Cap 04

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Chapter 4, Solution 1.

Free-Body Diagram:

(a)

ΣM B = 0:

− Ay ( 3.6 ft ) − (146 lb )(1.44 ft ) − ( 63 lb )( 3.24 ft ) − ( 90 lb )( 6.24 ft ) = 0

Ay = − 271.10 lb (b)

ΣM A = 0 :

or A y = 271 lb

or B y = 570 lb

By (3.6 ft ) − (146 lb)(5.04 ft ) − (63 lb)(6.84 ft ) − (90 lb)(9.84 ft ) = 0 By = 570.10 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 2.

Free-Body Diagram:

(a)

ΣM C = 0:

( 3.5 kips ) (1.6 + 1.3 + 19.5cos15o ) ft  − 2FB (1.6 + 1.3 + 14 ) ft  + ( 9.5 kips )(1.6 ft ) = 0 2FB = 5.4009 kips or FB = 2.70 kips

(b) ΣM B = 0:

( 3.5 kips ) (19.5cos15o − 14 ) ft  − ( 9.5 kips ) (14 + 1.3) ft  + 2 FC (14 + 1.3 + 1.6 ) ft  = 0 2FC = 7.5991 kips, or or FC = 3.80 kips

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 3.

Free-Body Diagram:

(a)

ΣM K = 0:

( 25 kN )( 5.4 m ) + ( 3 kN )( 3.4 m ) − 2FH ( 2.5 m ) + ( 50 kN )( 0.5 m ) = 0 2FH = 68.080 kN

(b)

ΣM H = 0:

or FH = 34.0 kN

( 25 kN )( 2.9 m ) + ( 3 kN )( 0.9 m ) − ( 50 kN )( 2.0 m ) + 2FK ( 2.5 m ) = 0 2FK = 9.9200 kN

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or FK = 4.96 kN

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Chapter 4, Solution 4.

Free-Body Diagram: (boom)

(a)

ΣM B = 0:

( 25 kN )( 2.6 m ) + ( 3 kN )( 0.6 m ) − ( 25 kN )( 0.4 m ) − TCD ( 0.7 m ) = 0 TCD = 81.143 kN

(b)

ΣFx = 0:

Bx = 0 so that B = By

ΣFy = 0:

( −25 − 3 − 25 − 81.143) kN + B = 0 B = 134.143 kN

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or TCD = 81.1 kN

or B = 134.1 kN

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Chapter 4, Solution 5.

Free-Body Diagram:

a1 = ( 20 in.) sin α − ( 8 in.) cos α a2 = ( 32 in.) cos α − ( 20 in.) sin α b = ( 64 in.) cos α From free-body diagram of hand truck

ΣM B = 0: P ( b ) − W ( a2 ) + W ( a1 ) = 0

(1)

ΣFy = 0: P − 2w + 2B = 0

(2)

α = 35°

For

a1 = 20sin 35° − 8cos 35° = 4.9183 in. a2 = 32 cos 35° − 20sin 35° = 14.7413 in.

b = 64cos 35° = 52.426 in. (a)

From Equation (1)

P ( 52.426 in.) − 80 lb (14.7413 in.) + 80 lb ( 4.9183 in.) = 0 ∴ P = 14.9896 lb (b)

or P = 14.99 lb

From Equation (2)

14.9896 lb − 2 ( 80 lb ) + 2 B = 0 ∴ B = 72.505 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or

B = 72.5 lb

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Chapter 4, Solution 6.

a1 = ( 20 in.) sin α − ( 8 in.) cos α

Free-Body Diagram:

a2 = ( 32 in.) cos α − ( 20 in.) sin α b = ( 64 in.) cos α From free-body diagram of hand truck

ΣM B = 0: P ( b ) − W ( a2 ) + W ( a1 ) = 0 (1) ΣFy = 0: P − 2w + 2B = 0

(2)

α = 40°

For

a1 = 20sin 40° − 8cos 40° = 6.7274 in. a2 = 32 cos 40° − 20sin 40° = 11.6577 in.

b = 64cos 40° = 49.027 in. (a)

From Equation (1)

P ( 49.027 in.) − 80 lb (11.6577 in.) + 80 lb ( 6.7274 in.) = 0 P = 8.0450 lb or P = 8.05 lb (b)

From Equation (2)

8.0450 lb − 2 (80 lb ) + 2 B = 0 B = 75.9775 lb or B = 76.0 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 7.

Free-Body Diagram:

(a) a = 2.9 m

ΣFx = 0:

ΣM B = 0:

Ax = 0

− (12 m ) Ay + (12 − 2.9 ) m  ( 3.9 kN ) + (12 − 2.9 − 2.6 ) m  ( 6.3 kN ) + ( 2.8 + 1.45 ) m  ( 7.9 kN ) + (1.45 m )( 7.3 kN ) = 0

or

ΣFy = 0: or

Ay = 10.0500 kN

or A = 10.05 kN

or B = 15.35 kN

10.0500 kN − 3.9 kN − 6.3 kN − 7.9 kN − 7.3 kN + By = 0 By = 15.3500 kN

(b) a = 8.1 m

ΣM B = 0:

− (12 m ) Ay + (12 − 8.1) m  ( 3.9 kN ) + (12 − 8.1 − 2.6 ) m  ( 6.3 kN ) + ( 2.8 + 4.05 ) m  ( 7.9 kN ) + ( 4.05 m )( 7.3 kN ) = 0

or

ΣFy = 0: or

Ay = 8.9233 kN

or A = 8.92 kN

8.9233 kN − 3.9 kN − 6.3 kN − 7.9 kN − 7.3 kN + By = 0 By = 16.4767 kN

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or B = 16.48 kN

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Chapter 4, Solution 8.

Free-Body Diagram:

(a)

ΣFx = 0:

ΣM B = 0:

Ax = 0

− (12 m ) Ay + (12 m − a )( 3.9 kN ) + (12 − 2.6 ) m − a  ( 6.3 kN ) a a  +  2.8 m +  ( 7.9 kN ) + ( 7.3 kN ) = 0 2 2 

or

(12 m ) Ay

= 128.14 kN ⋅ m − (10.2 kN ) a + (15.2 kN ) a 2

(12 m ) Ay

= 128.14 kN ⋅ m − ( 2.6 kN ) a

Thus Ay is maximum for the smallest possible value of a:

a =0 (b) The corresponding value of Ay is

( Ay )max = 10.6783 kN, and ΣFy = 0:

or A = 10.68 kN

or B = 14.72 kN

10.6783 kN − 3.9 kN − 6.3 kN − 7.9 kN − 7.3 kN + By = 0 By = 14.7217 kN

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 9.

Free-Body Diagram:

For (TC )max , TB = 0

ΣM O = 0:

(TC )max ( 4.8 in.) − (80 lb )( 2.4 in.) = 0 (TC ) = 40 lb  > [Tmax = 36 lb ] max 

(TC )max

= 36.0 lb

For (TC )min , TB = Tmax = 36 lb ΣM O = 0:

(TC )min ( 4.8 in.) + ( 36 lb )(1.6 in.) − (80 lb )( 2.4 in.) = 0 (TC )min

= 28.0 lb

Therefore:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

28.0 lb ≤ TC ≤ 36.0 lb

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Chapter 4, Solution 10.

Free-Body Diagram:

For Qmin , TD = 0 ΣM B = 0:

( 7.5 kN )( 0.5 m ) − Qmin ( 3 m ) = 0 Qmin = 1.250 kN

For Qmax , TB = 0 ΣM D = 0:

( 7.5 kN )( 2.75 m ) − Qmax ( 0.75 m ) = 0 Qmax = 27.5 kN

Therefore:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

1.250 kN ≤ Q ≤ 27.5 kN

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Chapter 4, Solution 11.

Free-Body Diagram:

ΣM D = 0:

( 7.5 kN )( 2.75 m ) − TB ( 2.25 m ) + ( 5 kN )(1.5 m ) − Q ( 0.75 m ) = 0 Q = ( 37.5 − 3TB ) kN

ΣM B = 0:

(1)

( 7.5 kN )( 0.5 m ) − ( 5 kN )( 0.75 m ) + TD ( 2.25 m ) − Q ( 3 m ) = 0 Q = ( 0.75 TD ) kN

(2)

For the loading to be safe, cables must not be slack and tension must not exceed 12 kN. Thus, making 0 ≤ TB ≤ 12 kN in. (1), we have 1.500 kN ≤ Q ≤ 37.5 kN

(3)

And making 0 ≤ TD ≤ 12 kN in. (2), we have 0 ≤ Q ≤ 9.00 kN (3) and (4) now give:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

(4) 1.500 kN ≤ Q ≤ 9.00 kN

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Chapter 4, Solution 12.

Free-Body Diagram:

For (WA )min , E = 0 ΣM F = 0:

(WA )min ( 7.5 ft ) + ( 9 lb )( 4.8 ft ) + ( 28 lb )( 3 ft ) − ( 90 lb )(1.8 ft ) = 0 (WA )min

= 4.6400 lb

For (WA ) max , F = 0 ΣM E = 0:

(WA )max (1.5 ft ) − ( 9 lb )(1.2 ft ) − ( 28 lb )( 3 ft ) − ( 90 lb )( 7.8 ft ) = 0

(WA )max

= 531.20 lb

Thus

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

4.64 lb ≤ WA ≤ 531 lb

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Chapter 4, Solution 13.

Free-Body Diagram:

ΣM D = 0:

( 750 N )( 0.1 m − a ) − ( 750 N )( a + 0.075 m − 0.1 m ) − (125 N )( 0.05 m ) + B ( 0.2 m ) = 0  87.5 N + 0.2 B  a=  1500 N  

(1)

Using the bounds on B:

B = − 250 N (i.e. 250 N downward) in (1) gives amin = 0.0250 m B = 500 N (i.e. 500 N upward) in (1) gives amax = 0.1250 m

Therefore:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

25.0 mm ≤ a ≤ 125.0 mm

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Chapter 4, Solution 14.

Free-Body Diagram:

Note that W = mg is the weight of the crate in the free-body diagram, and that

0 ≤ E y ≤ 2.5 kN ΣFx = 0: ΣM A = 0: or ΣFy = 0: or

Ax = 0 − (1.2 m )(1.2 kN ) − ( 2.0 m )(1.6 kN ) − ( 3.8 m ) E y + ( 6 m )W = 0

6W = 4.64 kN + 3.8E y

(1)

Ay − 1.2 kN − 1.6 kN − E y + W = 0 Ay = 2.8 kN + E y − W

(2)

Considering the smallest possible value of E y : For

E y = 0, W = Wmin = 0.77333 kN

From (2) the corresponding value of Ay is:

Ay = 2.02667 kN ≤ 2.5 kN, which satisfies the constraint on Ay .

For the largest allowable value of E y :

E y = 2.5 kN , W = Wmax = 2.3567 kN From (2) the corresponding value of Ay is:

Ay = 2.9433 kN ≥ 2.5 kN which violates the constraint on Ay .

Thus

( Ay )max = 2.5 kN. Solving (1) and (2) for W with ( Ay )max = 2.5 kN, W = Wmax = 1.59091 kN

Therefore:

773.33 N ≤ W ≤ 1590.91 N, or 773.33 N ≤ m(9.81 m/s 2 ) ≤ 1590.91 N, and 78.8 kg ≤ m ≤ 162.2 kg

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 15.

Free-Body Diagram:

Calculate lengths of vectors BD and CD:

BD = (11.2) 2 + (21.0) 2 ft = 23.8 ft CD = (a)

(11.2) + (8.4)2 ft = 14.0 ft

 11.2 ft   11.2 ft  (221 lb )(24 ft ) +  TCD (11.4 ft ) = 0 − (161 lb )(24 ft ) +   23.8 ft   14.0 ft 

ΣM A = 0 :

TCD = 150.000 lb

(b)

ΣFx = 0:

 11.2 ft   11.2 ft  161 lb −   ( 221 lb ) −   (150 lb ) + Ax = 0  23.8 ft   14.0 ft  Ax = 63.000 lb

ΣFy = 0:

or

A x = 63.000 lb

 21.0 ft   11.2 ft  (221 lb) −  Ay −    (150 lb) = 0  23.8 ft   14.0 ft 

Ay = 285.00 lb

A=

TCD = 150.0 lb

Ax2 + Ay2 =

or

A y = 285.00 lb

(63)2 + (285) 2 = 291.88 lb

( 63 )

θ = tan −1 285 = 77.535° Therefore

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

A = 292 lb

77.5°

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Chapter 4, Solution 16.

Free-Body Diagram:

(a)

Equilibrium for ABCD:

ΣM C = 0:

( A cos 60° )(1.6 in.) − ( 6 lb )(1.6 in.) + ( 4 lb )( 0.8 in.) = 0 A = 8.0000 lb

(b)

ΣFx = 0:

or

C x = 8.0000 lb

C y − 6 lb + ( 8 lb ) sin 60° = 0

or C y = −0.92820 lb C =

60°

Cx + 4 lb + ( 8 lb ) cos 60° = 0 or C x = − 8.0000 lb

ΣFy = 0:

A = 8.00 lb

C x2 + C y2 =

or

(8)2 + ( 0.92820 )2

C y = 0.92820 lb = 8.0537 lb

 − 0.92820   = 6.6182° −8  

θ = tan −1  Therefore:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

C = 8.05 lb

6.62°

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Chapter 4, Solution 17.

Free-Body Diagram:

Equations of equilibrium:

− ( 330 N )( 0.25 m ) + B sin α ( 0.3 m ) + B cos α ( 0.5 m ) = 0

ΣΜ Α = 0:

(1)

Ax − B sin α = 0

ΣFx = 0:

(2)

Ay − ( 330 N ) + B cos α = 0

ΣFy = 0:

(3)

(a) Substitution α = 0 into (1), (2), and (3) and solving for A and B:

B = 165.000 N, Ax = 0, Ay = 165.0 N or A = 165.0 N , B = 165.0 N

(b) Substituting α = 90° into (1), (2), and (3) and solving for A and B:

B = 275.00 N, Ax = 275.00 N, Ay = 330.00 N

A=

Ax2 + Ay2 =

θ = tan −1

Ay Ax

(275)2 + (330) 2 = 429.56 N

= tan −1

330 = 50.194° 275

∴ A = 430 N

50.2°, B = 275 N

(c) Substituting α = 30° into (1), (2), and (3) and solving for A and B:

B = 141.506 N, Ax = 70.753 N, Ay = 207.45 N, ⇒

A=

Ax2 + Ay2 =

θ = tan −1

Ay Ax

(70.753) 2 + (207.45) 2 = 219.18 N

= tan −1

207.45 = 71.168° 70.753 ∴ A = 219 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

71.2°, B = 141.5 N

60°

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Chapter 4, Solution 18.

Free-Body Diagram:

Equations of equilibrium:

ΣΜ Α = 0 :

− (82.5 N ⋅ m ) + B sin α (0.3 m ) + B cos α (0.5 m ) = 0

(1)

ΣFx = 0:

Ax − B sin α = 0

(2)

ΣFy = 0:

Ay + B cos α = 0

(3)

(a) Substituting α = 0 into (1), (2), and (3) and solving for A and B:

B = 165.000 N, Ax = 0, Ay = −165.0 N or A = 165.0 N , B = 165.0 N

∴ A = 275 N

(b) Substituting α = 90° into (1), (2), and (3) and solving for A and B:

B = 275.00 N, Ax = 275.00 N, Ay = 0 , B = 275 N

(c) Substituting α = 30° into (1), (2), and (3) and solving for A and B:

B = 141.506 N, Ax = 70.753 N, Ay = −122.548 N A = Ax2 + A y2 = (70.753) 2 + (−122.548) 2 = 141.506 N

θ = tan −1

Ay Ax

= tan −1

122.548 = 60.000° 70.753 ∴ A = 141.5 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

60.0°, B = 141.5 N

60°

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Chapter 4, Solution 19.

Free-Body Diagram: (a) From free-body diagram of lever BCD

ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 ∴ TAB = 300 (b) From free-body diagram of lever BCD

ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0 ∴ C x = −380 N

or

C x = 380 N

ΣFy = 0: C y + 0.8 ( 300 N ) = 0

∴ C y = −240 N C x2 + C y2 =

Then

C =

and

θ = tan −1 

or

C y = 240 N

( 380 )2 + ( 240 )2

= 449.44 N

 Cy  − 240   = tan −1   = 32.276°   − 380   Cx 

or C = 449 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

32.3°

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Chapter 4, Solution 20.

Free-Body Diagram: From free-body diagram of lever BCD

ΣM C = 0: TAB ( 50 mm ) − P ( 75 mm ) = 0 ∴ TAB = 1.5P

(1)

ΣFx = 0: 0.6TAB + P − C x = 0 ∴ C x = P + 0.6TAB

(2)

Cx = P + 0.6 (1.5P ) = 1.9 P

From Equation (1)

ΣFy = 0: 0.8TAB − C y = 0 ∴ C y = 0.8TAB

(3)

C y = 0.8 (1.5P ) = 1.2 P

From Equation (1) From Equations (2) and (3)

C = C x2 + C y2 =

(1.9 P )2 + (1.2 P )2

= 2.2472 P

Since Cmax = 500 N,

∴ 500 N = 2.2472Pmax or

Pmax = 222.49 lb or P = 222 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 21.

Free-Body Diagram:

(a)

ΣΜ Βx = 0 : or Fsp =

 2.4 in.  −  A − (0.9 in.)Fsp = 0  cosα 

8 lb = kx = k (1.2 in.) cos 30°

Solving for k:

k = 7.69800 lb/in.

k = 7.70 lb/in.

(b)

8 lb  =0  cos30° 

( 3 lb ) sin 30° + Bx + 

ΣFx = 0:

Bx = −10.7376 lb

or

− ( 3 lb ) cos 30° + B y = 0

ΣFy = 0:

By = 2.5981 lb

or B=

( −10.7376 )2 + ( 2.5981)2

θ = tan −1

= 11.0475 lb, and

2.5981 = 13.6020° 10.7376

Therefore:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

B = 11.05 lb

13.60°

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Chapter 4, Solution 22.

Free-Body Diagram:

(a)

ΣΜ Βx = 0: or

 2.4 in.    ( 3.6 lb ) − ( 0.9 in.)(12 lb ) = 0  cosα 

α = 36.9°

cosα = 0.80000, or α = 36.870°

(b)

ΣFx = 0:

( 3 lb ) sin 36.870° + Bx + (12 lb ) = 0

or

Bx = −14.1600 lb

ΣFy = 0:

− ( 3.6 lb ) cos 36.870° + By = 0

or

By = 2.8800 lb B=

( −14.1600 )2 + ( 2.8800 )2

θ = tan −1

= 14.4499 lb, and

2.8800 = 11.4966° 14.1600

Therefore:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

B = 14.45 lb

11.50°

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Chapter 4, Solution 23.

Free-Body Diagram:

From free-body diagram for (a):

− B ( 0.2 m ) − (100 N )( 0.3 m ) + ( 50 N )( 0.05 m ) − 10 N ⋅ m = 0

ΣΜ A = 0:

Β = −187.50 Ν

ΣFx = 0:

or B = 187.5 N

−187.5 N − 50 N + Ax = 0 Ax = 237.50 N

ΣFy = 0:

Ay − 100 N = 0 Ay = 100.000 N A=

and:

Ax2 + Ay2 =

θ = tan −1

Ay Ax

( 237.5)2 + (100 )2

= tan −1

= 257.69 N

100 = 22.834° 237.5

∴ A = 258 N

22.8°

From For (b)

ΣΜ A = 0:

− B cos 45° ( 0.2 m ) − (100 N )( 0.3 m ) + ( 50 N )( 0.05 m ) − 10 N ⋅ m = 0 Β = 265.17 Ν

or B = 265.17 N

45°

− ( 265.17 N ) cos 45° − 50 N + Ax = 0

ΣFx = 0:

Ax = 237.50 N

Ay + ( 265.17 ) sin 45° − 100 N = 0

ΣFy = 0:

Ay = −87.504 N and:

A=

Ax2 + Ay2 =

θ = tan −1

Ay Ax

(237.50)2 + (−87.504)2 = 253.11 N

= tan −1

87.504 = 20.226° 237.50

∴ A = 253 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

20.2°

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 24.

Free-Body Diagram:

From free-body diagram for (a):

− (100 N )( 0.3 m ) + A ( 0.2 m ) − ( 50 N )( 0.15 m ) − 10 N ⋅ m = 0

ΣΜ B = 0:

A = 237.50 Ν ΣFx = 0:

or A = 238 N

Bx + 237.5 N − 50 N = 0 Bx = −187.50 N

ΣFy = 0:

By − 100 N = 0 By = 100.000 N

and:

B=

Bx2 + By2 =

( −187.5)2 + (100 )2

By

100 = 28.072° 187.5

θ = tan −1

= tan −1

Bx

= 212.50 N

∴ B = 213 N

28.1°

From free-body diagram or (b):

− (100 N )( 0.3 m ) + A cos 45° ( 0.2 m ) − ( 50 N )( 0.15 m ) − 10 N ⋅ m = 0

ΣΜ B = 0:

A = 335.88 Ν ΣFx = 0:

or A = 336 N

45°

Bx + ( 335.88 N ) cos 45° − 50 N = 0 Bx = −187.503 N

ΣFy = 0:

B y + ( 335.88 N ) sin 45° − 100 N = 0

By = −137.503 N and:

B=

Bx2 + B y2 =

θ = tan −1

By Bx

(−187.503) 2 + (−137.503)2 = 232.52 N

= tan −1

137.503 = 36.254° 187.503

∴ B = 233 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

36.3°

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Chapter 4, Solution 25.

