections: 0; -57.5(4) + portion FED(1.5) 40(2) 0 Method Using the left of +the free = - body diagram, Fig. b. F = 100+ kN (C) -ED 57.5(2) FGH (1.5) = 0

Ans.

a + ©MH = 0;

3 (T) =F 76.7 (kN F GH 57.5 ) - 40 = 0 EH 5 -57.5(4) + FED(1.5) + 40(2) = 0 FEH = 29.2 kN (T) FED = 100 kN (C)

+ c ©Fy = 0;

3 57.5 - FEH ( ) - 40 = 0 5

a + ©ME = 0; + c ©Fy = 0;

Ans. Ans.

Ans.

FEH = 29.2 kN (T)

Ans: FGH = 76.7 kN (T) FED = 100 kN (C) FEH = 29.2 kN (T) 521

6–29. Determine the force in members HG, HE, and DE of the truss and state if the members are in tension or compression.

4 ft A

SOLUTION

3 ft

Method of Sections: The forces in members HG, HE, and DE are exposed by cutting the truss into two portions through section a–a and using the upper portion of the free-body diagram, Fig. a. From this free-body diagram, FHG and FDE can be obtained by writing the moment equations of equilibrium about points E and H, respectively. FHE can be obtained by writing the force equation of equilibrium along the y axis.

3 ft

F 3 ft

1500 lb 1500 lb 1500 lb 1500 lb 1500 lb

Joint D: From the free-body diagram in Fig. a, a + ©ME = 0;

FHG(4) - 1500(3) = 0 Ans.

FHG = 1125 lb (T) a + ©MH = 0;

FDE(4) - 1500(6) - 1500(3) = 0 Ans.

FDE = 3375 lb (C) + c ©Fy = 0;

4 FHE a b - 1500 - 1500 = 0 5 Ans.

FEH = 3750 lb (T)

Ans: FHG = 1125 lb (T) FDE = 3375 lb (C) FEH = 3750 lb (T) 522

6–30. Determine the force in members CD, HI, and CH of the truss and state if the members are in tension or compression.

4 ft A

SOLUTION

3 ft

Method of Sections: The forces in members HI, CH, and CD are exposed by cutting the truss into two portions through section b–b on the right portion of the free-body diagram, Fig. a. From this free-body diagram, FCD and FHI can be obtained by writing the moment equations of equilibrium about points H and C, respectively. FCH can be obtained by writing the force equation of equilibrium along the y axis. a + ©MH = 0;

3 ft

1500 lb 1500 lb 1500 lb 1500 lb 1500 lb

Ans.

FHI(4) - 1500(3) - 1500(6) - 1500(9) = 0 Ans.

FHI = 6750 lb (T) + c ©Fy = 0;

3 ft

FCD(4) - 1500(6) - 1500(3) = 0 FCD = 3375 lb (C)

a + ©Mc = 0;

3 ft

4 FCH a b - 1500 - 1500 = 0 5 Ans.

FCH = 5625 lb (C)

Ans: FCD = 3375 lb (C) FHI = 6750 lb (T) FCH = 5625 lb (C) 523

6–31. Determine the force in members CE, FE, and CD and state if the members are in tension or compression. Hint: The force acting at the pin G is directed along member GD. Why?

G 53

10 ft

D 10 ft

1000 lb

A F 2500 lb

SOLUTION a + ΣMA = 0;

- 2500(10) - 1000(20) - 1000(30) -

E 1000 lb

4 3 F (10) + FGD (30) = 0 5 GD 5

FGD = 7500 lb + c ΣFy = 0;

Ay +

3 (7500) - 2500 - 1000 - 1000 = 0 5

Ay = 0 + ΣFx = 0; S a + ΣME = 0;

a + ΣMC = 0;

+ c ΣFy = 0;

Ax =

4 (7500) = 6000 lb 5

4 3 - 1000(10) - 7500 a b(10) + 7500 a b(10) + FCD(10) = 0 5 5

FCD = 2500 lb = 2.5 kip (T)

Ans.

3 - 1000(10) + 7500 a b(10) - FFE (10) = 0 5

FFE = 3500 lb = 3.50 kip (T)

Ans.

3 7500 a b - 1000 - 1000 - FCE = 0 5

Ans.

FCE = 2500 lb = 2.50 kip (C)

Ans: FCD = 2.5 kip (T) FFE = 3.50 kip (T) FCE = 2.50 kip (C) 524

*6–32. Determine the force in members BC, FC, and FE and state if the members are in tension or compression. Hint: The force acting at the pin G is directed along member GD. Why?

G 53

10 ft

D 10 ft

1000 lb

A F 2500 lb

SOLUTION a + ΣMA = 0;

- 2500(10) - 1000(20) - 1000(30) -

E 1000 lb

4 3 FGD (10) + FGD (30) = 0 5 5

FGD = 7500 lb + c ΣFy = 0;

Ay +

3 (7500) - 2500 - 1000 - 1000 = 0 5

Ay = 0 4 (7500) = 6000 lb 5

+ ΣFx = 0; S

Ax =

a + ΣMF = 0;

FBC = 0

a + ΣMC = 0;

- 6000(10) + 2500(10) + FFE(10) = 0

Ans.

FFE = 3500 lb = 3.50 kip (T) + c ΣFy = 0;

FFC sin 45° - 2500 = 0 Ans.

FFC = 3535.53 lb = 3.54 kip (T)

Ans: FBC = 0 FFE = 3.50 kip (T) FFC = 3.54 kip (T) 525

6–33. Determine the force in members BC, HC, and HG of the bridge truss, and state if the members are in tension or compression.

3m A

B 3m

SOLUTION

C 3m

12 kN

Support Reactions: a + ©ME = 0;

D 3m

14 kN 18 kN

18132 + 14162 + 12192 - Ay 1122 = 0

Ay = 20.5 kN

Method of Sections: a + ©MC = 0;

FHG132 + 12132 - 20.5162 = 0 FHG = 29.0 kN 1C2

a + ©MH = 0;

Ans.

FBC132 - 20.5132 = 0 FBC = 20.5 kN 1T2

+ c ©Fy = 0;

Ans.

20.5 - 12 - FHC sin 45° = 0 FHC = 12.0 kN 1T2

Ans.

Ans: FHG = 29.0 kN (C) FBC = 20.5 kN (T) FHC = 12.0 kN (T)

526

6–34. Determine the force in members DE, DF, and GF of the cantilevered truss and state if the members are in tension or compression.

E 3 ft

I 4 ft

H 4 ft

G 4 ft

4 ft

1500 lb 5

SOLUTION

+ c ©Fy = 0;

4 3 F - 115002 = 0 5 DF 5

a + ©MD = 0;

3 4 1150021122 + 115002132 - FGF 132 = 0 5 5

a + ©MF = 0;

4 1150021162 - FDE 132 = 0 5

FDF = 2000 lb = 2.0 kip 1C2

Ans.

FGF = 5700 lb = 5.70 kip 1C2

Ans.

FDE = 6400 lb = 6.40 kip 1T2

Ans.

Ans: FDF = 2.0 kip (C) FGF = 5.70 kip (C) FDE = 6.40 kip (T) 527

6–35. Determine the force in members GF, FC, and CD of the bridge truss and state if the members are in tension or compression.

G 15 ft

30 ft E

A C

B 40 ft

40 ft 15 k

D 40 ft

40 ft 10 k

SOLUTION Support Reaction: a + ΣMA = 0;

Ey(160) - 5(120) - 10(80) - 15(40) = 0

Ey = 12.5 kip

+ ΣFx = 0; Ex = 0 S Method of Sections: 3 a + ΣMO = 0; FFC (120) + 5(80) - 12.5(40) = 0 5 Ans.

FFC = 1.389 kip = 1.39 kip (T) a + ΣMC = 0; 12.5(80) - 5(40) - FGF sin 20.556°(120) = 0

Ans.

FGF = 18.99 kip = 19.0 kip (C) a + ΣMF = 0; 12.5(40) - FCD (30) = 0

Ans.

FCD = 16.67 kip = 16.7 kip (T)

Ans: FFC = 1.39 kip (T) FGF = 19.0 kip (C) FCD = 16.7 kip (T)

528

*6–36. For the given loading, determine the force in members CD, CJ, and KJ of the Howe roof truss and state if the members are in tension or compression.

400 N 300 N

3m G

A L

12 m, 6 @ 2 m

SOLUTION Support Reactions: a + ΣMG = 0;

- Ay(12) + 300(8) + 400(6) = 0

+ ΣFx = 0; S

Ax = 0

Ay = 400 N

Method of Sections: a + ΣMJ = 0; FCD sin 26.56°(6) + 300(2) - 400(6) = 0 Ans.

FCD = 671 N (C) a + ΣMC = 0; FKJ(2) - 400(4) = 0

FKJ = 800 N (T)

a + ΣMA = 0; FCJ sin 45°(6) - 300(4) = 0

FCJ = 283 N (C)

Ans. Ans.

Ans: FCD = 671 N (C) FKJ = 800 N (T) FCJ = 283 N (C) 529

6–37. Determine the force in members EF, BE, BC and BF of the truss and state if the members are in tension or compression. Set P1 = 9 kN, P2 = 12 kN, and P3 = 6 kN.

3m D

C 3m

3m P1

3m P2

SOLUTION Support Reactions: Referring to the FBD of the entire truss shown in Fig. a, ND can be determined directly by writing the moment equation of equilibrium about point A. a+ΣMA = 0; ND(9) - 12(6) - 9(3) - 6(3) = 0 ND = 13.0 kN Method of Sections: Referring to the FBD of the right portion of the truss sectioned through a–a shown in Fig. b, FEF and FBC can be determined directly by writing the moment equation of equilibrium about points B and E, respectively. a+ΣMB = 0; 13.0(6) - 12(3) - FEF(3) = 0 FEF = 14.0 kN (C)

Ans.

a+ΣME = 0; 13.0(3) - FBC (3) = 0 FBC = 13.0 kN (T)

Ans.

Also, FBE can be determined by writing the force equation of equilibrium along the y axis. + c ΣFy = 0; 13.0 - 12 - FBE a

1 22

b = 0 FBE = 22 kN (T) = 1.41 kN (T) Ans.

Method of Joints: Using the result of FBE, the equilibrium of joint B, Fig. c, requires + c ΣFy = 0; FBF + a 22ba

1 22

b - 9 = 0 FBF = 8.00 kN (T)

Ans.

Ans: FEF = 14.0 kN (C) FBC = 13.0 kN (T) FBE = 1.41 kN (T) FBF = 8.00 kN (T) 530

6–38. Determine the force in members BC, BE, and EF of the truss and state if the members are in tension or compression. Set P1 = 6 kN, P2 = 9 kN, and P3 = 12 kN.

3m D

C 3m

3m P1

3m P2

SOLUTION Support Reactions: Referring to the FBD of the entire truss shown in Fig. a, ND can be determined directly by writing the moment equation of equilibrium about point A. a+ΣMA = 0; ND(9) - 12(3) - 6(3) - 9(6) = 0 ND = 12.0 kN Method of Sections: Referring to the FBD of the right portion of the truss s ectioned through a–a shown in Fig. b, FEF and FBC can be determined directly by writing the moment equation of equilibrium about points B and E, respectively. a+ΣMB = 0; 12.0(6) - 9(3) - FEF(3) = 0 FEF = 15.0 kN (C)

Ans.

a+ΣME = 0; 12.0(3) - FBC(3) = 0 FBC = 12.0 kN (T)

Ans.

Also, FBE can be obtained directly by writing the force equation of equilibrium along the y axis + c ΣFy = 0; 12.0 - 9 - FBE a

1 22

b = 0

Ans.

FBE = 322 kN = 4.24 kN (T)

Ans: FEF = 15.0 kN (C) FBC = 12.0 kN (T) FBE = 4.24 kN (T) 531

6–39. Determine the force in members FE, FB, and BC of the truss and state if the members are in tension or compression. After sectioning the truss, solve for each force directly using a single equilibrium equation to obtain each force. Assume that all joints are pin connected.

1000 lb

1000 lb 500 lb

500 lb G

10 ft

10 ft E

15 ft B A

SOLUTION a + ΣMC = 0; FFE (5) - 500(10) = 0 Ans.

FFE = 1000 lb = 1.0 kip (T) a + ΣMF = 0; FBC sin 26.56°(20) - 1000(10) - 500(20) = 0

Ans.

FBC = 2236 lb = 2.24 kip (C) Joint B:

Ans.

+ rΣFy′ = 0; FFB sin u = 0 FFB = 0

Ans: FFE = 1.0 kip (T) FBC = 2.24 kip (C) FFB = 0 532

*6–40. Determine the force in members CF, FG, and GC of the truss and state if the members are in tension or compression.

10 kN

5 kN H

308

B 2m

C 2m

D 2m

SOLUTION Support Reaction: a + ΣMA = 0; Ey(8) - 10(6) - 5 a

2 b = 0 Ey = 8.943 kN cos 30°

Method of Sections: a + ΣME = 0; - FCF sin 30°(4) + 10(2) = 0

FCF = 10 kN (C)

Ans.

b + ΣMC = 0; FFG sin 30°(4) + 10(2) - 8.943(4) = 0 Ans.

FFG = 7.887 kN (C) Joint G: + ΣFx = 0; S

FGH cos 30° - 7.887 cos 30° = 0

FGH = 7.887 kN

+ c ΣFy = 0;

2(7.887 sin 30°) - FGC = 0

FGC = 7.89 kN (T)

Ans.

Ans: FCF = 10 kN (C) FFG = 7.89 kN (C) FGC = 7.89 kN (T) 533

6–41. Determine the force members BC, FC, and FE and state if the members are in tension or compression.

6 kN B 6 kN

SOLUTION

Entire Truss: a + ©MA = 0;

- 6 (3) - 6(6) + Dy (9) = 0

Dy = 6 kN 3m

Section: a + ©MC = 0;

Ans.

- 6 (3) + 6 (6) - FBC sin 45° (3) = 0 Ans.

FBC = 8.49 kN (C) + ©F = 0; : x

-FFE cos 45° (3) + 6 (3) = 0 FFE = 8.49 kN (T)

a + ©MF = 0;

8.49 cos 45° - 8.49 cos 45° + FFC = 0 Ans.

FFC = 0

Ans: FFE = 8.49 kN (T) FBC = 8.49 kN (C) FFC = 0 534

6–42. Determine the force in members BC, HC, and HG and state if the members are in tension or compression.

12 kN 6 kN

9 kN

4 kN

6 kN

J 2m

E B 1.5 m

SOLUTION

C 1.5 m

D 1.5 m

1.5 m

Support Reactions: Referring to the FBD of the entire truss shown in Fig. a, NA can be determined directly by writing the moment equation of equilibrium about point E. a+ΣME = 0;

9(1.5) + 12(3) + 6(4.5) + 4(6) - NA(6) = 0

NA = 16.75 kN

Method of Sections: Referring to the FBD of the left portion of the truss sectioned through a–a shown in Fig. b, FHG, FHC and FBC can be determined directly by writing the moment equations of equilibrium about points C, A, and H, respectively. a+ΣMC = 0; FHG a

2 213

b (3) + 6(1.5) + 4(3) - 16.75(3) = 0

FHG = 4.875213 kN (C) = 17.6 kN (C)

a+ΣMA = 0; FHC a

2 213

b (3) - 6(1.5) = 0

Ans.

FHC = 1.5213 kN (C) = 5.41 kN (C)

a+ΣMH = 0; FBC(1) + 4(1.5) - 16.75(1.5) = 0 Ans.

FBC = 19.125 kN (T) = 19.1 kN (T)

Ans: FHG = 17.6 kN (C) FHC = 5.41 kN (C) FBC = 19.1 kN (T) 535

6–43. Determine the force in members CD, CJ, GJ, and CG, and state if the members are in tension or compression.

12 kN 6 kN

9 kN

4 kN

6 kN

J 2m

E B 1.5 m

SOLUTION

C 1.5 m

D 1.5 m

1.5 m

Support Reactions: Referring to the FBD of the entire truss shown in Fig. a, Ey can be determined directly by writing the moment equation of equilibrium about point A. a+ΣMA = 0; Ey(6) - 6(6) - 9(4.5) - 12(3) - 6(1.5) = 0 + ΣFx = 0; Ex = 0 S

Ey = 20.25 kN

Method of Sections: Referring to the FBD of the right portion of the truss sectioned through a–a shown in Fig. b, FGJ , FCJ and FCD can be determined directly by writing moment equations of equilibrium about point C, E and J, respectively. a+ΣMC = 0; 20.25(3) - 6(3) - 9(1.5) - FGJa

2 213

b(3) = 0

FGJ = 4.875 213 kN (C) = 17.6 kN (C)

a+ΣME = 0; 9(1.5) - FCJ a

213

b(3) = 0

Ans.

FCJ = 2.25213 kN (C) = 8.11 kN (C)

a+ΣMJ = 0; 20.25(1.5) - 6(1.5) - FCD(1) = 0 Ans.

FCD = 21.375 kN (T) = 21.4 kN (T)

Method of Joints: Using the result of FGJ to consider the equilibrium of joint G, Fig. c, + ΣFx = 0; FHG a 3 b - (4.875213)a 3 b = 0 FHG = 4.875213 kN (C) S 213 213

2 + c ΣFy = 0; 2 a4.875213ba b - 12 - FCG = 0 FCG = 7.50 kN (T) 213 Ans.

Ans: FGJ = 17.6 kN (C) FCJ = 8.11 kN (C) FCD = 21.4 kN (T) FCG = 7.50 kN (T) 536

*6–44. Determine the force in members DE, JI, and DO of the K truss and state if the members are in tension or compression. Hint: Use sections aa and bb.

15 ft 15 ft

A J

1500 lb 20 ft

20 ft

1800 lb 20 ft 20 ft

20 ft

SOLUTION a+ΣMI = 0;

- FDE(30) + 1400(40) = 0

FDE = 1867 lb = 1.87 kip (C)

Ans.

+ ΣFx = 0; S

Ans.

FJI = 1867 lb = 1.87 kip (T)

a+ΣMJ = 0; - 1867(30) + 1400(60) -

4 F (30) = 0 5 DO Ans.

FDO = 1167 lb = 1.17 kip (C)

Ans: FDO = 1.17 kip (C) FDE = 1.87 kip (C) FJI = 1.87 kip (T) 537

6–45. Determine the force in members GD and KJ of the K truss and state if the members are in tension or compression.

15 ft 15 ft

A J

1500 lb 20 ft

20 ft

1800 lb 20 ft 20 ft

20 ft

SOLUTION a+ΣMK = 0;

- 1800(20) + 1400(80) - FCD(30) = 0

FCD = 2533 lb = 2.53 kip (C)

Ans.

+ ΣFx = 0; S

Ans.

FKJ = 2533 lb = 2.53 kip (T)

Ans: FCD = 2.53 kip (C) FKJ = 2.53 kip (T)

538

6–46. The truss supports the vertical load of 600 N. Determine the force in members BC, BG, and HG as the dimension L varies. Plot the results of F (ordinate with tension as positive) versus L (abscissa) for 0 … L … 3 m.

B L

C L

SOLUTION + c ΣFy = 0;

FBG = sin u =

600 N

- 600 - FBG sin u = 0 600 sin u

3 2L2 + 9

FBG = - 200 2L2 + 9 a+ ΣMG = 0; a+ ΣMB = 0;

Ans.

- FBC (3) - 600 (L) = 0 FBC = -200L

Ans.

FHG (3) - 600(2L) = 0

Ans. FHG = 400L

Ans: FBG = -200 2L2 + 9 FBC = -200L FHG = 400L 539

6–47. Determine the force in members GF, CF, and CD of the roof truss and indicate if the members are in tension or compression.

1.5 kN C 1.70 m

2 kN 0.8 m

SOLUTION a + ©MA = 0;

Ey 142 - 210.82 - 1.512.502 = 0

1.5 m

Ey = 1.3375 kN

Method of Sections: a + ©MC = 0;

1.3375122 - FGF11.52 = 0 FGF = 1.78 kN 1T2

a + ©MF = 0;

Ans.

3 1.3375112 - FCD a b 112 = 0 5 FCD = 2.23 kN 1C2

a + ©ME = 0;

FCF

1.5 3.25

1 = 0

Ans. Ans.

FCF = 0

Ans: FGF = 1.78 kN (T) FCD = 2.23 kN (C) FCF = 0

540

*6–48. Determine the force in members JK, CJ, and CD of the truss and state if the members are in tension or compression.

K 3m

I H

A 2m

4 kN

SOLUTION

5 kN

D 2m

8 kN

6 kN

Method of Joints: Applying the equations of equilibrium to the free - body diagram of the truss, Fig. a, + ©F = 0; : x

Ax = 0

a + ©MG = 0

6(2) + 8(4) + 5(8) + 4(10) - Ay(12) = 0 Ay = 10.33 kN

Method of Sections: Using the left portion of the free - body diagram, Fig. a. a + ©MC = 0;

FJK(3) + 4(2) - 10.33(4) = 0 Ans.

FJK = 11.111 kN = 11.1 kN (C) a + ©MJ = 0;

FCD(3) + 5(2) + 4(4) - 10.33(6) = 0 Ans.

FCD = 12 kN (T) + c ©Fy = 0;

10.33 - 4 - 5 - FCJ sin 56.31° = 0 Ans.

FCJ = 1.602 kN = 1.60 kN (C)

Ans: FJK = 11.1 kN (C) FCD = 12 kN (T) FCJ = 1.60 kN (C) 541

6–49. Determine the force in members HI, FI, and EF of the truss and state if the members are in tension or compression.

K 3m

I H

A 2m

4 kN

SOLUTION

5 kN

D 2m

8 kN

6 kN

Support Reactions: Applying the moment equation of equilibrium about point A to the free - body diagram of the truss, Fig. a, a + ©MA = 0;

NG(2) - 4(2) - 5(4) - 8(8) - 6(10) = 0 NG = 12.67 kN

Method of Sections: Using the right portion of the free - body diagram, Fig. b. a + ©MI = 0;

12.67(4) - 6(2) - FEF(3) = 0 Ans.

FEF = 12.89 kN = 12.9 kN (T) a + ©MG = 0;

- FFI sin 56.31°(2) + 6(2) = 0 Ans.

FFI = 7.211 kN = 7.21 kN (T) a + ©MF = 0;

3 12.67(2) - FHI a b (2) = 0 5 Ans.

FHI = 21.11 kN = 21.1 kN (C)

Ans: FEF = 12.9 kN (T) FFI = 7.21 kN (T) FHI = 21.1 kN (C) 542

6–50. Determine the force in each member of the three-member three-member space truss truss that andsupports state if the theloading members are lb in and tension of 1000 state or if compression. the members are in tension or compression.

1000 lb B

5 ft

10 ft 10 ft

SOLUTION Joint D:

5 10 10 i + j + kb FAD = FAD a 15 15 15 FCD = FCD a FBD = FBD a

10 ft

5 ft 5 ft

2 ft y

5 10 2 i j + kb 11.358 11.358 11.358

10 5 10 i + j + kb 15 15 15

P = - 1000 k 10 2 10 b + FCD a b + FBD a b = 0 15 11.358 15

©Fx = 0;

FAD a -

©Fy = 0;

FAD a

5 5 5 b + FCD a b + FBD a b = 0 15 11.358 15

©Fz = 0;

FAD a

10 10 10 b + FCD a b + FBD a b - 1000 = 0 15 11.358 15

Solving, FAD = 300 lb C

Ans.

FBD = 450 lb C

Ans.

FCD = 568 lb C

Ans.

Ans: FAD = 300 lb (C) FBD = 450 lb (C) FCD = 568 lb (C) 543

6–51. Determine the force in each member of the space truss and state if the members are in tension or compression.The truss is supported by short links at A, B, and C.

z D

{500j} lb

8 ft

SOLUTION ©Mx¿ = 0;

-Az (6) - 500(8) = 0

Az = -666.67 lb

©My¿ = 0;

-( -666.67)(5) -Bz (5) = 0

©Mz¿ = 0;

Ay (5) + 500(2) = 0

©Fx¿ = 0;

Bx = 0

©Fy¿ = 0;

-200 + Cy + 500 = 0

©Fz¿ = 0;

- 666.67 + 666.67 + Cz = 0

2 ft

Bz = 666.67 lb

4 ft x

Ay = -200 lb

2 ft

Cy = - 300 lb Cz = 0

Joint B: ©Fz = 0; ©Fx = 0; ©Fy = 0;

3 277 2 277

FBD = 731.2 lb = 731 lb (C)

Ans.

(731.2) - FBC = 0

FBC = 250 lb (T)

Ans.

(731.2) - FBA = 0

FBA = 166.67 lb = 167 lb (T)

Ans.

666.67 -

277

FBD = 0

Joint A: ©Fz = 0;

8 289

FAD - 666.67 = 0 Ans.

FAD = 786.2 lb = 786 lb (T) ©Fx = 0;

5 261

FAC -

3 289

(786.2) = 0 Ans.

FAC = 390.5 lb = 391 lb (C) ©Fy = 0;

4 289

(786.2) + 166.7 - 200 -

6 261

(390.5) = 0

(Check)

Joint C: ©Fz = 0; ©Fy = 0; ©Fx = 0;

8 272 6

261 5 261

FCD = 0

Ans.

FCD = 0

(390.5) - 300 = 0

(Check)

(390.5) - 250 = 0

(Check)

Ans: FBD = 731 lb (C) FBC = 250 lb (T) FBA = 167 lb (T) FAD = 786 lb (T) FAC = 391 lb (C) FCD = 0 0 (Check) 0 (Check) 544

3 ft

*6–52. The truss supports the vertical load of 600 N. Determine the force in members BC, BG, and HG as the dimension L varies. Plot the results of F (ordinate with tension as positive) versus L (abscissa) for 0 … L … 3m.

SOLUTION Ans.

FFA = 0

ΣFx = 0;

FFD cos 60° - FFE cos 60° = 0 FFD = FFE = F

ΣFz = 0;

2(F sin 60°) - 3 = 0 F = 1.7320 kN (T) Ans.

FFD = FFE = F = 1.73 kN (T) Joint E: 0.8660 23.25

FEA - 3 - 1.7320 sin 60° = 0

FEA = 9.3675 kN = 9.37 kN (C) ΣFy = 0;

1.5 23.25

(9.3675) - FEB = 0

FEB = 7.7942 kN = 7.79 kN (T) ΣFx = 0;

FED + 1.7320 cos 60° -

0.5 23.25

(9.3675) = 0

FED = 1.7320 kN = 1.73 kN (T) Joint D: ΣFz = 0;

0.8660 23.25

FDA - 3 - 1.7320 sin 60° = 0

FDA = 9.3675 kN = 9.37 kN (C) ΣFy = 0;

1.5 23.25

(9.3675) - FDC = 0

FDC = 7.7942 kN = 7.79 kN (T) ΣFx = 0;

D L

600 N

Joint F:

ΣFz = 0;

ΣFy = 0;

0.5 23.25

(9.3675) - 1.7320 - 1.7320 cos 60° = 0

545

(Check !)

*6–52. Continued

Joint A: ΣFx = 0;

FAC cos 60° +

0.5 23.25

(9.3675) -

FAC = FAB = F ΣFz = 0;

2 (F sin 60°) - 2c

0.866 23.25

0.5 23.25

(9.3675) - FAB cos 60° = 0

(9.3675) d = 0 F = 5.1961 kN (T)

FAC = FAB = 5.20 kN (T) ΣFy = 0; Joint B:

Ay - 2 c

1.5 23.25

Ans.

(9.3675) d = 0 Ay = 15.59 kN

ΣFx = 0;

5.1961 cos 60° - FBC = 0 FBC = 2.60 kN (C)

ΣFy = 0;

7.7942 - By = 0 By = 7.794 kN

ΣFz = 0;

Bz - 5.1961 sin 60° = 0 Bz = 4.50 kN

Ans.

Ans: FFA = 0 F = 1.73 kN (T) FEA = 9.37 kN (C) FEB = 7.79 kN (T) FED = 1.73 kN (T) FDA = 9.37 kN (C) FDC = 7.79 kN (T) FAC = 5.20 kN (T) FBC = 2.60 kN (C) 546

6–53. The space truss supports a force F = 5 -500i + 600j + 400k6 lb. Determine the force in each member and state if the members are in tension or compression.

8 ft 6 ft D 6 ft

SOLUTION

Method of Joints: In this case, there is no need to compute the support reactions. We will begin by analyzing the equilibrium of joint C, and then that of joints A and D. x

Joint C: From the free-body diagram, Fig. a, 3 ©Fx = 0; FCA a b - 500 = 0 5

©Fz = 0;

6 ft

Ans.

FCA = 833.33 lb = 833 lb(T)

©Fy = 0;

6 ft

3 3 FCB a b - FCD a b + 600 = 0 5 5

(1)

4 4 4 400 - 833.33 a b - FCD a b - FCB a b = 0 5 5 5

(2)

Solving Eqs. (1) and (2) yields

FCB = - 666.67 lb = 667 lb(C)

Ans.

FCD = 333.33 lb = 333 lb(T)

Ans.

Joint A: From the free-body diagram, Fig. b, ©Fx = 0; FAD cos 45° - FAB cos 45° = 0 FAD = FAB = F 3 ©Fy = 0; F sin 45° + F sin 45° - 833.33 a b = 0 5 F = 353.55 lb

Ans.

Thus, FAD = FAB = 353.55 lb = 354 lb(C) ©Fz = 0;

4 833.33 a b - Az = 0 5

Az = 666.67 lb

Joint D: From the free-body diagram, Fig. c, ©Fy = 0;

3 FDB + 333.33 a b - 353.55 cos 45° = 0 5

Ans.

FDB = 50 lb(T)

©Fx = 0;

Dx - 353.55 sin 45° = 0 Dx = 250 lb

©Fz = 0;

4 333.33 a b - Dz = 0 5

Dz = 266.67 lb

Note:The equilibrium analysis of joint B can be used to determine the components of support reaction of the ball and socket support at B. 547

Ans: FCA = 833 lb (T) FCB = 667 lb (C) FCD = 333 lb (T) FAD = FAB = 354 lb (C) FDB = 50 lb (T)

6–54. The space truss supports a force F = 5600i + 450j 750k6 lb. Determine the force in each member and state if the members are in tension or compression.

8 ft 6 ft D 6 ft

SOLUTION

Method of Joints: In this case, there is no need to compute the support reactions. We will begin by analyzing the equilibrium of joint C, and then that of joints A and D. x

Joint C: From the free-body diagram, Fig. a,

6 ft 6 ft

3 ©Fx = 0; 600 + FCA a b = 0 5 Ans.

FCA = - 1000 lb = 1000 lb(C) ©Fy = 0;

3 3 FCB a b - FCD a b + 450 = 0 5 5

(1)

©Fz = 0;

4 4 4 -FCB a b - FCD a b -(- 1000) a b - 750 = 0 5 5 5

(2)

Solving Eqs. (1) and (2) yields FCD = 406.25 lb = 406 lb(T)

Ans.

FCB = - 343.75 lb = 344 lb(C)

Ans.

Joint A: From the free-body diagram, Fig. b, ©Fy = 0;

FAB cos 45° - FAD cos 45° = 0 FAB = FAD = F

©Fx = 0;

3 1000 a b - F sin 45° - F sin 45° = 0 5 F = 424.26 lb Ans.

Thus, FAB = FAD = 424.26 lb = 424 lb (T) 4 ©Fz = 0; Az - 1000a b = 0 5 Az = 800 lb Joint D: From the free-body diagram, Fig. c, ©Fy = 0;

3 406.25 a b + 406.25 cos 45° - FDB = 0 5 Ans.

FDB = 543.75 lb = 544 lb(C) ©Fx = 0; 424.26 sin 45° - Dx = 0 Dx = 300 lb 4 ©Fz = 0; 406.25 a b - Dz = 0 5 Dz = 325 lb

Note: The equilibrium analysis of joint B can be used to determine the components of support reaction of the ball and socket support at B. 548

Ans: FCA = 1000 lb (C) FCD = 406 lb (T) FCB = 344 lb (C) FAB = FAD = 424 lb (T) FDB = 544 lb (C)

6–55. Determine the force in each member of the space truss and state if the members are in tension or compression. The truss is supported by a ball-and-socket joint at A and short links at B and C.

z C B

D 8 ft A x

SOLUTION Σ(MAB)z = 0;

Cy = 0

ΣMx = 0;

By(8) - 600(8) = 0

By = 600 lb

ΣFy = 0;

Ay = 600 lb

ΣFz = 0;

Az = 600 lb

ΣMy = 0;

600(4) - 600 (8) + Bx (8) = 0

Bx = 300 lb

ΣFx = 0;

Ax = 300 lb

ΣFy = 0;

8 F - 600 = 0 12 BE

FBE = 900 lb (T)

Joint B:

ΣFx = 0; ΣFz = 0;

300 - FBC -

Ans.

4 (900) = 0 12 Ans.

FBC = 0 FAB -

8 (900) = 0 12 Ans.

FAB = 600 lb (C)

Ans.

Joints C and D: FDE = FCD = FAD = FCE = FAC = FBC = 0 Joint E: ΣFx = 0; ΣFz = 0; ΣFy = 0;

4 4 (900) FAE = 0 12 280

Ans.

FAE = 670.8 = 671 lb (C) 8 (900) - 600 = 0 12 8 280

(670.8) -

Check!

8 (900) = 0 12

Check!

549

F = {2600k} lb E

8 ft

4 ft

*6–56. Determine the force in each member of the space truss and state if the members are in tension or compression.

E D 2m

Determine Determinethe theforce forceininmembers membersBE, BE,BC, BC,BF, BF,and andCE CEofofthe the space space truss, truss,and and state state ifif the the members members are are inin tension tension oror compression. compression.

Method MethodofofJoints: Joints:InInthis thiscase, case,there thereisisno noneed needtotocompute computethe thesupport supportreactions.We reactions.Wewill will begin beginby byanalyzing analyzingthe theequilibrium equilibriumofofjoint jointC,C,and andthen thenthat thatofofjoints jointsEEand andB.B. xx

Joint JointC:C:From Fromthe thefree-body free-bodydiagram, diagram,Fig. Fig.a,a,we wecan canwrite write 1.5 1.5 FF bb --600 600==00 CE CEaa 13.25 13.25 FF 721.11NN==721 721NN(T) (T) CE CE==721.11 11 bb --FF 721.11 721.11aa BC BC==00 13.25 13.25 FF 400NN(C) (C) BC BC==400

Ans. Ans.

Joint ,F , and JointE:E:From Fromthe thefree-body free-bodydiagram, diagram,Fig, Fig,b,b,notice noticethat thatFF ,F , andFF lieininthe the EF ED EC EF ED EClie same sameplane plane(shown (shownshaded), shaded),and andFF theonly onlyforce forcethat thatacts actsoutside outsideofofthis thisplane. plane. BE BEisisthe IfIfthe thexx axis axisisisperpendicular perpendiculartotothis thisplane planeand andthe theforce forceequation equationofofequilibrium equilibriumisis written writtenalong alongthis thisaxis, axis,we wehave have ©F ©F x¿x¿==0;0;

FF cosuu==00 BE BE==cos Ans. Ans.

FF BE BE==00 Joint JointB:B:From Fromthe thefree-body free-bodydiagram, diagram,Fig. Fig.c,c, ©F ©F z z==0;0;

{22k} kN

DD {22k} kN

SOLUTION SOLUTION

©F ©F x x==0;0;

A 3m

©F ©F z z==0;0;

1.5 1.5 FF bb--900 900==00 BF BFaa 3.5 3.5 FF 2100NN==2.10 2.10kN kN(T) (T) BF BF==2100

550

Ans. Ans.

3 3mm

1.5 1.5mm

CC y y 1 1mm BB 1 1mm 600 600NN 900 900NN

first. From the free-body diagram of the truss, Fig. a, and writing the equations of y x equilibrium, weHibbeler. have Published by Pearson Education, Inc., Hoboken, NJ. All rights reserved. C laws © 2022 by R. C. This material is protected under all copyright 3m m publisher. as they currentlythe exist. No portion of this AF, material reproduced, Determine force in members AB, may AD,be ED, FD, and in any form or by any means, without permission inzwriting from1the B 600 N 1m BD theF space and state if the members are in tension ©M - 900(3) - 600(3) = 0 Fy = 3000 N x =of0; y(1.5)truss, or compression. A z = 900 N ©M 900(2) - A z(2) = 0 y = 0; Continued *6–56. F 900 N ©Mz = 0;

A y(2) - 3000(1) = 0

©Fx = 0;

A = 0

A y = 1500 N

x SOLUTION

Dy = 1500 N ©Fy = 0; Dy + 1500 - 3000 = 0 Support Reactions: In this case, it will be easier to compute the support reactions Dz writing ©Ffirst. Dzthe + 900 - 900 diagram - 600 = of 0 the truss, Fig. a, and = 600 Nthe equations of From free-body z = 0; x equilibrium, we have Method of Joints: Using the above results, we will begin by analyzing the equilibrium ©Mx A, = and 0; then Fy(1.5) 600(3) Fy = 3000 N of joint that-of900(3) joints C and D. = 0 ©M = 0; the 900(2) - A z(2) = 0 Fig. b, we can write Joint A:y From free-body diagram,

A z = 900 N

= 0; ©F©M y = z 0;

A y = 1500 N

©Fx = 0;

A y(2)1500 - 3000(1) - FAB == 00 Ax = 0

3m B

1.5 m

C 1m 600 N

900 N

Ans.

FAB = 1500 N = 1.50 kN (C)

Dy + 1500 - 3000 =1.50 Dy = 1500 N b = 0 900 - FAF a 13.25 Dz = 600 N Dz + 900 - 900 - 600 = 0 Ans. FAF = 1081.67 N = 1.08 kN (C) Method of Joints: Using the above results, we will begin by analyzing the equilibrium 1 ©Fof = 0; A, and then 1081.67a b and - FAD that of joints C D. = 0 x joint 13.25 Joint A: From the free-body diagram, Fig.Nb,(T) we can write Ans. F = 600 ©F = 0; ©Fz =y 0; ©Fz = 0;

©FC: 0; the free-body 1500 -diagram FAB = 0of the joint in Fig. c, notice that FCE, FCB, and y =From Joint the 600-N force lie in the x–z plane shaded). FAB (shown = 1500 N = 1.50 Thus, kN (C)if we write the force Ans. equation of equilibrium along the y axis, we have 1.5 900 z = 0; ©F©F =0 FAF a 13.25 b = 0 FDC y = 0; Ans. FAF = 1081.67 N = 1.08 kN (C) Joint D: From the free-body diagram, Fig. d, 1 1 ©F = 0; b - F1 = 0 1081.67 2a (1) FBC - a ©Fx =x 0; b + FFD a ADb + FFD a b + 600 = 0 13.25 3.5 113 13.25 FAD = 600 Ans. 3 3 N (T) (2) FBD a ©Fy = 0; b + FED a b + 1500 = 0 3.5 113 Joint C: From the free-body diagram of the joint in Fig. c, notice that FCE, FCB, and 1.5 x–z plane 1.5 600-N force F lie in the shaded). force ©Fthe b + FED a (shown b + 600 = 0 Thus, if we write the(3) z = 0; FD a 3.5we have equation of equilibrium113 along the y axis, Solving Eqs. (1) through (3) yields ©Fy = 0; FDC = 0 FFD = 0 FED = -1400N = 1.40 kN (C) Ans. Joint D: From the free-body diagram, Fig. d, FBD = -360.56 N = 361 N (C) Ans. 2 1 1 (1) FBC - a b + FFD a ©Fx = 0; b + FFD a b + 600 = 0 3.5 113 13.25 3 3 (2) FBD a ©Fy = 0; b + 1500 = 0 b + FED a 3.5 113 1.5 1.5 FFD a ©Fz = 0; b + 600 = 0 b + FED a (3) 3.5 113 Solving Eqs. (1) through (3) yields FFD = 0

Ans.

FED = - 1400N = 1.40 kN (C)

Ans.

FBD = -360.56 N = 361 N (C)

551

Ans: FCE = 721 N (T) FBC = 400 N (C) FBE = 0 FBF = 2.10 kN (T) FAB = 150 kN (C) FAF = 1.08 kN (C) FAD = 600 N (T) FFD = 0 FED = 1.40 kN (C) FBD = 361 N (C)

Determine the force in members AFthe of the Determine the force in members BD, BD, AD, AD, and and AF of 6–57. if members the members aretension in tension spacespace trusstruss and and statestate if the are in or or Determine the inis BD, AF ofat compression. The truss is supported byand short links at DetermineThe theforce force inmembers members BD, AD, and AF ofthe the compression. truss supported byAD, short links space truss A, D, andand F. state space truss and stateif ifthe themembers membersare areinintension tensionoror A, B, D,B,and F. compression. compression.The Thetruss trussisissupported supportedbybyshort shortlinks linksatat A,A,B,B,D,D,and andF.F.

z D

{ 250k} lb { 250k} lb

D zz

{ 250k} { 250k} Elblb E

SOLUTION SOLUTION Entire Entire truss:truss: SOLUTION SOLUTION

©M =truss: 0; 250 (3) 250 -(3)Bz-(6) Bz =(6)0 = 0 ©M Entire = 0; y truss: yEntire = lb 125 lb B =B 125 z (3) ©M ©M 250 (3)- -BB y y= =0;0; z 250 z (6) z (6)= =0 0 + F 250 = 0 ©Fz ©F = 0; + F 250 z = 0; 125B125 z z 125 lblb = 0 B z z= =125 F125 = lb 125 lb Fz 125 =125 z + ©F ©F +FF 250= =0 0 z z= =0;0; z z - -250 ©M = 0; 125 (6) 250(6 sin 60°) - 250(6) - Dy(660°) sin 60°) ©Mx = 0; 125F(6) 250(6 = 0= 0 x lblb sin 60°) - 250(6) - Dy(6 sin F = =-125 125

x xx

lb {250j}{250j} lb EE 6 ft 6 ft ft 6 ft 6{250j} lblb {250j} C

6 ft 6 ft

6 ft 6 ft F F 6 ftF F6 ft

BB 6 ft 6 ft 6 ft 6 ft

yy 6 ft 6 ft

z z

- 394.3 lb D =D -=(6) 394.3 lb sin y(6) ©M - -250(6 sin ©M 125 250(6 sin60°) 60°)- -250(6) 250(6)- -DD sin60°) 60°)= =0 0 x x= =0;0; y 125 y(6 y(6 0 ©Fx©F = 0; =A 0 x = 0; Ax D x== D 394.3lblb y y = -394.3 ©M = 0; B (6) 394.3 + (3) 250 =(3)0 = 0 ©M = 0; B (6) 394.3 (3) +(3)250 z y z©F ©F x x= =0;0; y AA x x= =0 0 By = 72.17 lb lb394.3 ©M (6) ©M = =0;0;By B=B72.17 (6)- 394.3(3) (3)+ +250 250(3) (3)= =0 0 z z

y y

- 394.372.17 + 72.17 + Fy + 250 ©Fy ©F = 0; = 0= 0 y = 0; - 394.3 BB = =+72.17 72.17lblb+ Fy + 250 y y

lb F =F72.17 y = 72.17 ©F ©F 394.3+lb+72.17 72.17+ +FF 250= =0 0 y y= =0;0; y - -394.3 y y+ +250

JointJoint B: B:

FF 72.17lblb y y= =72.17

©F - F sin = 60°0 = 0 ©FJoint = 0; - FBD sin zB:= BD60° B: 0; 125 125 z Joint Ans.Ans.

F=BD144.34 = 144.34 = lb 144 lb (C) FBD =sin 144 ©F 125 - -FF sin 60° = =(C) 00 ©F 125 60° z z= =0;0; BD BD 72.17©Fy ©F = 0; FBA F144.34 =BA0 ==0=144 y = 0; 72.17FF 144lblb(C) (C) BD BD= =144.34 F = 72.17 = 72.2 lb (C) =BA72.17 ©F 72.17FF ==72.2 ©F = =0;0; FBA 72.17=0 0 lb (C) y y

JointJoint E: E:

Ans. Ans.

BA BA

FF 72.17= =72.2 72.2lblb(C) (C) BA BA= =72.17

©F = 0; FAE F=AEFEF = FEF ©FJoint = 0; xE:E: xJoint ©F 2(F sin 60°) - 250 ©F©F = 0; sin - 250 = 0= 0 z = 0; 2(FF AE60°) z ©F AE F x x= =0;0; AE EF AE= =FF EF F=AEFEF = F=EF144.3 = 144.3 lb (C) lb250 (C) ©F 2(F sin = =0 0 ©F = =0;0; FAE 2(F sin60°) 60°)- -250 z z

JointJoint A: A:

AE AE

FF 144.3lblb(C) (C) AE EF AE= =FF EF= =144.3

sin 60° 3 3 6 6 6 sin660° Joint JointA:A: = F FAD F=ADFAD a AD - a - i - i- j + j + kb kb 272272 272272 272272 33 66 6 6sin sin60° 60° FAD -i j +j + k bk b FAD= =FF a- -660° sin i60° AD AD 6asin ©F = 0; F 144.3 sin = 60°0 = 0 272 272 272 272 272 ©Fz = 0; FAD 272 -AD144.3 sin 60° z 272272 6 6sin sin60° 60° ©F -=-144.3 sin 60° ©F F 144.3 sin 60°= =0 0 F 204.1 =F 204.1 204 lb (T) z z= =0;0; F AD =AD 204 lb (T) Ans.Ans. AD =AD 272 272 6 6 204.1 1 (T)1 F72.17 == 204 Ans. 204 Ans. - (204.1) (204.1) + FAC F=AC0 = 0 AD AD ©Fy ©F = 0; - = =204.1 +lblb(T) y = 0; 72.17F 272272 22 22 66 11 ©F -=-102.1 (204.1) + +(C) FF ©F 72.17 F =(204.1) 102 lb y y= =0;0; F 72.17 AC AC= =0 0 102.1 = 102 lb (C) AC =AC 272 22 272 22 3 3 1 1 F102.1 = =102 F =144.3 102.1 102lb lb(C) (C) F+=AF + 144.3 cos 60° - (204.1) (204.1) (102.1) AC AC ©Fx©F = 0; cos 60° + + (102.1) = 0= 0 x = 0; - FAF 272272 22 22 33 11 ©F 144.3 (204.1) (102.1) ©F 144.3 cos60° 60°- (204.1)+ + (102.1)= =0 0 Ans.Ans. FFF =+ + 72.2 lbcos (T) x x= =0;0; F - AF AF 72.2 lb (T) AF =AF 272 22 272 22 Ans. Ans.

FF 72.2lblb(T) (T) AF AF= =72.2 552

Ans: FBD = 144 lb (C) FAD = 204 lb (T) FAF = 72.2 lb (T)

6–58. Determine the force in members CF, EF, and DF of the space truss and state if the members are in tension or compression. The truss is supported by short links at A, B, D, and F.

z { 250k} lb D E {250j} lb 6 ft C

SOLUTION

Entire truss: ©My = 0;

6 ft

250 (3) - Bz (6) = 0

6 ft 6 ft

Bz = 125 lb ©Fz = 0;

125 + Fz - 250 = 0 Fz = 125 lb

©Mx = 0;

125 (6) - 250 (6 sin 60°) - 250 (6) - Dy (6 sin 60°) = 0 Dy = - 394.3 lb

©Fx = 0;

Ax = 0

©Mz = 0;

By (6) - 394.3 (3) + 250 (3) = 0 By = 72.17 lb

©Fy = 0;

- 394.3 + 72.17 + Fy + 250 = 0 Fy = 72.17 lb

Joint E: ©Fx = 0;

- FAE sin 30° + FEF sin 30° = 0 FAE = FEF

©Fz = 0;

2(FEF cos 30°) - 250 = 0 Ans.

FEF = 144.3 = 144 lb (C) Joint F: FDF = FDF ¢

3 272

i -

6 272

j +

©Fy = 0;

- FCF - FDF ¢

©Fz = 0;

125 +

6 sin 60° 272

6 sin 60° 272 6 272

k≤

≤ + 72.17 = 0

FDF - 144.34 sin 60° = 0

FDF = 0

Ans.

FCF = 72.2 lb (T)

Ans. Ans: FEF = 144 lb (C) FDF = 0 FCF = 72.2 lb (T) 553

6–59. Determinethe theforce forceinineach eachmember memberof ofthe thespace spacetruss truss Determine andstate stateififthe themembers membersare areinintension tensionor orcompression. compression.The The and The truss truss isis supported supported by by ball-and-socket ball-and-socketjoints jointsatatA, A,B, B, The and E. E.Set SetFF == 5800j6 5800j6NN. .Hint: Hint:The Thesupport supportreaction reactionatatEE and actsalong alongmember memberEC. EC.Why? Why? acts

DD FF

11mm

22mm CC

55mm

22mm

BB 1.5mm 1.5

SOLUTION Joint D: ΣFx = 0; ΣFy = 0;

1 5 1 - FAD + FBD + FCD = 0 3 231.25 27.25

2 1.5 1.5 - FAD + FBD + FCD + 800 = 0 3 231.25 27.25

2 2 2 - FAD + FBD + FCD = 0 3 231.25 27.25

FAD = 686 N (T)

Ans.

FBD = 0

Ans.

FCD = 615.4 = 615 N (C)

ΣFz = 0;

Ans.

Joint C: ΣFx = 0; ΣFy = 0; ΣFz = 0;

FBC -

1 27.25

(615.4) = 0 Ans.

FBC = 229 N (T) 1.5 27.25

(615.4) - FAC = 0 Ans.

FAC = 343 N (T) FBC -

2 27.25

(615.4) = 0 Ans.

FBC = 457 N (C)

Ans: FAD = 686 N (T) FBD = 0 FCD = 615 N (C) FBC = 229 N (T) FAC = 343 N (T) FEC = 457 N (C) 554

mine the force in each member of the space truss and f the members *6–60. are in tension or compression. The s supported by ball-and-socket joints at A, B, and E. Determine the force in each member = 5 -200i + 400j6 N. Hint: The support reaction at E of the space truss and state if the members are in tension or compression. The ong member EC. Why? truss is supported by ball-and-socket joints at A, B, and E. Set F = 5 -200i + 400j6 N. Hint: The support reaction at E acts along member EC. Why?

D F

SOLUTION

C 5m

UTION

1 5 1 - FD: + FBD + FCD - 200 = 0 Joint 3 AD 231.25 27.25

1 1.5 5 1 2 1.5 FBD =+ 0 FCD - 200 = 0 ©F - x F=AD0; + FBD- 3-FAD + FCD + 400 27.25 3 231.25 27.25 231.25

2 1.5 1.5 2 2 2 FCD + 400 = 0 ©F - y F=AD0; FBD- 3+FAD + FCD = 0FBD 27.25 3 231.25 27.25 231.25 2 2 2 Ans. - FAD FBD + FCD = 0 3 231.25 27.25 FBD = 186 N (T) Ans. FAD = 343 N (T) FCD = 397.5 = 397 N (C) Ans. FBD = 186 N (T)

A E

2m C

5m E

1.5 m x

B 1.5 m

FAD==0;343 N (T) ©F z

C: 0;

FBC C: Joint

1 27.25

1.5 27.25

(397.5) = 0

FEC -

FBC -

1 27.25

(397.5) = 0

Ans.

(397.5) - FACF= 0= 148 N (T)

Ans.

2 27.25

1.5

(397.5) - FAC = 0 27.25 (397.5) F= 0 = 221 N (T)

Ans.

FAC == 0;221 N (T) ©F y 0;

Ans. Ans.

FCD = 397.5 = 397 N (C)

FBC == 0;148 N (T) ©F x 0;

Ans.

FEC==0;295 N (C) ©F z

Ans.

FEC -

27.25

(397.5) = 0

Ans.

FEC = 295 N (C)

Ans: FAD = 343 N (T) FBD = 186 N (T) FCD = 397 N (C) FBC = 148 N (T) FAC = 221 N (T) FEC = 295 N (C) 555

6–61. Determine forceforce P required hold the 100-lb weightthe in Determine thethe P to required to hold equilibrium. 100-lb weight in equilibrium. D

SOLUTION Equations of Equilibrium: Applying the force equation of equilibrium along the y axis of pulley A on the free - body diagram, Fig. a, + c ©Fy = 0;

2TA - 100 = 0

B A

TA = 50 lb

Applying ©Fy = 0 to the free - body diagram of pulley B, Fig. b, + c ©Fy = 0;

2TB - 50 = 0

TB = 25 lb

From the free - body diagram of pulley C, Fig. c, + c ©Fy = 0;

2P - 25 = 0

P = 12.5lb

Ans.

Ans: P = 12.5 lb 556

6–62. Determine the force P needed to suspend the 100-lb weight. Each pulley has a weight of 10 lb. Also, what are the cord reactions at A and B?

2 in.

2 in. B

2 in.

SOLUTION

For the bottom pulley: + c ΣFy = 0; 2P + 2P + 10 - 110 = 0 P = 25 lb

Ans.

At A,

RA = P = 25 lb

Ans.

At B,

RB = 2P + 10 = 60 lb

Ans.

Ans: P = 25 lb RA = 25 lb RB = 60 lb 557

6–63. If each cord can support a maximum tension of 500 lb determine the largest weight that can be supported by the pulley system. Each pulley has a weight of 10 lb.

2 in.

2 in. B

2 in.

SOLUTION

2P + 10 = 500 P =

490 = 245 lb 2

+ c ΣFy = 0; 2P + 10 + 2P - W - 10 = 0 W = 4P = 4(245) = 980 lb Ans.

Maximum Weight = 980 lb

Ans: Wmax = 980 lb 558

*6–64. Determine the force P required to hold the 50-kg mass in equilibrium.

C B A

SOLUTION

Equations of Equilibrium: Applying the force equation of equilibrium along the y axis of each pulley. + c ©Fy = 0;

R - 3P = 0;

R = 3P

+ c ©Fy = 0;

T - 3R = 0;

T = 3R = 9P

+ c ©Fy = 0;

2P + 2R + 2T - 50(9.81) = 0

Ans.

Substituting Eqs.(1) and (2) into Eq.(3) and solving for P, 2P + 2(3P) + 2(9P) = 50(9.81) Ans.

P = 18.9 N

Ans: P = 18.9 N 559

6–65. Determine the horizontal and vertical components of force that pins A and B exert on the frame.

C 2 kN/m

SOLUTION Free-Body Diagram. The frame will be dismembered into members BC and AC. The solution will be very much simplified if one recognizes that member AC is a two force member. The FBDs of member BC and pin A are shown in Figs. a and b, respectively.

Equations of Equilibrium. Consider the equilibrium of member BC, Fig. a, 3 a+ΣMB = 0; 2(4)(2) - FAC a b(4) = 0 5 a+ΣMC = 0; Bx (4) - 2(4)(2) = 0

4 + c ΣFy = 0; 6.6667 a b - By = 0 5

FAC = 6.6667 kN

Bx = 4.00 kN

Ans.

By = 5.333 kN = 5.33 kN

Ans.

Ax = 4.00 kN

Ans.

Ay = 5.333 kN = 5.33 kN

Ans.

Then, the equilibrium of pin A gives

+ ΣFx = 0; Ax - 6.6667 a 3 b = 0 S 5

4 + c ΣFy = 0; Ay - 6.6667 a b = 0 5

Ans: Bx = 4.00 kN By = 5.33 kN Ax = 4.00 kN Ay = 5.33 kN 560

6–66. Determine the horizontal and vertical components of force at pins A and D.

0.3 m C A

1.5 m

1.5 m E

SOLUTION Free-Body Diagram. The assembly will be dismembered into member AC, BD and pulley E. The solution will be very much simplified if one recognizes that member BD is a two force member. The FBD of pulley E and member AC are shown in Fig. a and b respectively.

12 kN

Equations of Equilibrium. Consider the equilibrium of pulley E, Fig. a, + c ΣFy = 0; 2T - 12 = 0

T = 6.00 kN

Then, the equilibrium of member AC gives 4 a+ΣMA = 0; FBD a b(1.5) + 6(0.3) - 6(3) - 6(3.3) = 0 5 FBD = 30.0 kN + ΣFx = 0; Ax - 30.0 a 3 b - 6 = 0 S 5

4 + c ΣFy = 0; 30.0 a b - 6 - 6 - Ay = 0 5

Ax = 24.0 kN

Ans.

Ay = 12.0 kN

Ans.

Thus,

FA = 2A2x + A2y = 224.02 + 12.02 = 26.83 kN = 26.8 kN FB = FBD = 30.0 kN Dx =

3 (30) = 18.0 kN 5

Ans.

Dy =

4 (30) = 24.0 kN 5

Ans.

Ans: Ax = 24.0 kN Ay = 12.0 kN Dx = 18.0 kN Dy = 24.0 kN 561

6–67. Determine the horizontal and vertical components of force acting at the pins A, B, and C of the frame.

B 4 ft

4 ft

8 ft

4 ft 458 A

100 lb 100 lb

308

SOLUTION From FBD (a) a + ΣMA = 0; Bx(8 sin 45°) + By(8 cos 45°) - 100(4 cos 45°) = 0 [1] From FBD (b) a + ΣMC = 0; By(12 cos 30°) - Bx(12 sin 30°) + 100(8 cos 30°) = 0 [2] Solving Eqs. [1] and [2] yields:

From FBD (a)

By = - 23.96 lb = - 24.0 lb

Ans.

Bx = 73.96 lb = 74.0 lb

Ans.

+ ΣFx = 0; S

Ax - 73.96 = 0

+ c ΣFy = 0;

Ay - 100 + ( - 23.96) = 0

Ans.

Ax = 73.96 lb = 74.0 lb Ay = 123.96 lb = 124 lb

Ans.

From FBD (b) + ΣFx = 0; S

73.96 - Cx = 0

+ c ΣFy = 0;

Cy - 100 - ( - 23.96) = 0

Ans.

Cx = 73.96 lb = 74.0 lb Cy = 76.04 lb = 76.0 lb

Ans.

Ans: By = - 24.0 lb Bx = 74.0 lb Ax = 74.0 lb Ay = 124 lb Cx = 74.0 lb Cy = 76.0 lb 562

*6–68. The bridge frame consistsframe of three partsofwhich bewhich can be The bridge consists threecan parts considered pinned at A, pinned D, and at E, A, rocker supported at supported C considered D, and E, rocker at C and F, and roller at B. Determine the horizontalthe horizontal and supported F, and roller supported at B. Determine and vertical components reaction atofallreaction these supports and vertical of components at all these supports. due to the loading due toshown. the loading shown.

A 15 ft

15 ft30 ft

C 5 ft

a + ©MD = 0;a + ©M 2(30)(15) 2(30)(15) (30) y (30) = 0- By B y = =300 kip By = 30 kip D = 0; - B

Ans.

+ ©F = 0; : + ©F = 0; D = 0 : x x x

Ans.

a + ©MA = 0;a + ©M Ans. Cy (5) = 0 - 30(15) Cy = =135kip Cy (5) --30(15) 2(15)(7.5) 0 Cy = 135kip A = -0;2(15)(7.5)

Ans.

+ ©F = 0; : + ©F = 0; A = 0 : x x x

Ans.

+ c ©Fy = 0; + c ©F -A 2(15) 30-=2(15) 0 135 -Ay30= =750 kip Ay = 75 kipAns. y y= +0;135 --A y + -

Ans.

+ c ©Fy = 0; + c ©F Dyy += 30 0 - 2(30) Dy = =300 kip Dy = 30 kip 0; - 2(30) Dy +=30

E D

30 ft 15 ft

15 ft

20 ft

SOLUTION SOLUTION

Dx = 0

2 kip/ft

A B

20 ft

For segment BD: For segment BD:

2 kip/ft

F 5 ft

5 ft

For segment ABC: For segment ABC:

Ax = 0

For segment DEF: For segment DEF: a + ©Mg = 0;a + ©M Ans. -Fgy(5) = 0+ 30(15) Fy = =135 = 0;+ 2(15)(7.5) - Fy(5) + 30(15) 2(15)(7.5) 0 kip Fy = 135 kip

Ans.

+ ©F = 0; : + ©F = 0; E = 0 : x x x

Ans.

+ c ©Fy = 0; + c ©F -E 2(15) 30-=2(15) 0 135 -Ey30= =750 kip Ey = 75 kipAns. y y=+0;135 --E y + -

Ans.

Ex = 0

Ans: By = 30 kip Dx = 0 Dy = 30 kip Cy = 135 kip Ay = 75 kip Fy = 135 kip Ex = 0 Ey = 75 kip 563

6–69. Determine the reactions at supports A and B.

700 lb/ ft B D

500 lb/ ft

9 ft C

A 6 ft

SOLUTION

6 ft

8 ft

Member DB: a + ©MB = 0;

3 3.15 (6) - FCD a b (9) = 0 5 FCD = 3.50 kip

+ ©F = 0; : x

4 -Bx + 3.50a b = 0 5 Ans.

Bx = 2.80 kip + c ©Fy = 0;

3 By - 3.15 + 3.50 a b = 0 5 Ans.

By = 1.05 kip Member AC: + ©F = 0; : x

4 Ax - 3.50 a b = 0 5 Ans.

Ax = 2.80 kip + c ©Fy = 0;

3 Ay - 3 - 3.50 a b = 0 5 Ans.

Ay = 5.10 kip a + ©MA = 0;

3 MA - 3(6) - 3.50 a b (12) = 0 5 MA = 43.2 kip # ft

Ans.

Ans: Bx = 2.80 kip By = 1.05 kip Ax = 2.80 kip Ay = 5.10 kip MA = 43.2 kip # ft 564

6–70. Determine the horizontal and vertical components of force at pins B and C. The suspended cylinder has a mass of 75 kg.

0.3 m B

C 1.5 m

SOLUTION

Free-Body Diagram: The solution will be very much simplified if one realizes that member AB is a two force member. Also, the tension in the cable is equal to the weight of the cylinder and is constant throughout the cable. Equations of Equilibrium: Consider the equilibrium of member BC by referring to its FBD, Fig. a, 3 a+ΣMC = 0; FAB a b(2) + 75(9.81)(0.3) - 75(9.81)(2.8) = 0 5

0.5 m

FAB = 1532.81 N

a+ΣMB = 0; Cy (2) + 75(9.81)(0.3) - 75(9.81)(0.8) = 0 Ans.

Cy = 183.94 N = 184 N + ΣFx = 0; 1532.81a 4 b - 75(9.81) - Cx = 0 S 5 Cx = 490.5 N

Ans.

Thus, FB = FAB = 1532.81 N Bx =

4 (1532.81) = 1226.25 N = 1.23 kN 5

Ans.

By =

3 (1532.81) = 919.69 N = 920 kN 5

Ans.

Ans: Cy = 184 N Cx = 490.5 N Bx = 1.23 kN By = 920 kN 565

6–71. Determine the horizontal and vertical components of force that the pins C and D exert on member ECD of the frame.

4 ft

2 ft

D 2 ft

1 ft B 3 ft

SOLUTION

800 lb

a + ΣMD = 0; 800(6) - FBC sin 26.57°(2) - FBC cos 26.57°(2) = 0 FBC = 1788.85 lb Cx = 1788.85 cos 26.57° = 1600 lb

Ans.

Cy = 1788.85 sin 26.57° = 800 lb

Ans.

+ ΣFx = 0; S

Dx + 800 - 1788.85 cos 26.57° = 0

Dx = 800 lb

Ans.

+ c ΣFy = 0;

1788.85 sin 26.57° - Dy = 0

Dy = 800 lb

Ans.

Ans: Cx = 1600 lb Cy = 800 lb Dx = 800 lb Dy = 800 lb 566

*6–72. Determine the horizontal and vertical components of force that the pins at B and D exert on member ABD of the frame.

4 ft

2 ft

D 2 ft

1 ft B 3 ft

SOLUTION

800 lb

Equilibrium of the Entire Frame: a + ΣMD = 0; 800(6) - Ax (6) = 0

Ax = 800 lb

For member ABD: a + ΣMD = 0; FBC cos 26.57°(3) - 800(6) = 0

FBC = 1788.85 lb

Bx = 1788.85 cos 26.57° = 1600 lb

Ans.

By = 1788.85 sin 26.57° = 800 lb

Ans.

+ ΣFx = 0; d

Dx + 800 - 1788.85 cos 26.57° = 0

Dx = 800 lb

Ans.

+ c ΣFy = 0;

Dy - 1788.85 sin 26.57° = 0

Dy = 800 lb

Ans.

Ans: Bx = 1600 lb By = 800 lb Dx = 800 lb Dy = 800 lb

567

6–73. Determine the compressive force exerted on the stone by a vertical load of 50 N applied to the toggle press.

50 N

90 mm

400 mm

SOLUTION Equilibrium of Joint B: Since members AB and BC are two-force members, method of joints can be used to analyze the member forces. + ΣFx = 0; S

FAB cos 12.68° - FBC cos 12.68° = 0

+ c ΣFy = 0;

2(F sin 12.68°) - 50 = 0

FAB = FBC = F

F = 113.89 N

Force on Specimen: + ΣFx = 0; S

113.89 cos 12.68° - Fspec = 0

Fspec = 111 N

Ans.

Ans: Fspec = 111 N 568

6–74. The two ends of the spanner wrench fit loosely into the smooth slots of the bolt head. Determine the required force P on the handle in order to develop a torque of M = 50 N # m on the bolt. Also, what is the resultant force on the pin at B?

P A B

250 mm

M 70 mm

30 mm C

SOLUTION Bolt: ΣF = 0; FA = FB FB a

7 258

b(0.06) = 50

Ans.

FB = 906.64 = 907 N Wrench segment: c + ΣMA = 0;

P(0.320) - 906.64 a P = 156 N

7 258

b(0.03) - 906.64 a

3 258

b(0.07) = 0

Ans.

Ans: FB = 907 N P = 156 N 569

6–75. The two ends of the spanner wrench fit loosely into the smooth slots of the bolt head. Determine the torque M on the bolt and the resultant force on the pin at B when a force of P = 80 N is applied to the handle.

P A B

250 mm

M 70 mm

30 mm C

SOLUTION Wrench segment: a + ΣMA = 0;

- 80(0.320) + FB a

7 258

FB = 464.20 = 464 N

b(0.03) + FB a

3 258

b(0.07) = 0

Ans.

Bolt: a + ΣMA = 0;

462.20 a

7 258

b(0.06) - M = 0

M = 25.6 N # m

Ans.

Ans: FB = 464 N M = 25.6 N # m 570

*6–76. Determine the horizontal and vertical components of force which the pins at A and B exert on the frame.

400 N/m D

1.5 m E 3m

SOLUTION

Free-Body Diagram: The frame will be dismembered into members AD, EF, CD and BC. The solution will be very much simplified if one realizes that members CD and EF are two force member. Therefore, only the FBD of members AD and BC, Fig. a and b respectively, need to be drawn

1.5 m A

Equations of Equilibrium: Write the moment equations of equilibrium about point A for member AD, Fig. a, and point B for member BC, Fig. b. 4 a+ΣMA = 0; FEF a b(3) - FCD (4.5) - 400 (4.5)(2.25) = 0 5 4 a+ΣMB = 0; -FEF a b(1.5) + FCD (4.5) = 0 5

(1) (2)

Solving Eqs. (1) and (2)

FEF = 3375 N

FCD = 900 N

Write the force equation of equilibrium for member AD, Fig. a, + ΣFx = 0; Ax + 400(4.5) + 900 - 3375 a 4 b = 0 S 5

3 + c ΣFy = 0; 3375 a b - Ay = 0 5

Ax = 0

Ans.

Ay = 2025 N = 2.025 kN

Ans.

Also, for member BC, Fig. b

+ ΣFx = 0; 3375a 4 b - 900 - Bx = 0 S 5 + c ΣFy = 0;

3 By - 3375a b = 0 5

Bx = 1800 N = 1.80 kN

Ans.

By = 2025 N = 2.025 kN

Ans.

Ans: Ax = 0 Ay = 2.025 kN Bx = 1.80 kN By = 2.025 kN 571

6–77. Determine the reactions at the supports. The pin, attached to member BCD, passes through a smooth slot in member AB.

4 ft

200 lb/ft

2 ft B

1 ft C D

3 ft

SOLUTION

Member AB: a + ©MA = 0;

NB (5) - 1000 (2.5) = 0 NB = 500 lb

+ ©F = 0; : x

- Ax +

3 3 (1000) - (500) = 0 5 5 Ans.

Ax = 300 lb + c ©Fy = 0;

Ay -

4 4 (1000) + (500) = 0 5 5 Ans.

Ay = 400 lb Member BCD: a + ©MD = 0;

4 500 a b (3) - Cy (1) = 0 5 Ans.

Cy = 1200 lb + ©F = 0; : x

3 - Dx + 500 a b = 0 5 Ans.

Dx = 300 lb + c ©Fy = 0;

4 -500 a b + 1200 - Dy = 0 5 Ans.

Dy = 800 lb

Ans: Ax = 300 lb Ay = 400 lb Cy = 1200 lb Dx = 300 lb Dy = 800 lb 572

6–78. The compound beam is pin supported at B and supported by rockers at A and C. There is a hinge (pin) at D. Determine the reactions at the supports.

2 kN/m A

C D

B 6m

SOLUTION Free-Body Diagram: The compound beam will be dismembered into members ABD and CD of which their respective FBD are shown in Fig. a and b. Equations of Equilibrium: First, consider the equilibrium of member CD, Fig. b, NC = 3.00 kN

Ans.

a+ΣMC = 0; 2(3)(1.5) - Dy (3) = 0 Dy = 3.00 kN

Ans.

a+ΣMD = 0; NC (3) - 2(3)(1.5) = 0 + ΣFx = 0; Dx = 0 S

Next, the equilibrium of member ABD gives, a+ΣMB = 0; 2(9)(1.5) - 3.00(3) - NA(6) = 0

NA = 3.00 kN

Ans.

a+ΣMA = 0; By (6) - 2(9)(4.5) - 3.00(9) = 0

By = 18.0 kN

Ans.

+ ΣFx = 0; Bx = 0 S

Ans.

Ans: NC = 3.00 kN NA = 3.00 kN By = 18.0 kN Bx = 0 573

6–79. When force of of 22 lb lb is is applied applied to to the the handles handles of of the the brad brad When aa force squeezer, it pulls pulls in in the the smooth smooth rod rod AB. AB.Determine Determine the the force force squeezer, it P P exerted exerted on on each each of of the the smooth smooth brads brads at at C C and and D. D.

2 lb 0.25 in. P 1.5 in.

SOLUTION

+ ©ME = 0;

E C

1.5 in. B D P

Equations of Equilibrium: Applying the moment equation of equilibrium about point E to the free-body diagram of the lower handle in Fig. a, we have

2 in.

1 in. A 2 in.

2(2) - FAB(1) = 0

2 lb

FAB = 4 lb Using the result of FAB and considering the free-body diagram in Fig. b, + ©MB = 0; + ©F = 0; : x

NC(1.5) - ND(1.5) = 0 NC = ND

(1)

4 - NC - ND = 0

(2)

Solving Eqs. (1) and (2) yields Ans.

NC = ND = 2 lb

Ans: NC = ND = 2 lb 574

*6–80. A force of 4 lb is applied to the handle of the toggle clamp. Determine the normal force F that the bolt exerts on the clamp.

1.5 in. 4 in. A F

D 308

B C

6 in.

1.5 in.

SOLUTION From FBD (a) a+ΣMC = 0; FBE sin 30°(1.5) - 4(7.5) = 0

FBE = 40 lb

From FBD (b) a+ΣMD = 0; 40 sin 30°(1.5) - F(5.5) = 0

Ans.

F = 5.45 lb

Ans: F = 5.45 lb 575

6–81. The hoist supports the 125-kg engine.Determine the force this load creates along member DB and along member FB, which contains the hydraulic cylinder H.

1m G

2m E

SOLUTION

Free-Body Diagrams: The solution for this problem will be simplified if one realizes that members FB and DB are two-force members.

1226.25(3) - FFB ¢

1938.87 ¢

3 210

≤ (2) = 0

Ans.