Free-Body Diagram:

Geometry:

x AC = ( 8 in.) cos 20° = 7.5175 in. y AC = ( 8 in.) sin 20° = 2.7362 in. ⇒ yDA = 9.6 in. − 2.7362 in. = 6.8638 in.

 yDA  −1  6.8638   = tan   = 42.397° x  7.5175   AC 

α = tan −1 

β = 90° − 20° − 42.397° = 27.603° Equilibrium for lever: (a)

TAD cos 27.603° ( 8 in.) − ( 60 lb ) (12 in.) cos 20° = 0

ΣM C = 0:

TAD = 95.435 lb

(b)

TAD = 95.4 lb

Cx + ( 95.435 lb ) cos 42.397° = 0

ΣFx = 0:

C x = −70.478 lb

C y − 60 lb − ( 95.435 lb ) sin 42.397° = 0

ΣFy = 0:

C y = 124.348 lb

Cx2 + C y2 =

Thus:

C =

and

θ = tan −1

Cy Cx

(−70.478) 2 + (124.348) 2 = 142.932 lb

= tan −1

124.348 = 60.456° 70.478 ∴ C = 142.9 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

60.5°

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 26.

Free-Body Diagram:

(a) a = 2 in.

ΣΜ A = 0:

( 2 in.)( 20 lb ) − (1.5 in.)(16 lb ) − 32 lb ⋅ in. − ( 2 in.) E sin 60° − ( 5.5 in.) E cos 60° = 0 E = −3.5698 lb

ΣFx = 0:

or E = 3.57 lb

60.0°

Ax − 16 lb + ( 3.5698 lb ) cos 60° = 0 Ax = 14.2151 lb

ΣFy = 0:

Ay − 20 lb − ( 3.5698 lb ) sin 60° = 0

Ay = 23.092 lb A=

(14.2151)2 + ( 23.092 )2

θ = tan −1

= 27.117 lb

23.092 = 58.384° 14.2151

A = 27.1 lb

Therefore:

58.4°

(b) a = 7.5 in.

ΣΜ A = 0:

( 7.5 in.)( 20 lb ) − (1.5 in.)(16 lb ) − 32 lb ⋅ in. − ( 2 in.) E sin 60° − ( 5.5 in.) E cos 60° = 0 E = 20.973 lb

ΣFx = 0:

or E = 21.0 lb

60.0°

A = 26.6 lb

3.97°

Ax − 16 lb − ( 20.973 lb ) cos 60° = 0 Ax = 26.487 lb

ΣFy = 0:

Ay − 20 lb + ( 20.973 lb ) sin 60° = 0

Ay = 1.83685 lb A=

( 26.487 )2 + (1.83685)2

θ = tan −1

= 26.551 lb

1.83685 = 3.9671° 26.487

Therefore:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 27.

Free-Body Diagram:

Geometry:

( 2.52 )2 + ( 0.39 )2

Distance BC =

= 2.55 m

Equilibrium for mast: (a)

 2.52    TBC  ( 0.75 m ) − (135 N )( 2.16 m ) − ( 225 N )( 0.66 m ) = 0   2.55  

ΣΜ A = 0:

TBC = 593.79 N

(b)

or TBC = 594 N

 2.52  Ax −   ( 593.79 N ) − 225 N − 135 Ν = 0  2.55 

ΣFx = 0:

Ax = 586.80 N

 0.39  Ay +   ( 593.79 N ) − 225 N − 135 Ν = 0  2.55 

ΣFy = 0:

Ay = 269.19 N Ax2 + Ay2 =

Thus:

A=

and

θ = tan −1

Ay Ax

( 586.80 )2 + ( 269.19 )2

= tan −1

= 645.60 N

269.19 = 24.643° 586.80 ∴ A = 646 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

24.6°

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Chapter 4, Solution 28.

Free-Body Diagram:

Geometry:

( 2.52 )2 + ( 0.462 )2

Distance BC =

= 2.562 m

Equilibrium for mast: (a)

 2.52    TBC  ( 0.75 m ) − ( 90 N )( 2.16 m ) − (135 N )( 0.66 m ) = 0   2.562  

ΣΜ A = 0:

TBC = 384.30 N

(b)

or

TBC = 384 N

 2.52  Ax −   ( 384.30 N ) = 0  2.562 

ΣFx = 0:

Ax = 378.00 N

ΣFy = 0:

 0.462  Ay +   ( 384.30 N ) − 135 N − 90 Ν = 0  2.562 

Ay = 155.700 N Ax2 + Ay2 =

( 378.00 )2 + (155.700 )2

Ay

155.700 = 22.387° 378.00

Thus:

A=

and

θ = tan −1

Ax

= tan −1

= 408.81 N

∴ A = 409 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

22.4°

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Chapter 4, Solution 29.

Free-Body Diagram:

Geometry: AB =

Distance

(0.3)2 + (0.125)2

= 0.325 m

Equilibrium for bracket:

0.3   0.125  T  ( 0.175 m ) −  T  ( 0.225 m ) + T ( 0.075 m ) = 0  0.325   0.325 

(150 N )( 0.225 m ) − 

ΣΜ C = 0:

T = 195.000 N

T = 195.0 N

 0.3  Cx +  T  (195 N ) = 0  0.325 

ΣFx = 0:

C x = −180.000 N

ΣFy = 0:

 0.125  C y − 150 N +  T  (195 N ) + 195 N = 0  0.325 

C y = −120.000 N C x2 + C y2 =

Thus:

C =

and

θ = tan −1

Cy Cx

( −180 )2 + ( −120 )2

= tan −1

= 216.33 N

120 = 33.690° 180 ∴ C = 216 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

33.7°

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 30.

Free-Body Diagram:

Geometry: Distance BC =

( 4 ) 2 + ( 3) 2

Distance CD =

(14 )2 + ( 3)2

= 5 in. = 14.3178 in.

Equilibrium for bracket:

ΣΜ A = 0:

4  3 3   14    T  ( 4 in.) −  T  ( 9 in.) +  T  ( 4 in.) +  T  ( 9 in.) = 0 5 5 14.3178 14.3178        

(10 lb )( 9 in.) − 

T = 32.108 lb

or T = 32.1 lb

4  14  Ax +   ( 32.108 lb ) −   ( 32.108 lb ) = 0 5  14.3178 

ΣFx = 0:

Ax = 5.7089 lb

ΣFy = 0:

Ay +

3 3  ( 32.108 lb ) +   ( 32.108 lb ) + 10 lb = 0 5  14.3178 

Ay = −35.992 lb Ax2 + Ay2 =

Thus:

A=

and

θ = tan −1

Ay Ax

( 5.7089 )2 + ( −35.992 )2

= tan −1

= 36.442 lb

35.992 = 80.987° 5.7089 ∴ A = 36.4 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

81.0°

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 31.

Free-Body Diagram:

Geometry: Distance BC = (7.2)2 + (3)2 = 7.8 in.

Distance CD = (10.8)2 + (3)2 = 11.2089 in.

Equilibrium for bracket:

7.2   3   10.8  T  ( 4 in.) −  T  ( 9 in.) +  T  ( 4 in.)  7.8   7.8   11.2089 

(10 lb )( 9 in.) − 

ΣM A = 0:

3   + T  ( 9 in.) = 0  11.2089  or T = 101.0 lb

T = 101.014 lb  7.2   10.8  Ax +   (101.014 lb ) −   (101.014 lb ) = 0 7.8    11.2089 

ΣFx = 0:

Ax = 4.0853 lb

ΣFy = 0:

3  3    Ay +   (101.014 lb ) +   (101.014 lb ) + 10 lb = 0  7.8   11.2089 

Ay = − 75.887 lb Ax2 + Ay2 =

Thus:

A=

and

θ = tan −1

Ay Ax

( 4.0853)2 + ( − 75.887 )2

= tan −1

= 75.997 lb

75.887 = 86.919° 4.0853 ∴ A = 76.0 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

86.9°

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Chapter 4, Solution 32.

Free-Body Diagram:

Geometry: Distance AD = (0.9)2 + (0.375)2 = 0.975 m Distance BD = (0.5)2 + (0.375)2 = 0.625 m Equilibrium for beam: (a)

ΣM C = 0:

0.375   0.375  T  ( 0.9 m ) −  T  ( 0.5 m ) = 0  0.975   0.625 

(135 N )( 0.7 m ) − 

T = 146.250 N (b)

ΣFx = 0:

or T = 146.3 N

 0.9   0.5  Cx +  T  (146.250 N ) +  T  (146.250 N ) = 0 0.975    0.625 

C x = − 252.00 N

ΣFy = 0:

 0.375   0.375  Cy +  T  (146.250 N ) +  T  (146.250 N ) − 135 N = 0  0.975   0.625 

C y = − 9.0000 N Thus:

C = C x2 + C y2 =

and

θ = tan −1

Cy Cx

( − 252 )2 + ( − 9 )2

= tan −1

= 252.16 N

9 = 2.0454° 252

C = 252 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

2.05°

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Chapter 4, Solution 33.

Free-Body Diagram:

For both parts (a) and (b)

ΣM D = 0:

− RP − RC x = 0 Cx = −P

ΣFx = 0:

B cosθ − P = 0 B=

ΣFy = 0:

(1)

P cosθ

(2)

 P  Cy −   sin θ + P = 0  cosθ 

C y = P ( tanθ − 1)

(a)

(3)

The magnitudes of the forces at B and C are equal:

B = C x2 + C y2 2

2 2  P    = ( − P ) +  P ( tan θ − 1)   cosθ 

or

1 = 1 + tan 2 θ − 2 tan θ + 1 cos 2 θ

(

) continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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1 , this gives cos 2 θ 1 1 =1+ − 2 tan θ , or 2 cos θ cos 2 θ 1 tan θ = , so 2

tan 2 θ + 1 =

Noting that

θ = 26.565°

θ = 26.6°

(b) Using (2)

B=

P , or 2/ 5

∴ B=

5 P 2

26.6°

∴ C=

5 P 2

26.6°

and using (1) and (3)

C x = − P, C =

Cy = −

P 2 2

P 5 P  = 2  2

( −P )2 +  −

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 34.

Free-Body Diagram:

For both parts (a) and (b)

ΣM D = 0:

− RP − RC x = 0 Cx = −P B cosθ − P = 0

ΣFx = 0:

P cosθ

B=

ΣFy = 0:

(1)

 P Cy −   cosθ

(2)

  sin θ + P = 0 

C y = P ( tanθ − 1)

(3)

(a) The magnitude of the reaction at C:

C =

C x2 + C y2

C =

( −P )2 +  P ( tan θ

− 1) 

2

which is smallest when tan θ = 1 , or

θ = 45.0° (b) Using (2)

B=

P cos 45°

or B = 2P

45.0o

and

C x = − P,

Cy = 0

or C = P

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 35.

Free-Body Diagram:

Equilibrium for bracket:

ΣM C = 0:

− T ( a ) − P ( a ) + (T sin 40° )( 2a sin 40° ) + (T cos 40° )( a + 2a cos 40° ) = 0

T = 0.56624P ΣFx = 0:

or T = 0.566P

Cx + ( 0.56624 P ) sin 40° = 0

C x = 0.36397 P

ΣFy = 0:

C y + 0.56624 P − P + ( 0.56624 P ) cos 40° = 0

Cy = 0

or C = 0.364P

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 36.

Free-Body Diagram:

(a)

ΣM C = 0:

 2a  − aTABD − aP +  + a  TABD cosθ = 0  cosθ  or with TABD = 3P/4 :

3  2 3 − a P − aP +  + 1 P cosθ = 0 4  cosθ 4 cosθ = (b)

ΣFx = 0:

Cx −

θ = 70.5°

83   P = 0 3 4 

Cx =

ΣFy = 0:

1 , and θ = 70.529° 3

2 P 2

3 1 3  P + Cy − P +  P  = 0 4 3 4 

Cy = 0 Therefore:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

C=

1 P 2

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Chapter 4, Solution 37.

Free-Body Diagram:

Equilibrium for bracket: (a)

ΣFy = 0:

− 600 N + T = 0

T = 600 N (b)

ΣM C = 0:

− ( 600 N )( 0.6 m ) + A ( 0.09 m ) = 0

A = 4000 N ΣFx = 0:

or T = 600 N

or A = 4 kN

or B = 4 kN

B − 4000 N = 0

B = 4000 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 38.

Free-Body Diagram:

( ) = − ( 52 kg ) ( 9.81 m/s ) = 510.12 N

WC = − ( 80 kg ) 9.81 m/s 2 = 784.80 N

Note that

WD

2

(a)

ΣM A = 0:

− ( 784.80 N )( 0.8 m ) cos 30° − ( 510.12 N )( 2.2 m ) cos 30° + B ( 3.5 m ) cos 5° = 0

B = 434.69 N

or B = 435 N

55°

(b)

ΣFy = 0:

A cos10° − 784.80 N − 510.12 N + (434.69 N)cos35° = 0 A = 953.33 N

ΣFx = 0:

or A = 953 N

80°

P − ( 953.33 N ) sin10° − (434.69 N)sin35° = 0 P = 414.87 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or P = 415 N

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Chapter 4, Solution 39.

Free-Body Diagram:

Equilibrium for rod: (a)

ΣM E = 0:

( 6 lb ) cos 60° ( dOE ) − (T cos 45°) ( dOE ) = 0 T = 4.2426 lb

ΣFx = 0:

(b)

T = 4.24 lb

( 4.2426 lb ) cos 45° − ( 6 lb ) cos 60° − N A sin 45° + N D cos 45° = 0 N A = ND

ΣFy = 0:

(1)

− ( 6 lb ) sin 60° − ( 4.2426 lb ) sin 45° + N A cos 45° + N D cos 45° = 0 N A + N D = 11.5911 lb

(2)

Solving (1) and (2) gives:

N A = N D = 5.7956 lb Therefore:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

N A = 5.80 lb

45°

N D = 5.80 lb

45°

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Chapter 4, Solution 40.

Free-Body Diagram:

Equilibrium for rod: (a)

ΣM E = 0:

( 6 lb ) cos 60° ( dOE ) − (T cosθ ) ( dOE ) = 0 T =

3 lb cosθ

(1)

Thus T is minimum when cos θ is maximum: (b) (c)

With θ = 0°, (1) gives: T = 3 lb

ΣFx = 0:

θ = 0° T = 3.00 lb

3 lb − ( 6 lb ) cos 60° − N A sin 45° − N D sin 45° = 0 N A = ND

ΣFy = 0:

(2)

− ( 6 lb ) sin 60° + N A cos 45° + N D cos 45° = 0 N A + N D = 7.3485 lb

(3)

Solving (2) and (3) gives: N A = N D = 3.6742 lb

Therefore:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

N A = 3.67 lb

45°

N D = 3.67 lb

45°

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Chapter 4, Solution 41.

Free-Body Diagram:

Equilibrium for bracket:

ΣFy = 0:

T sin 20° − 270 N = 0 T = 789.43 N Tx = ( 789.43 N ) cos 20° = 741.82 N, and

Note that:

Ty = ( 789.43 N ) sin 20° = 270 N Thus Ty and the 270-N force form a couple:

270 N ( 0.25 m ) = 67.5 N ⋅ m clockwise ΣM B = 0:

( 741.82 N )( 0.125 m ) − 67.5 N ⋅ m + FCD ( 0.2 m ) = 0 FCD = −126.138 N

ΣFy = 0:

or

FCD = 126.138 N

FAB − 126.138 N − 741.82 N = 0 FAB = 867.96 N

or

FAB = 867.96 N

Thus, FCD acts to the left, while FAB acts to the right, i.e. these forces are exerted by rollers B and C, respectively. Rollers A and B exert no force. The forces exerted on the post are the opposites of the forces exerted by the rollers:

A = D=0

B = 868 N C = 126.1 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 42.

Free-Body Diagram:

Equilibrium for bracket:

ΣFy = 0:

T sin 30° − 270 N = 0 T = 540 N Tx = ( 540 N ) cos 30° = 467.65 N, and

Note that:

Ty = ( 540 N ) sin 30° = 270 N Thus Ty and the 270-N force form a couple:

270 N ( 0.25 m ) = 67.5 N ⋅ m clockwise ΣM B = 0:

( 467.65 N )( 0.125 m ) − 67.5 N ⋅ m + FCD ( 0.2 m ) = 0 FCD = 45.219 N or FCD = 45.219 N

ΣFy = 0:

FAB + 45.219 N − 467.65 N = 0

FAB = 422.43 N or FAB = 422.43 N Thus, both FCD and FAB act to the right, i.e. these forces are exerted on the bracket by rollers D and B, respectively. Rollers A and C exert no force. The forces exerted on the post are the opposites of the forces exerted on the bracket: A =C=0

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

B = 422 N

D = 45.2 N

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Chapter 4, Solution 43.

Free-Body Diagram:

Geometry: Equation of the slot: y =

x2 4

 2x   dy  = 1.20000   = slope of slot at C =  4  dx   ( x = 2.4 in.)  C It follows for the angles that:

α = tan −1 (1.2 ) = 50.194°

θ = 90° − α = 90° − 50.194° = 39.806°  4.8 − 2.64   = 12.6804° 9.6  

β = tan −1  Coordinates for C, D, and E:

( 2.4 )2

xC = 2.4 in.,

yC =

xD = 2.4 in.,

yD = 1.84 in. + (1.6 in.) tan β

4

= 2.44 in.

= 1.84 in. + (1.6 in.) tan12.6804° = 2.20000 in. xE = 0,

yE = yC + ( 2.4 in.) tan θ = 1.44 in. + ( 2.4 in.) tan 39.806° = 3.4400 in. continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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With P = 1 lb:

ΣM E = 0:

P ( yE ) − ( Q sin β )( yE − yD ) − ( Q cos β )( 2.4 in.) = 0

(1 lb )( 3.44 in.) − ( Q sin12.6804° )(1.24 in.) − ( Q cos12.6804° )( 2.4 in.) = 0 Q = 1.31616 lb ΣFx = 0:

P − N C cosθ − Q sin β = 0

1 lb − NC cos 39.806° − (1.31616 lb ) sin12.6804° = 0 NC = 0.92563 lb

ΣFy = 0:

N B + NC sin θ − Q cos β = 0 N B + ( 0.92563 lb ) sin 39.806° − (1.31616 lb ) cos12.6804° = 0 N B = 0.69148 lb (a) N = 0.691 lb , N = 0.926 lb B C

39.8°

Q = 1.316 lb

77.3°

(b)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 44.

Free-Body Diagram:

Geometry: Equation of the slot: y =

x2 4

 2x   dy  = 1.20000   = slope of slot at C =  4    ( x = 2.4 in.)  dx C It follows for the angles that:

α = tan −1 (1.2 ) = 50.194° θ = 90° − α = 90° − 50.194° = 39.806°  4.8 − 2.64   = 12.6804° 9.6  

β = tan −1  Coordinates for C, D, and E:

( 2.4 )2

xC = 2.4 in.,

yC =

xD = 2.4 in.,

yD = 1.84 in. + (1.6 in.) tan β

4

= 2.44 in.

= 1.84 in. + (1.6 in.) tan12.6804° = 2.20000 in. xE = 0,

yE = yC + ( 2.4 in.) tan θ = 1.44 in. + ( 2.4 in.) tan 39.806° = 3.4400 in. continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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With Q = 2 lb: (a)

ΣM E = 0:

P ( yE ) − ( Q sin β )( yE − yD ) − ( Q cos β )( 2.4 in.) = 0 P ( 3.44 in.) − ( 2 lb ) sin12.6804° (1.24 in.) − ( 2 lb ) cos12.6804° ( 2.4 in.) = 0

P = 1.51957 lb (b)

ΣFx = 0:

or P = 1.520 lb

P − NC cosθ − Q sin β = 0

1.51957 lb − NC cos 39.806° − ( 2 lb ) sin12.6804° = 0 NC = 1.40656 lb

ΣFy = 0:

or NC = 1.407 lb

39.8°

N B + NC sin θ − Q cos β = 0 N B + (1.40656 lb ) sin 39.806° − ( 2 lb ) cos12.6804° = 0 N B = 1.05075 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or N B = 1.051 lb

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Chapter 4, Solution 45.

Note: Weight of block is W = (10 kg)(9.81 m/s2) = 98.1 N (a) Free-Body Diagram:

ΣFx = 0:

Ax = 0

ΣFy = 0:

Ay − 98.1 N = 0 Ay = 98.1 N or A = 98.1 N

Therefore:

ΣM A = 0:

M A − ( 98.1 N )( 0.45 m ) = 0 M A = 44.145 N ⋅ m

or M A = 44.1 N ⋅ m

(b) Free-Body Diagram:

ΣFx = 0:

Ax − 98.1 N = 0

Ax = 98.1 N

ΣFy = 0:

Ay − 98.1 N = 0

Ay = 98.1 N continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Thus:

A=

Ax2 + Ay2 =

( 98.1)2 + ( 98.1)2

= 138.734 N

or A = 138.7 N

ΣM A = 0:

45°

M A + ( 98.1 N )( 0.45 m + 0.1 m ) = 0 M A = 44.145 N ⋅ m

or M A = 44.1 N ⋅ m

(c) Free-Body Diagram:

ΣFx = 0:

Ax = 0

ΣFy = 0:

Ay − 98.1 N − 98.1 N = 0 Ay = 196.2 N

ΣM A = 0:

or A = 196.2 N

M A − ( 98.1 N )( 0.45 m − 0.1 m ) − ( 98.1 N )( 0.45 m + 0.1 m ) = 0 M A = 88.290 N ⋅ m

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or M A = 88.3 N ⋅ m

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Chapter 4, Solution 46.

Free-Body Diagram:

With M = 0 and Ti = T0 = 12 lb ΣFx = 0:

C x − 12 lb = 0

Cx = 12 lb

ΣFy = 0:

C y − 12 lb = 0 CY = 12 lb

Thus: C =

C x2 + C y2 =

ΣM C = 0:

(12) 2 + (12)2 = 16.9706 lb or

C = 16.97 lb

or

M C = 2.40 lb ⋅ in.

45°

MC – (12 lb)[(1.8 – 1) in.] + (12 lb)[(2 + 1 − 2.4) in.] = 0 M C = 2.40 lb ⋅ in.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 47.

Free-Body Diagram:

With M = 8 lb.in. and Ti = 16 lb, To = 8 lb

ΣFx = 0:

Cx − 16 lb = 0 Cx = 16 lb

ΣFy = 0:

Cy – 8 lb = 0 Cy = 8 lb

Thus: C =

C x2 + C y2 =

and θ = tan −1

Cy Cx

(16)2 + (8)2 = 17.8885 lb

 8 = tan −1   = 26.565°  16  ∴

ΣM C = 0:

C = 17.89

26.6°

MC – (16 lb)[(1.8 – 1) in.] + (8 lb)[(2 + 1 –2.4) in.] – 8 lb ⋅ in. = 0 MC = 16.00 lb ⋅ in.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or

MC = 16.00 lb ⋅ in.

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Chapter 4, Solution 48. (a)

Free-Body Diagram:

ΣFx = 0:

Ax = 0

ΣFy = 0:

Ay − 2 lb − 1lb = 0 Ay = 3 lb or A = 3 lb

ΣM A = 0:

MA – (2 lb)(8 in.) – (1 lb)(12 in.) = 0

M A = 28 lb ⋅ in. (b)

or M A = 28 lb ⋅ in.

Free Body Diagram:

ΣFx = 0: ΣFy = 0:

Ax = 0

Ay − 2 lb − 1 lb + 1.2 lb = 0 Ay = 1.8 lb or A = 1.8 lb

ΣM A = 0:

MA – (2 lb)(8 in.) – (1 lb)(12 in.) + (1.2 lb)(16 in.) = 0

M A = 8.8 lb ⋅ in.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or M A = 8.8 lb ⋅ in.

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Chapter 4, Solution 49.

Free-Body Diagram:

Set M A = 20 lb ⋅ in. counter-clockwise to find Fmin:

ΣM A = 0:

20 lb ⋅ in. − (2 lb)(8 in.) – (1 lb)(12 in.) + Fmin (16 in.) = 0 Fmin = 0.5 lb

Set M A = 20 lb ⋅ in. clockwise to find Fmax :

ΣM A = 0:

− 20 lb ⋅ in. − (2 lb)(8 in.) – (1 lb)(12 in.) + Fmin (16 in.) = 0 Fmax = 3 lb

Therefore:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

0.5 lb ≤ FE ≤ 3 lb

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Chapter 4, Solution 50. (a)

Free-Body Diagram:

ΣFx = 0:

Ex = 0

ΣFy = 0:

E y − 16.2 kN − 5.4 kN − 18 kN = 0 Ey = 39.6 kN

ΣM E = 0:

M E + (16.2 kN)(4.8 m) + (5.4 kN)(2.6 m) − (18 kN)(1.5 m) = 0 M E = − 64.8 kN ⋅ m

(b)

or E = 39.6 kN

or ME = 64.8 kN. m

Free-Body Diagram:

ΣFx = 0:

Ex = 0

ΣFy = 0:

E y − 16.2 kN − 5.4 kN = 0 Ey = 21.6 kN

ΣM E = 0:

or E = 21.6 kN

ME + (16.2 kN)(4.8 m) + (5.4 kN)(2.6 m) = 0

M E = − 91.8 kN ⋅ m Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or ME = 91.8 kN ⋅ m

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Chapter 4, Solution 51.

Free-Body Diagram:

ΣM E = 0:

ME + (16.2 kN)x + (5.4 kN)(2.6 m) – T(1.5 m) = 0 ME = (1.5 T − 16.2 x – 14.04) kN ⋅ m

(1)

For x = 0.6 m, (1) gives: (ME )1 = (1.5 T − 23.76) kN ⋅ m For x = 7 m, (1) gives: (ME )2 = (1.5 T − 127.44) kN ⋅ m (a)

The maximum absolute value of ME is obtained when (ME )1 = − (ME ) and 1.5 T − 23.76 kN = − (1.5 T – 127.44 kN) T = 50.400 kN

(b)

or T = 50.4 kN

For this value of T: ME = 1.5(50.400) kN ⋅ m − 23.76 kN ⋅ m

= 51.84 kN ⋅ m

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or ME = 51.8 kN ⋅ m

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Chapter 4, Solution 52.

Free-Body Diagram:

Geometry: Distance BD =

(1.8)2 + (4)2 = 4.3863 m

Note also that: W = mg = (160 kg)(9.81 m/s2) = 1569.60 N With MA = 360 N ⋅ m clockwise: (i.e. corresponding to Tmax )

ΣM A = 0:

 1.8   − 360 N ⋅ m – [(540 N) cos 15o](5.6 m) +   Tmax  (4 m) = 0  4.3863   Tmax = 1998.79 N

or Tmax = 1.999 kN

With MA = 360 N ⋅ m counter-clockwise: (i.e. corresponding to Tmin ) ΣM A = 0:

 1.8   360 N ⋅ m – [(540 N) cos 15o](5.6 m) +   Tmin  (4 m) = 0  4.3863   Tmin = 1560.16 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or Tmin = 1.560 kN

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Chapter 4, Solution 53.

Free-Body Diagram:

(a) Using W = mg:

ΣFx = 0:

− A cos 45° + B sin 45° = 0

B= A ΣFy = 0:

(1)

A sin 45° + B sin 45° − mg = 0 A+ B =

2mg

(2)

From (1) and (2) it follows that

2A =

ΣM B = 0:

2 mg

and

A=

1 mg 2

 l    1  mg   cosθ  + M −  mg  [l cos(45° − θ )] = 0 2 2     

(3)

Using that cos(α − β ) = cos α cos β + sin α sin β , (3) gives

 mgl   mgl    cosθ + M −   ( cosθ + sin θ ) = 0  2   2   mgl  M −  sin θ = 0, and  2  sin θ =

(b)

2M mgl



 2(2.7 N ⋅ m)  = 20.122° 2  (2kg)(9.81 m/s )(0.8 m) 

θ = sin −1 

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

 2M  or θ = sin −1    mgl  or θ = 20.1°

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Chapter 4, Solution 54.

Free-Body Diagram:

For both parts (a) and (b)

(

)

o   l cos (θ + 30° ) cos 30° W − l cos θ + 30  T

ΣM D = 0:

(

)

(

)

+ l sin θ + 30o cos 60o  W + l sin θ + 30o  T = 0     or W cos(θ + 60°) + T sin (θ + 30° ) − cos (θ + 30° )  = 0 (a)

For T = 0, (1) gives

cos (θ + 60° ) = 0 (b)

(1)

or θ = 30.0°

For T = W , (1) gives: cosθ cos 60° − sin θ sin 60° + sin θ cos 30° + cosθ sin 30° − cosθ cos 30° + sin θ sin 30° = 0

or tan θ sin 30° + ( cos 60° + sin 30° − cos 30° ) = 0 Solving for θ :

tan θ = 2 ( cos 30° − 1) or θ = −150000°

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or θ = −15.00°

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Chapter 4, Solution 55.

Free-Body Diagram:

Using W = mg : (a)

P( R cosθ + R cosθ ) − mg ( R sin θ ) = 0

ΣM C = 0:

2P = mg tan θ tan θ = (b)

2P mg

or

 2P    mg 

θ = tan −1 

With m = 0.7 kg and P = 3 N:



 2(3 N) 2   (0.7 kg)(9.81 m/s ) 

θ = tan −1 

= 41.145°

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or

θ = 41.1°

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Chapter 4, Solution 56.

Free-Body Diagram:

Using W = mg, and h =

l tan α 2 M − (mg )(h sin θ ) = 0

ΣM C = 0:

or sin θ =

=

M mgh

M 2   cot α  mg  l 

 cot α  or θ = sin −1  2M  mgl   Note: θ ≤ 90° − α for cord BC to remain taut, (i.e. for TBC > 0).

With l = 1 m, m = 2 kg, and M = 3 N ⋅ m:



2(3 N ⋅ m) cot 30°   2  (2 kg)(9.81 m/s )(1 m) 

θ = sin −1 

= 31.984°

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or θ = 32.0°

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Chapter 4, Solution 57.

Free-Body Diagram: First note

T = tension in spring = ks s = elongation of spring

where

( )θ − ( AB )θ

= AB

= 90°

θ   90°  = 2l sin   − 2l sin   2  2 

  θ   1  = 2l sin   −     2   2    θ   1  ∴ T = 2kl sin   −     2   2 

(1)

(a) From free-body diagram of rod BC

  θ  ΣM C = 0: T l cos    − P ( l sin θ ) = 0  2   Substituting T From Equation (1)

  θ   1    θ  2kl sin   −   l cos    − P ( l sin θ ) = 0   2    2   2     θ   1   θ  θ   θ  2kl 2 sin   −   cos   − Pl  2sin   cos    = 0  2 2  2     2   2  Factoring out

θ  2l cos   , leaves 2 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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  θ   1  θ  kl sin   −   − P sin   = 0  2   2   2  or

1  kl  θ  sin   =   2  kl − P  2

  kl ∴ θ = 2sin −1    2 ( kl − P )  (b) P =

kl 4 

kl  2 kl −

θ = 2sin −1 

(

kl 4

)

  kl  4   −1  4   = 2sin −1     = 2sin    3 2   2  3 kl  

= 2sin −1 ( 0.94281) = 141.058°

or θ = 141.1°

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 58.

Free-Body Diagrams:

(

)

Note that WE = mg = (10 kg ) 9.81 m/s 2 = 98.1 N

ΣM A = 0:

M − rAT = 0 T =

( 58 N ⋅ m )φ M = rA 0.035 m

Since the torsion spring is unstretched when θ = 0:

( 70 mm )φ = ( 35 mm )θ φ =

1 θ 2

Therefore:

T =

( 58 N ⋅ m )θ

2(0.035 m)

ΣM B = 0:

rBT − lWE cosθ = 0

( 0.070 m )

( 58 N⋅ m )θ

2(0.035 m)

− ( 0.090 m )( 98.1N ) cosθ = 0

θ = 0.60890 cosθ Solving for θ numerically:

θ = 0.52645 rad

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or θ = 30.2°

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Chapter 4, Solution 59.

Free-Body Diagram:

Geometry: Triangle ABC is isosceles. Thus distance CD = l cos

θ 2

,

Elongation of spring is equal to distance AB:

x = 2l sin

θ 2

,

and T = kx = 2kl sin (a)

θ 2

.

Equilibrium for rod:

ΣM C = 0:

θ  P ( l cosθ ) − T  l cos  = 0 2  Pl cosθ − kl 2 (2sin

θ 2

θ

cos ) = 0 2

P cosθ − kl sin θ = 0

tan θ = (b)

P kl

P or θ = tan −1    kl 

For p = 2kl :

 2kl  −1  = tan (2) = 63.435°  kl 

θ = tan −1 

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or θ = 63.4°

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Chapter 4, Solution 60.

Free-Body Diagram:

Spring force: Fs = ks = k ( l − l cosθ ) = kl (1 − cosθ ) (a)

l  Fs ( l sin θ ) − W  cosθ  = 0 2 

ΣM D = 0:

kl (sin θ − cosθ sin θ ) −

kl (tan θ − sin θ ) − (b)

W cosθ = 0 2

W =0 2

or tan θ − sin θ =

W

2kl

For given values of W = 4 lb, l = 30 in., k = 1.8 lb/ft = 0.15 lb/in.

tan θ − sin θ = Solving numerically:

4 lb = 0.44444 2(0.15 lb/in.)(30 in.)

θ = 50.584°

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or θ = 50.6°

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Chapter 4, Solution 61.

Free-Body Diagram: 1. Three non-concurrent, non-parallel reactions (a)

Completely constrained

(b)

Determinate

(c)

Equilibrium

From free-body diagram of bracket:

ΣM A = 0: B (1 m ) − (100 N )( 0.6 m ) = 0 ∴ B = 60.0 N

ΣFx = 0: Ax − 60 N = 0 ∴ A x = 60.0 N

ΣFy = 0: Ay − 100 N = 0 ∴ A y = 100 N

( 60.0 )2 + (100 )2

Then

A=

= 116.619 N

and

θ = tan −1   = 59.036°  60.0 

 100 

∴ A = 116.6 N

59.0°

2. Four concurrent reactions through A (a)

Improperly constrained

(b)

Indeterminate

(c)

No equilibrium

3. Two reactions (a)

Partially constrained

(b)

Determinate

(c)

Equilibrium

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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From free-body diagram of bracket

ΣM A = 0: C (1.2 m ) − (100 N )( 0.6 m ) = 0 ∴ C = 50.0 N

ΣFy = 0: A − 100 N + 50 N = 0 ∴ A = 50.0 N

4. Three non-concurrent, non-parallel reactions (a)

Completely constrained

(b)

Determinate

(c)

Equilibrium

From free-body diagram of bracket

 1.0   = 39.8°  1.2 

θ = tan −1  BC =

(1.2 )2 + (1.0 )2

= 1.56205 m

 1.2   ΣM A = 0:   B  (1 m ) − (100 N )( 0.6 m ) = 0  1.56205   39.8°

∴ B = 78.1 N

ΣFx = 0: C − ( 78.102 N ) cos 39.806° = 0 ∴ C = 60.0 N

ΣFy = 0: A + ( 78.102 N ) sin 39.806° − 100 N = 0 ∴ A = 50.0 N

5. Four non-concurrent, non-parallel reactions (a)

Completely constrained

(b)

Indeterminate

(c)

Equilibrium

From free-body diagram of bracket

ΣM C = 0:

(100 N )( 0.6 m ) − Ay (1.2 m ) = 0 ∴ Ay = 50 N

or A y = 50.0 N

6. Four non-concurrent non-parallel reactions (a)

Completely constrained

(b)

Indeterminate

(c)

Equilibrium continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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From free-body diagram of bracket

ΣM A = 0: − Bx (1 m ) − (100 N )( 0.6 m ) = 0 ∴ Bx = −60.0 N or B x = 60.0 N

ΣFx = 0: − 60 + Ax = 0 ∴ Ax = 60.0 N or A x = 60.0 N

7. Three non-concurrent, non-parallel reactions (a)

Completely constrained

(b)

Determinate

(c)

Equilibrium

From free-body diagram of bracket

ΣFx = 0: Ax = 0 ΣM A = 0: C (1.2 m ) − (100 N )( 0.6 m ) = 0 ∴ C = 50.0 N

or C = 50.0 N

ΣFy = 0: Ay − 100 N + 50.0 N = 0 ∴ Ay = 50.0 N ∴ A = 50.0 N

8. Three concurrent, non-parallel reactions (a)

Improperly constrained

(b)

Indeterminate

(c)

No equilibrium

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 62.

Free-Body Diagram: 1. Three non-concurrent, non-parallel reactions Completely constrained

(a) (b)

Determinate

(c)

Equilibrium

From free-body diagram of plate

ΣM A = 0: C ( 30 in.) − 50 lb (15 in.) = 0 C = 25.0 lb

ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 50 lb + 25 lb = 0

Ay = 25 lb

A = 25.0 lb

2. Three non-current, non-parallel reactions Completely constrained

(a) (b)

Determinate

(c)

Equilibrium

From free-body diagram of plate B = 0

ΣFx = 0:

ΣM B = 0:

( 50 lb )(15 in.) − D ( 30 in.) = 0 D = 25.0 lb

ΣFy = 0: 25.0 lb − 50 lb + C = 0 C = 25.0 lb continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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3. Four non-concurrent, non-parallel reactions Completely constrained

(a) (b)

Indeterminate

(c)

Equilibrium

From free-body diagram of plate

ΣM D = 0: Ax ( 20 in.) − ( 50 lb )(15 in.) ∴ A x = 37.5 lb

∴ D x = 37.5 lb

ΣFx = 0: Dx + 37.5 lb = 0 4. Three concurrent reactions Improperly constrained

(a) (b)

Indeterminate

(c)

No equilibrium

5. Two parallel reactions (a)

Partial constraint

(b)

Determinate

(c)

Equilibrium

From free-body diagram of plate

ΣM D = 0: C ( 30 in.) − ( 50 lb )(15 in.) = 0 C = 25.0 lb

ΣFy = 0: D − 50 lb + 25 lb = 0 D = 25.0 lb

6. Three non-concurrent, non-parallel reactions (a)

Completely constrained

(b)

Determinate

(c)

Equilibrium

From free-body diagram of plate

ΣM D = 0: B ( 20 in.) − ( 50 lb )(15 in.) = 0 B = 37.5 lb ΣFx = 0: Dx + 37.5 lb = 0

ΣFy = 0: Dy − 50 lb = 0

D x = 37.5 lb

D y = 50.0 lb or D = 62.5 lb

53.1° continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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7. Two parallel reactions (a)

Improperly constrained

(b)

Reactions determined by dynamics No equilibrium

(c) 8. Four non-concurrent, non-parallel reactions

Completely constrained

(a) (b)

Indeterminate

(c)

Equilibrium

From free-body diagram of plate

ΣM D = 0: B ( 30 in.) − ( 50 lb )(15 in.) = 0 B = 25.0 lb

ΣFy = 0: Dy − 50 lb + 25.0 lb = 0

D y = 25.0 lb ΣFx = 0: Dx + C = 0

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 63.

Free-Body Diagram:

Note that the wheel is a three-force body, and let point D be the intersection of the three forces.

With a = 75 mm, it follows from the force triangle that

A P 90 N = = 125 100 75 Then:

P = 100 ( 90 N ) = 120 N 75

or P = 120.0 N

A = 125 ( 90 N ) = 150 N, and 75

(100 )

θ = tan −1 75 = 36.870°

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or A = 150.0 N

36.9°

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Chapter 4, Solution 64.

Free-Body Diagram:

Note that the wheel is a three-force body, and let point D be the intersection of the three forces.

From the force triangle it follows that

90 N = a

A = 90

A (100) + ( a ) 2

(100)

2

(a ) 2

2

+1

(1)

Setting A = 180 N and solving for a: 90 N = a

180 N

( )

2

2

a =

a=

( )

(100)2 + a 2

(100) + a

2

4 (100) 3

2

= 57.735 mm

From (1) it follows that A will decrease as a increases. Therefore the value of a calculated is a lower limit:

a ≼ 57.7 mm

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell Š 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 65.

Free-Body Diagram:

Geometry:

EF = ( 2.4 tan 30° + 0.9 ) in.

AF = EF tan 30°

tan φ =

0.9 ( 2.4 tan30° + 0.9 ) tan 30° + 2.4

φ = 13.6019° Equilibrium: force triangle

Using the law of sines on the force triangle:

Fsp 3 lb B = = sin120° sin ( 60° − φ ) sinφ

B = 11.05 lb Fsp = 9.2376 lb (a) Fsp = kx

9.2376 lb = k (1.2 in.)

Solving for k:

k = 7.698 lb/in.

(b)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or k = 7.70 lb/in.

B = 11.05 lb

13.60°

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Chapter 4, Solution 66.

Free-Body Diagram:

Note that the bent rod is a three-force body. D is the point where the lines of action of the three forces intersect. (a) The requirement B = C means that the force triangle must be isosceles. Therefore θ = φ .

Which leads to the force triangle shown.

From the geometry it follows that tan θ = 1 2

or θ = 26.6°

θ = 26.565° (b) From the force triangle:

2B sin θ = P, or with sin θ = B=C =

P =  1  2   5

1 5

5P 2

Therefore:

B= C=

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

5P 2 5P 2

26.6° 26.6°

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Chapter 4, Solution 67. (a) Free-Body Diagram: (α = 90° )

The bracket is a three-force body and A is the intersection of the lines of action of the three forces.