FFB = 1938.87 N = 1.94 kN + c ©Fy = 0;

1m C

Equations of Equilibrium: For FBD(a), a + ©ME = 0;

≤ - 1226.25 - Ey = 0

Ey = 613.125N + ©F = 0; : x

Ex - 1938.87 ¢

1 210

≤ = 0

Ex = 613.125 N From FBD (b), a + ©MC = 0;

613.125(3) - FBD sin 45°(1) = 0 Ans.

FBD = 2601.27 N = 2.60 kN

Ans: FFB = 1.94 kN FBD = 2.60 kN 576

6–82. The compound beam is fixed at E and supported by rollers at A and B. Determine the reactions at these supports. There are pins at C and D.

1200 lb 800 lb

600 lb

5 ft

10 ft

C 5 ft

4 ft

D 6 ft

E 5 ft

SOLUTION For segment CD: a + ΣMC = 0; Dy(10) - 600(4) = 0

Dy = 240 lb

+ ΣFx = 0; S

Cx - Dx = 0 [1]

+ c ΣFy = 0;

Cy + 240 - 600 = 0

Cy = 360 lb

For segment ABC: + ΣFx = 0; S

Cx = 0

From Eq. [1]

Dx = 0

a + ΣMA = 0; By(10) + 800(5) - 1200(10) - 360(15) = 0 Ans.

By = 1340 lb = 1.34 kip + c ΣFy = 0;

Ay + 1340 - 800 - 1200 - 360 = 0 Ans.

Ay = 1020 lb = 1.02 kip For segment DE: + ΣFx = 0; S + c ΣFy = 0;

Ey - 240 = 0

a + ΣME = 0;

240(5) - ME = 0

Ex = 0

Ans.

Ey = 240 lb

Ans.

ME = 1200 lb.ft = 1.20 kip # ft

Ans.

Ans: By = 1.34 kip Ay = 1.02 kip Ex = 0 Ey = 240 lb ME = 1.20 kip # ft 577

6–83. Determine the horizontal and vertical components of force that the pins exert on member ABD.

50 lb

1 ft

D 5 ft

SOLUTION

4 ft

From FBD (a)

+ ΣFx = 0; S

Dx - 50 = 0

Dx = 50 lb

Ans.

+ c ΣFy = 0;

Dy - 50 = 0

Dy = 50 lb

Ans.

a + ΣMA = 0; 50(9) - Bx(4) = 0

Bx = 112.5 lb = 112 lb

Ans.

+ ΣFx = 0; 112.5 - 50 - A = 0 S x

Ax = 62.5 lb

Ans.

C 3 ft

From FBD (b)

+ c ΣFy = 0;

Ay - By - 50 = 0 [1]

From FBD (c) a + ΣMC = 0;

- By(3) + 112.5(4) - 50(2) = 0 Ans.

By = 116.7 lb = 117 lb From Eq. [1]

Ay - 116.7 - 50 = 0

Ans.

Ay = 167 lb

Ans: Dx = 50 lb Dy = 50 lb Bx = 112 lb Ax = 62.5 lb By = 117 lb Ay = 167 lb 578

*6–84. Determine the horizontal and vertical components of force acting on pins A and C.

150 lb D

1 ft

10 ft B

308

9 ft

SOLUTION From FBD (a): Ans.

a+ΣMB = 0; 150(19) - Cy(10) = 0 Cy = 285 lb From FBD (b): a+ΣMA = 0; Cx (10 tan 30°) + 285(10) - 150(20 tan 30°) - 150(20) = 0 Cx = 325.98 lb = 326 lb

Ans.

+ ΣFx = 0; Ax + 150 - 325.98 = 0 Ax = 176 lb S

Ans.

+ c ΣFy = 0; 285 - 150 - Ay = 0

Ans.

Ay = 135 lb

Ans: Cy = 285 lb Cx = 326 lb Ax = 176 lb Ay = 135 lb 579

6–85. The picture frame is glued together at its corners and held in place by the 4-corner clamp. If the tension in the adjusting screw is 14 N, determine the horizontal and vertical components of the clamping force that member AD exerts on on the thesmooth smoothjoints jointsat at B and B and C. C. All connections are pins.

120 mm

B 160 mm 40 mm

SOLUTION + c ©Fy = 0;

40 mm

4 4 T a b - T¿ a b = 0 5 5 T = T¿

+ ©F = 0; : x

3 2 T a b - 14 = 0 5 T = 11.67 N

+ ©F = 0; : x

3 NB - 11.67 a b = 0 5 Ans.

NB = 7 N + c ©Fy = 0;

4 NC - 11.67 a b = 0 5 Ans.

NC = 9.33 N

Ans: NB = 7 N Nc = 9.33 N 580

6–86. 2m

The double tree AB is used to support the loadings applied to each of the single trees. Determine the total load that must be supported by the chain EG and its placement d for AB to remain horizontal.

E d

Equations of Equilibrium: From FBD (a), + c ©Fy = 0;

SOLUTION 0.4 m

FAC - 400 - 400 = 0

FAC = 800 N

FBD - 300 - 300 = 0

FBD = 600 N

400 N

0.4 m 400 N

B D

0.6 m 300 N

From FBD (b), + c ©Fy = 0; From FBD (c), + c ©Fy = 0;

FEG - 800 - 600 = 0 Ans.

FEG = 1400 N = 1.40 kN a + ©MG = 0;

800(d) - 600(2 - d) = 0 Ans.

d = 0.857 m

Ans: FEG = 1.40 kN d = 0.857 m 581

6–87. The pillar crane is subjected to the crate having a mass of 500 kg. Determine the force in the tie rod AB and the horizontal and vertical reactions at the pin support C when the boom is held in the position shown.

2.4 m 108 208

1.8 m

A C

SOLUTION Forces of pin on pulley at B are shown on pulley free-body diagram. Pin B: + ΣFx = 0; S + c ΣFy = 0;

4 FCB a b - FAB cos 20° - 250(9.81) cos 10° = 0 5

3 FCB a b - FAB sin 20° - 500(9.81) - 250(9.81) sin 10° = 0 5 FCB = 14 415 N = 14.4 kN

Ans.

FAB = 9702 N = 9.70 kN

Ans.

Thus 4 Cx = 14 415 a b = 11.5 kN 5

Ans.

3 Cy = 14 415 a b = 8.65 kN 5

Ans.

Ans: FCB = 14.4 kN FAB = 9.70 kN Cx = 11.5 kN Cy = 8.65 kN 582

*6–88. The machine shown is used for forming metal plates. It consists of two toggles ABC and DEF, which are operated by the hydraulic cylinder H.The toggles push the movable bar G forward, pressing the plate p into the cavity. If the force which the plate exerts on the head is P = 12 kN, determine the force F in the hydraulic cylinder when u = 30°.

200 mm

F P 12 kN

H G

SOLUTION

200 mm

Member EF:

a + ©ME = 0;

u 30

200 mm

u 30

- Fy (0.2 cos 30°) + 6 (0.2 sin 30°) = 0 Fy = 3.464 kN

Joint E: + ©F = 0; : x

FDE cos 30° - 6 = 0 FDE = 6.928 kN

+ c ©Fy = 0;

F - 3.464 - 6.928 sin 30° = 0 Ans.

F = 6.93 kN

Ans: F = 6.93 kN 583

6–89. Determine the horizontal and vertical components of force which pin C exerts on member ABC. The 600-N force is applied to the pin.

3m A

B 1.5 m

SOLUTION

600 N

Free-Body Diagram: The solution will be very much simplified if one determines the support reactions first and then considers the equilibrium of two of its three members after they are dismembered. The FBDs of the entire assembly, member DBF and member ABC are shown in Figs. a, b and c, respectively.

300 N

Equations of Equilibrium: Consider the equilibrium of the entire assembly, Fig. a, a+ΣME = 0; NA(3) - 300(4.5) - 600(4) = 0

NA = 1250 N

Next, write the moment equation of equilibrium about point D for member DBF, Fig. b. a+ΣMD = 0; Bx(1.5) - 300(3) = 0

Bx = 600 N

Finally, consider the equilibrium of member ABC, Fig. c + ΣFx = 0; 1250 - 600 - Cx = 0 S

Ans.

Cx = 650 N

Ans.

a+ΣMB = 0; Cy = 0

Ans: Cx = 650 N Cy = 0 584

6–90. The pipe cutter is clamped around around the the pipe pipe P. If the wheel at A exerts a normal force of F FAA = 80 N on the pipe, determine the forces of wheels B and B C on theCpipe. thenormal normal forces of wheels and on Also findThe thethree pin reaction on the at C. the pipe. wheels each havewheel a radius of The 7 mmthree and wheels a radius of 7ofmm the pipeeach has have an outer radius 10 and mm.the pipe has an outer radius of 10 mm.

C 10 mm B

10 mm

SOLUTION u = sin -1 a

10 b = 36.03° 17

Equations of Equilibrium: + c ©Fy = 0;

NB sin 36.03° - NC sin 36.03° = 0 N B = NC

+ ©F = 0; : x

80 - NC cos 36.03° - NC cos 36.03° = 0 NB = NC = 49.5 N

Ans.

Ans: NB = NC = 49.5 N 585

6–91. the hydraulic cylinderscylinders EF and AD Determine the force in created in the hydraulic EF in order to order hold the shovel the position shown. The and AD in to hold theinshovel in equilibrium. load has a mass 1.25 center of gravity at G. shovel load has a of mass of Mg 1.25and Mg aand a center of gravity at AllAll joints areare pinpin connected. G. joints connected.

0.25 m

1.5 m

E C

30 H

D 10 60

0.5 m G

SOLUTION Assembly FHG: a + ©MH = 0;

- 1250(9.81) (0.5) + FEF (1.5 sin 30°) = 0 Ans.

FEF = 8175 N = 8.18 kN (T) Assembly CEFHG: a + ©MC = 0;

FAD cos 40° (0.25) - 1250(9.81) (2 cos 10° + 0.5) = 0 Ans.

FAD = 158 130 N = 158 kN (C)

Ans: FEF = 8.18 kN (T) FAD = 158 kN (C) 586

*6–92. The clamp is used to hold the smooth strut S in place. If the tensile force in the bolt GH is 300 N, determine the force exerted at the smooth surface at A and B.

C S

458

20 mm

250 mm

H 20 mm 608

SOLUTION Clamp: a+ΣMC = 0;

- 300(0.250) + FDE cos 60°(0.25) + FDE sin 60°(0.05) = 0

250 mm 50 mm

FDE = 445.63 N Strut: + ΣFx = 0; S

FB - 445.63 cos 60° = 0 Ans.

FB = 223 N + c ΣFy = 0;

- FA + 445.63 sin 60° = 0 Ans.

FA = 386 N Note that ED is a two-force member and AEB is a three-force member.

Ans: FB = 223 N FA = 386 N 587

6–93. The gin-pole derrick is used to lift the 300-kg stone with constant velocity. If the derrick and the block and tackle are in the position shown, determine the horizontal and vertical components of force at the pin support A and the orientation u and tension in the guy cable BC.

B 13 12

3m 408 A

C 308

SOLUTION Pulley C: + ΣFx = 0; S

- 300(9.81) cos 30° + T sin u = 0

+ c ΣFy = 0;

- 300(9.81) - 300(9.81) sin 30°+ T cos u = 0 T = 5097.4 N = 5.10 kN

Ans.

u = 30°

Ans.

AB is a two-force member. a+ΣMA = 0;

12 5 b (3 cos 40°) - R a b (3 sin 40°) 13 13

+ 5097.4 sin 30° (3 cos 40°) - 5097.4 cos 30° (3 sin 40°) = 0

R = 1924.7 N = 1.92 kN + ΣFx = 0; S

Ax - 1924.7 a

12 b - 5097.4 sin 30° = 0 13

Ay - 1924.7 a

5 b - 5097.4 cos 30° = 0 13

Ans.

Ax = 4.33 kN + c ΣFy = 0;

Ans.

Ay = 5.15 kN

Ans: T = 5.10 kN u = 30° Ax = 4.33 kN Ay = 5.15 kN 588

6–94. Five coins are stacked in the smooth plastic container shown. If each coin weighs 0.0235 lb, determine the normal reactions of the bottom coin on the container at points A and B.

5 3 5 4 3

5 3 5

SOLUTION

All coins: + c ©Fy = 0;

NB - 5 (0.0235) = 0 Ans.

NB = 0.1175 lb

Bottom coin: + c ©Fy = 0;

4 0.1175 - 0.0235 - N a b = 0 5 N = 0.1175 lb

+ ©F = 0; : x

3 NA - 0.1175 a b = 0 5 Ans.

NA = 0.0705 lb

Ans: NB = 0.1175 lb NA = 0.0705 lb 589

6–95. The nail cutter consists of the handle and the two cutting blades. Assuming the blades are pin connected at B and the surface at D is smooth, determine the normal force on the fingernail when a force of 1 lb is applied to the handles as shown. The pin AC slides through a smooth hole at A and is attached to the bottom member at C.

0.25 in.

1.5 in.

SOLUTION

Handle: a + ©MD = 0;

1 lb

0.25 in.

C 1 lb

FA (0.25) - 1(1.5) = 0 FA = 6 lb

+ c ©Fy = 0;

ND - 1 - 6 = 0 ND = 7 lb

Top blade: a + ©MB = 0;

7(1.5) - FN (2) = 0 Ans.

FN = 5.25 lb Or bottom blade: a + ©MB = 0;

FN (2) - 6 (1.75) = 0 Ans.

FN = 5.25 lb

Ans: FN = 5.25 lb 590

*6–96. A man having a weight of 175 lb attempts to hold himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at at C. Neglect the weight of the platform. Do the results seem realistic?

B A

SOLUTION

(a)

(b)

Bar: + c ©Fy = 0;

2(F>2) - 2(87.5) = 0 Ans.

F = 175 lb Man: + c ©Fy = 0;

NC - 175 - 2(87.5) = 0 Ans.

NC = 350 lb (b) Bar: + c ©Fy = 0;

2(43.75) - 2(F>2) = 0 Ans.

F = 87.5 lb Man: + c ©Fy = 0;

NC - 175 + 2(43.75) = 0 Ans.

NC = 87.5 lb

Ans: F = 175 lb Nc = 350 lb F = 87.5 lb Nc = 87.5 lb 591

6–97. A woman having a weight 175lblbattempts attemptstotohold hold himself herself man having a weight ofof175 using one of the two methods shown. Determine the total force she mustexert exerton on bar bar AB AB in in each each case case and the normal he must reaction she exertson onthe the platform platform at at C. C.The The platform platform has a he exerts weight of 30 lb. Do the results seem realistic?

B A

SOLUTION

(a)

(b)

Bar: + c ©Fy = 0;

2(F>2) - 102.5 - 102.5 = 0 Ans.

F = 205 lb Woman: + c ©Fy = 0;

NC - 175 - 102.5 - 102.5 = 0 Ans.

NC = 380 lb (b) Bar: + c ©Fy = 0;

2(F>2) - 51.25 - 51.25 = 0 Ans.

F = 102 lb Woman: + c ©Fy = 0;

NC - 175 + 51.25 + 51.25 = 0 Ans.

NC = 72.5 lb

Ans: F = 205 lb NC = 380 lb F = 102 lb NC = 72.5 lb 592

6–98. Determine the couple moment M needed to create a force of F = 200 N on the slider block at C.

B 80 mm

F 5 200 N

M C

300 mm

200 mm

SOLUTION Geometry: Referring to Fig. a u = tan - 1 a

0.08 b = 21.80° 0.2

f = tan - 1 a

a = u + f = 21.80° + 14.93° = 36.73°

0.08 b = 14.93° 0.3

LAB = 20.22 + 0.082 = 0.2154 m Equations of Equilibrium: First consider the equilibrium of the piston, Fig. b. Here, F, N, and FBC are concurrent. Thus, moment equation of equilibrium is automatically satisfied. Write the force equation of equilibrium along x axis, + ΣFx = 0; S

200 - FBC cos 14.93° = 0 FBC = 206.99 N

Then using the result of FBC to write the moment equation of equilibrium about point A by referring to the FBD of the crank, Fig. c, a+ΣMA = 0; M - (206.99 sin 36.73°)(0.2154) = 0

M = 26.67 N # m = 26.7 N # m

Ans.

Ans: M = 26.7 N # m 593

6–99. If the 300-kg drum has a center of mass at point G, determine the horizontal and vertical components of force acting at pin A and the reactions on the smooth pads C and D. The grip at B on member DAB resists both horizontal and vertical components of force at the rim of the drum.

600 mm E

SOLUTION

60 mm 60 mm

Equations of Equilibrium: From the free - body diagram of segment CAE in Fig. a, a + ©MA = 0;

100 mm

Ans.

Ax - 12 743.56 = 0 Ans.

Ax = 12 743.56 N = 12.7 kN + c ©Fy = 0;

390 mm

300(9.81)(600 cos 30°) - NC(120) = 0 NC = 12 743.56 N = 12.7 kN

+ ©F = 0; : x

300(9.81) - Ay = 0 Ans.

Ay = 2943 N = 2.94 kN

Using the results for Ax and Ay obtained above and applying the moment equation of equilibrium about point B on the free - body diagram of segment BAD, Fig. b, a + ©MB = 0;

12 743.56(60) - 2943(100) - ND(450) = 0 Ans.

ND = 1045.14 N = 1.05 kN

Ans: NC = 12.7 kN Ax = 12.7 kN Ay = 2.94 kN ND = 1.05 kN 594

*6–100. Determine the horizontal and vertical components of force that the pins exert on member ABCD. The pin at C is attached to member ABCD and passes through the smooth slot in member ECF.

D 1 ft

1 ft

3 ft C 3 ft B

80 lb

E 3 ft

SOLUTION

4 ft

From FBD (a)

4 ft

a+ΣME = 0; 80(9) - 80(7) - FC (5) = 0 FC = 32 lb Cx =

3 (32) = 19.2 lb 5

Cy =

4 (32) = 25.6 lb 5

Ans.

From FBD (b) a+ΣMA = 0; 80(10) +

3 (32)(6) - FB (3) = 0 5

Bx = 305.1 lb = 305 lb

FB = 305.1 lb Ans.

By = 0

+ ΣFx = 0; 305.1 - 3 (32) - 80 - Ax = 0 Ax = 206 lb S 5 4 + c ΣFy = 0; Ay - (32) = 0 Ay = 25.6 lb 5

Ans. Ans.

Ans: Cx = 19.2 lb Cy = 25.6 lb Bx = 305 lb By = 0 Ax = 206 lb Ay = 25.6 lb 595

6–101. Determine the force that must be developed in the hydraulic cylinder AB in order to develop a normal force of 4 MN in the grip. Also, determine the magnitude of the force developed in the pin at C.

0.1 m B

0.2 m

1.2 m

A 0.6 m 1.5 m

SOLUTION Equilibrium: a+ΣMC = 0;

0.1 21.45

FAB (0.6) +

21.45

1.2

0.1 21.45

21.45

FAB (0.3) - 4(0.75) = 0

(8.6011) = 0

0.75 m

4 MN

FAB = 8.6011 MN = 8.60 MN

+rΣFy′ = 0; - Cy′ + 4 + p+ΣFx′ = 0; Cx′ -

1.2

4 MN

Ans.

Cy′ = 4.7143 MN Cx′ = 8.5714 MN

FC = 2C2x′ + C 2y′ = 28.57142 + 4.71432 = 9.78 MN

Ans.

Ans: FAB = 8.60 MN FC = 9.78 MN 596

6–102. If a force of F = 350 N is applied to the handle of the toggle clamp, determine the resulting clamping force at A.

235 mm

70 mm

30 mm C 30 mm A

SOLUTION

30 B

275 mm

D 30

Equations of Equilibrium: First, we will consider the free-body diagram of the handle in Fig. a. a + ©MC = 0;

FBE cos 30°(70) - FBE sin 30°(30) - 350 cos 30°(275 cos 30° + 70) -350 sin 30°(275 sin 30°) = 0 FBE = 2574.81 N

+ ©F = 0; : x

Cx - 2574.81 sin 30° + 350 sin 30° = 0 Cx = 1112.41 N

Subsequently, the free-body diagram of the clamp in Fig. b will be considered. Using the result of Cx and writing the moment equation of equilibrium about point D, a + ©MD = 0;

1112.41(60)-NA (235) = 0 Ans.

NA = 284.01 N = 284 N

Ans: NA = 284 N 597

6–103.

The double link grip is used to lift the beam. If the beam weighs 4 kN, determine the horizontal and vertical components of force acting on the pin at A and the horizontal and vertical components of force that the flange of the beam exerts on the grip at B. Assume B and F are pins.

4 kN 160 mm

160 mm

D E

SOLUTION

45 C

Free-Body Diagram: The solution for this problem will be simplified if one realizes that members ED and CD are two force members.

+ c ©Fy = 0;

4 - 2F sin 45° = 0

280 mm F

Equations of Equilibrium: Using method of joint, [FBD (a)],

120 mm

280 mm 280 mm

F = 2.828 kN

From FBD (b), + c ©Fy = 0;

2By - 4 = 0

Ans.

By = 2.00 kN

From FBD (c), a + ©MA = 0;

Bx 12802 - 2.0012802 - 2.828 cos 45°11202 - 2.828 sin 45°11602 = 0 Bx = 4.00 kN

+ c ©Fy = 0;

Ans.

Ay + 2.828 sin 45° - 2.00 = 0 Ans.

Ay = 0 + ©F = 0; : x

4.00 + 2.828 cos 45° - Ax = 0 Ans.

A x = 6.00 kN

Ans: By = 2.00 kN Bx = 4.00 kN Ay = 0 Ax = 6.00 kN 598

*6–104. The foot pump is used to fill the tire. If the force required to develop the necessary pressure in the cylinder is 70 lb, determine the vertical force P that must be applied to the pedal. The assembly is pin connected at A, B, and C.

308

18 in.

308 9 in.

9 in.

SOLUTION

308 70 lb

Geometry:

Equilibrium:

u = tan - 1 a

9 cos 30° b = 21.8° 24 - 9 sin 30°

C 24 in.

a+ΣMB = 0; 70 cos 8.213°(9) - P(18 sin 30° + 18 cos 30°) = 0 Ans.

P = 25.4 lb

Ans: P = 25.4 lb 599

6–105. Determine the reactions at pins A and B of the frame needed to support the 200-lb load. The pulley at C has a radius of 0.5 ft. The pulley at D has a radius of 0.25 ft.

C 4 ft 608

308 A

3 ft B

308

SOLUTION From FBD (a) + c ΣFy = 0;

2T - 200 = 0

T = 100 lb

From FBD (b): Notice that members AC and BC are two-force members. + ΣFx = 0; S

FAC cos 30° - FBC cos 30° - 100 = 0 [1]

+ c ΣFy = 0;

FAC sin 30° + FBC sin 30° - 100 - 100 = 0 [2]

Solving Eqs. [1] and [2] yields: FAC = 257.74 lb = 258 lb

FBC = 142.26 lb = 142 lb

From FBD (c) FA = 258 lb

Ans.

FB = 142 lb

Ans: FA = 258 lb FB = 142 lb 600

6–106. If d = 0.75 ft and the spring has an unstretched length of 1 ft, determine the force F required for equilibrium.

B 1 ft

1 ft d

SOLUTION

Spring Force Formula: The elongation of the spring is x = 2(0.75) - 1 = 0.5 ft. Thus, the force in the spring is given by

1 ft

150 lb/ft 1 ft

Fsp = kx = 150(0.5) = 75 lb Equations of Equilibrium: First, we will analyze the equilibrium of joint B. From the free-body diagram in Fig. a, + ©F = 0; : x

FAB cos 48.59° - FBC cos 48.59° = 0 FAB = FBC = F¿

+ c ©Fy = 0;

2F¿ sin 48.59° - 75 = 0 F¿ = 50 lb

From the free-body diagram in Fig. b, using the result FBC = F¿ = 50 lb, and analyzing the equilibrium of joint C, we have + c ©Fy = 0;

+ ©F = 0; : x

FCD sin 48.59° - 50 sin 48.59° = 0

FCD = 50 lb

2(50 cos 48.59°) - F = 0 Ans.

F = 66.14 lb = 66.1 lb

Ans: F = 66.1 lb 601

6–107. If a force of F = 50 lb is applied to the pads at A and C, determine the smallest dimension d required for equilibrium if the spring has an unstretched length of 1 ft.

F A

SOLUTION

1 ft

150 lb/ft 1 ft

Geometry: From the geometry shown in Fig. a, we can write sin u = d

1 ft

cos u = 21 - d2

Spring Force Formula: The elongation of the spring is x = 2d - 1. Thus, the force in the spring is given by Fsp = kx = 150(2d - 1) Equations of Equilibrium: First, we will analyze the equilibrium of joint B. From the free-body diagram in Fig. b, + ©F = 0; : x

FAB cos u - FBC cos u = 0

FAB = FBC = F¿

+ c ©Fy = 0;

2F¿(d) - 150(2d - 1) = 0

F¿ =

150d - 75 d

From the free-body diagram in Fig. c, using the result FBC = F¿ =

150d - 75 , and d

analyzing the equilibrium of joint C, we have + c ©Fy = 0; + ©F = 0; : x

FCD sin u - a

2c a

150d - 75 b sin u = 0 d

FCD =

150d - 75 d

150d - 75 b a 11 - d2 b d - 50 = 0 d

Ans.

Solving the above equation using a graphing utility, we obtain d = 0.6381 ft = 0.638 ft or d = 0.9334 ft = 0.933 ft

Ans: d = 0.638 ft 602

*6–108. The clamp has a rated load capacity of 1500 lb. Determine the compressive force this creates along segment AB of the screw and the magnitude of force exerted at pin C. The screw is pin connected at its end B and passes through the pin-connected (swivel) block at A.

3 in.

3 in. B

3.5 in. 1500 lb

2.5 in.

4 in.

SOLUTION a+ΣMC = 0;

1500(4) - FAB (3.5) = 0 FAB = 1714.3 lb = 1.71 kip

+ n ΣFx = 0;

1714.3 - Cx = 0

Cx = 1714.3 lb

+ c ΣFy = 0;

1500 - Cy = 0

Cy = 1500 lb

FC = 2C 2y + C 2x = 21714.32 + 15002 = 2278 lb = 2.28 kip

Ans.

Ans: FAB = 1.71 kip FC = 2.28 kip 603

6–109. P

If the vertical force P is applied to the two-bar mechanism, determine the equilibrium force F on the block. Plot this magnitude as a function of u, where 0° … u … 90°.

A 3 4 l

SOLUTION Equations of Equilibrium: First, consider the equilibrium of joint A, Fig. a. + ΣFx = 0; S

FAB cos f - FAC cos u = 0

+ c ΣFy = 0;

FAB sin f + c a

FAB =

FAC = a

cos f b FAB cos u

cos f b FAB d sin u - P = 0 cos u P sin f + tan u cos f

Then consider the equilibrium of the block, Fig. b. + ΣFx = 0; S

F - a

F =

P b cos f = 0 sin f + tan u cos f

P sin f + tan u cos f cos f

P (1) tan u + tan f

604

6–109. Continued

From the geometry shown in Fig. c, 3 l sin u 4 3 sin u tan f = = 9 216 - 9 sin2 u l 1 sin2 u A 16

Substitute this result into Eq. (1),

F = tan u + F = a

3 sin u 216 - 9 sin2 u

216 - 9 sin2 u

tan u 216 - 9 sin2 u + 3 sin u

b P

Ans.

The values of u and the corresponding values of F are tabulated below. u°

∞

2.15 P

1.02 P

0.615 P

0.387 P

0.209 P

The plot of F vs u is shown in Fig. d

Ans: F = a 605

216 - 9 sin2 u

tan u 216 - 9 sin2 u + 3 sin u

6–110. The spring has an unstretched length of 0.3 m. Determine the angle u for equilibrium if the uniform bars each have a mass of 20 kg.

k 150 N/ m

SOLUTION

u u

Free-Body Diagram: The assembly is being dismembered into members AB and BC of which their respective FBD are shown in Fig. b and a. Here, the spring stretches x = 2(2 sin u) - 0.3 = 4 sin u - 0.3. Thus, FSP = kx = 150 (4 sin u - 0.3) = 600 sin u - 45. Equations of Equilibrium: Considered the equilibrium of member BC, Fig. a, a+ΣMC = 0; By (2 cos u) - Bx (2 sin u) - 20(9.81) cos u = 0

(1)

Also, member AB, Fig. b a+ΣMA = 0; - By (2 cos u) - Bx (2 sin u) - 20(9.81) cos u = 0

(2)

+ c ΣFy = 0; (600 sin u - 45) - 20(9.81) - By = 0

(3)

Solving Eq. (1) and (2) 9.81 cos u sin u Substitute the result of By = 0 into Eq. (3) By = 0

Bx = -

600 sin u - 45 - 20(9.81) = 0 sin u = 0.402 Ans.

u = 23.7°

Ans: u = 23.7° 606

6–111. The spring has an unstretched length of 0.3 m. Determine the mass m of each uniform bar if u = 30° for equilibrium.

k 150 N/ m

SOLUTION

u u

Free-Body Diagram: The assembly is being dismembered into members AB and BC of which their respective FBD are shown in Fig. b and a. Here, the spring stretches x = 2(2 sin 30°) - 0.3 = 1.7 m. Thus, FSP = kx = 150(1.7) = 255 N. Equations of Equilibrium: Consider the equilibrium of member BC, Fig. a, a+ΣMC = 0; Bx (2 sin 30°) + By (2 cos 30°) - m(9.81) cos 30° = 0

(1)

Also, member AB, Fig. b a+ΣMA = 0; Bx (2 sin 30°) - By (2 cos 30°) - m(9.81) cos 30° = 0

(2)

+ c ΣFy = 0; 255 - m(9.81) - By = 0

(3)

Solving Eqs. (1) and (2) Bx = 8.4957 m

By = 0

Substitute the result of By = 0 into Eq. (3) 255 - m(9.81) = 0 Ans.

m = 26.0 kg

Ans: m = 26.0 kg 607

*6–112. The piston C moves vertically between the two smooth walls. If the spring has a stiffness of k = 15 lb>in., and is unstretched when u = 0°, determine the couple M that must be applied to AB to hold the mechanism in equilibrium when u = 30°.

A 8 in.

M u

SOLUTION

12 in.

Geometry: sin c sin 30° = 8 12

c = 19.47°

k = 15 lb/in.

f = 180° - 30° - 19.47 = 130.53° l¿ AC 12 = sin 130.53° sin 30°

l¿ AC = 18.242 in.

Free-Body Diagram: The solution for this problem will be simplified if one realizes that member CB is a two force member. Since the spring spring force is stretches x = lAC - l¿ AC = 20 - 18.242 = 1.758 in. the Fsp = kx = 15 (1.758) = 26.37 lb. Equations of Equilibrium: Using the method of joints, [FBD (a)], + c ©Fy = 0;

FCB cos 19.47° - 26.37 = 0 FCB = 27.97 lb

From FBD (b), a+ ©MA = 0;

27.97 cos 40.53° (8) - M = 0 M = 170.08 lb # in = 14.2 lb # ft

Ans.

Ans: M = 14.2 lb # ft 608

6–113. The scale consists consists of of aa combination combinationof ofthird thirdand andfirst first The platform platform scale class so that that the the load load on on one one lever lever becomes becomes the theeffort effort class levers levers so that moves the the next next lever. lever.Through Through this thisarrangement, arrangement,aasmall small that moves weight a heavy object. Ifobject. x = 450 weight can canbalance balance a massive If mm, x = determine 450 mm, the required of themass counterweight S requiredStorequired balance determine themass required of the counterweight the load having a mass to balance a 90-kg load,ofL.90 kg.

100 mm

250 mm

150 mm H

E C

150 mm

350 mm B

x L

SOLUTION Equations of Equilibrium: Applying the moment equation of equilibrium about point A to the free - body diagram of member AB in Fig. a, a + ©MA = 0;

FBG (500) - 90(9.81)(150) = 0 FBG = 264.87 N

Using the result of FBG and writing the moment equation of equilibrium about point F on the free - body diagram of member EFG in Fig. b, a + ©MF = 0;

FED (250) - 264.87(150) = 0 FED = 158.922 N

Using the result of FED and writing the moment equation of equilibrium about point C on the free - body diagram of member CDI in Fig. c, a + ©MC = 0;

158.922(100) - mS(9.81)(950) = 0 Ans.

mS = 1.705 kg = 1.71 kg

Ans: mS = 1.71 kg 609

6–114. The platform scale consists of a combination of third and first class levers so that the load on one lever becomes the effort that moves the next lever. Through this arrangement, a small small weight weight can can balance balanceaamassive massiveobject. object.IfIfxx == 450 450mm, mm and and,the the mass mass of of the the counterweight counterweight SS is is 22 kg, kg, determine the mass of the load L required required to to maintain maintain the the balance. balance.

100 mm

250 mm

150 mm H

E C

150 mm

350 mm B

x L

SOLUTION Equations of Equilibrium: Applying the moment equation of equilibrium about point A to the free - body diagram of member AB in Fig. a, a + ©MA = 0;

FBG (500) - ML(9.81)(150) = 0 FBG = 2.943 lb

Using the result of FBG and writing the moment equation of equilibrium about point F on the free - body diagram of member EFG in Fig. b, a + ©MF = 0;

FED (250) - 2.943mL(150) = 0 FED = 1.7658mL

Using the result of FED and writing the moment equation of equilibrium about point C on the free - body diagram of member CDI in Fig. c, a + ©MC = 0;

1.7658mL(100) - 2(9.81)(950) = 0 Ans.

mL = 105.56 kg = 106 kg

Ans: mL = 106 kg 610

6–115. The four-member “A” frame is supported at A and E by smooth collars and at G by a pin. All the other joints are ball-and-sockets. If the pin at G will fail when the resultant force there is 800 N, determine the largest vertical force P that can be supported by the frame. Also, what are the x, y, z force components which member BD exerts on members EDC and ABC? The collars at A and E and the pin at G only exert force components on the frame.

300 mm 300 mm

E 600 mm

A B

600 mm

F C

SOLUTION

GF is a two - force member, so the 800 - N force acts along the axis of GF. Using FBD (a), ©Mx = 0;

- P(1.2) + 800 sin 45°(0.6) = 0 - Ay (0.3) + Ey (0.3) = 0

©Fy = 0;

- Ay - Ey + 800 sin 45° = 0

Ans.

P = 283 N ©Mz = 0;

Ay = Ey = 283 N ©Mx = 0;

Az (0.6) + Ez (0.6) -283((0.6) = 0

©My = 0;

Az(0.3)- Ez (0.3) = 0 Az = Ez = 118 N

Using FBD (b), ©Fy = 0;

- By - Dy + 800 sin 45° = 0

©Mz = 0;

Dy(0.3) - By (0.3) = 0 Ans.

By = Dy = 283 N ©Fz = 0;

-Bz - Dz + 800 cos 45° = 0

©My = 0;

- Dz (0.3) + Bz (0.3) = 0 Ans.

Bz = Dz = 283 N ©Fx = 0;

- Bx + Dx = 0

Using FBD (c), ©Mz = 0;

- By (0.6) + 283(0.15) - 283(0.3) = 0 Ans.

Bx = Dx = 42.5 N

Ans: P = 283 N Bx = Dx = 42.5 N By = Dy = 283 N Bz = Dz = 283 N 611

*6–116. Member AB isis supported bythe a ball-and-socket A and The structure subjected to loadings shown.at Member smooth collar at B.by Member CD is supported a pin at C. AB is supported a ball-and-socket at A by and smooth Determine theMember x, y, z components of reaction andat C. C. collar at B. CD is supported byataApin

z 250 N

Determine the x, y, z components of reaction at A and C.

D A

SOLUTION

4m 800 N m

From FBD (a) x

©My = 0;

MBy = 0

©Mx = 0;

- MBx + 800 = 0

©Mz = 0;

By (3) -Bx (2) = 0

©Fz = 0;

Az = 0

©Fx = 0;

- Ax + Bx = 0

(2)

©Fy = 0;

- Ay + By = 0

(3)

MBx = 800 N # m

2m 1.5 m

(1) Ans.

From FBD (b) ©Mg = 0;

By(1.5) + 800 - 250 cos 45°(5.5) = 0

From Eq.(1)

114.85(3) - Bx(2) = 0

From Eq.(2)

Ax = 172 N

Ans.

From Eq.(3)

Ay = 115 N

Ans.

©Fx = 0;

Cx + 250 cos 60° - 172.27 = 0

Cx = 47.3 N

Ans.

©Fy = 0;

250 cos 45° - 114.85 - Cy = 0

Cy = 61.9N

Ans.

©Fz = 0;

250 cos 60° - Cz = 0

Cz = 125 N

Ans.

©My = 0;

MCy - 172.27(1.5) + 250 cos 60°(5.5) = 0

©Mz = 0;

By = 114.85 N

Bx = 172.27 N

MCy = - 429 N # m

Ans.

MCz = 0

Ans.

Negative sign indicates that MCy acts in the opposite sense to that shown on FBD.

Ans: Az = 0 Ax = 172 N Ay = 115 N Cx = 47.3 N Cy = 61.9 N Cz = 125 N MCy = - 429 N # m MCz = 0 612

6–117. Member AD isissupported bythe cable AB and a roller at C,AD and The structure subjected to loading shown. Member fits through a smooth circular hole at D. Member ED is supported by a cable AB and roller at C and fits throughis supported by a roller a poleED thatis fits in a smooth a smooth circular holeat at D D.and Member supported by a snug hole at E.that Determine the x, y,snug z components of rollercircular at D and a pole fits in a smooth circular hole reaction at E andthe thex,tension in cable AB. at E. Determine y, z components of reaction at E and

z B E

the tension in cable AB.

0.8 m

SOLUTION ©My = 0;

4 - FAB (0.6) + 2.5(0.3) = 0 5

0.3 m

Ans. Ans.

FAB = 1.5625 = 1.56 kN ©Fz = 0;

0.5 m

0.4 m

4 (1.5625) - 2.5 + Dz = 0 5

A 0.3 m

{ 2.5 } kN

Dz = 1.25 kN ©Fy = 0;

Dy = 0

©Fx = 0;

Dx + Cx -

©Mx = 0;

MDx +

3 (1.5625) = 0 5

(1)

4 (1.5625)(0.4) - 2.5(0.4) = 0 5

MDx = 0.5 kN # m 3 (1.5625)(0.4) - Cx (0.4) = 0 5

(2)

©Mz = 0;

MDz +

©Fz = 0;

Dz¿ = 1.25 kN

©Mx = 0;

MEx = 0.5 kN # m

Ans.

©My = 0;

MEy = 0

Ans.

©Fy = 0;

Ey = 0

Ans.

©Mz = 0;

E x (0.5) - MDz = 0

(3)

Solving Eqs. (1), (2) and (3): Cx = 0.938 kN MDz = 0 Ans.

Ex = 0

Ans: FAB = 1.56 kN Ex = 0 Ey = 0 MEx = 0.5 kN # m MEy = 0 613

6–118. The Thethree threepin-connected pin-connectedmembers membersshown shownininthe thetop topview view support forceofof6060 at If G.only If only vertical supportaa downward downward force lb lb at G. vertical forces forces are supported at the connections B,C,E at theA, are supported at the connections B, C, E and padand supports pad supports A, D, determine at each pad. D, F, determine theF,reactions at the eachreactions pad.

D 6 ft

2 ft

SOLUTION

Equations of Equilibrium: From FBD (a), a + ©MD = 0; + c ©Fy = 0;

2 ft

6 ft

60182 + FC162 - FB1102 = 0

(1)

FB + FD - FC - 60 = 0

(2)

FE162 - FC1102 = 0

(3)

FC + FF - FE = 0

(4)

FE1102 - FB162 = 0

(5)

C G

4 ft E

4 ft

6 ft

From FBD (b), a + ©MF = 0; + c ©Fy = 0; From FBD (c), a + ©MA = 0; + c ©Fy = 0;

FA + FE - FB = 0

(6)

Solving Eqs. (1), (2), (3), (4), (5) and (6) yields, FE = 36.73 lb

FC = 22.04 lb

FD = 20.8 lb

FF = 14.7 lb

FB = 61.22 lb FA = 24.5 lb

Ans.

Ans: FD = 20.8 lb FF = 14.7 lb FA = 24.5 lb 614

7–1.

Determine the shear force and moment at points C and D.

500 lb

300 lb

200 lb B

C 6 ft

4 ft

D 4 ft

6 ft

2 ft

SOLUTION Support Reactions: FBD (a). a + ©MB = 0;

500(8) - 300(8) - Ay (14) = 0

Ay = 114.29 lb Internal Forces: Applying the equations of equilibrium to segment AC [FBD (b)], we have + ©F = 0 : x + c ©Fy = 0;

NC = 0

114.29 - 500 - VC = 0

a + ©MC = 0;

Ans. VC = -386 lb

Ans.

MC + 500(4) - 114.29 (10) = 0 MC = -857 lb # ft

Ans.

Applying the equations of equilibrium to segment ED [FBD (c)], we have + ©F = 0; : x + c ©Fy = 0; a + ©MD = 0;

ND = 0

VD - 300 = 0

-MD - 300 (2) = 0

Ans.

VD = 300 lb

Ans.

MD = -600 lb # ft

Ans.

Ans: NC = 0 VC = - 386 lb MC = - 857 lb # ft ND = 0 VD = 300 lb MD = - 600 lb # ft 611

7–2. Determine the normal force, shear force, and moment in the shaft at points B and D. These points lie just to the right of the 150-lb force and the bearing at C, respectively. There is a thrust bearing at A and a journal bearing at C.

150 lb

75 lb

15 in.

10 in.

SOLUTION Support Reactions: a + ΣMA = 0;

Cy(30) - 150(15) - 75(40) = 0

+ ΣFx = 0; S

Cy = 175 lb

Ax = 0

+ c ΣFy = 0; Ay + 175 - 150 - 75 = 0

Ay = 50 lb

For Segment AB: + ΣFx = 0; NB = 0 S

Ans.

+ c ΣFy = 0;

50 - 150 - VB = 0

Ans.

a + ΣMB = 0;

MB - 50(15) = 0

VB = - 100 lb

MB = 750 lb # in

Ans.

For Segment CD: + ΣFx = 0; ND = 0 S

Ans.

+ c ΣFy = 0;

VD - 75 = 0

Ans.

a + ΣMD = 0;

-MD - 75(10) = 0

VD = 75 lb

MD = - 750 lb # in

Ans.

Ans: NB = 0 VB = -100 lb MB = 750 lb # in ND = 0 VD = 75 lb MD = -750 lb # in 612

7–3. The structural connections transmit the loads shown to the column. Determine the normal force, shear force, and moment acting in the column at a section passing horizontally through point A.

30 mm

40 mm 250 mm

16 kN 6 kN 23 kN 185 mm 6 kN

SOLUTION

+ ΣFx = 0; 6 - 6 - VA = 0 S Ans.

VA = 0 + c ΣFy = 0; - NA - 16 - 23 = 0

Ans.

NA = - 39 kN a + ΣMA = 0;

125 mm

- MA + 16(0.155) - 23(0.165) - 6(0.185) = 0

MA = - 2.42 kN # m

Ans.

Ans: NA = - 39 kN VA = 0 MA = - 2.425 kN # m 613

*7–4. The beam-column is fixed to the floor and supports the load shown. Determine the normal force, shear force, and moment at points A and B.

5 kN 0.8 kN ? m 2.5 kN 0.6 m A

0.4 m

0.2 m

SOLUTION

From FBD (a): + ΣFx = 0; VA = 0 S

Ans.

+ c ΣFy = 0;

Ans.

NA - 5 = 0 NA = 5 kN

a + ΣMA = 0; 0.8 - MA = 0 MA = 0.8 kN # m

Ans.

From FBD (b): + ΣFx = 0; VB = 0 S

Ans.

+ c ΣFy = 0;

Ans.

NB - 5 - 2.5 = 0 NB = 7.5 kN

a + ΣMB = 0; 0.8 - 2.5(0.4) - MB = 0 MB = -0.2 kN # m

Ans.

Ans: NA = 5 kN VA = 0 MA = 0.8 kN # m NB = 7.5 kN VB = 0 MB = -0.2 kN # m 614

7–5. The pliers are used to grip the tube at B. If a force of 20 lb is applied to the handles,determine the shear force and moment at point C. Assume the jaws of the pliers exert only normal forces on the tube.

20 lb 40

10 in.

0.5 in. 1 in. C B

SOLUTION + ©MA = 0;

20 lb

-20(10) + RB (1.5) = 0 RB = 133.3 lb

Segment BC: +Q ©Fy = 0;

VC + 133.3 = 0 Ans.

VC = - 133 lb a + ©MC = 0;

- MC + 133.3 (1) = 0 MC = 133 lb # in.

Ans.

Ans: VC = - 133 lb MC = 133 lb # in. 615

7–6.

Determine the distance a as a fraction of the beam’s length L for locating the roller support so that the moment in the beam at B is zero.

SOLUTION a + ©MA = 0;

-P a Cy =

a + ©M = 0;

L/3 L

2L - a b + Cy1L - a2 + Pa = 0 3 2P A L3 - a B L - a

M =

2P A L3 - a B L - a

2PL a a =

L b = 0 3

L - ab = 0 3

L 3

Ans.

Ans: a = 616

L 3

7–7. Determine the normal force, shear force, and moment at points C and D in the beam.

300 lb>ft 6000 lb ? ft A C 12 ft 15 ft

D 3 ft 6 ft

SOLUTION Support Reaction: FBD (a) a + ΣMA = 0; By (15) - 300(21)(10.5) - 6000 = 0

By = 4810 lb

From FBD (b) + ΣFx = 0; ND = 0 S

Ans.

+ c ΣFy = 0;

Ans.

VD - 300(3) = 0

VD = 900 lb

a + ΣMD = 0; - MD - 300(3)(1.5) - 6000 = 0

MD = - 7350 lb # ft = -7.35 kip # ft

Ans.

From FBD (c) + ΣFx = 0; NC = 0 S

Ans.

+ c ΣFy = 0; VC + 4810 - 300(9) = 0 Ans.

VC = - 2110 lb = - 2.11 kip a + ΣMC = 0; - MC - 300(9)(4.5) - 6000 + 4810(3) = 0

MC = - 3720 lb # ft = - 3.72 kip # ft

Ans.

Ans: ND = 0 VD = 900 lb MD = -7.35 kip # ft NC = 0 VC = -2.11 kip MC = -3.72 kip # ft 617

*7–8. Determine the internal shear force and moment acting at point C in the beam.

500 lb/ ft

900 lb ft

A 3 ft

900 lb ft

C 6 ft

B 6 ft

3 ft

SOLUTION Support Reactions: Referring to the FBD of the entire beam shown in Fig. a a+ ΣMB = 0; 500(12)(6) + 900 - 900 - Ay(12) = 0 Ay = 3000 lb + ΣFx = 0; S

Ax = 0

Internal Loadings: Referring to the FBD of the left segment of beam sectioned through C, Fig. b, + c ΣFy = 0; 3000 - 500(6) - VC = 0 VC = 0

Ans.

a+ ΣMC = 0; MC + 500(6)(3) + 900 - 3000(6) = 0

MC = 8100 lb # ft = 8.10 kip # ft

Ans.

Ans: VC = 0 MC = 8.10 kip # ft 618

7–9. Determine the normal force, shear force, and moment at point C. T ake P = 8 kN.

0.1 m

0.5 m C

0.75 m

SOLUTION a + ©MA = 0;

0.75 m

-T(0.6) + 8(2.25) = 0 T = 30 kN

+ ©F = 0; : x

Ax = 30 kN

+ c ©Fy = 0;

Ay = 8 kN

+ ©F = 0; : x

-NC - 30 = 0 Ans.

NC = - 30 kN + c ©Fy = 0;

VC + 8 = 0 Ans.

VC = - 8 kN a + ©MC = 0;

- MC + 8(0.75) = 0 MC = 6 kN # m

Ans.

Ans: NC = - 30 kN VC = - 8 kN MC = 6 kN # m 619

7–10. The cable will fail when subjected to a tension of 2 kN. Determine the largest vertical load P the frame will support and calculate the normal force, shear force, and moment at point C for this loading.

0.1 m

0.5 m C

0.75 m

-2(0.6) + P(2.25) = 0 Ans.

P = 0.533 kN + ©F = 0; : x

Ax = 2 kN

+ c ©Fy = 0;

Ay = 0.533 kN

+ ©F = 0; : x

-NC - 2 = 0 Ans.

NC = - 2 kN + c ©Fy = 0;

VC - 0.533 = 0 Ans.

VC = -0.533 kN a + ©M C = 0;

0.75 m

SOLUTION a + ©MA = 0;

0.75 m

- MC + 0.533(0.75) = 0 MC = 0.400 kN # m

Ans.

Ans: P = 0.533 kN NC = -2 kN VC = -0.533 kN MC = 0.400 kN # m 620

7–11. Determine the normal force, shear force, and moment acting at point C.

8 kN 5 kN>m 18 kN ? m A

C 1.25 m

0.75 m

0.5 m

SOLUTION Segment AC: + ΣFx = 0; NC = 0 S

Ans.

+ c ΣFy = 0;

- VC - 8 - 5 (0.75) = 0

a + ΣMC = 0;

MC + 8(2) + 18 + 5(0.75)(0.375) = 0

VC = -11.8 kN

MC = - 35.4 kN # m

Ans. Ans.

Ans: NC = 0 VC = -11.8 kN MC = -35.4 kN # m 621

*7–12. Determine the distance a between the bearings in terms of the shaft’s length L so that the moment in the symmetric shaft is zero at its center.

a L

SOLUTION Due to symmetry, Ay = By + c ©Fy = 0;

a + ©M = 0;

Ay + By -

w(L - a) w(L - a) - wa = 0 4 4

Ay = By =

w (L + a) 4

-M -

w(La) a wa a L a w a a b a + - b + (L + a) a b = 0 2 4 4 2 6 6 4 2

Since M = 0; 3a2 + (L - a)(L + 2a) - 3a (L + a) = 0 2a2 + 2a L - L2 = 0 Ans.

a = 0.366 L

Ans: a = 0.366 L 622

7–13. The 9-kip force is supported by the floor panel DE, which in turn is simply supported at its ends by floor beams. These beams transmit their loads to the girder AB. Determine the shear and moment acting at point C in the girder.

9 kip 6 ft

6 ft

2 ft

4 ft

6 ft

8 ft

24 ft

SOLUTION a + ΣMD = 0;

- 9(2) + Ey (6) = 0 Ey = 3 kip

+ c ΣFy = 0;

Dy - 9 + 3 = 0 Dy = 6 kip

a + ΣMB = 0;

- Ay (24) + 6(12) + 3(6) = 0 Ay = 3.75 kip

+ ΣFx = 0; S

Ax = 0

+ c ΣFy = 0;

3.75 - 6 - 3 + By = 0 By = 5.25 kip

+ c ΣFy = 0;

3.75 - VC = 0 Ans.

VC = 3.75 kip a + ΣMC = 0;

- 3.75(8) + MC = 0

MC = 30 kip # ft

Ans.

Ans: VC = 3.75 kip MC = 30 kip # ft 623

7–14. Determine the normal force, shear force, and moment at points F and E.

50 lb>ft

E 3 ft

2 ft

B 6 ft

4 ft

F D 2 ft

SOLUTION Support Reactions: FBD (a) a + ΣMB = 0; Cy (6) - 50(6)(3) = 0

Cy = 150 lb

+ ΣFx = 0; Bx = 0 S + c ΣFy = 0; By + 150 - 50(6) = 0

By = 150 lb

From FBD (b) + ΣFx = 0; NF = 0 S

Ans.

+ c ΣFy = 0; - 150 - VF = 0

Ans.

a + ΣMF = 0; MF + 150(4) = 0

VF = - 150 lb

MF = - 600 lb # ft

Ans.

From FBD (c) + ΣFx = 0; NE = 0 S

Ans.

1 + c ΣFy = 0; VE - 30(2) - (20)(2) - 150 = 0 VE = 230 lb 2 1 -ME - 30(2)(1) - (20)(2)(1.333) - 150(2) = 0 a + ΣME = 0; 2 ME = - 387 lb # ft

Ans.

Ans: NF = 0 VF = -150 lb MF = -600 lb # ft NE = 0 VE = 230 lb ME = -387 lb # ft 624

7–15. Determine the normal force, shear force, and moment at point C.

4 kN>m

C 3m

B 6m

SOLUTION For Segment BC: + ΣFx = 0; NC = 0 S

Ans.

1 (4)(6) = 0 2

VC = 12 kN

Ans.

1 (4)(6)(2) = 0 2

MC = -24 kN # m

Ans.

+ c ΣFy = 0;

VC =

a + ΣMC = 0;

-MC -

Ans: NC = 0 VC = 12 kN MC = - 24 kN # m 625

*7–16. Determine the normal force, shear force, and moment at point C of the beam.

400 N/m

200 N/m A

C 3m

SOLUTION Beam: a + ©MB = 0;

600 (2) + 1200 (3) - Ay (6) = 0 Ay = 800 N

+ ©F = 0; : x

Ax = 0

Segment AC: + ©F = 0; : x

NC = 0

+ c ©Fy = 0;

800 - 600 - 150 - VC = 0

Ans.

VC = 50 N a + ©MC = 0;

- 800 (3) + 600 (1.5) + 150 (1) + MC = 0 MC = 1350 N # m = 1.35 kN # m

Ans.

Ans: NC = 0 VC = 50 N MC = 1.35 kN # m 626

7–17. The cantilevered rack is used to support each end of a smooth pipe that has a total weight of 300 lb. Determine the normal force, shear force, and moment that act in the arm at its fixed support A along a vertical section.

6 in. B

SOLUTION

Pipe: + c ©Fy = 0;

NB cos 30° - 150 = 0 NB = 173.205 lb

Rack: + ©F = 0; : x

- NA + 173.205 sin 30° = 0 Ans.

NA = 86.6 lb + c ©Fy = 0;

VA - 173.205 cos 30° = 0 Ans.

VA = 150 lb a + ©MA = 0;

MA - 173.205(10.3923) = 0 MA = 1.80 lb # in.

Ans.

Ans: NA = 86.6 lb VA = 150 lb MA = 1.80 kip # in. 627

7–18. Determine the normal force, shear force, and moment in the beam at points C and D. The beam is supported at A and B.

250 lb> ft

3 ft

4 kip 4 ft

SOLUTION

4 ft

6 ft

4 ft

Support Reactions: FBD (a) a + ΣMB = 0;

0.25(18)(9) + 4(3) - Ay(10) = 0

Ay = 5.25 kip

+ ΣFx = 0; S

- Bx + 4 = 0

Bx = 4 kip

+ c ΣFy = 0;

- By + 5.25 - 0.25(18) = 0

By = 0.75 kip

+ ΣFx = 0; S

NC + 4 = 0

NC = -4 kip

Ans.

+ c ΣFy = 0;

- VC - 0.25(4) = 0

VC = -1.0 kip

Ans.

a + ΣMC = 0;

MC + 0.25(4)(2) + 4(3) = 0

MC = -14.0 kip # ft

Ans.

From FBD (b)

From FBD (c) + ΣFx = 0; S

- ND - 4 = 0

ND = -4 kip

Ans.

+ c ΣFy = 0;

VD - 0.25(4) - 0.75 = 0

VD = 1.75 kip

Ans.

a + ΣMD = 0;

- MD - 0.25(4)(2) - 0.75(4) = 0

MD = -5.0 kip # ft Ans.

Ans: NC = -4 kip VC = -1.0 kip MC = -14.0 kip # ft ND = -4 kip VD = 1.75 kip MD = -5.0 kip # ft 628

7–19. Determine the internal normal force, shear force, and moment at points F and E in the frame. The crate weighs 300 lb.

1.5 ft

1.5 ft 0.4 ft

4 ft

SOLUTION With reference to Fig. a, a + ©MA = 0;

4 FBC a b (3) + 300(0.4) -300(6.4) = 0 5

FBC = 750 lb

Referring to Fig. b, + ©F = 0; : x

- NE - 300 = 0

NE = -300 lb

Ans.

+ c ©Fy = 0;

VE - 300 = 0

VE = 300 lb

Ans.

a + ©ME = 0;

- ME + 300(0.4)- 300(1.9) = 0

ME = - 450 lb # ft

Ans.

Using the result of FBC and referring to Fig. c, + ©F = 0; : x

3 750a b -300- NF = 0 5

NF = 150 lb

Ans.

+ c ©Fy = 0;

4 VF + 750 a b -300 = 0 5

VF = - 300 lb

Ans.

a + ©MF = 0;

4 750a b (1.5) + 300(0.4)-300(4.9) -MF = 0 5

MF = -450 lb # ft

Ans.

The negative sign indicates that NE, VF, and MF act in the opposite sense to that shown in the free-body diagram.

Ans: NE = - 300 lb VE = 300 lb ME = - 450 lb # ft NF = 150 lb VF = - 300 lb MF = - 450 lb # ft

629

*7–20. Rod AB is fixed to a smooth collar D, which slides freely along the vertical guide. Determine the normal force, shear force, and moment at point C, which is located just to the left of the 60-lb concentrated load.

60 lb 15 lb/ft A

C 3 ft

SOLUTION

1.5 ft

With reference to Fig. a, we obtain + c ©Fy = 0;

FB cos 30° -

1 1 (15)(3) - 60 - (15)(1.5) = 0 2 2

FB = 108.25 lb

Using this result and referring to Fig. b, we have + ©Fx = 0; :

-NC - 108.25 sin 30° = 0

+ c ©Fy = 0;

VC - 60 -

NC = -54.1 lb

1 (15)(1.5) + 108.25 cos 30° = 0 2 Ans.

VC = - 22.5 lb a + ©MC = 0;

Ans.

108.25 cos 30°(1.5) -

MC = 135 lb # ft

1 (15)(1.5)(0.5) - MC = 0 2 Ans.

The negative signs indicates that NC and VC act in the opposite sense to that shown on the free-body diagram.

Ans: NC = - 54.1 lb VC = - 22.5 lb MC = 135 lb # ft 630

7–21. Determine the normal force, shear force, and moment at points E and F of the compound beam. Point E is located just to the left of 800-N force.

800 N

1200 N 400 N/m

4 3

A E 1.5 m

B 2m

F 1.5 m

1.5 m

SOLUTION Support Reactions: Referring to the FBD of member BC shown in Fig. a, 4 a+ ΣMB = 0; Cy(3) - 1200 a b(2) = 0 Cy = 640 N 5 4 a+ ΣMC = 0; 1200 a b(1) - By(3) = 0 By = 320 N 5 3 + ΣFx = 0; S 1200 a b - Bx = 0 Bx = 720 N 5 Internal Loadings: Referring to the right segment of member AB sectioned through E, Fig. b + ΣFx = 0; S

720 - NE = 0

NE = 720 N

Ans.

+ c ΣFy = 0;

VE - 800 - 320 = 0

VE = 1120 N = 1.12 kN

Ans.

a+ ΣME = 0; - ME - 320(1) = 0 ME = -320 N # m

Ans.

Referring to the left segment of member CD sectioned through F, Fig. c, + ΣFx = 0; S

NF = 0

+ c ΣFy = 0;

- VF - 640 - 400(1.5) = 0

Ans. VF = - 1240 N = - 1.24 kN Ans.

a+ ΣMF = 0; MF + 400(1.5)(0.75) + 640(1.5) = 0 MF = - 1410 N # m = -1.41 kN # m

Ans.

Ans: NE = 720 N VE = 1.12 kN ME = - 320 N # m NF = 0 VF = - 1.24 kN MF = - 1.41 kN # m 631

7–22. Determine the shear force and moment acting at point D of the beam.

2 kip/ ft

D 6 ft 9 ft

9 ft

SOLUTION + ©F = 0; : x

ND = 0

+ c ©Fy = 0;

9 - 4 - VD = 0

Ans.

VD = 5 kip a + ©MD = 0;

MD - 9 (6) + 4 (2) = 0 MD = 46 kip # ft

Ans.

Ans: ND = 0 VD = 5 kip MD = 46 kip # ft 632

7–23. Determine the normal force, shear force, and moment at point C.

0.5 ft B 2 ft

100 lb ? ft

A C 3 ft

SOLUTION

8 ft

4 ft 50 lb

Support Reactions: FBD (a) a + ΣMA = 0;

T(2.5) - 50(12) - 100 = 0

T = 280 lb

+ ΣFx = 0; S

Ax - 280 = 0

Ax = 280 lb

+ c ΣFy = 0;

Ay - 50 = 0

Ay = 50 lb

+ ΣFx = 0; S

NC + 280 = 0

NC = -280 lb

Ans.

+ c ΣFy = 0;

50 - VC = 0

VC = 50 lb

Ans.

a + ΣMC = 0;

MC - 50(3) = 0

MC = 150 lb # ft

Ans.

From FBD (b)

Ans: NC = - 280 lb VC = 50 lb MC = 150 lb # ft 633

*7–24. w

Determine the shear force and moment acting at point C of the beam. For the calculation use Simpson’s rule to evaluate the integrals.

11.58 kip>ft 1/

2 w 5 (2x3 1 x 1 4)

2 kip>ft

SOLUTION

C 2 ft

2 ft

Support Reactions: FBD (a) FR = 3 a2x3 + 2x + 4b dx = 21.4899 kip 4

1 2

3 2 3 x(2x + 2x + 4) dx 4

x =

56.0634 = 2.6088 ft 21.4899

a+ ΣMB = 0; 21.4899 (4 - 2.6088) - Ay (4) = 0 Ay = 7.4740 kip From FBD (b) FR1 = 3 a2x + 2x + 4b dx = 5.7862 kip 2

1 2

3 3 xa2x + 2x + 4b dx 2

x1 =

1 2

FR1

6.6255 = 1.1451 ft 5.7862

+ c ΣFy = 0; 7.4740 - 5.7862 - VC = 0 VC = 1.69 kip

Ans.

a+ ΣMC = 0; MC + 5.7862(2 - 1.1451) - 7.4740(2) = 0

MC = 10.0 kip # ft

Ans.

Ans: VC = 1.69 kip MC = 10.0 kip # ft 634

7–25. Determine the normal force, shear force, and moment in the beam at points D and E. Point E is just to the right of the 3-kip load.

3 kip 1.5 kip/ ft

6 ft

SOLUTION a + ©MB = 0;

B 6 ft

4 ft

1 2 (1.5)(12)(4) - Ay (12) = 0

Ay = 3 kip + ©F = 0; : x

Bx = 0

+ c ©Fy = 0;

By + 3 - 12 (1.5)(12) = 0 By = 6 kip

+ ©F = 0; : x

ND = 0

+ c ©Fy = 0;

3 - 12 (0.75)(6) - VD = 0

Ans.

VD = 0.75 kip a ©MD = 0;

MD + 12 (0.75) (6) (2) - 3 (6) = 0 MD = 13.5 kip # ft

Ans.

+ ©F = 0; : x

NE = 0

Ans.

+ c ©Fy = 0;

- VE - 3 - 6 = 0 Ans.

VE = -9 kip ©ME = 0;

ME + 6 (4) = 0 ME = - 24.0 kip # ft

Ans.

Ans: ND = 0 VD = 0.75 kip MD = 13.5 kip # ft NE = 0 VE = - 9 kip ME = - 24.0 kip # ft 635

7–26. Determine the distance a in terms of the beam’s length L between the symmetrically placed supports A and B so that the moment at the center of the beam is zero.

B a –– 2

a –– 2 L

SOLUTION In this problem, it is required that the internal moment at point C be equal to zero. With reference to Fig. a, a + ©MA = 0;

1 L 1 L L L a a By (a) - w0 a b B a + ¢ - ≤ R + w0 a b a - b = 0 2 2 3 2 2 2 3 2 By =

1 wL 4 0

Using this result and referring to Fig. b, a + ©MC = 0;

1 1 a L L w L a b - w0 a b a b = 0 4 0 2 2 2 3 a =

2 L 3

Ans.

Ans: a = 636

2 L 3

7–27. Determine the normal force, shear force, and moment at points F and G in the compound beam. Point F is located just to the right of the 500-lb force,while point G is located just to the right of the 600-lb force.

500 lb 2 ft

2 ft

A F

600 lb B

1.5 ft E G

SOLUTION

2 ft

With reference to Fig. b, + ©F = 0; : x

Dx = 0

Using this result and writing the moment equation of equilibrium about point A, Fig. a, and about point E, Fig. b, we have a + ©MA = 0;

Dy(6)- FBC (4)- 500(2) = 0

(1)

a + ©ME = 0;

600(2) -Dy (4)- FBC (6) = 0

(2)

Solving Eqs. (1) and (2) FBC = 560 lb

Dy = 540 lb

Using these results and referring to Fig. b, + c ©Fy = 0;

Ey - 600- 540 + 560 = 0

Ey = 580 lb

Again, using the results of Dx, Dy, and FBC, the force equation of equilibrium written along the x and y axes, Fig. a, + ©F = 0; : x

Ax = 0

+ c ©Fy = 0;

Ay - 500-560 +540 = 0

Ay = 520 lb

Using these results and referring to Fig. c, + ©F = 0; : x

NF = 0

+ c ©Fy = 0;

520- 500 -VF = 0

VF = 20 lb

Ans.

a + ©MF = 0;

MF - 520(2) = 0

MF = 1040 lb # ft

Ans.

Using the result for Ey and referring to Fig. d + ©F = 0; : x

NG = 0

+ c ©Fy = 0;

VG + 580 = 0

VG = - 580 lb

Ans.

a + ©MG = 0;

580(2) -MG = 0

MG = 1160 lb # ft

Ans.

The negative sign indicates that VG acts in the opposite sense to that shown in the free-body diagram. Ans: NF = 0 VF = 20 lb MF = 1040 lb # ft NG = 0 VG = - 580 lb MG = 1160 lb # ft 637

*7–28. Determine the normal force, shear force, and moment at points C and D.

A C

2 ft

2 kip>ft 6 ft

458

SOLUTION

3 ft

Support Reactions: FBD (a) a + ΣMA = 0;

By (6 + 6 cos 45°) - 2(6)(3 + 6 cos 45°) = 0

+ ΣFx = 0; S

Ax = 0

+ c ΣFy = 0;

Ay + 8.4853 - 2(6) = 0

Ay = 3.5147 kip

+R ΣFx′ = 0;

NC - 3.5147 cos 45° = 0

NC = 2.49 kip

Ans.

Q+ ΣFy′ = 0;

3.5147 sin 45° - VC = 0

VC = 2.49 kip

Ans.

a + ΣMC = 0;

MC - 3.5147 (2 sin 45°) = 0

MC = 4.97 kip # ft

Ans.

By = 8.4853 kip

From FBD (b)

From FBD (c) + ΣFx = 0; S

ND = 0

+ c ΣFy = 0;

VD + 8.4853 - 2(3) = 0

VD = -2.49 kip

a + ΣMD = 0;

- MD - 2(3)(1.5) + 8.4853(3) = 0

MD = 16.5 kip # ft Ans.

Ans. Ans.

Ans: NC = 2.49 kip VC = 2.49 kip MC = 4.97 kip # ft ND = 0 VD = -2.49 kip MD = 16.5 kip # ft 638

7–29. Determine the normal force, shear force, and moment acting at point C.

700 lb

800 lb 3 ft 1.5 ft

2 ft

600 lb 1 ft

3 ft

1.5 ft A

30 B

SOLUTION a + ©M A = 0;

- 800 (3) - 700(6 cos 30°) - 600 cos 30°(6 cos 30° + 3cos 30°) + 600 sin 30°(3 sin 30°) + By (6 cos 30° + 6 cos 30°) = 0 B y = 927.4 lb

+ ©F = 0; : x

800 sin 30° - 600 sin 30° - Ax = 0 Ax = 100 lb

+ c ©Fy = 0;

Ay - 800 cos 30° - 700 - 600 cos 30° + 927.4 = 0 Ay = 985.1 lb

Q + ©Fx = 0;

NC - 100 cos 30° + 985.1 sin 30° = 0 Ans.

N C = - 406 lb + a©Fy = 0;

100 sin 30° + 985.1 cos 30° - VC = 0 Ans.

V C = 903 lb a + ©M C = 0;

-985.1(1.5 cos 30°) - 100(1.5 sin 30°) + MC = 0 MC = 1355 lb # ft = 1.35 kip # ft

Ans.

Ans: NC = - 406 lb VC = 903 lb MC = 1.35 kip # ft 639

7–30. Determine the normal force, shear force, and moment acting at point D.

700 lb

800 lb 3 ft 1.5 ft

2 ft

600 lb 1 ft

3 ft

1.5 ft

SOLUTION a+ ©M A = 0;

30 B

- 800(3) - 700(6 cos 30°) - 600 cos 30°(6 cos 30° + 3 cos 30°) + 600 sin 30°(3 sin 30°) + By (6 cos 30° + 6 cos 30°) = 0 B y = 927.4 lb

+ ©F = 0; : x

800 sin 30° - 600 sin 30° - Ax = 0 Ax = 100 lb

+ c ©Fy = 0;

Ay - 800 cos 30° - 700 - 600 cos 30° + 927.4 = 0 Ay = 985.1 lb

+a©Fx = 0;

ND - 927.4 sin 30° = 0 Ans.

ND = - 464 lb Q + ©Fy = 0;

VD - 600 + 927.4 cos 30° = 0 Ans.

VD = - 203 lb a + ©M D = 0;

- MD - 600(1) + 927.4(4 cos 30°) = 0 MD = 2612 lb # ft = 2.61 kip # ft

Ans.

Ans: ND = -464 lb VD = -203 lb MD = 2.61 kip # ft 640

7–31. Determine the normal force, shear force, and moment in the beam at points C and D. Point D is just to the right of the 5-kip load.

5 kip

0.5 kip/ft

A 6 ft

C 6 ft

6 ft

SOLUTION Entire Beam: a + ©MB = 0;

5 (6) + 6 (18) - A y (24) = 0 Ay = 5.75 kip

+ ©F = 0; : x

Ax = 0

Segment AC: + ©F = 0 : x

NC = 0

+ c ©Fy = 0;

5.75 - 3 - VC = 0

Ans.

VC = 2.75 kip a + ©MC = 0;

MC + 3 (3) - 5.75 (6) = 0 MC = 25.5 kip # ft

Ans.

+ ©F = 0; : x

ND = 0

Ans.

+ c ©Fy = 0;

5.75 - 6 - 5 - VD = 0

Segment AD:

Ans.

VD = - 5.25 kip a + ©MD = 0;

MD + 6 (12) - 5.75 (18) = 0 MD = 31.5 kip # ft

Ans.

Ans: NC = 0 VC = 2.75 kip MC = 25.5 kip # ft ND = 0 VD = - 5.25 kip MD = 31.5 kip # ft 641

*7–32. Determine the normal force, shear force, and moment at points D and E of the frames.

0.25 m C

0.75 m

1m E

0.75 m

B 400 N>m

SOLUTION

608 A

Support Reactions: FBD (a) a + ΣMA = 0;

FBC cos 15°(1.5) - 400(1.5)(0.75) = 0

FBC = 310.58 N

Q+ ΣFx′ = 0;

310.58 sin 15° - Ax′ = 0

Ax′ = 80.38 N

+a ΣFy′ = 0;

Ay′ + 310.58 cos 15° - 400(1.5) = 0

Ay′ = 300 N

From FBD (b) + ΣFx = 0; S

310.58 cos 45° + ND = 0

ND = -220 N

Ans.

+ c ΣFy = 0;

- VD - 310.58 sin 45° = 0

VD = -220 N

Ans.

a + ΣMD = 0;

MD + 310.58 sin 45° (0.25) = 0

MD = -54.9 N # m

Ans.

From FBD (c) Q+ ΣFx′ = 0;

NE - 80.38 = 0

NE = 80.4 N

Ans.

+a ΣFy′ = 0;

- VE - 400(0.75) + 300 = 0

VE = 0

Ans.

a + ΣME = 0;

ME + 400(0.75)(0.375) - 300(0.75) = 0 ME = 112 N # m Ans.