 6

θ = tan −1   = 26.565°  12  From the force triangle:

A = (75 lb) cot θ = (75 lb) cot 26.565° = 150.000 lb

C =

75 lb 75 lb = = 167.705 lb sin θ sin 26.565° or A = 150.0 lb or C = 167.7 lb

63.4° continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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(b) Free-Body Diagram: (α = 45° ) Let E be the intersection of the lines of action of the three forces acting on the bracket.

Triangle ABE is isosceles and therefore

AE = AB = 16 in.

From triangle CEF

 CF  −1  6   = tan   = 12.0948°  EF   28 

θ = tan −1 

From force triangle:

β = 180° − 135° − θ = 180° − 135° − 12.0948° = 32.905° Using the law of sines:

A C 75 lb = = sin 32.905° sin135° sin12.0948°

Solving for A and C:

A = 194.452 lb C = 253.10 lb or A = 194.5 lb or C = 253 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

77.9°

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Chapter 4, Solution 68.

Free-Body Diagram: Let C be the intersection of the lines of action of the three forces acting on the girder From triangle BCD:

h = ( 40 ft ) cot 30° = 69.282 ft From triangle ACD:

 69.282 ft   = 73.8979°  20 ft 

α = tan −1 

or α = 73.9° From force triangle:

β = 90° − α = 90° − 73.8979° = 16.1021° γ = 180° − 30° − β = 180° − 30° − 16.1021° = 133.898° Using the law of sines:

TA TB 6000 lb = = sin 30° sin16.1021° sin133.898° Solving for TA and TB :

TA = 4163.3 lb,

TB = 2309.4 lb or TA = 4160 lb and TB = 2310 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 69.

Free-Body Diagram:

From the free-body diagram:

 (900 mm)sin 50°   = 82.726° 88 mm  

θ = tan −1 

From the force triangle:

FN = (130 N ) tan θ = (130 N ) tan 82.726° = 1018.48 N Force on nail is therefore

RB =

or FN = 1.018 kN

130 N 130 N = = 1026.74 N cosθ cos82.726° or R B = 1.027 kN

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

82.7°

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Chapter 4, Solution 70.

Free-Body Diagram:

From force triangle:  2600 N   = 83.636°  290 N 

θ = tan −1 

From free-body diagram:

tan θ = l =

(900 mm)sin50° l

( 900 mm ) sin 50° tan 83.636°

= 76.894 mm or l = 76.9 mm

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 71.

Free-Body Diagram:

We note from the free-body diagram that the ladder is a three-force body. Point C in the free-body diagram is the intersection between the lines of action of the three forces.

It then follows that:

sin θ =

1.75 m ( 9.2 − 1.8) m

θ = 13.6793° Also:

 9.2  − 1.8  m = 2.8 m AG =   2   9.2  BD =  m  cosθ = ( 4.6 m ) cosθ  2  CD = CG + GD

= =

AG sin θ 2.8

sin θ

+ BG sin θ +

9.2 2

sin θ continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Then

2.8 + 4.6sin13.6793° CD tan φ = = sin13.6793° BD 4.6cos13.6793°

φ = 70.928° Now using the law of sines on the force triangle:

FW B W = = sin(90° − φ ) sin θ sin [φ + (90° − θ )]

FW B W = = cos φ sin θ sin(φ − θ ) (a) From the law of sines and noting that W = ( 53 kg ) (9.81 m/s 2 ) = 519.93 N

FW 519.93 N = cos 70.928° cos(70.928° − 13.6793°) FW = 314.03 N

or FW = 314 N

76.3°

or B = 227 N

70.9°

(b) In the same way

B 519.93 N = sin13.6793° cos(70.928° − 13.6793°)

B = 227.28 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 72.

Free-Body Diagram:

We note from the free-body diagram that the ladder is a three-force body. Point D in the free-body diagram is the intersection between the lines of action of the three forces.

It then follows that:

BD =

9.2 m cosθ = ( 4.6 m ) cosθ =, but also that 2

1.75  BD = 1.75 tan θ + tan θ 

 m 

Therefore

BD =

9.2 m cosθ = ( 4.6 m ) cosθ 2

This implies:

4.6 cosθ = 1.75 tan θ +

1.75 tan θ

92sin θ = 35(1 + tan 2 θ ) = 35 sec2 θ 92sin θ cos 2 θ = 35 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Solving numerically for the smallest possible root:

θ = 31.722° Then

sin 31.722° =

1.75 9.2 − a

a = 5.8717 m

or a = 5.87 m

(b) From the force triangle

FW =

W cos 31.722°

FW = 611.24 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or FW = 611 N

58.3°

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Chapter 4, Solution 73.

Free-Body Diagram: Let D be the intersection of the lines of action of the three forces acting on the tool.

From the free-body diagram:

yDC =

(14.4 in.) cos 35° = 32.409 in. xBC = tan 20° tan 20°

yBC = (14.4 in.) sin 35° = 8.2595 in. 

α = tan −1 

 yDC

 3.6 in.  − yBC − 1.8 in. 

  3.6 in. = tan −1    ( 32.409 − 8.2595 − 1.8 ) in.  = 9.1505°

From the force triangle, using the law of sines:

20 lb A = sin α sin 20° or A = 43.0 lb

80.8° on tool, and A = 43.0 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

80.8° on rim of can.

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Chapter 4, Solution 74.

Free-Body Diagram: Let E be the intersection of the lines of action of the three forces acting on the tool.

From the free-body diagram, using law of sines:

6 in. + ( 0.76 in.) tan 35° L = sin 95° sin 30°

L = 13.0146 in.

Also: yBD = L − y AE − 0.88 in.

= 13.0146 in. −

0.76 in. − 0.88 in. cos 35°

= 11.2068 in.

And

 1.8 in.    yBD 

α = tan −1 

 1.8 in.  = tan −1   = 9.1247°  11.2086 in.  Then from the force triangle and using the law of sines:

B 14 lb = sin150° sin 9.1247° Solving for B:

B = 44.141 lb, or on the member B = 44.141 lb

80.9°, and on the lid B = 44.1 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

80.9°

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Chapter 4, Solution 75.

Free-Body Diagram: Based on the roller having impending motion to the left, the only contact between the roller and floor will be at the edge of the tile.

First note

(

)

W = mg = ( 20 kg ) 9.81 m/s 2 = 196.2 N

From the geometry of the three forces acting on the roller

 92 mm 

α = cos −1   = 23.074°  100 mm  and

θ = 90° − 30° − α = 60° − 23.074 = 36.926°

Applying the law of sines to the force triangle,

W P = sin θ sin α or

196.2 N P = sin 36.926° sin 23.074° ∴ P = 127.991 N

or P = 128.0 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

30°

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Chapter 4, Solution 76.

Free-Body Diagram: Based on the roller having impending motion to the right, the only contact between the roller and floor will be at the edge of the tile.

First note

(

W = mg = ( 20 kg ) 9.81 m/s 2

)

= 196.2 N

From the geometry of the three forces acting on the roller

 92 mm   = 23.074°  100 mm 

α = cos −1  and

θ = 90° + 30° − α = 120° − 23.074° = 96.926°

Applying the law of sines to the force triangle,

W P = sin θ sin α or

196.2 N P = sin 96.926° sin 23.074 ∴ P = 77.460 N

or P = 77.5 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

30°

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Chapter 4, Solution 77.

Free-Body Diagram: Note that the clamp is a three-force body. D is the intersection of the lines of action of the three forces.

From the free-body diagram it follows that:

y AD = (4.2 in.) tan 78o = 19.7594 in. yBD = y AD − 2.8 in.

= (19.7594 − 2.8) in. = 16.9594 in.

Then

 yBD    7.8 in. 

θ = tan −1 

 16.9594 in.  = tan −1   = 65.3013° , and  7.8 in. 

α = 90° − θ − 12° = 90° − 65.3013° − 12° = 12.6987° (a) Using the maximum allowable compressive force on the clamp:

( RB ) y = RB sin θ = 40 lb or RB =

40 lb = 44.028 lb sin 65.301°

or R B = 44.0 lb (b)

65.3°

Using the law of sines for the force triangle:

RB NA T = = sin12° sin α sin(90° + θ )

44.028 lb NA T = = sin12° sin12.6987° sin155.301° which gives:

N A = 46.551 lb (c)

T = 88.485 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or N A = 46.6 lb

or T = 88.5 lb

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Chapter 4, Solution 78.

Free-Body Diagram: (for hoist AD) Note that the hoist AD is a three-force body.

E is the intersection between the lines of action of the three forces acting on the hoist.

From the free-body diagram:

x AE = (48 in.) cos 30° = 41.5692 in. y AD = (48 in.)sin 30° = 24 in. yBE = x AD tan 75° = (41.5692 in.)tan75°

= 155.1384 in.

Then:

 yBE − 16 in.  −1  139.1384   = tan   x  41.5692  AD  

α = tan −1 

= 73.36588°

β = 75° − α = 75° − 73.36588° = 1.63412° θ = 180° − 15° − β = 165° − 1.63412° = 163.366° From the force triangle and using the law of sines:

260 lb B A = = sinβ sin θ sin15° 260 lb B A = = sin 1.63412° sin 163.366° sin 15° Solving for A and B: (a)

(b)

B = 2609.9 lb or B = 2.61kips

75.0°

or A = 2.36 kips

73.4°

A = 2359.8 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 79.

Free-Body Diagram:

Note that the member is a three-force body. In the free-body diagram, D is the intersection between the lines of action of the three forces. (a)

From the force triangle:

T − 110 N 3 = T 4 3T = 4T − 440 N

T = 440 N (b)

From the force triangle:

C 5 = T 4 C =

5 5 T = (440 N) = 550 N 4 4

or C = 550 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

36.9°

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Chapter 4, Solution 80.

Free-Body Diagram:

Note that the member is a three-force body. In the free-body diagram, E is the intersection between the lines of action of the three forces.

From the free-body diagram:

 15 

α = tan −1   = 61.928°  8  8

β = tan −1   = 33.690°  12  From the force triangle:

α − β = 61.928° − 33.690° = 28.238° 180° − α = 180° − 61.928° = 118.072°

Using the law of sines: T − 18 lb T C = = sin(α − β ) sin β sin(180° − α )

T − 18 lb T C = = sin ( 22.238° ) sin ( 33.690° ) sin(118.072°) Then:

(T

− 18 lb ) sin ( 33.690° ) = T sin ( 28.238° )

T = 122.414 lb and

or T = 122.4 lb

(122.414 lb ) sin (118.072°) = C sin ( 33.690° ) C = 194.723 lb or C = 194.7 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

33.7°

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Chapter 4, Solution 81.

Free-Body Diagram: Note that the peavey is a three-force body.

In the free-body diagram, D is the intersection of the lines of action of the three forces acting on the peavey.

It then follows:



 44 in.  = 40.2364°  44 in. + 8 in. 

β = tan −1 

α = 45° − β = 45° − 40.2364° = 4.7636°

From the force triangle, using the law of sines:

W C A = = sin β sin α sin135°

80 lb C A = = sin 40.236° sin 4.7636° sin135° Solving for C and A: (a)

(b)

C = 10.2852 lb or C = 10.29 lb

45.0°

or A = 87.6 lb

85.2°

A = 87.576 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 82.

Free-Body Diagram: Note that the peavey is a three-force body.

In the free-body diagram, D is the intersection of the lines of action of the three forces acting on the peavey.

It then follows: 

44 in.    DC + 8 in. 

β = tan −1 

where DC = ( 44 in. + a ) tan 30°

 R  a= −R  tan 30°   4 in.  =  − 4 in.  tan 30°  = 2.9282 in.

Then:

DC = ( 46.9282 in.) tan 30° = 27.0940 in. and



44 in.   = 51.4245°  35.0940 in. 

β = tan −1 

α = 60° − β = 60° − 51.4245° = 8.5755° Now from the force triangle, using the law of sines:

W C A = = sin β sin α sin120°

80 lb C A = = sin 51.424° sin 8.5755° sin120°

Solving for C and A: (a)

C = 15.2587 lb or C = 15.26 lb

(b)

A = 88.621 lb or A = 88.6 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

81.4°

30.0°

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Chapter 4, Solution 83.

Free-Body Diagram: From the free-body diagram, the member AB is a threeforce body. Let D be the intersection of the lines of action of the three forces acting on AB. Then, using triangle BCD:

CD = ( 250 mm ) tan 60° = 433.01 mm Also:

AF = AE + EF = AE + CD = (300 + 433.01) mm = 733.01 mm

FD 250 = tan −1 AF 733.01 = 18.8324°

θ = tan −1

From the force triangle

α = 180° − 30° − 18.8324° = 131.168° Using the law of sines

A B 330 N = = sin 30° sin18.8324° sin131.168° Solving for A and B:

A = 219.19 N,

B = 141.507 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or A = 219 N

71.2°

or B = 141.5 N

60.0°

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Chapter 4, Solution 84.

Free-Body Diagram:

From the free-body diagram it follows that

tan θ =

9.6 − 8sin20° 8cos20°

θ = 42.397° Also:

BE = 20sin 20° + ( 20cos 20° ) tan 42.397° = 12sin 20° (12cos 20° ) tan φ

φ = 60.456° Then using the law of sines on the force triangle:

TAD C 60 lb = = sin ( 90° − φ ) sin ( 90° + θ ) sin (φ − θ ) TAD C 60 lb = = cos 60.456° cos 42.397° sin18.059° (a)

TAD = 95.438 lb

(b)

C = 142.935 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or TAD = 95.41 lb or C = 142.935 lb

60.5°

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Chapter 4, Solution 85.

Free-Body Diagram:

(a)

Using the law of cosines on triangle ABC: 2

2

R 2 = ( 2R ) + ( 2R ) − 2 ( 2R )( 2R ) cosθ

1 = 8 − 8cosθ cosθ =

7 8

θ = 28.955° Also,

2R cos (θ + α ) = R cos α 2R ( cosθ cos α − sin θ sin α ) = R cos α tan α =

2 cosθ − 1 2cos 28.955° − 1 = 2 sin θ 2 sin 28.955° or α = 37.8°

α = 37.761° (b)

From the free-body diagram:

2R R = sin φ sin θ

sin φ = 2sin θ = 2 sin 28.955°

φ = 75.522° Now using the law of sines on the force triangle:

NA NB W = = sin [90° − (φ − α )] sin 90° − (θ + α )  sin (θ + α ) + (φ − α ) 

NA NB mg = = cos(φ − α ) cos (θ + α ) sin (θ + φ ) NA NB mg = = cos 37.762° cos 66.716° sin104.478°

Solving for N A and N B :

N A = 0.816 mg N B = 0.408 mg Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

66.7° 37.8°

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Chapter 4, Solution 86.

Free-Body Diagram:

(a)

Note that the rod is a three-force body. Using the law of cosines on triangle ABC: 2

R 2 = ( 2R ) + L2 − 2 ( 2 R ) L cosθ cosθ =

3R 2 + L2 4RL

(1)

Also,

L cos 45° = 2 R cos(θ + 45°) 2 L cos 45° = 2R(cosθ cos 45° − sin θ sin 45°) 2 L = cosθ − sin θ 4R Using (1) and that sin θ = 1 − cos 2 θ

sin θ =

( 4 RL )2 − ( 3R 2 + L2 ) 4RL

2

, this gives continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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L 3R 2 + L2 = − 4R 4RL

(

( 4RL )2 − ( 3R 2 + L2 )

2

4RL

)

( 4RL )2 − ( 3R 2 + L2 )

L2 − 3R 2 + L2 = −

2

Squaring both sides and simplifying:

(

)

9R 4 = 16R 2 L2 − 9R 4 + 6R 2 L2 + L4 , or

L4 − 10R 2 L2 + 18R 4 = 0 Solving for L2 :

(

)

(

)

L2 = 5 ± 7 R 2 , and taking the largest root L2 = 5 + 7 R 2 (b)

or L = 2.77 R

Using the value of L obtained in (a) and (1)

cosθ =

(

)

3R 2 + 5 + 7 R 2

(

4R 5 + 7

)

1/2

R

θ = 15.7380° Now using the law of sines on triangle ABC in the free body diagram: 2R R = sin φ sin θ

sin φ = 2sin θ = 2 sin15.7380°

φ = 32.852° Now using the law of sines on the force triangle:

NA NB mg = = cos (135° − 32.852° ) sin ( 45° − 15.7380° ) sin (15.7380° + 32.852° ) Solving for N A and N B :

N A = 1.303 mg N B = 0.652 mg

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

60.7° 12.15°

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Chapter 4, Solution 87.

Free-Body Diagram: Note that the wheel is a two-force body and therefore the force at C is directed along CA and perpendicular to the incline.

The wheelbarrow is a three-force body. Let D be the intersection of the lines of action of the three forces acting on the wheelbarrow. Then, using the triangle DEG

DE = EG tan 72° = ( 8 in.) tan 72° = 24.6215 in.

DF = DE − EF = 24.6215 in. − 9 in. = 15.6215 in.

Using triangle DFB:

φ = tan −1

FB  40  = tan −1   = 68.667° DF  15.6215 

From the force triangle: α = φ − 18° = 68.667° − 18° = 50.667°

β = 180° − 50.667° − 18° = 111.333° Using the law of sines:

B C 120 lb = = sin18° sin 50.667° sin111.333° B = 39.809 lb, C = 9.644 lb (a)

(b)

Noting that the force on each handle is B/2:

1 B = 19.90 lb 2

39.3°

C = 99.6 lb

72.0°

Reaction at C:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 88.

Free-Body Diagram:

From the free-body diagram:

a = ( 40 in.) tan θ − 32 in. =

8 in. − ( 23 in.) tan18° tan18°

θ = 40.048° Then,

φ = 90° − 18° − 40.048° = 31.952° Now, using the law of sines on the force triangle:

2( B/2) C 120 lb = = sin18° sin 31.952° sin (162° − 31.952° ) and solving for B/2 and C:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

(a)

1 B = 24.2 lb 2

58.0°

(b)

C = 83.0 lb

72.0°

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Chapter 4, Solution 89.

Free-Body Diagram: Note that the rod is a three-force body. In the free-body diagram, E is the intersection between the lines of action of the three forces.

Using triangle ACF in the free-body diagram:

yCF = d tan θ From triangle CEF:

xFE = yCF tan θ = d tan 2 θ and from triangle AGE:

cosθ = Noting that 1 + tan 2 θ = sec 2 θ =

d + xFE

( L2 )

=

d + d tan 2 θ

( L2 )

(1)

1 cos 2 θ

(1) gives

cosθ =

2d  1 L  cos 2 θ

cos3 θ =

  , or 

2d L

Using the given values of d = 2.8 in., and L = 10 in.

cos3 θ =

2(2.8 in.) = 0.56 10 in.

cosθ = 0.82426

θ = 34.486°

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or θ = 34.5°

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Chapter 4, Solution 90.

Free-Body Diagram: As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force geometry

tan β =

xGB y AB

where

y AB = L cosθ

xGB =

and

∴ tan β =

1 2

1 L sin θ 2

L sin θ

L cosθ

=

1 tan θ 2 or tan θ = 2 tan β

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 91.

Free-Body Diagram

(a) As shown in the free-body diagram. of the slender rod AB, the three forces intersect at C. From the geometry of the forces

xCB yBC

tan β = where

xCB =

1 L sin θ 2

yBC = L cosθ

and

∴ tan β =

1 tan θ 2

or

tan θ = 2 tan β

For

β = 25° tan θ = 2 tan 25° = 0.93262 ∴ θ = 43.003°

or θ = 43.0°

(

)

W = mg = (10 kg ) 9.81 m/s 2 = 98.1 N

(b)

From force triangle

A = W tan β = ( 98.1 N ) tan 25° = 45.745 N or A = 45.7 N and

B=

W 98.1 N = = 108.241 N cos β cos 25° or B = 108.2 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

65.0°

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Chapter 4, Solution 92.

Free-Body Diagram: Note that the rod is a three-force body. In the free-body diagram, D is the intersection between the lines of action of the three forces.

Using triangle BCE: a = BE = BC sin θ

and from triangle BCD BC = BD sin θ

Then a = BD sin 2 θ Also from triangle ABD BD = L sin θ , so

a = L sin 3 θ 1

a 3 or θ = sin   L −1 

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 93.

Free-Body Diagram:

Note that the athlete is a three-force body. From the free-body diagram

1 a tan θ 4a 2 = tan θ tan (θ + φ )

3a +

tan (θ + φ ) =

4 tan θ 1 3 + tan θ 2

(1)

From the force triangle, using the law of sines

FH W = or, using FH = 0.8W sin θ sin 180° − (θ + φ ) 

sin (θ + φ ) = 1.25sin θ Now, using (1)



−1 

4 tan θ θ + φ = tan  1  3 + tan θ 2 

  −1  = sin (1.25sin θ )  

Solving numerically for θ

θ = 15.04°

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 94.

Free-Body Diagram: Note that the rod is a three-force body. In the free-body diagram, E is the intersection of the lines of action of the three forces. (a)

Using triangle DBC which is isosceles DB = a

and using triangle BDE ED = DB tan 2θ = a tan 2θ

From triangle GED

ED =

a tan 2θ =

( L − a) , tan θ

and therefore

L−a , or tan θ

a (tan θ tan 2θ + 1) = L

(1)

From triangle BCD:

a=

L a

1.25 L 2

cos θ

= 1.6 cos θ

(2)

Using (2) in (1): 1.6 cos θ = 1 + tan θ tan 2θ

(3)

(

)

2 1 − cos 2 θ sin θ sin 2θ sin θ 2sin θ cosθ , (3) gives = = Noting that tan θ tan 2θ = cosθ cos 2θ cosθ 2cos 2 θ − 1 2cos 2 θ − 1

1.6 cosθ = 1 +

(

2 1 − cos 2 θ 2

) , or

2 cos θ − 1

3

3.2 cos θ − 1.6 cos θ − 1 = 0 or θ = 23.5°

Solving numerically, θ = 23.515° (b) Substituting into (2) for L = 8 in.,

a=

5 ( 8 in.) = 5.4528° 8 cos 23.515° or a = 5.45 in.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 95.

Free-Body Diagram:

The forces acting on the three-force member intersect at D. (a) From triangle ACO

 r  −1  1   = tan   = 18.4349°  3r  3

θ = tan −1 

tan θ =

(b) From triangle DCG

∴ DC = and

or θ = 18.43°

r DC

r r = = 3r tan θ tan18.4349°

DO = DC + r = 3r + r = 4r

 yDO    x AG 

α = tan −1  where

yDO = ( DO ) cosθ = ( 4r ) cos18.4349° = 3.4947r

and

x AG = ( 2r ) cosθ = ( 2r ) cos18.4349° = 1.89737r

 3.4947r  ∴ α = tan −1   = 63.435°  1.89737r  where

90° + (α − θ ) = 90° + 45° = 135.00°

Applying the law of sines to the force triangle,

mg R = A sin 90° + (α − θ )  sin θ ∴ RA = ( 0.44721) mg Finally,

P = RA cos α

= ( 0.44721mg ) cos 63.435° = 0.20000mg

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or P =

mg

5

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Chapter 4, Solution 96.

Free-Body Diagram:

ΣM A = 0:

( 450 mm ) i  × D + (150 mm ) i  × ( −180 N ) j + ( 250 mm ) i  × ( − 300 N ) k  = 0

( 450 mm ) Dyk − (450 mm)Dz j − (150 mm)(180 N)k + (250 mm)(300 N) j = 0 Setting the coefficients of the unit vectors equal to zero:

k:

Dy (450 mm) − (180 N )(150 mm ) = 0,

or

Dy = 60.000 N

i:

( 300 N )( 250 mm ) − Dz ( 450 mm ) = 0,

or

Dz = 166.667 N D = ( 60.0 N ) j + (166.7 N ) k

ΣFx = 0:

Ax = 0

ΣFy = 0:

Ay + 60.000 N − 180 N = 0, or Ay = 120.000 N

ΣFz = 0:

Az + 166.667 N − 300 N = 0, or Az = 133.333 N

A = (120.0 N ) j + (133.3 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 97.

Free-Body Diagram:

ΣFx = 0: ΣM D = 0:

Dx = 0 ( − 7 in.) i  × C + ( 2 in.) i + ( 3 in.) k  × ( 530 lb ) j + ( −192 lb ) k  + ( −3 in.) i + ( 6 in.) j × ( −96 lb ) j + ( 265 lb ) k  = 0

or − ( 7 in.) C yk + ( 7 in.) C z j + ( 2 in.)( 530 lb ) k + ( 2 in.)(192 lb ) j − ( 3 in.)( 530 lb ) i + ( 3 in.)( 96 lb ) k + ( 3 in.)( 265 lb ) j + ( 6 in.)( 265 lb ) i = 0 Setting the coefficients of the unit vectors to zero:

k:

− C y ( 7 in.) + ( 96 lb )( 3 in.) + ( 530 lb )( 2 in.) = 0,

or

C y = 192.571 lb

j:

C z ( 7 in.) + ( 265 lb )( 3 in.) + (192 lb )( 2 in.) = 0,

or

C z = −168.429 lb C = (192.6 lb ) j − (168.4 lb ) k

Then: ΣFy = 0:

192.571 lb − 96 lb + Dy + 530 lb = 0,

ΣFz = 0:

−168.429 lb + 265 lb + Dz − 192 lb = 0,

or

Dy = −626.57 lb or

Dz = 95.429 lb

D = − ( 626 lb ) j + ( 95.4 lb ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 98.

Free-Body Diagram:

ΣFx = 0:

Dx = 0

ΣM D = 0:

( − 7 in.) i  × C + ( 2 in.) i + ( 3 in.) k  × ( 530 lb ) j + ( − 192 lb ) k  + ( − 3 in.) i + ( 6 in.) j × ( 265 lb ) j + ( − 96 lb ) k  = 0

or − ( 7 in.) C yk + ( 7 in.) C z j + ( 2 in.)( 530 lb ) k + ( 2 in.)(192 lb ) j − ( 3 in.)( 530 lb ) i

− ( 3 in.)( 265 lb ) k − ( 3 in.)( 96 lb ) j − ( 6 in.)( 96 lb ) i = 0 Setting the coefficients of the unit vectors equal to zero:

k:

j:

− C y ( 7 in.) − ( 265 lb )( 3 in.) + ( 530 lb )( 2 in.) = 0, C z ( 7 in.) + ( 96 lb )( 3 in.) + (192 lb )( 2 in.) = 0, or

or

C y = 37.857 lb

C z = − 96.000 lb

C = ( 37.9 lb ) j − ( 96.0 lb ) k Then:

ΣFy = 0:

37.857 lb + 265 lb + Dy + 530 lb = 0, or

Dy = − 832.86 lb

ΣFz = 0:

− 96 lb + 96 lb + Dz − 192 lb = 0,

Dz = 192.000 lb

or

D = − ( 833 lb ) j + (192.0 lb ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 99.

Free-Body Diagram:

(

)

Note that W = mg = (18 kg ) 9.81 m/s 2 = 176.580 N Moment equilibrium: ΣM A = 0:

(

)

rB/ A × By j + Bz k + rC/ A × C zk + rG/ A × ( −176.580 N ) j = 0

or

i j k i j k i j k 1.5 0 0 + 1.2 1.2sin 60° −1.2cos 60° + 0.6 0.6sin 60° − 0.6cos 60° = 0 Cz 0 B y Bz 0 0 0 −176.580 0 or

(

)

1.2C z sin 60° − (105.948 N ) cos 60° i + ( −1.5Bz − 1.2C z ) j + 1.5By − 105.948 N k = 0

Solving the equation one component at a time: From i component:

1.2C z sin 60° − (105.948 N ) cos 60° = 0,

From j component:

−1.5Bz − 1.2C z = 0,

From k component:

1.5By − 105.948 N = 0,

or

or

C z = 50.974 N

Bz = − 0.8 ( 50.974 N ) = − 40.779 N or

By = 70.632 N

Force equations:

ΣFy = 0:

Ay − 176.580 N + 70.632 N = 0,

or

Ay = 105.948 N

ΣFz = 0:

Az + 50.974 N − 40.779 N = 0,

or

Az = −10.195 N

Therefore:

A = (105.9 N ) j − (10.20 N ) k B = ( 70.6 N ) j − ( 40.8 N ) k C = ( 51.0 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 100.

Free-Body Diagram:

(a)

( 250 mm ) i  × D + (150 mm ) i + ( 50 mm ) j × (150 N ) k

ΣM C = 0:

+ (150 mm ) i + ( − 50 mm ) j × T k + ( 325 mm ) i + ( 55 mm ) j × (− FE ) j = 0

or

( 250 mm ) Dyk − ( 250 mm ) Dz j + (150 mm )(150 N ) j − ( 50 mm )(150 N ) i − (150 mm ) T j − ( 50 mm ) T j − ( 325 mm ) FE k = 0 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Setting the coefficients of the unit vectors equal to zero:

i:

(150 N )( 50 mm ) − T ( 50 mm ) = 0 ,

j:

( −150 N )(150 mm ) − (150 N )(150 mm ) − Dz ( 250 mm ) = 0,

k:

( Dy ) ( 250 mm ) − FE ( 325 mm ) = 0 ,

or

or

T = 150 N

Dy =

T = 150 N

or

Dz = −180 N

325 FE 250

Spring force FE = kx, where

k = 366 N/m elongation of spring x = ( yE )θ =180° − ( yE )θ = 0°

= ( 300 + 55 ) mm − ( 300 − 55 ) mm = 110 mm So,

FE = ( 366 N/m )( 0.110 m ) = 40.26 N

Substituting into expression for Dy:

Dy = 52.338 N Force equations:

ΣFx = 0:

Dx = 0

ΣFy = 0:

C y + 52.338 N − 40.26 N = 0,

ΣFz = 0:

C z + 150 N + 150 N − 180

or

C y = −12.078 N

N = 0,

or

C z = −120.000 N

Therefore:

C = − (12.08 N ) j − (120.0 N ) k D = ( 52.3 N ) j − (180.0 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 101.

Free-Body Diagram:

Start by determining the spring force, FE : FE = − FE cosθ j + FE sin θ k

The magnitude FE = kx, where

k = 366 N/m, and elongation of spring x = ( yE )θ = 90° − ( yE )θ = 0°

=

( 300 mm )2 + ( 55 mm )2 − ( 300 mm − 55mm )

= 305 mm − 245 mm = 60 mm

So, FE = ( 366 N/m )( 0.06 m ) = 21.96 N. Note that the length of the spring at θ = 90° is therefore 305 mm.

Then  300   55  FE = − ( 21.96 N )   j + ( 21.96 N )  k 305    305 

or FE = − ( 21.6 N ) j + ( 3.96 N ) k continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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( 250 mm ) i  × D + (150 mm ) i + ( 50 mm ) j × (150 N ) k + (150 mm ) i + ( −50 mm ) j × T k

ΣM C = 0:

+ ( 325 mm ) i + ( −55 mm ) k  ×  − ( 21.6 N ) j + ( 3.96 N ) k  = 0

( 250 mm ) Dyk − ( 250 mm ) Dz j + (150 mm )(150 N ) j

or

− ( 50 mm )(150 N ) i − (150 mm ) T j − ( 50 mm ) T j − ( 325 mm )( 21.6 N ) k − ( 325 mm )( 3.96 N ) j − ( 55 mm )( 21.6 N ) i = 0 Setting the coefficients of the unit vectors equal to zero: (a)

i:

(150 N )( 50 mm ) − T ( 50 mm ) − ( 21.6 N )( 55 mm ) = 0,

or

T = 126.240 N T = 126.2 N

j:

( −150 N )(150 mm ) − (126.24 N )(150 mm ) − ( 3.96 N )( 325 mm ) − Dz ( 250 mm ) = 0, or Dz = −170.892 N

k:

( Dy ) ( 250 mm ) − ( 21.6 N )( 325 mm ) = 0,

Dy = 28.080 N

or

Force equations:

ΣFx = 0:

Dx = 0

ΣFy = 0:

C y + 28.080 N − 21.6 N = 0,

ΣFz = 0:

C z + 126.240 N + 150 N − 170.892 N + 3.96 N = 0,

or

C y = −6.4800 N or

C z = −109.308 N

Therefore:

C = − ( 6.48 N ) j − (109.3 N ) k D = ( 28.1 N ) j − (170.9 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 102.

Free-Body Diagram:

(

)

The weight W is W = mg = (170 kg ) 9.81 m/s 2 = 1667.7 N

ΣM C = 0: or

rCA × N A + rCB × N B + rCG × W = 0

( −0.3 m ) i + (1.2 m ) k  × N A j + (1.8 m ) i + ( 0.9 m ) k  × N B j + ( 0.6 m ) i + ( 0.6 m ) k  × ( −W ) j = 0

or

− ( 0.3 m ) N Ak − (1.2 m ) N Ai + (1.8 m ) N Bk − ( 0.9 m ) N Bi − ( 0.6 m )Wk + ( 0.6 m )Wi = 0

Equating the coefficients of the unit vectors to zero: i:

−1.2 N A − 0.9 N B + 0.6W = 0

4 N A + 3 N B 0 = 2W

j:

(1)

− 0.3 N A + 1.8 N B − 0.6W = 0 − N A + 6 N B 0 = 2W

− 2 ×  Eq. (1)  + Eq. ( 2 )

(2)

gives

−9 N A = 2W

NA =

2 W = 370.60 N 9 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Now (2) gives

NB =

1 2  10 W  2W + W  = 6 9  27

N B = 617.67 N,

and from (1)

ΣFy = 0:

N A + N B + NC − W = 0 370.60 N + 617.67 N + NC − 1667.7 N = 0 NC = 679.43 N

Therefore the forces on the blocks are:

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

N A = 371 N

N B = 618 N

NC = 679 N

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Chapter 4, Solution 103.

Free-Body Diagram:

The location of the bucket of sand will be ( xS , c, zS ).

(

)

The weight W is W = mg = (170 kg ) 9.81 m/s 2 = 1667.7 N

ΣFy = 0: N = ΣM O = 0:

or

3N − W − WS = 0

1 (W + WS ) 3

(1)

rOA × N + rOB × N + rOC × N + rOG × W − rOS × WS = 0

( 0.3 m ) i + (1.2 m ) k  × Nj + ( 2.4 m ) i + ( 0.9 m ) k  × Nj

+ ( 0.6 m ) i × Nj + (1.2 m ) i + ( 0.6 m ) k  × ( −W ) j + ( xS i + zS k ) × ( −WS ) j = 0 or

( 0.3 m ) N k − (1.2 m ) N i + ( 2.4 m ) N k − ( 0.9 m ) N i + ( 0.6 m ) N k − (1.2 m )W k + ( 0.6 m )W i − xSWS k + zSWS i = 0

Equating the coefficients of the unit vectors to zero:

i:

− (1.2 ) N − ( 0.9 ) N + ( 0.6 )W + zSWS = 0 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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or, using (1)

1  −2.1  (W + WS )  + 0.6W + zSWS = 0 3  WS = k:

0.1W zS − 0.7 m

0.3 N + 2.4 N + 0.6 N − 1.2W − xSWS = 0

or, using (1)

1  3.3  (W + WS )  − 1.2W − xSWS = 0 3   WS =

0.1W 1.1 m − xS

(2)

For (WS )min , (1) and (2) imply that xS, should be chosen as small as possible and that zS should be chosen as large as possible with the constraint that

(1.1 m − xS ) = ( zS

− 0.7 m )

xS + zS = 1.8 m.

or

The smallest xS and the largest zS that satisfy this condition are xS = 0.6 m

zS = 1.2 m The corresponding value of WS is:

WS =

0.1(1667.7 N ) = 333.54 N 1.1 m − 0.6 m

Therefore the smallest mass of the bucket of sand is

( mS )min

=

333.54 N = 34.000kg 9.81 m/s 2

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or

( mS )min

= 34.0 kg

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Chapter 4, Solution 104.

Free-Body Diagram:

WAB = ( 5 lb/ft )( 2 ft ) = 10 lb

First note

WBC = ( 5 lb/ft )( 4 ft ) = 20 lb W = WAB + WBC = 30 lb

To locate the equivalent force of the pipe assembly weight rG/B × W = Σ ( ri × Wi ) = rG ( AB ) × WAB + rG ( BC ) × WBC

( xGi + zGk ) × ( −30 lb ) j = (1 ft ) k × ( −10 lb ) j + ( 2 ft ) i × ( −20 lb ) j

or

∴

− ( 30 lb ) xGk + ( 30 lb ) zG i = (10 lb ⋅ ft ) i − ( 40 lb ⋅ ft ) k

From i-coefficient k-coefficient

zG =

10 lb ⋅ ft 1 = ft 30 lb 3

xG =

40 lb ⋅ ft 1 = 1 ft 30 lb 3

From free-body diagram. of piping

ΣM x = 0:

W ( zG ) − TA ( 2 ft ) = 0 1  1  ∴ TA =  ft  30 lb  ft  = 5 lb 2  3 

ΣFy = 0:

or

TA = 5.00 lb

5 lb + TD + TC − 30 lb = 0

∴ TD + TC = 25 lb

(1) continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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ΣM z = 0:

4  TD (1.25 ft ) + TC ( 4 ft ) − 30 lb  ft  = 0 3  ∴ 1.25TD + 4TC = 40 lb ⋅ ft

−4  Equation (1) 

−4TD − 4TC = −100

(3)

−2.75TD = −60

Equation (2) + Equation (3)

∴ TD = 21.818 lb From Equation (1)

(2)

or

TD = 21.8 lb

TC = 25 − 21.818 = 3.1818 lb

or

Results:

TC = 3.18 lb TA = 5.00 lb TC = 3.18 lb TD = 21.8 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 105.

Free-Body Diagram:

First note

W AB = (5 lb/ft )(2 ft ) = 10 lb WBC = ( 5 lb/ft )( 4 ft ) = 20 lb

From free-body diagram. of pipe assembly

ΣFy = 0: TA + TC + TD − 10 lb − 20 lb = 0 ∴ TA + TC + TD = 30 lb

(1)

ΣM x = 0: (10 lb )(1 ft ) − TA ( 2 ft ) = 0 or

TA = 5.00 lb TC + TD = 25 lb

From Equations (1) and (2)

(2) (3)

ΣM z = 0: TC ( 4 ft ) + TD ( amax ) − 20 lb ( 2 ft ) = 0 or

( 4 ft ) TC

+ TD amax = 40 lb ⋅ ft

(4) continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Using Equation (3) to eliminate TC

4 ( 25 − TD ) + TD amax = 40 amax = 4 −

or

60 TD

By observation, a is maximum when TD is maximum. From Equation (3), (TD )max occurs when TC = 0.

Therefore, (TD )max = 25 lb and

amax = 4 −

60 25

= 1.600 ft Results: (a) amax = 1.600 ft

(b)

TA = 5.00 lb TC = 0 TD = 25.0 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 106.

Free-Body Diagram:

The free-body diagram indicates the forces on the camera and tripod slid along their lines of action to the plane ABCD. Note that the x-coordinate of the center of mass of the camera is:

xCAM = − ( 2.4 in. − 1 in.) = −1.4 in. ΣM B = 0:

or

( − 3 in.) k × C y j + ( −1.5 in.) k − (1.4 in.) i  × ( − 0.44 lb ) j + ( −1.5 in.) k − ( 2.8 in.) i  × ( − 0.53 lb ) j + ( −1.5 in.) k − ( 3.2 in.) i  × Ay j = 0 ( 3 in.) C yi − (1.5 in.)( 0.44 lb ) i + (1.4 in.)( 0.44 lb ) k − (1.5 in.)( 0.53 lb ) i + ( 2.8 in.)( 0.53 lb ) k + (1.5 in.) Ay i − ( 3.2 in.) Ayk = 0

Setting the coefficients of the unit vectors equal to zero: k:

− Ay ( 3.2 in.) + ( 0.53 lb )( 2.8 in.) + ( 0.44 lb )(1.4 in.) = 0

Ay = 0.65625 lb

or A y = 0.656 lb

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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i:

( 3 in.) C y − (1.5 in.)( 0.44 lb ) − (1.5 in )( 0.53 lb ) + (1.5 in.) (0.65625 lb) = 0 C y = 0.156875 lb

or C y = 0.1569 lb

or B y = 0.1569 lb

ΣFx = By − 0.53 lb − 0.44 lb + 0.156875 lb + 0.156875 lb = 0

By = 0.156875 lb (b) Free-Body Diagram:

Condition for no tipping: By > 0

ΣM A = 0:

( 3.2 in.) i + ( −1.5 in.) k  × C y j + ( 3.2 in.) i + (1.5 in.) k  × B y j

{

}

+ (1.8 in.) i × ( − 0.44 lb ) j + 1.8 in. − (1.4 in.) cosθ  i − (1.4 in.) sinθ k × ( − 0.53 lb ) j = 0 or

( 3.2 in.) C yk + (1.5 in.) C yi + ( 3.2 in.) By k − (1.5 in.) Byi − (1.8 in.)( 0.44 lb ) k + 1.8 in. − (1.4 in.) cosθ  ( − 0.53 lb ) k + (1.4 in.) sinθ ( − 0.53 lb ) i = 0

Setting the coefficients of the unit vectors equal to zero: i:

C y (1.5 in.) − By (1.5 in.) − ( 0.53 lb )(1.4 in.) sin θ = 0

C y = B y + ( 0.53 lb ) k:

1.4 sin θ 1.5

( By + C y ) ( 3.2 in.) − ( 0.44 lb )(1.8 in.) − ( 0.53 lb ) 1.8 in. − (1.4 in ) cosθ  = 0 or

 1.4   2By ( 3.2 in.) + ( 0.53 lb )   ( 3.2 in.) sin θ + (1.4 in.) cosθ  − ( 0.44 lb + 0.53 lb )(1.8 in.) = 0  1.5   continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Solving for By : By =

  1.4    1 ( 0.44 lb + 0.53 lb )1.8 in. − 0.53 lb   ( 3.2 in.) sin θ + (1.4 in.) cosθ   2 ( 3.2 in.)   1.5   

and By > 0 2.3531 > 2.1333sin θ + cosθ

To solve for θ :

2.1333sin θ + cosθ = A cos (θ + α ) = A ( cosθ cos α − sin θ sin α ) where A=

( 2.1333)2 + (1)2

= 2.3561, and

 − 2.1333   , which (noting that cos α > 0) gives 1  

α = tan −1 

α = − 64.885° The inequality for By becomes:

2.3531 > 2.3561cos(θ − 64.885°) or cos (θ − 64.885° ) < 0.99873 or θ − 64.885° < cos −1 ( 0.99873) or θ < 64.885° ± 2.8879°

θ max = 62.00°

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 107.