Ans: ND = -220 N VD = -220 N MD = -54.9 N # m NE = 80.4 N VE = 0 ME = 112 N # m 642

7–33. 1.5 m

Determine the normal force, shear force, and moment at points D of the two-member frame.

1.5 m

1.5 kN/m

1.5 m

SOLUTION

1.5 m

Member BC: a + ©MC = 0;

4.5 (1.5) - Bx (3) = 0

+ ©F = 0; : x

2 kN/m

Bx = 2.25 kN

2.25 + Cx - 4.5 = 0 Cx = 2.25 kN

Member AB: a + ©MA = 0;

2.25 (3) - 3 (1) - By (3) = 0 By = 1.25 kN

Segment DB: + ©F = 0; : x

- ND - 2.25 = 0 Ans.

ND = - 2.25 kN + c ©Fy = 0;

VD - 1.25 = 0 Ans.

VD = 1.25 kN a + ©MD = 0;

- MD - 1.25 (1.5) = 0 MD = - 1.88 kN # m

Ans.

Ans: ND = - 2.25 kN VD = 1.25 kN MD = - 1.88 kN # m 643

7–34. 1.5 m

Determine the normal force, shear force, and moment at point E.

1.5 m

1.5 kN/m

1.5 m

SOLUTION

1.5 m

Member BC: a + ©MC = 0;

4.5 (1.5) - Bx (3) = 0

+ ©F = 0; : x

2 kN/m

Bx = 2.25 kN

2.25 + Cx - 4.5 = 0 Cx = 2.25 kN

Member AB: a + ©MA = 0;

2.25 (3) - 3 (1) - By (3) = 0 By = 1.25 kN

Segment BE: + c ©Fy = 0;

1.25 - NE = 0 Ans.

NE = 1.25 kN + ©F = 0; : x

VE + 2.25 - 2.25 = 0 Ans.

VE = 0 + ©Mg = 0;

Mg - 2.25 (0.75) = 0 Mg = 1.6875 kN # m = 1.69 kN # m

Ans.

Ans: NE = 1.25 kN VE = 0 MB = 1.69 kN # m 644

7–35. The man has a weight of 250 lb and a center of gravity at G. If the bench upon which he is sitting is fixed connected to the center support, determine the x, y, z components of loading at sections passing through point A and the base B.

G 1.5 ft A

SOLUTION

1 ft 1.5 ft

0.75 ft

A: ΣFx = 0;

VAx = 0

Ans.

ΣFy = 0;

NAy = 0

Ans.

ΣFz = 0;

VAz = 250 lb

Ans.

ΣMAx = 0;

MAx - 0.75(250) = 0;

ΣMAy = 0;

MAy - 0.75(250) = 0;

ΣMAz = 0;

MAz = 0

Ans.

ΣFx = 0;

VBx = 0

Ans.

ΣFy = 0;

VBy = 0

Ans.

ΣFz = 0;

NBz = 250 lb

Ans.

ΣMBx = 0;

MBx - 1.75(250) = 0;

ΣMBy = 0;

MBy - 0.75(250) = 0;

ΣMBz = 0;

MBz = 0

MAx = 187.5 lb # ft MAy = 187.5 lb # ft

Ans. Ans.

MBx = 437.5 lb # ft MB = 187.5 lb # ft y

Ans. Ans. Ans.

Ans: VAx = 0 NAy = 0 VAz = 250 lb MAx = 187.5 lb # ft MAy = 187.5 lb # ft MAz = 0 VBx = 0 VBy = 0 NBz = 250 lb MBx = 437.5 lb # ft MBy = 187.5 lb # ft MBz = 0 645

*7–36. Determine the normal force, shear force, and moment acting at points B and C on the curved member.

A C 458 2 ft 500 lb 308

4 3

SOLUTION From FBD (a) Q+ ΣFx′ = 0;

400 cos 30° + 300 cos 60° - VB = 0

VB = 496 lb

Ans.

+a ΣFy′ = 0;

NB + 400 sin 30° - 300 sin 60° = 0

NB = 59.8 lb

Ans.

a + ΣMO = 0;

300 (2) - 59.80 (2) - MB = 0

MB = 480 lb # ft Ans.

Q+ ΣFx′ = 0;

400 cos 45° + 300 cos 45° - NC = 0

NC = 495 lb

Ans.

+a ΣFy′ = 0;

- VC + 400 sin 45° - 300 sin 45° = 0

VC = 70.7 lb

Ans.

a + ΣMO = 0;

300 (2) + 495 (2) - MC = 0 MC = 1590 lb # ft = 1.59 kip # ft

Ans.

From FBD (b)

Ans: VB = 496 lb NB = 59.8 lb MB = 480 lb # ft NC = 495 lb VC = 70.7 lb MC = 1.59 kip # ft 646

7–37. Determine the normal force, shear force, and moment at point D of the two-member frame.

250 N/m B

D 2m

1.5 m C

SOLUTION

300 N/m

Member AB: a + ©MA = 0;

By (4) - 1000 (2) = 0 By = 500 N

Member BC: a + ©MC = 0;

- 500 (4) + 225 (0.5) + Bx (1.5) = 0 Bx = 1258.33 N

Segment DB: + ©F = 0; : x

-ND + 1258.33 = 0 Ans.

ND = 1.26 kN + c ©Fy = 0;

VD - 500 + 500 = 0 Ans.

VD = 0 a + ©MD = 0;

-MD + 500 (1) = 0 MD = 500 N # m

Ans.

Ans: ND = 1.26 kN VD = 0 MD = 500 N # m 647

7–38. Determine the normal force, shear force, and moment at point E of the two-member frame.

250 N/m B

D 2m

1.5 m C

SOLUTION

300 N/m

Member AB: a + ©MA = 0;

By (4) - 1000 (2) = 0 By = 500 N

Member BC: a + ©MC = 0;

- 500 (4) + 225 (0.5) + Bx (1.5) = 0 Bx = 1258.33 N

Segment EB: + ©F = 0; : x

- NE - 1258.33 - 225 = 0 Ans.

NE = - 1.48 kN + c ©Fy = 0;

VE - 500 = 0 Ans.

VE = 500 N a + ©ME = 0;

- ME + 225 (0.5) + 1258.33 (1.5) - 500 (2) = 0 ME = 1000 N # m

Ans.

Ans: NE = - 1.48 kN VE = 500 N ME = 1000 N # m 648

7–39. w = w0 sin θ

The distributed loading w = w0 sin u, measured per unit length, acts on the curved rod. Determine the normal force, shear force, and moment in the rod at u = 45°.

SOLUTION w = w0 sin u Resultants of Distributed Loading: u

FRx =

w0 sin u(r du) cos u = rw0

L0 u

FRy =

sin u cos u du =

1 r w0 sin2 u 2

w0 sin u(r du) sin u = rw0

Q+ ©Fx = 0;

1 l sin2 u du = rw0 c u - sin 2u d 2 4 L0

-V + FRx cos 45° + FRy sin 45° = 0 1 1 p 1 - sin 90°b sin 45° V = a r w0 sin2 45° b cos 45° + w0 a 2 2 4 4 Ans.

V = 0.278 w0r +a©Fy = 0;

- N - FRy cos 45° + FRx sin 45° = 0 1 p 1 1 N = - r w0 c a b - sin 90° d cos 45° + a r w0 sin2 45°b sin 45° 2 4 4 2 Ans.

N = 0.0759 w0 r a + ©MO = 0;

M - (0.0759 r w0)(r) = 0 M = 0.0759

w0 r2

Ans.

Ans: V = 0.278 w0 r N = 0.0759 w0 r M = 0.0759 w0 r 2 649

*7–40. w = w0 sin θ

Solve Prob. 7–39 for u = 120°.

SOLUTION Resultants of Distributed Load: FRx =

w0 sin u(r du) cos u = rw0

sin u cos u =

L0 u

FRy =

1 rw sin2 u 2 0

u 1 1 sin2 u du = rw0 c u sin 2u drw0 (sin u) 2 = r w0 (sin u) 2 4 L0 0 u

w0 sin u(r du) sin u = rw0

FRx = 12 r w0 sin2 120° = 0.375 r w0 1 120° 1 b - sin 240° d = 1.2637 r w0 FRy = r w0 c (p)a 2 180° 4 +

b©Fx¿ = 0;

N + 0.375 rw0 cos 30° + 1.2637 r w0 sin 30° = 0 Ans.

N = - 0.957 r w0 +a©Fy¿ = 0;

- V + 0.375 rw0 sin 30° - 1.2637 r w0 cos 30° = 0 Ans.

V = - 0.907 rw0 a + ©MO = 0;

- M - 0.957 r w0 (r) = 0 M = - 0.957 r2w0

Ans.

Ans: N = - 0.957 r w0 V = - 0.907 rw0 M = - 0.957 r 2w0 650

7–41. 3m

Determine the normal force, shear force, and moment at point D of the two-member frame. Neglect the thickness of the member.

SOLUTION Support Reactions: Member AB is a two force member. From FBD (a). + ©MC = 0;

450 N

4 3 450 a b (1.5) - FAB a b (4) = 0 5 5

FAB = 225 N

1.5 m A

Internal Forces: Applying the equations of equilibrium to segment AD [FBD (b)], we have + ©F = 0; : x

3 225 a b - VD = 0 5

VD = 135 N

Ans.

+ c ©Fy = 0;

4 225a b - ND = 0 5

ND = 180 N

Ans.

MD = 270 N # m

Ans.

a + ©ME = 0;

225

3 (2) - MD = 0 5

Ans: VD = 135 N ND = 180 N MD = 270 N # m 651

7–42. Determine the x, y, z components of force and moment at point C in the pipe assembly. Neglect the weight of the pipe. The load acting at (0, 3.5 ft, 3 ft) is F1 = 5-24i - 10k6 lb and M = {-30k} lb # ft and at point (0, 3.5 ft, 0) F2 = {-80i} lb.

F1 F1 z

3 ft

SOLUTION

3 ft

Free-Body Diagram: The support reactions need not be computed.

C A

Internal F orces: Applying the equations of equilibrium to segment BC, we have Ans.

©Fx = 0;

(VC)x - 24 - 80 = 0

(VC)x = 104 lb

©Fy = 0;

NC = 0

©Fz = 0;

(VC)z - 10 = 0

(VC)z = 10.0 lb

Ans.

©Mx = 0;

(MC)x - 10(2) = 0

(MC)x = 20.0 lb # ft

Ans.

©My = 0;

(MC)y - 24 (3) = 0

(MC)y = 72.0 lb # ft

Ans.

©Mz = 0;

(MC)z + 24 (2) + 80 (2) - 30 = 0

1.5 ft x x

1.5 ft

y C

2 ft 2 ft

F2 F2

Ans.

(MC)z = -178 lb # ft

Ans.

Ans: NC = 0 (VC)x = 104 lb (VC)z = 10 lb (MC)x = 20 lb # ft (MC)y = 72 lb # ft (MC)z = - 178 lb # ft 652

7–43. Determine the x, y, z components of internal loading at a section passing through point B in the pipe assembly. Neglect the weight of the pipe. Take F1 = 5 200i - 100j - 400k6 N and F2 = 5 300i - 500k 6 N.

A B 1m

1m x 1.5 m

SOLUTION

Internal Loadings: Referring to the FBD of the free end segment of the pipe assembly sectioned through B, Fig. a, ΣFx = 0;

Nx + 300 + 200 = 0

Nx = -500 N

Ans.

ΣFy = 0;

Vy - 100 = 0

Vy = 100 N

Ans.

ΣFz = 0;

Vz - 500 - 400 = 0

Vz = 900 N

Ans.

ΣMx = 0;

Mx - 400(1.5) = 0

ΣMy = 0;

My + 500(1) + 400(1) = 0

ΣMz = 0

Mz - 200(1.5) - 100(1) = 0

Mx = 600 N # m My = -900 N # m Mz = 400 N # m

Ans. Ans. Ans.

The negative signs indicate that Nx and My act in the opposite sense to those shown in FBD.

Ans: Nx = - 500 N Vy = 100 N Vz = 900 N Mx = 600 N # m My = - 900 N # m Mz = 400 N # m 653

*7–44. Determine the x, y, z components of internal loading at a section passing through point B in the pipe assembly. Neglect the weight of the pipe. Take F1 = 5100i - 200j - 300k 6 N and F2 = 5 100i + 500j 6 N.

A B 1m

1m x 1.5 m

SOLUTION

Internal Loadings: Referring to the FBD of the free end segment of the pipe assembly sectioned through B, Fig. a ΣFx = 0;

Nx + 100 + 100 = 0

Nx = - 200 N

Ans.

ΣFy = 0;

Vy + 500 - 200 = 0

Vy = - 300 N

Ans.

ΣFz = 0;

Vz - 300 = 0

Vz = 300 N

Ans.

ΣMx = 0;

Mx - 300(1.5) = 0

ΣMy = 0;

My + 300(1) = 0

ΣMz = 0;

Mz + 500(1) - 100(1.5) - 200(1) = 0

Mx = 450 N # m My = - 300 N # m

Ans. Ans.

Mz = - 150 N # m

Ans.

The negative signs indicates that Nx, Vy, My and Mz act in the senses opposite to those shown in FBD.

Ans: Nx = -200 N Vy = - 300 N Vz = 300 N Mx = 450 N # m My = - 300 N # m Mz = - 150 N # m 654

7–45. Draw the shear and moment diagrams for the shaft (a) in terms of the parameters shown; (b) set P = 9 kN,a = 2 m, L = 6 m. There is a thrust bearing at A and a journal bearing at B.

B a L

SOLUTION (a) c + ©MB = 0;

(Ay)(L) - P(L - a) = 0 Ay = a

L - a bP L

Ay = a 1 + c ©Fy = 0;

a bP L

Ay + By - P = 0 By = P - Ay = a

+ ©F = 0; : x

a bP L

Ax = 0

For 0 … x … a + c ©Fy = 0;

a bP - V = 0 L

a1 -

V = a1 + ©F = 0; : x

A = 0

c + ©M = 0;

a1 -

a bP L

Ans.

a b Px - M = 0 L

M = a1 -

a b Px L

Ans.

For a 6 x 6 L + c ©Fy = 0;

a1 -

a bP - P - V = 0 L

V = -a c + ©M = 0;

a1 -

a bP L

Ans.

a b Px - P(x - a) - M = 0 L

M = Px - a

a bPx - Px + Pa L

655

7–45. Continued

(b) M = Paa c + ©MB = 0;

a xb L

Ans.

Ay (6) - 9(4) = 0 Ay = 6 kN

+ c ©Fy = 0;

By = 3 kN

For 0 … x … 2 m + c ©Fy = 0;

6 - V = 0 Ans.

V = 6 kN c + ©M = 0;

6x - M = 0 M = 6x kN # m

Ans.

For 2 m 6 x … 6 m + c ©Fy = 0;

6 - 9 - V = 0 Ans.

V = - 3 kN c + ©M = 0;

6x - 9(x - 2) - M = 0 M = 18 - 3x kN # m

Ans.

Ans: 0 … x 6 a: V = a1 M = a1 -

a bPx L

a bP L

a a 6 x … L: V = - a bP L a M = P aa - xb L

0 … x 6 2 m: V = 6 kN, M = {6x} kN # m # 2 m 6 x … 6 m:V = - 3 kN M = {18 - 3x} kN # m 656

7–46.

Draw the shear and moment diagrams for the beam (a) in terms of the parameters shown; (b) set P = 800 lb, a = 5 ft, L = 12 ft.

a L

SOLUTION (a)

For 0 … x 6 a + c ©Fy = 0;

V = P

Ans.

a + ©M = 0;

M = Px

Ans.

+ c ©Fy = 0;

V = 0

Ans.

a + ©M = 0;

-Px + P(x - a) + M = 0

For a 6 x 6 L- a

M = Pa

Ans.

+ c ©Fy = 0;

V = -P

Ans.

a + ©M = 0;

- M + P(L - x) = 0

For L- a 6 x … L

M = P(L - x)

Ans.

657

7–46. Continued

(b)

Set P = 800 lb,

a = 5 ft,

L = 12 ft

For 0 … x 6 5 ft + c ©F y = 0;

V = 800 lb

Ans.

a + ©M = 0;

M = 800x lb # ft

Ans.

+ c ©Fy = 0;

V = 0

Ans.

a + ©M = 0;

- 800x + 800(x - 5) + M = 0

For 5 ft 6 x 6 7 ft

M = 4000 lb # ft

Ans.

+ c ©F y = 0;

V = -800 lb

Ans.

a + ©M = 0;

- M + 800(12 - x) = 0

For 7 ft 6 x … 12 ft

M = (9600 - 800x) lb # ft

Ans.

Ans: For 0 … x 6 a, V = P, M = Px For a 6 x 6 L - a, V = 0, M = Pa For L - a 6 x … L, V = - P, M = P(L - x) For 0 … x 6 5 ft, V = 800 lb M = 800x lb # ft For 5 ft 6 x 6 7 ft, V = 0 M = 4000 lb # ft For 7 ft 6 x … 12 ft, V = - 800 lb M = (9600 - 800x) lb # ft 658

7–47. Draw the shear and moment diagrams for the beam (a) in terms of the parameters shown; (b) set P = 600 lb, a = 5 ft, b = 7 ft.

B a

SOLUTION (a) For 0 … x 6 a Pb - V = 0 a + b

+ c ©Fy = 0;

V = a + ©M = 0;

M -

Pb a + b

Ans.

Pb x = 0 a + b

M =

Pb x a + b

Ans.

For a 6 x … 1a + b2 Pb - P - V = 0 a + b

+ c ©Fy = 0;

V = a + ©M = 0;

Pa a + b

Ans.

Pb x + P1x - a2 + M = 0 a + b M = Pa -

Pa x a + b

Ans.

(b) For P = 600 lb, a = 5 ft, b = 7 ft

659

7–47. Continued

(b) c + ©MB = 0;

A y(12) - 600(7) = 0 A y = 350 lb

+ c ©Fy = 0;

By = 250 lb

For 0 … x … 5 ft + c ©Fy = 0;

350 - V = 0 Ans.

V = 350 lb c + ©M = 0;

350x - M = 0

M = 350x lb # ft

Ans.

For 5 ft 6 x … 12 ft + c ©Fy = 0;

350 - 600 - V = 0 Ans.

V = -250 lb c + ©M = 0;

350x - 600(x - 5) - M = 0

M = {3000 – 250x} lb # ft

Ans.

Ans: Pb Pb ,M = x a + b a + b Pa For a 6 x … a + b, V = a + b Pa M = Pa x a + b For 0 … x 6 5 ft, V = 350 lb M = 350x lb # ft For 5 ft 6 x … 12 ft, V = -250 lb M = 53000 - 250x6 lb # ft

For 0 … x 6 a, V =

660

*7–48.

Draw the shear and moment diagrams for the cantilevered beam.

100 lb 800 lb ft C

A B 5 ft

5 ft

SOLUTION

Ans: V = 100 lb Mmax = - 1800 lb # ft 661

7–49.

Draw the shear and moment diagrams for the beam (a) in terms of the parameters shown; (b) set M0 = 500 N # m, L = 8 m.

L/3

L/ 3

SOLUTION (a) For 0 … x …

L 3

+ c ©Fy = 0;

V = 0

Ans.

a + ©M = 0;

M = 0

Ans.

+ c ©Fy = 0;

V = 0

Ans.

a + ©M = 0;

M = M0

Ans.

+ c ©Fy = 0;

V = 0

Ans.

a + ©M = 0;

M = 0

Ans.

For

L 2L 6 x 6 3 3

2L 6 x … L 3

(b) Set M0 = 500 N # m, L = 8 m For 0 … x 6

8 m 3

+ c ©Fy = 0;

V = 0

Ans.

c + ©M = 0;

M = 0

Ans.

For

8 16 m 6 x 6 m 3 3

+ c ©F y = 0;

V = 0

Ans.

c + ©M = 0;

M = 500 N # m

Ans.

For

Ans: L : V = 0, M = 0 3 L 2L 6 x 6 : V = 0, M = M0 3 3 2L 6 x … L: V = 0, M = 0 3 8 0 … x 6 m: V = 0, M = 0 3 8 16 m 6 x 6 m: V = 0, M = 500 N # m 3 3 16 m 6 x … 8 m: V = 0, M = 0 3 0 … x 6

16 m 6 x … 8m 3

+ c ©Fy = 0;

V = 0

Ans.

c + ©M = 0;

M = 0

Ans.

662

7–50.

If L = 9 m, the beam will fail when the maximum shear force is Vmax = 5 kN or the maximum bending moment is Mmax = 2 kN # m. Determine the magnitude M0 of the largest couple moments it will support.

L/3

L/ 3

SOLUTION See solution to Prob. 7–48 a. Mmax = M0 = 2 kN # m

Ans.

Ans: Mmax = 2 kN # m 663

7–51. Draw the shear and moment diagrams for the beam.

10 kN 3 kN/m

A 6m

SOLUTION + c ©Fy = 0;

28 - 3 x - V = 0 Ans.

V = 28 - 3 x a + ©M = 0;

x 114 - 28 x + 3 x a b + M = 0 2 M = 28 x - 1.5 x2 - 114

Ans.

Ans: V = 28 - 3x M = 28x - 1.5x2 - 114 664

*7–52. w

Draw the shear and moment diagrams for the beam. A

L –– 2 L

SOLUTION Support Reactions: From FBD (a), wL 3L a b = 0 2 4

Cy =

3wL 8

wL 3wL = 0 8 2

Ay =

wL 8

a + ©MA = 0;

Cy 1L2 -

+ c ©Fy = 0;

Ay +

L Shear and Moment Functions: For 0 ◊ x< [FBD (b)], 2 + c ©Fy = 0; a + ©M = 0;

wL 8

Ans.

wL x 8

Ans.

wL - V = 0 8

V =

wL 1x2 = 0 8

M =

M -

L For <x ◊ L [FBD (c)], 2 + c ©Fy = 0;

V +

3wL - w1L - x2 = 0 8 V =

a + ©M = 0;

w 15L - 8x2 8

Ans.

L - x 3wL 1L - x2 - w1L - x2a b - M = 0 8 2 M =

w 1- L2 + 5Lx - 4x22 8

Ans.

Ans: wL 8 wL M = x 8 w V = (5L - 8x) 8 w M = ( -L2 + 5Lx - 4x2 ) 8 V =

665

7–53. Draw the shear and moment diagrams for the beam.

20 kN 40 kN/ m A

B 8m

C 3m

150 kN m

SOLUTION 0 … x 6 8 + c ©Fy = 0;

133.75 - 40x - V = 0 Ans.

V = 133.75 - 40x a + ©M = 0;

x M + 40x a b - 133.75x = 0 2 M = 133.75x - 20x2

Ans.

8 6 x … 11 + c ©Fy = 0;

V - 20 = 0 Ans.

V = 20 c + ©M = 0;

M + 20(11 - x) + 150 = 0 Ans.

M = 20x - 370

Ans: V = 133.75 - 40x M = 133.75x - 20x2 V = 20 M = 20x - 370 666

7–54. w

The shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. Draw the shear and moment diagram for the shaft (a) in terms of the parameters shown; (b) set w = 500 lb>ft, L = 10 ft.

B L

SOLUTION (a) For 0 … x … L + c ©Fy = 0;

wL - wx - V = 0 2 V = - wx + V =

a + ©M = 0;

(b) Set w = 500 lb/ft,

wL 2

w (L - 2x) 2

Ans.

wL x x + wx a b + M = 0 2 2

M =

wL wx2 x 2 2

M =

w a Lx - x2 b 2

Ans.

L = 10 ft

For 0 … x … 10 ft + c ©Fy = 0;

2500 - 500x - V = 0 Ans.

V = (2500 - 500x) lb a + ©M = 0;

- 2500x + 500

x2 + M = 0 2

M = (2500x - 250x2) lb # ft

Ans.

Ans: w (L - 2x) 2 w M = (Lx - x2) 2 V = (2500 - 500x) lb M = (2500x - 250x2) lb # ft V =

667

7–55. w

Draw the shear and moment diagrams for the compound beam. The beam is pin connected at E and F.

SOLUTION

Support Reactions: From FBD (b), wL L L b a b = 0 3 3 6

Fy =

wL wL = 0 6 3

Ey =

a + ©ME = 0;

Fy a

+ c ©Fy = 0;

Ey +

wL 6

From FBD (a), a + ©MC = 0;

wL L 4wL L a b a b = 0 6 3 3 3

Dy =

7wL 18

4wL L wL L a b a b - A y 1L2 = 0 3 3 6 3

Ay =

7wL 18

Dy 1L2 +

From FBD (c), a + ©MB = 0; + c ©Fy = 0;

By +

4wL wL 7wL = 0 18 3 6

By =

10wL 9

Shear and Moment Functions: For 0 ◊ x<L [FBD (d)], + c ©Fy = 0;

7wL - wx - V = 0 18 V =

a + ©M = 0;

w 17L - 18x2 18

Ans.

7wL x x = 0 M + wx a b 2 18 M =

w 17Lx - 9x 22 18

Ans.

668

L –– 3

C L

7–55. Continued

For L ◊ x<2L [FBD (e)], + c ©Fy = 0;

10wL 7wL + - wx - V = 0 18 9 V =

a + ©M = 0;

w 13L - 2x2 2

Ans.

7wL 10wL x 1x - L2 = 0 x M + wxa b 2 18 9 M =

w 127Lx - 20L2 - 9x 22 18

Ans.

For 2L<x ◊ 3L [FBD (f)], + c ©Fy = 0;

V +

7wL - w13L - x2 = 0 18

V = a + ©M = 0;

w 147L - 18x2 18

7wL 3L - x - w 3L - x 18 M =

3L - x 2

Ans. - M = 0

w 147Lx - 9x 2 - 60L 22 18

Ans.

Ans: For 0 … x 6 L w V = (7L - 18x) 18 w ( 7Lx - 9x2 ) M = 18 For L 6 x 6 2L w V = (3L - 2x) 2 w (27Lx - 20L2 - 9x2 ) M = 18 For 2L 6 x … 3L w V = (47L - 18x) 18 w ( 47Lx - 9x2 - 60L2 ) M = 18 669

*7–56. Draw the shear and moment diagrams for the beam.

1.5 kN/ m A

2m 4m

SOLUTION 0 … x … 2 m: + c ©Fy = 0;

0.75 - V = 0 Ans.

V = 0.75 kN a + ©M = 0;

M - 0.75 x = 0 M = 0.75 x kN # m

Ans.

2 m 6 x 6 4 m: + c ©Fy = 0;

0.75 - 1.5 (x - 2) - V = 0 Ans.

V = 3.75 - 1.5 x kN a + ©M = 0;

M +

1.5 (x - 2)2 - 0.75 x = 0 2

M = - 0.75 x2 + 3.75 x - 3 kN # m

Ans.

Ans: V = 0.75 kN M = 0.75 x kN # m V = 3.75 - 1.5 x kN M = -0.75 x2 + 3.75 x - 3 kN # m 670

7–57. Draw the shear and moment diagrams for beam ABC. There is a pin at B. Solve the problem (a) in terms of the parameters shown; (b) set w = 5 kN>m, L = 12 m.

SOLUTION (a) a + ΣMB = 0;

wL L L a b + Cy a b = 0 4 8 4

+ ΣFx = 0; S

wL 8 Bx = 0

+ c ΣFy = 0;

By =

a + ΣMA = 0;

Cy =

wL 8

3wL 3L wL 3L a b a b + MA = 0 4 8 8 4

MA =

3 2 wL 8

+ ΣFx = 0; S

Ax = 0

+ c ΣFy = 0;

Ay -

3wL wL = 0 4 8

Ay =

7 wL 8

For 0 … x … L + c ΣFy = 0; (a)

V =

w (7L - 8x) 8

a + ΣM = 0; (b) Set w = 5 kN/m,

7 wL - wx - V = 0 8 Ans. 7 3 wx2 wLx - wL2 - M = 0 8 8 2 w M = - (4x2 - 7Lx + 3L2) 8

Ans.

L = 12 m

For 0 … x … 12 m + c ΣFy = 0; (b)

V = 5(10.5 - x) a + ΣM = 0;

52.5 - 5x - V = 0 Ans. x2 - 270 + 52.5x - 5a b - M = 0 2 M = - 2.5(x2 - 21x + 108)

Ans. Ans: w V = (7L - 8x) 8 w M = - (4x2 - 7Lx + 3L2) 8 V = 5(10.5 - x) M = -2.5(x2 - 21x + 108) 671

7–58. The two segments of the girder are pin connected at B. Draw the shear and moment diagrams for the girder.

15 kip 500 lb>ft A

B C 3 ft

4 ft

2 ft

8 kip ? ft 12 ft

SOLUTION

Ans: Vmax = -11.89 kip Mmax = -106.7 kip · ft 672

7–59. Draw the shear and moment diagrams for the beam. 120 lb/ft 30 lb/ft A 12 ft

SOLUTION + c ©Fy = 0;

360 - 30 x -

1 (7.5 x) (x) - V = 0 2

V = - 3.75 x2 - 30 x + 360 V = 0 = -3.75x2 - 30x + 360 x = 6.58 ft a + ©M = 0;

1 x 1 - 360 x + (30 x)a b + (7.5 x) (x)a xb + M = 0 2 2 3 M = - 1.25 x3 - 15 x2 + 360 x 3 2 Mmax max = - 1.25(6.58) - 15(6.58) + 360(6.58)

= 1363 lb # ft = 1.36 kip # ft

Ans.

Ans: Vmax = - 540 lb Mmax = 1.36 kip # ft 673

*7–60. The shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. Draw the shear and moment diagrams for the shaft.

300 lb/ft A

B 6 ft

6 ft

SOLUTION Since the loading is discontinuous at the midspan, the shear and moment equations must be written for regions 0 … x 6 6 ft and 6 ft 6 x … 12 ft of the beam. The free-body diagram of the beam’s segment sectioned through the arbitrary points within these two regions are shown in Figs. b and c. Region 0 … x 6 6 ft, Fig. b + c ©Fy = 0; a + ©M = 0;

1 (50x)(x) - V = 0 2 x 1 M + (50x)(x) ¢ ≤ - 600(x) = 0 2 3 600 -

V = {600 - 25x2} lb

M = {600x - 8.333x3} lb # ft

(1)

(2)

Region 6 ft 6 x … 12 ft, Fig. c + c ©Fy = 0;

V + 300 = 0

a + ©M = 0;

300(12 - x) - M = 0

V = -300 lb

M = {300(12 - x)} lb # ft

(3) (4)

The shear diagram shown in Fig. d is plotted using Eqs. (1) and (3). The location at which the shear is equal to zero is obtained by setting V = 0 in Eq. (1). 0 = 600 - 25x2

x = 4.90 ft

The moment diagram shown in Fig. e is plotted using Eqs. (2) and (4). The value of the moment at x = 4.90 ft (V = 0) is evaluated using Eq. (2). M ƒ x = 4.90 ft = 600(4.90) - 8.333(4.903) = 1960 lb # ft The value of the moment at x = 6 ft is evaluated using either Eq. (2) or Eq. (4). M ƒ x = 6 ft = 300(12 - 6) = 1800 lb # ft

Ans: Vmax = 600 lb M 0 x = 4.90 ft = 1960 lb # ft M 0 x = 6 ft = 1800 lb # ft 674

7–61. Draw the shear and moment diagrams for the beam.

20 kip

20 kip 4 kip/ ft

A 15 ft

B 30 ft

15 ft

SOLUTION

Ans: x = 15 V = -20 M = - 300 x = 30 + V = 0 M = 150 x = 45 V = -60 M = - 300 675

7–62. Determine the placement a of the roller support B so that the maximum moment within the span AB is equivalent to the moment at the support B.

B a L

SOLUTION Since the loading is discontinuous at support B, the shear and moment equations must be written for regions 0 … x 6 a and a 6 x … L. The free-body diagram of the beam’s segment sectioned through the arbitrary points within these two regions are shown in Figs. b and c. Region 0 … x 6 a, Fig. b w0 (2aL - L2) - w0x - V = 0 2a

+ c ©Fy = 0;

V =

w0 x (2aL - L2)x = 0 a + ©M = 0; M + w0x ¢ ≤ 2 2a

M =

w0 (2aL - L2 - 2ax) 2a

w0 C (2aL - L2)x - ax2 D 2a

(1)

(2)

Region a 6 x … L, Fig. c + c ©Fy = 0;

V - w0(L - x) = 0

(3)

V = w0(L - x)

w0 1 a + ©M = 0; -M - w0(L - x)c (L - x) d = 0 M = - (L - x)2 2 2

(4)

The location at which the shear is equal to zero is obtained by setting V = 0 in Eq. (1). 0 =

w0 (2aL - L2 - 2ax) 2a

x =

2aL - L2 2a

The maximum span moment occurs at the position at which V = 0. Thus, using Eq. (2), we obtain

A Mspan B max =

w0 w0 2aL - L2 2aL - L2 2 C (2aL - L2) ¢ ≤ - a¢ ≤ S = 2 c A 2aL - L2 B 2 d 2a 2a 2a 8a

The support moment at B is evaluated using Eq. (2). Msuppport =

w0 w0 w0 (2aL - L2 - a2) = - (L - a)2 C (2aL - L2)a - a3 D = 2a 2 2

The support moment at B can also be computed from Eq. (4). Msupport = -

w0 (L - a)2 2

Here, we require (Mmax)span = Msupport Thus, w0 8a2

(2aL - L2)2 = a =

w0 (L - a)2 2

Ans.

Ans: a =

676

L 22

7–63. The jib crane supports a load of 750 lb. If the boom AB has a uniform weight of 60 lb>ft, draw the shear and moment diagrams for the boom.

7 ft

3 ft

SOLUTION Support Reactions: FBD (a) Shear and Moments Functions: For 0 … x 6 7 ft + c ΣFy = 0;

1350 - 60x - V = 0

a + ΣM = 0;

x M + 8250 + 60x a b - 1350x = 0 2

V = {1350 - 60x} lb

Ans.

M = 5 1350x - 30x2 - 8250 6 lb # ft Ans. For 7 6 x … 10 ft + c ΣFy = 0;

1350 - 60 (7) - 60 (x - 7) - 750 - V = 0 V = {600 - 60x} lb

a + ΣM = 0;

Ans.

x - 7 b 2 + 750 (x - 7) - 1350x = 0

M + 8250 + 60 (7) (x - 3.5) + 60 (x - 7) a

M = 5 600x - 30x2 - 3000 6 lb # ft

Ans.

Ans: V = {1350 - 60x} lb M = 5 1350x - 30x2 - 8250 6 lb # ft V = {600 - 60x} lb M = 5 600x - 30x2 - 3000 6 lb # ft 677

*7–64. The semicircular rod is subjected to a distributed loading w = w0 sin u. Determine the normal force, shear force. and moment in the rod as a function of u.

w0 w 5 w0 sin u

r u

SOLUTION FRx = 3 w cos u r du = 3 w0 sin u cos u r d u u

1 1 = w0 r c sin2 u d = w0 r sin2 u d 2 2 0 u

u 1 1 FRy = 3 w sin u r du = 3 w0 sin2 u r du = w0 r c u - sin 2 u d 2 4 0 0 0 u

1 1 = w0 r a u sin 2 u b T 2 4 +QΣFx′ = 0;

1 1 1 w r sin2 u (cos u) - w0 r a u - (sin u cos u) b sin u = 0 2 0 2 2 1 V = w0 r u sin u Ans. 2 V -

+ aΣFy′ = 0;

N +

1 1 1 w r sin2 u (sin u) - w0 r a u - (sin u cos u) b cos u = 0 2 0 2 2

N =

1 w r (sin u - u cos u) 2 0

a + Σ MO = 0;

N (r) - M = 0 M =

1 w r 2 (sin u - u cos u) 2 0

Ans.

Ans: 1 w0 r u sin u 2 1 N = w0 r (sin u - u cos u) 2 1 M = w0 r 2 (sin u - u cos u) 2 V =

678

7–65. Draw the shear and moment diagrams for the beam.

12 kN/ m A

C 3m 6m

SOLUTION Support Reactions: Referring to the FBD of the entire beam shown in Fig. a, 1 a+ ΣMB = 0; 12(3)(1.5) + (12)(3)(4) - Ay(6) = 0 Ay = 21.0 kN 2 1 a+ ΣMA = 0; By(6) - (12)(3)(2) - 12(3)(4.5) = 0 By = 33.0 kN 2 + ΣFx = 0; S Bx = 0 Shear and Moment Functions: The beam will be sectioned at two arbitrary distances x in region AC (0 … x 3 m) and region CB (3 m 6 x … 6 m). For region 0 … x 6 3 m, Fig. b 1 +c ΣFy = 0; 21.0 - (4x)(x) - V = 0 V = 5 21.0 - 2x2 6 kN Ans. 2 1 x a+ ΣMO = 0 M + c (4x)(x) d a b - 21.0x = 0 2 3 M = e 21.0x -

2 3 x f kN # m 3

679

Ans.

7–65. Continued

For region 3 m 6 x … 6 m, Fig. c +c ΣFy = 0;

V + 33.0 - 12(6 - x) = 0 V = {39.0 - 12x} kN Ans.

1 a+ ΣMO = 0 33.0 (6 - x) - [12(6 - x)] c (6 - x) d - M = 0 2 M = 5 - 6x2 + 39x - 18 6 kN # m

Ans.

Plotting the shear and moment functions obtained, the shear and moment diagram shown in Fig. d and e resulted.

Ans: For 0 … x 6 3 m V = 5 21.0 - 2x2 6 kN M = e 21.0x -

2 3 x f kN # m 3

For 3 m 6 x … 6 m V = 5 39.0 - 12x6 kN

M = 5 -6x2 + 39x - 18 6 kN # m

680

7–66. Express the shear and moment acting in the pipe as a function of y, where 0 … y … 4 ft.

4 lb>ft

2 ft y

4 ft

SOLUTION +c ΣFy = 0;

16 - 4y - V = 0

a+ ΣM = 0;

M + 4y ( 2 ) + 40 - 16y = 0

V = {16 - 4y} lb

Ans.

M = { - 2y2 + 16y - 40} lb # ft

Ans.

Ans: V = {16 - 4y} lb M = { -2y2 + 16y - 40} lb # ft 681

7–67. Determine the normal force, shear force, and m oment in the curved rod as a function of u. The force P acts at the constant angle f.

r u f P

SOLUTION Support Reactions: Not required Internal Loadings: Referring to the FBD of the free and segment of the sectioned curved rod, Fig. a, + aΣFy = 0; N - P sin (u + f) = 0 N = P sin (u + f)

Ans.

+ QΣFx = 0; P cos (u + f) + V = 0 V = -P cos (u + f)

Ans.

a+ ΣMO = 0; P cos f(r sin u) - P sin f[r (1 - cos u)] - M = 0 M = Pr(sin u cos f + cos u sin f - sin f) Using the identity sin (u + f) = sin u cos f + cos u sin f, Ans.

M = Pr[sin (u + f) - sin f]

Ans: N = P sin (u + f) V = - P cos (u + f) M = Pr[sin (u + f) - sin f] 682

*7–68. Determine the normal force, shear force, and moment in the curved rod as a function of u.

SOLUTION For 0 … u … 180° Q + ©Fx = 0;

+R©Fy = 0;

a + ©M = 0;

N -

3 4 P cos u - P sin u = 0 5 5

N =

P (4 cos u + 3 sin u) 5

V -

3 4 P sin u + P cos u = 0 5 5

V =

P (4 sin u) - 3 cos u) 5

Ans.

3 4 P(r - r cos u) + P(r sin u) + M = 0 5 5

M =

Pr (4 - 4 cos u - 3 sin u) 5

Ans.

Also, a + ©M = 0;

4 - Pa b (r) + N(r) + M = 0 5 M =

Pr (4 - 4 cos u - 3 sin u) 5

Ans: P (4 cos u + 3 sin u) 5 P V = (4 sin u + 3 cos u) 5 Pr M = (4 - 4 cos u - 3 sin u) 5 N =

683

7–69. Determine the distance a between the bearings in terms of the shaft’s length L so that the moment in the symmetric shaft is zero at its center.

a L

SOLUTION Due to symmetry, +c ΣFy = 0;

Ay = By

Ay + By Ay = By =

a+ ΣM = 0;

-M -

w(L - a) 4

- wa -

w(L - a) 4

= 0

w (L + a) 4

w(L - a) a wa a L a w a a b a + - b + (L + a)a b = 0 2 4 4 2 6 6 4 2

Since M = 0; 3a2 + (L - a)(L + 2a) - 3a(L + a) = 0 2a2 + 2aL - L2 = 0 Ans.

a = 0.366 L

Ans: a = 0.366 L 684

7–70. Draw the shear and moment diagrams for the beam.

800 lb

A B 12 ft

12 ft

SOLUTION

Ans: Vmax = 1.20 kip Mmax = 19.2 kip · ft 685

7–71. Draw the shear and moment diagrams for the cantilever beam.

50 lb

60 lb ? ft

4 ft

8 ft

SOLUTION

Ans: Vmax = 50 lb Mmax = - 660 lb # ft 686

*7–72. Draw the shear and moment diagrams for the beam. There is a pin at C and A is a fixed support.

400 lb

500 lb? ft A

C 5 ft

8 ft

12 ft

SOLUTION

Ans: Vmax = 358 lb Mmax = - 1458 lb · ft 687

7–73. Draw the shear and moment diagrams for the beam.

400 lb

200 lb

B 6 ft

SOLUTION

6 ft

Ans: Vmax = -550 lb Mmax = 4200 lb · ft 688

7–74. Draw the shear and moment diagrams for the beam. The supports at A and B are a thrust bearing and journal bearing, respectively.

600 N 1200 N/ m

300 N

A 0.5 m

0.5 m

SOLUTION Support Reactions: Referring to the FBD of the shaft shown in Fig. a, a+ ΣMA = 0;

NB(1) + 300(0.5) - 1200(1)(0.5) - 600(1.5) = 0

a+ ΣMB = 0;

1200(1)(0.5) + 300(1.5) - 600(0.5) - Ay(1) = 0 Ay = 750 N

+ ΣFx = 0; S

NB = 1350 N

Ax = 0

Ans: x = 0.5 + V = 450 N M = - 150 N # m x = 1.5 V = - 750 N M = - 300 N # m 689

7–75.

Draw the shear and moment diagrams for the beam.

SOLUTION

250 N/m

Support Reactions: A

a + ©MA = 0;

3 FC a b142 - 500122 - 500112 = 0 5

FC = 625 N

+ c ©Fy = 0;

3 A y + 625a b - 500 - 500 = 0 5

A y = 625 N

500 N

Ans: x = 2+ V = -375 N M = 750 N # m 690

*7–76. Draw the shear and moment diagrams for the beam.

12 kN>m 30 kN ? m A

SOLUTION

Ans: Vmax = 48 kN Mmax = 96 kN · m 691

7–77. Draw the shear and moment diagrams for the beam.

2 kip/ ft 50 kip ft

50 kip ft

A 10 ft

B 20 ft

10 ft

SOLUTION Support Reactions. Referring to the FBD of the beam shown in Fig. a, a+ ΣMA = 0; By(20) + 50 - 2(20)(10) - 50 = 0 By = 20.0 kip a+ ΣMB = 0; 2(20)(10) + 50 - 50 - NA(20) = 0 NA = 20.0 kip + ΣFx = 0; S

Bx = 0

Ans: x = 10 + V = 20.0 kip M = -50.0 kip # ft 692

7–78. Draw the shear and moment diagrams for the beam.

8 kN

15 kN/ m

20 kN m

SOLUTION

Ans: Vmax = 45 kN Mmax = - 67.5 kN · m 693

7–79. Draw the shear and moment diagrams for the beam.

50 lb>ft 200 lb ? ft A

B 20 ft

C 10 ft

SOLUTION

Ans: Vmax = -510 lb Mmax = 2401 lb · ft 694

*7–80. Draw the shear and moment diagrams for the beam.

8 kN

15 kN/m

20 kN · m A

B 1m

0.75 m

0.25 m

SOLUTION Support Reactions:

a+ ΣMA = 0;

Dy(3) - 8(1) - 8(2) - 15.0(3.5) - 20 = 0 Dy = 32.167 kN

+ c ΣFy = 0;

32.167 - 8 - 8 - 15.0 - Ay = 0 Ay = 1.167 kN

Ans: Vmax = -17.2 kN Mmax = 16.5 kN · m 695

7–81. Draw the shear and moment diagrams for the beam.

200 lb>ft

A 6 ft

6 ft

SOLUTION

Ans: Vmax = 800 lb Mmax = -1200 lb · ft 696

7–82. Draw the shear and moment diagrams for the beam.

4 kN/ m 2 kN/ m B A 3m

1.5 m

SOLUTION Support Reactions: Referring to the FBD of the cantilevered beam shown in Fig. a. a+ ΣMA = 0; MA - 2(3)(1.5) + c ΣFy = 0; Ay - 2(3) -

1 (4)(1.5)(3.5) = 0 MA = 19.5 kN # m 2

1 (4)(1.5) = 0 Ay = 9.00 kN 2

Ans: Vmax = 9.00 kN Mmax = - 19.5 kN · m 697

7–83. 3 kip/ft

Draw the shear and moment diagrams for the beam.

3 kip/ft

B 6 ft

6 ft

SOLUTION Support Reactions: Shown on FBD (a) From FBD (b) + c ©Fy = 0;

- V6- - 12 132162 = 0 V6- = - 9 kip

©M = 0;

M6 + 12 132162142 = 0 M6 = - 36 kip # ft

Ans: Vmax = 9 kip Mmax = -36 kip · ft 698

*7–84. Draw the shear and moment diagrams for the beam.

6 kN/m 3 kN/ m C

A B 3m

SOLUTION Support Reactions: Referring to the FBD of the simply supported beam shown in Fig. a, a+ ΣMA = 0; NC(6) - 3(6)(3) -

1 (3)(3)(5) = 0 NC = 12.75 kN 2

1 a+ ΣMC = 0; (3)(3)(1) + 3(6)(3) - Ay(6) = 0 2 + S ΣFx = 0; Ax = 0

Ay = 9.75 kN

Shear and Moment Functions: Referring to the FBD of the left segment of the beam sectioned at a distance x within region BC (3 m 6 x … 6 m), + c ΣFy = 0;

9.75 - 3x -

a+ ΣMO = 0;

M +

1 (x - 3)(x - 3) - V = 0 2

V = e 5.25 =

1 2 x f kN 2

1 1 x (x - 3)(x - 3) c (x - 3) d + 3x a b - 9.75x = 0 2 3 2 1 M = e - x3 + 5.25x + 4.50 f kN # m 6

Set V = 0, we obtain

1 2 x x = 110.5 m 2 The corresponding moment is 1 M = - 1 110.5 2 3 + 5.25110.5 + 4.50 = 15.8 kN # m 6 0 = 5.25 -

Ans: Vmax = - 12.75 kN Mmax = 15.8 kN # m 699

7–85. 30 lb ft

Draw the shear and moment diagrams for the beam.

180 lb · ft A

B 9 ft

C 4.5 ft

SOLUTION 0 … x 6 9 ft: + c ©Fy = 0;

25 -

1 (3.33 x) (x) - V = 0 2

V = 25 - 1.667 x2 V = 0 = 25 - 1.667 x2 x = 3.87 ft a + ©M = 0;

M +

x 1 (3.33 x)(x) a b - 25 x = 0 2 3

M = 25 x - 0.5556 x3 Mmax = 25 (3.87) - 0.5556 (3.87)3 = 64.5 lb # ft 9 ft 6 x 6 13.5 ft: + c ©Fy = 0;

25 - 135 + 110 - V = 0 V = 0

a + ©M = 0;

- 25 x + 135 (x - 6) - 110 (x - 9) + M = 0 M = - 180

Ans: Vmax = -110 lb Mmax = -180 lb # ft 700

7–86. 2 kN m

Draw the shear and moment diagrams for the beam.

2 kN m

B 3m

SOLUTION Support Reactions: From FBD (a), a+ ΣMA = 0;

Cy (6) - 3.00(1) - 3.00(5) = 0

Cy = 3.00 kN

+ c ΣFy = 0;

Ay + 3.00 - 3.00 - 3.00 = 0

Ay = 3.00 kN

Shear and Moment Diagrams: The peak value of the moment diagram can be evaluated using the method of sections. The maximum moment occurs at the midspan (x = 3 m) where V = 0. From FBD (b), a+ ΣM = 0;

M - 3.00(1) = 0

M = 3.00 kN # m

Ans: Vmax = 3 kN Mmax = 3 kN # m 701

7–87. Draw the shear and moment diagrams for the beam.

9 kN/m

9 kN/ m

B 3m

SOLUTION Support Reactions. Referring to the FBD of the simply supported beam shown in Fig. a, 1 1 a+ ΣMA = 0; NB(6) - (9)(3)(2) - (9)(3)(5) = 0 NB = 15.75 kN 2 2 1 1 a+ ΣMB = 0; (9)(3)(1) + (9)(3)(4) - Ay(6) = 0 Ay = 11.25 kN 2 2 + ΣFx = 0; S

Ax = 0

Shear And Moment Functions. Referring to the FBD of the left segment of the beam sectioned at a distance x within region 0 6 x … 3 m. 1 3 (3x)(x) - V = 0 V = e 11.25 - x2 f kN 2 2 1 x a+ ΣMO = 0; M + c (3x)(x) d a b - 11.25x = 0 2 3 1 M = e 11.25x - x3 f kN # m 2 Set V = 0; 3 0 = 11.25 - x2 x = 27.5 m 2 The corresponding moment is 1 M = 11.25 ( 17.5) - ( 17.5 ) 3 = 20.5 kN # m 2 The moment at x = 3 m is 1 M = 11.25(3) - (33) = 20.25 kN # m 2 + c ΣFy = 0; 11.25 -

Ans: x = 3 V = - 2.25 kN M = 20.25 kN # m 702

*7–88. Draw the shear and moment diagrams for the beam.

3 kN

6 kN/ m

A 1.5 m

1.5 m

SOLUTION Support Reactions: Referring to the FBD of the cantilevered beam shown in Fig. a. 1 (6)(3)(1.5) - 3(3) = 0 MA = 22.5 kN # m 2 1 + c ΣFy = 0; Ay - (6)(3) - 3 = 0 Ay = 12.0 kN 2 a+ ΣMA = 0; MA -

+ ΣFx = 0 Ax = 0 S Internal Loadings: Referring to the FBD of the right segment of the beam sectioned at x = 1.5 m, the internal moment at this section is a+ ΣMO = 0; - M -

1 (6)(1.5)(0.5) - 3(1.5) = 0 M = -6.75 kN # m 2

Ans: Vmax = 12 kN Mmax = - 22.5 kN # m 703

7–89. Draw the shear and moment diagrams for the shaft. The support at A is a thrust bearing and at B it is a journal bearing.

2 kN 3 kN>m A

500 mm

300 mm

200 m

SOLUTION

Ans: Vmax = 1.98 kN Mmax = 0.395 kN # m 704

7–90. 9 kN/ m

Draw the shear and moment diagrams for the beam.

6 kN m

A 3m

SOLUTION Support Reactions: Referring to the FBD of the cantilivered beam shown in Fig. a, 1 a+ ΣMA = 0; MA - (9)(3)(1) - 6 = 0 MA = 19.5 kN # m 2 + c ΣFy = 0; Ay -

1 (9)(3) = 0 Ay = 13.5 kN 2

+ ΣFx = 0; Ax = 0 S

Ans: Vmax = 13.5 kN Mmax = - 9.5 kN # m 705

7–91. 700 lb

The beam consists of two segments pin connected at B. Draw the shear and moment diagrams for the beam.

150 lb/ft

A 8 ft

4 ft

800 lb · ft

6 ft

SOLUTION

Ans: Vmax = 1017 lb Mmax = -9.40 lb # ft 706

*7–92. Draw the shear and moment diagrams for the beam.

50 lb>ft 200 lb? ft A

B 9 ft

C 4 ft

SOLUTION

Ans: Vmax = - 172 lb Mmax = - 200 lb # ft 707

7–93.

Draw the shear and moment diagrams for the beam.

2 kip/ ft

A 1 kip/ ft 15 ft

SOLUTION Shear and Moment Functions: For 0 … x 6 15 ft + c ©Fy = 0;

1x - x2>15 - V = 0 V = {x - x2>15} N

a + ©M = 0;

Ans.

x M + (x2>15) a b - 1x(x>2) = 0 3 M = {x2>2 - x3>45} N # m

Ans.

Ans: Vmax = 3.75 kip Mmax = 37.5 kip # ft 708

7–94. The cable supports the three loads shown. Determine the sags yB and yD of B and D. Take P1 = 800 N, P2 = 500 N.

E A

D C

P2 3m

SOLUTION Support Reactions: Referring to the FBD of the cable system sectioned through cable AB shown in Fig. a, a+ ΣME = 0;

500(3) + 800(9) + 500(15) - FAB a

yB 2y2B + 9

- FABa FABa

18yB + 3 2y2B + 9

b(15) 3

2y2B + 9

b = 16200

b(yB + 1) = 0 (1)

Also, referring to the FBD of the cable segment sectioned through cables AB and CD, shown Fig. b, yB 3 a+ ΣMC = 0; 500(6) + FAB a b(4 - yB) - FABa b(6) = 0 2 2yB + 9 2y2B + 9 FABa

9yB - 12

2y2B + 9

b = 3000

(2)

Divide Eq. (1) by (2) yB = 2.216 m = 2.22 m

Ans.

Substituting this result into Fig. (1) FAB c

18(2.216) + 3 22.2162 + 9

d = 16200

FAB = 1408.93 N

709

P1 6m

7–94. Continued

Method of Joints: Perform the joint equilibrium analysis first for joint B and then joint C. Joint B. Fig. c + ΣFx = 0; S

FBC cos 16.56° - 1408.93 cos 36.45° = 0 FBC = 1182.39 N

+ c ΣFy = 0; 1408.93 sin 36.45 - 1182.39 sin 16.56° - 500 = 0

(Check!)

Joint C. Fig. d + ΣFx = 0; S

FCDa

6 2(4 - yD)2 + 36 6

2(4 - yD)2 + 36 + c ΣFy = 0;

FCDa

2(4 - yD)2 + 36

(3)

FCD = 1133.33

4 - yD

b - 1182.39 cos 16.56° = 0

b + 1182.39 sin 16.56° - 800 = 0 (4)

FCD = 462.96

Divide Eq (4) by (3)

4 - yD = 0.4085 yD = 1.549 m = 1.55 m 6

Ans.

Ans: yB = 2.22 m yD = 1.55 m 710

7–95. The cable supports the three loads shown. Determine the magnitude of P1 if P2 = 600 N and yB = 3 m. Also find sag yD.

E A

D C

P2 3m

SOLUTION Support Reactions: Referring to the FBD of the cable system sectioned through cable BC, Fig. a a+ ΣME = 0;

600(3) + P1(9) - FBCa

6 237

b(5) - FBC a

39 FBC - 9P1 = 1800 237

1 237

b(9) = 0 (1)

Method of Joints: Perform the joint equilibrium analysis for joint B first, Fig. b, + ΣFx = 0; FBCa 6 b - FABa 1 b = 0 (2) S 237 22 + c ΣFy = 0; FABa Solving Eqs (2) and (3),

b - FBC a

237

b - 600 = 0

(3)

FBC = 120237 N FAB = 72022 N Substitute the result of FBC into Fig. (1), Ans.

P1 = 320 N

711

P1 6m

7–95. Continued

Next. Consider the equilibrium of joint C, Fig. c, + ΣFx = 0; FCDa S

6 2

2(4 - yD) + 36 6

2(4 - yD)2 + 36 + c ΣFy = 0; FCDa

b - 1 120237 2 a

2(4 - yD) + 36

4 - yD

2(4 - yD)2 + 36

237

b = 0 (4)

FCD = 720

4 - yD 2

b + 1 120237 2 a

1 237

b - 320 = 0 (5)

FCD = 200

Divide Eq (5) by (4)

4 - yD 5 = yD = 2.3333 m = 2.33 m Ans. 6 18

Ans: P1 = 320 N yD = 2.33 m 712

*7–96. Determine the tension in each segment of the cable and the cable’s total length.

A 4 ft D

7 ft

B C

SOLUTION a+ ΣMD = 0;

- TAB cos 60.26°(3) - TAB sin 60.26°(8) + 100(8) + 200(3) = 0 Ans.

TAB = 165.99 lb = 166 lb + ΣFx = 0; S

4 ft

100 lb 5 ft

3 ft 200 lb

Dx - 165.99 cos 60.26° = 0 Dx = 82.34 lb

+ c ΣFy = 0; Dy + 165.99 sin 60.26° - 100 - 200 = 0 Dy = 155.87 lb

Joint D: + ΣFx = 0; S

- TCD cos f + 82.34 = 0 [1]

+ c ΣFy = 0;

- TCD sin f + 155.87 = 0 [2]

Solving Eqs. [1] and [2] yields: f = 62.15° Ans.

TCD = 176 lb Joint B: + ΣFx = 0; S

TBC cos u - 165.99 cos 60.26° = 0 [3]

+ c ΣFy = 0;

165.99 sin 60.26° - TBC sin u - 100 = 0 [4]

Solving Eqs. [3] and [4] yields: u = 28.18° Ans.

TBC = 93.4 lb 7 3 5 + cos 62.15° + cos 28.18° = 20.2 ft Total lenght of the cable lT = sin 60.26°

Ans.

Ans: TAB = 166 lb TCD = 176 lb TBC = 93.4 lb lT = 20.2 ft 713

7–97. xB

The cable supports the loading shown. Determine the distance xB the force at B acts from A. Set P = 800 N. A 4m

SOLUTION Support Reactions: Referring to the FBD of the cable system sectioned through 1 m cable CD, Fig. a a+ ΣMA = 0; 800(4) + 600(10) - FCDa

25 FCD = 935.08 N

+ ΣFx = 0; 800 + 600 - 935.08 a S + c ΣFy = 0; Ay - 935.08 a

1 25

2 25

b(11) = 0

C D

600 N

b - Ax = 0 Ax = 563.64 N

b = 0 Ay = 418.18 N

Method of Joints: Consider the equilibrium of joint A, Fig. b + ΣFx = 0; FABa S FABa

xB 2x2B + 16 xB

2x2B + 16

+ c ΣFy = 0; 418.18 - FABa FABa Divide Eq (1) by (2)

b - 563.64 = 0 b = 563.64 4 2x2B + 16

2x2B + 16

(1)

b = 0

b = 418.18

(2)

Ans.

xB = 5.3913 m = 5.39 m

Ans: xB = 5.39 m 714

7–98. The cable supports the loading shown. Determine the magnitude of the horizontal force P so that xB = 5 m.

xB A 4m B

6m C

SOLUTION

Support Reactions: Referring to the FBD of the cable system sectioned through cable AB, Fig. a 5 FAB = a b(11) - P(7) - 600(1) = 0 a+ ΣMD = 0; 241 55 FAB - 7P = 600 (1) 241

600 N

Method of Joints: Consider the equilibrium of joint B, Fig. b, + c ΣFy = 0; FABa

4 241

+ ΣFx = 0; P - FABa S FAB =

b - FBCa 5

241

2 25

b - a

241 P 7

b = 0 FBC =

225 241

FAB ba

1 25

225 241

FAB

b = 0 (2)

Substituting Eq. (2) into (1) 241 P b - 7P = 600 7 241 55

Ans.

P = 700 N

Ans: P = 700 N 715

7–99. The cable supports the three loads shown. Determine the sags yB and yD of points B and D. Take P1 = 400 lb, P2 = 250 lb.

4 ft

E A

14 ft

D C

SOLUTION At B + ©F = 0; : x + c ©Fy = 0;

12 ft

20 2

2(14 - yB) + 400 -

TBC -

14 - yB 2

2(14 - yB) + 400 32yB - 168

2y2B + 144

TBC +

20 ft

15 ft

12 ft

TAB = 0

yB 2y2B + 144

TAB - 250 = 0 (1)

TBC = 3000

2(14 - yB)2 + 400

At C + ©F = 0; : x

+ c ©Fy = 0;

15 2

2(4 - yD) + 225

TCD -

14 - yD 2

2(14 - yD) + 225

2(14 - yB)2 + 400

TCD +

-20yD + 490 - 15yB 2(14 - yB)2 + 400 - 20yD + 490 - 15yB 2(14 - yD)2 + 225

TBC = 0

14 - yB 2(14 - yB)2 + 400

TBC - 400 = 0

TBC = 6000

(2)

TCD = 8000

(3)

At D + ©F = 0; : x + c ©Fy = 0;

12 2

2(4 + yD) + 144

TDB -

4 - yD 2

2(4 + yD) + 144

15 2(14 - yD)2 + 225

TDE -

-108 + 27yD 2(14 - yD)2 + 225

TCD = 0

14 - yD 2(14 - yD)2 + 225

TCD - 250 = 0 (4)

TCD = 3000

Combining Eqs. (1) & (2) 79yB + 20yD = 826 Combining Eqs. (3) & (4) 45yB + 276yD = 2334 yB = 8.67 ft

Ans.

yD = 7.04 ft

Ans. Ans: yB = 8.67 ft yD = 7.04 ft 716

*7–100. The cable supports the three loads shown. Determine the magnitude of P1 if P2 = 300 lb and yB = 8 ft. Also find the sag yD.

4 ft

E A

14 ft

D C

SOLUTION At B

12 ft

+ ©F = 0; : x

20 2436 -6

+ c ©Fy = 0;

2436

TBC TBC +

12 2208 8 2208

P1 20 ft

15 ft

12 ft

TAB = 0 TAB - 300 = 0

TAB = 983.3 lb TBC = 854.2 lb At C + ©F = 0; : x

-20 2436 6

+ c ©Fy = 0;

2436

(854.2) +

15 2(14 - yD)2 + 225 14 - yD 2(14 - yD)2 + 225

TCD = 0

(1)

TCD - P1 = 0

(2)

At D + ©F = 0; : x

+ c ©F ΣFyy = = 0; 0; From Eq. 1,

12 2

2(4 + yD) + 144

TDE -

15 2(14 - yD)2 + 225

TCD = 0

4 + yD4 + yD 14 - yD 14 - yD TDE - TDE TCD - T 300 0 (4)= 0 - 300 CD = 2 2 2 2 2(4 +2(4 yD) ++ y144 yD) -+ y225 2(14 D) + 144 2(14 D) + 225

TCD 2(14 - yD)2 + 225

3600 2225 + (14 - yD)2 TCD = = 54.545 lb 27yD - 108

Substitute into Eq. (1): and then from Eq. 3, Ans.

yD = 6.44 ft = 68.182 lb 2(4 + yD)2 + 144 TCD = 916.1 lb TDE

Now Eq. 2 simplifies to

Ans.

P1 = 658 lb

(5)

P1 + 54.545yD = 1009.09 and Eq. 4 simplifies to 68.182(4 + yD) - 54.545(14 - yD) = 300

Ans.

yD = 6.444 = 6.44 ft Substituting into Eq. 5,

Ans.

P1 = 658 lb 717

Ans: yD = 6.44 ft P1 = 658 lb

7–101. Cable ABCD supports the 120-kg uniform beam. Determine the maximum tension in this cable and the sag of point B.

D 1m

yB C B 1m 0.5 m

2m E

1m F

0.5 m

SOLUTION a+ ΣMB = 0; TC (2) - 1177.2(0.5) = 0 TC = 294.30 N + c ΣFy = 0; TB + 294.30 - 1177.2 = 0 TB = 882.90 N a+ ΣMA = 0; TCD sin 45° (4) - 882.90 (1) - 294.30 (3) = 0 TCD = 624.30 N Joint C: + ΣFx = 0; 624.30 cos 45° - TCB cos u = 0 S + c ΣFy = 0; 624.30 sin 45° - TCB sin u - 294.30 = 0 u = 18.43° TCB = 465.33 N Joint B: + ΣFx = 0; 465.33 cos 18.43° - TBA cos f = 0 S + c ΣFy = 0; TBA sin f + 465.33 sin 18.43° - 882.90 = 0 TBA = 858.06 N f = 59.0° Maximum tension: T = 858 N

Ans.

yB = 1 tan 59.0° = 1.67 m

Ans.

Ans: T = 858 N yB = 1.67 m 718

7–102.

Determine the maximum uniform loading w, measured in lb>ft, that the cable can support if it is capable of sustaining a maximum tension of 3000 lb before it will break.

50 ft 6 ft

SOLUTION y =

1 a wdxbdx FH L L

At x = 0,

dy = 0 dx

At x = 0, y = 0 C1 = C2 = 0 y =

w 2 x 2FH

At x = 25 ft, y = 6 ft

FH = 52.08 w

dy w 2 2 = tan umax = x dx max FH x = 25 ft umax = tan - 1 (0.48) = 25.64° Tmax =

FH = 3000 cos umax

FH = 2705 lb Ans.

w = 51.9 lb/ft

Ans: w = 51.9 lb>ft 719

7–103. The cable is subjected to a uniform loading of w = 250 lb>ft. Determine the maximum and minimum tension in the cable.

50 ft 6 ft

SOLUTION From Example 7–12: FH =

250 (50)2 w0 L2 = = 13 021 lb 8h 8 (6)

umax = tan - 1 a Tmax =

250 (50) w0 L b = tan - 1 a b = 25.64° 2FH 2(13 021)

FH 13 021 = = 14.4 kip cos umax cos 25.64°

Ans.

The minimum tension occurs at u = 0°. Ans.

Tmin = FH = 13.0 kip

Ans: Tmax = 14.4 kip Tmin = 13.0 kip 720

*7–104. The cable AB is subjected to a triangular loading. If the angles with the tangents at points A and B are 45° and 80°, respectively, determine the deflection curve of the cable and the maximum tension developed in the cable.

y B 808

SOLUTION

458

From Eq. 7–12: y =

1 1 w(x) dx 2 dx; w = 2.5 x FH 3 1

A 20 ft

1 y = 1 2.5 x dx 2 dx FH 3 1 =

1 1 1.25x2 + C1 2 dx FH

1 1 0.417x3 + C1 x + C2 2 FH

50 lb>ft

x = 0, y = 0, 6 C2 = 0 dy 1 = (1.25x2 + C1) dx FH dy = tan 45° = 1 at x = 0, 6 C1 = FH dx At x = 20, 5.671 =

dy = tan 80° = 5.671 dx

1 (1.25(20)2 + FH); FH = 107.04 lb FH

Thus, y =

1 (0.417x3 + 107.04x) 107.04

y = 3.89 (10-3)x3 + x

Ans.

Maximum tension occurs at point of umax (pt. B), T =

FH 107.04 = = 616 lb cos u cos 80°

Ans.

Ans: y = 3.89 (10-3)x3 + x T = 616 lb 721

7–105. If x = 2 ft and the crate weighs 300 lb, which cable segment AB, BC, or CD has the greatest tension? What is this force and what is the sag yB?

2 ft

3 ft

yB A

D 3 ft

SOLUTION The forces FB and FC exerted on joints B and C will be obtained by considering the equilibrium on the free-body diagram, Fig. a. + ©ME = 0;

FC(3) - 300(2) = 0

FC = 200 lb

+ ©MF = 0;

300(1) - FB(3) = 0

FB = 200 lb

Referring to Fig. b, we have + ©MA = 0;

TCD sin 45°(8) - 200(5) - 100(2) = 0 TCD = 212.13 lb = 212 lb (max)

Using these results and analyzing the equilibrium of joint C, Fig. c, we obtain + ©F = 0; : x

212.13 cos 45° - TBC cos u = 0

+ c ©Fy = 0;

TBC sin u + 212.13 sin 45° - 200 = 0 Ans.

TAB = TCD = 212 lb (max) Solving, u = 18.43°

TBC = 158.11 lb

Using these results to analyze the equilibrium of joint B, Fig. d, we have + : ©Fx = 0;

158.11 cos 18.43° - TAB cos f = 0

+ c ©Fy = 0;

TAB sin f - 100 - 158.11 sin 18.43° = 0

Solving, f = 45° TAB = 212.13 lb = 212 lb (max) Thus, both cables AB and CD are subjected to maximum tension. The sag yB is given by yB = tan f = tan 45° 2 Ans.

yB = 2 ft

Ans: TAB = TCD = 212 lb (max), yB = 2 ft 722

7–106. If yB = 1.5 ft, determine the largest weight of the crate and its placement x so that neither cable segment AB, BC, or CD is subjected to a tension that exceeds 200 lb.

2 ft

3 ft

yB A

D 3 ft

SOLUTION The forces FB and FC exerted on joints B and C will be obtained by considering the equilibrium on the free-body diagram, Fig. a. a + ©ME = 0;

FC(3) - w(x) = 0

a + ©MF = 0;

w(3 - x) - FB(3) = 0

3 ft

wx 3 w FB = (3 - x) 3 FC =

Since the horizontal component of tensile force developed in each cable is constant, cable CD, which has the greatest angle with the horizontal, will be subjected to the greatest tension. Thus, we will set TCD = 200 lb. First, we will analyze the equilibrium of joint C, Fig. b. + ©Fx = 0; :

200 cos 45° - TBC cos 26.57° = 0

+ c ©Fy = 0;

200 sin 45° + 158.11 sin 26.57° -

TBC = 158.11 lb wx = 0 3 wx = 212.13 3

(1)

Using the result of TBC to analyze the equilibrium of joint B, Fig. c, we have + ©Fx = 0; :

4 158.11 cos 26.57° - TAB a b = 0 5

+ c ©Fy = 0;

3 w 176.78 a b - 158.11 sin 26.57° - (3 - x) = 0 5 3

TAB = 176.78 lb

w (3 - x) = 35.36 3

(2)

Solving Eqs. (1) and (2) x = 2.57 ft

Ans.

W = 247 lb

Ans: x = 2.57 ft W = 247 lb 723

7–107. The cable supports a girder which weighs 850 lb>ft. Determine the tension in the cable at points A, B, and C.

100 ft A 40 ft

C B

SOLUTION y = y =

1 ( w dx) dx FH L L 0

1 (425x2 + C1x + C2) FH dy C1 850 = x + dx FH FH

At x = 0,

dy = 0 C1 = 0 dx

At x = 0 , y = 0

C2 = 0 y =

425 2 x FH

20 =

425(x¿)2 FH

At y = 20 ft, x = x¿

At y = 40 ft, x = (100- x¿) 40 =

425(100 - x¿)2 FH

2(x¿)2 = (x¿)2 - 200x¿ + 1002 (x¿)2 + 200x¿ - 1002 = 0 x¿ =

-200 < 22002 + 4(100)2 = 41.42 ft 2 FH = 36 459 lb

At A, 2(425)x dy = tan uA = = 1.366 ` dx FH x = - 58.58 ft uA = 53.79° TA =

FH 36 459 = = 61 714 lb cos uA cos 53.79° Ans.

TA = 61.7 kip

724

20 ft

7–107. Continued

At B, Ans.

TB = FH = 36.5 kip At C, dy 2(425)x = tan uC = = 0.9657 ` dx FH x = 41.42 ft uC = 44.0° TC =

FH 36 459 = = 50 683 lb cos u C cos 44.0° Ans.

TC = 50.7 kip

Ans: TA = 61.7 kip TB = 36.5 kip TC = 50.7 kip 725

*7–108. The cable is subjected to the loading parabolic loading The cable is subjected to the parabolic w 200(1 2 2) lb w 200(1 where x is in the ft. Determine (x =50) ft, (x>50) where x) lb>ft, is in ft. Determine equation ythe equation = f(x) which defines cable AB, and f(x) whichy defines the cable shapethe AB, andshape the maximum the maximum tension in slope the cable. The slope of isthe cable at tension in the cable. The of the cable at A zero. A is zero.

y 50 ft 10 ft

SOLUTION y =

1 ( w(x) dx) dx FH L 1

y =

x 2 1 a 200 a1 - a b b dx b dx FH L L 50

y =

x3 1 a 200 a x b + C1 b dx FH L 3 (50)2

y =

1 x4 a 100 x2 + C1x + C2 b FH 150

200 lb / ft

dy C1 200 x x3 = + dx FH 37.5 FH FH dy = 0, dx

C1 = 0

At x = 0, y = 0,

C2 = 0

At x = 0,

y =

1 x4 a 100 x2 b FH 150

At x = 50 ft, y = 10 ft, y =

FH = 20 833 lb

1 x4 c 100 x2 d ft 20 833 150

Ans.

dy 1 x3 = c 200 x d = tan umax dx 20 833 37.5 x = 50 umax = 17.74° Tmax =

FH 20 833 = = 21 874 lb cos umax cos 17.74° Ans.