Free-Body Diagram:

ΣM B = 0:

or

rBA × TA + rBC × TC + rBO × WAB + rBF × WBD + rBG × WAD = 0

( )

3   L L 3  1  − Li × TA j +  − 4 i + Lk   × TC j − i ×  − Wj    4 2  tan 60°   3   

 L   1   3L   1  3 3 +  − i + Lk  ×  − Wj  +  − i + Lk  ×  − Wj  = 0 4 4  4   3   4   3 

or

− LTAk −

3 3 L L 3 L 3 LTC k − LTC i + Wk + Wk + LWi − Wk + LWi = 0 4 4 6 12 12 4 12

Equating the coefficients of the unit vectors to zero:

i:

W 3 W 3 3 L+ L − TC L = 0 3 4 3 4 4

TC = k:

2W 3 3

− TA L + TA =

W 3L W L W L 2W 3 L + + − =0 3 4 3 4 3 2 3 34 3

W 3 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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ΣFy = 0:

TB + TB =

W 2W + −W = 0 3 3 3 2 1  1 − W 3 3

Therefore:

TA = TB =

2 1  1 − W 3 3

TC =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

W

3

2W

3 3

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Chapter 4, Solution 108.

Free-Body Diagram:

Note that:

xC =

L 1 − zC 2 3

(a) Setting: TA = TB = T

ΣM O = 0: or

rOA × TA + rOB × TB + rOC × TC + rOH × W + rOF × WBD + rOG × WAD = 0

−

L L i × TA j + i × TB j + 2 2

( )

3  L L  3   − 4  i + Lk  × TC j 4   2 tan 60°    

L   1   L   1  3 3 Lk  ×  − Wj  +  − i + Lk  ×  − Wj  + ( xC i + zC k ) × ( − Wj) = 0 +  i + 4 4 4   3   4   3 

or

( )

3 L  L L L 3  T k + LTC i − TAk + TAk +  − 4  2 tan 60°  C 2 2 4  

−

1L 1 3 1L 1 3 Wk + LWi + Wk + LWi − xCWk + zCWi = 0 34 3 4 34 3 4 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Equating the coefficients of the unit vectors to zero:

i:

W 3 W 3 3L L+ L − TC + WzC = 0 3 4 3 4 4

(1)

or, using the relation between xC, and zC:

 L  3 3 − LTC +  +  − 3 xC + 4 2  2 3  k:

 L   W = 0  

(2)

1 L L 3 1 1 LW + LW − xCW = 0 − T + T + L  −  TC − 2 2 2 4 12 12  

1 3 L  −  TC = xCW 4  2

(3)

Substituting (3) into (2)

 1  1 3 3 3  − LTC +  +  LW − 3  L  −  TC  = 0  4 2  4     2 2 3

4 W 3 4 ΣFy = 0: 2T + W − 2W = 0 3 W T = 3 TC =

TA = TB =

Therefore:

TC = (b) Using TC in (1):

W

3

4 W 3

W 3 W 3 4 3 L+ L −  W  L + WzC = 0 3 4 3 4 3 4 1   zC = 1 − L 2 3  and from geometry

xC =

L 1 1 zC = 2− 3 L − 2 3 3

(

)

Therefore:

xC =

1 2− 3 L 3

(

)

 1  zC = 1 − L 2 3 

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 109.

Free-Body Diagram:

(a) ΣFy = 0:

3 ( R ) − 135 N = 0

R = 45.0 N

ΣM A = 0:

rAB × R B + rAC × R C + rAG × FW = 0

or

( 0.9 m ) i + l k  × R j + l i + ( 0.9 m ) k  × R j + ( 0.450 m ) i + ( 0.450 m ) k  × ( − FW j) = 0

or

( 0.9 m ) R k − lR i + lR k − ( 0.9 m ) R i − ( 0.450 m ) FW k + ( 0.450 m ) FW i = 0

Equating the coefficients of the i unit vector to zero:

i:

− lR − ( 0.9 m ) R + ( 0.45 m ) FW = 0

Using that FW = 3R

− l − ( 0.9 m ) + ( 0.45 m )( 3) = 0 l = 0.450 m

or l = 450 mm continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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(b) Free-Body Diagram:

ΣM A = 0: rAB × R B + rAC × R C + rAG × FW = 0 or ( 0.9 m ) i + ( 0.5 m ) k  × R j + ( 0.5 m ) i + ( 0.9 m ) k  × R j + ( 0.450 m ) i + ( 0.450 m ) k  × ( −135 N ) j = 0

( 0.9 m ) R k − ( 0.5 m ) R i + ( 0.5 m ) R k − ( 0.9 m ) R i − ( 0.450 m ) FW k + ( 0.450 m ) FW i = 0 Equating the coefficients of the unit vectors to zero i:

− 0.5RB − 0.9 RC + 60.75 = 0

(1)

k:

0.9 RB + 0.5RC − 60.75 = 0

(2)

0.5 × [ Eq. (1)] + 0.9 × [ Eq. (2) ]

gives

 − 0.5 ( 0.5 ) + ( 0.9 )( 0.9 )  RB + ( 0.5 − 0.9 ) 60.75 = 0 RB = 43.393 N

Now using (1)

− 0.5 ( 43.393) − 0.9 RC + 60.67 = 0 RC = 43.393 N

ΣFy = 0:

RA + 43.393 + 43.393 − 135 = 0 RA = 48.2 N RB = 43.4 N

RC = 43.4 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 110.

Free-Body Diagram:

ΣM O = 0:

( 0.75 ft ) i × ( Ay j + Azk ) + ( 3.75 ft ) i × ( By j + Bzk ) + ( 2.25 ft ) i + ( 3 ft ) k  × ( − 27 lb ) j + ( 4.5 ft ) i + ( 5.25 ft ) k  × C y j = 0

or

( 0.75 ft ) Ayk − ( 0.75 ft ) Az j + ( 3.75 ft ) Byk − ( 3.75 ft ) Bz j − ( 2.25 ft )( 27 lb ) k + ( 3 ft )( 27 lb ) i + ( 4.5 ft ) C yk − ( 5.25 ft ) C y i = 0

Setting the coefficients of the unit vectors equal to zero:

i:

( 27 lb )( 3 ft ) − C y ( 5.25 ft ) = 0 C y = 15.4286 lb

k:

( 27 lb )(1.5 ft ) + By ( 3 ft ) − (15.4286 lb )( 3.75 ft ) = 0 By = − 5.78575 lb

ΣFy = 0:

Ay − 5.78575 lb − 27 lb + 15.4286 lb = 0 Ay = 17.3572 lb

Therefore: (a) (b) (c)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

A y = 17.36 lb

B y = 5.79 lb C y = 15.43 lb

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Chapter 4, Solution 111.

Free-Body Diagram:

ΣM O = 0:

( 0.75 ft ) i × ( Ay j + Azk ) + ( 3.75 ft ) i × ( By j + Bzk ) + ( 2.25 ft ) i + ( 3 ft ) k  × ( − 27 lb ) j + ( 3.75 ft ) i + ( 6 ft ) k  × C y j = 0

or

( 0.75 ft ) Ayk − ( 0.75 ft ) Az j + ( 3.75 ft ) Byk − ( 3.75 ft ) Bz j − ( 2.25 ft )( 27 lb ) k + ( 3 ft )( 27 lb ) i + ( 3.75 ft ) C yk − ( 6.0 ft ) C y i = 0

Setting the coefficients of the unit vectors equal to zero: ΣM x = 0:

( 27 lb )( 3 ft ) − C y ( 6 ft ) = 0 C y = 13.5000 lb

ΣM z = 0:

− ( 27 lb )(1.5 ft ) + By ( 3 ft ) − (13.5000 lb )( 3 ft ) = 0 By = 0

ΣFy = 0:

Ay + 13.5000 lb − 27 lb = 0 Ay = 13.5000 lb

Therefore:

A y = 13.50 lb

(a) (b) (c)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

By = 0 C y = 13.50 lb

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Chapter 4, Solution 112 Free-Body Diagram:

Express all forces in terms of rectangular components:

rE = ( 3 ft ) i + ( 3 ft ) j rB = ( 3 ft ) sin 30° j + ( 3 ft ) cos 30°k = (1.5 ft ) j + ( 2.598 ft ) k rD = − ( 3 ft ) i + ( 3 ft ) j

or

or

rA = (10 ft ) sin 30° j − (10 ft ) cos 30°k = ( 5 ft ) j + (8.66 ft ) k uuur BE = rE − rB = ( 3 ft ) i + ( 3 ft ) j − (1.5 ft ) j − ( 2.598 ft ) k uuur BE = ( 3 ft ) i + (1.5 ft ) j − ( 2.598 ft ) k, and BE = 4.243 ft uuur BD = rD − rB = − ( 3 ft ) i + ( 3 ft ) j − (1.5 ft ) j − ( 2.598 ft ) k uuur BD = − ( 3 ft ) i + (1.5 ft ) j − ( 2.598 ft ) k, and BD = 4.243 ft

Then

uuur ur BD T BD = TBD = TBD ( − 0.707i + 0.3535j − 0.6123k ) BD uuur ur BE T BE = TBE = TBE ( 0.707i + 0.3535j − 0.6123k ) BE continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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rB × TBD + rB × TBE + ( 5 ft ) j + ( 8.66 ft ) k  × ( − 75 lb ) j

ΣM C = 0:

i j k i j k 0 1.5 2.598 TBD + 0 1.5 2.598 + 649.5 i = 0 or − 0.707 0.3535 − 0.6123 0.707 0.3535 − 0.6123

Equating the coefficients of the unit vectors to zero:

j:

−1.837 TBD + 1.837 TBE = 0

i:

−1.837 TBD + 1.837 TBE + 649.5 lb = 0

TBD = 176.8 lb

TBE = 176.8 lb Force equations:

C x + (176.8 )( − 0.707 ) + (176.8 )( 0.707 ) = 0,

or

Cx = 0

C y + (176.8 )( 0.3535 ) + (176.8 )( 0.3535 ) − 75 lb = 0,

or

C z + (176.8 )( − 0.6123) + (176.8 )( − 0.6123) = 0,

C z = 216.5 lb

or

C y = − 50 lb

C = − ( 50 lb ) j + ( 216.5 lb ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 113.

Free-Body Diagram:

Express the forces in terms of their rectangular components: uuur FB 12i − 18j + 36k 2 3 6 TFB = TFB = TFB = TFB i − TFB j + TFBk 2 2 2 FB 7 7 (12 ) + ( −18) + ( 36 ) 7

TFC = TFC

uuur FC = TFC FC

−12i − 18j + 36k

( −12 )

2

2

+ ( −18 ) + ( 36 )

2

2 3 6 = − TFBi − TFB j + TFBk 7 7 7

From the free-body diagram, note that

zE 12 = 27 18 z E = 18.00 ft continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Then, using TED = 2720 lb

TED = 2720

− 27 j + 57.75k

( − 27 )2 + (18 + 39.75)2

TED = ( 32 lb )( − 36j + 77k ) (a)

ΣM A = 0:

rAD × TED + rAF × TFB + rAE × W = 0

i j k i j k i j k i j k TFB TFC or 32 0 0 39.75 + 0 18 −12 + 0 18 −12 + 2720 0 27 −18 = 0 7 7 0 − 36 77 2 −3 6 0 −1 0 − 2 −3 6 Equating the coefficients of the unit vectors to zero

i:

32 ( 39.75 )( 36 ) +

TFB T ( 72 ) + FC ( 72 ) + 2720 ( −18) = 0 7 7

72 72 TFB + TFC − 3168 = 0 7 7 j:

(1)

TFB T ( − 24 ) + FC ( 24 ) = 0 7 7 TFB = TFC

(2)

Substituting Eq. (2) in (1) gives

 72  2   TFB − 3168 = 0  7 

or TFB = 154.0 lb TFC = 154.0 lb (b) ΣFx = 0:

Ax +

2 2 (154.0) − (154.0 ) = 0 7 7

Ax = 0

ΣFy = 0:

Ay −

3 3 (154.0) − (154.0 ) − ( 32 )( 36 ) − 2720 = 0 7 7

Ay = 4004 lb ΣFz = 0:

Az +

6 6 (154.0) + (154.0 ) + ( 32 )( 77 ) = 0 7 7

Az = − 2728 lb

Therefore:

A = ( 4.00 kips ) j − ( 2.73 kips ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 114.

Free-Body Diagram

First express tensions in terms of rectangular components: uuur BE − 8i − 8j + 4k 2 2 1 TBE = TBE = TBE = − TBE i − TBE j + TBE k 2 2 2 BE 3 3 3 ( − 8) + ( − 8) + ( 4 )

TBF = TBF

uuur BF = TBF BF

8i − 8j + 14k

(8)

2

2

+ ( − 8 ) + (14 )

2

=

4 4 7 TBF i − TBF j + TBF k 9 9 9

TCD = TCD ( cos φ sin θ i − sin φ j − cos φ cosθ k ) ΣM A = 0:

or

rAB × TBE + rAB × TBF + rAC × TCD = 0

8j ×

TBE T ( − 2i − 2 j + k ) + 8j × BF ( 4i − 4 j + 7k ) 3 9

+ 10 j × TCD ( cos φ sin θ i − sin φ j − cos φ cosθ k ) = 0 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Equating the coefficients of the unit vectors to zero:

(a)

i:

8 56 TBE + TBF − 10TCD cos φ cosθ = 0 3 9

(1)

k:

16 32 TBE − TBF − 10TCD cos φ sin θ = 0 3 9

(2)

− 2 ×  Eq. (1)  + Eq. ( 2 ) gives:

  56  32  − 2   −  TBF − 10 ( 600 ) cos10° ( − 2 cos 30° + sin 30° ) = 0   9  9 TBF = 455.00 N

Using this in Eq. (1),

8 56 TBE + ( 455.00 ) − 10 ( 600 ) cos10° cos 30° = 0 3 9 TBE = 857.29 N

Therefore:

TBE = 857 N TBF = 455 N (b)

ΣFx = 0:

ΣFy = 0:

Ax −

2 4 (857.29 N ) + ( 455.00 N ) + ( 600 N ) cos10° sin 30° = 0 3 9

Ax = 73.9 N 2 4 Ay − ( 857.29 N ) − ( 455.00 N ) + ( 600 N ) sin10° = 0 3 9

Ay = 878 N ΣFz = 0:

Az +

1 7 (857.29 N ) + ( 455.00 N ) − ( 600 N ) cos10° cos 30° = 0 3 9

Az = −127.9 N Therefore:

A = ( 73.9 N ) i + ( 878 N ) j − (127.9 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 115.

Free-Body Diagram:

First express tensions in terms of rectangular components: uuur BE − 8i − 8j + 4k 2 2 1 = TBE = − TBE i − TBE j + TBE k TBE = TBE 2 2 2 BE 3 3 3 ( − 8) + ( − 8) + ( 4 )

TBF = TBF

uuur BF = TBF BF

8i − 8j + 14k

(8)

2

2

+ ( − 8 ) + (14 )

2

=

4 4 7 TBF i − TBF j + TBF k 9 9 9

TCD = TCD ( cos φ sin θ i − sin φ j − cos φ cosθ k ) ΣM A = 0: or

rAB × TBE + rAB × TBF + rAC × TCD = 0

8j ×

TBE T (− 2i − 2 j + k ) + 8j × BF (4i − 4 j + 7k ) 3 9

+ 10 j × TCD ( cos φ sin θ i − sin φ j − cos φ cosθ k ) = 0 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Equating the coefficients of the unit vectors to zero:

8 56 TBE + TBF − 10TCD cos φ cosθ = 0 3 9

i:

8 56 (840 N ) + ( 450 N ) = 10TCD cos φ cosθ 3 9

(1)

16 32 TBE − TBF − 10TCD cos φ sin θ = 0 3 9

k:

16 32 (840 N ) − ( 450 N ) = 10TCD cos φ sin θ 3 9 (a)

(2)

 Eq. ( 2)  gives:  Eq. (1)  16 (840) − 32 ( 450) 10TCD cos φ sin θ 3 9 = 8 10TCD cos φ cos θ (840) + 56 ( 450) 3 9

tan θ =

1 1.75

θ = 29.7°

θ = 29.745° (b) Substituting into (1) gives:

8 56 (840 N ) + ( 450 N ) − 10TCD cos8° cos 29.745° = 0, 3 9

or

TCD = 586.19 N

or TCD = 586 N (c)

ΣFx = 0:

Ax −

2 4 (840 N ) + ( 450 N ) + ( 586.19 N ) cos8° sin 29.745° = 0 3 9

Ax = 72.0 N

ΣFy = 0:

Ay −

2 4 (840 N ) − ( 450 N ) − ( 586.19 N ) sin 8° = 0 3 9

Ay = 842 N ΣFz = 0:

Az +

1 3 (840 N ) + ( 450 N ) − ( 586.19 N ) cos8° cos 29.745° = 0 3 9

Az = −126.0 N Therefore:

A = ( 72.0 N ) i + ( 842 N ) j − (126.0 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 116.

Free-Body Diagram:

Express all forces in terms of rectangular components:

rA = ( 2.4 m ) i rB = (1.8 m ) j uuur AD = − ( 2.4 m ) i + ( 0.3 m ) j + (1.2 m ) k uuur BE = − (1.8 m ) i + ( 0.6 m ) j − ( 0.9 m ) k W = − ( 880 N ) j Then

uuur ur AD = TAD T AD = TAD AD uuur ur BE T BE = TBE = TBE BE

− 2.4i + 0.3j + 1.2k

( − 2.4 )

2

2

+ ( 0.3) + (1.2 )

2

−1.8i + 0.6 j − 0.9k

( −1.8)

2

2

+ ( 0.6 ) + ( − 0.9 )

2

8 1 4 = − TADi + TAD j + TADk 9 9 9 6 2 3 = − TADi + TAD j − TADk 7 7 7 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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ΣM C = 0:

or

rA × TAD + rB × TBE + rA × W = 0

i 2.4 8 − 9

j 0 1 9

k i j k 0 TAD + 1.8 0 0 TBE + ( 2.4 ) i × ( − 880 ) j = 0 4 6 2 3 − − 9 7 7 7

Equating the coefficients of the unit vectors to zero:

−

j:

9.6 5.4 TAD + TBE = 0 9 7

2.4 3.6 TAD + TBE − 2112 = 0 9 7

k:

or TAD = 2160 N

TBE = 2990 N Force equations:

Cx −

8 6 ( 2160.0 N ) − ( 2986.7 N ) = 0, 9 7

Cy +

1 2 ( 2160.0 N ) + ( 2986.7 N ) − 880 N = 0, or 9 7

Cz +

4 3 ( 2160.0 N ) − ( 2986.7 N ) = 0, 9 7

or

or

C x = 4480.0 N

C y = − 213.34 N

C z = 320.01 N

C = ( 4480 N ) i − ( 213 N ) j + ( 320 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 117.

Free-Body Diagram:

Express all forces in terms of rectangular components:

rA = ( 2.4 m ) i rB = (1.8 m ) j uuur AD = − ( 2.4 m ) i + ( 0.3 m ) j + (1.2 m ) k uuur BE = − (1.8 m ) i + ( 0.6 m ) j − ( 0.9 m ) k WA = − ( 440 N ) j WB = − ( 440 N ) j Then

uuur ur AD T AD = TAD = TAD AD uuur ur BE = TBE T BE = TBE BE

− 2.4i + 0.3j + 1.2k

( − 2.4 )

2

2

+ ( 0.3) + (1.2 )

2

−1.8i + 0.6 j − 0.9k

( −1.8)

2

2

+ ( 0.6 ) + ( − 0.9 )

2

8 1 4 = − TADi + TAD j + TADk 9 9 9 6 2 3 = − TADi + TAD j − TADk 7 7 7 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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ΣM C = 0:

or

rA × TAD + rB × TBE + rA × WA + rB × WB = 0

i 2.4 8 − 9

j 0 1 9

k i j k 0 TAD + 1.8 0 0 TBE + ( 2.4 ) i × (− 440) j + (1.8 ) i × (− 440) j = 0 4 6 2 3 − − 9 7 7 7

Equating the coefficients of the unit vectors to zero:

j:

−

9.6 5.4 TAD + TBE = 0 9 7

2.4 3.6 TAD + TBE − 1848 = 0 9 7

k:

or TAD = 1890 N

TBE = 2610 N Force equations:

Cx −

8 6 (1890.00 N ) − ( 2613.3 N ) = 0, 9 7

Cy +

1 2 (1890.00 N ) + ( 2613.3 N ) − 440 N − 440 N = 0, 9 7

Cz +

4 3 (1890.00 N ) − ( 2613.3 N ) = 0, 9 7

or

or

C x = 3920.0 N

or

C y = −76.657 N

C z = 279.99 N

C = ( 3920 N ) i − ( 76.7 N ) j + ( 280 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 118.