Tmax = 21.9 kip

Ans:

1 x4 c 100x2 d ft 20 833 150 Tmax 21.9 kip y =

726

7–109. If the pipe has a mass per unit length of 1500 kg> m, determine the maximum tension developed in the cable.

30 m A

SOLUTION As shown in Fig. a, the origin of the x, y coordinate system is set at the lowest point of the cable. Here, w(x) = w0 = 1500(9.81) = 14.715(103) N>m. Using Eq. 7–12, we can write y =

1 a w0dxb dx FH L L 1 14.715(103) 2 x + c1x + c2 b a FH 2

dy = 0 at x = 0, results in c1= 0. dx Applying the boundary condition y = 0 at x = 0 results in c2 = 0. Thus, Applying the boundary condition

y =

7.3575(103) 2 x FH

Applying the boundary condition y = 3 m at x = 15 m, we have 3 =

7.3575(103) (15)2 FH

FH = 551.81(103) N

Substituting this result into Eq. (1), we have dy = 0.02667x dx The maximum tension occurs at either points at A or B where the cable has the greatest angle with the horizontal. Here, umax = tan - 1 a

dy ` b = tan-1 [0.02667(15)] = 21.80° dx 15 m

Thus, Tmax =

551.8(103) FH = = 594.32(103) N = 594 kN cos umax cos 21.80°

Ans: Tmax = 594 kN 727

7–110. If the pipe has a mass per unit length of 1500 kg> m, determine the minimum tension developed in the cable.

30 m A

SOLUTION As shown in Fig. a, the origin of the x, y coordinate system is set at the lowest point of the cable. Here, w(x) = w0 = 1500(9.81) = 14.715(103) N>m. Using Eq. 7–12, we can write y =

1 a w0dxb dx FH L L 3

1 14.715(10 ) 2 x + c1x + c2 b a FH 2

dy = 0 at x = 0, results in c1= 0. dx Applying the boundary condition y = 0 at x = 0 results in c2 = 0. Thus, Applying the boundary condition

y =

7.3575(103) 2 x FH

Applying the boundary condition y = 3 m at x = 15 m, we have 3 =

7.3575(103) (15)2 FH

FH = 551.81(103) N

Substituting this result into Eq. (1), we have dy = 0.02667x dx The minimum tension occurs at the lowest point of the cable, where u = 0°. Thus, Tmin = FH = 551.81(103) N = 552 kN

Ans: Tmin = 552 kN 728

7–111. The power line has a mass of 500 g/m. If it has a length of 32 m between the poles, determine the maximum tension in the line and its sag.

29 m

SOLUTION w0 = 0.5(9.81) = 4.905 N/m From Example 7–15, s =

w0 FH sinha xb w0 FH

At s = 16 m, x = 14.5 m 78.48 = FH sinha

71.12 b FH

FH = 91.64 N dy w0 s ` = tan umax = ` = 0.856 dx max FH s = 16 m umax = tan - 1 (0.856) = 40.58° Tmax =

FH 91.64 = = 121 N cos umax cos 40.58°

Ans.

At x = 14.5 m y =

w0 FH c cosha xb - 1 d w0 FH

h =

91.64 4.905 c cosha (14.5) b - 1 d = 5.91 m 4.905 91.64

Ans.

Ans: Tmax = 121 N h = 5.91 m 729

*7–112. The cable will break when the maximum tension reaches Tmax = 10 kN. Determine the minimum sag h if it supports the uniform distributed load of w = 600 N>m.

25 m h

SOLUTION

600 N/m

The Equation of the Cable: y = =

1 ( w(x)dx)dx FH L L 1 w0 2 ¢ x + C1x + C2 ≤ FH 2

[1]

dy 1 (w x + C1) = dx FH 0

[2]

Boundary Conditions: 1 (C ) FH 2

C2 = 0

dy 1 (C ) = 0 at x = 0, then from Eq.[2] 0 = dx FH 1

C1 = 0

y = 0 at x = 0, then from Eq.[1] 0 =

Thus, w0 2 x 2FH

[3]

dy w0 = x dx FH

[4]

y =

y = h, at x = 12.5m, then from Eq.[3] h =

w0 (12.52) 2FH

FH =

78.125 w0 h

u = umax at x = 12.5 m and the maximum tension occurs when u = umax. From Eq.[4] tan umax =

dy w0 2 = 18.125 x = 0.0128h(12.5) = 0.160h dx x - 12.5m w0 h

Thus, cos umax =

1 20.0256h2 + 1

The maximum tension in the cable is Tmax =

10 =

FH cos umax

18.125 h (0.6)

1 20.0256h2 + 1 Ans.

h = 7.09 m

730

Ans: h = 7.09 m

7–113. The chain has a weight of 3 lb lb>ft. tension at at ft. Determine the tension points A, B, and C. C necessary for equilibrium.

12 ft A

SOLUTION

Performing the Integration Yields:

8 ft

FH 1 e sinh - 1 c (3s + C1) d + C2 f x = 3 FH From Eq. 7–14

At s = 0;

10 ft

dy 1 w ds = dx FH L 0 dy 1 (3s + C1) = dx FH

dy = 0 dx

hence C1 = 0 dy 3s = tan u = dx FH

(2)

Applying boundary conditions at x = 0; s = 0 to Eq. (1) and using the result C1 = 0 yields C2 = 0. Hence s =

FH 3 sinh a xb 3 FH

(3)

Substituting Eq. (3) into (2) yields: dy 3x = sinh a b dx FH

(4)

FH 3 cosh a x b + C3 3 FH

(5)

Performing the integration y =

FH Applying boundary conditions at x = 0; y = 0 yields C3 = . Therefore 3 FH 3 y = B cosh a x b - 1 R 3 FH At x = 10 ft; y = 12 ft. 12 =

FH = 16.40 lb

By trial and error At point A

FH 3 (10) b -1 R B cosh a 3 FH

x = - 8 ft

From Eq. (4)

dy 3(- 8) = sinh a b uA = - 63.94° dx 16.40 FH 16.40 = 37.3 lb TA = = cos uA cos (- 63.94°)

tan uA =

At point B

x = 10 ft

Ans.

From Eq. (4)

3(10) dy = sinh a b uB = 71.76° dx 16.40 FH 16.40 = = 52.4 lb TB = cos uB cos 71.76°

tan uB =

At point C,

dy = tan uC = 0 dx

uC = 0°

TC =

Ans.

FH 16.40 = = 16.4 lb Ans. cos uC cos 0° 731

Ans: TA = 37.3 lb TB = 52.4 lb TC = 16.4 lb

7–114. If the side loading on the suspension bridge in Example 7.12 is w0 = 0.75 kN>m and L = 190 m, h = 20 m, determine the maximum tension in the cable.

SOLUTION The Equation of the Cable: y =

1 1 w dx 2 dx FH 3 1 0

y =

1 1 0.75 dx 2 dx FH 3 1

1 1 0.375x2 + C1 x + C2 2 [1] FH

dy 1 = (0.75x + C1) [2] dx FH Boundary Conditions: At

x = 0;

y = 0

From Eq. [1]

x = 0;

dy = 0 dx

From Eq. [2]

Hence

y =

1 (C ) FH 2 1 0 = (C ) FH 1 0 =

C2 = 0 C1 = 0

0.375 2 x [3] FH

dy 0.75 = tan u = x [4] dx FH At x = 95 m; y = 20 m

From Eq. [3] 20 =

0.375 (95)2 FH

FH = 169.22 kN At x = 95 m; u = umax and the maximum tension occurs when u = umax. From Eq. [4] tan umax = Tmax =

0.75 (95) umax = 22.834° 169.22 FH 169.22 = = 184 kN cos umax cos 22.834°

Ans.

Ans: Tmax = 184 kN 732

7–115. The cable will break when the maximum tension reaches Tmax = 12 kN. Determine the uniform distributed load w required to develop this maximum tension. Set h = 6 m.

7.5 m

SOLUTION

The Equation of the Cable: y =

1 1 w dx 2 dx FH 3 1 0

y =

1 1 wdx 2 dx FH 3 1

1 w 2 a x + C1 x + C2 b [1] FH 2

dy 1 = (wx + C1) [2] dx FH Boundary Conditions: At

x = 0;

y = 0

From Eq. [1]

x = 0;

dy = 0 dx

From Eq. [2]

1 (C ) FH 2 1 0 = (C ) FH 1 0 =

C2 = 0 C1 = 0

w 2 x [3] 2FH

Hence y =

At x = 7.5 m; y = 6 m

From Eq. [3] 6 =

w (7.5)2 FH = 4.6875w 2FH

dy w w = x = x = 0.2133x dx FH 4.6875w At x = 7.5 m; u = umax Hence

tan umax =

dy 2 = 0.2133 (7.5) = 1.6 dx x = 7.5 m umax = 57.994°

Maximum tension occurs when u = umax Tmax = 12(10)3 =

FH cos umax 4.6875w cos 57.994° Ans.

w = 1357 N/m = 1.36 kN/m

Ans: w = 1.36 kN/m 733

*7–116. The uniform beam weighs 80 lb>ft and is held in the horizontal position by means of cable AB, which has a weight of 10 lb>ft. If the slope of the cable at A is 15°, determine the length of the cable.

SOLUTION Equilibrium: Tension in the cable at point A can be found by summing moments about point C. a + ΣMC = 0;

80(10)(5) - TA sin 15°(10) = 0

FH = TA cos u = 1545.5 cos 15° = 1492.8 lb x= 3

158

10 ft

TA = 1545.5 lb

ds 1 1 e 1 + 2 (w0ds)2 f 2 FH

Performing the Integration Yields: x =

FH 1 e sinh-1 c (10s + C1) d + C2 f [1] 10 FH

From Eq. 7-13 dy 1 = w ds dx FH 3 0 dy 1 = (10s + C1) dx FH

s = 0;

dy = tan 15° dx

hence C1 = FH tan 15°

Applying boundary conditions at x = 0; s = 0 to Eq. [1] and using the result C1 = FH tan 15° yields C2 = - sinh-1 (tan 15°). Hence x = At x = 10 ft; 10 =

FH 1 e sinh-1 c (10s + FH tan 15°) d - sinh-1 (tan 15°) f [2] 10 FH s = ℒ

From Eq. [2]

1492.8 1 e sinh-1 c (10ℒ + 1492.8 tan 15°) d - sinh-1 (tan 15°) f 10 1492.8 ℒ = 10.4 ft

Ans.

Ans: x = 10.4 ft 734

7–117. The cable has a weight of 8 N>m and length of 10 m. If the tension in the cable at C is 300 N, determine the distance from A to C.

308 A

SOLUTION FH = TA cos u = 300 cos 30° = 259.81 N x= 3

ds 1 1 e 1 + 2 (w0ds)2 f 2 FH

Perform the integration x =

FH 1 e sinh-1 c (8s + C1) d + C2 f [1] 8 FH

From Eq. 7-13 dy 1 = w ds dx FH 3 0 dy 1 = (8s + C1) dx FH At

s = 0;

dy = tan 30°. dx

Hence C1 = FH tan 30°

Apply boundary conditions at x = 0; s = 0 to Eq. [1] and using the result C1 = FH tan 30° yields C2 = - sinh-1 (tan 30°). Hence x =

FH 1 e sinh-1 c (8s + FH tan 30°) d - sinh-1 (tan 30°) f 8 FH

259.81 1 e sinh-1 c (8(10) + 259.81 tan 30°) d - sinh-1 (tan 30°) f 8 259.81 Ans.

= 8.07 m

Ans: x = 8.07 m 735

7–118. Show that the deflection curve of the cable discussed in Example 7–13 reduces to Eq. 4 in Example 7–12 when the hyperbolic cosine function is expanded in terms of a series and only the first two terms are retained. (The answer indicates that the catenary may be replaced by a parabola in the analysis of problems in which the sag is small. In this case, the cable weight is assumed to be uniformly distributed along the horizontal.)

SOLUTION cosh x = 1 +

x2 + Á 21

Substituting into y =

FH w0 B cosh ¢ x ≤ - 1 R w0 FH

FH w 20x2 + Á - 1R B1 + w0 2F 2H

w0x2 2FH

Using Eq. (3) in Example 7–12, FH = We get

y =

w0L2 8h

4h 2 x L2

QED

736

7–119. The power line is supported at A by the tower. If the cable weighs 0.75 lb>ft, and the sag s = 3 ft, determine the resultant horizontal force the cable exerts at A.

A s

C 7 ft

SOLUTION From Example 7–15, y =

w0 FH c cosh a xb - 1 d w0 FH

Line AB At x = 75 ft, 3 =

y = 3 ft

0.75(75) FH c cosh a b - 1d 0.75 FH

FH = 703.5 lb Line AC At x = 87.5 ft, 7 =

y = 7 ft

0.75(87.5) FH c cosh a b - 1d 0.75 FH

FH = 411.0 lb FRx = 411.0 - 703.5 = 292 lb

Ans.

Ans: FRx = 292 lb 737

*7–120. The power line is supported at A by the tower. If the cable weighs 0.75 lb>ft, determine the total length of the cable, BAC. Set s = 3 ft.

A s

C 7 ft

SOLUTION From Prob. 7–99 Line AB FH = 703.5 lb From Example 7–15, s =

w0 FH c sinh a xb d w0 FH

At x = 75 ft,

s = 75.08 ft

LAB = 2s = 150.2 ft Line AC At x = 87.5 ft,

s = 87.75 ft

LAC = 2s = 175.7 ft Ans.

Total length BAC = 150.2 + 175.7 = 326 ft

Ans: l = 326 ft 738

7–121. The power line is supported at A by the tower. If the cable weighs 0.75 lb>ft, determine the required sag s so that the resultant horizontal force the cable exerts at A is zero.

A s

C 7 ft

SOLUTION From Prob. 7–99, Line AC, FH = 411.0 lb Line AB, y = At x = 75 ft, s =

w0 FH c cosh a xb - 1 d w0 FH

FH = 411.0 lb

0.75(75) 411.0 c cosh a b - 1 d = 5.14 ft 0.75 411.0

Ans.

Ans: s = 5.14 ft 739

7–122. The man picks up the 52-ft chain and holds it just high enough so it is completely off the ground. The chain has points of attachment A and B that are 50 ft apart. If the chain has a weight of 3 lb/ft, and the man weighs 150 lb, determine the force he exerts on the ground. Also, how high h must he lift the chain? Hint: The slopes at A and B are zero.

25 ft

SOLUTION Deflection Curve of The Cable: x =

where w0 = 3 lb>ft

1 2 L 31 + 11>F H 21 1 w0 ds2242

Performing the integration yields x =

FH 1 b sinh-1 B 13s + C12 R + C2 r 3 FH

(1)

From Eq. 7–14 dy 1 1 13s + C12 w0 ds = = dx FH L FH

(2)

Boundary Conditions: dy = 0 at s = 0. From Eq. (2) dx

0 =

1 10 + C12 FH

C1 = 0

Then, Eq. (2) becomes dy 3s = tan u = dx FH

(3)

s = 0 at x = 0 and use the result C1 = 0. From Eq. (1) x =

FH 1 b sinh-1 B 10 + 02 R + C2 r 3 FH

C2 = 0

Rearranging Eq. (1), we have s =

FH 3 x≤ sinh ¢ 3 FH

(4)

Substituting Eq. (4) into (3) yields dy 3 x≤ = sinh ¢ dx FH Performing the integration y =

FH 3 x cosh 3 FH

y = 0 at x = 0. From Eq. (5) 0 =

(5)

+ C3

FH FH cosh 0 + C3 ,thus, C3 = 3 3

740

25 ft

7–122. Continued

Then, Eq. (5) becomes 3 26 ft at

cosh

(6)

25 ft. From Eq. (4) 26

sinh

154.003 lb By trial and error at

25 ft. From Eq. (6) 154.003 3 25 cosh 3 154.003

Ans.

6.21 ft

From Eq. (3) tan 26 ft

The vertical force

3 26

26.86°

0.5065

154.003

that each chain exerts on the man is tan

154.003 tan 26.86°

78.00 lb

Equation of Equilibrium: By considering the equilibrium of the man, 0;

150

2 78.00

Ans.

306 lb

Ans: h = 6.21 ft Nm = 306 lb 741

8–1. P 2 P 2

Determine the maximum force P the connection can support so that no slipping occurs between the plates. There are four bolts used for the connection and each is tightened so that it is subjected to a tension of 4 kN. The coefficient of static friction between the plates is ms = 0.4.

SOLUTION Free-Body Diagram: The normal reaction acting on the contacting surface is equal to the sum total tension of the bolts. Thus, N = 4(4) kN = 16 kN. When the plate is on the verge of slipping, the magnitude of the friction force acting on each contact surface can be computed using the friction formula F = msN = 0.4(16) kN. As indicated on the free-body diagram of the upper plate, F acts to the right since the plate has a tendency to move to the left. Equations of Equilibrium: + ©Fx = 0; :

0.4(16) -

P = 0 2

p = 12.8 kN

Ans.

Ans: P = 12.8 kN 742

8–2. If the coefficient of static friction between the incline and the 200-lb crate is ms = 0.3, determine the force P that must be applied to the rope to begin moving the crate up the incline.

SOLUTION Q + ΣFy′ = 0; N - 200 cos 30° = 0 N = 173.2 lb 308

+ R ΣFx′ = 0; 200 sin 30° + 0.3(173.2) - 2P = 0 Ans.

P = 76.0 lb

Ans: P = 76.0 lb 743

8–3. 10 kN

The mine car and its contents have a total mass of 6 Mg and a center of gravity at G. If the coefficient of static friction between the wheels and the tracks is ms = 0.4 when the wheels are locked, find the normal force acting on the front wheels at B and the rear wheels at A when the brakes at both A and B are locked. Does the car move?

0.9 m

SOLUTION

0.15 m

1.5 m

NA 11.52 + 1011.052 - 58.8610.62 = 0 Ans.

NA = 16.544 kN = 16.5 kN + c ©Fy = 0;

A 0.6 m

Equations of Equilibrium: The normal reactions acting on the wheels at (A and B) are independent as to whether the wheels are locked or not. Hence, the normal reactions acting on the wheels are the same for both cases. a + ©MB = 0;

NB + 16.544 - 58.86 = 0 Ans.

NB = 42.316 kN = 42.3 kN

When both wheels at A and B are locked, then 1FA2max = msNA = 0.4116.5442 = 6.6176 kN and 1FB2max = msNB = 0.4142.3162 = 16.9264 kN. Since 1FA2max + FB max = 23.544 kN 7 10 kN, the wheels do not slip. Thus, the mine car does not move. Ans.

Ans: NA = 16.5 kN NB = 42.3 kN It does not move. 744

*8–4. The winch on the truck is used to hoist the garbage bin onto the bed of the truck. If the loaded bin has a weight of 8500 lb and center of gravity at G, determine the force in the cable needed to begin the lift. The coefficients of static friction at A and B are mA = 0.3 and mB = 0.2, respectively. Neglect the height of the support at A.

G A

10 ft

12 ft

SOLUTION a + ©MB = 0;

8500(12) - NA(22) = 0 NA = 4636.364 lb

+ ©F = 0; : x

T cos 30° - 0.2NB cos 30° - NB sin 30° - 0.3(4636.364) = 0 T(0.86603) - 0.67321 NB = 1390.91

+ c ©Fy = 0;

4636.364 - 8500 + T sin 30° + NB cos 30° - 0.2NB sin 30° = 0 T(0.5) + 0.766025 NB = 3863.636

Solving: Ans.

T = 3666.5 lb = 3.67 kip NB = 2650.6 lb

Ans: T = 3.67 kip 745

8–5. The automobile has a mass of 2 Mg and center of mass at G. Determine the towing force F required to move the car if the back brakes are locked, and the front wheels are free to roll. Take ms = 0.3.

F G

0.6 m

0.3 m

0.75 m

A 1m

1.50 m

SOLUTION Equations of Equilibrium: Referring to the FBD of the car shown in Fig. a, + ΣFx = 0; FB - F cos 30° = 0 S

(1)

+ c ΣFy = 0; NA + NB + F sin 30° - 2000(9.81) = 0

(2)

a + ΣMA = 0; F cos 30°(0.3) - F sin 30°(0.75) + (3)

NB (2.5) - 2000(9.81)(1) = 0 Friction: It is required that the rear wheels are on the verge to slip. Thus

(4)

FB = ms NB = 0.3 NB Solving Eqs. (1) to (4),

Ans.

F = 2,762.72 N = 2.76 kN NB = 7975.30 N

NA = 10, 263.34 N

FB = 2392.59 N

Ans: F = 2.76 kN 746

8–6. The automobile has a mass of 2 Mg and center of mass at G. Determine the towing force F required to move the car. Both the front and rear brakes are locked. Take ms = 0.3.

F G

0.6 m

0.3 m

0.75 m

A 1m

1.50 m

SOLUTION Equations of Equilibrium: Referring to the FBD of the car shown in Fig. a, + ΣFx = 0; FA + FB - F cos 30° = 0 S

(1)

+ c ΣFy = 0; F sin 30° + NA + NB - 2000(9.81) = 0

(2)

a + ΣMA = 0; F cos 30°(0.3) - F sin 30°(0.75) + NB (2.5) - 200(9.81)(1) = 0

(3)

Friction: It is required that both the front and rear wheels are on the verge to slip. Thus FA = ms NA = 0.3 NA

(4)

FB = ms NB = 0.3 NB

(5)

Solving Eqs. (1) to (5), Ans.

F = 5793.16 N = 5.79 kN NB = 8114.93 N

NA = 8608.49 N

FA = 2582.55 N

FB = 2434.48 N

Ans: F = 5.79 kN 747

8–7. The 100-lb cylinder rests between the two inclined planes. When P = 15 lb, the cylinder is on the verge of impending motion. Determine the coefficient of static friction between the surfaces of contact and the cylinder.

458

SOLUTION Equilibrium: + ΣFx = 0; S + c ΣFy = 0;

FA cos 45° + NA cos 45° + FB cos 45° - NB cos 45° = 0 FA + NA + FB - NB = 0 [1] 15 - 100 - FA sin 45° + NA sin 45° + FB sin 45° + NB sin 45° = 0

NA + FB + NB - FA = 120.21 [2]

a + ΣMO = 0;

FB (r) + FA (r) - 15(r) = 0 FB + FA = 15 [3] FA = mNA and FB = mNB [4]

Substituting Eq. [4] into [1], [2], and [3] and solving yields: NA = 51.66 lb

NB = 66.65 lb Ans.

m = 0.127

Ans: m = 0.127 748

*8–8. The bracket can move freely along the pole until a force P is placed on it. If the coefficient of static friction at A and B is ms = 0.6, determine the smallest distance d at which the force can be applied so that the bracket does not slip. Neglect the weight of the bracket.

8 in.

B d

SOLUTION + ΣFx = 0; NA - NB = 0 [1] S + c ΣFy = 0; 0.6NA + 0.6NB - P = 0 [2]

0.5 in.

a + ΣMB = 0; P(d - 0.25) - NA (8) + 0.6NA(0.5) = 0 [3] Solving Eqs. [1] to [3] yields: NA = NB = 0.8333P Ans.

d = 6.67 in.

Ans: d = 6.67 in. 749

P all copyright laws up the inclined ofPublished by Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under © 2022 by R. plane C. Hibbeler. as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. n angle f, show that for u), where u is the angle

8–9. The pipe of weight W is to be pulled up the inclined plane of slope a using a force P. If P acts at an angle f, show that for slipping P = W sin(a + u)>cos(f - u), where u is the angle of static friction; u = tan - 1 ms.

α α

SOLUTION N +cos P sinaf - W = 0 cos a = +a©F 0 y¿ =N0; = W Pcos sina f +Q©Fx¿ = 0;

N = W cos a - P sin f

P cos f - W sin a - tan u(W cos a - P sin f) = 0

- tan u(W cos a - P sin f) = 0 W(sin a + tan u cos a)

an u cos a) n u sin f

P = =

cos f + tan u sin f W sin(a + u) W(cos u sin a + sin u cos a) = cos f cos u + sin f sin u cos(f - u)

W sin(a + u) + sin u cos a) = cos(f - u) + sin f sin u

Q.E.D.

750

Q.E.D.

P he applied force P should all as possible pulling © 2022 by for R. C. Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. s the corresponding value φ slope a is known. Express 8–10. angle of kinetic friction, Determine the angle f at which the applied force P should act on the pipe so that P is as small as possible for pulling the pipe up the incline. What is the corresponding value of P? The pipe weighs W and the slope a is known. Express the answer in terms of the angle of kinetic friction, u = tan - 1 mk.

α α

W cos a =SOLUTION 0 N = W cos a - P sin f +a©Fy¿ = 0;

N + P sin f - W cos a = 0

+Q©Fx¿ = 0;

P cos f - W sin a - tan u (W cos a - P sin f) = 0

n a - tan u (W cos a - P sin f) = 0 P =

N = W cos a - P sin f

W(sin a + tan u cos a) cos f + tan u sin f

W(cos u sin a + sin u cos a) cos f cos u + sin f sin u

W sin (a + u) cos (f - u)

W sin (a + u) sin (f - u) dP = 0 = df cos2(f - u)

= 0

W sin (a + u) sin (f - u) = 0

W sin (a + u) = 0

sin (f - u) = 0

f = u

P =

f - u = 0

Ans.

W sin (a + u) = W sin (a + u) cos (u - u)

Ans.

W sin (a + u) = 0 f = u

Ans.

Ans: f = u P = W sin (a + u) 751

8–11. The man has a mass of 40 kg. He plans to scale the vertical crevice using the method shown. If the coefficient of static friction between his shoes and the rock is ms = 0.4 and between his backside and the rock, mb = 0.3, determine the smallest horizontal force his body must exert on the rocks in order to do this.

SOLUTION + c ΣFy = 0; 0.4F + 0.3F - 40(9.81) = 0

Ans.

F = 561 N

Ans: F = 561 N 752

*8–12. The block brake is used to stop the wheel from rotating when the wheel is subjected to a couple moment M0. If the coefficient of static friction between the wheel and the block is ms, determine the smallest force P that should be applied.

P a b C c

SOLUTION a + ©MC = 0;

Pa - Nb - ms Nc = 0 N =

c + ©MO = 0;

Pa (b + ms c)

ms Nr - M0 = 0 ms P a P =

a b r = M0 b + ms c

M0 (b + ms c) ms ra

Ans.

Ans: P = 753

M0 (b + ms c) msra

8–13. The brake is to be designed to be self locking, that is, it will not rotate when no load P is applied to it when the disk is subjected to a clockwise couple moment M 0. Determine the distance d of the lever that will allow this to happen. The coefficient of static friction at B is ms = 0.5.

A P d 1.5 ft

1.5 ft B M0

SOLUTION a + ©MO = 0;

M0 - 0.5 NB (1) = 0

1 ft

NB = 2M0 a + ©MA = 0;

P (3) - NB (1.5) + 0.5 NB (d) = 0 P =

2M0 (1.5 - 0.5 d) = 0 3

1.5 - 0.5d = 0 Ans.

d = 3 ft

Ans: d = 3 ft 754

8–14. The car has a mass of 1.6 Mg and center of mass at G. If the coefficient of static friction between the shoulder of the road and the tires is ms = 0.4, determine the greatest slope u the shoulder can have without causing the car to slip or tip over if the car travels along the shoulder at constant velocity.

2.5 ft G

B 5 ft

SOLUTION

Tipping: a + ©MA = 0;

- W cos u12.52 + W sin u12.52 = 0 tan u = 1 u = 45°

Slipping: Q + ©Fx = 0;

0.4 N - W sin u = 0

a + ©Fy = 0;

N - W cos u = 0 tan u = 0.4 u = 21.8°

Ans. (car slips before it tips)

Ans: u = 21.8° 755

8–15. l

The log has a coefficient of static friction of ms = 0.3 with the ground and a weight of 40 lb/ft. If a man can pull on the rope with a maximum force of 80 lb, determine the greatest length l of log he can drag.

80 lb A

SOLUTION Equations of Equilibrium: + c ©Fy = 0;

N - 40l = 0

N = 40l

+ ©F = 0; : x

4(80) - F = 0

F = 320 lb

Friction: Since the log slides, F = (F)max = ms N 320 = 0.3 (40l) Ans.

l = 26.7 ft

Ans: l = 26.7 ft 756

*8–16. Determine the torque M that must be applied by the engine to the rear wheels of the tractor to cause them to slip. The front wheels are free to roll. The tractor weighs 3500 lb and has a center of gravity at G. The coefficient of static friction between the rear wheels and the ground is ms = 0.5.

G 2 ft

M 2 ft A

5 ft

3 ft

SOLUTION For the tractor: + ΣFx = 0; T - 0.5NB = 0 [1] S a + ΣMA = 0; NB (8) - 3500(5) - T(2) = 0 [2] Solving Eqs. [1] and [2] yields: NB = 2500 lb

T = 1250 lb

Check if wheel A lifts off the ground. + c ΣFy = 0;

NA + 2500 - 3500 = 0

NA = 1000 lb

NA is a positive value, which indicates that wheel A still has contact with the ground. For the rear Wheels: a + ΣMO = 0; M - 0.5(2500)(2) = 0

M = 2500 lb # ft = 2.50 kip # ft

Ans.

Ans: M = 2.50 kip # ft 757

8–17. If the coefficient of static friction between the rear wheels of the tractor and the ground is ms = 0.6, determine if the rear wheels slip or the front wheels lift off the ground as the engine provides a torque M to the rear wheels. What is the torque needed to cause the motion? The front wheels are free to roll. The tractor weighs 2500 lb and has a center of gravity at G.

G 2 ft

M 2 ft A

5 ft

3 ft

SOLUTION For the tractor: + ΣFx = 0; T - 0.6NB = 0 [1] S a + ΣMA = 0; NB (8) - 2500(5) - T(2) = 0 [2] Solving Eqs. [1] and [2] yields: NB = 1838.2 lb

T = 1102.9 lb

Check if wheel A lifts off the ground. + c ΣFy = 0;

NA + 1838.2 - 2500 = 0

NA = 661.8 lb

NA is a positive value, which indicates that wheel A still has contact with the ground. For the rear Wheels: a + ΣMO = 0; M - 0.6(1838)(2) = 0

M = 2206 lb # ft = 2.21 kip # ft

Ans.

Ans: M = 2.21 kip # ft 758

8–18. The spool of wire having a weight of 300 lb rests on the ground at B and against the wall at A. Determine the force P required to begin pulling the wire horizontally off the spool. The coefficient of static friction between the spool and its points of contact is ms = 0.25.

3 ft O 1 ft

A P

SOLUTION Equations of Equilibrium: Referring to the FBD of the spool shown in Fig. a, + ΣFx = 0; P - NA - FB = 0 [1] S + c ΣFy = 0; NB - FA - 300 = 0 [2] a+ ΣMO = 0; P(1) - FB(3) - FA(3) = 0

[3]

Frictions: It is required that slipping occurs at A and B. Thus, FA = m NA = 0.25 NA [4] FB = m NB = 0.25 NB [5] Solving Eqs. [1] to [5], Ans.

P = 1350 lb NA = 1200 lb

NB = 600 lb

FA = 300 lb

FB = 150 lb

Ans: P = 1350 lb 759

8–19. The spool of wire having a weight of 300 lb rests on the ground at B and against the wall at A. Determine the normal force acting on the spool at A if P = 300 lb. The coefficient of static friction between the spool and the ground at B is ms = 0.35. The wall at A is smooth.

3 ft O 1 ft

A P

SOLUTION Equations of Equilibrium: Referring to the FBD of the spool shown in Fig. a, a + ΣMB = 0; NA(3) - 300(2) = 0 NA = 200 lb a + ΣMO = 0; 300(1) - FB(3) = 0

Ans.

FB = 100 lb

+ c ΣFy = 0; NB - 300 = 0 NB = 300 lb Friction: Since FB 6 (FB)max = ms NB = 0.35(300) = 105 lb, slipping will not occur at B. Thus, the spool will remain at rest.

Ans: NA = 200 lb 760

*8–20. The ring has a mass of 0.5 kg and is resting on the surface of the table. To move the ring, a normal force P from the finger is exerted on it. Determine its magnitude when the ring is on the verge of slipping at A. The coefficient of static friction at A is mA = 0.2 and at B, mB = 0.3.

B 60

75 mm A

SOLUTION FA = FB P cos 60° - FB cos 30° - FA = 0 NA - 0.5(9.81) - P sin 60° - FB sin 30° = 0 FA = 0.2 NA NA = 19.34 N FA = FB = 3.868 N Ans.

P = 14.4 N

(FB)max = 0.3(14.44) = 4.33 N 7 3.868 N (O.K!)

Ans: P = 14.4 N 761

8–21. The fork lift has a weight of 2400 lb and a center of gravity at G. If the rear wheels are powered, whereas the front wheels are free to roll, determine the maximum number of 300-lb crates the fork lift can push forward. The coefficient of static friction between the wheels and the ground is ms = 0.4, and between each crate and the ground msœ = 0.35.

G 2.5 ft A

B 3.50 ft

SOLUTION

1.25 ft

Fork lift: a + ©MB = 0;

2400 (3.50) - NA (4.75) = 0 NA = 1768.4 lb

+ ©F = 0; : x

0.4 (1768.4) - P = 0 P = 707.37 lb

Crate: + c ©Fy = 0;

NC - 300 = 0 NC = 300 lb

+ ©F = 0; : x

P¿ - 0.35 (300) = 0 P¿ = 105 lb

Thus 105 n = 707.37 Fork lift can push n = 6.74 crates Ans.

n = 6 crates

Ans: n = 6 crates 762

8–22. The cylinder is confined by the brake, where ms = 0.4. Determine the required compression in the spring in order to resist a torque of 800 N · m on the cylinder. The spring has a stiffness of k = 3 MN>m and an unstretched length of 60 mm. End C of member ABC slides freely along the smooth vertical guide.

k C B

250 mm

400 mm A

SOLUTION Equations of Equilibrium: Since the cylinder is required to be on the verge to rotate, FB = mBNB = 0.4NB. For the wheel, Fig. a a+ ΣMO = 0;

800 - 0.4NB(0.25) = 0

200 mm

NB = 8000N

Then FB = 0.4(8000) = 3200N. For member ABC, Fig. b, a+ ΣMA = 0;

8000(0.4) + 3200(0.05) - FSP(0.2) = 0

FSP = 16800N

The required compression of the spring is given by FSP = kd;

16800 = 3(106)d d = 5.60(10-3) m = 5.60 mm

Ans.

Ans: d = 5.60 mm 763

8–23. Determine the minimum force P needed to push the tube E up the incline. The tube has a mass of 75 kg and the roller D has a mass of 100 kg. The force acts parallel to the plane, and the coefficients of static friction at the contacting surfaces are mA = 0.3, mB = 0.25, and mC = 0.4. The roller and tube each have a radius of 150 mm.

E A

P 30

SOLUTION For roller D +Q ©Fx¿ = 0;

P - NA - FC - 100(9.81) sin 30° = 0

(1)

a + ©Fy¿ = 0;

Nc + FA - 100(9.81) cos 30° = 0

(2)

FA(0.15) - FC(0.15) = 0

(3)

+Q ©Fx¿ = 0;

NA - FB - 75(9.81) sin 30° = 0

(4)

+a©Fy¿ = 0;

NB - FA - 75(9.81) cos 30° = 0

(5)

a + ©MO = 0;

FA(0.15) - FB(0.15) = 0

(6)

a + ©MO = 0; For tube E

Assuming slipping occurs only at A. (7)

FA = 0.3NA Solving Eqs. (1) to (7) yields: NA = 525.54 N

NC = 691.91 N

NB = 794.84 N

FA = FB = FC = 157.66 N Ans.

P = 1173.7 N = 1.17 kN Since FB = 157.66 N 6 mBNB = 0.25(794.84) = 198.71 N and FC = 157.66 N 6 mCNC = 0.4(691.91) = 276.76 N, the assumption is OK.

Ans: P = 1.17 kN 764

*8–24.

The uniform thin pole has a weight of 30 lb and a length of 26 ft. If it is placed against the smooth wall and on the rough floor in the position d = 10 ft, will it remain in this position when it is released? The coefficient of static friction is ms = 0.3.

26 ft

SOLUTION SOLUTION a + ©MA = 0;

30 (5) - NB (24) = 0 NB = 6.25 lb

+ ©F = 0; : x

6.25 - FA = 0 d

FA = 6.25 lb + c ©Fy = 0;

NA - 30 = 0 NA = 30 lb

(FA)max = 0.3 (30) = 9 lb 7 6.25 lb Ans.

Yes, the pole will remain stationary.

Ans: Yes, the pole will remain stationary. 765

8–25. The uniform pole has a weight of 30 lb and a length of 26 ft. Determine the maximum distance d it can be placed from the smooth wall and not slip. The coefficient of static friction between the floor and the pole is ms = 0.3.

26 ft

SOLUTION + c ©Fy = 0;

NA - 30 = 0 A

NA = 30 lb FA = (FA)max = 0.3 (30) = 9 lb + ©F = 0; : x

NB - 9 = 0 NB = 9 lb

a + ©MA = 0;

30 (13 cos u) - 9 (26 sin u) = 0 u = 59.04° Ans.

d = 26 cos 59.04° = 13.4 ft

Ans: d = 13.4 ft 766

8–26. The truck has a mass of 1.25 Mg and a center of mass at G. Determine the greatest load it can move if (a) the truck has rear-wheel drive while the front wheels are free to roll, and (b) the truck has four-wheel drive. The coefficient of static friction between the wheels and the ground is mw = 0.5, and between the crate and the ground, it is mc = 0.4.

800 mm G 600 mm

B 1.5 m

SOLUTION For crate: + c ΣFy = 0; N - W = 0 + ΣFx = 0; T - 0.4W = 0 S

N=W T = 0.4W

(a) For two-wheel-drive truck + c ΣFy = 0; NA + NB - 1250(9.81) = 0 [1] + ΣFx = 0; 0.5NA - 0.4W = 0 [2] S a+ ΣMA = 0; NB(2.5) - 1250(9.81)(1.5) + 0.4W(0.6) = 0 [3] Solving Eqs. [1] to [3] yields: Ans.

W = 6967 N = 6.97 kN NB = 6689 N NA = 5574 N (b) For four-wheel-drive truck

+ c ΣFy = 0; NA + NB - 1250(9.81) = 0 [4] + ΣFx = 0; 0.5NB + 0.5NA - 0.4W = 0 [5] S a+ ΣMA = 0; NB(2.5) - 1250(9.81)(1.5) + 0.4W(0.6) = 0 [6] Solving Eqs. [4] to [6] yields: Ans.

W = 15328 N = 15.3 kN NB = 5886 N NA = 6376.5 N

Ans: W = 6.97 kN W = 15.3 kN 767

8–27. Solve Prob. 8–26 if the truck and crate are resting on an upward 10° incline.

800 mm G 600 mm

B 1.5 m

SOLUTION For crate: + a ΣFy′ = 0; N - W cos 10° = 0

N = 0.9848W

Q + ΣFx′ = 0; T - 0.4(0.9848W) - W sin 10° = 0

T = 0.5676W

(a) For two-wheel-drive truck + a ΣFy′ = 0; NA + NB - 1250(9.81) cos 10° = 0 [1] Q + ΣFx′ = 0; 0.5NA - 0.5676W -1250(9.81) sin 10° = 0 [2] a+ ΣMA = 0; NB(2.5) - 1250(9.81) cos 10° (1.5) + 1250(9.81) sin 10°(0.8) + 0.5676W(0.6) = 0 [3] Solving Eqs. [1] to [3] yields: Ans.

W = 1254 N = 1.25 kN NB = 6393 N NA = 5683 N (b) For four-wheel-drive truck

+ a ΣFy′ = 0; NA + NB - 1250(9.81) cos 10° = 0 [4] Q + ΣFx′ = 0; 0.5NA + 0.5NB - 0.5676W - 1250(9.81) sin 10° = 0 [5] a+ ΣMA = 0; NB(2.5) - 1250(9.81) cos 10° (1.5) + 1250(9.81) sin 10°(0.8) + 0.5676W(0.6) = 0 [6] Solving Eqs. [4] to [6] yields: Ans.

W = 6886 N = 6.89 kN NB = 5626 N NA = 6450 N

Ans: W = 1.25 kN W = 6.89 kN 768

*8–28.

0.5 m

The uniform crate resting on the dolly has a mass of 500 kg and mass center at G. If the front casters contact a high step, and the coefficient of static friction between the crate and the dolly is ms = 0.45, determine the greatest force P that can be applied without causing motion of the crate. The dolly does not move.

P 0.6 m

0.3 m 0.1 m A

SOLUTION

0.4 m

Slipping of crate on dolley: + ΣFx = 0; S

B 0.3 m

P = F = 0.45(4905) = 2207.25 N

Tipping of crate on dolley: a+ Σ MC = 0; - P(0.6) + 4905(0.20) = 0 Thus,

P = 1635 N Ans.

P = 1.64 kN

Ans: P = 1.64 kN 769

8–29. The friction pawl is pinned at A and rests against the wheel at B. It allows freedom of movement when the wheel is rotating counterclockwise about C. Clockwise rotation is prevented due to friction of the pawl which tends to bind the wheel. If 1ms2B = 0.6, determine the design angle u which will prevent clockwise motion for any value of applied moment M. Hint: Neglect the weight of the pawl so that it becomes a two-force member.

θ B M

SOLUTION

20°

Friction: When the wheel is on the verge of rotating, slipping would have to occur. Hence, FB = mNB = 0.6NB . From the force diagram (FAB is the force developed in the two force member AB) tan120° + u2 =

0.6NB = 0.6 NB Ans.

u = 11.0°

Ans: u = 11.0° 770

8–30.

Two blocks A and B have a weight of 10 lb and 6 lb, respectively. They are resting on the incline for which the coefficients of static friction are mA = 0.15 and mB = 0.25. Determine the incline angle u for which both blocks begin to slide. Also find the required stretch or compression in the connecting spring for this to occur. The spring has a stiffness of k = 2 lb>ft.

2 lb/ft

A u

2x + FA - 10 sin u = 0

(1)

a+ ©Fy¿ = 0;

NA - 10 cos u = 0

(2)

+ Q©Fx¿ = 0;

FB - 2x - 6 sin u = 0

(3)

a+ ©Fy¿ = 0;

NB - 6 cos u = 0

(4)

From FBD (b),

Friction: If block A and B are on the verge to move, slipping would have to occur at point A and B. Hence. FA = msA NA = 0.15NA and FB = msB NB = 0.25NB. Substituting these values into Eqs. (1), (2),(3) and (4) and solving, we have u = 10.6° NA = 9.829 lb

Ans.

x = 0.184 ft NB = 5.897 lb

Ans: u = 10.6° x = 0.184 ft 771

8–31. Two blocks A and B have a weight of 10 lb and 6 lb, respectively. They are resting on the incline for which the coefficients of static friction are mA = 0.15 and mB = 0.25. Determine the angle u which will cause motion of one of the blocks. What is the friction force under each of the blocks when this occurs? The spring has a stiffness of k = 2 lb>ft and is originally unstretched.

2 lb/ft

A u

SOLUTION Equations of Equilibrium: Since neither block A nor block B is moving yet, the spring force Fsp = 0. From FBD (a), +Q©Fx¿ = 0;

FA - 10 sin u = 0

(1)

a+ ©Fy¿ = 0;

NA - 10 cos u = 0

(2)

+Q©Fx¿ = 0;

FB - 6 sin u = 0

(3)

a+ ©Fy¿ = 0;

NB - 6 cos u = 0

(4)

From FBD (b),

Friction: Assuming block A is on the verge of slipping, then (5)

FA = mA NA = 0.15NA Solving Eqs. (1),(2),(3),(4), and (5) yields u = 8.531°

NA = 9.889 lb

FB = 0.8900 lb

FA = 1.483 lb

NB = 5.934 lb

Since (FB)max = mB NB = 0.25(5.934) = 1.483 lb 7 FB, block B does not slip. Therefore, the above assumption is correct. Thus u = 8.53°

FA = 1.48 lb

Ans.

FB = 0.890 lb

Ans: u = 8.53° FA = 1.48 lb FB = 0.890 lb 772

*8–32. Determine the smallest force P that must be applied to begin moving the 150-lb uniform crate. The coefficent of static friction between the crate and the floor is ms = 0.5.

2 ft P

3 ft

SOLUTION Equations of Equilibrium: Referring to the FBD of the crate shown in Fig. a, + ΣFx = 0; F - P = 0 S + c ΣFy = 0; N - 150 = 0

(1) N = 150 lb

a+ ΣMO = 0; P(3) - 150x = 0

(2)

Friction: Assuming that the crate slides before tipping. Thus F = m N = 0.5(150) = 75 lb Substitute this value into Eq. (1) P = 75 lb Then Eq. (2) gives 75(3) - 150x = 0

x = 1.5 ft

Since x > 1 ft, the crate tips before sliding. Thus, the assumption was wrong. Substitute x = 1 ft into Eq. (2), P(3) - 150(1) = 0 Ans.

P = 50 lb

Ans: 50 lb 773

8–33. The man having a weight of 200 lb pushes horizontally on the crate. If the coefficient of static friction between the 450-lb crate and the floor is ms = 0.3 and between his shoes and the floor m′s = 0.6, determine if he can move the crate.

2 ft P

3 ft

SOLUTION Equations of Equilibrium: Referring to the FBD of the crate shown in Fig. a, + ΣFx = 0; FC - P = 0 S + c ΣFy = 0; NC - 450 = 0

(1) NC = 450 lb

a+ ΣMO = 0; P(3) - 450(x) = 0

(2)

Also, from the FBD of the man, Fig. b, + ΣFx = 0; P - Fm = 0 S + c ΣFy = 0; Nm - 200 = 0

(3) Nm = 200 lb

Friction: Assuming that the crate slides before tipping. Thus FC = ms NC = 0.3(450) = 135 lb Using this result to solve Eqs. (1), (2) and (3) Fm = P = 135 lb

x = 0.9 ft

Since x < 1 ft, the crate indeed slides before tipping as assumed. Also, since Fm > (Fm)max = ms ′NC = 0.6(200) = 120 lb, the man slips. Thus he is not able to move the crate.

Ans: No 774

8–34. The uniform hoop of weight W is subjected to the horizontal force P. Determine the coefficient of static friction between the hoop and the surface of A and B if the hoop is on the verge of rotating.

B B

SOLUTION Equations of Equilibrium: Referring to the FBD of the hoop shown in Fig. a, + ΣFx = 0; P + FA - NB = 0 S

(1)

+ c ΣFy = 0; NA + FB - W = 0

(2)

a+ ΣMA = 0; NB(r) + FB (r) - P(2r) = 0

(3)

Friction: It is required that slipping occurs at point A and B. Thus FA = ms NA

(4)

FB = ms NB

(5)

Substituting Eq. (5) into (3), NB r + ms NB r = 2Pr

NB =

2P 1 + ms

(6)

Substituting Eq. (4) into (1) and Eq. (5) into (2), we obtain NB - ms NA = P

(7)

NA + ms NB = W

(8)

Eliminate NA from Eqs. (7) and (8), NB =

P + msW 1 + ms2

(9)

Equating Eq. (6) and (9) P + msW 2P = 1 + ms 1 + m2s 2P(1 + m2s ) = (P + msW)(1 + ms) 2P + 2m2s P = P + Pms + msW + m2s W (2P - W)m2s - (P + W)ms + P = 0

If P =

1 W, the quadratic term drops out, and then 2 P ms = P + W = 1

1 2W

2W + W

Ans.

= 0.333

775

8–34. Continued

If P ≠

1 W, then 2 ms =

(P + W) { 2[ - (P + W)]2 - 4(2P - W)P

ms =

(P + W) { 2W 2 + 6PW - 7P2

ms =

(P + W) { 2(W + 7P)(W - P)

2(2P - W)

In order to have a solution, (W + 7P)(W - P) 7 0 Since W + 7P > 0 then W - P>0

W>P

Also, P > 0. Thus 0 6 P 6 W Choosing the smaller value of ms, ms =

(P + W) - 2(W + 7P)(W - P) 2(2P - W)

for 0 6 P 6 W and P ≠

W Ans. 2

1 1 W and P ≠ W, are continuous. 2 2 Note: Choosing the larger value of ms in the quadratic solution leads to NA, FA < 0, which is nonphysical. Also, (ms)max = 1. For ms > 1, the hoop will tend to climb the wall rather than rotate in place. The two solutions, for P =

Ans: ms = 0.333 776

8–35. Determine the maximum horizontal force P that can be applied to the 30-lb hoop without causing it to rotate. The coefficient of static friction between the hoop and the surfaces A and B is ms = 0.2. Take r = 300 mm.

B B

SOLUTION Equations of Equilibrium: Referring to the FBD of the hoop shown in Fig. a, + ΣFx = 0; P + FA - NB = 0 S

(1)

+ c ΣFy = 0; NA + FB - 30 = 0

(2)

a+ ΣMA = 0; FB(0.3) + NB(0.3) - P(0.6) = 0

(3)

Friction: Assuming that the hoop is on the verge to rotate due to the slipping occur at A and B. Then FA = ms NA = 0.2 NA

(4)

FB = ms NB = 0.2 NB

(5)

Solving Eq. (1) to (5) NA = 27.27 lb

NB = 13.64 lb

FA = 5.455 lb

FB = 2.727 lb Ans.

P = 8.182 lb = 8.18 lb

Since NA is positive, the hoop will be in contact with the floor. Thus, the assumption was correct.

Ans: P = 8.18 lb 777

*8–36. Determine the minimum force P needed to push the tube E up the incline. The coefficients of static friction at the contacting surfaces are mA = 0.2, mB = 0.3, and mC = 0.4. The 100-kg roller and 40-kg tube each have a radius of 150 mm.

E A

SOLUTION Equations of Equilibrium: Referring to the FBD of the roller, Fig. a, + ΣFx = 0; P - NA cos 30° - FA sin 30° - FC = 0 S

(1)

+ c ΣFy = 0; NC + FA cos 30° - NA sin 30° - 100(9.81) = 0

(2)

a + ΣMD = 0; FA(0.15) - FC (0.15) = 0

(3)

Also, for the FBD of the tube, Fig. b, +QΣFx = 0; NA - FB - 40(9.81) sin 30° = 0

(4)

+a ΣFy = 0; NB - FA - 40(9.81) cos 30° = 0

(5)

a + ΣME = 0; FA(0.15) - FB(0.15) = 0

(6)

Friction: Assuming that slipping is about to occur at A. Thus (7)

FA = mA NA = 0.2 NA Solving Eqs. (1) to (7)

Ans.

P = 285.97 N = 286 N NA = 245.25 N

NB = 388.88 N

NC = 1061.15 N

FA = FB = FC = 49.05 N

Since FB 6 (FB)max = mB NB = 0.3(388.88) = 116.66 N and FC < (FC)max = mC NC = 0.4(1061.15) = 424.46 N, slipping indeed will not occur at B and C. Thus, the assumption was correct.

Ans: 286 N 778

8–37.

0.2 m C

The double-block brake mechanism is used to prevent the wheel from turning when the wheel is subjected to the torque of M = 5 N # m. If the coefficient of static friction between the blocks and the wheel is ms = 0.8, determine the smallest vertical force P applied to the handle needed to stop the wheel.

0.1 m D

M 0.1 m

0.4 m

0.2 m I O

0.4 m

J L

A 0.5 m

SOLUTION

0.15 m

Wheel: a + ΣMO = 0;

5 - (0.8NH + 0.8NI)0.2 = 0 (1)

NH + NI = 31.25 Member BK: a + ΣMB = 0;

-NI(0.4) + FKC(0.8) = 0

FKC = 0.5 NI Member AD: a + ΣMA = 0;

NH(0.4) - 0.7 Dx = 0

Dx = 0.57143 NH Member EDC: + ΣFx = 0; 0.5NI - 0.57143 NH = 0 S a + ΣMD = 0; FEJ(0.2) - 0.5NI(0.1) = 0 Using Eq.(1) and solving: NI = 16.667 N NH = 14.5833 N FEJ = 4.1667 N Member LJ: a + ΣMC = 0; P(0.65) - 4.1667(0.15) = 0 Ans.

P = 0.962 N

Ans: P = 0.962 N 779

8–38. Solve Prob. 8–37 if the torque is M = 5 N # m clockwise. 0.2 m C

0.1 m D

M 0.1 m

0.4 m

0.2 m I O

0.4 m

J L

A 0.5 m

0.15 m

SOLUTION Wheel: a + ΣMO = 0;

0.8NH(0.2) + 0.8NI(0.2) - 5 = 0 (1)

NH + NI = 31.25 Member BK: a + ΣMB = 0;

- NI(0.4) + FKC(0.8) = 0

FKC = 0.5 NI Member AD: a + ΣMA = 0;

NH(0.4) + 0.7 Dx = 0

Dx = 0.57143 NH Member EDC: + ΣFx = 0; 0.5NI - 0.57143 NH = 0 S a + ΣMD = 0; FEJ(0.2) - 0.5NI(0.1) = 0 Using Eq. (1) and solving: NI = 16.667 N NH = 14.5833 N FEJ = 4.1667 N Member LJ: a + ΣMj = 0; P(0.65) - 4.1667(0.15) = 0 Ans.

P = 0.962 N

Ans: P = 0.962 N 780

8–39. The car has a weight of 4000 lb and a center of gravity at G. If it pulls off the side of the road, determine the greatest angle of tilt u it can have without slipping or tripping over. The coefficient of static friction between its wheels and the ground is ms = 0.4.

2 ft

u 4 ft 4 ft

SOLUTION Equilibrium: + R ΣFx′ = 0; 4000 sin u - FA - FB = 0

[1]

Q + ΣFy′ = 0; NA + NB - 4000 cos u = 0

[2]

a+ ΣMA = 0; NB(8) - 4000 sin u(2) - 4000 cos u(4) = 0

[3]

Assume slipping occurs. Therefore FA = 0.4NA and FB = 0.4NB

[4]

Substituting Eq. [4] into [1], [2], and [3] and solving yields: NA = 1485.6 lb

NB = 2228.3 lb Ans.

u = 21.8°

Since NA is a positive value, the tire indeed has the contact at A. Therefore, the assumption is right.

Ans: u = 21.8° 781

*8–40. The 800-lb concrete pipe is being lowered from the truck bed when it is in the position shown. If the coefficient of static friction at the points of support A and B is ms = 0.4, determine where it begins to slip first: at A or B, or at both A and B.

15 in. 308 A

908

B 30 in.

SOLUTION

5 in.

ΣFx = 0; NA + FB - 800 sin 30° = 0 ΣFy = 0; FA + NB - 800 cos 30° = 0 a+ ΣMO = 0; FB(15) - FA(15) = 0 FA = FB Assume slipping at A: FA = 0.4 NA Thus, NA = 285.71 lb NB = 578.53 lb FA = FB = 114.29 lb At B: (O.K!)

(FB)max = 0.4 NB = 0.4(578.53) = 231.4 lb 7 114.29 lb

Ans

Thus, slipping occurs at A.

782

18 in.

8–41. If the coefficient of static friction at A and B is ms = 0.6, determine the maximum angle u so that the frame remains in equilibrium, regardless of the mass of the cylinder. Neglect the mass of the rods.

SOLUTION Free-Body Diagram : Due to the symmetrical loading and system, ends A and B of the rod will slip simultaneously. Since end B is on the verge of sliding to the right, the friction force FB must act to the left such that FB = msNB = 0.6NB as indicated on the free-body diagram shown in Fig. a.

NB - FBC cos u = 0

+ ©Fx = 0; :

FBC sin u - 0.6(FBC cos u) = 0

NB = FBC cos u

tan u = 0 .6 Ans.

u = 31.0°

Ans: u = 31.0° 783

8–42. The 100-kg disk rests on a surface for which mB = 0.2. Determine the smallest vertical force P that can be applied tangentially to the disk which will cause motion to impend.

A 0.5 m B

SOLUTION Equations of Equilibrium: Referring to the FBD of the disk shown in Fig. a, + c ΣFy = 0; NB - P - 100(9.81) = 0

(1)

a + ΣMA = 0; P(0.5) - FB(1) = 0

(2)

Friction: It is required that slipping impends at B.Thus, (3)

FB = mB NB = 0.2 NB Solving Eqs. (1), (2) and (3)

Ans.

P = 654 N NB = 1635 N

FB = 327 N

Ans: P = 654 N 784

8–43. The tongs are used to suspend the 80-kg plate. Determine the smallest coefficient of static friction to prevent slipping of the plate.

150 mm

150 mm A

SOLUTION

200 mm

Equation of Equilibrium. The required friction can be determined by considering the equilibrium of the plate, Fig. a + c ΣFy = 0;

2Ff - 80(9.81) = 0

Ff = 392.4 N

20 mm

Next, consider the equilibrium of the hook, Fig. b. Here the suspending force must be equal to the weight of the plate for equilibrium, ie;

B D

150 mm 150 mm

P = 80(9.81) = 784.8N + ΣFx = 0; FAB a 3 b - FAC a 3 b = 0 S 5 5 4 + c ΣFy = 0; 784.8 - 2c F a b d = 0 5

FAB = FAC = F F = 490.5 N

Finally, write the moment equation of equilibrium about point D by referring to the FBD of the tong, Fig. c, a+ ΣMD = 0;

4 3 490.5 a b(0.15) + 490.5 a b(0.15) + 392.4(0.02) - N(0.15) = 0 5 5 N = 739.02N

Friction: For the plate not to slip, Ff … msN 392.4 … ms(739.02) Ans.

Therefore, ms = 0.531

Ans: ms = 0.531 785

*8–44.

The 50-lb board is placed across the channel and a 100-lb girl attempts to walk across. If the coefficient of static friction at A and B is ms = 0.4, determine if she can make the crossing; and if not, how far will she get from A before the board slips? A

d 10 ft 5 3 4

4 3 0.4 NA + 0.4 NB a b - NB a b = 0 5 5

(1)

+ c ©Fy = 0;

4 3 NA - 100 - 50 + NB a b +0.4 NB a b = 0 5 5

(2)

a + ©MB = 0;

(3)

50(5) + 100 (10 - d) - NA (10) = 0

Solving, NA = 60.3 lb NB = 86.2 Ib d = 6.47 ft 6 10 ft Board will slip.

Ans.

Ans: Board will slip. 786

8–45. The man has a mass of 60 kg and the crate has a mass of 100 kg. If the coefficient of static friction between his shoes and the ground is ms = 0.4 and between the crate and the ground is mc = 0.3, determine if the man is able to move the crate using the rope-and-pulley system shown. 458 A 308

SOLUTION For the crate: + ΣFx = 0; T cos 30° - 0.3NC = 0 [1] S + c ΣFy = 0; NC + T sin 30° - 100(9.81) = 0 [2] Solving Eqs. [1] and [2] yields: T = 289.66 N

NC = 836.17 N

For the man: + ΣFx = 0; 289.66 cos 45° - FM = 0 S

FM = 204.82 N

+ c ΣFy = 0; NM + 289.66 sin 45° - 60(9.81) = 0

NM = 383.78 N

Since FM = 204.82 N 7 0.4(383.78) = 153.51 N, the man slips. Consequently, he can not move the crate.

Ans: He can not move the crate. 787

8–46. The uniform cylindrical tank has a mass of 250 kg. If the coefficient of static friction is ms = 0.15, determine the force P applied to the rope needed to move the tank.

458 3m P

SOLUTION

Equations of Equilibrium: Assuming that slipping occurs before tipping, then Ff = msN = 0.15N. Refering to the FBD of the tank, Fig. a + ΣFx = 0; P sin 45° - 0.15 N = 0 (1) S + c ΣFy = 0; N - P cos 45° - 250(9.81) = 0 (2) a+ ΣMA = 0; [250 (9.81)]x - P sin 45°(3) - P cos 45°(1 - x) = 0 (3) Solving Eqs. 1 through 3, N = 2885.29 N

x = 0.8571 m

P = 612.06 N = 0.6 m

Ans.

Since x < 1 m, the tank slips before it tips as assumed.

Ans: P = 0.6 m 788

8–47. A chain having a length of 20 ft and weight of 8 lb>ft rests on a street for which the coefficient of static friction is ms = 0.2. If a crane is used to hoist the chain, determine the force P it applies to the chain if the length of chain remaining on the ground begins to slip when the horizontal component of P becomes 10 lb. What length of chain remains on the ground?

- 10 + 0.2 NC = 0 NC = 50 lb

+ c ©Fy = 0;

Py - 160 + 50 = 0 Py = 110 lb P = 2(10)2 + (110)2 = 110 lb

Ans.

The length on the ground is supported by NC = 50 lb, thus L =

50 = 6.25 ft 8

Ans.

Ans: P = 110 lb L = 6.25 ft 789

*8–48. Determine the smallest couple moment which can be applied to the 20-lb wheel that will cause impending motion. The cord is attached to the 30-lb block, and the coefficients of static friction are mB = 0.2 and mD = 0.3.

1.5 ft C M 1.5 ft B

SOLUTION

3 ft

NB - 20 = 0

+ ©F = 0; : x

T - 0.2(20) = 0

NB = 20 lb T = 4 lb

a + ©MB = 0;

M - 4(3) = 0

M = 12 lb # ft

+ c ©Fy = 0;

ND - 30 = 0

ND = 30 lb

+ ©F = 0; : x

FD - 4 = 0

FD = 4 lb

Ans.

For block

a + ©MO = 0;

4(3) - 30(x) = 0

x = 0.4 ft

Since FD = 4 lb 6 mND = 0.3(30) = 9 lb and x = 0.4 ft 6 0.75 ft, neither slipping not tipping of the block occurs.

Ans: M = 12 lb # ft 790

8–49. The crate has a weight of 200 lb and a center of gravity at G. Determine the horizontal force P required to tow it. Also, find the location of the resultant normal force measured from A. Take h = 4 ft and ms = 0.4.

3.5 ft P G

P - FO = 0

+ c ©Fy = 0;

NO = 200 lb

3 ft

- P (4) + 200(x) = 0 2 ft

FO = 0.4 NO

2 ft

Ans.

P = 80 lb x = 1.6 ft The distance of NO from A is

Ans.

2 - 1.6 = 0.4 ft

Ans: P = 80 lb d = 0.4 ft 791

8–50. The crate has a weight of 200 lb and a center of gravity at G. Determine the height h of the tow rope so that the crate slips and tips at the same time. What horizontal force P is required to do this? Take ms = 0.4.

3.5 ft P G

NA = 200 lb

+ ©F = 0; : x

P - FA = 0

Fs = ms N;

FA = 0.4 (200) = 80 lb

Ans.

3 ft

2 ft

- 80 h + 200 (2) = 0 Ans.

h = 5 ft

Ans: P = 80 lb h = 5 ft 792

8–51. Beam AB has a negligible mass and thickness, and supports the 200-kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the minimum force P needed to move the post. The coefficients of static friction at B and C are mB = 0.4 and mC = 0.2, respectively.

B 1.5 m

1.5 m 1m 0.75 m

SOLUTION Equations of Equilibrium: Referring to the FBD of member AB shown in Fig. a, a + ΣMA = 0; NB (3) - 200(9.81)(1.5) = 0

NB = 981 N

Then consider the FBD of member BC shown in Fig. b, 3 + c ΣFy = 0; NC + P a b - 981 - 20(9.81) = 0 5

(1)

4 a + ΣMC = 0; FB (1.75) - P a b(0.75) = 0 5

(2)

4 a + ΣMB = 0; P a b(1) - FC (1.75) = 0 5

(3)

Friction: Assuming that slipping occurs at C. Then (4)

FC = mC NC = 0.2 NC Solving Eqs. (1) to (4)

Ans.

P = 407.94 N = 408 N NC = 932.44 N

FC = 186.49 N

FB = 139.87

Since FB 6 (FB)max = mB NB = 0.4(981) N = 392.4. Indeed slipping will not occur at B. Thus, the assumption is correct.

Ans: P = 408 N 793

*8–52. Beam AB has a negligible mass and thickness, and supports the 200-kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to P = 300 N, the post slips at both B and C simultaneously.

B 1.5 m

1.5 m 1m 0.75 m

SOLUTION Equations of Equilibrium: Referring to the FBD of member AB shown in Fig. a, a+ ΣMA = 0; NB(3) - 200(9.81)(1.5) = 0 NB = 981 N Then consider the FBD of member BC shown in Fig. b, 3 + c ΣFy = 0; NC + 300 a b - 981 - 20(9.81) = 0 NC = 997.2 N 5

4 a + ΣMC = 0; FB (1.75) - 300 a b(0.75) = 0 FB = 102.86 N 5

4 a + ΣMB = 0; 300 a b(1) - FC (1.75) = 0 5

FC = 137.14 N

Friction: It is required that slipping occurs at B and simultaneously. Then FB = mB NB; 102.86 = mB (981)

mB = 0.1048 = 0.105

Ans.

FC = mC NC; 137.14 = mC (997.2)

mC = 0.1375 = 0.138

Ans.

Ans: mB = 0.105 mC = 0.138 794

8–53. Determine the smallest couple moment that can be applied to the 150-lb wheel that will cause impending motion. The uniform concrete block has a weight of 300 lb. The coefficients of static friction are mA = 0.2, mB = 0.3, and between the concrete block and the floor, m = 0.4.

1 ft

5 ft

M 1.5 ft A

SOLUTION Equations of Equilibrium: Referring to the FBD of the concrete block, Fig. a. + ΣFx = 0; FC - NB = 0 S

(1)

+ c ΣFy = 0; NC - FB - 300 = 0

(2)

a + ΣMO = 0; NB (1.5) - 300x - FB (0.5 + x) = 0

(3)

Also, from the FBD of the wheel, Fig. b. + ΣFx = 0; NB - FA = 0 S

(4)

+ c ΣFy = 0; NA - FB - 150 = 0

(5)

a + ΣMA = 0; M - NB(1.5) - FB(1.5) = 0

(6)

Friction: Assuming that the impending motion is caused by the rotation of wheel due to the slipping at A and B. Thus, FA = mANA = 0.2NA

(7)

FB = mBNB = 0.3NB

(8)

Solving Eqs. (1) to (8), NA = 141.51 lb

FA = 28.30 lb

NB = 28.30 lb

NC = 308.49 lb

FC = 28.30 lb

x = 0.1239 ft

FB = 8.491 lb

M = 55.19 lb # ft = 55.2 lb # ft

Ans.

Since FC 6 (FC)max = mC NC = 0.4(308.49) = 123.40 lb, and x < 0.5 ft, the c oncrete block will not slide or tip. Also, NA is positive, so the wheel will be in contact with the floor. Thus, the assumption was correct.

Ans: M = 55.2 lb # ft 795

8–54. Two blocks A and B, each having a mass of 6 kg, are connected by the linkage shown. If the coefficients of static friction at the contacting surfaces are mB = 0.8 and mA = 0.2, determine the largest vertical force P that may be applied to pin C without causing the blocks to slip. Neglect the weight of the links.

P C

308

308 A

308

SOLUTION Joint C: + ΣFx = 0; FAC cos 30° = FBC sin 30° S + c ΣFy = 0; FAC sin 30° + FBC cos 30° - P = 0 FAC = 0.5 P FBC = 0.866 P Block B: Assume slipping at B,

FB = 0.8 NB

Q + ΣFx = 0;

NB - 0.866P cos 60° - 6(9.81) cos 30° = 0

+ a ΣFy = 0;

FB - 6(9.81) sin 30°- 0.866P sin 60° = 0

Block A:

P = 28.1 N

+ ΣFx = 0; FA - 0.5(28.12) cos 30° = 0 S FA = 12.18 N + c ΣFy = 0; NA - 6(9.81) - 0.5(28.12) sin 30° = 0 NA = 65.89 N (FB)max = 0.2(65.89) = 13.18 N > 12.18 N Thus,

(O.K!) Ans.

P = 28.1 N

Ans: P = 28.1 N 796

8–55. Two blocks A and B, each having a mass of 6 kg, are connected by the linkage shown. If the coefficient of static friction at the contacting surfaces is ms = 0.5, determine the largest vertical force P that may be applied to pin C without causing the blocks to move. Neglect the weight of the links.

FAC sin 30° - P = 0;

+ ©F = 0; : x

2P (cos 30°) - FBC = 0;

FAC = 2P FBC = 1.732 P

FA - 2 P cos 30° = 0

+ c ©Fy = 0;

NA - 6(9.81) - 2 P sin 30° = 0

FB - 1.732 P cos 30° + 6(9.81) sin 30° = 0

+a ©Fy = 0;

NB - 6(9.81) cos 30° - 1.732 P sin 30° = 0

Assume slipping occurs at A, then FA = 0.5 NA Solving, FA = 41.4 N, FB = 6.40 N NA = 82.7 N, NB = 71.7 N Ans.

P = 23.9 N Since (FB)max = 0.5 (71.6) = 35.8 N 7 6.40 N Block A will slip first, B remains stationary.

Ans: P = 23.9 N 797

*8–56. The uniform 5-kg rod rests on the ground at B and against the wall at A. If the rod does not slip at B, determine the smallest coefficient of static friction at A which will prevent slipping at A.

z A 4m

3m y

1m x

SOLUTION Σ(MB)Z = 0; FA a

4 217

(mNAx) a

b(3) - NAx(1) = 0 Since slipping occurs FA = mNAx 4

217 m = 0.344

b(3) - NAx(1) = 0

Ans.

Ans: m = 0.344 798

ic friction is ms = 0.3 at C, eWpole start toC.slip. 2022 by R. Hibbeler. andwill is©lowered slowlyPublished by Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected underBall copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 4m ward the horizontal using ic friction is ms = 0.3 at C, 8–57. A θ e pole will start to slip. The uniform pole has a weight W and is lowered slowly from a vertical position u = 90° toward the horizontal using cable AB. If the coefficient of static friction is ms = 0.3 at C, determine the angle u at which the pole will start to slip.

+ T sin f (2) = 0

f = 0 u

f = 0

fu = 0

W cos u sin f

ms N - T cos f = 0 W cos u tan f

N - W - T sin f = 0

os u)

N = W(1 + cos u)

f = 0 tan f = cos u

ms (1 + cos u) tan f = cos u

os u)

- W (2) cos u + T sin f (2) = 0

CA 2m

θ 2m

SOLUTION

From geometry tan f =

4 sin u 2 + 4 cos u

tan f = cos u 4 ms sin u (1 + cos u) = cos u (2 + 4 cos u) nu cos u Set ms = 0.3, solving for u,

ncos u u) = cos u (2 + 4 cos u) cos u

Ans.

u = 65.2°

cos u) = cos u (2 + 4 cos u) = 65.2°

Ans.

= 65.2°

Ans.

Ans: u = 65.2° 799

8–58. The blocks each have a weight of 50 lb. If the coefficient of static friction at A is ms = 0.2 and between each block m′s = 0.4, determine how many blocks can be stacked as shown before they begin to topple.

208

208 A

SOLUTION + ΣFx = 0; FB cos 20° - NB sin 20° = 0 S + c ΣFy = 0; NB cos 20° + FB sin 20° - 50n = 0 NB = 46.98n FB = 17.10n (FB)max = 0.4(46.98n) = 18.79n > 17.10n

(O.K!)

For the whole stack No slipping can occur at A. Ans.

Thus, any number of blocks can be stacked.

Ans: Any number of blocks can be stacked. 800

8–59. If the coefficient of static friction between the axe and the wood is ms = 0.2, determine the smallest angle u of the blade which will cause the axe to be self-locking. Neglect the weight of the axe.