Free-Body Diagram:

Express all forces in terms of rectangular components:

rA = (140 in.) i rC = ( 72 in.) i uuur EG = − (120 in.) i + (126 in.) k uuuur FH = − (120 in.) i − ( 90 in.) k uur CI = − ( 72 in.) i + ( 44.8 in.) j W = − (140 lb ) j Then

uuur ur EG = TEG T EG = TEG EG

−120i + 126k

( −120 )

uuuur ur FH T FH = TFH = TFH FH uur ur CI T CI = TCI = TCI CI

2

+ (126 )

2

=−

−120i − 90k

( −120 )2 + ( − 90 )2 − 72i + 44.8j

( − 72 )

2

+ ( 44.8 )

2

=−

20 21 TEG i + TEGk 29 29

= −0.8TFH i − 0.6TFH k 45 28 TCI i + TCI j 53 53 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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ΣM D = 0:

or

rE × TEG + rF × TFH + rC × TCI + rA × W = 0

i j k i j k i j k TEG TFH T 120 0 10 + 120 0 −10 + 72 0 0 CI − 19600 lb ⋅ in. = 0 29 5 53 − 20 0 21 −4 0 −3 − 45 28 0

Noting that TCI = TFH and equating the coefficients of the unit vectors to zero:

j: k:

− 93.793TEG + 80TCI = 0 38.038TCI − 19600 lb ⋅ in. = 0

or TCI = TFH = 515 lb or TEG = 440 lb Force equations:

ΣFx = 0:

Dx −

20 45 4 ( 439.50 lb ) − ( 515.28 lb ) − ( 515.28 lb ) = 0, 29 53 5

ΣFy = 0:

Dy +

28 ( 515.28 lb ) − 140 lb = 0, 53

ΣFz = 0:

Dz +

21 3 ( 439.50 lb ) − ( 515.28) lb = 0, 29 5

or

or

Dx = 1152.83 lb

Dy = −132.223 lb or

Dz = − 9.0906 lb

D = (1153 lb ) i − (132.2 lb ) j − ( 9.09 lb ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 119.

Free-Body Diagram:

Express all forces in terms of rectangular components:

rB = (120 in.) i rC = ( 72 in.) i uuur EG = − (120 in.) i + (126 in.) k uuuur FH = − (120 in.) i − ( 90 in.) k uur CI = − ( 72 in.) i + ( 44.8 in.) j W = − (140 lb) j Then

uuur ur EG T EG = TEG = TEG EG

−120i + 126k

uuuur ur FH T FH = TFH = TFH FH uur ur CI T CI = TCI = TCI CI

( −120 )

2

+ (126 )

2

=−

−120i − 90k

( −120 )2 + ( − 90 )2 − 72i + 44.8j

( − 72 )

2

+ ( 44.8 )

2

=−

20 21 TEG i + TEGk 29 29

= − 0.8TFH i − 0.6TFH k 45 28 TCI i + TCI j 53 53 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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ΣM D = 0: or

rE × TEG + rF × TFH + rC × TCI + rA × W = 0 i j k i j k i j k TEG TFH T 120 0 10 + 120 0 −10 + 72 0 0 CI − 19600 lb ⋅ in. = 0 29 5 53 − 20 0 21 −4 0 −3 − 45 28 0

Noting that TCI = TFH and equating the coefficients of the unit vectors to zero:

j: k:

− 93.793TEG + 80TCI = 0 38.038TCI − 16800 lb ⋅ in. = 0 or TCI = TFH = 442 lb

TEG = 377 lb Force equations:

ΣFx = 0:

Dx −

20 45 4 ( 376.72 lb ) − ( 441.67 lb ) − ( 441.67 lb ) = 0, 29 53 5

ΣFy = 0:

Dy +

28 ( 441.67 lb ) − 140 lb = 0, 53

ΣFz = 0:

Dz +

21 3 ( 376.72 lb ) − ( 441.67 lb ) = 0, 29 5

or

or

Dx = 988.15 lb

Dy = − 93.335 lb or

Dz = − 7.7952 lb

D = ( 998 lb ) i − ( 93.3 lb ) j − ( 7.80 lb ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 120.

Free-Body Diagram:

Geometry: Using triangle ACD and the law of sines

sin α sin 50° or α = 20.946° = 15 in.

7 in.

β = 50° + 20.946° = 70.946° Expressing FCD in terms of its rectangular coordinates:

FCD = FCD sin β j + FCD cos β k

= FCD sin 70.946° j + FCD cos 70.946°k FCD = 0.94521FCD j + 0.32646 FCDk

ΣM B = 0:

( − 26 in.) i × A + ( −13 in.) i + (16 in.) sin 50°j + (16 in.) cos 50°k  × ( − 75 lb ) j + ( − 26 in.) i + ( 7 in.) k  × FCD = 0

or

− ( 26 in.) Ayk + ( 26 in.) Az j + (13 in.)( 75 lb ) k + (16 in.)( 75 lb ) cos 50°i − ( 26 in.) ( 0.94521FCD ) k + ( 26 in.) ( 0.32646 FCD ) j − ( 7 in.) ( 0.94521FCD ) i = 0 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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(a) Setting the coefficients of the unit vectors to zero:

i:

( 75 lb ) (16 in.) cos 50° − ( 0.94521FCD ) ( 7 in.) = 0 FCD = 116.6 lb

(b) ΣFx = 0:

k:

Ax = 0

− 0.94521(116.580 lb )  ( 26 in.) + ( 75 lb )(13 in.) − Ay ( 26 in.) = 0

Ay = − 72.693 lb j:

 0.32646 (116.580 lb )  ( 26 in.) + Az ( 26 in.) = 0 Az = − 38.059 lb

ΣFy = 0:

− 72.693 lb + 0.94521(116.580 lb ) − 75 lb + By = 0

By = 37.500 lb ΣFz = 0:

− 38.059 lb + 0.32646 (116.580 lb ) + Bz = 0

Bz = 0 Therefore:

A = − ( 72.7 lb ) j − ( 38.1 lb ) k B = ( 37.5 lb ) j

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 121.

Free-Body Diagram:

Express tension in terms of rectangular components: uuuur DG = − ( 4.8 in.) j − ( 9 in.) k uuuur − 4.7 j − 9k DG 8 15 TDG = TDG = TDG = − TDG j − TDG k 2 2 DG 17 17 ( − 4.8) + ( − 9 ) Equilibrium:

ΣM F = 0:

( 6.4 in.) i + ( − 2.4 in.) j × TDG + ( 5.2 in.) j × E + ( 7.6 in.) j + ( 9.6 in.) k  × ( − 55 lb ) i = 0

or

15   8   15   TDG i − ( 6.4 in.)   TDG k + ( 6.4 in.)   TDG j  17   17   17 

( 2.4 in.) 

− ( 5.2 in.) Ex k + ( 5.2 in.) Ez i + ( 7.6 in.) (55 lb)k − ( 9.6 in.) (55 lb)j = 0 Setting the coefficients of the unit vectors equal to zero: (a)

j:

15 TDG ( 6.4 in.) − ( 55 lb )( 9.6 in.) = 0 17 TDG = 93.5 lb continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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(b)

k:

8  (93.500 lb)  ( 6.4 in.) = 0 17  

( 55 lb )( 7.6 in.) − Ex ( 5.2 in.) −  E x = 26.231 lb

i:

 15  Ez ( 5.2 in.) +  (93.500 lb)  ( 2.4 in.) = 0 17  E z = − 38.077 lb

ΣFx = 0:

− 55 lb + 26.231 lb + Fx = 0 Fx = 28.796 lb

ΣFy = 0:

Fy −

8 ( 93.500 lb ) = 0 17

Fy = 44.000 lb

ΣFz = 0:

− 38.077 lb + Fz −

15 ( 93.500 lb ) = 0 17

Fz = 120.577 lb Therefore:

E = ( 26.2 lb ) i − ( 38.1 lb ) k F = ( 28.8 lb ) i + ( 44.0 lb ) j + (120.6 lb ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 122.

Free-Body Diagram:

(

)

W = mg = (15 kg ) 9.81 m/s 2 = 147.15 N

First note (a)

 ( 0.08 m ) i + ( 0.25 m ) j − ( 0.2 m ) k  T TEF = λ EF TEF =  TEF = EF ( 0.08i + 0.25j − 0.2k )  0.33 ( 0.08 )2 + ( 0.25 )2 + ( 0.2 )2 m   From free-body diagram of rectangular plate ΣM x = 0:

(147.15 N )( 0.1 m ) − (TEF ) y ( 0.2 m ) = 0  0.25   14.715 N ⋅ m −   TEF  ( 0.2 m ) = 0  0.33  

or

TEF = 97.119 N

or

or TEF = 97.1 N

(b)

ΣFx = 0:

Ax + (TEF ) x = 0

 0.08  Ax +   ( 97.119 N ) = 0  0.33  ∴ Ax = −23.544 N

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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ΣM B( z -axis ) = 0:

− Ay ( 0.3 m ) − (TEF ) y ( 0.04 m ) + W ( 0.15 m ) = 0

 0.25   − Ay ( 0.3 m ) −   97.119 N  ( 0.04 m ) + 147.15 N ( 0.15 m ) = 0  0.33  

or

∴ Ay = 63.765 N ΣM B( y -axis ) = 0:

Az ( 0.3 m ) + (TEF ) x ( 0.2 m ) + (TEF ) z ( 0.04 m ) = 0  0.08    0.2   Az ( 0.3 m ) +   TEF  ( 0.2 m ) −   TEF  ( 0.04 m ) = 0 0.33 0.33       ∴ Az = −7.848 N

and A = − ( 23.5 N ) i + ( 63.8 N ) j − ( 7.85 N ) k ΣFy = 0:

Ay − W + (TEF ) y + By = 0

 0.25  63.765 N − 147.15 N +   ( 97.119 N ) + By = 0  0.33  ∴ By = 9.81 N ΣFz = 0:

Az − (TEF ) z + Bz = 0

 0.2  −7.848 N −   ( 97.119 N ) + Bz = 0  0.33  ∴ Bz = 66.708 N

and B = ( 9.81 N ) j + ( 66.7 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 123.

Free-Body Diagram:

(

(a)

)

W = mg = (15 kg ) 9.81 m/s 2 = 147.15 N

First note

TEH = λ EH TEH

  − ( 0.3 m ) i + ( 0.12 m ) j − ( 0.2 m ) k  T  = T = EH  − ( 0.3) i + ( 0.12 ) j − ( 0.2 ) k    EH 2 2 2 0.38 ( 0.3) + ( 0.12 ) + ( 0.2 ) m  

From free-body diagram of rectangular plate

ΣM x = 0:

(147.15 N )( 0.1 m ) − (TEH ) y ( 0.2 m ) = 0  0.12    TEH  ( 0.2 m ) = 0  0.38  

or

(147.15 N )( 0.1 m ) − 

or

TEH = 232.99 N or TEH = 233 N

(b)

ΣFx = 0:

Ax + (TEH ) x = 0

 0.3  Ax −   ( 232.99 N ) = 0  0.38  ∴ Ax = 183.938 N continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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ΣM B( z -axis ) = 0:

− Ay ( 0.3 m ) − (TEH ) y ( 0.04 m ) + W ( 0.15 m ) = 0  0.12 − Ay ( 0.3 m ) −  ( 232.99 N ) ( 0.04 m ) + (147.15 N )( 0.15 m ) = 0  0.38 

or

∴ Ay = 63.765 N ΣM B( y -axis ) = 0:

Az ( 0.3 m ) + (TEH ) x ( 0.2 m ) + (TEH ) z ( 0.04 m ) = 0

 0.3   Az ( 0.3 m ) −   ( 232.99 N )  ( 0.2 m ) −  0.38  

or

 0.2    ( 232.99 )  ( 0.04 m ) = 0   0.38  

∴ Az = 138.976 N

and A = (183.9 N ) i + ( 63.8 N ) j + (139.0 N ) k ΣFy = 0:

Ay + By − W + (TEH ) y = 0

 0.12  63.765 N + By − 147.15 N +   ( 232.99 N ) = 0  0.38  ∴ By = 9.8092 N ΣFz = 0:

Az + Bz − (TEH ) z = 0

 0.2  138.976 N + Bz −   ( 232.99 N ) = 0  0.38  ∴ Bz = −16.3497 N and B = ( 9.81 N ) j − (16.35 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 124.

Free-Body Diagram:

Express tension, weight in terms of rectangular components: uuur EF = ( 300 mm ) i + (1350 mm ) j − ( 700 mm ) k uuur EF 300 i + 1350 j − 700 k T=T =T EF ( 300 )2 + (1350 )2 + ( − 700 )2

=

6 27 14 Ti + T j− Tk 31 31 31

(

)

W = − (mg ) j = − ( 7 kg ) 9.81 m s 2 j = − (68.67 N)j ΣM B = 0:

− ( 750 mm ) i × A +  − ( 375 mm ) i + ( 350 mm ) k  × ( − 68.7 N ) j + ( −100 mm ) i + ( 700 mm ) k  × T = 0

or

− ( 750 mm ) Ayk + (125 mm ) Az j + ( 375 mm )( 68.7 N ) k + ( 350 mm )( 68.7 N ) i i j k T + −100 0 700 ( mm ) = 0 31 6 27 −14 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Setting the coefficients of the unit vectors equal to zero: (a)

i:

−

27 T ( 700 mm ) + ( 68.67 N )( 350 mm ) = 0 31 or T = 39.4N

T = 39.422 N (b)

k:

 27  − Ay ( 750 mm ) + ( 68.67 N )( 375 mm ) −  ( 39.422 N )  (100 mm ) = 0 31   Ay = 29.757 N

j:

14  6  Az ( 750 mm ) −  ( 39.422 N )  (100 mm ) +  ( 39.422 N )  ( 700 mm ) = 0  31   31  Az = − 4.7476 N

ΣFx = 0:

Bx +

6 ( 39.422 N ) = 0 31

Bx = − 7.6301 N

ΣFy = 0:

29.757 N + B y − 68.67 N +

27 ( 39.422 N ) = 0 31

By = 4.5777 N ΣFz = 0:

− 4.7476 N + Bz −

14 ( 39.422 N ) = 0 31

Bz = 22.551 N Therefore:

A = ( 29.8 N ) j − ( 4.75 N ) k B = − ( 7.63 N ) i + ( 4.58 N ) j + ( 22.6 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 125.

Free-Body Diagram:

Express tension, weight in terms of rectangular components: uur IF = ( 75 mm ) i + (1350 mm ) j − ( 250 mm ) k

uur IF T=T =T IF =

75i + 1350 j − 250 k

( 75)2 + (1350 )2 + ( − 250 )2

3 54 10 Ti + Tj− Tk 55 55 55

(

)

W = − (mg ) j = − ( 7 kg ) 9.81 m/s 2 j = − (68.67 N)j ΣM B = 0:

− ( 750 mm ) i × A +  − ( 375 mm ) i + ( 350 mm ) k  × ( − 68.7 N ) j + (125 mm ) i + ( 250 mm ) k  × T = 0

or

− ( 750 mm ) Ay k + (125 mm ) Az j + ( 375 mm )( 68.7 N ) k + ( 350 mm )( 68.7 N ) i i j k T + 125 0 250 ( mm ) = 0 55 3 54 −10 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Setting the coefficients of the unit vectors equal to zero: (a)

i:

−

54 T ( 250 mm ) + ( 68.67 N )( 350 mm ) = 0 55

T = 97.918 N (b)

k:

or T = 97.9 N

 54  − Ay ( 750 mm ) + ( 68.67 N )( 375 mm ) −  ( 97.918 N )  (125 mm ) = 0  55 

Ay = 50.358 N j:

 10  Az ( 750 mm ) −  ( 97.918 N )  (125 mm ) + 55  

3   55 ( 97.918 N )  ( 250 mm ) = 0  

Az = − 4.7475 N

ΣFx = 0:

Bx +

3 ( 97.918 N ) = 0 55

Bx = − 5.3410 N

ΣFy = 0:

50.358 N + B y − 68.67 N +

54 ( 97.918 N ) = 0 55

By = − 77.826 N ΣFz = 0:

− 4.7475 N + Bz −

10 ( 97.918 N ) = 0 55

Bz = 22.551 N

Therefore:

A = ( 50.4 N ) j − ( 4.75 N ) k B = − ( 5.34 N ) i − ( 77.8 N ) j + ( 22.6 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 126.

Free-Body Diagram:

Express forces, weight in terms of rectangular components: uuur CE = ( 3 ft ) i + ( 4 ft ) j − ( 2 ft ) k

FCE = FCE

uuur CE = FCE CE

3i + 4 j + 2k

( 3) 2 + ( 4 )2 + ( 2 ) 2

= 0.55709 FCE i + 0.74278 FCE j + 0.37139 FCE k

W = − (mg ) j = − (300 lb)j ΣM B = 0:

or

( 4 ft ) k × A + (1.5 ft ) i + ( 2 ft ) k  × ( − 300 lb ) j + ( 3 ft ) i + ( 4 ft ) k  × FCE = 0 − ( 4 ft ) Ay i + ( 4 ft ) Az j − (1.5 ft )( 300 lb ) k + ( 2 ft )( 300 lb ) i i j k + 3 0 4 FCE ( ft ) = 0 0.55709 0.74278 0.37139 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Setting the coefficients of the unit vectors equal to zero:

k:

( 0.74278 FCE ) ( 3 ft ) − ( 300 lb )(1.5 ft ) = 0 FCE = 201.94 lb

or FCE = 202 lb j:

Ax ( 4 ft ) + 0.55709 ( 201.94 lb )  ( 4 ft ) − 0.37139 ( 201.94 lb )  ( 3 ft ) = 0 Ax = − 56.250 lb

i:

− Ay ( 4 ft ) −  0.74278 ( 201.94 lb )  ( 4 ft ) + ( 300 lb )( 2 ft ) = 0

Ay = 0 ΣFx = 0:

− 56.250 lb + Bx + 0.55709 ( 201.94 lb ) = 0 Bx = − 56.249 lb

ΣFy = 0:

0 + By − 300 lb + 0.74278 ( 201.94 lb ) = 0

By = 150.003 lb ΣFz = 0:

Bz + 0.371391( 201.94 lb ) = 0 Bz = − 74.999 lb

Therefore:

A = − ( 56.3 lb ) i B = − ( 56.2 lb ) i + (150.0 lb ) j − ( 75.0 lb ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 127.

Free-Body Diagram:

Express forces, weight in terms of rectangular components: uuur CA = − (1.2 m ) i + (1.2 m ) j − ( 0.6 m ) k

uuur CB = (1.2 m ) i + (1.2 m ) j − ( 0.6 m ) k By symmetry FCA = FCB , and at the load corresponding to buckling FCA = FCB = 1.8 kN

FCA

uuur CA = FCA = (1.8 kN ) CA

−1.2 i + 1.2 j − 0.6 k

( −1.2 )2 + (1.2 )2 + ( − 0.6 )2

FCA = − (1.2 kN ) i + (1.2 kN ) j − ( 0.6 kN ) k FCB

uuur CB = FCB = (1.8 kN ) CB

1.2 i + 1.2 j − 0.6 k

(1.2 )2 + (1.2 )2 + ( − 0.6 )2

FCB = (1.2 kN ) i + (1.2 kN ) j − ( 0.6 kN ) k continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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ΣM D = 0:

( 2.4 m ) i × E + ( 2.4 m ) i + (1.2 m ) j × FCB + (1.2 m ) j × FCA + (1.2 m ) i + ( 0.6 m ) j × Pk = 0 i

or

( 2.4 m ) Ex

j

k

i

j

k

0 0 + 2.4 1.2 0 kN ⋅ m E y Ez 1.2 1.2 − 0.6

i j k i + 0 1.2 0 kN ⋅ m + (1.2 m ) −1.2 1.2 − 0.6 0

j k ( 0.6 m ) 0 = 0 0 P

Setting the coefficient of the unit vector i equal to zero: (a)

i : P(0.6 m) − ( 0.6 )(1.2 ) kN ⋅ m − ( 0.6 )(1.2 ) kN ⋅ m = 0 P = 2.4000 kN or P = 2.40 kN

(b)

By symmetry, Dz = Ez

ΣFz = 0:

Dz + Dz + 2.4 kN − 0.6 kN − 0.6 kN = 0 Dz = Ez = − 0.60000 kN

Therefore:

E z = − ( 0.600 kN ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 128.

Free-Body Diagram:

Notice that the forces in the belts can be equivalently moved to the center of the pulley because their net moment about this point is zero. (a) ΣFx = 0:

3 lb + ( 3 lb ) cos 30° − T = 0 T = 5.5981 lb or T = 5.60 lb

(b) ΣFy = 0:

ΣFz = 0:

Dy = 0 Dz − ( 3 lb ) sin 30° = 0 D = (1.500 lb ) k

ΣM D = 0:

M D + ( 0.72 in.) j − (1.2 in.) k  × ( −T ) i + ( 0.88 in.) j − ( 3 in.) k  × ( 3 lb )(1 + cos 30° ) i − ( 3 lb ) sin 30°k  = 0

i or M Dx i + M Dy j + M Dz k + 0 −T

(

)

j ( 0.72 in.)