SOLUTION u u + c ΣFy = 0; 2 aN sin b - 2 a0.2N cos b = 0 2 2 u tan = 0.2 u = 22.6° 2

Ans.

Ans: u = 22.6° 801

*8–60. Each block has a weight of 400 lb. Determine how far the force P can compress the spring until block B slips on block A. What is P for this to occur?

P k = 800 lb>ft

m9s 5 0.5

308

B A m s 5 0.2

SOLUTION Wedge B: + c ΣFy = 0;

12 5 N (0.5NB) - 400 - P sin 30° = 0 13 B 13

+ ΣFx = 0; 5 NB + 12 (0.5NB) - P cos 30° = 0 S 13 13 NB = 1651 lb Ans.

P = 1613 lb = 1.61 kip FB = 825.6 lb Wedge A: + c ΣFy = 0; NA + 825.6 a

5 12 b - 1651 a b - 400 = 0 13 13

NA = 1607 lb

+ ΣFx = 0; F + 0.2(1607) - 1651 a 5 b - 825.6 a 12 b = 0 S 13 13 F = 1076 lb x=

1076 = 1.34 ft 800

Ans.

Ans: P = 1.61 kip x = 1.34 ft 802

8–61. If the coefficient of static friction between all the surfaces of contact is ms, determine the force P that must be applied to the wedge in order to lift the brace that supports the load F. Neglect friction between the wall and the brace.

SOLUTION

(1)

P - NB - ms NA = 0 NA - ms NB - F = 0

(2)

+ ©F = 0; : x

P - ms NA - ms N¿ cos a - N¿ sin a = 0

(3)

+ c ©Fy = 0;

NA - N¿ cos a + ms N¿ sin a = 0

(4)

A a

Wedge A:

From Eqs. (3) and (4): N¿ = NA =

P - ms NA (ms cos a + sin a) P(1 - ms tan a)

(5)

(2 ms + (1 - m2s) tan a)

From Eqs. (1) and (2): NB = P - msNA NA =

Pms + F

(6)

(1 + m2s )

Combining Eqs. (5) and (6): P = F¢

(1 - m2s ) tan a + 2 ms 1 - 2 ms tan a - m2s

≤

Ans.

Ans: P = Fa 803

(1 - m2s ) tan a + 2ms 1 - 2ms tan a - m2s

8–62. If P = 250 N, determine the required minimum compression in the spring so that the wedge will not move to the right. Neglect the weight of A and B. The coefficient of static friction for all contacting surfaces is ms = 0.35. Neglect friction at the rollers.

15 kN/m

SOLUTION

Free-Body Diagram: The spring force acting on the cylinder is Fsp = kx = 15(103)x. Since it is required that the wedge is on the verge to slide to the right, the frictional force must act to the left on the top and bottom surfaces of the wedge and their magnitude can be determined using friction formula. (Ff)1 = mN1 = 0.35N1

(Ff)2 = 0.35N2

N1 - 15(103)x = 0

N1 = 15(103)x

N2 cos 10° - 0.35N2 sin 10° - 15(103)x = 0 N2 = 16.233(103)x

+ ©Fx = 0; :

250 - 5.25(103)x - 0.35316.233(103)x4cos 10° - 316.233(103)x4sin 10° = 0 Ans.

x = 0.01830 m = 18.3 mm

Ans: x = 18.3 mm 804

8–63. Determine the minimum applied force P required to move wedge A to the right. The spring is compressed a distance of 175 mm. Neglect the weight of A and B. The coefficient of static friction for all contacting surfaces is ms = 0.35. Neglect friction at the rollers.

k = 15 kN/m B

SOLUTION Equations of Equilibrium and Friction: Using the spring formula, Fsp = kx = 1510.1752 = 2.625 kN. If the wedge is on the verge of moving to the right, then slipping will have to occur at both contact surfaces. Thus, FA = msNA = 0.35NA and FB = msNB = 0.35NB. From FBD (a), + c ©Fy = 0;

NB - 2.625 = 0

10°

NB = 2.625 kN

NA cos 10° - 0.35NA sin 10° - 2.625 = 0 NA = 2.841 kN

+ ©F = 0; : x

P - 0.3512.6252 - 0.3512.8412 cos 10° - 2.841 sin 10° = 0 Ans.

P = 2.39 kN

Ans: P = 2.39 kN 805

*8–64. The wedge has a negligible weight and a coefficient of static friction ms = 0.35 with all contacting surfaces. Determine the angle u so that it is self-locking. This requires no slipping for any magnitude of the force P applied to the joint.

–– 2

SOLUTION Friction: When the wedge is on the verge of slipping, then F = mN = 0.35N. From the force diagram (P is the ‘locking’ force.), tan

0.35N u = = 0.35 2 N Ans.

u = 38.6°

Ans: u = 38.6° 806

8–65. If it takes a vertical force of 60,000 lb to separate the two parts at A and C, determine the force P the hydraulic cylinder must exert on the wedge in order to push it forward. The coefficient of static friction between the wedge and members AB and CD is ms = 0.2.

60,000 lb

7.58

C 60,000 lb

SOLUTION From FBD (a): + c ΣFy = 0;

N cos 3.75° - 0.2N sin 3.75° - 60000 = 0

N = 60927.4 lb

From FBD (b): + ΣFx = 0; S

2 3 60927.4 sin 3.75° + 0.2 (60927.4) cos 3.75° 4 - P = 0 P = 32288 lb = 32.3 kip

Ans.

Ans: P = 32.3 kip 807

8–66. The load bar of negligible weight is screwed in place between the walls of the truck bed in order to keep loads from shifting. Determine the torsional moment M that must be applied to the center of the bar so that it produces a compression in the bar of 80 lb. The square-threaded screw at A has a lead of 0.2 in. and a radius of 0.5 in., and the coefficient of static friction between the bar and the screw is ms = 0.3. Also, what amount of horizontal force F must be applied to the center of the bar to pull it out? The coefficient of static friction between the walls of the truck and the end pads is ms = 0.7.

B M

SOLUTION Since friction at two screws must be overcome, this requires

M = 2 3 W r tan (u + f) 4

Where W = 80 lb, f = f s = tan - 1 ms = tan - 1 (0.3) = 16.70°, u = tan - 1 (L/2pr) = tan - 1 a

0.2 b = 3.643°. 2p(0.5)

M = 2 3 80 (0.5) tan (3.643° + 16.70°) 4 = 29.7 lb.in

Ans.

F = 2 3 0.7 (80) 4 = 112 lb

Ans.

Ans: M = 29.7 lb.in F = 112 lb 808

8–67. If the clamping force at G is 900 N, determine the horizontal force F that must be applied perpendicular to the handle of the lever at E. The mean diameter and lead of both single square-threaded screws at C and D are 25 mm and 5 mm, respectively. The coefficient of static friction is ms = 0.3.

200 mm G

200 mm A

SOLUTION

D E

Referring to the free-body diagram of member GAC shown in Fig. a, we have FCD = 900 N ©MA = 0; FCD(0.2) - 900(0.2) = 0 L b = Since the screw is being tightened, Eq. 8–3 should be used. Here, u = tan-1 a 2pr 5 -1 tan c d = 3.643°; 2p(12.5)

125 mm

fs = tan - 1 ms = tan - 1(0.3) = 16.699°; and M = F(0.125). Since M must overcome the friction of two screws, M = 2[Wr tan(fs + u)] F(0.125) = 2 [900(0.0125)tan(16.699° + 3.643°)] Ans.

F = 66 .7 N Note: Since fs 7 u, the screw is self-locking.

Ans: F = 66.7 N 809

*8–68. If a horizontal force of F = 50 N is applied perpendicular to the handle of the lever at E, determine the clamping force developed at G. The mean diameter and lead of the single square-threaded screw at C and D are 25 mm and 5 mm, respectively. The coefficient of static friction is ms = 0.3.

200 mm G

200 mm A

SOLUTION

D E

Since the screw is being tightened, Eq. 8–3 should be used. Here, u = tan-1 a tan-1 c

5 d = 3.643°; 2p(12.5)

L b = 2pr

125 mm

fs = tan-1ms = tan-1(0.3) = 16.699°; and M = 50(0.125). Since M must overcome the friction of two screws, M = 2[Wr tan(fs + u)] 50(0.125) = 2[FCD(0.0125)tan(16.699° + 3.643°)] Ans.

FCD = 674.32 N

Using the result of FCD and referring to the free-body diagram of member GAC shown in Fig. a, we have ©MA = 0; 674.32(0.2) - FG(0.2) = 0 Ans.

FG = 674 N Note: Since fs 7 u, the screws are self-locking.

Ans: FCD = 674 N FG = 674 N 810

8–69. The hand clamp is constructed using a square-threaded screw having a mean diameter of 36 mm, a lead of 4 mm, and a coefficient of static friction at the screw of ms = 0.3. To tighten the screw, a force of F = 20 N is applied perpendicular to the handle. Determine the clamping force in the board AB.

50 mm

SOLUTION f = tan-1 (0.3) = 16.70° u = tan-1 a

4 b = 2.026° 2p (18)

M = Wr tan (f + u) 20 (0.05) = FAB (0.018) tan (16.70° + 2.026°) Ans.

FAB = 164 N

Ans: FAB = 164 N 811

8–70. The hand clamp is constructed using a square-threaded screw having a mean diameter of 36 mm, a lead of 4 mm, and a coefficient of static friction at the screw of ms = 0.3. If the clamping force in the board AB is 300 N, determine the reversed force -F that must be applied perpendicular to the handle in order to loosen the screw.

50 mm

SOLUTION f = tan-1 (0.3) = 16.70° u = tan-1 a

4 b = 2.026° 2p(18)

F (0.05) = 300(0.018) tan (16.70° - 2.026°) Ans.

F = 28.3 N

Ans: F = 28.3 N 812

8–71. If couple forces of F = 10 lb are applied perpendicular to the lever of the clamp at A and B, determine the clamping force on the boards. The single square-threaded screw of the clamp has a mean diameter of 1 in. and a lead of 0.25 in. The coefficient of static friction is ms = 0.3.

6 in. A

6 in. B

SOLUTION Since the screw is being tightened, Eq. 8–3 should be used. Here, M = 10(12) = 120 lb # in; u = tan - 1 ¢

L 0.25 ≤ = tan - 1 B R = 4.550°; 2pr 2p(0.5)

fs = tan - 1ms = tan - 1(0.3) = 16.699°. Thus M = Wr tan (fs + u) 120 = P(0.5) tan (16.699° + 4.550°) Ans.

P = 617 lb Note: Since fs 7 u, the screw is self-locking.

Ans: P = 617 lb 813

*8–72. If the clamping force on the boards is 600 lb, determine the required magnitude of the couple forces that must be applied perpendicular to the lever AB of the clamp at A and B in order to loosen the screw. The single square-threaded screw has a mean diameter of 1 in. and a lead of 0.25 in. The coefficient of static friction is ms = 0.3.

6 in. A

6 in. B

SOLUTION Since the screw is being loosened, Eq. 8–5 should be used. Here, M = F(12); u = tan - 1 ¢

L 0.25 ≤ = tan - 1 B R = 4.550°; 2pr 2p(0.5)

fs = tan - 1ms = tan - 1(0.3) = 16.699°; and W = 600 lb. Thus M = Wr tan (fs - u) F(12) = 600(0.5) tan (16.699° - 4.550°) Ans.

F = 5.38 lb

Ans: F = 5.38 lb 814

8–73. Prove that the lead l must be less than 2prms for the jack screw shown in Fig. 8–15 to be self-locking.

-" 1/ " For self–locking, fs 7 uP or tan fs 7 tan up; ms 7

l ; 2p r

Q.E.D.

l 6 2prms

815

8–74. The square-threaded bolt is used to join two plates together. If the bolt has a mean diameter of d = 20 mm and a lead of l = 3 mm, determine the smallest torque M required to loosen the bolt if the tension in the bolt is T = 40 kN. The coefficient of static friction between the threads and the bolt is ms = 0.15. M

SOLUTION f = tan-1 0.15 = 8.531° u = tan-1

3 = 2.734° 2p(10)

M = r W tan (f - u) = (0.01)(40 000) tan (8.531° - 2.734°) M = 40.6 N # m

Ans.

Ans: M = 40.6 N # m 816

8–75. The shaft has a square-threaded screw with a lead of 9 mm and a mean radius of 15 mm. If it is in contact with a plate gear having a mean radius of 20 mm, determine the resisting torque M on the gear which can be overcome if a torque of 7 N # m is applied to the shaft. The coefficient of static friction between the gear and the screw is ms = 0.2. Neglect friction at the bearings located at A and B.

7N?m

15 mm A

B M 20 mm

SOLUTION ΣMO = 0;

M - (0.02)F = 0

F = 50M -1

f s = tan

(0.2) = 11.31°

up = tan-1 c

9 d = 5.455° 2p(15)

M′ = F(0.015) tan(f s + up)

7 = 50M(0.015) tan(11.31° + 5.455°) M = 31.0 N # m

Ans.

Ans: M = 31.0 N # m 817

*8–76. If couple forces of F = 35 N are applied to the handle of the machinist’s vise, determine the compressive force developed in the block. Neglect friction at the bearing A. The guide at B is smooth. The single square-threaded screw has a mean radius of 6 mm and a lead of 8 mm, and the coefficient of static friction is ms = 0.27.

F 125 mm A

SOLUTION f = tan - 1 (0.27) = 15.11° u = tan - 1 a

125 mm

8 b = 11.98° 2p(6)

M = Wr tan (u + f) 35 (0.250) = P (0.006) tan (11.98° + 15.11°) Ans.

P = 2851 N = 2.85 kN

Ans: P = 2.85 kN 818

8–77. Determine the horizontal force P applied perpendicular to the handle of the jack screw necessary to start lifting the 3-kN load. The square-threaded screw has a lead of 5 mm and a mean diameter of 60 mm. The coefficient of static friction for the screw is ms = 0.2.

3 kN

400 mm

SOLUTION up = tan-1 c

5 d = 1.520° 2p(30)

f s = tan-1 (0.2) = 11.31° M = W(r) tan(f s + up)

P(0.4) = 3000(0.03) tan(11.31° + 1.520°) Ans.

P = 51.2 N

Ans: P = 51.2 N 819

8–78. automobile jack is subjected to a vertical load of The automobile jackhaving is subjected a vertical load of screw, a lead of 5tomm 8 kN. If a square-threaded F = 8 kN. having a lead of 5 mm a mean diameter of If 10a square-threaded mm, is used inscrew, the jack, and that a mean diameter 10 mm, is to used in the jack, rmine the force must be appliedof perpendicular force applied handle to (a)determine raise thethe load, andthat (b)must lowerbethe load; perpendicular to the handle to exerts (a) raise load,forces and (b) plate onlythe vertical at lower the load; 0.2. The supporting ms = 0.2. The supporting plate exerts only vertical forces at nd B. A and B.

LUTION

F F P P 400 mm

SOLUTION

ations of Equilibrium: From FBD (a), Equations of Equilibrium: From FBD (a), 81x2 - Dy 12x2 = 0 ©ME = 0; Dy = 4.00 kN 81x2 - Dy 12x2 = 0 Dy = 4.00 kN a + ©ME = 0; m FBD (b), From FBD (b), FB 12x2 - 81x2 = 0 ©MA = 0; FB = 4.00 kN FB 12x2 - 81x2 = 0 a + ©MA = 0; FB = 4.00 kN m FBD (c), From FBD (c), ©MC = 0; Dx 10.1 sin 30°2 - 4.0010.2 cos 30°2 = 0 a + ©MC = 0; Dx 10.1 sin 30°2 - 4.0010.2 cos 30°2 = 0 Dx = 13.86 kN Dx = 13.86 kN mber DF is a two force member. Analysing the forces that act on pin D[FBD (d)], Member DF is a two force member. Analysing the forces that act on pin D[FBD (d)], ave we have ©Fy = 0; FDF = 8.00 kN FDF sin 30° - 4.00 = 0 + c ©Fy = 0; FDF = 8.00 kN FDF sin 30° - 4.00 = 0 ©Fx = 0; P¿ - 13.86 - 8.00 cos 30° = 0 P¿ = 20.78 kN + ©F = 0; : P¿ - 13.86 - 8.00 cos 30° = 0 P¿ = 20.78 kN x l 5 b = tan-1 c tional Forces on Screw: Here, u = tan-1 a l d = 9.043°, 5 2pr u = tan-12p152 b = tan-1 c d = 9.043°, a Frictional Forces on Screw: Here, -1 -1 2pr = P¿ = 20.78 kN, M = 0.4P and fs = tan ms = tan 10.22 = 11.310°.2p152 = P¿ = is 20.78 kN,the M = we 0.4P lying Eq. 8–3W if the jack raising load, haveand fs = tan-1ms = tan-1 10.22 = 11.310°. Applying Eq. 8–3 if the jack is raising the load, we have M = Wr tan1u + f2 M = Wr tan1u + f2 0.4P = 20.7810.0052 tan19.043° + 11.310°2 0.4P = 20.7810.0052 tan19.043° + 11.310°2 P = 0.09638 kN = 96.4 N Ans. P = 0.09638 kN = 96.4 N Ans. lying Eq. 8–5 if the jack is lowering the load, we have Applying Eq. 8–5 if the jack is lowering the load, we have M– = Wr tan1f - u2 M– = Wr tan1f - u2 0.4P = 20.7810.0052 tan 111.310° - 9.043°2 0.4P = 20.7810.0052 tan 111.310° - 9.043°2 P = 0.01028 kN = 10.3 N Ans. P = 0.01028 kN = 10.3 N Ans. e: Since fs 7 u, the screw is self-locking. It will not unscrew even if force P is Note: Since fs 7 u, the screw is self-locking. It will not unscrew even if force P is oved. removed.

400 mm

C 30°

E C 30°

10 mm

Ans: P = 96.4 N raising P = 10.3 N lowering 820

8–79. If the required clamping force at the board A is to be 50 N, determine the force F that must be applied to the handle to tighten the clamp. The single square-threaded screw has a mean radius of 10 mm, a lead of 3 mm, and the coefficient of static friction is ms = 0.35.

150 mm

–F

SOLUTION

M = W r tan (u + f)

Where f = f s = tan - 1 ms = tan - 1 (0.35) = 19.29°, W = 50 N, M = 0.15 F, u = tan - 1 (L/2pr) = tan - 1 a

3 b = 2.733°. 2p(10)

0.15 F = 50 (0.01) tan (2.733° + 19.29°) Ans.

F = 1.35 N

Ans: F = 1.35 N 821

*8–80. Determine the clamping force on the board A if the screw of the hold-down clamp is tightened with a force of F = 400 N. The single square-threaded screw has a mean radius of 8 mm, a lead of 2 mm, and the coefficient of static friction is ms = 0.3.

150 mm

–F

SOLUTION

M = W r tan (u + f)

Where f = f s = tan - 1 ms = tan - 1 (0.3) = 16.670°, M = 400 (0.15) = 60 N.m, W = P, u = tan - 1 (L/2pr) = tan - 1 a

2 b = 2.279°. 2p(8)

60 = P(0.008) tan (2.279° + 16.670°) Ans.

P = 21809 N = 21.8 kN

Ans: P = 21.8 kN 822

8–81. If a horizontal force of P = 100 N is applied perpendicular to the handle of the lever at A, determine the compressive force F exerted on the material. Each single squarethreaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction at all contacting surfaces of the wedges is ms = 0.2, and the coefficient of static friction at the screw is msœ = 0.15.

A 15

C 15

250 mm

SOLUTION Since the screws are being tightened, Eq. 8–3 should be used. Here, L 7 .5 b = tan -1 c d = 5 .455°; u = tan -1 a 2pr 2p(12.5)

fs = tan - 1ms = tan - 1(0.15) = 8.531°; M = 100(0.25) = 25 N # m; and W = T, where T is the tension in the screw shank. Since M must overcome the friction of two screws, M = 2[Wr, tan(fs + u)4 25 = 23T(0.0125) tan (8 .531° + 5 .455°)4 Ans.

T = 4015.09 N = 4.02 kN

4015 .09 - 0 .2N¿ - 0 .2N cos 15° - N sin 15° = 0

(1)

N¿ + 0 .2N sin 15° - N cos 15° = 0

(2)

Solving, N = 6324.60 N

N¿ = 5781.71 N

2(6324 .60) cos 15° - 230 .2(6324.60) sin 15°4 - F = 0 F = 11563 .42 N = 11 .6 kN

Ans.

Ans: T = 4.02 kN F = 11.6 kN 823

8–82. Determine the horizontal force P that must be applied perpendicular to the handle of the lever at A in order to develop a compressive force of 12 kN on the material. Each single square-threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction at all contacting surfaces of the wedges is ms = 0.2, and the coefficient of static friction at the screw is msœ = 0.15.

A 15

C 15

250 mm

2N cos 15° - 230 .2N sin 15°4 - 12000 = 0 N = 6563.39 N

N¿ - 6563 .39 cos 15° + 0 .2(6563 .39) sin 15° = 0 N¿ = 6000 N

+ ©Fx = 0; :

T - 6563 .39 sin 15° - 0 .2(6563 .39) cos 15° - 0 .2(6000) = 0 T = 4166 .68 N

Since the screw is being tightened, Eq. 8–3 should be used. Here, u = tan - 1 c

L 7 .5 d = tan-1 c d = 5.455°; 2pr 2p(12.5)

fs = tan-1ms = tan-1(0 .15) = 8.531°; M = P(0 .25); and W = T = 4166.68N. Since M must overcome the friction of two screws, M = 23Wr tan (fs + u)4 P(0.25) = 234166 .68(0.0125) tan (8.531° + 5.455°)4 P = 104 N

Ans.

Ans: P = 104 N 824

8–83. Determine the minimum tension in the rope at points A and B that is necessary to maintain equilibrium. Take ms = 0.3 between the rope and the fixed post D.

B 608 D

300 lb

SOLUTION T2 = T1 e mb 300 = TA e

p 0.3a 2 b

Ans.

TA = 187 lb

300 = TB e 0.3 6 p TB = 137 lb

Ans.

Ans: TA = 187 lb TB = 137 lb 825

*8–84. Determine the maximum tension in the rope at points A and B that is necessary to maintain equilibrium. Take ms = 0.3 between the rope and the fixed post D.

B 608 D

300 lb

SOLUTION T2 = T1 e mb TA = 300 e

p 0.3a 2 b

Ans.

TA = 481 lb

TB = 300 e 0.3 6 p TB = 658 lb

Ans.

Ans: TA = 481 lb TB = 658 lb 826

8–85. A180-lb farmer tries to restrain the cow from escaping by wrapping the rope two turns around the tree trunk. If the cowexerts a force of 250 lb on the rope, determine if the farmer can successfully restrain the cow. The coefficient of static friction between the rope and the tree trunk is ms = 0.15, and between the farmer’s shoes and the ground mœs = 0.3.

SOLUTION Since the cow is on the verge of moving, the force it exerts on the rope is T2 = 250 lb and the force exerted by the man on the rope is T1. Here, b = 2(2p) = 4p rad. Thus, T2 = T1ems b 250 = T1e0.15(4p) T1 = 37.96 lb Using this result and referring to the free - body diagram of the man shown in Fig. a, + c ©Fy = 0;

N - 180 = 0

N = 180 lb

+ ©F = 0; : x

37.96 - F = 0

F = 37.96 lb

Since F 6 Fmax = ms ¿N = 0.3(180) = 54 lb, the man will not slip, and he will successfully restrain the cow.

Ans: He will successfully restrain the cow. 827

8–86. The 100-lb boy at A is suspended from the rope that passes over the quarter circular cliff rock. Determine if it is possible for the 185-lb woman to hoist him up; and if this is possible, what smallest force must she exert on the rope? The coefficient of static friction between the rope and the rock is ms = 0.2, and between the shoes of the woman and the ground msœ = 0.8.

SOLUTION b =

p 2

0.2 p2

= 136.9 lb

N - 185 = 0 N = 185 lb

+ ©F = 0; : x

136.9 - F = 0 F = 136.9 lb

Fmax = 0.8 (185) = 148 lb 7 136.9 lb Yes, just barely.

Ans.

Ans: Yes, it is possible. F = 137 lb 828

8–87. The 100-lb boy at A is suspended from the rope that passes over the quarter circular cliff rock. What force must the woman exert on the rope in order to let the boy descend at constant velocity? The coefficients of static and kinetic friction between the rope and the rock are ms = 0.4 and mk = 0.35, respectively.

SOLUTION b =

p 2

T2 = T1 e mb;

100 = T1 e

0.35 p2

Ans.

T1 = 57.7 lb

Ans: T1 = 57.7 lb 829

*8–88. If the force T1 is applied to the rope at A, determine the force T2 at B needed to pull the rope over the two fixed drums having the angles of contact and the coefficients of static friction shown.

µ1 A

β1 T1

µ2

β2

SOLUTION T¿ = T1 e m1 b1 T2 = T¿ e

m2 b2

Thus, T2 = T1 e m1 b1 e m2 b2 or T2 = T1 e (m1 b1 + m2 b2)

Ans.

Ans: T2 = T1 e (m1b1 + m2b2) 830

8–89. Determine the minimum tension in the rope at points A and B which is necessary to maintain equilibrium. The coefficient of static friction between the rope and the fixed post D is ms = 0.3. The rope is wrapped only once around the post.

5 kN

SOLUTION T2 = T1 e mb

Where

T2 = 5 kN, T1 = TA, b = p rad

5 = TA e 0.3 (p) Ans.

TA = 1.95 kN

T2 = T1 e mb 5 = TB e

Where

T2 = 5 kN, T1 = TB, b = 2p rad

0.3 (2p)

Ans.

TB = 0.759 kN

Ans: TA = 1.95 kN TB = 0.759 kN 831

8–90. The choker sling is used to lift the smooth pipe that has a mass of 600 kg. If the coefficient of static friction between the loop at the end A of the sling and the rope is ms = 0.3, determine the angle u at the connection.

SOLUTION u u + c ΣFy = 0; T - P cos - F cos = 0 2 2 + ΣFx = 0; - P sin u + F sin u = 0 S 2 2 P=F T = 2P cos

u 2

Also, T = Pe mu 2P cos 2 cos

u u a b = Pe m 2 2

u u a b = e 0.3 2 2

Solving, u = 0.866 rad. = 49.6° 2 Ans.

u = 99.2°

Ans: u = 99.2° 832

8–91. Two 8–kg blocks are attached to a cord that passes over two fixed drums. If ms = 0.3 at the drums, determine the equilibrium angle u the cord makes with the horizontal when a vertical force P = 200 N is applied to the cord to pull it downward.

0.1 m

2 T2 sin u = 200 T2 sin u = 100

T2 = T1 e mb;

T2 = 8 (9.81)e 0.3A 2 + uB p

Thus, p 100 = (8) (9.81) a e 0.3A 2 B b A e 0.3 u B sin u

0.795 = sin u A e 0.3 u B Solving, Ans.

u = 0.701 rad. = 40.1°

Ans: u = 40.1° 833

*8–92. Determine the force P that must be applied to the handle of the lever so that the wheel is on the verge of turning if M = 300 N # m. The coefficient of static friction between the belt and the wheel is ms = 0.3.

P 300 mm 25 mm B C 60 mm

700 mm

SOLUTION Frictional Force on Flat Belt: Here b = 270° =

3p rad. 2

TD = TAe ms b TD = TAe 0.3 ( 2 ) 3p

(1)

TD = 4.1112 TA Equations of Equilibrium: Referring to the FBD of the wheel shown in Fig. a, a+ ΣMB = 0; 300 + TA (0.3) - TD (0.3) = 0

(2)

Solving Eqs. (1) and (2), TA = 321.42 N

TD = 1321.42 N

Subsequently, from the FBD of the lever, Fig. b a+ ΣMC = 0; 1321.42(0.025) - 321.42(0.06) - P(0.7) = 0 Ans.

P = 19.64 N = 19.6 N

Ans: 19.6 N 834

8–93. If a force of P = 30 N is applied to the handle of the lever, determine the largest couple moment M that can be resisted so that the wheel does not turn. The coefficient of static friction between the belt and the wheel is ms = 0.3.

P 300 mm 25 mm B C 60 mm

700 mm

SOLUTION Frictional Force on Flat Belt: Here b = 270° =

3p rad. 2

TD = TAe mb TD = TAe 0.3 ( 2 ) 3p

(1)

TD = 4.1112 TA Equations of Equilibrium: Referring to the FBD of the wheel shown in Fig. a, a+ ΣMB = 0; M + TA (0.3) - TD (0.3) = 0

(2)

Solving Eqs. (1) and (2) TA = 1.0714 m

TD = 4.4047 m

Subsequently, from the FBD of the lever, Fig. b a+ ΣMC = 0; 4.4047 M(0.025) - 1.0714 M(0.06) - 30(0.7) = 0 M = 458.17 N # m = 458 N # m

Ans.

Ans: M = 458 N # m 835

8–94. 5 ft

The 50-lb cylinder is attached to a cord which passes over the fixed drum. If the coefficient of static friction at the drum is ms = 0.3, determine the maximum force P that can be applied to the roller when u = 30° without causing the cylinder to move.

θ 1 ft

SOLUTION T2 = T1e mb T2 = 50e 0.3A 2 + b B p

T2 sin 30° - P = 0 Ans.

P = 93.723 sin 30° = 46.9 lb

Ans: P = 46.9 lb 836

8–95. 5 ft

The 50-lb cylinder is attached to a cord which passes over the fixed drum. If the coefficient of static friction between the cord and the drum is m s = 0.3, determine the angle u of of the cord if the maximum vertical load that can be applied to the roller without causing the cylinder to move is P = 100 lb.

θ 1 ft

SOLUTION Rope friction: T2 = T1e mb

where

T1 = 50 lb, b = p2 + u

T2 = 50e0.3(p>2 + 0)

(1)

(2)

T2 sin u - 100 = 0

Substituting Eq. (1) into (2) yields: 50e 0.3(p/2 + u) sin u = 100 50e 0.3(p/2)e0.3 u sin u = 100 e0.3 u sin u = 1.2485 By trial and error

Ans.

u = 1.1089 rad = 63.5°

Ans: u = 63.5° 837

*8–96. The load can be supported by two boys when the cord is suspended over the pipe (a half-turn). If each boy can pull with a force of 125 lb, determine the maximum weight of the load. Can one boy support the same load when the cord is wrapped one-and-a-half times around the pipe? The coefficient of static friction is ms = 0.3.

SOLUTION For two boys: T2 = T1 e mb Where T2 = W, T1 = 2 (125) = 250 lb, b = p rad W = 250 e 0.3 (p) Ans.

= 641.6 lb = 642 lb For one boy: T2 = T1 e mb Where T2 = W = 641.6 lb, T1 = T, b = 3p rad 641.6 = Te 0.3 (3p)

Ans.

T = 37.96 lb

Since the required pulling force T = 37.96 lb 6 125 lb, one boy can support the load.

Ans: W = 642 lb T = 37.96 lb 838

8–97. A girl weighing 100 lb attempts to pull herself up a tree using a rope which is draped over the limb at B. If ms = 0.4, determine the smallest force at which she must pull on the rope to lift herself. Do you think this is possible?

SOLUTION For the girl: FBD (a) + c ΣFy = 0; T1 + T2 - 100 = 0 T1 = 100 - T2 Rope friction: T2 = T1 e mb Where T1 = 100 - T2, b = p rad T2 = (100 - T2) e 0.4 (p) Ans.

= 77.8 lb

Ans: T2 = 77.8 lb 839

8–98. Show that the frictional relationship between the belt tensions, the coefficient of friction m, and the angular contacts a and b for the V-belt is T2 = T1emb>sin(a>2).

Impending motion

a b

SOLUTION

- (T + dT) cos

du du + T cos + 2 dF = 0 2 2

©Fy = 0;

-(T + dT) sin

du a du - T sin + 2 dN sin = 0 2 2 2

Replace

sin

du du by , 2 2

cos

du by 1, 2

dF = m dN Using this and (dT)(du) : 0, the above relations become dT = 2m dN T du = 2 adN sin

a b 2

Combine dT du = m T sin a2 Integrate from u = 0, T = T1 to u = b, T = T2 we get, ¢

a≤

T2 = T1 e sin 2

Q.E.D

840

8–99. The wheel is subjected to a torque of M = 50 N # m. If the coefficient of kinetic friction between the band brake and the rim of the wheel is mk = 0.3, determine the smallest horizontal force P that must be applied to the lever to stop the wheel from rotating.

400 mm

SOLUTION

A 150 mm

Wheel: a + ΣMO = 0; - T2 (0.150) + T1 (0.150) + 50 = 0 T2 = T1 e mb;

50 mm 25 mm

100 mm

a 3p b

T2 = T1 e 0.3 2

T1 = 107.14 N Link: + ΣMB = 0; Lever:

107.14 (0.05) - F (0.025) = 0 F = 214.28 N

a + ΣMA = 0; - P (0.4) + 214.28 (0.1) = 0 Ans.

P = 53.6 N

Ans: P = 53.6 N 841

*8–100. Blocks A and B have a mass of 7 kg and 10 kg, respectively. Using the coefficients of static friction indicated, determine the largest vertical force P which can be applied to the cord without causing motion.

µD = 0.1

300 mm

µB = 0.4

400 mm P

µC = 0.4 A

SOLUTION

µA = 0.3

Frictional Forces on Flat Belts: When the cord pass over peg D, b = 180° = p rad and T2 = P. Applying Eq. 8–6, T2 = T1 e mb, we have P = T1 e 0.1 p

T1 = 0.7304P

When the cord pass over peg C, b = 90° = Applying Eq. 8–6, T2 ′ = T1 ′e mb, we have 0.7304P = T1 ′e 0.4(p>2)

p rad and T2 ′ = T1 = 0.7304P. 2

T1 ′ = 0.3897P

Equations of Equilibrium: From FDB (b), + c ΣFy = 0;

NB - 98.1 = 0

+ ΣFx = 0; S

FB - T = 0

(1)

a+ ΣMO = 0 ;

T(0.4) - 98.1(x) = 0

(2)

NB = 98.1 N

From FDB (b), + c ΣFy = 0;

NA - 98.1 - 68.67 = 0

+ ΣFx = 0; S

0.3897P - FB - FA = 0

NA = 166.77 N (3)

Friction: Assuming the block B is on the verge of tipping, then x = 0.15 m. A1 for motion to occur, block A will have slip. Hence, FA = (ms)ANA = 0.3(166.77) = 50.031 N. Substituting these values into Eqs. (1), (2) and (3) and solving yields Ans.

P = 222.81 N = 223 N FB = T = 36.79 N

Since (FB)max = (ms)B NB = 0.4(98.1) = 39.24 N 7 FB, block B does not slip but tips. Therefore, the above assumption is correct.

Ans: P = 223 N 842

8–101. The uniform bar AB is supported by a rope that passes over a frictionless pulley at C and a fixed peg at D. If the coefficient of static friction between the rope and the peg is mD = 0.3, determine the smallest distance x from the end of the bar at which a 20-N force may be placed and not cause the bar to move.

- 20 (x) + TB (1) = 0

;

20 N

TA + TB - 20 = 0 TA = TB e

0.3(p2 )

= 1.602 TB

Solving,

B 1m

TA = 12.3 N TB = 7.69 N Ans.

x = 0.384 m

Ans: x = 0.384 m 843

B. If the© 2022 motor develop# by R. C.can Hibbeler. Published by Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. the M as=they0.80 Nm, smallest nt the belt from slipping. tween the8–102. belt and the drum The belt on the portable dryer wraps around the drum D, idler pulley A, and motor pulley B. If the develop# 0motor can 1uP a maximum torque of determine the M = 0.80 Nm, smallest spring tension required to prevent the belt from slipping. The coefficient of static friction between the belt and the drum % and motor pulley is ms = 0.3.

10.022 - 0.8 = 0

-" 1/ "

-T1 10.022 + T2 10.022 - 0.8 = 0

T2 = T1 emb;

PP $ $

1uP

$ $

T2 = T1 e10.321p2 = 2.5663T1

-Fs10.052 + 125.537 + 25.537 sin 30°210.1 cos 45°2 + 25.537 cos 30°10.1 sin 45°2 = 0

5.537 + 25.537 sin 30°210.1 cos 45°2 + 25.537 cos 30°10.1 sin 45°2 = 0 Ans.

Fs = 85.4 N

Ans.

Ans: Fs = 85.4 N 844

8–103. Two boys have a tug-of-war using a rope. In the process the rope rubs against a post such that it is deflected 3° on each side of the post. If boy A weighs 60 lb and boy B weighs 50 lb, determine if it possible for one of them to pull the rope toward him. The coefficient of static friction between the rope and the post is mp = 0.3, and the coefficient of static friction between the boys’ shoes and the ground is ms = 0.4.

38 A

SOLUTION The maximum force which boy A can exert on the rope is TA 3 FBD (a) 4 . + c ΣFy = 0;

NA - 60 = 0

NA = 60 lb

+ ΣFx = 0; S

TA - 0.4 (60) = 0

TA = 24 lb

Rope friction: T2 = T1 e mb Where T1 = TB, T2 = TA = 24 lb, b =

6° p = 0.1047 rad 180°

24 = TB e 0.3 (0.1047) TB = 23.257 lb For boy B: FBD (b) + c ΣFy = 0;

NB - 50 = 0

NB = 50 lb

+ ΣFx = 0; S

FB - 23.257 = 0

FB = 23.257 lb

Since FB = 23.257 lb 7 mNB = 0.4 (50) = 20 lb, boy B slips. Therefore, it is possible for boy A to pull boy B towards him.

Ans: Therefore, it is possible for boy A to pull boy B towards him. 845

*8–104. Two boys have a tug-of-war using a rope. In the process the rope rubs against a post such that it is deflected 3º on each side of the post. If boy A weighs 60 lb and boy B weighs 50 lb, determine if it is possible for one of them to pull the rope toward him. The coefficient of static friction between the rope and the post is mp = 0.3, and the coefficient of static friction between the shoes of the boy at A and the ground is mA = 0.4, whereas for boy B it is mB = 0.5.

38 A

SOLUTION The maximum force which boy B can exert on the rope is TB 3 FBD (a) 4 . + c ΣFy = 0;

NB - 50 = 0

NB = 50 lb

+ ΣFx = 0; S

- TB + 0.5 (50) = 0

TB = 25 lb

Rope friction: T2 = T1 e mb Where T1 = TA, T2 = TB = 25 lb, b =

6° p = 0.1047 rad 180°

25 = TA e 0.3 (0.1047) TA = 24.227 lb For boy A: FBD (b) + c ΣFy = 0;

NA - 60 = 0

NA = 60 lb

+ ΣFx = 0; S

24.227 - FA = 0

FA = 24.227 lb

Since FA = 24.227 lb 7 mNA = 0.4 (60) = 24 lb, boy A slips. Therefore, it is possible for boy B to pull boy A towards him.

Ans: Therefore, it is possible for boy B to pull boy A towards him. 846

8–105. A 10-kg cylinder D, which is attached to a small pulley B, is placed on the cord as shown. Determine the largest angle uX so that the cord does not slip over the peg at C. The cylinder at E also has a mass of 10 kg, and the coefficient of static friction between the cord and the peg is ms = 0.1.

B E

SOLUTION

Since pully B is smooth, the tension in the cord between pegs A and C remains constant. Referring to the free-body diagram of the joint B shown in Fig. a, we have + c ©Fy = 0;

2T sin u - 10(9.81) = 0

T =

49.05 sin u

49.05 In the case where cylinder E is on the verge of ascending, T2 = T = and sin u p T1 = 10(9.81) N. Here, + u, Fig. b. Thus, 2 T2 = T1e msb p 49.05 = 10(9.81) e 0.1 a 2 + ub sin u

p 0.5 = 0.1 a + u b sin u 2

Solving by trial and error, yields u = 0.4221 rad = 24.2° In the case where cylinder E is on the verge of descending, T2 = 10(9.81) N and 49.05 p T1 = . Here, + u. Thus, sin u 2 T2 = T1e m s b 10(9.81) =

49.05 0.1 a p + ub e 2 sin u

ln (2 sin u) = 0.1 a

p + ub 2

Solving by trial and error, yields u = 0.6764 rad = 38.8° Thus, the range of u at which the wire does not slip over peg C is 24.2° 6 u 6 38.8° Ans.

umax = 38.8°

Ans: umax = 38.8° 847

8–106. A conveyer belt is used to transfer granular material and the frictional resistance on the top of the belt is F = 500 N. Determine the smallest stretch of the spring attached to the moveable axle of the idle pulley B so that the belt does not slip at the drive pulley A when the torque M is applied. What minimum torque M is required to keep the belt moving? The coefficient of static friction between the belt and the wheel at A is ms = 0.2.

0.1 m

M A

F = 500 N

0.1 m B k = 4 kN>m

SOLUTION Frictional Force on Flat Belt: Here, b = 180° = p rad and T2 = 500 + T and T1 = T. Applying Eq. 8–6, we have T2 = T1 emb 500 + T = Te0.2p T = 571.78 N Equations of Equilibrium: From FBD (a), M + 571.7810.12 - 1500 + 578.1210.12 = 0

a + ©MO = 0;

M = 50.0 N # m

Ans.

Fsp - 21578.712 = 0

Fsp = 1143.57 N

Thus, the spring stretch is x =

Fsp k

1143.57 = 0.2859 m = 286 mm 4000

Ans.

Ans: M = 50.0 N # m x = 286 mm 848

8–107. The collar bearing uniformly supports an axial force of P = 5 kN. If the coefficient of static friction is ms = 0.3, determine the smallest torque M required to overcome friction.

P M

150 mm 200 mm

SOLUTION Bearing Friction: With R2 = 0.1 m, R1 = 0.075 m, P = 5 ( 103 ) N, and ms = 0.3, M = =

R23 - R13 2 ms P a 2 b 3 R2 - R12

0.13 - 0.0753 2 (0.3) 3 5 ( 103 )4 a 2 b 3 0.1 - 0.0752

= 132 N # m

Ans: M = 132 N # m 849

*8–108. The collar bearing uniformly supports an axial force of P = 8 kN. If a torque of M = 200 N # m is applied to the shaft and causes it to rotate at constant velocity, determine the coefficient of kinetic friction at the surface of contact.

P M

150 mm 200 mm

SOLUTION

Bearing Friction: With R2 = 0.1 m, R1 = 0.075 m, M = 300 N # m, and P = 8 ( 103 ) N, M = 200 =

R23 - R13 2 mk P a 2 b 3 R2 - R12

0.13 - 0.0753 2 mk 3 8 ( 103 )4 a 2 b 3 0.1 - 0.0752

Ans.

mk = 0.284

Ans: mk = 0.284 850

8–109. The floor-polishing machine rotates at a constant angular velocity. If it has a weight of 80 lb, determine the couple forces F the operator must apply to the handles to hold the machine stationary. The coefficient of kinetic friction between the floor and brush is mk = 0.3. Assume the brush exerts a uniform pressure on the floor.

1.5 ft

SOLUTION M =

2 mPR 3 2 ft

2 F(1.5) = (0.3) (80)(1) 3 Ans.

F = 10.7 lb

Ans: F = 10.7 lb 851

8–110. The double-collar bearing is subjected to an axial force P = 4 kN. Assuming that collar A supports 0.75P and collar B supports 0.25P, both with a uniform distribution of pressure, determine the maximum frictional moment M that may be resisted by the bearing. Take ms = 0.2 for both collars.

P M

20 mm B

SOLUTION

R32 - R31 2 M = ms P ¢ 2 ≤ 3 R2 - R21 M =

(0.03)3 - (0.01)3 (0.02)3 - (0.01)3 2 (0.75) (4000) + (0.25) (4000) ≤ (0.2) ¢ 5 (0.03)2 - (0.01)2 (0.02)2 - (0.01)2

= 16.1 N # m

10 mm 30 mm

Ans.

Ans: M = 16.1 N # m 852

8–111. The double-collar bearing is subjected to an axial force P = 16 kN. Assuming that collar A supports 0.75P and collar B supports 0.25P, both with a uniform distribution of pressure, determine the smallest torque M that must be applied to overcome friction. Take ms = 0.2 for both collars.

P 100 mm

M A B

50 mm

75 mm 30 mm

SOLUTION Bearing Friction: Here (RA)2 = 0.1 m, (RA)1 = 0.05 m, PA = 0.75 316 ( 103 ) N4 = 12 ( 103 ) N, (RB)2 = 0.075 m, (RB)1 = 0.05 m and PB = 0.25 316 ( 103 ) N4 = 4 ( 103 ) N. m = =

(RA)23 - (RA)13 (RB)23 - (RB)13 2 2 ms PA c d + m P c d 3 3 s B (RB)22 - (RB)12 (RA)22 - (RA)12

0.13 - 0.053 0.0753 - 0.053 2 2 3 ( ) (0.2) 312 ( 103 )4 a 2 b + (0.2) 3 4 10 4 a b 3 3 0.1 - 0.052 0.0752 - 0.052

= 237.33 N # m = 237 N # m

Ans.

Ans: M = 237 N # m 853

which varies as shown. If the M on is m,© determine thePublished torque M Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright 2022 by R. C. Hibbeler. by Pearson laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. tion if the shaft supports an axial *8–112. The pivot bearing is subjected to a pressure distribution at its surface of contact which varies as shown. If the coefficient of static friction is m, determine the torque M required to overcome friction if the shaft supports an axial force P. For the solution it is necessary to determine the peak pressure p0 in terms of P and the bearing radius R.

pr b dA 2R SOLUTION

R r

r dr du

pr b dA dF = m dN = m p0 cos a 2R

pr rm p0 cos a M = b r dr du 2R 2p LA

r bdrb du R L0 = m p0

A pr b + R - 2d

a r2 cos a

pr du b drb 2R L0

p 2 2 A 2R RB r - 2 pr pr R pr cos a sin a b + b d 12p2 2R 0 A p B 2sin a2R b d 12p2 A p B3

p 3 A 2R B 16R

= 0.7577m p0 R3 2p

dN =

= p0 B

p0 a cos a

2p cos a

p A 2R BThus,

M = 0.521 PmR

p R

b R 12p2

pr0 R2 = 1.454p

sin a

pr b rdrb du 2R L0

r pr pr R b + b R 12p2 sin a p 2R 2R 0 A B

pr A B b rdrb du 2R 02 a1 - 2 b = 4p L R p 2 2R

b +

p 2 = mp0 ¢ 2 ≤ c a b - 2 d 2 p

P =

cos a

r p = p0 cos π 2R

L0 p 2 2 - 22r 2R = r m p0 B

r p = p0 cos π 2R

Ans.

M = 0.521 PmR

Ans.

Ans: M = 0.521 PmR 854

8–113. Because of wearing at the edges, the pivot bearing is subjected to a conical pressure distribution at its surface of contact. Determine the torque M required to overcome friction and turn the shaft, which supports an axial force P. The coefficient of static friction is ms. For the solution it is necessary to determine the peak pressure p0 in terms of P and the bearing radius R.

P M

SOLUTION Equations of Equilibrium and Bearing Friction: Using similar triangles, p =

p0 p = , R - r R

p0 (R - r).Also, dA = 2prdr, dN = pdA and dF = ms dN = ms pdA. R

+ c ©Fy = 0;

pdA - P = 0

p0 (R - r)(2prdr) - P = 0 L R R 2pp0 r(R - r)dr - P = 0 R L0

3P pR 2

[1]

(ms pdA)r - M = 0

ms p0 (R - r)(2prdr)r - M = 0 L0 R 2pms p0 R 2 r (R - r)dr - M = 0 R L0 pms R 3 p0 6

[2]

ms PR pms R 3 3P a b = 2 6 2 pR

Ans.

M = Substituting Eq. [1] into [2] yields M =

Ans: M = 855

ms PR 2

8–114. A tube has a total weight of 200 lb, length l = 8 ft, and radius = 0.75 ft. If it rests in sand for which the coefficient of static friction is ms = 0.23, determine the torque M needed to turn it. Assume that the pressure distribution along the length of the tube is defined by p = p0 sin u. For the solution it is necessary to determine p0, the peak pressure, in terms of the weight and tube dimensions.

SOLUTION Equations of Equilibrium and Friction: Here, dN = plrdu = p0 lr sin udu. Since the tube is on the verge of slipping, dF = ms dN = p0 ms lr sin udu.

+ c ΣFy = 0;

p 2

p = p0 sin p0

dN sin u - W = 0 p

p0 lr sin2 udu = W

P0 lr

( 1 - cos 2u ) du = W p p0 lr a b = W 2 p0 =

2W plr

(1)

a+ ΣMO = 0;

M = 2

L0 p 2

dF(r) - M = 0

p0 ms lr 2 sin udu = 2p0 ms lr 2

(2)

Substituting Eq. (1) into (2) yields M =

4Wms r p

However, W = 200 lb, ms = 0.23 and r = 0.75 ft, then M =

4(200)(0.23)(0.75) = 43.9 lb # ft p

Ans.

Ans: M = 43.9 lb # ft 856

8–115. The corkscrew is used to remove the 15-mm-diameter cork from the bottle. Determine the smallest vertical force P that must be applied to the handle if the gauge pressure in the bottle is p = 175 kPa and the cork pushes against the sides of the bottle’s neck with a uniform pressure of 90 kPa. The coefficient of static friction between the bottle and the cork is ms = 0.4. Hint: The force exerted on the bottom of the cork is F = pA, where A is the surface area of the cork’s bottom and p is the gauge pressure.

P 20 mm A

180 mm B

30 mm

SOLUTION a + ΣMA = 0; P(200) - TB(20) = 0

TB = 10 P

F = PbAb = 175(103)(p)(0.0075)2 = 30.925 N Fr = ms(ps As) = 0.4(90)(103)(2p)(0.0075)(0.03) = 50.89 N + c ΣFy = 0; 10P - 50.89 + 30.925 = 0 Ans.

P = 2.00 N

Ans: P = 2.00 N 857

*8–116. The collar fits loosely around a fixed shaft that has a radius of 2 in. If the coefficient of kinetic friction between the shaft and the collar is mk = 0.3, determine the force P on the horizontal segment of the belt so that the collar rotates counterclockwise with a constant angular velocity. Assume that the belt does not slip on the collar; rather, the collar slips on the shaft. Neglect the weight and thickness of the belt and collar. The radius, measured from the center of the collar to the mean thickness of the belt, is 2.25 in.

P 2.25 in. 2 in.

SOLUTION

20 lb

Ry - 20 = 0

Ry = 20 lb

+ ©F = 0; : x

P - Rx = 0

Rx = P

- a 2P2 + 202 b (0.5747) + 20(2.25) - P(2.25) = 0 Ans.

P = 13.8 lb

Ans: P = 13.8 lb 858

8–117. The collar fits loosely around a fixed shaft that has a radius of 2 in. If the coefficient of kinetic friction between the shaft and the collar is mk = 0.3, determine the force P on the horizontal segment of the belt so that the collar rotates clockwise with a constant angular velocity. Assume that the belt does not slip on the collar; rather, the collar slips on the shaft. Neglect the weight and thickness of the belt and collar. The radius, measured from the center of the collar to the mean thickness of the belt, is 2.25 in.

P 2.25 in. 2 in.

SOLUTION

20 lb

Ry - 20 = 0

Ry = 20 lb

+ ©F = 0; : x

P - Rx = 0

Rx = P

a 2P2 + 202 b(0.5747) + 20(2.25) - P(2.25) = 0 Ans.

P = 29.0 lb

Ans: P = 29.0 lb 859

8–118. If the coefficient of static friction at the cone clutch is ms, determine the smallest force P that should be applied to the handle in order to transmit the torque M.

SOLUTION

dr b dA = 2p ra sin u

Since dN = p dA, then ro

©Fx = 0;

P¿ = =

dN sin u =

(sin u)p(2p r)a

dr b sin u

ro 2pp r dr 2 Lp

= p p A r2o - r2i B p =

P¿ pA

r2o - r2i

B ro

©M = 0;

M = = =

+ ©MC = 0;

r dF =

rms dN =

2pr ms p a

r dr b sin u

2 ms p p (r3o - r3i ) 3

sin u

2 ms P¿ (r3o - r3i ) 3 sin u A r2o - r2i B

P l - P¿ b = 0 l P¿ = Pa b b

Thus P =

3Mb sin u(r 2o - r 2i )

Ans.

2msl r3o - r3i

Ans: P = 860

3Mb sin u(ro2 - ri2) 2ms l(ro3 - ri3)

8–119. A disk having an outer diameter of 120 mm fits loosely over a fixed shaft having a diameter of 30 mm. If the coefficient of static friction between the disk and the shaft is ms = 0.15 and the disk has a mass of 50 kg, determine the smallest vertical force F acting on the rim which must be applied to the disk to cause it to slip over the shaft.

SOLUTION

Frictional Force on Journal Bearing: Here, fs = tan-1ms = tan-10.15 = 8.531°. Then the radius of friction circle is rf = r sin fs = 0.015 sin 8.531° = 2.225110 -32 m Equation of Equilibrium: a + ©MP = 0;

490.512.2252110-32 - F30.06 - 12.2252110-324 = 0 Ans.

F = 18.9 N

Ans: F = 18.9 N 861

*8–120. The 4-lb pulley has a diameter of 1 ft and the axle has a diameter of 1 in. If the coefficient of kinetic friction between the axle and the pulley is mk = 0.20, determine the vertical force P on the rope required to lift the 20-lb block at constant velocity.

6 in.

SOLUTION Frictional Force on Journal Bearing: Here f k = tan - 1 mk = tan - 1 0.2 = 11.3099°. Then the radius of the friction circle is rf = r sin f k = 0.5 sin 11.3099° = 0.09806 in. Equations of Equilibrium: Referring to the FBD of the pulley shown in Fig. a, a + ΣMP = 0; P(6 - 0.09806) - 4(0.09806) - 20(6 + 0.09806) = 0 Ans.

P = 20.73 = 20.7 lb

Ans: P = 20.7 lb 862

8–121. Solve Prob. 8–120 if the force P is applied horizontally to the left.

6 in.

SOLUTION Frictional Force on Journal Bearing: Here f k = tan - 1mk = tan - 1 0.2 = 11.3099°. Then the radius of the friction circle is rf = r sin f k = 0.5 sin 11.3099° = 0.09806 in. Equations of Equilibrium: Referring to the FBD of the pulley shown in Fig. a. a + ΣMO = 0; P(6) - 20(6) - R(0.09806) = 0 (1)

R = 61.1882 P - 1223.76 + ΣFx = 0; Rx - P = 0 Rx = P S + c ΣFy = 0; Ry - 4 - 20 = 0

Ry = 24 lb

Thus, the magnitude of R is R = 2Rx2 + Ry2 = 2P2 + 242

(2)

Equating Eqs. (1) and (2)

61.1882 P - 1223.76 = 2P2 + 242

3743.00 P2 - 149,760.00 P + 1,497,024.00 = 0 P2 - 40.01 P + 399.95 = 0 chose the root P 7 20 lb, Ans.

P = 20.52 lb = 20.5 lb

Ans: P = 20.5 lb 863

8–122. Determine the tension T in the belt needed to overcome the tension of 200 lb created on the other side. Also, what are the normal and frictional components of force developed on the collar bushing? The coefficient of static friction is ms = 0.21.

2 in. 1.125 in.

SOLUTION Frictional Force on Journal Bearing: Here, fs = tan-1ms = tan-10.21 = 11.86°. Then the radius of friction circle is rf = r sin fk = 1 sin 11.86° = 0.2055 in.

200 lb

20011.125 + 0.20552 - T11.125 - 0.20552 = 0 Ans.

T = 289.41 lb = 289 lb + c Fy = 0;

R - 200 - 289.41 = 0

R = 489.41 lb

Thus, the normal and friction force are N = R cos fs = 489.41 cos 11.86° = 479 lb

Ans.

F = R sin fs = 489.41 sin 11.86° = 101 lb

Ans.

Ans: T = 289 lb N = 479 lb F = 101 lb 864

8–123. If a tension force T = 215 lb is required to pull the 200-lb force around the collar bushing, determine the coefficient of static friction at the contacting surface. The belt does not slip on the collar. 2 in. 1.125 in.

20011.125 + rf2 - 21511.125 - rf2 = 0

200 lb

rf = 0.04066 in. Frictional Force on Journal Bearing: The radius of friction circle is rf = r sin fk 0.04066 = 1 sin fk fk = 2.330° and the coefficient of static friction is Ans.

ms = tan fs = tan 2.330° = 0.0407

Ans: ms = 0.0407 865

*8–124. If the smallest tension required to pull the belt downward at A over the shaft S is TA = 500 N, determine the coefficient of kinetic friction between the loosely fitting collar bushing B and the shaft. Assume that the belt does not slip on the collar; rather, the collar slips on the shaft.

20 mm S B A 46 mm

SOLUTION 400 N

a + ΣMP = 0; 400(23 + rf) - 500(23 - rf) = 0

rf = 2.5556 mm rf = 2.5556 = 20 sin f s; f s = 7.341° Ans.

ms = tan f s = tan 7.341° = 0.129 Also, by approximation, d ≈ ms(20) d A = 23 + 20 ms d B = 23 - 20 ms + ΣMP = 0; - 500(23 - 20ms) + 400(23 + 20ms) = 0 ms = 0.128 Ans. (Approximately)

Ans: ms = 0.129 ms = 0.128 866

8–125. The 5-kg skateboard rolls down the 5° slope at constant speed. If the coefficient of kinetic friction between the 12.5-mm-diameter axles and the wheels is mk = 0.3, determine the radius of the wheels. Neglect rolling resistance of the wheels on the surface. The center of mass for the skateboard is at G.

75 mm G 5 250 mm

300 mm

Fs - 5(9.81) sin 5° = 0

Fs = 4.275 N

©Fy¿ = 0;

N - 5(9.81) cos 5° = 0

N = 48.86 N

The effect of the forces acting on the wheels can be represented as if these forces are acting on a single wheel as indicated on the free-body diagram shown in Fig. b.We have ©Fx¿ = 0;

Rx¿ - 4.275 = 0

Rx¿ = 4.275 N

©Fy¿ = 0;

48.86 - Ry¿ = 0

Ry¿ = 48.86 N

Thus, the magnitude of R is R = 2Rx¿ 2 + Ry¿ 2 = 24.2752 + 48.862 = 49.05 N fs = tan-1 ms = tan-1(0.3) = 16.699°. Thus, the moment arm of R from point O is (6.25 sin 16.699°) mm. Using these results and writing the moment equation about point O, Fig. b, we have a + ©MO = 0;

4.275(r) - 49.05(6.25 sin 16.699° = 0) Ans.

r = 20.6 mm

Ans: r = 20.6 mm 867

8–126. The bell crank fits loosely into a 0.5-in.-diameter pin. Determine the required force P which is just sufficient to rotate the bell crank clockwise. The coefficient of static friction between the pin and the bell crank is ms = 0.3.

12 in.

50 lb

10 in.

P cos 45° - Rx = 0

Rx = 0.7071P

+ c ©Fy = 0;

Ry - P sin 45° - 50 = 0

Ry = 0.7071P + 50

Thus, the magnitude of R is R = 2Rx 2 + Ry 2 = 2(0.7071P)2 + (0.7071P + 50)2 = 2P2 + 70.71P + 2500 We find that fs = tan - 1 ms = tan - 1(0.3) = 16.699°. Thus, the moment arm of R from point O is (0.25 sin 16.699°) mm. Using these results and writing the moment equation about point O, Fig. a, a + ©MO = 0;

50(10) + 2P2 + 70.71P + 2500(0.25 sin 16.699°) - P(12) = 0

Choosing the larger root, Ans.

P = 42.2 lb

Ans: P = 42.2 lb 868

8–127. The bell crank fits loosely into a 0.5-in.-diameter pin. If P = 41 lb, the bell crank is then on the verge of rotating counterclockwise. Determine the coefficient of static friction between the pin and the bell crank.

12 in.

50 lb

10 in.

41 cos 45° - Rx = 0

Rx = 28.991 lb

+ c ©Fy = 0;

Ry - 41 sin 45° - 50 = 0

Ry = 78.991 lb

Thus, the magnitude of R is R = 2Rx 2 + Ry 2 = 228.9912 + 78.9912 = 84.144 lb We find that the moment arm of R from point O is 0.25 sin fs. Using these results and writing the moment equation about point O, Fig. a, a + ©MO = 0;

50(10) - 41(12) - 84.144(0.25 sin fs) = 0 fs = 22.35°

Thus, Ans.

ms = tan fs = tan 22.35° = 0.411

Ans: ms = 0.411 869

*8–128. The lawn roller has a mass of 80 kg. If the arm BA is held at an angle of 30° from the horizontal and the coefficient of rolling resistance for the roller is 25 mm, determine the force P needed to push the roller at constant speed. Neglect friction developed at the axle, A, and assume that the resultant force P acting on the handle is applied along arm BA.

P B

250 mm A

SOLUTION u = sin - 1 a

25 b = 5.74° 250

a + ©MO = 0;

-25(784.8) - P sin 30°(25) + P cos 30°(250 cos 5.74°) = 0

Solving, Ans.

P = 96.7 N

Ans: P = 96.7 N 870

8–129. The tractor has a weight of 16, 000 lb and the coefficient of rolling resistance is a = 2 in. Determine the force P needed to overcome rolling resistance at all four wheels and push it forward.

P 2 ft

SOLUTION Applying Eq. 8–11 with W = 16 000 lb, a = a Wa P L = r

16000 a 2

3 ft

2 b ft and r = 2 ft, we have 12

2 b 12

6 ft

2 ft

Ans.

= 1333 lb

Ans: P = 1333 lb 871

8–130. The cylinder is subjected to a load that has a weight W. If the coefficients of rolling resistance for the cylinder’s top and bottom surfaces are aA and aB, respectively, show that a horizontal force having a magnitude of P = [W(aA + aB)]>2r is required to move the load and thereby roll the cylinder forward. Neglect the weight of the cylinder.

W P A

SOLUTION

+ ©F = 0; : x

(RA)x - P = 0

(RA)x = P

+ c ©Fy = 0;

(RA)y - W = 0

(RA)y = W

a + ©MB = 0;

P(r cos fA + r cos fB) - W(aA + aB) = 0

(1)

Since fA and fB are very small, cos fA - cos fB = 1. Hence, from Eq. (1) P =

W(aA + aB) 2r

(QED)

872

8–131. A large stone having a mass of 500 kg is moved along the incline using a series of 150-mm-diameter rollers for which the coefficient of rolling resistance is 3 mm at the ground and 4 mm at the bottom surface of the stone. Determine the force T that will cause the stone to descend the plane at a constant speed.

308

SOLUTION Q + ΣFy = 0; N - 4905 cos 30° = 0 + a ΣFx = 0; T - 4905 sin 30° + FR = 0

T = 4905 sin 30° - FR

FR =

FR = 198.2 N

T = 4905 sin 30° - 198.2

4247.85(0.003 + 0.004) 2(0.075)

Ans.

T = 2.25 kN

Ans: T = 2.25 kN 873

9–1. Determine the distance x to the center of gravity of the rod bent in the form of a parabola. If the rod has a weight per unit length of 0.5 lb>ft, determine the components of reaction at the fixed support A.

y 1 ft

y 5 2x2 2 ft

SOLUTION The Differential Element:

dL = 2(dx)2 + (dy)2 = a dL = a 21 + 16x2 b dx

1 + a

dy 2 b b dx dx

Since y = 2x2 then

dy = 4x dx

Centroid: x = x LL

x dL = LL

L =

dL =

1L x dL 1L dL ~

x =

x a 31 + 16x2 b dx = 1.4394

a 31 + 16x2 b dx = 2.3234 ft

1.4394 = 0.6195 ft = 0.620 ft 2.3234

Ans.

Equilibrium: + ΣFx = 0; S

Ax = 0

+ c ΣFy = 0;

Ay - 2.3234(0.5) = 0

a+ΣMA = 0;

MA - 2.3234(0.5)(0.6195) = 0

Ans. Ay = 1.16 lb

MA = 0.720 lb # ft

Ans.

Ans: x = 0.620 ft Ax = 0 Ay = 1.16 lb MA = 0.720 lb # ft 873

9–2. Locate the center of gravity ( x , y) of the homogeneous rod. What is its weight if it has a mass per unit length of 3 kg m? Solve the problem by evaluating the integrals using Simpson’s rule.

π x) y = 4 sin (–– 2

SOLUTION y = 4 sin a dL = B

1 + a

L =

x xb ; 2

L0 A

dy p = 2p cos a x b dx 2

2 p dy 2 b R dx = = B 1 + a 2p cos a xb b R dx dx A 2

1 + a 2p cos a

x 2m

p x b b dx = 8.377 m 2

y = y 2

y dL =

2 p p 2 4 sin a x b 1 + a2p cos a x b b dx = 17.23 m 2 A 2 L0

Thus, W = 8.377 (3) (9.81) = 247 N

Ans.

x = 1m

Ans.

Due to symmetry,

y =

y dL

= dL

17.23 = 2.06 m 8.377

Ans.

Ans: W = 247 N x = 1m y = 2.06 m 874

9–3. y

Locate the center of gravity x of the homogeneous rod. If the rod has a weight per unit length of 100 N>m, determine the vertical reaction at A and the x and y components of reaction at the pin B.

1m B

y x2 A

SOLUTION Length and Moment Arm: The length of the differential element is dy dy 2 dL = 2dx2 + dy2 = c 1 + a b d dx and its centroid is ~ x = x. Here = 2x. A dx dx Perform the integration L =

dL =

= 2

= cx

21 + 4x2 dx

x2 +

= 1.4789 m LL

x~ dL =

= 2

1 dx 4

1 1 1 1m + ln ax + x2 + b d 4 4 A 4 0

x21 + 4x2 dx

x2 +

1 dx 4

2 1 3>2 1 m = c ax2 + b d 3 4 0 = 0.8484 m2 Centroid: x =

~ 0.8484 m2 1L x dL = 0.5736 m = 0.574 m = 1.4789 m 1L dL

Ans.

Equations of Equilibrium: Refering to the FBD of the rod shown in Fig. a + ΣFx = 0; S a+ΣMB = 0;

Ans.

Bx = 0 100(1.4789) (0.4264) - Ay(1) = 0

Ans.

Ay = 63.06 N = 63.1 N a+ΣMA = 0;

By(1) - 100(1.4789) (0.5736) = 0 Ans.

By = 84.84 N = 84.8 N

Ans: x = 0.574 m Bx = 0 By = 84.8 N Ay = 63.1 N 875

*9–4. y

Locate the center of gravity y of the homogeneous rod.

1m B

y x2 A

SOLUTION Length and Moment Arm: The length of the differential element is dy dy 2 dL = 2dx2 + dy2 = c 1 + a b d dx and its centroid is ~ y = y. Here = 2x. A dx dx Perform the integration, L =

dL =

= 2

21 + 4x2 dx

= cx

x2 +

= 1.4789 m LL

y~ dL =

= 2

1 dx 4

1 1 1 1m + ln ax + x2 + b d 4 4 A 4 0

x2 21 + 4x2 dx

= 2c

x2 +

1 dx 4

x 1 1 1 1 1m 1 3 2 x x2 + ln ax + x2 + b d ax + b 4A 32 A 4 128 A 4 0 4

= 0.6063 m2 Centroid:

0.6063 m2 1L y dL y = = 0.40998 m = 0.410 m = 1.4789 m 1L dL ~

Ans.

Ans: y = 0.410 m 876

9–5. Locate the center of gravity x of the homogeneous rod. If the rod has a weight per unit length of 0.5 lb>ft, determine the vertical reaction at A and the x and y components of reaction at the pin B.

y 1 ft B

SOLUTION dL = B

2 ft

dy = 4x dx

y = 2x2 ;

1 + a

y 2x2

dy b R dx = = C 21 + (4)2 x2 D dx dx 2

L = 4

1 1 2 1 2 1 1 2 a b + x2 dx = 2 B x x2 + a b + ln ¢ x + a b + x2 ≤ R B 4 16 B 4 0 L0 B 4

= 2.323 ft ' x = x

' xdL = 4

1 2 4 1 2 3 1 a b + x2 R dx = x2 + a b ≤ ` ¢ B 4 3B 4 0 = 1.4394 ft2

x =

' xdL

= dL

1.4394 = 0.6195 = 0.620 ft 2.323

Ans.

(1.162)(0.3805) - A y (1) = 0 Ans.

Ans.

Bx = 0 0.442 - 1.162 + By = 0

Ans.

By = 0.720 lb

Ans: x = 0.620 ft Ay = 0.442 lb Bx = 0 By = 0.720 lb 877

9–6.

Locate the centroid y of the area.

1 1 – – x2 4

1m x 2m

SOLUTION Area and Moment Arm: The area of the differential element is y 1 1 1 ' = a1 - x 2 b . dA = ydx = a1 - x2 b dx and its centroid is y = 4 2 2 4 Centroid: Due to symmetry Ans.

x = 0 Applying Eq. 9–4 and performing the integration, we have 2m

' ydA

y =

1 2 1 1 ¢ 1 - x ≤ ¢ 1 - x2 ≤ dx 2 4 4 L- 2m 2m

L- 2m

¢1 -

1 2 x ≤ dx 4

x3 x5 2m x + ≤` 2 12 160 - 2m x ¢x ≤` 12 - 2m 3

2 m 5

Ans.

Ans: y = 878

2 m 5

9–7. y

Locate the centroid x of the shaded area.

h y = ahn xn

SOLUTION Area

and

Moment

Arm: The area of the differential xn ' dA = 1h - y2dx = h ¢ 1 - n ≤ dx and its centroid is x = x. a

element

Centroid: Applying Eq. 9-6 and performing the integration, we have

x =

' xdA

xBh¢1 -

dA L0

h¢

h¢1 -

xn ≤ dx an

a xn + 2 x2 n≤ ` 2 1n + 22a 0

h x =

xn ≤ dx R an

a xn + 1 n ` 1n + 12a 0

n + 1 a 2 n + 2

Ans.

Ans: x = 879

n + 1 a 2(n + 2)

*9–8. Determine the location (x, y) of the centroid of the area.

y3 5 a x 8

a 2

SOLUTION Area of the differential element dA = x dy =

~ x =

~ y =

~ 1A x dA

a 2

~ 1A y dA

a 2

8 3 x 8 3 y dy and ~ y = y, ~ x = = y. 2 2 a 2a2

8 3 8 3 y a 2 y dyb 2 a 2 L0 2a = a = a 7 8 dA 2 3 1A y dy 2 L0 a

1A dA

Ans.

8 y a 2 y3dyb a 2 = a a 5 8 2 y3 dy 2 L0 a

Ans.

Ans:

2 ∼ x = a 7 2 ∼ y = a 5

880

9–9. y

Locate the centroid x of the area.

4m y

1 2 x 4 x

SOLUTION Area and Moment Arm: The area of the differential element shown shaded in Fig. a 1 is dA = x dy and its centroid is at ~ x = x . Here, x = 2y1>2 2 Centroid: Perform the integration

x =

~ 1A x dA

1A dA

1 a2y1>2 ba2y1>2 dyb 2 L0

2y1>2 dy

3 m 2

Ans.

Ans: x = 881

3 m 2

9–10. y

Locate the centroid y of the area.

4m y

1 2 x 4 x

SOLUTION Area and Moment Arm: The area of the differential element shown shaded in Fig. a y = y. Here, x = 2y1>2. is dA = x dy and its centroid is at ∼ Centroid: Perform the integration

y =

~ 1A y dA

1A dA

y a2y1>2 dyb

2y1>2 dy

4m 4 a y 5>2 b ` 5 0 4m 4 a y3>2 b ` 3 0

12 m 5

Ans.

Ans: y = 882

12 m 5

9–11. y

Locate the centroid x of the area.

h y —2 x2 b

SOLUTION

dA = y dx ' x = x

x =

' x dA

= dA

h 3 x dx 2 L0 b b

h 2 x dx 2 L0 b

h 4 x R 4b2 0

h B 2 x3 R 3b 0

3 b 4

Ans.

Ans: x = 883

3 b 4

*9–12. y

Locate the centroid y of the area.

SOLUTION

h 2 — x b2

dA = y dx y ' y = 2

y =

' y dA

= dA

h 4 x dx 4 L0 2b b

h 2 x dx 2 b L0

h2 5 b x R B 10b4 0

h 3 x R 3b2 0

3 h 10

Ans.

Ans: y = 884

3 h 10

9–13. y

Locate the centroid x of the area.

1 y 5 4 2 –– x2 16

SOLUTION

1 dA = 14 - y2dx = a x 2 b dx 16 ' x = x

x =

' xdA

LA x = 6m

= dA

L0 L0

xa a

1 2 –– x 16 8m

x2 b dx 16

1 2 x b dx 16 Ans.

Ans: x = 6m 885

9–14. y

Locate the centroid y of the area.

SOLUTION

dA = 14 - y2dx = a y =

y =

1 2 x b dx 16

1 2 –– x 16 8m

4 + y 2

' ydA

= dA

x2 x2 1 b a b dx ¢8 2 L0 16 16 8

1 2 x b dx 16 Ans.

y = 2.8 m

Ans: y = 2.8 m 886

9–15. y

Locate the centroid x of the area. Solve the problem by evaluating the integrals using Simpson’s rule.

y 5 0.5ex2

SOLUTION 1m

A = 3 0.5 e x dx = 0.7313 1

∼ x2 3 x dA = 3 x(0.5 e ) dx = 0.4296 1

x =

0.4296 = 0.587 m 0.7313

Ans.

Ans: x = 0.587 m 887

*9–16. y

Locate the centroid y of the area. Solve the problem by evaluating the integrals using Simpson’s rule.

y 5 0.5ex2

SOLUTION 1m

A = 3 0.5 e x dx = 0.7313 1

3 y dA = 3

1 2 (0.5)2(e 2x ) dx = 0.2956 2 0

y =

0.2956 = 0.404 m 0.7313

Ans.

Ans: y = 0.404 m 888

9–17. y

Locate the centroid of the area.

y = a sin π–x L a x

SOLUTION dA = y dx ∼

x = x

∼

y =

y 2 px

dA =

ydA =

ydA

LA x =

L 2

2px

1 px a2 sin L c - 4p a2 sin2 dx = 2 L0 L 2 L

∼

y =

a cos L L px 2aL dx = c d = p p L 0 L

∼

a sin

a2L 4

= 2aL = p

x L a2L d = 2 0 4

ap 8

Ans.

(By symmetry)

Ans: ap 8 L x = 2 y =

889

9–18. x of shaded area. Solve theproblem problem by x the of the area. Solve the Locate the thecentroid centroid evaluating the evaluating theintegrals integralsusing usingSimpson’s Simpson’srule. rule.

y y = 0.5ex2

SOLUTION SOLUTION At x = 1 m 5/3 y = 0.5e1 A==1.359 m1/2 3 2x + 2x dx = 3.396 2

2 dA = (1.359 - y) dx = a 1.359 = 0.5 ex b dx = 0.6278 m2 ∼ L0= x 2x1/2 + 2x5/3 dx L0 = 4.1607 LA x dA 3 3 0 x = x 4.1607 x =1 = 1.23 ft 3.396 2 x dA = x a 1.359 - 0.5 ex b dx LA L0

Ans.

= 0.25 m3

x =

x dA LA LA

0.25 = 0.398 m 0.6278

Ans.

Ans: x = 1.23 ft 890

9–19. Locate the centroid y of the area. Solve the problem by evaluating the integrals using Simpson’s rule.

y y = 0.5ex2

SOLUTION 1 SOLUTION

(1.359 - y) dx = a 1.359 - 0.5ex b dx = 0.6278 m 2 0 L0 L LA A = 3 2x1/2 + 2x5/3 dx = 3.396 1.359 + y 0 y = 2 2 2 3 1 1/2 1 2 3/2 5/3 8/3 y dA = (x + 2x ) dx = c x + 2a b(x ) d = 3.3239 3 3 2 3 8 2 10 2 0 1.359 + 0.5 ex 2 y dA = a b A 1.359 - 0.5 ex B dx 0 3.3239 2 LA L y = = 0.980 ft 3.396 1 1 2 a 1.847 - 0.25 e2x b dx = 0.6278 m3 = 2 L0 dA =

y =

y dA LA dA

0.6278 = 1.00 m 0.6278

x 1m

Ans.

Ans: y = 0.980 ft 891

*9–20. Locate the centroid (x, y) of the exparabolic segment of area.

a x

SOLUTION

y 5 2b x2 a2

b 2 b 1 3 1 3 dA = 3 a2 (x ) dx = a2 a 3 a b = 3 ab 0 A a

x = -x

b 3 b 1 4 1 2 3 x‾ dA = 3 - a2 (x ) dx = - a2 a 4 a b = - 4 ba 0 A a

3 x‾ dA

1 - ba2 4 3 x‾ = = = - a 1 4 ab 3 dA 3 A A

Ans.

1 1 b y = - y = - a 2 x2 b 2 2 a

1 b2 1 2 y dA = a 4 b (x4) dx = (b a) ‾ 3 3 2 10 a 0 A a

y‾ =

3 y‾ dA A

3 dA A

1 2 (b a) 10 3 = b 1 10 ab 3

Ans.

Ans: 3 x‾ = - a 4 3 y‾ = - a 10 892

9–21. y

Locate the centroid x of the area.

y (4 x2 )2

16 ft

4 ft x

4 ft

SOLUTION Area and Moment Arm: The area of the differential element shown shaded in Fig. a is dA = y dx = ( 4 - x1>2 ) 2 dx = (x - 8x1>2 + 16)dx and its centroid is at ~ x = x. Centroid: Perform the integration

x =

~ 1A x dA

1A dA

4 ft

L0 a

a = 1

x(x - 8x1>2 + 16)dx

4 ft

( x - 8x1>2 + 16) dx

4 ft x3 16 5>2 x + 8x2 b ` 3 5 0

4 ft x2 16 3>2 x + 16xb ` 2 3 0

3 ft 5

Ans.

Ans:

3 x = 1 ft 5

893

9–22. y

Locate the centroid y of the area.

y (4 x2 )2

16 ft

4 ft x

4 ft

SOLUTION Area and Moment Arm: The area of the differential element shown shaded in 1

Fig. a is dA = y dx = ( 4 - x 2 ) 2 dx = (x - 8x1>2 + 16)dx and its centroid is at y 1 = ( 4 - x1>2 ) 2. y~ = 2 2 Centroid: Perform the integration,

y =

~ 1A y dA

1A dA

4 ft

1 ( 4 - x1>2 ) 2 ( x - 8x1>2 + 16 ) dx 2 L0

4 ft

(x - 8x1>2 + 16)dx

1 a x2 - 8x3>2 + 48x - 128x1>2 + 128bdx 2 L0

4 ft

( x - 8x1>2 + 16 ) dx

4 ft x3 16 5>2 256 3>2 x + 24x2 x + 128xb ` 6 5 3 0

= 4.15 ft

4 ft 16 3>2 x2 x + 16xb ` 2 3 0

Ans.

Ans: y = 4.15 ft 894

9–23. y

Locate the centroid x of the area.

y h2 x2 h a h

SOLUTION Area and Moment Arm: The area of the differential element shown shaded in Fig. a h is dA = y dx = a - 2 x2 + hbdx and its centroid is at ~ x = x. a Centroid: Perform the integration,

x =

~ 1A x dA

1A dA

h x a - 2 x2 + hbdx a L0 a

h 2 a - 2 x + hbdx L0 a

h 4 h 2 a x + x b` 2 4a2 0 = a h 3 a - 2 x + hxb ` 3a 0 a-

3 a 8

Ans.

Ans: x = 895

3 a 8

*9–24. y

Locate the centroid y of the area.

y h2 x2 h a h

SOLUTION Area and Moment Arm: The area of the differential element shown shaded in Fig. a y h 1 h2 is dA = y dx = a - 2 x2 + hbdx and its centroid is at ~ y = = a - x2 + hb. 2 2 a a Centroid: Perform the integration,

y =

~ 1A y dA

1 h h a - 2 x2 + hba - 2 x2 + hbdx a a L0 2 = a h dA 2 1A a - 2 x + hbdx L0 a =

a 1 h2 5 2h2 3 2 a 4x x + h xb ` 2 5a 3a2 0

2 h 5

a h 3 x + hxb ` 2 3a 0

Ans.

Ans: y = 896

2 h 5

9–25. z

The plate has a thickness of 0.25 ft and a specific weight of g = 180 lb>ft3. Determine the location of its center of gravity. Also, find the tension in each of the cords used to support it.

SOLUTION 1 2

Area and Moment Arm: Here, y = x - 8x + 16. The area of the differential 1 ' element is dA = ydx = 1x - 8x 2 + 162dx and its centroid is x = x and 1 1 ' y = 1x - 8x2 + 162. Evaluating the integrals, we have 2 16 ft

A =

16 ft

dA =

1 y2

1 x2

C y

1x - 8x2 + 162dx 1

16 ft 1 16 3 = a x2 x 2 + 16x b ` = 42.67 ft2 2 3 0

' xdA =

16 ft

x31x - 8x2 + 162dx4

16 ft 1 16 5 = a x3 x 2 + 8x 2 b ` = 136.53 ft3 3 5 0

' ydA =

16 ft

1 1 1 1x - 8x2 + 16231x - 8x2 + 162dx4 2

16 ft 1 1 3 32 5 512 3 a x x 2 + 48x2 x2 + 256x b ` 2 3 5 3 0

= 136.53 ft3 Centroid: Applying Eq. 9–6, we have

x =

' xdA

y =

136.53 = 3.20 ft 42.67

Ans.

136.53 = 3.20 ft 42.67

Ans.

' ydA

192013.202 - TA1162 = 0

TA = 384 lb

Ans.

©My = 0;

TC1162 - 192013.202 = 0 TC = 384 lb

Ans.

©Fz = 0;

TB + 384 + 384 - 1920 = 0 Ans.

TB = 1152 lb = 1.15 kip 897

Ans: x = 3.20 ft y = 3.20 ft TA = 384 lb TC = 384 lb TB = 1.15 kip

9–26. y

Locate the centroid x of the area.

4 ft

1 2 x 4

SOLUTION Area and Moment Arm: The area of the differential element shown shaded in Fig. a 1 is dA = y dx = x2 dx and its centroid is at ~ x = x. 4 Centroid: Perform the integration

x =

~ 1A x dA

1A dA

4 ft

L0 a a

x 4 ft

1 x a x2 dxb 4

4 ft

1 2 x dx 4

1 4 4 ft x b` 16 0 1 3 4 ft x b` 12 0

= 3 ft

Ans.

Ans: x- = 3 ft 898

9–27. y

Locate the centroid y of the area.

4 ft

1 2 x 4

SOLUTION Area and Moment Arm: The area of the differential element shown shaded in Fig. a y 1 1 1 1 is dA = y dx = x2 dx and its centroid is located at ~ y = = a x2 b = x2. 2 2 4 8 4 Centroid: Perform the integration,

y =

~ 1A y dA

1A dA

4 ft

1 2 1 2 x a x dxb 8 4

L0 6 ft 5

4 ft

1 2 x dx 4

Ans.

Ans: y = 899

6 ft 5

*9–28. y

Locate the centroid x of the area.

y x

100 mm

y 1 x2 100 100 mm

SOLUTION

1 2 x . Thus the area of the 100 1 2 x ) dx differential element shown shaded in Fig. a is dA = ( y2 - y1 ) dx = ( x 100 ~ and its centroid is at x = x. Area and Moment Arm: Here, y2 = x and y1 =

Centroid: Perform the integration

x =

~ 1A x dA

1A dA

100 mm

L0 a a

x ax -

100 mm

ax -

1 2 x bdx 100

x3 1 4 100 mm x b` 3 400 0

x2 1 3 100 mm x b` 2 300 0

= 50.0 mm

Ans.

Ans: x = 50.0 mm 900

9–29. y

Locate the centroid y of the area.

y x

100 mm

y 1 x2 100 100 mm

SOLUTION Area and Moment Arm: Here, x2 = 10y 1>2 and x1 = y. Thus, the area of the differential element shown shaded in Fig. a is dA = ( x2 - x1 ) dy = ( 10y1>2 - y ) dy and its centroid is at ~ y = y. Centroid: Perform the integration,

y =

~ 1A y dA

1A dA

100 mm

y a10y1>2 - ybdy

100 mm

a4y 5>2 -

1>2

a10y

- ybdy

y3 100 mm b` 3 0

y2 100 mm 20 3>2 y - b` 3 2 0

= 40.0 mm

Ans.

Ans: y = 40.0 mm 901

9–30. y

Locate the centroid x of the area.

h y –– ax

y (

h )(x b) a b x

SOLUTION

a a - b y and x2 = a b y + b. Thus the area h h a a - b by + b - y d dy = of the differential element is dA = ( x2 - x1 ) dy = c a h h x -x b a b b 1 ( b - y ) dy and its centroid is at ~x = x1 + 2 1 = ( x2 + x1 ) = y - y + . h h 2h 2 2 2 Area and Moment Arm: Here x1 =

Centroid: Perform the integration,

x =

~ 1A x dA

1A dA

a b b b a y y + b c ab - ybdy d h 2h 2 h L0 L0

ab -

b ybdy h

h b b b2 (a - b)y2 + (b - 2a)y3 + yd ` 2 2h 2 6h 0

bh (a + b) 6 = bh 2 1 = (a + b) 3

aby -

b 2 h y b` 2h 0

Ans.

Ans: x = 902

1 (a + b) 3

9–31. y

Locate the centroid y of the area.

h y –– ax

y (

h )(x b) a b x

SOLUTION

a - b a by + b. Thus the area y and x2 = a h h a a - b by + b - y d dy = of the differential element is dA = ( x2 - x1 ) dy = c a h h b ab - ybdy and its centroid is at ~ y = y. h Centroid: Perform the integration,

Area and Moment Arm: Here, x1 =

y =

~ 1A y dA

1A dA

y ab -

ab -

b ybdy h b ybdy h

b b 3 h a y2 y b` 2 3h 0 aby -

b 2 h y b` 2h 0

1 2 bh h 6 = = 1 3 bh 2

Ans.

Ans: y = 903

h 3

*9–32. y

Locate the centroid x of the area.

SOLUTION

Area and Moment Arm: The area of the differential element is x dA = ydx = a sin dx and its centroid are x = x a

x =

a sin ax

' xdA

= dA

x aa sin

x dxb a

x dx a

c a3 sin

a sin

pa x x - x a a2 cos b d ` a a 0

a -a2 cos

x pa b` a 0

p a 2

Ans.

Ans: x = 904

p a 2

9–33. y

Locate the centroid y of the area.

SOLUTION

a sin ax

Area and Moment Arm: The area of the differential element is y a x x = sin . dA = ydx = a sin dx and its centroid are y = a a 2 2

pa 1 2x 1 2 x x a bd` c a a x - a sin sin aa sin dxb a 4 2 a a 0 pa L0 2 y = LA = = = pa pa 8 x x 2 dA a sin dx a -a cos b ` a LA L0 a 0 pa

ydA

Ans.

Ans: y = 905

pa 8

9–34. Locate the center of gravity of the homogeneous cantilever beam and determine the reactions at the fixed support. The material has a density of 8 Mg>m3.

4m y

1m z

SOLUTION 0

1 — x2 16

0.5 m x

1 dV = (0.5)|z| dx = 0.5 a x2 bdx = 0.667 m3 L-4 L0 16 LV ' x = -x |z| ' z = 2

' xdV =

x =

' z dV =

z =

1 3 x dx = - 2 m4 16

-2 = -3 m 0.667

y = 0.25 m LV

-x(0.5|z|dx) = 0.5

Ans. Ans.

(By symmetry) 0

1 z 1 4 x dx = -0.2 m4 a - b (0.5) (|z|dx) = - a b 2 4 256 L0 L-4

-0.2 = -0.3 m 0.667

Ans.

Ax = Ay = 0

+ c ©Fy = 0;

A z - 52.3 = 0

Ans.

MA - 52.3(1) = 0 MA = 52.3 kN # m

Ans.

Ans: x = -3 m y = 0.25 m z = -0.3 m Ax = 0 Az - 52.3 = 0 Az = 52.3 kN

MA = 52.3 kN # m 906

9–35. y

Locate the centroid x of the area.

h y h —n xn a h

h y h — a x x

SOLUTION

h n h x and y1 = h - x. Thus, the an a area of the differential element shown shaded in Fig. a is dA = ( y2 - y1 ) dx

Area and Moment Arm: Here, y2 = h -

= ch -

h n h h h x - ah - xb d dx = a x - n xn bdx and its centroid is ~ x = x. an a a a

Centroid: Perform the integration

x =

~ 1A x dA

1A dA

h h x a x - n xn bdx a a L0 L0

h h n x - n x bdx a a

a h 3 h x - n xn + 2 d ` 3a a (n + 2) 0

a h 2 h x - n xn + 1 b ` 2a a (n + 1) 0

ha2 (n - 1) =

3(n + 2) ha(n - 1) 2(n + 1)

= c

2(n + 1) 3(n + 2)

Ans.

Ans: x = c 907

2(n + 1) 3(n + 2)

*9–36. y

Locate the centroid y of the area.

h y h —n xn a h

h y h — a x x

SOLUTION

h n h x and y1 = h - x. Thus, the an a area of the differential element shown shaded in Fig. a is dA = ( y2 - y1 ) dx h h h h = c h - n xn - ah - xb d dx = a x - n xn bdx and its centroid is at a a a a

Area and Moment Arm: Here, y2 = h -

∼

y = y1 + a

y2 - y1 1 1 h h 1 h h b = ( y2 + y1 ) = ah - n xn + h - xb = a2h - n xn - xb. 2 2 2 a a 2 a a

Centroid: Perform the integration

y =

~ 1A y dA

h 1 h h h a2h - n xn - xba x - n xn bdx 2 a a a a L0 = a h h n dA 1A a x - n x bdx a a L0 =

a 1 h2 2 h2 2h2 h2 x2n + 1 d ` c x - 2 x3 - n xn + 1 + 2n 2 a a (n + 1) 3a a (2n + 1) 0

h2a c

= c

a h 2 h x - n xn + 1 b ` 2a a (n + 1) 0

(4n + 1) (n - 1)

6(n + 1)(2n + 1) hac

n - 1 d 2(n + 1)

(4n + 1)

3(2n + 1)

Ans.

d h

Ans: y = c 908

(4n + 1) 3(2n + 1)

d h

9–37. Determine the distance y‾ to the centroid of the cone.

z z 5 ha y

a y

SOLUTION a2 The volume of the differential thin disk element dV = pz2dy = p a 2 y2 b dy and h ~ y = y.

~ y =

~ 1V y dV

1V dV

a2 y c p a 2 y2 bdy d h L0 h

a p a 2 y2 bdy h L0

3 h 4

Ans.

Ans: 3 ~ y = h 4 909

9–38. Determine the distance y‾ to the center of mass of the cone. The density of the material varies linearly from zero at the origin to r0 at x = h.

z z 5 ha y

a y

SOLUTION The mass of the differential thin disk element dm = rpz2dy and ~ y = y. However, pr0a2 3 r0 r0 a2 2 r = y, then dm = a yb p a 2 y b dy = y dy h h h h3

~ y =

~ 1m y dm

1m dm

pr0 a2 h3

pr0 a 3

y3dy d 3

y dy

4 h 5

Ans.

Ans: 4 ~ y = h 5 910

9–39. Locate the centroid of the volume formed by rotating the shaded area about the aa axis.

z a 1m

z2 = 2y

SOLUTION z =

1 z = 22 2y 2 2

dV = 2p(3 - y)(z) dy = 222 p a 3y2 - y2 b dy 1

V = 2 22 p

z dV = 2p

z =

a 2

2 5 y2 d = 30.16 m3 5 0

1 222y (2p)(3 - y) A 22 2y B dy = 2p A 3y - y2 B dy 2

z dV =

a 3y2 - y2 b dy = 222 pc 2y2 1

z dV

= dV

A 3y - y2 B dy = 2pc y2 - y3 d = 20.94 m4 3 2

1 3

2 0

20.94 = 0.694 m 30.16

Ans.

Due to symmetry: y = 3m

Ans.

x = 0

Ans.

Ans: x = 0 y = 3m z = 0.694 m 911

*9–40. z

Locate the centroid y of the paraboloid.

z2 = 4y 4m y

SOLUTION Volume and Moment Arm: The volume of the thin disk differential element is ' dV = pz2dy = p14y2dy and its centroid y = y.

Centroid: Applying Eq. 9–3 and performing the integration, we have

y =

' ydV

= dV

y3p14y2dy4 4m

p14y2dy

L0 4p =

y3 3

y 4p 2

Ans.

= 2.67 m

Ans: y = 2.67 m 912

9–41. Locate the centroid of the solid.

z y2 5 a (a2–) 2

SOLUTION

Volume and Moment Arm: The volume of the thin disk differential element is z z z = z. dV = py2 dz = xc aaa - b d dz = paaa - bdz and its centroid is at ~ 2 2

Centroid: Due to symmetry about the z axis ~ x =~ y = 0

Ans.

Applying Eq.9–5 and performing the integration, we have

& z =

~ 1V z dV

1V dV

paa =

zc paaa -

az2 z3 2a - b` 2 6 0

z2 2a paaaz - b ` 4 0

paaa -

z bdz d 2

z bdz 2

2 a 3

Ans.

913

9–42. Locate the centroid x of the solid.

y y = x 3/2

8 in. C

SOLUTION

–x

dV =

p y2 dx =

p A x3dx B = p

1 424 x = 64p in3 4 0

x = x 4

x dV =

x =

x A p y2 dx B =

x dV

= dV

4 in.

1 px4 dx = pa x5 b 2 = 204.8p in4 5 0

204.8 p = 3.20 in. 64 p

Ans.

Ans: x = 3.20 in. 914

9–43. Locate the center of gravity for the homogeneous half-cone.

z (a/h)y

SOLUTION x

Due to symmetry:

Ans.

x = 0 dV =

1 2 1 a2 pz dy = p a 2 b y2 dy 2 2 h

' y = y Since the centroid of a semicircle is located at

4r , then 3p

4a y 4z ' = z = 3p 3hp

y =

z =

' ydV dV

= dV

1 a p a 2 b y2 dy L0 2 h h

' z dV

1 a2 3 p a 2 b y dy L0 2 h

3 h 4

4a y 1 a 2 2 b p a b y dy 3hp 2 h2 h

a 1 p a 2 b y2dy h L0 2

Ans.

a p

Ans.

Ans: x = 0 3 y = h 4 a z = p 915

*9–44. z

The king’s chamber of the Great Pyramid of Giza is located at its centroid. Assuming the pyramid to be a solid, prove that this point is at z = 14 h, Suggestion: Use a rectangular differential plate element having a thickness dz and area (2x)(2y).

SOLUTION dV = 12x212y2 dz = 4xy dz x = y =

a x

a 1h - z2 h

L L

dV =

' z dV =

4 a2 z3 h 4 a2 h 4 a2 1h - z22 dz = 2 B h2z - hz2 + R = 2 3 0 3 h L0 h h

4 a2 z2 4 a2 a 2 h2 z3 z4 h - 2h + R = 1h - z22z dz = 2 B h2 2 2 3 4 0 3 h L0 h

' L z =

' z dV

= dV

a 2 h2 3 4 a2 h 3

h 4

(QED)

Ans: z = 916

h 4

9–45. Determine the distance y to the centroid of the bell-shaped volume.

z 1 ft

z = y2

1 ft

SOLUTION The volume of the differential thin disk element dV = pz2dy = py4dy and ~ y = y. 1

y 1py dy2 ~ L0 1V y dV ~ y = = 0.833 ft = 1 1V dV py4dy L0 4

Ans.

Ans: ~ y = 0.833 ft 917

9–46. Determine the distance y to the centroid of the semiellipsoid.

y2 1 z2 5 1 a2 b2

SOLUTION

The volume of the differential thin-disk element dV = pz2dy = pb2 a1 -

and ~ y = y.

~ y =

~ 1V y dV

1V dV

yc pb2 a1 -

pb a1 -

y2 a2 2

bdy

3 a 8

y2 a2

b dy

Ans.

Ans: 3 ~ y = a 8 918

9–47. Locate the center of mass x of the hemisphere. The density of the material varies linearly from zero at the origin O to r0 at the surface. Hint: Choose a hemispherical shell element for integration.

SOLUTION

x ' x = 2

x r = r0 a b a dV = 2 p x2 dx dW = r dV = a

x =

2p b r0 x3 dx a a

' xdW = dW

L0 2

a a

2p r b x3 dx a 0

5 a

1 x c d 2 5 0 c

x4 a d 4 0

= 0.4 a

Ans.

Ans: x = 0.4 a 919

*9–48. z

Determine the distance: y to the center of gravity of the volume. The material is homogeneous.

z 1 y2 100 4 in. 1 in.

10 in.

SOLUTION Volume and Moment Arm: The volume of the thin disk differential element shown 1 2 2 p shaded in Fig. a is dV = pz2 dy = pa y b dy = y4 dy and its centroid is 100 10000 y = y. at ~ Centroid: Perform the integration

y =

~ 1V y dV

1V dV

20 in.

L10 in.

20 in.

p y4 dyb 10000

L10 in. 10000 a a

y4 dy

20 in. p y6 b ` 60000 10 in. 20 in. p y5 b ` 50000 10 in.

Ans.

= 16.94 in. = 16.9 in.

Ans: y = 16.9 in. 920

9–49. Locate the center of gravity y of the volume. The material is homogeneous.

y2 100z 100 mm 25 mm

SOLUTION y4 The volume of the differential thin disk element dV = pz2 dy = p a bdy and 10 000 ' y = y.

y =

100

' ydV

= dV

L50

y c pa

100

L50

y4 b dy d 10 000

y4 pa b dy 10 000

50 mm

Ans.

= 84.7 mm

Ans: y = 84.7 mm 921

9–50. Determine the location z of the centroid for the tetrahedron. Suggestion: Use a triangular “plate” element parallel to the x–y plane and of thickness dz.

b a

SOLUTION z = c a1 -

1 1 y b = ca 1 - xb a b c

dV =

c y

1 1 z abc z (x)(y)dz = a a 1 - b ba1 - b dz = c c 2 L0 6 L0 2

z a b c2 z 1 ' z a a1 - b ba1 - bdz = z dV = c c 2 L0 24 L a b c2 c 24 L = z = = abc 4 dV 6 L ' z dV

Ans.

Ans: z = 922

c 4

9–51. The truss is made from five members, each having a length of 4 m and a mass of 7 kg>m. If the mass of the gusset plates at the joints and the thickness of the members can be neglected, determine the distance d to where the hoisting cable must be attached, so that the truss does not tip (rotate) when it is lifted.

d B 4m

4m 4m

SOLUTION

Ans.

Ans: d = 3m 923

*9–52. z

Determine the location (x, y, z) of the centroid of the homogeneous rod.

200 mm x 30 600 mm 100 mm

SOLUTION

Centroid: Referring to Fig. a, the length of the segments and the locations of their respective centroids are tabulated below Segment L(mm)

x~(mm)

y~(mm)

z~(mm)

x~ L(mm2)

100

y~ L(mm2) z~ L(mm2) 0

20.0 ( 103 )

200

600

300 cos 30° 300 sin 30°

155.88 ( 103 ) 90.0 ( 103 )

100

600 cos 30° 600 sin 30°

–50

51.96 ( 103 ) 30.0 ( 103 ) -5.0 ( 103 )

900

207.85 ( 103 ) 120.0 ( 103 ) 15.0 ( 103 )

Thus, x =

207.85 ( 10 ) mm Σ~ xL = = 230.94 mm = 231 mm ΣL 900 mm

Ans.

y =

120.0 ( 103 ) mm2 Σ~ yL = = 133.33 mm = 133 mm ΣL 900 mm

Ans.

z =

15.0 ( 10 ) mm Σ~ zL = = 16.67 mm = 16.7 mm ΣL 900 mm2

Ans.

Ans: x = 231 mm y = 133 mm z = 16.7 mm 924

9–53. Determine the location y of the centroid C of the “rollformed” member. Neglect the thickness of the material and any slight bends at the corners.

SOLUTION

~ 2 a y L = 2 [20(150) + 150(75) + 40(0)] = 28.500 mm

150 mm y

a L = 2 [20 + 150 + 40] = 420 mm y =

x 40 mm

Σy~L 28.500 = = 67.9 mm ΣL 420

Ans.

40 mm 20 mm

Ans: y = 67.9 mm 925

9–54. y

Locate the centroid (x, y) of the metal cross section. Neglect the thickness of the material and slight bends at the corners. 50 mm

150 mm

SOLUTION

Centroid: The length of each segment and its respective centroid are tabulated below. Segment

L (mm)

' y (mm)

' y L (mm2)

50p

168.17

26415.93

180.28

13520.82

400

180.28

13520.82

917.63

53457.56

- = 0 Due to symmetry about y axis, x

y =

50 mm 100 mm 100 mm 50 mm

Ans.

Ans: x = 0 y = 58.3 mm 926

9–55. Determine the location (x, y) of the centroid of the sheet metal cross section. Neglect the thickness of the material and slight bends at the corners.

2 in.

SOLUTION

a L = 2p(2) + 8 + 3 + 6 = 29.57 in.

8 in.

~ 2 a x L = 3[2p(2)] + 3(8) + 1.5(3) + 3(6) = 84.20 in 3 in.

~ 2 a y L = 10[2p(2)] + 4(8) + 0(3) + 0(6) = 157.66 in

x =

Σx~L 84.20 = = 2.85 in. ΣL 29.57

Ans.

y =

Σy~L 157.66 = = 5.33 in. ΣL 29.57

Ans.

3 in.

Ans: x = 2.85 in. y = 5.33 in. 927

*9–56. The steel and aluminum plate assembly is bolted together and fastened to the wall. Each plate has a constant width in the z direction of 200 mm and thickness of 20 mm. If the density of A and B is rs = 7.85 Mg>m3, and for C, ral = 2.71 Mg>m3, determine the location x of the center of mass. Neglect the size of the bolts.

100 mm 200 mm A

SOLUTION

B 300 mm

©m = 2 C 7.85(10)3(0.3)(0.2)(0.02) D + 2.71(10)3(0.3)(0.2)(0.02) = 22.092 kg ' ©xm = 150{2 C 7.85(10)3(0.3)(0.2)(0.02) D }+350 C 2.71(10)3(0.3)(0.2)(0.02) D = 3964.2 kg.mm x =

Ans.

Ans: x = 179 mm 928

9–57. Determine the location (x, y) of the centroid of the rod. Neglect the rod’s thickness.

150 mm

100 mm x 100 mm

SOLUTION Due to symmetry y- = 0

x =

Ans.

150[2p(150)] + 0(200) Σx~L = = 124 mm ΣL 2p(150) + 200

150 mm

Ans.

Ans: y- = 0 x = 124 mm 929

9–58. The rectangular horn antenna is used for receiving microwaves. Determine the location x of its center of gravity G. The horn is made of plates having a constant thickness and density and is open at each end.

_ x

0.3 ft 30° G

0.1 ft 30°

SOLUTION

0.3 ft

0.4 ft

0.6 ft

1 0.6 2 (0.6) b a b (0.6)(0.6 tan 30°) + 2(0.4 + 0.3)(0.3) a b 3 2 cos 30°

= 0.7715 ft3 0.6 1 b ©A = 2(0.3)(0.4) + 2(0.1)(0.4) + 2(0.1)(0.6) + 4 a b (0.6)(0.6 tan 30°) + 2(0.3) a 2 cos 30° = 1.2714 ft2 x =

Ans.

Ans: x = 0.607 ft 930

9–59. The truss is made from five members, each having a length of 4 m and a mass of 7 kg>m. If the mass of the gusset plates at the joints and the thickness of the members can be neglected, determine the distance d to where the hoisting cable must be attached, so that the truss does not tip (rotate) when it is lifted.

d B 4m

4m 4m 608

SOLUTION

2b1h2 + b2h2 b2 - b1 h h Σ∼ yA = b1ha b + ha ba b = 2 2 3 6 ΣA = b1h + ha & Σy A = y- = ΣA

b1h + b2h b2 - b1 b = 2 2

2b1h2 + b2h2 h 2b1 + b2 6 = a b b1h + b2h 3 b1 + b2 2

Ans.

Ans: h 2b1 + b2 y= a b 3 b1 + b2 931

*9–60. Locate the centroid y for the beam’s cross-sectional area.

120 mm

240 mm

SOLUTION

x y

1 ©A = 2 c (120)(240) d + 120(240) + 600(120) = 129 600 mm2 2 1 2 ' ©yA = (240)c 2a b (120)(240) d + 120(120)(240) + 300(600) (120) = 29 664 000 mm3 3 2 y =

120 mm

Ans.

Ans: y = 229 mm 932

9–61. Determine the location y of the centroid for the beam’s cross-sectional area.

15 mm

300 mm

15 mm 400 mm y

15 mm

SOLUTION

200 mm

Centroid: The location of the centroid measuring from x axis for segments 1, 2 and 3 are indicated in Fig. a. Thus y =

422.5(15)(300) + 215(400)(15) + 7.5(15)(200) 15(300) + 400(15) + 15(200) Ans.

= 238.06 mm = 238 mm

Ans: y = 238 mm 933

9–62. y

Locate the centroid (x, y) of the area.

6 in. 3 in.

6 in. x

SOLUTION

6 in.

Centroid: Referring to Fig. a, the areas of the segments and the locations of their respective centroids are tabulated below Segment

A(in.2)

x~(in.)

y~(in.)

x~A(in.3)

y~A(in.3)

1 (6)(9) 2

54.0

162.0

1 (6)(3) 2

-2

-18.0

63.0

6(6)

-3

-108.0

108.00

72.0

-72.0

333.0

Thus, x =

Σx~A - 72.0 in.3 = = - 1.00 in. ΣA 72.0 in.2

Ans.

y =

Σy~A 333.0 in.3 = = 4.625 in. ΣA 72.0 in.2

Ans.

Ans: x = - 1.00 in. y = 4.625 in. 934

9–63. y

Locate the centroid y of the cross-sectional area of the beam. 8 in. 6 in.

8 in. 1 in.

C 1 in.

SOLUTION Centroid: The area of each segment and its respective centroid are tabulated below. Segment

A (in2)

' y (in.)

' y A(in3)

14(1)

15.5

217.0

6(2)

156.0

16(1)

128.0

42.0

10 in.

y x

1 in.

501.0

Thus, y =

Ans.

Ans: y = 11.9 in. 935

*9–64. y

Determine the location (x, y) of the centroid of the area.

6 in. 2 in. x 6 in.

3 in.

SOLUTION 1 1 6(12)(2) + 1a b(4)(3) + 11a b(4)(3) ~ 2 2 Σx A x = = = 6.00 in. ΣA 1 (12)(2) + 2a b(4)(3) 2

3 in.

Ans.

This result could also be determined by symmetry.

1 1(12)(2) + 2c 3.333a b(4)(3) d Σy~A 2 y = = = 1.78 in. ΣA 1 (12)(2) + 2a b(4)(3) 2

Ans.

Ans: x = 6.00 in. y = 1.78 in. 936

9–65. y

Determine the location y of the centroid of the area. a 2

a 2 a

SOLUTION 23a 23a 23a 1 a 23a ba ba ba b + a b(a) (2)a Σy~A 6 2 2 2 4 2 y = = = 0.385 a ΣA 1 a 23a 23a b + 2a ba ba (a) 2 2 2 2

Ans.

Ans: y = 0.385 a 937

9–66. x

The rectangular horn antenna is used for receiving microwaves. Determine the location xX of its center of gravity G. The horn is made of plates having a constant thickness and density and is open at each end.

0.3 ft 308 0.1 ft

308 0.3 ft 0.4 ft

SOLUTION

0.6 ft

Using symmetry about the y axis, Σ∼ yA = 150(25)(12.5) + 100(25)(75) = 234 375 mm3 ΣA = 150(25) + 25(100) = 6250 mm2 & Σy A 234 375 y = = = 37.5 mm ΣA 6250

Ans.

Ans: y = 37.5 mm 938

9–67. Locate the centroid y of the cross-sectional area of the beam constructed from a channel and a plate. Assume all corners are square and neglect the size of the weld at A.

20 mm y 350 mm C A

10 mm 70 mm

SOLUTION

325 mm

Centroid: The area of each segment and its respective centroid are tabulated below. Segment

A (mm2)

' y (mm)

' y A (mm3)

350(20)

175

1 225 000

630(10)

355

2 236 500

70(20)

385

539 000

14 700

4 000 500

Ans.

Ans: ∼ y = 272 mm 939

*9–68. Determine the distance y‾ to the centroid of the area.

150 mm 150 mm

600 mm

C 100 mm

y x

SOLUTION

300 mm

Centroid: Referring to Fig. a, the areas of the segments and the locations of their respective centroids are tabulated below Segment 1

A(mm2) 300(600)

y~(mm)

y~A(mm3)

300

54(10)6

200

18(10)6

- 127.32

-18(10)6

= 180(10)3 2

1 (300)(600) 2 = 90(10)3

1 (p)(300)2 2 = 141.37(10)3

- (p)(100)2 = - 31.41(10)3

54(10)6

379.96(10)3

y =

54(10)6 Σy~A = = 142 mm ΣA 379.96(10)3

Ans.

Ans: y = 142 mm 940

9–69. Determine the distance y‾ to the centroid of the beam’s cross-sectional area.

y 50 mm

75 mm

50 mm

25 mm C

100 mm

25 mm

SOLUTION y =

50[2(25)(100)] + 112.5[300(25)] Σy~A = = 87.5 mm ΣA 2(25)(100) + 300(25)

25 mm

Ans.

Ans: y = 87.5 mm 941

9–70. Determine the location x of the centroid of the solid made from a hemisphere, cylinder, and cone.

x x–

SOLUTION ©V =

1 2 p (30)2 (80) + p (30)2 (60) + p (30)3 = 301.593 A 103 B mm3 3 3

Ans.

Ans: x = 105 mm 942

9–71. Determine the location (x, y) of the centroid of the area.

SOLUTION ΣA =

b c(2a + b) 1 1 ac (a + b)c - (a) a b = 2 2 a + b 2(a + b) 2

b c(b c + 3abc + 3 a c) c 1 ac 1 ac ∼ Σ y A = c (a + b)c d c (a)a bd = 3 2 3(a + b) 2 a + b 6(a + b)2 2

b c(b + 3a) a + b 1 a 1 ac ∼ c (a + b)c d - ab + b c (a) a Σ xA = bd = 3 2 3 2 a + b 6(a + b) ∼ Σ yA y‾ = = ΣA

b c(b2c + 3 ab c + 3 a2c) 6(a + b)2 b c(2a + b)

b2 c + 3 ab c + 3 a2 c 3 (a + b)(2a + b)

Ans.

2(a + b) b2c(b + 3a) ∼ 6(a + b) b(b + 3a) Σ xA = x‾ = = bc(2a + b) ΣA 3(2a + b) 2(a + b)

Ans.

Ans:

943

x‾ =

∼ b(b + 3a) Σ xA = ΣA 3(2a + b)

y‾ =

b2 c + 3 ab c + 3 a2 c 3 (a + b)(2a + b)

*9–72. A toy skyrocket consists of a solid conical top, rt = 600 kg>m3, a hollow cylinder, rc = 400 kg>m3, and a stick having a circular cross section, rs = 300 kg>m3. Determine the length of the stick, x, so that the center of gravity G of the skyrocket is located along line aa.

a 3 mm

5 mm

100 mm

20 mm

10 mm a G

SOLUTION 20 1 x ' ©xm = a b c a b p (5)2 (20) d (600) - 50 C p A 52 - 2.52 B (100) D (400) - C (x) p (1.5)2 D (300) 4 3 2 = - 116.24 A 106 B - x2(1060.29) kg # mm4>m3 1 ©m = c p (5)2 (20) d (600) + p A 52 - 2.52 B (100)(400) + C xp (1.5)2 D (300) 3 = 2.670 A 106 B + 2120.58x kg # mm3/m3 x =

- 116.24 A 106 B - x2 (1060.29) = - 267.0 A 106 B - 212.058 A 103 B x 1060.29x2 - 212.058 A 103 B x - 150.80 A 106 B = 0

Solving for the positive root gives Ans.

x = 490 mm

Ans: x = 490 mm 944

9–73. Determine the location (x, y) of the centroid of the area.

2 in.

4 in.

2 in.

SOLUTION y =

4 in.

1[4(2)] + 4[8(2)] + 7[4(2)] Σy~A = = 4 in. ΣA 2(4)(2) + 8(2)

Ans.

y 2 in. x

1[8(2)] + 4[2(4)(2)] Σx~A x = = = 2.5 in. ΣA 2(4)(2) + 8(2)

Ans.

Ans: x‾ = 2.5 in. y‾ = 4 in. 945

9–74. Determine the distance x‾ to the center of gravity of the gas dehydrator assembly. The weight and the center of gravity of each of the various components are indicated. What are the vertical reactions at A and B needed to support the assembly?

G2 (2400 lb)

SOLUTION

G1 (900 lb)

ΣW = 2400 + 900 = 3300 lb xΣW = ΣMA; 3300x = 900(5) + 2400(8)

Ans.

x = 7.182 ft = 7.18 ft

3 ft

1 ft

Equilibrium: 5 ft

a+ ΣMA = 0; By(9) - 3300(7.182) = 0 Ans.

By = 2633.4 lb = 2633 lb + c ΣFy = 0; Ay + 2633.4 - 3300 = 0

x–

Ans.

Ay = 667 lb

Ans: x = 7.18 ft Ay = 667 lb By = 2633 lb 946

9–75. Determine the distance y‾ to the center of mass of the assembly, which has a hole bored through its center. The material has a density of 3 Mg>m3.

y 40 mm 15 mm C

100 mm

20 mm

30 mm

SOLUTION Segment

m(kg)

y~(m)

y~m(kg.m)

p r a b(0.04)2(0.06) 3

0.115

3.68(10)-6rp

r (p)(0.04)2(0.1)

0.05

8(10) - 6rp

p - r a b(0.02)2(0.03) 3

0.14

-0.56(10) - 6rp

- r (p)(0.015)2(0.13)

0.065

-1.90125(10) - 6rp

= 32(10)-6rp

= 160(10) - 6rp 3

= - 4(10) - 6rp

= 29.25(10) - 6rp Σ

158.75(10) - 6rp

y =

9.21875(10) - 6rp

9.21875(10) - 6rp Σy~m = = 0.0581 m = 58.1 mm Σm 158.75(10) - 6rp

~ ~ Due to symmetry x = z = 0

Ans. Ans.

Ans: ~ ~ x = z = 0 y = 58.1 mm 947

*9–76. The sheet metal part has the dimensions shown. Determine the location 1x, y, z2 of its centroid.

3 in. C

SOLUTION

©A = 4(3) +

4 in. B

1 (3)(6) = 21 in2 2

y 6 in.

Ans.

y =

Ans.

z =

Ans.

Ans: x = - 1.14 in. y = 1.71 in. z = - 0.857 in. 948

9–77. The sheet metal part has a weight per unit area area of of 22 lb>ft lb/ft 22 and is supported by the smooth rod and the cord at C. If supported by the smooth rod and at C. If the cordthe is cord is cut, willabout rotatethe about theuntil y axis until it cut, the partthe willpart rotate y axis it reaches reaches equilibrium. Determine the equilibrium of equilibrium. Determine the equilibrium angle angle of tilt, tilt, measured downward from negativex xaxis, axis, that that AD measured downward from thethe negative makes with the -x -xaxis. axis.

3 in. C

SOLUTION

Since the material is homogeneous, the center of gravity coincides with the centroid. See solution to Prob. 9-76. u = tan

4 in. B

y 6 in.

-1

1.14 b = 53.1° a 0.857

Ans.

Ans: u = 53.1° 949

9–78. Determine the distance x‾ to the center of gravity of the generator assembly. The weight and the center of gravity of each of the various components are indicated. What are the vertical reactions at blocks A and B needed to support the assembly?

G2(1000 lb)

G1(200 lb)

G3(250 lb)

SOLUTION

B 1 ft

3 ft

4 ft

2 ft

x–

ΣW = 200 + 1000 + 250 = 1450 lb xΣW = ΣMA; 1450x = 200(1) + 1000(4) + 250(8)

Ans.

x = 4.276 ft = 4.28 ft Equilibrium: a+ ΣMA = 0; By(10) - 1450(4.276) = 0 By = 620 lb

Ans.

+ c ΣFy = 0; Ay + 620 - 1450 = 0

Ans.

Ay = 830 lb

Ans: x = 4.28 ft By = 620 lb Ay = 830 lb 950

9–79. The buoy is made from two homogeneous cones each having a radius of 1.5 ft. If h = 1.2 ft, find the distance z to the buoy’s center of gravity G.

h 1.5 ft z 4 ft

1 1 p (1.5)2 (1.2) + p(1.5)2 (4) 3 3

Ans.

Ans: z = 0.70 ft 951

*9–80. The buoy is made from two homogeneous cones each having a radius of 1.5 ft. If it is required that the buoy’s center of gravity G be located at z = 0.5 ft, determine the height h of the top cone.

h 1.5 ft z 4 ft

1 1 p (1.5)2 (h) + p(1.5)2 (4) 3 3

= 2.3562 h + 9.4248 ' - 0.5890 h2 + 9.4248 © zV ' = = 0.5 z = ©V 2.3562 h + 9.4248 -0.5890 h2 + 9.4248 = 1.1781 h + 4.7124 Ans.

h = 2.00 ft

Ans: h = 2.00 ft 952

9–81. The assembly is made from a steel hemisphere, rst = 7.80 Mg>m3, and an aluminum cylinder, ral = 2.70 Mg>m3. Determine the mass center of the assembly if the height of the cylinder is h = 200 mm.

z 80 mm

G _ z

SOLUTION

160 mm

= 9.51425(10 - 3) Mg # m

= 77.7706(10 - 3) Mg z =

Ans.

Ans: z = 122 mm 953

9–82. The assembly is made from a steel hemisphere, and an aluminum cylinder, rst = 7.80 Mg>m3, ral = 2.70 Mg>m3. Determine the height h of the cylinder so that the mass center of the assembly is located at z = 160 mm.

z 80 mm

G _ z

SOLUTION

160 mm

= 66.91(10 - 3) + 54.29(10 - 3) h z =

Solving Ans.

h = 0.385 m = 385 mm

Ans: h = 385 mm 954

9–83. The car rests on four scales and in this position the scale readings of both the front and rear tires are shown by FA and FB. When the rear wheels are elevated to a height of 3 ft above the front scales, the new readings of the front wheels are also recorded. Use this data to compute the location x and y to the center of gravity G of the car. The tires each have a diameter of 1.98 ft.

_ y

A _ x

975 lb

SOLUTION In horizontal position

3.0 ft

984 lb

9.40 ft FA 1959 lb

1129 lb

1168 lb

2297(9.40) - 4256 x = 0

1269 lb

1307 lb

2576 lb

Ans.

x = 5.0733 = 5.07 ft u = sin - 1 a

2297 lb

3 - 0.990 b = 12.347° 9.40

2576(9.40 cos 12.347°) - 4256 cos 12.347°(5.0733) - 4256 sin 12.347° y¿ = 0 y¿ = 2.86 ft Ans.

y = 2.815 + 0.990 = 3.80 ft

Ans: x = 5.07 ft y = 3.80 ft 955

*9–84. The solid is formed by boring a conical hole into the cylinder. Determine the distance to the center of gravity.

h G

SOLUTION

p 2 ΣV = pa h - a2h = pa2h 3 3 2

h 3 p 2 pa2h2 2 Σz~V = (pa h) - h a a hb = 2 4 3 4 pa2h2 ~ Σz V 4 3 z~ = = = h ΣV 2 8 p a2h 3

Ans.

Ans: 3 z~ = h 8 956

9–85. Determine the distance z to the center of mass of the casting that is formed from a hollow cylinder having a density of 8 Mg>m3 and a hemisphere having a density of 3 Mg>m3.

z 20 mm

40 mm

SOLUTION Segment

m(kg)

~ z (m)

~ zm(kg.m)

8(10)3(p)(0.04)2(0.12)

0.06

0.09216p

3(0.04)

-0.00192p

120 mm

= 1.536p 2

2 3(10)3 a p(0.04)3 b 3

40 mm

= 0.128p

- 8(10)3(p)(0.02)2(0.12)

0.06

-0.02304p

= -0.384p Σ

1.28p

0.0672p

Σz~m 0.0672p z = = = 0.0525 m = 52.5 mm Σm 1.28p

Ans. Ans.

Due to symmetry x = y = 0

Ans: x = y = 0 z = 52.5 mm 957

9–86. Locate the center of mass z of the assembly. The material has a density of r = 3 Mg/m3. There is a 30-mm diameter hole bored through the center.

z 40 mm 20 mm 30 mm

100 mm

SOLUTION

–z

Centroid: Since the density is the same for the whole material, the centroid of the volume coincide with centroid of the mass. The volume of each segment and its respective centroid are tabulated below.

y x

Segment

V(mm )

z(mm)

zV(mm )

1 p140221602 3

115

3.68p11062

2 3 4

p1402211002

1 - p120221302 3 -p1152211302 158.75p11032

50 137.5 65

8.00p11062 -0.550p11062 -1.901 25p11062 9.228 75p11062

Thus, z =

Ans.

Ans: z = 58.1 mm 958

9–87. Major floor loadings in a shop are caused by the weights of the objects shown. Each force acts through its respective center of gravity G. Locate the center of gravity (x, y) of all these objects.

z y 450 lb 1500 lb G2

G1 9 ft 6 ft

SOLUTION Centroid: The floor loadings on the floor and its respective centroid are tabulated below. Loading

W (lb)

x (ft)

y (ft)

xW(lb # ft)

yW(lb # ft)

1 2 3 4

450 1500 600 280

6 18 26 30

7 16 3 8

2700 27000 15600 8400

3150 24000 1800 2240

2830

53700

31190

600 lb

7 ft

280 lb

4 ft

5 ft 3 ft

12 ft 8 ft

Thus, x =

Ans.

y =

Ans.

Ans: x = 19.0 ft y = 11.0 ft 959

*9–88. Determine the dimension h of the block so that the centroid C of the assembly lies at the base of the cylinder as shown.

6 in.

C 5 in. 2 in.

SOLUTION ~ xΣV = Σx V h36(6)(h) + p(2)2(5) 4 =

h 36(6)(h) 4 + (h + 2.5) 3p(2)2(5) 4 2

Ans.

h = 2.95 in.

Ans: h = 2.95 in. 960

9–89. The composite plate is made from both steel (A) and brass (B) segments. Determine the mass and location 1x, y, z2 of its mass center G. Take rst = 7.85 Mg>m3 and rbr = 8.74 Mg>m3.

225 mm

G 150 mm B 150 mm 30 mm

SOLUTION

1 1 ©m = ©rV = c 8.74 a (0.15)(0.225)(0.03)b d + c 7.85a (0.15)(0.225)(0.03) b d 2 2 + [7.85(0.15)(0.225)(0.03)] = C 4.4246 A 10 - 3 B D + C 3.9741 A 10 - 3 B D + C 7.9481 A 10 - 3 B D = 16.347 A 10 - 3 B = 16.4 kg

2 1 (0.150) b (4.4246) A 10 - 3 B + a 0.150 + (0.150) b (3.9741) A 10 - 3 B 3 3

Ans.

1 (0.150)(7.9481)(10 -3) = 2.4971(10 -3) kg # m 2

1 2 0.225 b(7.9481) A 10 - 3 B ©zm = a (0.225) b(4.4246) A 10 - 3 B + a (0.225)b (3.9471) A 10 - 3 B + a 3 3 2 = 1.8221 A 10 - 3 B kg # m

x =

Ans.

Due to symmetry: Ans.

y = - 15 mm z =

Ans.

Ans: m = 16.4 kg x = 153 mm y = - 15 mm z = 111 mm 961

9–90. The starter starter for foran anelectric electricmotor motorhas is the a full cylinder andarea has cross-sectional the cross-sectional areashas shown. If copper has a3 shown. If copper wiring a density of rcu =wiring 8.90 Mg>m 3 density rcu frame = 8.90has Mg/m and the has 3a, and the ofsteel a density of steel rst = frame 7.80 Mg>m 3 density rsttotal = 7.80 Mg/m , estimate the total estimateofthe mass of the starter. Neglect themass size of the starter. wire. Neglect the size of the copper wire. copper

50 mm Copper

30 mm

Steel

SOLUTION

60 mm

100 mm

= 986 A 10 B p mm 3

80 mm

103 b = 0.0395 Mg = 39.5 kg 109

Ans.

Ans: m = 39.5 kg 962

9–91. Determine the outside surface area of the storage tank.

15 ft 4 ft

30 ft

SOLUTION Surface Area: Applying the theorem of Pappus and Guldinus, Eq.9–7. with u = 2p, L1 = 2152 + 4 2 = 2241 ft, L2 = 30 ft, r1 = 7.5 ft and r2 = 15 ft, we have A = u©r~L = 2p C 7.5 A 2241 B + 15(30) D = 3.56 A 103 B ft2

Ans.

Ans: A = 3.56 ( 103 ) ft 2 963

*9–92. Determine the volume of the storage tank.

15 ft 4 ft

30 ft

SOLUTION Volume: Applying the theorem of Pappus and Guldinus, Eq. 9–8 with u = 2p, 1 r1 = 5 ft, r2 = 7.5 ft, A 1 = (15)(4) = 30.0 ft2 and A 2 = 30(15) = 450 ft2, we have 2 V = u©rA = 2p[5(30.0) + 7.5(450)] = 22.1 A 103 B ft3

Ans.

Ans: V = 22.1 ( 103 ) ft 3 964

9–93. The Gates Manufacturing Co. produces pulley wheels such as the one shown. Determine the weight of the wheel if it is made from steel having a specific weight of 490 lb>ft3.

1 in.

5 in. 0.5 in.

SOLUTION

1 in.

1 1 (1) b a b (0.375)(1) R = 10.60 in3 3 2

10.60 b = 3.01 lb (12)3

0.25 in.

Ans.

Ans: W = 3.01 lb 965

9–94. The Gates Manufacturing Co. produces pulley wheels such as the one shown. Determine the total surface area of the wheel in order to estimate the amount of paint needed to protect its surface from rust.

1 in.

5 in. 0.5 in.

SOLUTION

Ans.

1 in. 1 in. 0.25 in.

Ans: A = 70.0 in2 966

9–95. A ring is generated by rotating the quarter circular area about the x axis. Determine its volume. a

SOLUTION

Volume: Applying the theorem of Pappus and Guldinus, Eq. 9–10, with u = 2p, 6p + 4 4a p r = 2a + = a and A = a2, we have 3p 3p 4 V = urA = 2p

6p + 4 a 3p

p 2 a 4

p(6p + 4) 3 a 6

Ans.

Ans: V = 967

p(6p + 4) 3 a 6

*9–96.

A ring is generated by rotating the quarter circular area about the x axis. Determine its surface area. a

SOLUTION

Surface Area: Applying the theorem of Pappus and Guldinus, Eq. 9–11, with u = 2p , 2(p + 1) pa 5 , r1 = 2a, r 2 = L1 = L3 = a, L2 = a and r3 = a, we have p 2 2 A = u©rL = 2p 2a(a) +

2(p + 1) 5 pa b + a (a) a a p 2 2

= p(2p + 11)a2

Ans.

Ans: A = p(2p + 11)a2 968

9–97. Determine the volume of concrete needed to construct the curb. 100 mm 150 mm 30

150 mm 150 mm

SOLUTION p p 1 V = Σu A r = a b[(0.15)(0.3)(4.15)] + a b c a b(0.15)(0.1)(4.25)d 6 6 2 V = 0.114 m3

Ans.

Ans: V = 0.114 m3 969

9–98. Determine the surface area of the curb. Do not include the area of the ends in the calculation. 100 mm 150 mm 30

150 mm 150 mm

SOLUTION A = ΣurL =

p {4(0.15) + 4.075(0.15) + (4.15 + 0.075) ( 20.152 + 0.12 ) 6 + 4.3(0.25) + 4.15(0.3)}

A = 2.25 m2

Ans.

Ans: A = 2.25 m2 970

9–99. Determine the approximate amount of aluminum necessary to make the funnel. It consists of a full circular part having a thickness of 2 mm.

50 mm

80 mm 30 mm

SOLUTION

30 mm

V = a urX A

60 mm

= 2p[0.09(0.002)(0.05) + 0.06(0.002)(0.1) + 0.03(0.002)(0.03)] = 143(10)-6 m3

Ans.

Ans: V = 143(10)-6 m3 971

*9–100. Determine the approximate outer surface area of the funnel. It consists of a full circular part of negligible thickness.

50 mm

80 mm 30 mm

SOLUTION

30 mm

A = a urX L

60 mm

= 2p[0.09(0.05) + 0.06(0.1) + 0.03(0.03)] = 71.6(10)-3 m2

Ans.

Ans: A = 71.6(10)-3 m2 972

9–101. The elevated water storage tank has a conical top and hemispherical bottom and is fabricated using thin steel plate. Determine how many square feet of plate is needed to fabricate the tank.

8 ft 6 ft 10 ft 8 ft

SOLUTION Surface Area: The perpendicular distance measured from the z axis to the centroid of each line segment is indicated in Fig. a. A = 2p©rL = 2p B 4 ¢ 282 + 62 ≤ + 8(10) + ¢

2(8) p(8) ≤¢ ≤R p 2

= 2p(184) = 1156 ft2

Ans.

Ans: A = 1156 ft 2 973

9–102. The elevated water storage tank has a conical top and hemispherical bottom and is fabricated using thin steel plate. Determine the volume within the tank.

8 ft 6 ft 10 ft 8 ft

SOLUTION Volume: The peipendicular distance measured from the z axis to the centroid of each area segment is indicated in Fig. a. 4(8) p(82) 8 1 V = 2p©rA = 2p B a b a b (6)(8) + 4(10)(8) + a b¢ ≤R 3 2 3p 4 = 2p(554.67) = 3485 ft3

Ans.

Ans: V = 3485 ft 3 974

9–103. Determine the surface area and the volume of the ring formed by rotating the square about the vertical axis.

= 4pcba -

a a sin 45°b(a) d + 2 c 2p a b + sin 45°b(a) d 2 2

a2 a2 sin 45° + ba + sin 45° d 2 2 Ans.

Ans.

Ans: A = 8pba V = 2pba2 975

*9–104. Using integration, determine both the area and the distance y to the centroid of the shaded area. Then using the second theorem of Pappus–Guldinus, determine the volume of the solid generated by revolving the shaded area about the x axis.

y y2

2 ft x

SOLUTION Area of the differential element dA = a 1 + 2

A =

' ydA =

y =

dA =

ya 1 +

ydA

a1 +

y2 2

y2 b dy and y = y. 2

b dy = 3.333 ft2 = 3.33 ft2

Ans.

V = urA = 2p (1.2)(3.333) = 25.1 ft3

Ans.

= dA

2 ft

y2 b dy = 4 ft3 2 4 = 1.2 ft 3.333

1 ft

Volume:

Ans: V = 25.1 ft 3 976

9–105. The heat exchanger radiates thermal energy at the rate of 2500 kJ h for each square meter of its surface area. Determine how many joules (J) are radiated within a 5-hour period.

0.5 m

0.75 m

0.75 + 0.5 b 2(0.75)2 + (0.25)2 + (0.75)(1.5) + (0.5)(1) R 2

0.75 m

= 16.419 m2 Q = 2500 A 103 B a

1.5 m

J b A 16.416 m2 B (5 h) = 205 MJ h # m2

Ans. 0.5 m

Ans: Q = 205 MJ 977

9–106. Using integration, determine the area and the centroidal distance y of the shaded area. Then, using the second theorem of Pappus-Guldinus, determine the volume of a paraboloid formed by revolving the area about the x axis.

y y2 5 x

C y

1 ft x

SOLUTION 1 ft

2 A = 3 2x dx = ft 2 = 0.667 ft 2 3 0 1

yA = 3

Ans.

1 y 1 1 2x dx = 3 x dx = 4 0 2 0 2

1/4 3 = ft = 0.375 ft 2/3 8

Ans.

3 2 V = urX A = 2p a ba b = 1.57 ft 3 8 3

Ans.

Ans: A = 0.667 ft 2 Xy = 0.375 ft V = 1.57 ft 3 978

9–107. The suspension bunker is made from plates which are curved to the natural shape which a completely flexible membrane would take if subjected to a full load of coal. This curve may be approximated by a parabola, y = 0.2x2. Determine the weight of coal which the bunker would contain when completely filled. Coal has a specific weight of g = 50 lb>ft3, and assume there is a 20% loss in volume due to air voids. Solve the problem by integration to determine the cross-sectional area of ABC; then use the second theorem of Pappus–Guldinus to find the volume.

y 10 ft C B 20 ft

SOLUTION y 0.2x2

x x = 2

' y = y dA = x dy 20 y 3 2 dy = y2 2 = 133.3 ft2 0 L0 C 0.2 3 20.2 20

dA =

x dA =

x =

y y2 20 2 = 500 ft3 dy = 0.4 0.8 0

x dA LA LA

500 = 3.75 ft 133.3

V = u r A = 2p (3.75) (133.3) = 3142 ft3 W = 0.8 g V = 0.8(50)(3142) = 125 664 lb = 126 kip

Ans.

Ans: W = 126 kip 979

*9–108. Sand is piled between two walls as shown. Assume the pile to be a quarter section of a cone and that 26 percent of this volume is voids (air space). Use the second theorem of Pappus–Guldinus to determine the volume of sand.

3m 3m

SOLUTION p 1 ' V = ur A = c a b (1) a b (3)(2) d (0.74) = 3.49 m3 2 2

Ans.

Ans: V = 3.49 m3 980

9–109. Determine the surface area and volume of the wheel formed by revolving the cross-sectional area 360° about the z axis.

SOLUTION

1 in.

1.5 in.

Surface Area: The perpendicular distance measured from the z axis to the centroid of each line segment is indicated in Fig. a. 2(1) 2(1.5) A = 2p©rL = 2p B ¢ 2 ≤ p(1) + 2(1) + 4(2)(4) + 6(2) + ¢ 6 + ≤ p(1.5) R p p = 2p(83.0575) = 522 in2

2 in.

4 in.

Ans.

Volume: The perpendicular distance measured from the z axis to the centroid of each area segment is indicated in Fig. a. V = 2p©rA = 2p B a2 -

4(1) p(12) 4(1.5) p(1.52) b¢ b¢ ≤ + 4(4)(1) + a6 + ≤R 3p 2 3p 2

= 2p(41.9307) = 263 in3

Ans.

Ans: A = 522 in2 V = 263 in3 981

9–110. Determine the surface area of the silo which consists of a cylinder and hemispherical cap. Neglect the thickness of the plates.

10 ft 10 ft 10 ft

80 ft

2(10) 1 a (2p)(10)b + 10(80) d = 5.65 A 103 B ft2 p 4

Ans.

Ans: A = 5.65 (103) ft 2 982

9–111. Using integration, determine the area and the centroidal distance y of the shaded area. Then, using the second theorem of Pappus–Guldinus, determine the volume of a solid formed by revolving the area about the x axis.

xy 1

SOLUTION 2

A =

y dx =

1 dx = ln 2 - ln (0.5) = 1.386 = 1.39 ft2 x L0.5

Ans.

y ' y = 2

' y dA =

y =

0.5 ft 2

2 ft

1 2 1 y dx = dx = 0.75 ft3 2 L0.5 2 L0.5 2x LA LA

' ydA = dA

0.75 = 0.541 ft 1.386

Ans.

V = u r A = 2p (0.541)(1.386) = 4.71 ft3

Ans.

Ans: A = 1.39 ft 2 y = 0.541 ft V = 4.71 ft 3 983

*9–112. Using integration, determine the area and the centroidal distance y of the shaded area. Then, using the second theorem of Pappus–Guldinus, determine the volume of a paraboloid formed by revolving the area about the x axis.

y y2 = x + 4

2 ft

_ y

SOLUTION

4 ft

dA = - x dy y = y 2

A =

- x dy = -

(y2 - 4)dy = 5.33 ft2

Ans.

y dA

y =

= dA

(y)(y2 - 4)dy 5.33

Ans.

= 0.750 ft

V = 2p yA = 2p (0.75)(5.33) V = 25.1 ft3

Ans.

Ans: A = 5.33 ft 2 y = 0.750 ft V = 25.1 ft 3 984

9–113. Determine the volume of material needed to make the casting.

2 in.

6 in.

SOLUTION

6 in.

Side View

4 in.

Front View

Ans.

Ans: V = 1.40 ( 10 3 ) in3 985

9–114. Determine the weight of the wedge which is formed by rotating a right triangle of base 6 in. and height 3 in. through an angle of 30°. The specific weight of the material is g = 0.22 lb>in3.

308

3 in.

6 in.

SOLUTION V = u rA =

p 1 (1)a (6)(3) b = 1.5p in3 6 2

Ans.

W = gV = 0.22(1.5p) = 1.04 lb

Ans: W = 1.04 lb 986

9–115. y

The pressure loading on the plate is described by the function p = [ -240 > (x + 1) + 340] Pa. Determine the magnitude of the resultant force and the coordinates (x, y) of the point where the line of action of the source intersects the plate.

300 Pa

100 Pa 6m

5m x

SOLUTION p=

- 240 + 340 x+1

FR = 6 3 p dx = 6 3 a 5

- 240 + 340b dx x+1

= c - 1440 ln(x + 1) + 2040 x d

5 0

= 7619.8 N

Ans.

FR = 7.62 kN - 240x & 3 x p dx = 6 3 a x + 1 + 340xb dx 0 5

= c - 1440 {x - ln(x + 1)} + 2040 a

= 20 880 N # m

xX =

& 3 x p dx 3 p dx

y = 3m

x2 5 bd 2 0

20 880 = 2.74 m 7619.8

Ans.

(By symmetry)

Ans: FR = 7.62 kN xX = 2.74 m y = 3m 987

*9–116. p

The load over the plate varies linearly along the sides of the plate such that p = (12 - 6x + 4y) kPa. Determine the magnitude of the resultant force and the coordinates ( x, y ) of the point where the line of action of the force intersects the plate.

12 kPa 18 kPa

x 1.5 m

6 kPa 2m

SOLUTION Centroid: Perform the double integration. FR =

r(x, y)dA = = =

1.5 m

(12 -6x + 4y)dxdy 2m

1 12x - 3x2 + 4xy 2 `

(8y + 12)dy

= (4y2 + 12y) `

1.5 m 0

Ans.

= 27.0 kN LA

xr(x, y)dA = = =

1.5 m

1 12x - 6x2 + 4xy 2 dx dy

(6x2 - 2x3 + 2x2y) `

= 21.0 kN # m LA

= =

(8y + 8)dy

= 1 4y2 + 8y 2 ` yr(x, y)dA =

1.5 m

1.5 m 0

1 12y - 6xy + 4y2 2 dx dy

1 12xy - 3x2y + 4xy2 2 `

(8y2 + 12y)dy

1.5 m 8 = a y3 + 6y2 b ` 3 0

= 22.5 kN # m

988

*9–116. Continued

Thus,

x =

xp(x, y)dA

21.0 kN # m 7 = m = 0.778 m 27.0 kN 9

22.5 kN # m = 0.833 m 27.0 kN

p(x, y)dA

yp(x, y)dA

p(x, y)dA

Ans.

Ans: FR = 27.0 kN x = 0.778 m y = 0.833 m 989

9–117. Determine the magnitude and location of the resultant force of the parabolic loading acting on the plate.

p p5

y2 lb>ft2 25

4 lb>ft2 5 ft y 5 ft

10 ft

SOLUTION dF = dV = 10pdy = 0.4y2dy FR = LV

dV = 0.4

y~dV =

y =

y2dy = 133.33 lb = 133 lb

Ans.

y(0.4y2dy) = 1000 lb # ft

ydV

dV LV

1000 = 7.50 ft 133.33

Ans.

x = 0 Due to symmetry

Ans: FR = 133 lb xX = 0 y = 7.50 ft 990

9–118. p

The rectangular plate is subjected to a distributed load over its entire surface. The load is defined by the expression p = p0 sin (px>a) sin (py>b), where p0 represents the pressure acting at the center of the plate. Determine the magnitude and location of the resultant force acting on the plate.

y x

SOLUTION

a b

Resultant Force and its Location: The volume of the differential element is py px dy b . dxb asin dV = dFR = pdxdy = p0 asin a b a

FR =

dFR = p0

LFR

asin

= p0 B a =

py px dyb dxb a sin a b L0

a b px 2 a px 2 b b R cos b a - cos p a p b 0 0

4ab p0 p2

Ans.

Since the loading is symmetric, the location of the resultant force is at the center of the plate. Hence, x =

a 2

y =

b 2

Ans.

Ans: FR =

4ab p0 p2

a 2 b y = 2

x =

991

9–119. A wind loading creates a positive pressure on one side of the chimney and a negative (suction) pressure on the other side, as shown. If this pressure loading acts uniformly along the chimney’s length, determine the magnitude of the resultant force created by the wind.

p p0 cos u

SOLUTION

p p 2 2 FRx = 2l 2p (p0 cos u) cos u r du = 2rlp0 p cos u du L- 2 L- 2

p = 2rlp0 a b 2

FRx = plrp0 FRy = 2l

p 2 p (p0 cos u) sin u r du = 0 L- 2

Thus, Ans.

FR = plrp0

Ans: FR = plrp0 992

*9–120. When the tide water A subsides, the tide gate automatically swings open to drain the marsh B. For the condition of high tide shown, determine the horizontal reactions developed at the hinge C and stop block D. The length of the gate is 6 m and its height is 4 m. rw = 1.0 Mg/m3.

4m 3m

SOLUTION Fluid Pressure: The fluid pressure at points D and E can be determined using Eq. 9–13, p = rgz. pD = 1.01103219.812122 = 19 620 N>m2 = 19.62 kN>m2 pE = 1.01103219.812132 = 29 430 N>m2 = 29.43 kN>m2

Thus, wD = 19.62162 = 117.72 kN>m wE = 29.43162 = 176.58 kN>m Resultant Forces: FR1 =

1 1176.582132 = 264.87 kN 2

FR2 =

1 1117.722122 = 117.72 kN 2

264.87132 - 117.7213.3332 - Dx 142 = 0 Ans.

264.87 - 117.72 - 100.55 - Cx = 0 Ans.

Cx = 46.6 kN

Ans: Dx = 101 kN Cx = 46.6 kN 993

9–121. The tank is filled with water to a depth of d = 4 m. Determine the resultant force the water exerts on side A and side B of the tank. If oil instead of water is placed in the tank, to what depth d should it reach so that it creates the same resultant forces? ro = 900 kg>m3 and rw = 1000 kg>m3.

SOLUTION For water At side A: wA = b rw g d = 2(1000)(9.81) (4) = 78 480 N/m FRA =

1 (78 480)(4) = 156 960 N = 157 kN 2

Ans.

At side B: wB = b rw g d = 3(1000)(9.81)(4) = 117 720 N>m FRB =

1 (117 720)(4) = 235 440 N = 235 kN 2

Ans.

For oil At side A: wA = b ro g d = 2(900)(9.81)d = 17 658 d FRA =

1 (17 658 d)(d) = 156 960 N 2 Ans.

d = 4.22 m

Ans: For water: FRA = 157 kN FRB = 235 kN For oil: d = 4.22 m 994

9–122. Determine the magnitude and location of the resultant hydrostatic force acting on each of the cover plates A and B. rw =1.0 Mg/m3.

0.5 m 1m 0.5 m 1.25 m

SOLUTION

0.75 m B

wu = 1000(9.81)(0.75)(0.5) = 3678.75 N/m wD = 1000(9.81)(0.5)(0.5) = 2452.5 N/m wh = 1000(9.81)(2.25) = 22 073 N/m2 FRA = F1 + F2 1 = (2452.5)(0.5) + 3678.75(0.5) = 2453 N 2 Ans.

FRA = 2.45 kN a + (MR)B = ΣMB; 2453 d =

0.5 2 1 (3678.75)(0.5) + (0.5)a b(2452.5)(0.5) 2 3 2

d = 0.2708 m

Hence, the resultant force on plate A is located at Ans.

d′ = 0.75 + 0.2708 = 1.02 m from the liquid surface. FRB = 22 073 p a

0.75 2 b = 9.75 kN 2

Ans. Ans.

Located at the center of plate B.

Ans:

FRA = 2.45 kN d′ = 1.02 m FRB = 9.75 kN

Located at the center of plate B. 995

9–123. The tank is used to store a liquid having a specific weight of g = 80 lb>ft 3. If it is filled to the top, determine the magnitude of the resultant force the liquid exerts on side ABEF.

C E

B D 5 ft

F 1.5 ft

2 ft

1.5 ft

8 ft

SOLUTION dF = dV = p dA = g(5 - z)(2y dz) = (800y - 160 yz) dz Line AB: y = -0.3z + 2.5

F = V = 3 dV = 3 (48z - 640z + 2000) dz 5

= 16z3 - 320z2 + 2000z 50 = 4000 lb Ans.

F = 4 kip

Ans: F = 4 kip 996

*9–124. The tank is used to store a liquid having a specific weight of g = 80 lb > ft 3. If it is filled to the top, determine the magnitude of force the liquid exerts on side ABCD.

C E

B D 5 ft

F 1.5 ft

2 ft

1.5 ft

8 ft

SOLUTION w = 80(5)(8) = 3200 lb/ft 1 F = (3200) 252 + (1.5)2 = 8.35 kip 2

Ans.

Ans:

F = 8.35 kip

997

9–125. The trapezoidal back plate of the boat has the dimensions shown. If the water on the outside of the boat is 3 in. from the top, determine the resultant force on the plate and its location, measured from the top of the boat. gw = 62.4 lb > ft 3. 3 in. 16 in.

SOLUTION 3

8 in.

gw = 62.4 lb/ft = 0.03611 lb/in

20 in. 8 in.

For side triangles: d Ft = dV = p dA = gw z (2 x dz) = 0.07222 z x dz Since x = 8 - 0.5z Ft = V = 3 dV = 3 (0.07222)(z)(8 - 0.5z) dz 16

= 0.07222 3 (8z - 0.5 z2) dz 16

= 24.652 lb

zX =

3 z dV v

3 dV

0.07222 3

(8 z2 - 0.5 z3) dz 24.652

= 8 in.

For rectangle: Fr =

1 [(0.03611)(16)](16)(20) = 92.44 lb 2 Ans.

FR = 24.652 + 92.44 = 117.0963 = 117 lb 2 117.0963(d′) = 24.652(8) + 92.44 a b(16) 3

d′ = 10.105 in.

Ans.

d = 3 + 10.105 = 13.1 in.

Ans: FR = 117 lb d = 13.1 in. 998

9–126. The quarter circular “gravity” dam is held in place by its own weight. Determine the smallest possible density of the material composing the dam so that it will be prevented from overturning about its end A. rw = 1.0 Mg>m3.

r A

SOLUTION Per meter of length: a+ ΣMA = 0;

1 1 1 4 - c [(1000)(9.81)(r)] (r)a r b d + r(9.81)a p r 2 b(r) a1 b = 0 2 3 4 3p

r = 369 kg/m3

Ans.

Ans: r = 369 kg/m3 999

9–127. The 2-m-wide rectangular gate is pinned at its center A and is prevented from rotating by the block at B. Determine the reactions at these supports due to hydrostatic pressure. rw = 1.0 Mg>m3.

6m 1.5 m A B

SOLUTION

1.5 m

w1 = 1000(9.81)(3)(2) = 58 860 N/m w2 = 1000(9.81)(3)(2) = 58 860 N/m F1 =

1 (3)(58 860) = 88 290 2

88 290(0.5) - FB (1.5) = 0 Ans.

88 290 + 176 580 - 29 430 - FA = 0 Ans.

FA = 235 440 N = 235 kN

Ans: FB = 29.4 kN FA = 235 kN 1000

*9–128. 3 The tank withwith a liquid that has a density 900 kg/mof . tankis isfilled filled a liquid which has aof density Determine resultantthe force that it force exertsthat on the elliptical 900 kg>m3. the Determine resultant it exerts on end and end the location of the of pressure, the plate, elliptical plate, and thecenter location of the measured center of from the xmeasured axis. pressure, from the x axis.

y 1m

1m 4 y2

0.5 m x 0.5 m

SOLUTION Fluid Pressure: The fluid pressure at an arbitrary point along y axis can be determined using Eq. 9–13, p = g(0.5 - y) = 900(9.81)(0.5 - y) = 8829(0.5 - y). Resultant Force and its Location: Here, x = 21 - 4y2. The volume of the differential element is dV = dFR = p(2xdy) = 8829(0.5 - y)[221 - 4y2] dy. Evaluating integrals using Simpson’s rule, we have

the

0.5 m

FR =

LFR

d FR = 17658

L-0.5 m

(0.5 - y)(21 - 4y 2)dy Ans.

= 6934.2 N = 6.93 kN 0.5 m

LFR

yd FR = 17658

y(0.5 - y)(21 - 4y2)dy

= - 866.7 N # m ' ydFR

y =

L-0.5 m

LFR

= dFR

-866.7 = - 0.125 m 6934.2

Ans.

Ans: FR = 6.93 kN y = - 0.125 m 1001

9–129. The structure is used for temporary storage of oil at sea for later loading into ships. When it is empty the water level is at A (sea level). As oil is poured into it, the water is displaced through exit ports at D. If the riser EC is filled with oil, i.e., to a depth of C, determine the height h to B of the oil level above sea level. ro = 900kg/m3 and rw = 1020 kg/m3.

E B A

50 m 2m

SOLUTION pC = 1020(9.81)(50) = 500.31 kPa

Require,

8m D

500.31(103) = 900(9.81)(h + 50)

7m 1m

Ans.

h = 6.67 m

Ans: h = 6.67 m 1002

9–130. If the structure in Prob. 9–129 is totally filled with oil, i.e., until it reaches a depth of 58 m below sea level, how high h will the oil level extend above sea level?

E B A

50 m 2m

SOLUTION pD = 1020(9.81)(58) = 580.36 kPa

Require,

8m D

580.36(103) = 900(9.81)(h + 58)

7m 1m

Ans.

h = 7.73 m

Ans: h = 7.73 m 1003

10–1. Determine the moment of inertia of the area about the x axis.

y3 5 x 1 in. x 1 in.

SOLUTION Area of the differential element (shaded) dA = (1 - x)dy where x = y3, hence, dA = (1 - x)dy = (1 - y3)dy. Ix =

y2dA = =

y2 (1 - y3) dy (y2 - y5) dy

1 1 1 = c y3 - y6 d ` 3 6 0

= 0.167 in4

Ans.

Ans: Ix = 0.167 in4 1004

10–2. Determine the moment of inertia of the area about the y axis.

y3 5 x 1 in. x 1 in.

SOLUTION 1

Area of the differential element (shaded) dA = y dx where y = x3 , hence, 1 dA = y dy = x3 dx. Iy =

x2dA = =

= c

x2 ( x3 ) dx 7

x3 dx

3 10 1 x3 d ` 10 0

= 0.3 in4

Ans.

Ans: Iy = 0.3 in4 1005

10–3. Determine the moment of inertia of the triangular area about the x axis.

h (b x) y –– b h

SOLUTION Area of the differential element (shaded) dA = xdy where x = b - hb y, hence, dA = xdy = (b - hb y) dy. h

Ix =

y2dA =

y2 a b-

b y b dy h

a by2 -

x b

b 3 y b dy h

b 3 b 42h y y 3 4h 0

1 bh3 12

Ans.

Ans: Ix = 1006

1 3 bh 12

*10–4. Determine the moment of inertia of the triangular area about the y axis.

h (b x) y –– b h

SOLUTION Area of the differential element (shaded) dA = ydx where y = h - hbx, hence, dA = ydx = (h - hbx) dx. b

Iy =

x2dA = =

L0 L0

x2 a h -

h x bdx b

h 3 x bdx b

a hx2 -

h 3 h 42b x x 3 4b 0

1 hb3 12

x b

Ans.

Ans: Iy = 1007

1 3 hb 12

10–5. y

Determine the moment momentofofinertia inertiaof of area area about the Determine the thethe shaded about xthe axis. x axis.

y = 1 (400 – x 2 ) 5

80 mm

SOLUTION Differential Element: The area of the differential element parallel to y axis is dA = ydx. The moment of inertia of this element about x axis is dIx = dIx¿ + dAy

x 20 mm

y 2 1 (dx) y3 + ydx a b 12 2

3 1 1 c A 400 - x2 B d dx 3 5

1 c - x6 + 1200x4 - 480 A 103 B x2 + 64 A 106 B ddx 375

Moment of Inertia: Performing the integration, we have 20 mm

Ix =

dIx =

1 c - x6 + 1200x4 - 480 A 103 B x2 + 64 A 106 B ddx 375 L- 20 mm

20 mm 1 1 c - x7 + 240x5 - 160 A 103 B x3 + 64 A 106 B x d 375 7 - 20 mm

= 3.12 106 mm4

Ans.

Ans: Ix = 3.12 (106) mm4 1008

10–6. y

Determine the the moment momentofofinertia inertiaof of area area about the Determine thethe shaded about y axis. the y axis.

y = 1 (400 – x 2 ) 5

80 mm

SOLUTION Element: The area of the element parallel to y axis is 1 dA = ydx = A 400 - x2 B dx. 5

Differential

20 mm

Moment of Inertia: Applying Eq. 10–1 and performing the integration, we have 20 mm

Iy =

x2 dA =

1 x2 c A 400 - x2 B dx d 5 L- 20 mm

= a

20 mm 400 3 1 x - x5 b 3 5 - 20 mm

= 171 103 mm4

Ans.

Ans: Iy = 171(103) mm4 1009

10–7. Determine the moment of inertia of the area about the x axis.

π x) y 2 cos (–– 8 2 in. x 4 in.

4 in.

SOLUTION dIx = dIx ¿ + dA y2 =

y 2 1 1 dx y3 + y dx a b = y3 dx 12 2 3 4

Ix =

dIx =

8 p cos3 a x b dx 3 8 L-4

4 p p sin a x b sin3 a x b 8 8 8 256 = D = 9.05 in4 T = p 3 3p 9p 8 8 -4

Ans.

Ans: Ix = 9.05 in4 1010

*10–8. Determine the moment of inertia of the area about the y axis.

π x) y 2 cos (–– 8 2 in. x 4 in.

4 in.

SOLUTION 4

Iy =

x2dA =

L-4

x2 2 cos a

p x b dx 8

4 p p p x2 sin a x b 2x cos a x b 2 sina xb 8 8 8 = 2D + T p p 2 p 3 a b a b 8 8 8 -4

= 4a

128 1024 b = 30.9 in4 p p3

Ans.

Ans: Iy = 30.9 in4 1011

10–9. y

Determine the moment of inertia of the area about the x axis. Solve the problem in two ways, using rectangular differential elements: (a) having a thickness dx and (b) having a thickness of dy.

y = 2.5 – 0.1x2 2.5 ft 5 ft

SOLUTION (a) Differential Element: The area of the differential element parallel to y axis is dA = ydx. The moment of inertia of this element about x axis is ' dIx = dIx¿ + dAy 2 =

y 2 1 1dx2y 3 + ydx a b 12 2

1 12.5 - 0.1x 223 dx 3

1 1- 0.001x 6 + 0.075x 4 - 1.875x 2 + 15.6252 dx 3

Moment of Inertia: Performing the integration, we have 5 ft

Ix =

dIx =

1 1- 0.001x6 + 0.075x 4 - 1.875x2 + 15.6252 dx 3 L-5 ft

5 ft 0.001 7 1 0.075 5 1.875 3 ax + x x + 15.625xb ` 3 7 5 3 -5 ft

= 23.8 ft4

Ans.

(b) Differential Element: Here, x = 225 - 10y . The area of the differential element parallel to x axis is dA = 2xdy = 2 225 - 10y dy. Moment of Inertia: Applying Eq. 10–1 and performing the integration, we have Ix =

y 2dA 2.5 ft

= 2

y 2 225 - 10ydy

= 2c -

2.5 ft y2 2y 3 3 7 2 125 - 10y22 125 - 10y22 125 - 10y22 d ` 15 375 13125 0

= 23.8 ft4

Ans.

Ans: Ix = 23.8 ft 4 1012

10–10. Determine the radius of gyration ky of the parabolic area.

y = 0.1(1600 – x2)

SOLUTION

160 mm 40

A =

y dx = 0.1

= c 160x -

L- 40

A 1600 - x2 B dx

0.1 3 40 x d 3 - 40

x 40 mm

= 8533.3 mm2 40

Iy =

x2 dA = 0.1

= 0.1 c

L- 40

(1600x2 - x4) dx

40 1600 3 1 x - x5 d 3 5 - 40

= 2730.67 A 103 B mm4 ky =

2730.67(103) = 17.9 mm 8533.3

Ans.

Ans: ky = 17.9 mm 1013

10–11. Determine the forfor thethe shaded about themoment momentofofinertia inertia area area about the the x axis. x axis.

y2 x 2 in.

SOLUTION Ix =

4 y 2 3 1 1 1 1 2 5 4 (dx)(y)3 + y(dx) a b R = y3 dx = x2 dx = a b a bx2 R 2 3 LA 3 L0 3 5 0 LA 12

= 4.267 in4 = 4.27 in4

Ans.

x 4 in.

Also, 2

Ix =

y2 (4 - y2) dy =

4y3 y5 2 ` = 4.27 in4 3 5 0

Ans.

Ans: Ix = 4.27 in4 1014

*10–12. Determine the forfor thethe shaded about themoment momentofofinertia inertia area area about the the y axis. y axis.

y2 x 2 in.

SOLUTION 4

Iy =

x2 dA =

x2y dx =

5 2 7 4 x2 dx = a b x2 R 7 0 L0

= 36.5714 in4 = 36.6 in4

Ans.

4 in.

Ans: Iy = 36.6 in4 1015

10–13. Determine the forfor thethe shaded about themoment momentofofinertia inertia area area about the the x axis. x axis.

y2 4x

4 in.

SOLUTION 4

Ix =

y2 dA =

= =

a 4y2 -

y2 (4 -x) dx

4 in.

y4 bdy 4

4 3 1 5 4 y y R = 34.1 in4 3 20 0

Ans.

Ans: Ix = 34.1 in4 1016

10–14. forfor thethe shaded about themoment momentofofinertia inertia area area about the Determine the the y axis. y axis.

y2 4x

4 in.

SOLUTION 4

Iy =

x2 dA = 4

= 2

x2 dx =

x2 y dx =

x2 a 2x2 b dx

4 in.

4 7 4 x2 R = 73.1 in4 7 0

Ans.

Ans: Iy = 73.1 in4 1017

10–15. y

Determine the moment of inertia for the area about the x axis. 4 in.

y2 x x 16 in.

SOLUTION Differential Element: The area of the differential element parallel with the x axis shown shaded in Fig. a is dA = x dy = y2 dy. Moment of Inertia: Perform the integration, Ix =

y2dA = = =

4 in.

y2(y2dy) y4dy

y5 4 in. ` 5 0

= 204.8 in4 = 205 in4

Ans.

Ans: Ix = 205 in4 1018

*10–16. y

Determine the moment of inertia for the area about the y axis.

4 in.

y2 x x 16 in.

SOLUTION Differential Element: The moment of inertia of the differential element parallel to the x axis shown shaded in Fig. a about the y axis is ∼2 dIy = dIy + dAx = =

1 x 2 (dy)x3 + (xdy) a b 12 2 1 3 x dy 3

1 23 (y ) dy 3

1 6 y dy 3

Moment of Inertia: Perform the integration, Iy =

dIy = =

4 in.

1 6 y dy 3

1 y7 4 in. a b` 3 7 0

= 780.19 in4 = 780 in4

Ans.

Ans: Iy = 780 in4 1019

10–17. Determine the moment of inertia of the quarter circular area about the x axis.

x2 1 y2 5 r 2

SOLUTION

x 1

Area of the differential1 element (shaded) dA = x dy where x = (r 2 - y2)2, hence, dA = x dy = (r 2 - y2)2 dy. Ix =

y2dA =

= c=

y2 a(r 2 - y2)2 b dy

y y r r2 1 3(r 2 - y2)3 + ay3r 2 - y2 b + r 2 sin - 1 d ` 4 8 8 r 0

pr 4 16

Ans.

Ans: Ix = 1020

pr 2 16

10–18. Determine the moment of inertia for the area about the y axis.

y h x3 b3

x b

SOLUTION Differential Element: The area of the differential element parallel to the y axis h shown shaded in Fig. a is dA = ydx = 3 x3dx b Moment of Inertia: Perform the integration, Iy =

x2dA =

h x2 a 3 x3 bdx b L0 b

= = =

h x5dx b3 L0 h x6 6 a b` b3 6 0 b3h 6

Ans.

Ans: Iy = 1021

b3h 6

10–19. Determine the moment of inertia of the area about the x axis.

y 5 9 2 x2 9 in.

SOLUTION

Ix = =

L L0

3 in.

y2 dA 9

y2(9 - y)1/2 dy

= 333 in4

Ans.

Ans: Ix = 333 in4 1022

*10–20. Determine the moment of inertia of the area about the y axis.

y 5 9 2 x2 9 in.

SOLUTION

Iy = =

L L0

3 in.

x2 dA 3

x2(9 - x3) dx

= 32.4 in4

Ans.

Ans: Iy = 32.4 in4 1023

10–21. Determine the moment of inertia of the area about the x axis.

y 5 2x 2 x 2 1m x 2m

SOLUTION The moment of inertia of the element about the x axis dIx = y = 2x - x2.

Ix =

1 dx y3 where 3

dIx =

1 3 1 (2x - x2)3 dx y dx = 3 3 LA L0 2

1 ( -x6 + 6x5 - 12x4 + 8x3) dx 3 L0

2 1 1 12 5 = c a - x7 + x6 x + 2x4 b d ` 3 7 5 0

= 0.305 m4

Ans.

Ans: Ix = 0.305 m4 1024

10–22. Determine the moment of inertia of the area about the y axis.

y 5 2x 2 x 2 1m x 2m

SOLUTION Area of the differential element (shaded) dA = y dx where y = 2x - x2, hence, dA = y dx = (2x - x2)dx.

Iy =

x2 dA = =

x2(2x - x2) dx (2x3 - x4) dx

2 1 1 = c x4 - x5 b d ` 2 5 0

= 1.6 m4

Ans.

Ans: Iy = 1.6 m4 1025

10–23. y

Determine the moment of inertia for the area about the x axis.

b2 y2 —x a b x2 y — a2

SOLUTION

a 2 y . Thus, the area of the 2 b b differential element parallel to the x axis shown shaded in Fig. a is dA = (x2 - x1) dy

Differential Element: Here x2 =

1 2

y2 and x1 =

a a 1 = a 1 y2 - 2 y2 b dy. b 2 b

Moment of Inertia: Perform the integration, Ix =

y2dA =

a 1 a y2 a 1 y2 - 2 y2 bdy b L0 b2 b

a 5 a a 1 y2 - 2 y4 bdy b 2 L0 b

= a =

1 2

3ab3 35

y2 -

a 5 2b y b 5b2 0

Ans.

Ans: Ix = 1026

3ab3 35

*10–24. Determine the moment of inertia of the equilateral triangle about the x′ axis passing through its centroid.

3 ( a 2 x) 2

a C

x9 x

SOLUTION Ix = Ix = Ix =

L L0 L0

= 2c

y2 dA 23 2 a

y (2x) dy

23 a 2

2y2 a

y a b dy 2 23 23

a 2 ay3 y4 d 6 423 0

23 a4 32 @ Ix = Ix′ = + Ad 2 =

23 a4 1 23 23 2 @ = Ix′ + (a)a aba ab 32 2 2 6 23 a4 23 a4 @ Ix′ = 32 48 23 a4 @ Ix′ = 96

Ans.

Ans: 23 a4 @ Ix′ = 96 1027

10–25. Determine the moment of inertia of the composite area about the x axis.

y 3 in.

SOLUTION

3 in.

Composite Parts: The composite area can be subdivided into three segments as shown in Fig. a. The perpendicular distance measured from the centroid of each segment to the x axis is also indicated.

3 in.

6 in.

Moment of Inertia: The moment of inertia of each segment about the x axis can be determined using the parallel-axis theorem. Thus, -

Ix = Ix¿ + A(dy)2 = B

1 1 1 (3)(33) + (3)(3)(4)2 R + B (3)(33) + 3(3)(1.5)2 R 36 2 12

+ B

1 1 (6)(63) + (6)(6)(2)2 R 36 2

= 209 in4

Ans.

Ans: Ix = 209 in4 1028

10–26. y

Determine the moment of inertia of the composite area about the y axis.

3 in.

SOLUTION

3 in.

Composite Parts: The composite area can be subdivided into three segments as shown in Fig. a. The perpendicular distance measured from the centroid of each segment to the x axis is also indicated.

3 in.

6 in.

Moment of Inertia: The moment of inertia of each segment about the y axis can be determined using the parallel-axis theorem. Thus, -

Iy = Iy ¿ + A(dx)2 = B

1 1 1 (3)(33) + (3)(3)(2)2 R + B (3)(33) + 3(3)(1.5)2 R 36 2 12

+ B

1 1 (6)(63) + (6)(6)(5)2 R 36 2

= 533 in4

Ans.

Ans: Iy = 533 in4 1029

10–27. Determine the moment of inertia of the “Z” section with respect to the x and y centroidal axes.

1 in.

4 in. 1 in.

8 in.

SOLUTION Ix =

Iy =

1 1 (1)(10)3 + 2c (4)(1)3 + 1(4)(4.5)2 d = 246 in4 12 12 1 1 (10)(1)3 + 2c (1)(4)3 + 1(4)(2.5)2 d = 61.5 in4 12 12

Ans.

1 in. 4 in.

1 in.

Ans.

Ans: Ix = 246 in4 Iy = 61.5 in4 1030

*10–28. Determine the moment of inertia of the beam’s cross-sectional area with respect to the x and y centroidal axes.

y 2.5 in.

2.5 in. 1 in. 1 in.

6 in.

x 3 in. 1 in.

SOLUTION Moment of inertia about x axis: Ix =

1 1 (5)(8)3 (4)(6)3 = 141 in4 12 12

Ans.

Moment of inertia about y axis: Iy = 2c

1 1 (1)(5)3 d + (6)(1)3 = 21.3 in4 12 12

Ans.

Ans: Ix = 141 in4 Iy = 21.3 in4 1031

10–29. Determine the location y of the centroid of the channel’s cross-sectional area and then calculate the moment of inertia of the area about this axis.

50 mm

250 mm –y

SOLUTION Centroid: The area of each segment and its respective centroid are tabulated below.

50 mm 350 mm

Segment

A (mm2)

' y (mm)

' y A (mm3)

100(250)

125

3.12511062

250(50)

37.511032

0.312511062 3.437511062

Ans.

Moment of Inertia: The moment of inertia about the x¿ axis for each segment can be determined using the parallel-axis theorem Ix¿ = Ix¿ + Ad2y. Segment

Ai (mm2)

(dy)i (mm)

100(250)

33.33

250(50)

66.67

(Ix¿)i (mm4)

(Ad 2y)i (mm4)

1 3 12 12502150 2

55.55611062

1 3 12 110021250 2

27.77811062

(Ix¿)i (mm4)

157.9911062 58.1611062

Ans.

Ans: ∼ y = 91.7 mm Ix′ = 216(106) mm4 1032

10–30. y

Determine the moment of inertia of the cross-sectional area of the beam about the x axis.

1 in.

1 in. 8 in. 3 in. 1 in.

10 in.

SOLUTION Moment of Inertia: The moment of inertia about x axis for each segment can be determined using the parallel axis theorem, Ix = Ix′ + Ad 2y. Referring to Fig. a, Segment

Ai(in2)

(dy)i (in.)

1(8)

8(1)

0.5

1(3)

1.5

(Ix′)i (in4) 1 (1)(83) 12 1 (8)(13) 12 1 (1)(33) 12

(Ad 2y)i (in4)

(Ix)i (in4)

128

170.67

2.67

6.75

9.00

Thus, Ix = Σ(Ix)i = 182.33 in4 = 182 in4 Ans.

Ans: Ix = 182 in4 1033

10–31. y

Determine the moment of inertia of the cross-sectional area of the beam about the y axis.

1 in.

1 in. 8 in. 3 in. 1 in.

SOLUTION

10 in.

Moment of Inertia: The moments of inertia about the y axis for each segment can be determined using the parallel axis theorem, Ix = Ix′ + Ad 2y. Referring to Fig. a, Segment

Ai(in2)

(dx)i (in.)

8(1)

9.5

1(8)

3(1)

0.5

Thus,

(Iy')i (in4) 1 (8)(13) 12 1 (1)(83) 12 1 (3)(13) 12

(Adx2)i (in4)

(Iy)i (in4)

722

722.67

200

242.67

0.75

1.00

Iy = Σ(Iy)i = 966.33 in4 = 966 in4 Ans.

Ans: Iy = 966 in4 1034

*10–32. y

Determine the moment of inertia Ix of the area about the x axis.

100 mm

150 mm

SOLUTION

75 mm x

Moment of Inertia: The moment of inertia about the x axis for each segment can be determined using the parallel axis theorem, Ix = Ix′ + Ad 2y. Referring to Fig. a Segment

Ai(mm2)

(dy)i (mm)

200(300)

150

1 (150)(300) 2

100

- p(752)

150

Thus,

(Ix′)i (mm4)

(Ady)2i (mm4)

(Ix)i (mm4)

1 (200)(3003) 12 1 (150)(3003) 36 p(754) 4

1.35(109)

1.80(109)

0.225(109)

0.3375(109)

-0.3976(109)

-0.4225(109)

Ix = Σ(Ix)i = 1.715(109) mm4 = 1.72(109) mm4 Ans.

Ans: Ix = 1.72(109) mm4 1035

10–33. y

Determine the moment of inertia Iy of the shaded area about the y axis.

100 mm

150 mm

SOLUTION

75 mm x

Moment of Inertia: The moment of inertia about the y axis for each segment can be determined using the parallel-axis theorem, Iy = Iy′ + Ad 2x. Referring to Fig. a Segment

Ai(mm2)

(dx)i (mm)

200(300)

100

1 (150)(300) 2

250

- p(752)

100

Iy′(mm4) 1 (300)(2003) 12 1 (300)(1503) 36 p(754) 4

(Ad 2x)i (mm4)

(Iy)i (mm4)

0.6(109)

0.800(109)

1.40625(109) 1.434375(109) -0.1767(109) -0.20157(109)

Thus, Iy = Σ(Iy)i = 2.033(109) mm4 = 2.03(109) mm4 Ans.

Ans: Iy = 2.03 ( 109 ) mm4 1036

10–34. Determine inertia of the area Determine the the moment momentofof inertia of crosssectional the beam’s crossof the beam about the y axis. sectional area about the y axis.

150 mm 150 mm

50 mm

SOLUTION Moment of Inertia: The dimensions and location of centroid of each segment are shown in Fig. a. Since the y axis passes through the centroid of both segments, the moment of inertia about y axis for each segment is simply (Iy)i = (Iy¿)i. Iy = g (Iy)i =

250 mm

x¿ C x¿

1 1 (50)(3003) + (250)(503) 12 12 6

_ y 4

Ans.

= 115.10(10 ) mm = 115(10 ) mm

25 mm

Ans: Iy = 115 ( 106 ) mm4 1037

10–35. Determine y, which locates the centroidal axis x′ x¿ for the cross-sectional area of the T-beam, and then find the moment of inertia about the x¿ x′ axis. axis.

150 mm 150 mm

50 mm

SOLUTION y =

250 mm

x¿ C x¿ y

= 206.818 mm Ans.

y = 207 mm 1 Ix¿ = c (50)(250)3 + 50(250)(206.818 - 125)2 d 12 +c

25 mm

1 (300)(50)3 + 50(300)(275 - 206.818)2 d 12

I x¿ = 222(106) mm4

Ans.

Ans: y = 207 mm Ix′ = 222 ( 106 ) mm4 1038

*10–36. Determine the distance x to the centroid for the beam’s cross-sectional area, then find Iy

y9 x

2 in. 4 in.

6 in. C

SOLUTION 6(2)(1) + 2(1)(4)(4) x@ = = 2.20 in. 6(2) + 2(1)(4)

1 in. 1 in. 1 in. 1 in.

Ans.

1 @ Iy′ = (6)(2)3 + 6(2)(1 - 2.2)2 12 + 2c

1 (1)(4)3 + 1(4)(4 - 2.2)2 d 12

= 57.9 in4

Ans.

Ans: x = 2.20 in. Iy′ = 57.9 in4 1039

10–37. Determine the moment of inertia of the beam’s crosssectional area about the x axis.

y9 x

2 in. 4 in.

6 in. C

SOLUTION Ix =

1 1 (2)(6)3 + 2c (4)(1)3 + 4(1)(1.5)2 d 12 12

= 54.7 in4

1 in. 1 in. 1 in. 1 in.

Ans.

Ans: Ix = 54.7 in4 1040

10–38. y

Determine the moment of inertia of the area about the x axis. 3 in.

6 in. x 6 in.

6 in.

SOLUTION Moment of Inertia: The moment of inertia about the x axis for each segment can be determined using the parallel-axis theorem, Ix = Ix′ + Ad 2y. Referring to Fig. a, Segment

Ai(in2)

(dy)i (in.)

6(6)

1 (6)(3) 2

1 (6)(9) 2

Thus,

(Ix′)i (in4) 1 (6)(63) 12 1 (6)(33) 36 1 (6)(93) 36

(Ad 2y)i (in4)

(Ix)i (in4)

324

432.0

441

445.5

972

1093.5

Ix = Σ(Ix)i = 1971 in4 Ans.

Ans: Ix = 1971 in4 1041

10–39. y

Determine the moment of inertia of the area about the y axis. 3 in.

6 in. x 6 in.

6 in.

SOLUTION Moment of Inertia: The moment of inertia about the y axis for each segment can be determined using the parallel-axis theorem, Iy = Iy′ + Ad 2x. Referring to Fig. a, Segment

Ai(in2)

(dx)i (in.)

6(6)

1 (6)(3) 2

1 (9)(6) 2

(Iy′)i (in4) 1 (6)(63) 12 1 (3)(63) 36 1 (9)(63) 36

(Ad 2x)i (in4)

(Iy)i (in4)

324

432.0

144

162.0

1728

1782.0

Thus, Iy = Σ(Iy)i = 2376 in4 Ans.

Ans: Iy = 2376 in4 1042

*10–40. Determine the distance x to the centroid for the beam’s cross-sectional area, then find Iy.

y y9 x

200 mm C 50 mm 50 mm

SOLUTION

100 mm

100(200)(50) + 300(100)(250) x@ = = 170 mm 100(200) + 300(100)

300 mm

Ans.

1 @ Iy′ = (200)(100)3 + 200(100)(50 - 170)2 12

1 (100)(300)3 + 100(300)(250 - 170)2 12

= 722(106) mm4

Ans.

Ans: x = 170 mm Iy′ = 722(106) mm4 1043

10–41. Determine the moment of inertia of the beam’s crosssectional area about the x axis.

y y9 x

200 mm C 50 mm

SOLUTION

50 mm

1 1 @ Ix = (100)(100)3 + (300)(100)3 12 12

100 mm

= 91.7(106) mm4

300 mm

Ans.

Ans: Ix = 91.7 ( 106 ) mm4 1044

10–42. Determine the moment of inertia of the beam’s crosssectional area about the x axis.

y¿

30 mm 30 mm 70 mm 140 mm

SOLUTION

30 mm

1 (170)(30)3 + 170(30)(15)2 Ix = 12

C y

x¿

30 mm 170 mm

1 + (30)(170)3 + 30(170)(85)2 12 +

1 (100)(30)3 + 100(30)(185)2 12

Ix = 154(106) mm4

Ans.

Ans: Ix = 154 ( 106 ) mm4 1045

10–43. y

Determine the moment of inertia of the beam’s crosssectional area about the y axis.

y¿

30 mm 30 mm

SOLUTION Iy =

1 (30)(170)3 + 30(170)(115)2 12 +

140 mm

1 (170)(30)3 + 30(170)(15)2 12

30 mm

1 + (30)(100)3 + 30(100)(50)2 12

70 mm _ x C _ y

x¿

30 mm 170 mm

Iy = 91.3(106) mm4

Ans.

Ans: Iy = 91.3 ( 106 ) mm4 1046

*10–44. Determine the distance y to the centroid C of the beam’s cross-sectional area and then compute the moment of inertia Ix¿ about the x¿ axis.

y¿

30 mm 30 mm 70 mm 140 mm

SOLUTION

30 mm

170(30)(15) + 170(30)(85) + 100(30)(185) y = 170(30) + 170(30) + 100(30) Ans.

= 80.68 = 80.7 mm Ix¿ = c

x C y

x¿

30 mm 170 mm x

1 (170)(30)3 + 170(30)(80.68 - 15)2 d 12

+c +

1 (30)(170)3 + 30(170)(85 - 80.68)2 d 12

1 (100)(30)3 + 100(30)(185 - 80.68)2 12

Ix¿ = 67.6(106) mm4

Ans.

Ans: y = 80.7 mm Ix′ = 67.6(106) mm4 1047

10–45. Determine the distance x to the centroid C of the beam’s findcompute the moment inertia Iof cross-sectional area areaand andthen then the of moment y′ about the y′ axis.the y¿ axis. inertia Iy¿ about

y¿

30 mm 30 mm 70 mm 140 mm

SOLUTION

30 mm

170(30)(115) + 170(30)(15) + 100(30)(50) x = 170(30) + 170(30) + 100(30) Ans.

= 61.59 = 61.6 mm Iy¿ = c

x C y

x¿

30 mm 170 mm x

1 (30)(170)3 + 170(30)(115 - 61.59)2 d 12

+c +

1 (170)(30)3 + 30(170)(15 - 61.59)2 d 12

1 (30)(100)3 + 100(30)(50 - 61.59)2 12

Iy¿ = 41.2(106) mm4

Ans.

Ans: x = 61.6 mm Iy= = 41.2 ( 106 ) mm4 1048

10–46. Determine y which locates the centroid C of the crosssectional area of the wing channel, and then determine the moment of inertia Ix′ about the centroidal x′ axis. Neglect the effect of rounded corners. The material has a uniform thickness of 0.5 in.

y 1.5 in.

1.5 in.

2 in.

SOLUTION y =

3 in.

2(2)(0.5)(0.25) + 2(3)(0.5)(1.5) + 2(0.5)(2.75) 2(2)(0.5) + 2(3)(0.5) + 2(0.5)

= 1.292 = 1.29 in. Ix= = 2c

Ans.

1 (2)(0.5)3 + 2(0.5)(0.25 - 1.292)2 d 12

+ c

1 (3)3(0.5) + 3(0.5)(1.5 - 1.292)2 d 12

1 (2)(0.5)3 + 2(0.5)(2.75 - 1.292)2 12

= 6.74 in4

Ans.

Ans: y = 1.29 in Ix= = 6.74 in4 1049

10–47. Determine the moment of inertia of the cross-sectional area of the wing channel about the y axis.

y 1.5 in.

2 in.

SOLUTION Iy =

1.5 in.

1 1 1 (3)(3)3 (2.5)(2)3 + 2c (0.5)(2)3 + (2)(0.5)(2.5)2 d 12 12 12

Iy = 18.3 in4

3 in.

2 in.

Ans.

Ans: Iy = 18.3 in4 1050

*10–48. The assembly consists of two pipes and a plate. Determine the inner radii r of the pipes so that the moment of inertia of the assembly about the x axis is Ix = 527 in4.

r 2 in. 5 in. 0.5 in.

SOLUTION 1 1 Ix = 2c p(2)4 + p(2)2(7)2 d - 2c pr 4 + pr 2(7)2 d 4 4 +

x 5 in.

2 in.

1 (0.5)(103) 12

= 527 in

1 4 r + 49r 2 - 122.76 = 0 4 r 2 = 2.474 Ans.

r = 1.57 in

Ans: r = 1.57 in 1051

10–49. Determine the polar moment of inertia of the shaft’s crosssectional area about the center O

a O

SOLUTION Ix = Iy = =

1 1 d 4 pa b (a)(a)3 4 2 12

1 1 4 pd4 a 64 12

Jo = Ix + Iy =

1 1 pd4 - a4 32 6

Ans.

Ans: JO = 1052

1 1 pd4 - a4 32 6

10–50. Locate the centroid y of the cross section and determine the moment of inertia of the section about the x¿ axis. 0.4 m

–y

0.05 m

SOLUTION

0.2 m 0.2 m

0.3 m

0.2 m 0.2 m

Centroid: The area of each segment and its respective centroid are tabulated below. Segment

A (m2)

y (m)

yA (m3)

0.3(0.4)

0.25

0.03

1 2 10.4210.42

0.1833

0.014667

1.1(0.05)

0.025

0.001375

0.255

0.046042

Thus, y =

Ans.

Moment of Inertia: The moment of inertia about the x¿ axis for each segment can be determined using the parallel-axis theorem Ix¿ = Ix¿ + Ad2y. Segment

A i (m 2 )

(d y) i (m)

0.3(0.4)

0.06944

1 2 10.4 210.4 2

0.002778

1.1(0.05)

0.1556

(I x¿) i (m4 )

1 3 12 10.3 210.4 2 1 3 36 10.4 210.4 2 1 3 12 11.1 210.05 2

(Ad 2y) i (m4 )

(Ix¿) i (m 4 )

0.5787110 -32

2.1787110-32

1.3309110 -32

1.3423110-32

0.6173110 -62

0.7117110-32

Ans.

Ans: y = 0.181 m Ix′ = 4.23(10-3) m4 1053

10–51. the moment momentof of inertia for beam’s the beam’s Determine the inertia for the crosscrosssectional area about the passing x′ axis through that passes through sectional area about the x¿ axis the centroid theofcentroid of the cross section. C the crossCsection.

100 mm

25 mm 200 mm 200 mm 45

SOLUTION Ix¿ =

141.4 2 1 1 1 b d (200)(332.8)3 + 4 c (141.4)(141.4)3 + a (141.4)(141.4) b a 12 36 2 3

45 45

x¿

25 mm

1 1 p - 2 c (200)4 a - sin90° b d 4 4 2 = 520(106) mm4

Ans.

Ans: Ix= = 520 ( 106 ) mm4 1054

*10–52. Determine the moment of inertia of the area about the x axis.

y 3 in.

3 in.

6 in.

SOLUTION Ix = B

1 1 1 (6)(10)3 + 6(10)(5)2 R - B (3)(6)3 + a b(3)(6)(8)2 R 12 36 2

1 - B p (2)4 + p(2)2(4)2 R = 1.19(103) in4 4

2 in. 4 in.

Ans.

Ans: Ix = 1.19(103) in4 1055

10–53. y

Determine the moment of inertia of the area about the y axis.

3 in.

SOLUTION Iy = c

1 1 1 (10)(6)3 + 6(10)(3)2 d - c (6)(3)3 + a b 6(3)(5)2 d 12 36 2

6 in.

1 - c p(2)4 + p(2)2(3)2 d = 365 in4 4

Ans. 2 in. 4 in. x

Ans: Iy = 365 in4 1056

10–54.

y t

Determine the product of inertia of the thin strip of area with respect to the x and y axes. The strip is oriented at an angle u from the x axis. Assume that t V l. l

SOLUTION 1

lxy =

xydA =

(s cos u)(s sin u)tds = sin u cos ut

s ds

1 3 l t sin 2u 6

Ans.

Ans: Ixy = 1057

1 3 tl sin 2u 3

10–55. Determine with respect to Determine the the product productof ofinertia inertiaofofthe thearea shaded area with the x and y axes. respect to the x and y axes.

SOLUTION

3 in.

Differential Element: The area of the differential element parallel to the y axis 1 3 x dx. The coordinates of the centroid of 9

1 x3 9

shown shaded in Fig. a is dA = y dx =

y 1 3 = x . Thus, the product of inertia of this 2 18 element with respect to the x and y axes is

x 3 in.

this element are xc = x and yc = '' dIxy = dIx¿y¿ + dAx y

1 1 = 0 + a x3dxb (x)a x3 b 9 18 =

1 7 x dx 162

Product of Inertia: Performing the integration, we have 3 in

Ixy =

dIxy =

3 in. 1 1 7 x dx = (x 8) 2 = 5.06 in4 162 1296 0

Ans.

Ans: Ixy = 5.06 in4 1058

*10–56. Determine the the product productofofinertia inertia parabola with forfor thethe shaded portion of respect to thewith x and y axes.to the x and y axes. the parabola respect

y 100 mm

200 mm y

SOLUTION

1 2 x 50

Differential Element: Here, x = 250y2 . The area of the differential element 1 parallel to the x axis is dA = 2xdy = 2 250y2 dy. The coordinates of the centroid for this element are x = 0, y = y. Then the product of inertia for this element is

dIxy = dIx¿y¿ + dAx y = 0 + ¢ 2 250y2 dy ≤ (0)(y) 1

= 0 Product of Inertia: Performing the integration, we have Ixy =

Ans.

dIxy = 0

Note: By inspection, Ixy = 0 since the shaded area is symmetrical about the y axis.

Ans: Ixy = 0 1059

10–57. Determine the area with respect to the product productofofinertia inertiaofofthe the shaded area with the x and axes, and ythen use thethen parallel-axis theorem to respect to ythe x and axes, and use the parallel-axis find the product of the area with the theorem to find of theinertia product of inertia of respect the areatowith centroidal x′ and y′ axes.x¿ and y¿ axes. respect to the centroidal

y¿

2m C

SOLUTION

Differential Element: The area of the differential element parallel to the y axis shown shaded in Fig. a is dA = y dx = x 1>2 dx. The coordinates of the centroid of y ' ' 1 1>2 this element are x = x and y = Thus, the product of inertia of this = x 2 2 element with respect to the x and y axes is

~~ dIxy = dIx¿y¿ + dAx y 1 = 0 + A x1>2 dx B (x) a x 1>2 b 2 1 2 x dx 2 Product of Inertia: Performing the integration, we have =

Ixy =

x¿

dIxy =

4m 1 1 2 = 10.67 m4 = 10.7 m4 x dx = a x3 b 2 2 6 0

Ans.

Using the information provided on the inside back cover of this book, the location of 2 3 the centroid of the parabolic area is at x = 4 - (4) = 2.4 m and y = (2) = 0.75 m 5 8 2 and its area is given by A = (4)(2) = 5.333 m2. Thus, 3 Ixy = Ix¿y¿ + Adxdy 10.67 = Ix¿y¿ + 5.333(2.4)(0.75) Ix¿y¿ = 1.07 m4

Ans.

Ans: Ixy = 10.7 m4 Ix′y′ = 1.07 m4 1060

10–58. Determine the product of inertia for the parabolic area with respect to the x and y axes.

b 1/2 x a1/2

SOLUTION

' x = x y ' y = 2 dA = y dx dIxy = Ixy =

xy2 dx 2

d Ixy a

1 b2 2 a b x dx L0 2 a

a 1 b2 B a b x3 R a 6 0

1 2 2 a b 6

Ans.

Ans: Ixy = 1061

1 2 2 ab 6

10–59. Determine the product of inertia of the area with respect to the x and y axes.

y 1m

y 5 x3

SOLUTION The product of inertia of the element (shaded) is dIxy = dIx′y′ + dAx~ y~ 1 x = 0 + x (dy)a b(y) = x2 y dy Where x2 = y6 2 2 =

1 7 y dy 2

dIxy = Ixy =

Integrating 1

1 7 y dy 2 L0

1 1 1 = c a y8 b d ` 2 8 0

= 0.0625 m4

Ans.

Ans: Ixy = 0.0625 m4 1062

*10–60. Determine the product of inertia of the shaded area with respect to the x and y axes.

x2 + y2 = 4 2 in.

SOLUTION

Differential Element: Here, y = 24 - x2 . The area of the differential element

2 in.

parallel to the y axis is dA = ydx = 24 - x2dx. The coordinates of the centroid y 1 ' ' = 24 - x2 . Then the product of inertia for this for this element are x = x, y = 2 2 element is '' dIxy = dIx¿y¿ + dAx y 1 = 0 + A 24 - x2dx B 1x2 a 24 - x2 b 2 =

1 14x - x32 dx 2

Product of Inertia: Performing the integration, we have Ixy =

dIxy = =

1 2 L0

2in.

14x - x32 dx

1 x4 2x2 2 4

2in. 0

= 2.00 in4

Ans.

Ans: Ixy = 2.00 in4 1063

10–61. y

Determine the product of inertia of the shaded area with respect to the x and y axes.

1 in. y = 0.25x 2 x 2 in.

SOLUTION 1

Differential Element: Here, x = 2y 2. The area of the differential element parallel 1

to the x axis is dA = xdy = 2y 2 dy. The coordinates of the centroid for this element 1 x are x = = y 2, y = y. Then the product of inertia for this element is 2 dIxy = dIx′y′ + dAx y 1

= 0 + a2y 2 dyb ( y2 ) (y) = 2y2 dy Product of Inertia: Performing the integration, we have l in.

Ixy =

dIxy =

2y2 dy =

2 3 1 in. y ` = 0.667 in4 3 0

Ans.

Ans: Ixy = 0.667 in4 1064

10–62. Determine the product of inertia of the beam’s crosssectional area with respect to the x and y axes.

1 in.

3 in. 1 in.

12 in.

SOLUTION 4

Ixy = 8(1)(4)(0.5) + 10(1)(0.5)(6) + 3(1)(1.5)(11.5) = 97.8 in

Ans.

8 in.

1 in. x

Ans: Ixy = 97.8 in4 1065

10–63. Determine the moments of inertia for the shaded area with respect to the u and v axes.

y 0.5 in. v

SOLUTION Moment and Product of Inertia about x and y Axes: Since the shaded area is symmetrical about the x axis, Ixy = 0. Ix = Iy =

1 1 (1)(53) + (4)(13) = 10.75 in4 12 12

0.5 in. x 0.5 in.

5 in.

1 in.

4 in.

1 1 (1)(43) + 1(4)(2.5)2 + (5)(13) = 30.75 in4 12 12

Moment of Inertia About the Inclined u and v Axes: Applying Eq. 10-9 with u = 30°, we have Iu = =

I x + Iy 2

Ix - I y 2

cos 2u - Ixy sin 2u

10.75 + 30.75 10.75 - 30.75 + cos 60° - 0(sin 60°) 2 2

= 15.75 in4 Iv = =

Ix + I y 2

Ans. -

I x - Iy 2

cos 2u + Ixy sin 2u

10.75 - 30.75 10.75 + 30.75 cos 60° + 0(sin 60°) 2 2

= 25.75 in4

Ans.

Ans: Iu = 15.75 in4 Iu = 25.75 in4 1066

the(xcentroid (x, beam’s y) of the beam’s cross-sectional Locate theLocate centroid , y) of the cross-sectional *10–64. area,determine and then determine inertia area, and then the productthe of product inertia ofofthis areaof this area with respect to the x¿ centroidal x¿ the and beam’s y¿ axes. cross-sectional Locate the centroid (x, yy¿) axes. of with respect to the centroidal and area, and then determine the product of inertia of this area with respect to the centroidal x¿ and y¿ axes.

y¿

y x y¿ 10 mm 10 mm x 10 mm 100 mm 100 mm x

300 mm

SOLUTION SOLUTION

10 mm

10 mm 100 mm

300 mm

10 mm

C mm 300

x¿

y Parts: The cross-sectional composite cross-sectional thebe beam can be subdivided CompositeComposite Parts: The composite area of the area beamofcan subdivided SOLUTION into three segments as shown in Figs. a and b. C into three segments as shown in Figs. a and b. 10 mm 10 mm Composite Parts: The composite cross-sectional area of the beam can be subdivided into three The segments as shown in Figs. afrom and b. x Centroid: perpendicular distances measured from the of each segment Centroid: The perpendicular distances measured the centroid ofcentroid each segment 10 mm x and y axes areinindicated to the x andtoythe axes are indicated Fig. a. in Fig. a. 200 mm 200 mm Centroid: The perpendicular distances measured from the centroid of each segment ' ' 3 B + B 5+A 300(10) B a.+ 105 55©x + A 90(10) 5 Aare 300(10) 105 190(10) x and axes indicated in AFig. A 90(10) B A 190(10) A yB 55 264(10B3) 264(10 ) ©xA to the 200 mm x= = = Ans. == 45.52 3mm= =45.52 45.5mm mm= 45.5 mm Ans. x = = 3 ©A 90(10) + 300(10) + 190(10) 5.8(10 ) ©A 90(10) + 300(10) 190(10) 5.8(10 ) ' 3 55 A 90(10) B + 5 A 300(10) B + 105 A 190(10) B 264(10 ) ©x A ' 3 = 45.52 mm = 45.5 mm Ans. B ++150 B + B 5 A 190(10) 150 5 A 190(10) A'90(10) B + A 90(10) A 300(10) B +A 300(10) 3 ) ©y A = 295 725(10 ©yA x =295 725(10B3=) 5.8(10 ©A 90(10) 300(10) + 190(10) ) = 125 mm y= = = Ans. == 125 mm y = Ans. = 3 3 ©A 90(10) + 300(10) ©A + 300(10) 190(10) + 190(10) 5.8(10 ) 5.8(10 3) ' 90(10) 295 A 90(10) B + 150 A 300(10) B + 5 A 190(10) B 725(10 ) ©yA = 125 mm Ans. = = y = 3 ©ASince 90(10) 300(10) +axes 190(10) of Inertia: Since the+axes centroidal are axessegments of all theare segments are the 5.8(10 ) the Product of Product Inertia: the centroidal are the axes of the all the axes of then symmetry, Ix¿y¿ the . Thus, the ofeach inertia of each segment with = 0product axes of symmetry, Ix¿y¿ =then of product inertia of segment with 0. Thus, respect toofy¿ the x¿ and y¿ be determined using the of parallel-axis theorem. Product Inertia: Since the can centroidal axes the axes all the segments areThe the respect to the x¿ and axes can beaxes determined using theare parallel-axis theorem. The perpendicular distances measured fromthe the centroid each to the x¿ with and 0. centroid axesdistances of symmetry, then Ifrom Thus, product ofofinertia of segment perpendicular measured of each segment tosegment theeach x¿ and x¿y¿ =the y¿ axes are indicated in axes Fig. b.can be determined using the parallel-axis theorem. The to in the x¿ and y¿ axes are respect indicated Fig. b. y¿ perpendicular distances measured from the centroid of each segment to the x¿ and = x¿Iare Adx¿dy¿in=Fig. Adx¿dy¿ Ix¿y¿ = Ix¿y¿I + axes Ad dx¿y¿ =+ Adx¿dy¿ b. y¿x¿y¿ y¿ indicated

y x¿ y

6 = I 90(10)(9.483)(170) + 300(10)( - 40.52)(25) + 190(10)(59.48)( - x¿y¿ 120)= =-15.15(10 ©Ix¿y¿ = 6)-15.15(10 ) mm4 Ans. = 90(10)(9.483)(170) + x¿300(10)( - 40.52)(25) + 190(10)(59.48)( -120) = ©I mm4 Ans. Ix¿y¿ = dy¿ = Adx¿dy¿ x¿y¿ + Ad

Ans: x = 45.5 mm y = 125 mm Ix=y= = -15.15 ( 106 ) mm4 1067

x x

10–65. Determine the product of inertia for the area shaded area with with respect to the x to and respect they xaxes. and y axes.

y 2 in.

2 in.

SOLUTION

1 in.

4 in.

Ans. x

Ans: Ixy = 119 in4 1068

10–66. Determine the product of inertia of the cross-sectional area of therespect member to the x and y axes. with to with the xrespect and y axes.

100 mm 20 mm

SOLUTION

400 mm

Product of Inertia: The area for each segment, its centroid and product of inertia with respect to x and y axes are tabulated below. Segment

Ai (mm2)

(dx)i (mm)

(dy)i (mm)

100(20)

410

49.211062

840(20)

100(20)

- 60

-410

49.211062

(Ixy)i (mm4)

400 mm

20 mm 100 mm

20 mm

Ans.

Ans: Ixy = 98.4 ( 106 ) mm4 1069

10–67. Determine the location (x, y) y) of of the centroid C of the angle’s cross-sectional area, area,and andthen thendetermine compute the product of inertia with respect to the x′ and y′ axes.

18 mm

y¿

150 mm x¿

C y 150 mm

x 18 mm

SOLUTION Centroid: x =

y =

9(18)(150) + 84(18)(132) ΣxA = = 44.11 mm = 44.1 mm ΣA 18(150) + 18(132) ΣyA ΣA

75(18)(150) + 9(18)(132) 18(150) + 18(132)

= 44.11 mm = 44.1 mm

Ans.

Product of inertia about x′ and y′ axes: Ix′y′ = 18(150)( -35.11)(30.89) + 18(132)(39.89)( -35.11) = - 6.26(106) mm4

Ans.

Ans: x = y = 44.1 mm Ix=y= = - 6.26 ( 106 ) mm4 1070

*10–68. Locate the centroid, y, and determine the orientation of the principal centroidal axes for the composite area. What are the moments of inertia with respect to these axes?

y 8 in.

2 in.

8 in.

12 in. x

C y

SOLUTION Because of symmetry, x and y are principal axes. 1 1 ΣyA = 6(4)(12) + 2 a b(12)a b(8)(12) 3 2 = 672 in3

1 ΣA = 4(12) + 2 a b(8)(12) 2 = 144 in2

672 = 4.67 in. Ans. 144 1 1 1 Ix = c (4)(12)3 + 4(12)(6 - 4.667)2 d + 2c (8)(12)3 + (8)(12)(4.667 - 4)2 d 12 36 2 y =

= 1.47(103) in4

Iy = c

Ans.

2 1 1 1 8 (12)(4)3 d + 2c (12)(8)3 + (12)(8)a + 2b d 12 36 12 3

= 2.50(103) in4

Ans.

Ans: y = 4.67 in. Ix = 1.47(103) in4 Iy = 2.50(103) in4 1071

10–69. y

Determine the principal moments of inertia of the composite area with respect to a set of principal axes that have their origin located at the centroid C. Use the equations developed in Sec. 10.7. Ixy = - 15.84(106) mm4.

60 mm

80 mm

SOLUTION

Ixy = - 15.84 A 106 B mm4 Ix = B

1 1 1 (120)(160)3 R + B (120)(60)3 + (120)(60)(100)2 R 12 36 2

+ B

1 1 (120)(60)3 + (120)(60)(- 100)2 R = 114.4 A 106 B mm4 36 2

Iy = B

1 1 1 (160)(120)3 R + B (60)(120)3 + (60)(120)(20)2 R 12 36 2

+ B

1 1 (60)(120)3 + (60)(120)(- 20)2 R = 31.68 A 106 B mm4 36 2

tan 2u =

-2Ixy Ix - Iy

min min

140 mm 60 mm

60 mm 60 mm

- 2(15.84)(106) 114.4(106) - 31.68(106) Ans.

u = 10.5° Imax =

Ix + Iy 2

;

Ix - Iy 2

b + I2xy

114.4(10 ) + 31.68(106) 114.4(106) - 31.68(106) 2 ; a b + ( -15.84 (106))2 2 C 2

Imax = 117 117(A 1066)B mm4 max =

Ans.

IImin = 28.8 28.8( 10 1066) mm4 min =

Ans.

Ans: u = 10.5° Imax = 117 ( 106 ) mm4 Imin = 28.8 ( 106 ) mm4 1072

10–70. Locate the position x, y for the centroid C of the beam’s cross-sectional area, and then determine the product of inertia with respect to the x′ and y′ axes.

y9 x 50 mm

100 mm 50 mm

Σ~ xA ΣA 100(200)(50) + 25(50)(50) 200(50) + 50(50) Ans.

= 85.0 mm y =

x9 y

200 mm

SOLUTION x =

Σ~ yA ΣA 25(50)(200) + 75(50)(50) 200(50) + 50(50) Ans.

= 35.0 mm

@ I x′y′ = 0 - (100 - 85)(35 - 25)(200)(50) + 0 - (85 - 25)(75 - 35)(50)(50) = -7.50(106) mm4

Ans.

Ans: x = 85.0 mm y = 35.0 mm @ Ix′y′ = - 7.50(106) mm4 1073

10–71. Determine the product of inertia of the beam’s crosssectional area with respect to the x and y axes.

y9 x 50 mm

100 mm 50 mm

x9 y

200 mm

SOLUTION Ixy = 100(25)(200)(50) + 25(75)(50)(50) = 29.7(106) mm4

Ans.

Ans: Ixy = 29.7(106) mm4 1074

*10–72. Determine the moments of inertia Iu and Iv and the product of inertia Iuv for the rectangular area. The u and v axes pass through the centroid C. Take u = 30°.

3 in.

u C

8 in.

SOLUTION Moments of inertia Ix and Iy. Ix =

1 (12)(5)3 = 125 in4 12

1 (5)(12)3 = 720 in4 12 The section is symmetric about both x and y axes; therefore Ixy = 0. Iy =

Iu = =

Ix + Iy 2

Ix - Iy 2

cos 2u - Ixy sin 2u

125 + 720 125 - 720 + cos 60° - 0 sin 60° 2 2

= 274 in4 Iv = =

Ix + Iy 2

Ix - Iy 2

Ans.

cos 2u + Ixy sin 2u

125 + 720 125 - 720 + cos 60° + 0 sin 60° 2 2

= 571 in4 Iuv =

Ix - Iy 2

sin 2u + Ixy cos 2u

125 - 720 sin 60° + 0 cos 60° 2

= -258 in4

Ans: Iu = 274 in4 Iv = 571 in4 Iuv = - 258 in4 1075

10–73. Determine the moments of inertia Iu and Iv and the product of inertia Iuv for the rectangular area. The u and v axes pass through the centroid C. Take u = 20°.

3 in.

u C

8 in.

SOLUTION Ix =

1 (12)(5)3 = 125 in4 12

Iy =

1 (12)3(5) = 720 in4 12

Ixy = 0 (symmetry) Iu = Iu = Iv = Iv = Iuv = Iuv =

Ix + Iy 2

Ix - Iy 2

cos 2u - Ixy sin 2u

125 + 720 125 - 720 + cos 40° = 194.60 in4 = 195 in4 2 2 Ix + Iy 2

Ix - Iy 2

cos 2u + Ixy sin 2u

125 + 720 125 - 720 cos 40° = 650.39 in4 = 650 in4 2 2 Ix - Iy 2

Ans.

sin 2u + Ixy cos 2u

125 - 720 sin 40° = - 191.22 in4 = - 191 in4 2

Ans.

Ans: Iu = 195 in4 Iv = 650 in4 Iuv = - 191 in4 1076

10–74. Determine the principal moments of inertia of the composite area with respect to a set of principal axes that have their origin located at the centroid C. Use the equations developed in Sec.10.7. Ixy = 15.84(106) mm4.

60 mm 60 mm

x 60 mm

140 mm

SOLUTION

80 mm

60 mm

Moments of inertia Ix and Iy: Ix = Iy =

1 (220)(120)3 = 31.68(10)6 12 1 1 1 (120)(160)3 + 2c (120)(60)3 + (120)(60)(100)2 d = 114.4(10)6 mm4 12 36 2

Orientation:

tan 2 uP =

- Ixy Ix - Iy 2

2uP = 20.96°

- 15.84 = 0.3830 31.68 - 114.4 2 Ans.

uP = 10.5°

Principal moments of inertia: Imax max =

Ix + Iy 2

min min

= J

{

Ix - Iy 2 b + I 2xy B 2 a

31.68 + 114.4 31.68 - 114.4 2 { a b + 15.842 R 106 2 B 2

Imax = 117(10)6 mm4 Imin = 28.8(10)6 mm4

Ans.

Ans: uP = 10.5° Imax = 117(10)6 mm4 Imax = 28.8(10)6 mm4 1077

10–75. Solve Prob. 10–74 using Mohr’s circle.

60 mm 60 mm

x 60 mm

140 mm

SOLUTION

80 mm

60 mm

Moments of inertia Ix and Iy: Ix = Iy =

1 (220)(120)3 = 31.68(10)6 12 1 1 1 (120)(160)3 + 2c (120)(60)3 + (120)(60)(100)2 d = 114.4(10)6 mm4 12 36 2

Mohr’s circle:

OA = c 2(73.04 - 31.68)2 + 15.842 d (10)6 = 44.29(10)6 Imax = (73.04 + 44.29)(10)6 = 117(10)6 mm4 6

Ans.

Imin = (73.04 - 44.29)(10) = 28.8(10) mm

Ans.

15.84 73.04 - 31.68

Ans.

tan 2u =

u = 10.5°

Ans: Imax = 117 ( 10 ) 6 mm4 Imin = 28.8 ( 10 ) 6 mm4 u = 10.5° 1078

*10–76. Determine the moments of inertia and the product of inertia of the beam’s cross-sectional area with respect to the u and v axes.

y v 1.5 in.

3 in.

SOLUTION

Moment and Product of Inertia with Respect to the x and y Axes: Since the rectangular beam cross-sectional area is symmetrical about the x and y axes, Ixy = 0. Ix =

1.5 in.

1 (3)(63) = 54 in4 12

Iy =

3 in.

1 (6)(33) = 13.5 in4 12

Moment and Product of Inertia with Respect to the u and v Axes: With u = 30°, Iu = =

Ix + Iy 2

Ix - Iy 2

cos 2u - Ixy sin 2u

54 - 13.5 54 + 13.5 + cos 60° - 0 sin 60° 2 2

= 43.9 in4 Iv = =

Ix + Iy 2

Ans. -

Ix - Iy 2

cos 2u + Ixy sin 2u

54 + 13.5 54 - 13.5 cos 60° + 0 sin 60° 2 2

= 23.6 in4 Iuv = =

Ix - Iy 2

Ans. sin 2u + Ixy cos 2u

54 - 13.5 sin 60° + 0 cos 60° 2

= 17.5 in4

Ans.

Ans: Iu = 43.9 in4 Iv = 23.6 in4 Iuv = 17.5 in4 1079

10–77. Solve Prob. 10–76 Mohr’s circle. Use Mohr’s circle using to determine the moments of inertia and the product of inertia of the beam’s cross-sectional area with respect to the u and y axes.

y v 1.5 in.

3 in.

SOLUTION

Moment and Product of Inertia with Respect to the x and y Axes: Since the rectangular beam cross-sectional area is symmetrical about the x and y axes, Ixy = 0. Ix =

1.5 in.

1 (3)(63) = 54 in4 12

Iy =

3 in.

1 (6)(33) = 13.5 in4 12

Construction of Mohr’s Circle: The center C of the circle lies along the u axis at a distance Iavg =

Ix + Iy 2

54 + 13.5 = 33.75 in4 2

The coordinates of the reference point A are (54, 0) in4. The circle can be constructed as shown in Fig. a. The radius of the circle is R = CA = 54 - 33.75 = 20.25 in4 Moment and Product of Inertia with Respect to the u and v Axes: By referring to the geometry of the circle, Iu = 33.75 + 20.25 cos 60° = 43.9 in4

Ans.

Iv = 33.75 - 20.25 cos 60° = 23.6 in4

Ans.

Iuv = 20.25 sin 60° = 17.5 in4

Ans.

Ans: Iu = 43.9 in4 Iv = 23.6 in4 Iuv = 17.5 in4 1080

10–78. section of ansection airplane has thewing moments of area of the cross of wing an airplane has the The cross inertia and the product of the inertia x andthrough y axes following properties about x andabout y axesthe passing 4 of I = 450 in44, I = 1730 in4, passing through the centroid C:the Ix centroid = 450 inC , Iy = . x 1730 in , Iyxy = 138 Ixy = 138 in4the . Determine of axes the principal Determine orientationtheof orientation the principal and the axes and the principal moments of inertia. principal moments of inertia.

SOLUTION - 2Ixy

tan 2u =

Ix - Iy

- 2(138) 450 - 1730

Ans.

u = 6.08° Imax/min max/min = =

Ix + Iy 2

;

Ix - Iy 2 b + I2xy A 2 a

450 + 1730 450 - 1730 2 ; a b + 1382 2 A 2

3 4 Imax max = 1.74(10 ) in

Ans.

Imin = 435 in4

Ans.

Ans: u = 6.08° Imax = 1.74 ( 103 ) in4 Imin = 435 in4 1081

10–79. Solve Prob. 10–78 using Mohr’s circle.

SOLUTION tan 2u =

- 2Ixy Ix - Iy

- 2(138) 450 - 1730 Ans.

u = 6.08° Center of circle:

Ix + Iy 2

450 + 1730 = 1090 in4 2

R = 2(1730 - 1090)2 + (138)2 = 654.71 in4 Imax = 1090 + 654.71 = 1.74(103) in4

Ans.

Imin = 1090 - 654.71 = 435 in4

Ans.

Ans: u = 6.08° Imax = 1.74 ( 103 ) in4 Imin = 435 in4 1082

*10–80. Determine the directions of the principal axes with origin located at point O, and the principal moments of inertia for the area about these axes.

100 mm

SOLUTION

x 60 mm

Moments of inertia Ix and Iy: Ix =

1 (60)(100)3 = 20(10)6 mm4 3

Iy =

1 (100)(60)3 = 7.2(10)6 mm4 3

Product of inertia Ixy: Ixy = 60(100)(30)(50) = 9(10)6 mm4 Orientation: tan 2 uP =

- Ixy Ix - Iy 2

2uP = - 54.58°

-9 = -1.40625 20 - 7.2 2 Ans.

uP = - 27.3°

Principal moments of inertia: Imax =

Ix + Iy 2

min

= J

{

Ix - Iy 2 b + I 2xy B 2 a

20 + 7.2 20 - 7.2 2 { a b + 92 R 106 2 B 2

Imax = 24.6(10)6 mm4

Imin = 2.56(10)6 mm4

Ans.

Ans: uP = -27.3° Imax = 24.6(10)6 mm4 Imin = 2.56(10)6 mm4 1083

10–81. Solve Prob. 10–80 using Mohr’s circle.

100 mm

SOLUTION

x 60 mm

Moments of inertia Ix and Iy: Ix =

1 (60)(100)3 = 20(10)6 mm4 3

Iy =

1 (100)(60)3 = 7.2(10)6 mm4 3

Product of inertia Ixy: Ixy = 60(100)(30)(50) = 9(10)6 mm4 Mohr’s circle: OA = c 2(20 - 13.6)2 + 92 d (10)6 = 11.04(10)6 Imax = (13.6 + 11.04)(10)6 = 24.6(10)6 mm4 6

Ans.

Imin = (13.6 - 11.04)(10) = 2.56(10) mm tan 2u =

9 20 - 13.6

Ans.

u = - 27.3°

Ans: Imax = 24.6 ( 10 ) 6 mm4 Imin = 2.56 ( 10 ) 6 mm4 u = -27.3° 1084

10–82. Determine the directions of the principal axes with origin Determine the directions of the principal axes with origin located at point O and the principal moments of inertia of located at point O, and the principal moments of inertia of the area about these axes. the area about these axes.

50 mm

SOLUTION Ix = Iy = B -B

150 mm

1 (0.2)(0.2)3 + (0.2)(0.2)(0.1)2 R 12

1 (0.15)(0.15)3 + (0.15)(0.15)(0.075)2 R = 364.583 A 10-6 B m4 12

Ixy = C 0 + (0.1)(0.1)(0.2)(0.2) D - [0 + (0.075)(0.075)(0.15)(0.15)] = 273.4375 A 10 tan 2up =

-Ixy

Ix - Iy

= -

273.4375(10-6)

O 150 mm -6

50 mm

(364.583 - 364.583)(10-6) 2

Thus, up1 = - 45°; Imax = max min

Ix + Iy 2

Ans.

up2 = 45° ;

Ix - Iy 2

b + I2xy

= (364.583 ; 273.4375) A 10-6 B

Imax = 638 A 10-6 B m4

Ans.

-6 Imin m4 min = 91.1 10

Ans.

Ans: up1 = -45° up2 = 45° Imax = 638 (10 - 6) m4 Imin = 91.1 (10 - 6) m4 1085

10–83. SolveMohr’s Prob. 10–82 using circle. Use circle to Mohr’s determine the directions of the principal axes with origin located at point O, and the principal moments of inertia for the rectangular area about these axes.

50 mm

150 mm

SOLUTION See solution to Prob. 10–81. Ix = Iy = 364.58 ( 10-6 ) m4

150 mm

Ixy = 273.4375 ( 10-6 ) m4 Ix + Iy 2

O 50 mm

= 364.58 ( 10-6 ) m4

R = 273.4375 (10-6) m4 -6 -6 4 Imax max = ( 364.58 + 273.4375 )( 10 ) = 638(10 ) m

Ans.

Imin = ( 364.58 - 273.4375 )( 10-6 ) = 91.1(10-6) m4

Ans.

uP 1 =

90° = - 45° 2

Ans.

90° = 45° 2

Ans.

uP 2 = -

Ans: Imax = 638 (10 - 6) m4 Imin = 91.1 (10 - 6) m4 uP1 = - 45° uP2 = 45° 1086

*10–84. y

Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.

R x

SOLUTION 2p

Iz =

r A(R du)R2 = 2p r A R3

L0 2p

m =

r A R du = 2p r A R

Thus, Iz = m R 2

Ans.

Ans: Iz = m R2 1087

10–85. Determine the moment of inertia of the ellipsoid with respect to the x axis and express the result in terms of the mass m of the ellipsoid. The material has a constant density r.

y x2 a2

y2 b2

1 b x

SOLUTION a

dm= py2dx L d Ix =

y2dm 2

m =

r dV =

L-a

rp b a 1 2

x a

b dx =

4 2 prab 3

Ix =

x2 2 8 1 prab4 rpb4 a 1 - 2 b dx = 2 15 a L-a

Thus, Ix =

2 mb2 5

Ans.

Ans: Ix = 1088

2 mb2 5

10–86. The paraboloid formed revolving shaded area around the isradius of by gyration of the paraboloid. The kx the Determine the x axis. Determine of3. gyration kx of the density of the material isthe r =radius 5 Mg>m paraboloid. The density of the material is r = 5 Mg>m3.

y y2

50 x 100 mm x

SOLUTION

200 mm

Differential Disk Element: The mass of the differential disk element is dm = rdV = rpy2 dx = rp(50x) dx. The mass moment of inertia of this element rp 1 1 (2500x2) dx. is dIx = dmy2 = [rp(50x) dx](50x) = 2 2 2 Total Mass: Performing the integration, we have m =

dm =

200 mm

mm rp(50x)dx = rp(25x2)|200 = 1(106)rp 0

Mass Moment of Inertia: Performing the integration, we have Ix =

dIx = =

200 mm

rp (2500x2) dx 2

rp 2500x3 200 mm a b` 2 3 0

= 3.333(109)rp The radius of gyration is kx =

Ix 3.333(109)rp = = 57.7 mm Am A 1(106)rp

Ans.

Ans: kx = 57.7 mm 1089

10–87. The paraboloid is formed by revolving the shaded area around the x axis. Determine the moment of inertia about axis and andexpress expressthe theresult resultininterms termsofof the mass m of the xx axis the total mass m thethe paraboloid.The material hashas a constant density r. r. of paraboloid. The material a constant density

y 2

y2 = a–h x a x

SOLUTION

h 2

dm = r dV = r (p y dx)

d Ix =

1 1 dm y2 = r p y4 dx 2 2 h

Ix = =

1 a4 r p a 2 b x2 dx h L0 2 1 p ra4 h 6 h

m = = Ix =

1 a2 r p a bx dx h L0 2 1 r p a2 h 2 1 ma2 3

Ans.

Ans: Ix = 1090

1 ma2 3

*10–88. z

Determine the moment of inertia of the homogenous triangular prism with respect to the y axis. Express the result in terms of the mass m of the prism. Hint: For integration, use thin plate elements parallel to the x-y plane having a thickness of dz.

z = –h a (x – a)

SOLUTION Differential Thin Plate Element: Here, x = a a 1 -

z b. The mass of the h

z differential thin plate element is dm = rdV = rbxdz = rab a1 - b dz. The mass h moment of inertia of this element about y axis is dIy = dIG + dmr2 =

1 x2 + z2 ≤ dmx 2 + dm ¢ 12 4

1 2 x dm + z2 dm 3

= B =

z 2 z a2 a1 - b + z2 R B raba 1 - b dz R 3 h h

rab 2 a2 3a 2 3a 2 3z3 z - 3 z3 + 3z2 b dz ¢ a + 2 z2 3 h h h h

Total Mass: Performing the integration, we have h

m =

dm =

rab a 1 -

z z2 h 1 b dz = rab ¢ z ≤ ` = rabh h 2h 0 2

Mass Moment of Inertia: Performing the integration, we have h

Iy =

dIy =

rab 2 3a 2 3a2 3z3 a2 z - 3 z3 + 3z2 b dz ¢ a + 2 z2 h h h h L0 3

rab 2 a2 3a 2 2 a2 4 3z4 h 3 z z + z ¢ a z + 2 z3 ≤` 3 2h 4h 0 h 4h3

rabh 2 1a + h22 12

The mass moment of inertia expressed in terms of the total mass is Iy =

1 rabh 6 2

a2 + h2 =

m 2 a + h2 6

Ans.

Ans: Iy = 1091

m 2 (a + h2) 6

10–89. Determine the moment of inertia of the semi-ellipsoid with respect to the x axis and express the result in terms of the mass m of The material has a constant of the the semi-ellipsoid. semiellipsoid. The density r.

x2 a2

y2 b2

b x

SOLUTION

2 2 Differential Disk Element: Here, y = b a 1 -

x2 b . The mass of the differential disk element is a2

x2 b dx. The mass moment of inertia of this element is a2 rp b4 x4 1 1 x2 x2 2x2 dIx = dmy2 = c rp b2 a 1 - 2 b dx d c b2 a 1 - 2 b d = a 4 - 2 + 1b dx. 2 2 2 a a a a dm = rdV = rp y2 dx = rp b2 a 1 -

Total Mass: Performing the integration, we have a

m =

dm =

rp b2 a 1 -

x2 x3 a b dx = rp b2 ax b` 2 a 3a2 0 =

2 rpab2 3

Mass Moment of Inertia: Performing the integration, we have a

Ix =

dIx =

rp b4 x4 2x2 a 4 - 2 + 1 b dx a a L0 2

a rp b4 x5 2x3 a 4 + xb ` 2 2 5a 3a 0

4 rp ab4 15

The mass moment of inertia expressed in terms of the total mass is. Ix =

2 2 2 a rp ab2 b b2 = mb2 5 3 5

Ans.

Ans: Ix = 1092

2 mb2 5

10–90. y

Determine the radius of gyration kx of the solid formed by revolving the shaded area about the x axis. The density of the material is r.

yn h x a h x

SOLUTION Differential Disk Element: The mass of the differential disk element shown h 1 h 1 2 shaded in Fig. a is dm = rdv = rpy2dx. Here y = 1 x n . Thus, dm = rp a 1 xn b an an rph2 2 dx = xndx. The mass moment of Inertia of this element about the x axis is 2 an rxh4 4 1 1 rph2 n2 h 1 2 dIx = (dm)y2 = a 2>n x dxb a 2 xn b = a xn dxb 4 2 2 a an 2an Total Mass: Perform the integration, m =

dm =

= a = a

rph2 2

rph2 a

2 n

(x ndx)

a n b axn n+ 2 b 2 n + 2 0

n brpah2 n + 2

Mass Moment of Inertia: Perform the integration, Ix =

a rph4 4 dIx = (xn dx) 4 L L0 2an a

rph4

n n+4 = a 4>n ba bax b3 n n + 4 2a = c

n d rpah4 2(n + 4)

The radius of gyration is Ix kx = = Am

n d rp ah4 2(n + 4)

n b rp ah2 n + 4

n + 2 h A 2(n + 4)

Ans.

Ans: kx = 1093

n + 2 h A 2(n + 4)

10–91. y

The concrete shape is formed by rotating the shaded area about the y axis. Determine the moment of inertia Iy. The specific weight of concrete is g = 150 lb>ft3.

6 in.

2 2 8 in. y x 9

SOLUTION d Iy = =

1 1 (dm)(10)2 - (dm)x2 2 2 1 1 [pr (10)2 dy](10)2 - prx2 dyx2 2 2 8

Iy =

4 in.

1 9 2 pr B (10)4 dy a b y2 dy R 2 L0 L0 2 =

1 2 p (150) (10)4(8) 3 B

32.2(12)

9 2 1 a b a b (8)3 R 2 3

= 324.1 slug # in2 Iy = 2.25 slug # ft2

Ans.

Ans: Iy = 2.25 slug # ft 2 1094

*10–92. y

The solid is formed by revolving the shaded area around The solid is formed by revolving the shaded are around the the x axis. Determine the radius of gyration kx. The density x axis. Determine the radius of3 gyration kx. The density of of the material is r = 5 Mg>m . the material is r = 5 Mg>m3.

y2 = 1 – 50 x 1m

SOLUTION

Differential Disk Element: The mass of the differential disk element is dm = rdV = rpy2 dx = rp(1 - 0.5x) dx. The mass moment of inertia of this element is rp 1 1 dIx = dmy2 = [rp(1 - 0.5x) dx](1 - 0.5x) = A 0.25x2 - x + 1 B dx. 2 2 2 Total Mass: Performing the integration, we have 2m

m =

dm =

rp(1 - 0.5x) dx = rxa x -

0.5x2 2 2 m b = rp 2 0

Mass Moment of Inertia: Performing the integration, we have 2m

Ix =

dIx = =

rp A 0.25x2 - x + 1 B dx 2

2m rp 0.25x3 x2 a + xb 2 2 3 2 0

= 0.3333rp The radius of gyration is kx =

Ix = m

0.3333rp = 0.577 m rp

Ans.

Ans: kx = 0.577 m 1095

10–93. The right circular cone is formed by revolving the shaded area around the x axis. Determine the moment of inertia Ix and express the result in terms of the total mass m of the cone. The cone has a constant density r.

y y –hr x r x

SOLUTION dm = r dV = r(p y2 dx)

r2 r2 1 1 m = r(p) ¢ 2 ≤ x2 dx = rp ¢ 2 ≤ a b h3 = rp r2h 3 3 h h L0 dIx =

1 2 y dm 2

1 2 y (rp y2 dx) 2

1 r4 r(p)a 4 b x4 dx 2 h h

Ix =

1 r4 1 r(p)a 4 b x4 dx = rp r4 h 2 10 h L0

Thus, Ix =

3 m r2 10

Ans.

Ans: Ix = 1096

3 mr 2 10

10–94. y

The hemisphere is formed by rotating the shaded area about the y axis. Determine the moment of inertia Iy and express the result in terms of the total mass m of the hemisphere. The material has a constant density r.

x2 + y 2 = r 2

SOLUTION

x r

m =

r dV = r

= rp c r2y Iy = =

p x dy = rp

A r2 - y2 B dy

1 3 r 2 y d = r p r3 3 3 0

r r 4 rp r 2 1 (dm) x2 = px dy = A r - y2 B 2 dy 2 L0 2 L0 Lm 2 rp 4 y5 r 4rp 4 2 c r y - r2 y3 + d = r 2 3 5 0 15

Thus, Iy =

2 m r2 5

Ans.

Ans: Iy = 1097

2 m r2 5

10–95. Determine the moment of inertia of the wire triangle about an axis perpendicular to the page and passing through point O. Also, locate the mass center G and determine the moment of inertia about an axis perpendicular to the page and passing through point G.The wire has a mass of 0.3 kg>m. Neglect the size of the ring at O.

O –y

SOLUTION

60°

100 mm

Mass Moment of Inertia About an Axis Through Point O: The mass for each wire segment is mi = 0.310.12 = 0.03 kg. The mass moment of inertia of each segment about an axis passing through the center of mass can be determined using 1 ml2. Applying Eq. 10–16, we have 1IG2i = 12 IO = ©1IG2i + mi d 2i = 2c +

1 10.03210.122 + 0.0310.0522 d 12 1 10.03210.122 + 0.0310.1 sin 60°22 12

= 0.450110-32 kg # m2

Ans.

Location of Centroid: y =

= 0.05774 m = 57.7 mm

Mass Moment of Inertia About an Axis Through Point G: Using the result IO = 0.450110 -32 kg # m2 and d = y = 0.05774 m and applying Eq. 10–16, we have IO = IG + md 2 0.450110 -32 = IG + 310.03210.05774 22 IG = 0.150 10 -3 kg # m2

Ans.

Ans: IO = 0.450(10 -3) kg # m2 y = 57.7 mm IG = 0.150(10 -3) kg # m2 1098

*10–96. A

The pendulum consists of a 8-kg circular disk A, a 2-kg circular disk B, and a 4-kg slender rod. Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O.

0.4 m

0.5 m

0.2 m

SOLUTION Mass Moment of Inertia About An Axis Through O: The mass moment of inertia of each rod segment and disk segment shown in Fig. a about an axis passes through 1 1 their center of mass can be determined using (IG)i = m l 2 and (IG)i = mi r i2. 12 i i 2 IO = Σ c (IG)i + mi d 2i d

= c

1 1 (4) ( 1.52 ) + 4 ( 0.252 ) d + c (2) ( 0.12 ) + 2 ( 0.62 ) d 12 2 1 + c (8) ( 0.22 ) + 8 ( 1.22 ) d 2

= 13.41 kg # m2

The total mass is 8 kg + 2 kg + 4 kg = 14 kg The radius of gyration is kO =

13.41 kg # m2 IO = = 0.9787 m = 0.979 m Am C 14 kg

Ans.

Ans: kO = 0.979 m 1099

10–97. Determine the moment of inertia IIz of the frustum of the cone which has a conical depression. The material has a 3 3 density of 200 r = kg>m 200 kg>m . .

0.2 m

0.8 m 0.6 m

SOLUTION Iz =

3 1 [ p (0.4)2(1.6)(200)](0.4)2 10 3 -

3 1 [ p (0.2)2(0.8)(200)](0.2)2 10 3

3 1 [ p (0.4)2(0.6)(200)](0.4)2 10 3

0.4 m

Iz = 1.53 kg # m2

Ans.

Ans: Iz = 1.53 kg # m2 1100

10–98. The slender rods have a weight of 3 lb>ft. Determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through point A.

2 ft

SOLUTION

1.5 ft

3(3) 1 3(2) 1 3(3) IA = c d(2)2 + c d(3)2 + c d(2)2 = 1.58 slug # ft2 3 32.2 12 32.2 32.2

1.5 ft

Ans.

Ans: IA = 1.58 slug # ft 2 1101

10–99. Determine the the moment moment of of inertia inertia of of the the thin thin plate plate about about an an Determine axis perpendicular to the page and passing through the pin axis perpendicular to the page and passing through the pin at O. O. The The plate platehas hasa ahole holeininitsits center. thickness is at center. Its Its thickness is 50 3 3 50 mm, material a density = kg>m 50 kg>m mm, andand thethe material hashas a density of rof=r 50 . .

150 mm

SOLUTION 1.40 m

1 1 IG = C 50(1.4)(1.4)(0.05) D C (1.4)2 + (1.4)2 D - C 50(p)(0.15)2(0.05) D (0.15)2 12 2

1.40 m

= 1.5987 kg # m2 IO = IG + md2 m = 50(1.4)(1.4)(0.05) - 50(p)(0.15)2(0.05) = 4.7233 kg IO = 1.5987 + 4.7233(1.4 sin 45°)2 = 6.23 kg # m2

Ans.

Ans: IO = 6.23 kg # m2 1102

*10–100. The pendulum consists of a plate having a weight of 12 lb and a slender rod having a weight of 4 lb. Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O.

1 ft 1 ft

3 ft

2 ft

1 4 4 1 12 12 a b (5)2 + a b (0.5)2 + a b (12 + 12) + a b(3.5)2 12 32.2 32.2 12 32.2 32.2

= 4.917 slug # ft2 m = a kO =

4 12 b + a b = 0.4969 slug 32.2 32.2

IO 4.917 = = 3.15 ft Am A 0.4969

Ans.

Ans: kO = 3.15 ft 1103

10–101. If the large ring, small ring and each of the spokes weigh 100 lb, 15 lb, and 20 lb, respectively, determine the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A.

4 ft 1 ft O

SOLUTION Composite Parts: The wheel can be subdivided into the segments shown in Fig. a. The spokes which have a length of (4 - 1) = 3 ft and a center of mass located at a 3 distance of a1 + b ft = 2.5 ft from point O can be grouped as segment (2). 2

Mass Moment of Inertia: First, we will compute the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point O. IO = ¢

100 1 20 20 15 ≤ (42) + 8 B ¢ ≤ (32) + ¢ ≤ (2.52) R + ¢ ≤ (12) 32.2 12 32.2 32.2 32.2

= 84.94 slug # ft2 The mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A can be found using the parallel-axis theorem IA = IO + md2, 100 15 20 + 8¢ = 8.5404 slug and d = 4 ft.Thus, where m = ≤ + 32.2 32.2 32.2 IA = 84.94 + 8.5404(42) = 221.58 slug # ft2 = 222 slug # ft2

Ans.

Ans: IA = 222 slug # ft 2 1104

10–102. Determine the moment of inertia of the center crank about the x axis. The material is steel having a specific weight of gst = 490 lb>ft3.

0.5 in.

1 in.

0.5 in.

1 in. 0.5 in. 4 in.

SOLUTION ms =

490 p(0.25)2(1) a b = 0.0017291 slug 32.2 (12)3

mp =

490 (6)(1)(0.5) ¢ ≤ = 0.02642 slug 32.2 (12)3

Ix = 2 B

0.5 in.

0.5 in. x 1 in. 1 in.

1 in.

1 1 (0.02642) A (1)2 + (6)2 B + (0.02642)(2)2 R + 2 B (0.0017291)(0.25)2 R 12 2 +

1 (0.0017291)(0.25)2 + (0.0017291)(4)2 2

= 0.402 slug # ft2

Ans.

Ans: Ix = 0.402 slug # ft 2 1105

10–103. Each of the three slender rods has a mass m. Determine the moment of inertia of the assembly about an axis that is perpendicular to the page and passes through the center point O. a

a O

SOLUTION IO = 3 B

1 a sin 60° 2 1 ma2 + m ¢ ≤ R = ma2 12 3 2

Ans.

Ans: IO = 1106

1 ma2 2

*10–104. The thin plate has a mass per unit area of 10 kg>m2. Determine its mass moment of inertia about the y axis.

z 200 mm 100 mm

200 mm

SOLUTION

100 mm

Composite Parts: The thin plate can be subdivided into segments as shown in Fig. a. Since the segments labeled (2) are both holes, the y should be considered as negative parts.

200 mm 200 mm x

200 mm

200 mm 200 mm

Mass Moment of Inertia: The mass of segments (1) and (2) are m1 = 0.4(0.4)(10) = 1.6 kg and m2 = p(0.12)(10) = 0.1p kg. The perpendicular distances measured from the centroid of each segment to the y axis are indicated in Fig. a. The mass moment of inertia of each segment about the y axis can be determined using the parallel-axis theorem. Iy = © A Iy B G + md2 = 2c

1 1 (1.6)(0.42) + 1.6(0.22) d - 2 c (0.1p)(0.12) + 0.1p(0.22) d 12 4

= 0.144 kg # m2

Ans.

Ans: Iy = 0.144 kg # m2 1107

10–105. The thin plate has a mass per unit area of 10 kg>m2. Determine its mass moment of inertia about the z axis.

z 200 mm 100 mm

200 mm

SOLUTION

100 mm

Composite Parts: The thin plate can be subdivided into four segments as shown in Fig. a. Since segments (3) and (4) are both holes, the y should be considered as negative parts.

200 mm 200 mm x

200 mm

Mass Moment of Inertia: Here, the mass for segments (1), (2), (3), and (4) are m1 = m2 = 0.4(0.4)(10) = 1.6 kg and m3 = m4 = p(0.12)(10) = 0.1p kg. The mass moment of inertia of each segment about the z axis can be determined using the parallel-axis theorem. Iz = © A Iz B G + md2 =

1 1 1 1 (1.6)(0.42) + c (1.6)(0.42 + 0.42) + 1.6(0.22) d - (0.1p)(0.12) - c (0.1p)(0.12) + 0.1p(0.22) d 12 12 4 2

= 0.113 kg # m2

Ans.

Ans: Iz = 0.113 kg # m2 1108

10–106. Determine the moment of inertia of the wheel about the x axis that passes through the center of mass G. The material has a specific weight of g = 90 lb>ft3. 1 ft x

SOLUTION

0.25 ft

Mass Moment of Inertia About an Axis Through Point G: The mass moment of inertia of each disk about an axis passing through the center of mass can be 0.5 ft 1 determine using 1IG2i = mr2. Applying Eq. 10–16, we have 2

0.25 ft 2 ft

x¿

x¿ 1 ft

1 p12.5 21121902 1 p12 210.7521902 B R 12.522 - B R 1222 2 32.2 2 32.2 2

p10.25 210.2521902 1 p10.25 210.2521902 - 4b B R 10.2522 + B R 1122 r 2 32.2 32.2 2

= 118 slug # ft2

Ans.

Ans: IG = 118 slug # ft 2 1109

10–107. the moment moment of ofinertia inertiaof ofthe thewheel wheelabout aboutthe thex¿x' Determine the that passes passes through through point point O. O.The Thematerial materialhas hasaaspecific specific axis that = 90 90 lb>ft lb/ft3. weight of g = 1 ft x

SOLUTION

0.25 ft

0.25 ft 2 ft

x¿

x¿ 1 ft

1 p12.5 21121902 1 p12 210.7521902 B R 12.522 - B R 1222 2 32.2 2 32.2 2

1 p10.25 210.2521902 - 4b B R 10.2522 2 32.2 2

+ B

p10.252210.2521902 32.2

R 1122 r

= 118.25 slug # ft2 Mass Moment of Inertia About an Axis Through Point O: The mass of the wheel is m =

p12.5221121902 32.2

p122210.7521902 32.2

- 4B

p10.252210.2521902 32.2

= 27.989 slug Using the result IG = 118.25 slug # ft2 and applying Eq. 10–16, we have IO = IG + md2 = 118.25 + 27.98912.522 = 293 slug # ft2

Ans.

Ans: IO = 293 slug # ft 2 1110

*10–108. The pendulum consists of two slender rods AB and OC which have a mass of 3 kg>m. The thin plate has a mass of 12 kg>m2. Determine the location y of the center of mass G of the pendulum, then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.

0.4 m

B O

_ y G

1.5 m

SOLUTION y =

1.5(3)(0.75) + p(0.3)2(12)(1.8) - p(0.1)2(12)(1.8)

0.1 m

Ans.

= 0.8878 m = 0.888 m IG =

1.5(3) + p(0.3)2(12) - p(0.1)2(12) + 0.8(3)

0.3 m

1 (0.8)(3)(0.8)2 + 0.8(3)(0.8878)2 12 +

1 (1.5)(3)(1.5)2 + 1.5(3)(0.75 - 0.8878)2 12

1 + [p(0.3)2(12)(0.3)2 + [p(0.3)2(12)](1.8 - 0.8878)2 2 -

1 [p(0.1)2(12)(0.1)2 - [p(0.1)2(12)](1.8 - 0.8878)2 2

IG = 5.61 kg # m2

Ans.

Ans: y = 0.888 m IG = 5.61 kg # m2 1111

10–109. z

the moment moment of of inertia inertia IIzz of ofthe thefrustrum frustum of the Determine the cone which has a conical depression. The material has a kg>m33.. density of 200 kg>m

200 mm 600 mm

SOLUTION

400 mm

z + 1 z Mass Moment of Inertia About z Axis: From similar triangles, = , 0.2 0.8 z = 0.333 m. The mass moment of inertia of each cone about z axis can be 3 mr 2. determined using Iz = 10 Iz = ©1Iz2i =

800 mm

3 p c 10.82211.333212002 d10.822 10 3 -

3 p c 10.2 2210.333212002 d10.2 22 10 3

3 p c 10.2 2210.6212002 d10.2 22 10 3

= 34.2 kg # m2

Ans.

Ans: Iz = 34.2 kg # m2 1112

11–1. Determine the tension in cable The lampthe weighs 10 lb.in Use the method of virtual work AC. to determine tensions cable AC. The lamp weighs 10 lb.

B C

45°

SOLUTION

30°

Free-Body Diagram: The tension in cable AC can be determined by releasing cable AC. The system has only one degree of freedom defined by the independent coordinate u. When u undergoes a positive displacement du, only FAC and the weight of lamp (10 lb force) do work. Virtual Displacement: Force FAC and 10 lb force are located from the fixed point B using position coordinates yA and xA. xA = l cos u

dxA = - l sin udu

(1)

yA = l sin u

dyA = l cos udu

(2)

Virtual–Work Equation: When yA and xA undergo positive virtual displacements dyA and dxA, the 10 lb force and horizontal component of FAC, FAC cos 30°, do positive work while the vertical component of FAC, FAC sin 30°, does negative work. dU = 0;

10dyA - FAC sin 30°dyA + FAC cos 30°, dxA = 0

(3)

Substituting Eqs. (1) and (2) into (3) yields (10 cos u - 0.5FAC cos u - 0.8660FAC sin u)ldu = 0 Since ldu Z 0, then FAC =

10 cos u 0.5 cos u + 0.8660 sin u

At the equilibrium position u = 45°, FAC =

10 cos 45° = 7.32 lb 0.5 cos 45° + 0.8660 sin 45°

Ans.

Ans: FAC = 7.32 lb 1113

11–2. The scissors jack supports a load P. Determine the axial force in the screw necessary for equilibrium when the jack is in the position u. Each of the four links has a length L and is pin connected at its center. Points B and D can move horizontally.

SOLUTION x = L cos u,

dx = -L sin u du

y = 2L sin u,

dy = 2L cos u du

dU = 0;

- Pdy - Fdx = 0

B u

-P(2L cos u du) - F(- L sin u du) = 0 Ans.

F = 2P cot u

Ans: F = 2P cot u 1114

11–3. The disk has a weight of 10 lb and is subjected to a vertical force P = 8 lb and a couple moment M = 8 lb # ft. Determine the disk’s rotation u if the end of the spring wraps around the periphery of the disk as the disk turns. The spring is originally unstretched.

M 8 lb ft 1.5 ft

k 12 lb/ft

SOLUTION

P 8 lb

dyF = dyP = 1.5du PdyP + Mdu - FdyF = 0

dU = 0;

8(1.5 du) + 8 du - F(1.5 du) = 0 u(20 - 1.5F) = 0 Since du Z 0 20 -1.5F = 0

F = 13.33 lb

F = kx Where x = 1.5u 13.33 = 12(1.5u) Ans.

u = 0.7407 rad = 42.4°

Ans: u = 42.4° 1115

*11–4. If a force of P = 5 lb is applied to the handle of the mechanism, determine the force the screw exerts on the cork of the bottle. The screw is attached to the pin at A and passes through the collar that rests on the bottle neck at B.

5 lb

SOLUTION

Free - Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only the force in the screw Fs and force P do work when the virtual displacements take place.

u 3 in.

30°

Virtual Displacement: The position of the points of application for Fs and P are specified by the position coordinates yA and yD, measured from the fixed point B, respectively. yA = 2(3 sin u)

dyA = 6 cos udu

(1)

yD = 6(3 sin u)

dyD = 18 cos udu

(2)

Virtual–Work Equation: Since P acts towards the positive sense of its corresponding virtual displacement, it does positive work. However, the work of Fs is negative since it acts towards the negative sense of its corresponding virtual displacement. Thus, dU = 0;

PdyD + A - FSdyA B = 0

(3)

Substituting P = 5 lb, Eqs. (1) and (2) into Eq. (3), 5(18 cos udu)FS (6 cos udu) = 0 cos udu(90 - 6FS) = 0 Since cos udu Z 0, then 90 - 6FS = 0 Ans.

FS = 15 lb

Ans: FS = 15 lb 1116

11–5. A 5-kg uniform serving table is supported on each side by pairs of two identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg, determine the angle u where the table is in equilibrium. The springs each have a stiffness of k = 200 N>m and are unstretched when u = 90°. Neglect the mass of the links.

250 mm

150 mm

250 mm

SOLUTION

B 150 mm

Virtual Displacement: The position of points of application of Wb, Wt, and Fsp are specified by the position coordinates yGb, yGt, and xC, respectively. Here, yGb and yGt are measured from the fixed point B while xC is measured from the fixed point D. yGb = 0.25 sin u + b

dyGb = 0.25 cos udu

(1)

yGt = 0.25 sin u + a

dyGt = 0.25 cos udu

(2)

xC = 0.25 cos u

dxC = -0.25 sin udu

(3)

Virtual–Work Equation: Since Wb, Wt, and Fsp act towards the negative sense of their corresponding virtual displacement, their work is negative. Thus, dU = 0;

-WbdyGb + A - WtdyGt B + A - FspdxC B = 0

(4)

1 5 Substituting Wb = a b(9.81) = 4.905 N, Wt = a b(9.81) = 24.525 N, 2 2 Fsp = 50 cos u N, Eqs. (1), (2), and (3) into Eq. (4), we have - 4.905(0.25 cos udu) - 24.525(0.25 cos udu) - 50 cos u( -0.25 sin udu) = 0 du A - 7.3575 cos u + 12.5 sin u cos u B = 0 Since du Z 0, then -7.3575 cos u + 12.5 sin u cos u = 0 cos u(- 7.3575 + 12.5 sin u) = 0 Solving the above equation, cos u = 0

Ans.

u = 90°

- 7.3575 + 12.5 sin u = 0 Ans.

u = 36.1°

Ans: u = 90° u = 36.1° 1117

11–6. A 5-kg uniform serving table is supported on each side by pairs of two identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg and is in equilibrium when u = 45°, determine the stiffness k of each spring. The springs are unstretched when u = 90°. Neglect the mass of the links.

250 mm

150 mm

250 mm

SOLUTION

B 150 mm

dyGb = 0.25 cos udu

(1)

yGt = 0.25 sin u + a

dyGt = 0.25 cos udu

(2)

xC = 0.25 cos u

dxC = -0.25 sin udu

(3)

Virtual–Work Equation: Since Wb, Wt, and Fsp act towards the negative sense of their corresponding virtual displacement, their work is negative. Thus, dU = 0;

-WbdyGb + A - WtdyGt B + A - FspdxC B = 0

(4)

1 5 Substituting Wb = a b(9.81) = 4.905 N, Wt = a b(9.81) = 24.525 N, 2 2 Fsp = 0.25k cos u N, Eqs. (1), (2), and (3) into Eq. (4), we have - 4.905(0.25 cos udu) - 24.525(0.25 cos udu) - 0.25k cos u(-0.25 sin udu) = 0 du A - 7.3575 cos u + 0.0625k sin u cos u B = 0 Since du Z 0, then - 7.3575 cos u + 0.0625k sin u cos u = 0 117.72 k = sin u When u = 45°, then 117.72 k = = 166 N>m sin 45°

Ans.

Ans: k = 166 N>m 1118

11–7. When u = 20°, the 50-lb uniform block compresses the two vertical springs 4 in. If the uniform links AB and CD each weigh 10 lb, determine the magnitude of the applied couple moments M needed to maintain equilibrium when u = 20°.

2 lb/in.

2 lb/ in. 1 ft

1 ft

SOLUTION

Free-Body Diagram: The system has only one degree of freedom defined by the independent coordinate u. When u undergoes a positive displacement du, only the spring forces Fsp, the weight of the block (50 lb), the weights of the links (10 lb) and the couple moment M do work.

M A

1 ft 2 ft M

4 ft

Virtual Displacements: The spring forces Fsp, the weight of the block (50 lb) and the weight of the links (10 lb) are located from the fixed point C using position coordinates y3, y2 and y1 respectively. y3 = 1 + 4 cos u

dy3 = - 4 sin udu

(1)

y2 = 0.5 + 4 cos u

dy2 = -4 sin udu

(2)

y1 = 2 cos u

(3)

dy1 = - 2 sin udu

Virtual–Work Equation: When y1, y2 and y3 undergo positive virtual displacements dy1, dy2 and dy3, the spring forces Fsp, the weight of the block (50 lb) and the weights of the links (10 lb) do negative work. The couple moment M does negative work when the links undergo a positive virtual rotation du. dU = 0;

- 2Fspdy3 - 50dy2 - 20dy1 - 2Mdu = 0

(4)

Substituting Eqs. (1), (2) and (3) into (4) yields (8Fsp sin u + 240 sin u - 2M) du = 0 Since du Z 0, then 8Fsp sin u + 240 sin u - 2M = 0 M = sin u(4Fsp + 120) At the equilibrium position u = 20°, Fsp = kx = 2(4) = 8 lb. M = sin 20°[4(8) + 120] = 52.0 lb # ft

Ans.

Ans: M = 52.0 lb # ft 1119

*11–8. The bar is supported by the spring and smooth collar that allows the spring to be always perpendicular to the bar for any angle u. If the unstretched length of the spring is l0, determine the force P needed to hold the bar in the equilibrium position u. Neglect the weight of the bar.

a A

B u k C l

SOLUTION s = a sin u,

ds = a cos u du

y = l sin u,

dy = l cos u du

Fs = k(a sin u - l0) dU = 0;

Pdy - Fsds = 0

Pl cos du - k(a sin u - l0) a cos u du = 0 P =

ka (a sin u - l0) l

Ans.

Ans: P = 1120

ka (a sin u - l0) l

11–9. Determine the force P that must be applied perpendicular to the handle in order to hold the mechanism in equilibrium for any angle u of rod CD. There is a couple moment M applied to the link BA. A smooth collar attached at A slips along rod CD.

P D

L A

SOLUTION Virtual–Work Equation: dU = 0;

P(Ldu) -M(2 du) = 0 du(PL - 2M) = 0

since du ≠ 0

2M L

M B

L> 2

PL - 2M = 0 P =

L> 2

Ans.

Ans: P = 1121

2M L

11–10. The thin rod of weight W rest against the smooth wall and floor. Determine the magnitude of force P needed to hold it in equilibrium for a given angle u.

SOLUTION

Free-Body Diagram: The system has only one degree of freedom defined by the independent coordinate u. When u undergoes a positive displacement du, only the weight of the rod W and force P do work. Virtual Displacements: The weight of the rod W and force P are located from the fixed points A and B using position coordinates yC and xA, respectively yC =

1 sin u 2

dyC =

xA = l cos u

1 cos udu 2

dxA = - l sin udu

(1) (2)

Virtual–Work Equation: When points C and A undergo positive virtual displacements dyC and dxA, the weight of the rod W and force F do negative work. (3)

dU = 0; - WdyC - PdyA = 0 Substituting Eqs. (1) and (2) into (3) yields a Pl sin u -

Wl cos u b du = 0 2

Since du Z 0, then Pl sin u P =

Wl cos u = 0 2

W cot u 2

Ans.

Ans: P = 1122

W cot u 2

11–11. If each of the three links of the mechanism has a mass of 4 kg, determine the angle u for equilibrium. The spring, which always remains vertical, is unstretched when u = 0°.

M 30 N m u 200 mm

k 3 kN/m D

200 mm

SOLUTION

Free-Body Diagram: The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only the weights W1 = W2 = W3 = W, couple moment M, and spring force Fsp do work. Virtual Displacement: The positions of the weights W1, W2, W3 and spring force Fsp are measured from fixed point A using position coordinates y1, y2, y3 and y4 respectively y1 = 0.1 sin u

dy1 = 0.1 cos u du

(1)

y2 = 0.2 sin u + 0.1

dy2 = 0.2 cos u du

(2)

y3 = 0.1 sin u + 0.2

dy3 = 0.1 cos u du

(3)

y4 = 0.5 sin u

dy4 = 0.2 cos u du

(4)

Virtual–Work Equation: When all the weights undergo positive virtual displacement, all of them do positive work. However, Fsp does negative work when its undergoes positive virtual displacement. Also, M does positive work when it undergoes positive virtual angular displacement. dU = 0;

W1dy1 + W2dy2 + W3dy3 - Fspdy4 + Mdu = 0

(5)

Substitute Eqs (1), (2) and (3) into (5), using W1 = W2 = W3 = W. W(0.1 cos u du) + W(0.2 cos u du) + W(0.1 cos u du) - Fsp(0.2 cos u du) + Mdu = 0 (0.4 W cos u - 0.2 Fsp cos u + M)du = 0 Since d ≠ 0, then 0.4 W cos u - 0.2 Fsp cos u + M = 0

Here M = 30 N # m, W = 4(9.81)N = 39.24 N and Fsp = kx = 3000(0.2 sin u) = 600 sin u. Substitute these results into this equation, 0.4(39.24) cos u - 0.2(600 sin u)cos u + 30 = 0 15.696 cos u - 120 sin u cos u + 30 = 0

1123

u 200 mm

11–11. Continued

Solve numerically: Ans.

u = 23.8354° = 23.8° or

Ans.

u = 72.2895° = 72.3° Note: u = 23.8° is a stable equilibrium, while u = 72.3° is an unstable one.

Ans: u = 23.8° u = 72.3° 1124

*11–12. If the disk is subjected to a couple moment M, determine the disk’s rotation u required for equilibrium. The end of the spring wraps around the periphery of the disk as the disk turns. The spring is originally unstretched.

k 4 kN/ m

0.5 m

M 300 N m

SOLUTION Free-Body Diagram: The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only the spring force Fsp and couple moment M do work. Virtual–Work Equation: When the disk undergoes a positive angular displacement du, correspondingly point A undergoes a positive displacement of dxA. As a result couple moment M does positive work whereas spring force Fsp does negative work. dU = 0;

(1)

Mdu + ( -FspdxA) = 0

Here, Fsp = kxA = 4000(0.50) = 2000u and dxA = 0.5du. Substitute these results into Eq. (1) 300du - 2000u(0.5du) = 0 (300 - 1000u)du = 0 Since du ≠ 0, 300 - 1000u = 0 u = 0.3 rad = 17.19° = 17.2° Ans.

Ans: u = 17.2° 1125

11–13. Determine the horizontal force F required to maintain equilibrium of the slider mechanism when u = 60°. Set M = 6 N # m

D F 0.5 m B 0.5 m

0.5 m M u

SOLUTION dU = 0;

- F cos 30°(1du) sin 30° + 6du = 0 - F(0.866)(0.5) + 6 = 0 Ans.

F = 13.9 N

Ans: F = 13.9 N 1126

11–14. Determine the moment M that must be applied to the slider mechanism in order to maintain the equilibrium position u = 60° when the horizontal force F = 100 N is applied at D.

D F 0.5 m B 0.5 m

0.5 m M u

SOLUTION dU = 0;

- Mdf + F sin u(1du) = 0

But

0.5df cos 2 u = 0.5du

df =

du cos 2u

-Ma

du b + F sin u(1du) = 0 cos 2u

M = - F sin u cos 2u = -100(sin 60°) cos 2(60°) = 43.3 N # m

Ans.

Ans: M = 43.3 N # m 1127

11–15. The spring is unstretched when u = 0°. If P = 8 lb, determine the angle u for equilibrium. Due to the roller guide, the spring always remains vertical. Neglect the weight of the links.

k 5 50 lb>ft 2 ft 2 ft

SOLUTION y1 = 2 sin u,

4 ft

dy1 = 2 cos u du

y2 = 4 sin u + 4, dy2 = 4 cos u du

4 ft

Fs = 50(2 sin u) = 100 sin u dU = 0;

- Fs dy1 + Pdy2 = 0

- 100 sin u(2 cos u du) + 8(4 cos u du) = 0 200 sin u = 16 Choosing the smallest root: Ans.

u = 9.21°

Ans: u = 9.21° 1128

*11–16. If each of the three links of the mechanism has a weight of 20 lb, determine the angle u for equilibrium. The spring, which always remains vertical due to the roller guide, is unstretched when u = 0°. Set P = 0.

k 5 50 lb>ft 2 ft 2 ft

SOLUTION y1 = 2 sin u,

dy1 = 2 cos u du

y2 = 4 sin u,

dy2 = 4 cos u du

y3 = 2 sin u,

dy3 = 2 cos u du

4 ft

y4 = 4 sin u + 4, dy4 = 4 cos u du

Fs = 50(2 sin u) = 100 sin u dU = 0;

- Fs dy1 + W(dy1 + dy2 + dy3) + Pdy4 = 0

- 10 sin u(2 cos u du) + 20(8 cos u du) + 0 = 0 200 sin u = 160 and cos u = 0 Choosing the smallest root: Ans.

u = 53.1°

Ans: u = 53.1° 1129

11–17. If a force P = 40 N is applied perpendicular to the handle of the toggle press, determine the vertical compressive force developed at C; u = 30°.

50 mm

300 mm

P 5 40 N

350 mm

SOLUTION dU = 0;

- FC dy - 40ds = 0 ds = 300 du (350)2 = (50)2 + y2 - 2(50)(y)(cos u) 0 = 0 + 2y dy - 100 dy cos u + 100 y sin ud

For u = 30°;

y = 392.407 dy = - 28.10du - FC ( -28.10)du - 40(300)(du) = 0 (28.10FC - 12.000) du = 0 Ans.

FC = 427 N

Ans: FC = 427 N 1130

11–18. The “Nuremberg scissors” is subjected to a horizontal force of P = 600 N. Determine the angle u for equilibrium. The spring has a stiffness of k = 15 kN>m and is unstretched when u = 15°.

E u

200 mm A k

SOLUTION

B 200 mm C

Free - Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is a formed. We observe that only the spring force Fsp acting at points A and B and the force P do work when the virtual displacements take place. The magnitude of Fsp can be computed using the spring force formula, Fsp = kx = 15(103) C 2(0.2 sin u) - 2(0.2 sin 15°) D = 6000(sin u - 0.2588) N Virtual Displacement: The position of points A and B at which Fsp acts and point C at which force P acts are specified by the position coordinates yA, yB, and yC, measured from the fixed point E, respectively. yA = 0.2 sin u

dyA = 0.2 cos udu

(1)

yB = 3(0.2 sin u)

dyB = 0.6 cos udu

(2)

yC = 8(0.2 sin u)

dyB = 1.6 cos udu

(3)

Virtual–Work Equation: Since Fsp at point A and force P acts towards the positive sense of its corresponding virtual displacement, their work is positive. The work of Fsp at point B is negative since it acts towards the negative sense of its corresponding virtual displacement. Thus, Fsp dyA + A - FspdyB B + PdyC = 0

dU = 0;

(4)

Substituting Fsp = 6000(sin u - 0.2588), P = 600 N, Eqs. (1), (2), and (3) into Eq. (4), 6000(sin u - 0.2588)(0.2 cos udu - 0.6 cos udu) + 600(1.6 cos udu) = 0 cos udu C -2400(sin u - 0.2588) + 960 D = 0 Since cos udu Z 0, then - 2400(sin u - 0.2588) + 960 = 0 Ans.

u = 41.2°

Ans: u = 41.2° 1131

11–19. The “Nuremberg scissors” is subjected to a horizontal force of P = 600 N. Determine the stiffness k of the spring for equilibrium when u = 60°. The spring is unstretched when u = 15°.

E u

200 mm A k

SOLUTION

B 200 mm C

Free - Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only the spring force Fsp acting at points A and B and the force P do work when the virtual displacements take place. The magnitude of Fsp can be computed using the spring force formula. Fsp = kx = k C 2(0.2 sin u) - 2(0.2 sin 15°) D = (0.4)k(sin u - 0.2588) N Virtual Displacement: The position of points A and B at which Fsp acts and point C at which force P acts are specified by the position coordinates yA, yB, and yC, measured from the fixed point E, respectively. yA = 0.2 sin u

dyA = 0.2 cos udu

(1)

yB = 3(0.2 sin u)

dyB = 0.6 cos udu

(2)

yC = 8(0.2 sin u)

dyB = 0.6 cos udu

(3)

dU = 0;

(4)

Substituting Fsp = k(sin u - 0.2588), P = 600 N, Eqs. (1), (2), and (3) into Eq. (4), (0.4)k(sin u - 0.2588)(0.2 cos udu - 0.6 cos udu) + 600(1.6 cos udu) = 0 cos udu C -0.16k(sin u - 0.2588) + 960 D = 0 Since cos udu Z 0, then - 0.16k(sin u - 0.2588) + 960 = 0 k =

6000 sin u - 0.2588

When u = 60°, k =

6000 = 9881N>m = 9.88 kN>m sin 60° - 0.2588

Ans.

Ans: k = 9.88 kN>m 1132

*11–20. The spring has an unstretched length of 0.3 m. Determine the angle u for equilibrium if the uniform links each have a mass of 5 kg. Set P = 0.

0.5 m B

u D

k 5 200 N>m 0.5 m

SOLUTION

Virtual Displacement: y = 0.5 cos u

dy = - 0.5 sin u du

x1 = 0.5 sin u

dx1 = 0.5 cos u du

x2 = 1.5 sin u

dx2 = 1.5 cos u du

Virtual–Work Equation: dU = 0;

- 235(9.81)dy4 + Fsdx1 - Fsdx2 = 0

- 98.1( - 0.5 sin u du) + Fs(0.5 cos u du) - Fs(1.5 cos u du) = 0 du(49.05 sin u - Fs cos u) = 0 Since du ≠ 0 49.05 sin u - Fs cos u = 0 [1] Spring Force: Fs = kx where

x = (0.5 sin u + 0.5 sin u) - 0.3 = sin u - 0.3

Fs = 200(sin u - 0.3) = 200 sin u - 60 [2] Substituting Eq. [2] into [1] yields: 49.05 sin u - (200 sin u - 60)cos u = 0 49.05 sin u - 200 sin ucos u + 60 cos u = 0 Ans.

u = 24.2°

Ans: u = 24.2° 1133

11–21. The spring has an unstretched length of 0.3 m. Determine the angle u for equilibrium if P = 50 N. Neglect the weight of the links.

0.5 m B

u D

k 5 200 N>m 0.5 m

SOLUTION

Virtual Displacement: x1 = 0.5 sin u

dx1 = 0.5 cos u du

x2 = 1.5 sin u

dx2 = 1.5 cos u du

x3 = 1 sin u

dx3 = cos u du

Virtual–Work Equation: dU = 0;

Fsdx1 - Fsdx2 - Pdx3 = 0

Fs(0.5 cos u du) - Fs(1.5 cos u du) + 50(cos u du) = 0 du(50 cos u - Fs cos u) = 0 Since d ≠ 0 50 cos u - Fs cos u = 0 [1] Spring Force: Fs = kx where

x = (0.5 sin u + 0.5 sin u) - 0.3 = sin u - 0.3

Fs = 200(sin u - 0.3) = 200 sin u - 60 [2] Substituting Eq. [2] into [1] yields: 50 cos u - (200 sin u - 60)cos u = 0 cos u (110 - 200 sin u) = 0 cos u = 0

Ans.

u = 90°

Ans.

110 - 200 sin u = 0 u = 33.4°

Ans: u = 90° u = 33.4° 1134

11–22. The dumpster has a weight W and a center of gravity at G. Determine the force in the hydraulic cylinder needed to hold it in the general position u.

b a

SOLUTION

G d

s = 2a2 + c2 - 2a c cos (u + 90°) = 2a2 + c2 + 2a c sin u c

ds = (a2 + c2 + 2a c sin u)- 2 ac cos u du y = (a + b) sin u + d cos u dy = (a + b) cos u du - d sin u du dU = 0;

Fds - Wdy = 0 1

F(a2 + c2 + 2a c sin u)- 2 ac cos u du - W(a + b) cos u du + Wd sin u du = 0 F = a

W(a + b - d tan u) b 2a2 + c2 + 2a c sin u ac

Ans.

Ans: F = 1135

W(a + b - d tan u) ac

2a2 + c 2 + 2ac sin u

11–23. P1

If vertical forces P1 = 40 lb and P2 = 20 lb act at C and E as shown, determine the angle u for equilibrium. The spring is unstretched when u = 45°. Neglect the weight of the members.

C k 5 200 lb>ft 2 ft

SOLUTION

2 ft

y = 4 sin u A

x = 4 cos u

2 ft

dy = 4 cos u du dx = - 4 sin u du dU = 0;

- Fsdx - 60dy = 0

[ - Fs( -4 sin u) - 60(4 cos u)]du = 0 Fs = a

60 b tan u

Fs = k(4 cos u - 4 cos 45°) 60 = 200 tan u (4 cos u - 4 cos 45°) sin u - 0.707 tan u - 0.075 = 0 Ans.

Solving, u = 16.6°

Ans: u = 16.6° 1136

*11–24. P1

If vertical forces P1 = P2 = 30 lb act at C and E as shown, determine the angle u for equilibrium. The spring is unstretched when u = 45°. Neglect the weight of the members.

C k 5 200 lb>ft 2 ft

SOLUTION

2 ft

y = 4 sin u x = 4 cos u A

dy = 4 cos u du dx = - 4 sin u du

2 ft

- Fsdx - 30dy - 30dy = 0

dU = 0;

[ - Fs( -4 sin u) - 60(4 cos u)]du = 0 Fs = 60 a

cos u b sin u

Since Fs = k(4 cos u - 4 cos 45°) = 200(4 cos u - 4 cos 45°) 60 cos u = 800(cos u - cos 45°) sin u sin u - 0.707 tan u - 0.075 = 0 Ans.

u = 16.6°

Ans: u = 16.6° 1137

11–25. The members of the mechanism are pin connected. If a vertical force of 800 N acts at A, determine the angle u for equilibrium. The spring is unstretched when u = 0°. Neglect the mass of the links.

k 6 kN/ m

SOLUTION

800 N

Virtual Displacement: The positions of Fsp and P are measured from fixed point B using position coordinates yc and yA respectively. yc = 1 sin u dyc = cos u du

(1)

yA = 3(1 sin u) dyA = 3 cos u du

(2)

Virtual–Work Equation: When Fsp and P undergo their respective positive virtual displacement, P does positive work whereas Fsp does negative work. (3)

- Fspdyc + PdyA = 0

dU = 0; Substitute Eqs. (1) and (2) into (3)

- Fsp(cos udu) + P(3 cos udu) = 0 ( -Fsp cos u + 3P cos u)du = 0 Since du ≠ 0, and assuming u 6 90°, then - Fsp cos u + 3P cos u = 0 Fsp = 3P Here Fsp = kx = 6000(1 sin u) = 6000 sin u and P = 800 N, Then 6000 sin u = 3(800) sin u = 0.4 Ans.

u = 23.58° = 23.6°

Ans: u = 23.6° 1138

11–26. The potential energy of a one-degree-of-freedom system is defined by V = (20x3 - 10x2 - 25x - 10) ft # lb, where x is in ft. Determine the equilibrium positions and investigate the stability for each of these positions.

SOLUTION Equilibrium Configuration: Taking the first derivative of V, we have dV = 60x2 - 20x - 25 dx Equilibrium requires

dV = 0. Thus, dx 60x 2 - 20x - 25 = 0 x =

20 ; 3( - 20)2 - 4(60)( - 25) 2(60) Ans.

x = 0.833 ft and - 0.5 ft Stability: The second derivative of V is d2V = 120x - 20 dx2 At x = 0.8333 ft, d2V 2 = 120(0.8333) - 20 = 80 7 0 dx2 x = 0.8333 ft

Ans.

Thus, the equilibrium configuration at x = 0.8333 ft is stable. At x = -0.5ft, d 2V 2 = 120( -0.5) - 20 = -80 6 0 dx2 x = -0.5 ft

Ans.

Thus, the equilibrium configuration at x = -0.5 ft is unstable.

Ans: x = - 0.5 ft unstable x = 0.833 ft stable 1139

11–27. If the potential function for a one-degree-of-freedom system is V = (12 sin 2u + 15cos u) J, where 0° 6 u 6 180°, determine the positions for equilibrium and investigate the stability at each of these positions.

SOLUTION V = 12 sin 2u + 15 cos u dV = 0; du

24 cos 2u - 15 sin u = 0

2411 - 2 sin2 u2 - 15 sin u = 0 48 sin2 u + 15 sin u - 24 = 0 Choosing the angle 0° 6 u 6 180° u = 34.6°

Ans.

u = 145°

Ans.

and

d2V = - 48 sin 2u - 15 cos u du2 u = 34.6°,

d 2V = - 57.2 6 0 du2

Unstable

Ans.

u = 145°,

d2V = 57.2 7 0 du2

Stable

Ans.

Ans: Unstable at u = 34.6° stable at u = 145° 1140

*11–28. If the potential function for a one-degree-of-freedom system is V = (8x3 - 2x2 - 10) J, where x is given in meters, determine the positions for equilibrium and investigate the stability at each of these positions.

SOLUTION V = 8x3 - 2x2 - 10 dV = 24x2 - 4x = 0 dx 124x - 42x = 0 x = 0

Ans.

and x = 0.167 m

d2V = 48x - 4 dx2 x = 0,

d 2V = -4 6 0 dx2

x = 0.167 m,

Ans.

Unstable

d2V = 4 7 0 dx2

Ans.

Stable

Ans: x = 0.167 m d 2V = -4 6 0 Unstable dx2 d 2V = 4 7 0 Stable dx2 1141

11–29. If the potential function for a one-degree-of-freedom system is V = (10 cos3 - 2u + 25 sin u) J, where 0° 6 u 6 180°, determine the positions for equilibrium and investigate the stability at each of these positions.

SOLUTION V = 10 cos 2u + 25 sin u For Equilibrium: dV = - 20 sin 2u + 25 cos u = 0 du 1- 40 sin u + 252 cos u = 0 u = sin-1 a

25 b = 38.7° and 141° 40

Ans.

and u = cos-1 0 = 90°

Ans.

Stability: d2V = - 40 cos 2u - 25 sin u du2 u = 38.7°,

d2V = - 24.4 6 0, du2

Unstable

Ans.

u = 141°,

d2V = - 24.4 6 0, du2

Unstable

Ans.

u = 90°,

d 2V = 15 7 0, du2

Stable

Ans.

Ans: u = 38.7° unstable u = 90° stable u = 141° unstable 1142

11–30. If the potential energy for a one-degree-of-freedom system is expressed by the relation V = (4x3 - x2 - 3x + 10) ft # lb, where x is given in feet, determine the equilibrium positions and investigate the stability at each position.

SOLUTION V = 4x3 - x 2 - 3x + 10 Equilibrium P osition: dV = 12x2 - 2x - 3 = 0 dx x =

2 ; 2(- 2)2 - 4(12)( -3) 24

x = 0.590 ft

and

Ans.

- 0.424 ft

Stability: d 2V = 24x - 2 dx 2 At x = 0.590 ft

d2V = 24(0.590) - 2 = 12.2 7 0 dx 2

Stable

Ans.

At x = -0.424 ft

d 2V = 24( -0.424) - 2 = -12.2 6 0 dx2

Unstable

Ans.

Ans: x = - 0.424 ft unstable x = 0.590 ft stable 1143

11–31. The uniform link ABhas a mass of 3 kg. The rod BD, having negligible weight, passes through a swivel block at C. If the spring has astiffness of k = 100 N>m and is unstretched when u = 0°, determine the angle u for equilibrium and investigate the stability at the equilibrium position. Neglect the size of the swivel block.

400 mm

D k

100 N/m

C u

SOLUTION

400 mm

s = 2(0.4)2 + (0.4)2 - 2(0.4)2 cos u

= (0.4)22(1 - cos u) V = Vg + Ve = -(0.2)(sin u)3(9.81) +

1 (100) C (0.4)2(2)(1 - cos u) D 2

dV = -(5.886) cos u + 16(sin u) = 0 du

(1) Ans.

u = 20.2° d 2V = 5.886 sin u + (16) cos u = 17.0 7 0 du2

stable

Ans.

Ans: u = 20.2° stable 1144

*11–32. The spring of the scale has an unstretched length of a. Determine the angle u for equilibrium when a weight W is supported on the platform. Neglect the weight of the members. What value W would be required to keep the scale in neutral equilibrium when u = 0°?

L k

SOLUTION Potential Function: The datum is established at point A. Since the weight W is above the datum, its potential energy is positive. From the geometry, the spring stretches x = 2L sin u and y = 2L cos u. V = Ve + Vg

L u

1 2 kx + Wy 2

1 (k)(2 L sin u)2 + W(2L cos u) 2

= 2kL2 sin 2 u + 2WL cos u Equilibrium P osition: The system is in equilibrium if

dV = 0. du

dV = 4kL2 sin u cos u - 2WL sin u = 0 du dV = 2kL2 sin 2u - 2WLsin u = 0 du Solving, u = 0°

u = cos-1 a

Stability: To have neutral equilibrium at u = 0°,

W b 2kL

Ans.

d2V 2 = 0. du2 u - 0°

d 2V = 4kL2 cos 2 u - 2WL cos u du2 d 2V 2 = 4kL2 cos 0° - 2WLcos 0° = 0 du2 u - 0° Ans.

W = 2kL

Ans: u = cos-1a W = 2kL 1145

W b 2kL

11–33. The uniform bar has a mass of 80 kg. Determine the angle u for equilibrium and investigate the stability of the bar when it is in this position. The spring has an unstretched length when u = 90°. B 4m

k 400 N/ m

SOLUTION Potential Function: The Datum is established through point A, Fig. a. Since the center of gravity of the bar is above the datum, its potential energy is positive. Here, y = 2 sin u and the spring stretches x = 4(1 - sin u) m. Thus,

V = Ve + Vg 1 2 kx + Wy 2 1 = (400) 3 4(1 - sin u) 4 2 + 80(9.81)(2 sin u) 2 =

= 3200 sin2 u - 4830.4 sin u + 3200

Equilibrium Position: The bar is in equilibrium of

dV = 0 du

dV = 6400 sin u cos u - 4830.4 cos u = 0 du cos u(6400 sin u - 4830.4) = 0 Solving, Ans.

u = 90° or u = 49.00° = 49.0° Using the trigonometry identify sin 2u = 2 sin u cos u, dV = 3200 sin 2u - 4830.4 cos u du d 2V = 6400 cos 2u + 4830.4 sin u du 2

d 2V d 2V Stability: The equilibrium configuration is stable if 7 0, unstable if 6 0 du 2 du 2 d 2V and neutral if = 0. du 2 At u = 90°, d 2V 2 = 6400 cos [2(90°)] + 4830.4 sin 90° = -1569.6 6 0 du 2 u = 90° Thus, the bar is in unstable equilibrium at u = 90° At u = 49.00°, d 2V = 6400 cos [2(49.00°)] + 4830.4 sin 49.00° = 2754.26 7 0 du2 Thus, the bar is in stable equilibrium at u = 49.0°.

Ans: Unstable at u = 90° Stable at u = 49.0° 1146

11–34. Determine the force in the spring required tokeep the 6-kg rod in equilibrium when u = 30°. The spring remains horizontal due to the roller guide.

SOLUTION x = 0.5 cos u, y =

0.5 sin u, 2

dU = 0;

0.5 m

40 N · m

dx = - 0.5 sin u du dy =

200 N/ m

0.5 cos u du 2

- Fs dx - 6(9.81)dy - 40du = 0 - Fs ( -0.5 sin u du) - 58.86a Fs (0.5 sin 30°) - 58.86 a

0.5 cos u dub - 40du = 0 2

0.5 cos 30°b - 40 = 0 2 Ans.

Fs = 211 N

Ans: Fs = 211 N 1147

11–35. The uniform beam has a weight W. If the contacting surfaces are smooth, determine the angle u for equilibrium. The spring is uncompressed when u = 90°.

SOLUTION

L 1 L V = -W a sin u b + k(L cos u)2 2 2 2 = -

WL WL 1 + sin u + kL2 cos2 u 2 2 2

dV WL = cos u - kL2 sin u cos u = 0 du 2 L cos u a

W - kL sin u b = 0 2

cos u = 0 sin u =

W 2kL

u = 90° u = sin-1 a

W b 2kL

Ans.

Ans: u = sin-1 a 1148

W b 2kL

*11–36. Each of the three links of the mechanism has a weight of 10 lb. Determine the angle u for equilibrium. The spring is unstretched when u = 0°. Investigate the stability at the equilibrium position.

4 ft

k 5 40 lb>ft

2 ft u

4 ft

2 ft

SOLUTION Potential Function: The spring stretches s = 2 sin u V = Ve + Vg =

1 (40)(2 sin u)2 + 2310(2 cos u) 4 + 10(4 cos u) 2

= 80 sin2 u + 80 cos u Equilibrium Position:

dV = 0 du

dV = 160 sin u cos u - 80 sin u = 0 du

sin u(160 cos u - 80) = 0 sin u = 0 u = 0°

Ans.

160 cos u - 80 = 0 u = 60°

Ans.

Stability: d 2V = 160 cos 2u - 80 cos u du 2 At u = 0°,

d 2V = 160 cos 0° - 80 cos 0° = 80 7 0 (stable) du 2

At u = 60°,

d 2V = 160 cos 120° - 80 cos 60° = -120 6 0 (unstable) du 2

Ans: u = 0° stable u = 60° unstable 1149

The door has a uniform weight of 50 lb. It is hinged at A and 11–37. is held open by the 30-lb weight and the pulley. Determine Theu door has a uniform weight of 50 lb. It is hinged at A and the angle for equilibrium.

is held open by the 30-lb weight and the pulley. Determine the angle u for equilibrium.

B B

6 ft 6 ft

6 ft

SOLUTION

When u = 0°,

BC = 6 22

When u = u,

BC = 62(1 -sin u)2 + 2cos2u 2

When u = 0°,

BC = 6 22

When u = u,

BC = 62(1 - sin u) + cos u

= 6 22 21 - sin u

= 6 2221 - sin u

u) sin - 30(622)(1 - 21 - sin u) u) V = 50(3 V =sin50(3 u) - 30(6 22)(1 - 21 - sin dV 1 1 dV -1/2 -cos u) u) = 150 cos u + b(1 = 150 cos254.56a u + 254.56 a b-(1sin - u) sin u)(-1/2 ( -cos du 2 2 du dV dV = 0 requires = 0du requires du -1/2

150 - 127.28(1- -1/2 150 - 127.28(1sin u) sin u)= 0 = 0 Ans.

u = 16.3°

Ans.

u = 16.3°

Ans: u = 16.3° 1150

11–38. The uniform rod OA weighs 20 lb, and when the rod is in the vertical position, the spring is unstretched. Determine the position u for equilibrium. Investigate the stability at the equilibrium position.

3 ft A

SOLUTION

1 ft

Potential Function: The spring stretches s = 12(u) in., where u is in radians. V = Ve + Vs =

1 (2)(12u)2 + 20 C 1.5(12) cos u D 2

2 lb/in.

= 144u2 + 360 cos u Equilibrium Position:

dV = 0 du dV = 288u - 360 sin u = 0 du u = 1.1311 rad = 64.8°

Ans.

u = 0°

Ans.

Stability: d2V = 288 - 360 cos u du2 At u = 64.8°,

d 2V = 288 - 360 cos 64.8° = 135 7 0 du2

At u = 0°,

d 2V = 288 - 360 cos 0° = - 72 6 0 du2

stable

Ans.

unstable

Ans.

Ans: u = 64.8°, u = 0° At u = 64.8° stable At u = 0°

1151

unstable

11–39. A spring with a torsional stiffness k is attached to the hinge at B. It is unstretched when the rod assembly is in the vertical position. Determine the weight W of the block that results in neutral equilibrium. Hint: Establish the potential energy function for a small angle u, i.e., approximate sin u L 0, and cos u L 1 - u2>2.

L 2 A L 2

SOLUTION Potential Function: With reference to the datum, Fig. a, the gravitational potential

energy of the block is positive since its center of gravity is located above the datum. Here, 3 L the rods are tilted with a small angle u. Thus, y = cos u + L cos u = L cos u. 2 2 u2 However, for a small angle u, cos u 1 - .Thus, 2 3 u2 3WL u2 Vg = Wy = Wa L b a 1 b = a1 - b 2 2 2 2

L 2 C

The elastic potential energy of the torsional spring can be computed using 1 Ve = kb 2, where b = 2u. Thus, 2 Vg =

1 k(2u)2 = 2ku2 2

The total potential energy of the system is V = Vg + Ve =

u2 3WL a 1 - b + 2ku2 2 2

Equilibrium Configuration: Taking the first derivative of V, we have dV 3WL 3WL = u + 4ku = u a + 4k b du 2 2 dV = 0. Thus, du 3WL + 4k b = 0 ua 2

Equilibrium requires

u = 0° Stability: The second derivative of V is 3WL d2V + 4k = 2 2 du

To have neutral equilibrium at u = 0°, 3WL + 4k = 0 2 8k W = 3L

d2V 2 = 0. Thus, du2 u = 0°

Ans.

Note: The equilibrium configuration of the system at u = 0° is stable if 8k d2V 8k d 2V W< > 0 and is unstable if W > < 0 . 2 3L du 3L du2 1152

Ans: W =

8k 3L

*11–40. A homogeneous cone rests on top of the cylindrical surface. Determine a relationship between the radius r of the cylinder and the height h of the cone for neutral equilibrium. Hint: Establish the potential function for a small angle u of tilt of the cone, i.e., approximate sin u L u and cos u L 1 - u2>2.

SOLUTION

a r

h V = Vs = c ar + bcos u + r u sin u dW 4 For small u, sin u L u cos u L 1 -

u2 2

V L c ar +

u2 h b a1 - b + ru2 d W 4 2

dV h = c - ar + b u + 2ru d W = 0 du 4 Ans.

u = 0° For neutral equilibrium, d 2V h = 0 = r 4 du2 r =

h 4

Ans.

Ans: u = 0° h r = 4 1153

11–41. The uniform rod has a mass of 100 kg. If the spring is unstretched when u = 60°, determine the angle u for equilibrium and investigate the stability at the equilibrium position. The spring is always in the horizontal position due to the roller guide at B.

2m k 500 N/ m B

SOLUTION Potential Function: The Datum is established through point A, Fig. a. Since the center of gravity of the bar is below the datum, its potential energy is negative. Here y = 2 cos u, and the spring stretches x = 2 sin 60° - 2 sin u = 2(sin 60° - sin u). Thus V = Ve + Vg =

1 2 kx + Wy 2

1 (500)[2(sin 60° - sin u)]2 + [ - 100(9.81)(2 cos u)] 2

= 1000 sin2 u - 100013 sin u - 1962 cos u + 750 dV Equilibrium Position: The bar is in equilibrium if = 0. du dV = 2000 sin u cos u - 100013 cos u + 1962 sin u du Using the trigonometry identity sin 2u = 2 sin u cos u, dV = 1000 sin 2u - 100013 cos u + 1962 sin u = 0 du Solved numerically, Ans.

u = 24.62° = 24.6° d 2V = 2000 cos 2u + 100013 sin u + 1962 cos u du 2

d 2V d 2V 7 0, unstable if 6 0 Stability: The equilibrium configuration is stable if 2 du du 2 d 2V and neutral if = 0. du 2 At u = 24.62°, d 2V = 2000 cos[2(24.62°)] + 100013 sin 24.62° + 1962 cos 24.62° = 3811.12 7 0 du 2 Thus, the bar is in stable equilibrium at u = 24.6°.

Ans: u = 24.6° stable 1154

11–42. Each bar has a mass per length of m0. Determine the angles u and f for equilibrium. The contact at A is smooth, and both are pin connected at B.

l u

3l 2

SOLUTION

A l 2

Require G for system to be at its lowest point. 0 x =

l l 3 a b - (0.75 l sin 26.565°)a lb 4 2 2 = -0.20937 l l 3 l + + l 2 2

lx 1 3 - a b(l) - l a b - (0.75 l cos 26.565°)a lb 2 2 2 y = = -0.66874 l l 3 l + + l 2 2 f = tan-1a

0.20937 l b = 17.38° = 17.4° 0.66874 l

Ans. Ans.

u = 26.565° - 17.38° = 9.18°

Ans: f = 17.4° u = 9.18° 1155

11–43. The truck has a mass of 20 Mg and a mass center at G. Determine the steepest grade u along which it can park without overturning and investigate the stability in this position.

G 3.5 m

SOLUTION Potential Function: The datum is established at point A. Since the center of gravity for the truck is above the datum, its potential energy is positive. Here, y = (1.5 sin u + 3.5 cos u) m.

1.5 m u

1.5 m

V = Vg = Wy = W(1.5 sin u + 3.5 cos u) Equilibrium Position: The system is in equilibrium if

dV = 0 du

dV = W(1.5 cos u - 3.5 sin u) = 0 du Since W Z 0, 1.5 cos u - 3.5 sin u = 0 Ans.

u = 23.20° = 23.2° Stability: d2V = W(-1.5 sin u - 3.5 cos u) du2 d2V = W( - 1.5 sin 23.20° - 3.5 cos 23.20°) = -3.81W 6 0 du2 u = 23.20° Thus, the truck is in unstable equilibrium at u = 23.2°

Ans.

Ans: u = 23.2° unstable 1156

*11–44. The hemisphere has a conical cavity cut into it as shown. Determine the depth d of the cavity in terms of r so that the hemisphere balances on the pivot and remains in neutral equilibrium.

r d

SOLUTION y =

3r 2 3 d 1 b pr - a b pr 2d 8 3 4 3 3r 2 - d 2 = [1] 2 3 4(2r - d) 1 pr - pr 2d 3 3

Potential Function: -

V = W[y - d]cos u = W c Equilibrium Position:

3r 2 - d 2 3r 2 + 3d 2 - 8rd - d d cos u = W c d cos u 4(2r - d) 4(2r - d)

3r 2 + 3d 2 - 8rd dV = -Wc d sin u = 0 du 4(2r - d)

sin u = 0

u = 0°

Stability: To have neutral equilibrium at u = 0°,

d 2V 3r 2 + 3d 2 - 8rd d cos u = -c 2 4(2r - d) du

d 2V ` = 0. du 2 u = 0°

d 2V 3r 2 + 3d 2 - 8rd ` = -Wc d cos 0° = 0 2 4(2r - d) du u = 0°

3d 2 - 8rd + 3r 2 = 0 Ans.

Solve for the smallest root d = 0.451r

Note: B y substituting d = 0.451r into Eq. [1], one realizes that the fulcrum must be at the center of gravity for neutral equilibrium.

Ans: d = 0.451r 1157

11–45. A 3-lb weight is attached to the end of rod ABC. If the rod is supported by a smooth slider block at C and rod BD, determine the angle u for equilibrium. Neglect the weight of the rods and the slider.

6 in. B

D 4 in.

10 in.

SOLUTION x = 2(6)2 - (4 sin u)2 = 236 - 16 sin2 u

16 6 = x (x + 4 cos u) + y x + 4 cos u + y = 2.667x y = - 4 cos u + 1.667 236 - 16 sin2 u 1 1 dy = 4 sin u du + 1.667a b (36 - 16 sin2 u)- 2 ( -32 sin u cos u)du 2

dU = 0;

W dy = 0 1

W c 4 - 0.8333(36 - 16 sin2 u)- 2 (32 cos u) d sin u du = 0 Thus, sin u = 0 Ans.

u = 0° or, 1

A 36 - 16 sin2u B 2 = 6.667 cos u Ans.

u = 33.0°

Ans: u = 0° u = 33.0° 1158

11–46. A cone is attached to the hemisphere. If both pieces have the same density, determine the height h of the cone if the configuration is to be in neutral equilibrium.

3 ft

SOLUTION Potential Function: V = Wsys + Wcyc 3 1 2 1 = rg a pr 3 bar - r cos u b + rg a pr 2hb c r + h cos u d 3 8 3 4

rgpr 2 12

(h2 cos u - 3r 2 cos u + 4rh + 8r 2)

Equilibrium Position:

rgpr 2 dV = ( - h2 + 3r 2) sin u = 0 du 12

sin u = 0

u = 0°

Stability: To have neutral equilibrium at u = 0°,

rgpr 2 d 2V = ( - h2 + 3r 2) cos u 2 12 du

d 2V ` = 0. du 2 u = 0°

rgpr 2 d 2V ` = ( -h2 + 3r 2) cos 0° = 0 12 du 2 u = 0° 3r 2 - h2 = 0

h = a 23br set r = 3 ft Ans.

= a 23b3 = 5.20 ft

Ans: h = 5.20 ft 1159

11–47. The homogeneous block has a mass of 10 kg and rests on the smooth corners of two ledges. Determine the angle u for placement that will cause the block to be stable.

200 mm 200 mm

SOLUTION Since d(cos u sin u) =

d sin 2u 2

150 mm

d Then y = cos(45° - u) - sin 2u 2 22

V = Wy =

cos(45° - u) -

Wd sin 2u 2

dV Wa = sin(45° - u) - Wd cos 2u = 0 du 22 -W c

-a + d(cos u + sin u) d (sin u - cos u) = 0 2

Ans.

Thus, u = 45°

and cos u + sin u Since

a 2d

a 200 = = 0.6667 = cos u + sin u 2d 2(150)

(No solution for 0° … u … 90°) d 2V Wa = cos(45° - u) + 2Wd sin 2u 2 du 22

For u = 45° and problem data

10(9.81)(0.2) d 2V ` = cos 0° + 2(10)(9.81)(0.15) sin 90° 2 du u = 45° 22

= 15.6 7 0

Stable at u = 45° (O.K!)

Ans: u = 45° 1160

stable

*11–48. The block weighs W and is supported by links AB and BC. Determine the necessary spring stiffness k required to hold the system in neutral equilibrium. The springs are subjected to an initial compression F0 when the links are vertical as shown.

A l B k

k l C

SOLUTION 1 1 V = Wy + c k(x + △)2 + k(x - △)2 d 2 2 y = 2L cos u

x = L sin u 1 1 V = W(2L cos u)y + c k(L sin u + △)2 + k(L sin u - △)2 d 2 2

dV = -W(2L sin u) + k(L sin u + △)(L cos u) + k(L sin u - △)(L cos u) du = -W(2L sin u) + kL2 sin 2u

For equilibrium, u = 0° For neutral equilibrium, d 2V = - W(2L cos u) + kL2(2) cos 2u = 0 du 2 u = 0°; - W(2L) + kL2(2) = 0 k =

W L

Ans.

Ans: k = 1161

W L

11–49. The The60-lb 60-lbhemisphere hemispheresupports supportsaacylinder cylinderhaving havingaaspecific specific 33 weight Determinethe the height height hh ofofthe the weightofof gg==311 311lblb ft ft. .Determine the thecylinder cylinderwhich whichwill willproduce produceneutral neutralequilibrium equilibriumininthe the position positionshown. shown.

55in. in.

SOLUTION SOLUTION 3 3 ggCC==311 311lb/ft lb/ft3==0.180 0.180lb/in lb/in3

3(5) 3r 3(5) = 1.875 in. dd== 3r == 88 88 = 1.875 in. h VV==VV 5 ++ hcos u b A 0.180hp (5)2 2B g g==(5 (5--1.875 1.875cos cosu)(60) u)(60)++ aa5 22 cos u b A 0.180hp (5) B Equilibrium: Equilibrium: dV dV = 112.5 sin u - 7.068 h2 2sin u = 0 = 112.5 sin u - 7.068 h sin u = 0 du du uu==0°0° Neutral NeutralStability: Stability: 2 dd2V V = 112.5 - 7.068 h2 2 cos u = 0 2 2 = A A 112.5 - 7.068 hB B cos u = 0 du du

At Atuu==0°0°, require , require 112.5 hh== 112.5 ==3.99 3.99in. in. AA7.068 7.068

Ans. Ans.

Ans: h = 3.99 in. 1162