0

k i 0 ( −1.2 in.) + 0 1 + cos 30°

j k ( 0.88 in.) ( − 3 in.) ( 3 lb ) = 0 − sin 30° 0 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Setting the coefficients of the unit vectors equal to zero:

i:

M Dx − ( 3 lb ) sin 30° ( 0.88 in.) = 0 M Dx = 1.3200 lb ⋅ in.

j:

M Dy + ( 5.5981 lb )(1.2 in.) − ( 3 lb )( 3 in.)(1 + cos 30° ) = 0

M Dy = 10.0765 lb ⋅ in. k:

M Dz + ( 5.5981 lb )( 0.72 in.) − ( 3 lb )( 0.88 in.)(1 + cos 30° ) = 0 M Dz = 0.89568 lb ⋅ in.

or M D = (1.320 lb ⋅ in.) i + (10.08 lb ⋅ in.) j + ( 0.896 lb ⋅ in.) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 129.

Free-Body Diagram:

Express the tension in terms of its rectangular components: uuuur DG = − ( 4.8 in.) j − ( 9 in.) k

TDC = TDG

ΣM E = 0:

uuuur DG = TDG DG

− 4.8 j − 9 k

( − 4.8)

2

+ ( − 9)

2

=−

8 15 TDG j − TDG k 17 17

M E + ( 6.4 in.) i + ( − 7.6 in.) j × TDG + ( 2.4 in.) j + ( 9.6 in.) k  × ( − 44 lb ) i = 0

15   8   15   TDG i − ( 6.4 in.)   TDGk + ( 6.4 in.)   TDG j  17   17   17 

 or ( M Ex i + M Ey j + M Ez k ) + ( 7.6 in.) 

+ ( 2.4 in.)( 44 lb ) k − ( 9.6 in.)( 44 lb ) j = 0 Setting the coefficient of the unit vector j equal to zero: (a) j:

 15   TDG  ( 6.4 in.) − ( 44 lb )( 9.6 in.) = 0 17   TDG = 74.800 lb or TDG = 74.8 lb continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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(b)

ΣFx = 0:

E x − 44 lb = 0, or Ex = 44.000 lb

ΣFy = 0:

Ey −

8 ( 74.8 lb ) = 0, or E y = 35.200 lb 17

ΣFz = 0:

Ez −

15 ( 74.8 lb ) = 0, or Ez = 66.000 lb 17 or E = ( 44.0 lb ) i + ( 35.2 lb ) j + ( 66.0 lb ) k

Using the moment equation again and setting the coefficients of the unit vectors i and k equal to zero:

i:

 15  M Ex + ( 7.6 in.)   ( 74.800 lb ) = 0  17  M Ex = −501.60 lb ⋅ in.

k:

8  M Ez + ( 44 lb )( 2.4 in.) −  ( 74.8 lb )  ( 6.4 in.) = 0 17  M Ex = 119.680 lb ⋅ in. or M E = − ( 502 lb ⋅ in.) i + (119.7 lb ⋅ in.) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 130.

Free-Body Diagram:

Express forces and moments in terms of rectangular components:

FDE = FDE

− 40 i − 70 j + 40k

( − 40 )

2

2

+ ( − 70 ) + ( 40 )

2

=

FDE (− 4i − 7 j + 4k ) 9

FA = ( 24 N )( sin 20° i − cos 20° j) B = By j + Bzk , (a) ΣFx = 0:

M B = M By j + M Bzk

4 − FDE + 24sin 20° = 0 9 FDE = 18.4691 N

ΣM B = 0:

or

FCF

or FDE = 18.47 N

rBC × FCF + rBD × FDE + rBA × FA + M B = 0

i j k i j k i j k 18.4691 −1 +  M B y j + M B zk  0 − 48 36 + 0 − 80 60 + ( 80 )( 24 ) 0 0 9 −4 −7 4 0 −1 0 sin 20° − cos 20° 0 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Equating the coefficients of the unit vectors to zero:

i:

36FCF +

18.4691 (100 ) + (80 )( 24 )( − cos 20° ) = 0 9 FCF = 44.417 N

or FCF = 44.4 N

(b)

j:

18.4691 ( − 240 ) + (80 )( 24 )( − sin 20° ) + M By = 0 9 M By = 1149.19 N ⋅ mm

k:

18.4691 ( − 320 ) + M Bz = 0 9 M Bz = 656.68 N ⋅ mm

ΣFy = 0:

By − 44.417 −

7 (18.4691) − 24 cos 20° = 0 9

By = 81.3 N

ΣFz = 0:

Bz +

4 (18.4691) = 0 9

Bz = −8.21 N Therefore:

B = ( 81.3 N ) j − ( 8.21 N ) k M B = (1.149 N ⋅ m ) j + ( 0.657 N ⋅ m ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 131.

Free-Body Diagram:

Express tension, weight in terms of rectangular components: uuur EF = ( 300 mm ) i + (1350 mm ) j − ( 700 mm ) k

uuur EF =T T=T EF

=

300 i + 1350 j − 700 k

( 300 )2 + (1350 )2 + ( − 700 )2

6 27 14 Ti + T j− Tk 31 31 31

(

)

W = − ( mg ) j = − ( 7 kg ) 9.81 m/s 2 j = − ( 68.67 N ) j

ΣM B = 0:

M B +  − ( 375 mm ) i + ( 350 mm ) k  × ( − 68.7 N ) j

+ ( −100 mm ) i + ( 700 mm ) k  × T = 0 or

( M By j + M Bzk ) + ( 375 mm )( 68.7 N ) k + ( 350 mm )( 68.7 N ) i i j k T + −100 0 700 ( mm ) = 0 31 6 27 −14 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Setting the coefficients of the unit vector i equal to zero: (a)

i:

−

27 T ( 700 mm ) − ( 68.67 N )( 350 mm ) = 0 31 or T = 39.4 N

T = 39.422 N (b)

ΣFx = 0:

Bx +

6 ( 39.422 N ) = 0 31

Bx = − 7.6301 N

ΣFy = 0:

By − 68.67 N +

27 ( 39.422 N ) = 0 31

By = 34.335 N

ΣFz = 0:

Bz −

14 ( 39.422 N ) = 0 31

Bz = 17.8035 N

B = − ( 7.63 N ) i + ( 34.3 N ) j + (17.80 N ) k Using the moment equation again and setting the coefficients of the unit vectors j and k to zero:

ΣM B ( y − axis) = 0:

14  M By −  ( 39.422 N )  (100 mm ) + 31  

6   31 ( 39.422 N )  ( 700 mm ) = 0  

M By = − 3.5607 N ⋅ m

ΣM B ( z − axis) = 0:

 27  M Bz + ( 68.67 N )( 375 mm ) −  ( 39.422 N )  (100 mm ) = 0  31  M Bz = − 22.318 N ⋅ m,

Therefore:

M B = − ( 3.56 N ⋅ m ) j − ( 22.3 N ⋅ m ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 132.

Free-Body Diagram:

Express tensions, load in terms of rectangular components: uuur BD = − ( 60 in.) i + ( 25 in.) k

uuur BE = − ( 60 in.) i + ( 25 in.) j uuur CF = − ( 60 in.) i + ( 25 in.) j BD = BE = CF =

TBD TBE TCF

= 65 in.

uuur BD 12 5 = TBD = − TBD i + TBD k BD 13 13 uuur BE 12 5 = TBE = − TBE i + TBE j BE 13 13 uuur CF 12 5 = TCF = − TCF i + TCF j CF 13 13

ΣM A = 0: or

( −60 )2 + ( 25)2

rB × TBD + rB × TBE + rC × TCF + rG × W = 0

i j k i j k i j k i j k TBD TBE TCF 60 0 0 in. + 60 0 0 in. + 60 0 − 30 in. + 60 0 −15 lb ⋅ in. = 0 13 13 13 −12 0 5 −12 5 0 −12 5 0 0 − 500 0 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Equating the coefficients of the unit vectors to zero:

i:

150 TCF − 7500 = 0 13 TCF = 650.00 lb

j:

−

300 360 TBD + ( 650 lb ) = 0 13 13 TBD = 780.00 lb

k:

or TCF = 650 lb

or TBD = 780 lb

300 300 TBE − 30000 + ( 650.00 lb ) = 0 13 13 TBE = 650.00 lb

ΣFx = 0:

Ax −

or TBE = 650 lb

12 12 12 ( 780 lb ) − ( 650 lb ) − ( 650 lb ) = 0 13 13 13

Ax = 1920.00 lb

ΣFy = 0:

Ay +

5 5 ( 780 lb ) + ( 650 lb ) − 500 lb = 0 13 13

Ay = 0

ΣFz = 0:

Az +

5 ( 780 lb ) = 0 13

Az = −300.00 lb Therefore,

A = (1920 lb ) i − ( 300 lb ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 133.

Free-Body Diagram:

Express tensions, load in terms of rectangular components: uuur BD = − ( 60 in.) i + ( 25 in.) k

uuur BE = − ( 60 in.) i + ( 25 in.) j uuur CF = − ( 60 in.) i + ( 25 in.) j BD = BE = CF =

( −60 )2 + ( 25)2

= 65 in.

uuur BD 12 5 = − TBD i + TBD k BD 13 13 uuur BE 12 5 = TBE = − TBE i + TBE j BE 13 13 uuur CF 12 5 = TCF = − TCF i + TCF j CF 13 13

TBD = TBD

TBE TCF

WG = − ( 500 lb ) j WC = − ( 800 lb ) j

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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ΣM A = 0:

rB × TBD + rB × TBE + rC × TCF + rG × WG + rC × WC = 0

i j k i j k i j k i j k TBD TBE TCF 60 0 0 in. + 60 0 0 in. + 60 0 − 30 in. + 60 0 −15 lb ⋅ in.

13 13 13 −12 0 5 −12 5 0 −12 5 0 0 − 500 0

or

i j k + 60 0 −30 lb ⋅ in. = 0 0 −800 0 Equating the coefficients of the unit vectors to zero:

i:

150 TCF − 7500 − 24000 = 0 13 TCF = 2730 lb

j:

−

or TCF = 2.73 kips

300 360 TBD + ( 2730 lb ) = 0 13 13 TBD = 3276 lb

k:

or TBD = 3.28 kips

300 300 TBE − 30000 + ( 2730 lb ) − ( 60 )(800 lb ) = 0 13 13 TBE = 650.00 lb

ΣFx = 0:

Ax −

or TBE = 650 lb

12 12 12 ( 3276 lb ) − ( 650 lb ) − ( 2730 lb ) = 0 13 13 13

Ax = 6144 lb

ΣFy = 0:

Ay +

5 5 ( 2730 lb ) + ( 650 lb ) − 500 lb − 800 lb = 0 13 13

Ay = 0

ΣFz = 0:

Az +

5 ( 3276 lb ) = 0 13

Az = −1260.00 lb Therefore,

A = ( 6.14 kips ) i − (1.260 kips ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 134.

First note

Free-Body Diagram:

TDI = λ DI TDI =

= TEH = λ EH TEH =

= TFG = λ FGTFG =

=

− ( 0.65 m ) i + ( 0.2 m ) j − ( 0.44 m ) k

( 0.65)2 + ( 0.2 )2 + ( 0.44 )2 m

TDI

TDI ( − 0.65i + 0.2j − 0.44k ) 0.81 − ( 0.45 m ) i + ( 0.24 m ) j

( 0.45)2 + ( 0.24 )2 m

TEH

TEH ( − 0.45 i ) + ( 0.24 j) 0.51

− ( 0.45 m ) i + ( 0.2 m ) j + ( 0.36 m ) k

( 0.45

)

2

2

2

+ ( 0.2 ) + ( 0.36 ) m

TFG

TFG ( −0.45i + 0.2j + 0.36k ) 0.61

From free-body diagram of frame ΣM A = 0: rD/ A × TDI + rC/ A × ( −280 N ) j + rH / A × TEH + rF / A × TFG + rF / A × ( −360 N ) j = 0 or

i j k i j k i j k i j k  TDI   TEH   TFG  0.65 0.2 0  0.32 0   + 0.65 0 0 ( 280 N ) + 0  + 0.45 0 0.06    0.81   0.51   0.61  −0.65 0.2 −0.44 −0.45 0.24 0 −0.45 0.2 0.36 0 −1 0 i j k + 0.45 0 0.06 ( 360 N ) = 0 0 −1 0

or

( − 0.088 i + 0.286 j + 0.26 k ) + ( − 0.012 i − 0.189 j + 0.09 k )

TDI T + ( − 0.65 k ) 280 N + ( 0.144 k ) EH 0.81 0.51

TFG + ( 0.06 i − 0.45 k )( 360 N ) = 0 0.61 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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From i-coefficient

T  T  −0.088  DI  − 0.012  FG  + 0.06 ( 360 N ) = 0  0.81   0.61 

∴ 0.108642TDI + 0.0196721TFG = 21.6 From j-coefficient

(1)

T  T  0.286  DI  − 0.189  FG  = 0 0.81    0.61 

∴ TFG = 1.13959TDI

(2)

From k-coefficient

T  T  T  0.26  DI  − 0.65 ( 280 N ) + 0.144  EH  + 0.09  FG  0.81 0.51      0.61  − 0.45 ( 360 N ) = 0 ∴ 0.32099TDI + 0.28235TEH + 0.147541TFG = 344 N

(3)

Substitution of Equation (2) into Equation (1)

0.108642TDI + 0.0196721(1.13959TDI ) = 21.6 ∴ TDI = 164.810 N TDI = 164.8 N

or Then from Equation (2)

TFG = 1.13959 (164.810 N ) = 187.816 N TFG = 187.8 N

or And from Equation (3)

0.32099 (164.810 N ) + 0.28235TEH + 0.147541(187.816 N ) = 344 N ∴ TEH = 932.84 N TEH = 933 N

or The vector forms of the cable forces are:

TDI =

164.810 N ( −0.65i + 0.2j − 0.44k ) 0.81

= − (132.25 N ) i + ( 40.694 N ) j − ( 89.526 N ) k TEH =

932.84 N ( − 0.45i + 0.24 j) = − (823.09 N ) i + ( 438.98 N ) j 0.51

187.816 N ( − 0.45i + 0.2 j + 0.36k ) 0.61 = − (138.553 N ) i + ( 61.579 N ) j + (110.842 N ) k

TFG =

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Then, from free-body diagram of frame

ΣFx = 0: Ax − 132.25 − 823.09 − 138.553 = 0 ∴ Ax = 1093.89 N ΣFy = 0: Ay + 40.694 + 438.98 + 61.579 − 360 − 280 = 0 ∴ Ay = 98.747 N ΣFz = 0: Az − 89.526 + 110.842 = 0 ∴ Az = −21.316 N or

A = (1094 N ) i + ( 98.7 N ) j − ( 21.3 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 135.

Free-Body Diagram: First note

TDI = λ DI TDI =

= TEH = λ EH TEH =

=

TFG = λ FGTFG =

− ( 0.65 m ) i + ( 0.2 m ) j − ( 0.44 m ) k

( 0.65)2 + ( 0.2 )2 + ( 0.44 )2 m TDI ( −65i + 20 j − 44k ) 81 − ( 0.45 m ) i + ( 0.24 m ) j

( 0.45)2 + ( 0.24 )2 m

TDI

TEH

TEH ( −15i + 8j) 17 − ( 0.45 m ) i + ( 0.2 m ) j + ( 0.36 m ) k

( 0.45)2 + ( 0.2 )2 + ( 0.36 )2 m

TFG

TFG ( −45i + 20j + 36k ) 61 From free-body diagram of frame

=

ΣM A = 0: rD/ A × TDI + rC/ A ×  − ( 280 N ) j + ( 50 N ) k  + rH / A × TEH + rF / A × TFG + rF / A × ( −360 N ) j or

i j k i j k i j k  TDI  0.65 0.2 0  0 + 0 0.32 0  + 0.65 0  81  −65 20 −44 −15 8 0 0 −280 50 i j k T + 0.45 0 0.06  FG  61 −45 20 36

and

 TEH     17 

i j k   + 0.45 0 0.06 ( 360 N ) = 0  0 −1 0

TDI   TEH   + ( −32.5j − 182k ) + ( 4.8k )    81   17 

( −8.8i + 28.6 j + 26k ) 

T  + ( −1.2i − 18.9 j + 9.0k )  FG  + ( 0.06i − 0.45k ) ( 360 ) = 0  61  continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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T  T −8.8  DI  − 1.2  FG  81   61

From i-coefficient

  + 0.06 ( 360 ) = 0 

∴ 0.108642TDI + 0.0196721TFG = 21.6

(1)

T  T  From j-coefficient 28.6  DI  − 32.5 − 18.9  FG  = 0 81    61 

∴ 0.35309TDI − 0.30984TFG = 32.5

(2)

From k-coefficient T  T  T  26  DI  − 182 + 4.8  EH  + 9.0  FG  − 0.45 ( 360 ) = 0  81   17   61  ∴ 0.32099TDI + 0.28235TEH + 0.147541TFG = 344

−3.25 × Equation (1) Add Equation (2)

(3)

−0.35309TDI − 0.063935TFG = −70.201 0.35309TDI − 0.30984TFG =

−0.37378TFG

32.5

= −37.701

∴ TFG = 100.864 N TFG = 100.9 N

or Then from Equation (1)

0.108642TDI + 0.0196721(100.864 ) = 21.6 ∴ TDI = 180.554 N

TDI = 180.6 N

or and from Equation (3)

0.32099 (180.554 ) + 0.28235TEH + 0.147541(100.864 ) = 344 ∴ TEH = 960.38 N TEH = 960 N

or The vector forms of the cable forces are:

TDI =

180.554 N ( −65i + 20j − 44k ) 81

= − (144.889 N ) i + ( 44.581 N ) j − ( 98.079 N ) k TEH =

960.38 N ( −15i + 8j) = − (847.39 N ) i + ( 451.94 N ) j 17

100.864 N ( −45i + 20j + 36k ) 61 = − ( 74.409 N ) i + ( 33.070 N ) j + ( 59.527 N ) k

TFG =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Then from free-body diagram of frame

ΣFx = 0: Ax − 144.889 − 847.39 − 74.409 = 0 ∴ Ax = 1066.69 N ΣFy = 0: Ay + 44.581 + 451.94 + 33.070 − 360 − 280 = 0 ∴ Ay = 110.409 N ΣFz = 0: Az − 98.079 + 59.527 + 50 = 0 ∴ Az = −11.448 N

Therefore,

A = (1067 N ) i + (110.4 N ) j − (11.45 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 136.

Free-Body Diagram:

ΣFx = 0:

Bx + C x = 0, or Bx = − C x

(1)

ΣFy = 0:

Ay + By + C y = 0

(2)

ΣFy = 0:

Az − P = 0, or Az = P = 40.0 lb

(3)

ΣM O = 0:

or

rOA × A + rOB × B + rOC × C + M Ai − M C k = 0

(

)

(

)

(

)

ai × Ay j + Azk + bj × Bxi + By j + ck × C x i + C y j + M Ai − M C k = 0 or

a Ayk − a Az j − bBx j + cCx j − cC y i + M Ai − M C k = 0

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Equating the coefficients of the unit vectors to zero:

i : − cC y + M A = 0 Cy = j:

(4)

− a Az + cC x = 0, or using (3) Cx =

k:

MA 36 lb ⋅ ft = = 36.0 lb c 1 ft

a 9 in.

P= ( 40 lb ) = 30.0 lb c 12 in.

(5)

a Ay − b Bx − M C = 0, or, using (1) and (5) b M 6 in.

0 Ay = − P + C = − = − 20.0 lb ( 40 lb ) + c a 12 in.

9 in.

(6)

Finally substituting into (1) and (2) gives: Βx = −30.0 lb

By = − Ay − C y = 20.0 lb − 36.0 lb = −16.00 lb Therefore:

A = − ( 20.0 lb ) j + ( 40.0 lb ) k

B = − ( 30.0 lb ) i − (16.00 lb ) j C = ( 30.0 lb ) i + ( 36.0 lb ) j

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 137.

Free-Body Diagram:

ΣFx = 0:

Bx + C x = 0, or Bx = − C x

(1)

ΣFy = 0:

Ay + By + C y = 0

(2)

ΣFz = 0:

Az − P = 0, or Az = P = 60.0 N

(3)

ΣM O = 0:

rOA × A + rOB × B + rOC × C + M Ai − M C k = 0

(

)

(

)

(

)

or

ai × Ay j + Azk + bj × Bxi + By j + ck × C x i + C y j + M Ai − M C k = 0

or

a Ayk − a Az j − bBx j + cCx j − cC y i + M Ai − M C k = 0 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Equating the coefficients of the unit vectors to zero:

i:

− cC y + M A = 0 Cy =

j:

(4)

− a Az + cC x = 0, or using (3) Cx =

k:

MA 6.3 N ⋅ m = = 35.0 N c 0.180 m

a 0.240 m P= ( 60 N ) = 80.0 N c 0.180 m

(5)

a Ay − b Bx − M C = 0, or, using (1) and (5) b M 0.200 m 13 N ⋅ m Ay = − P + C = − = − 12.50 N ( 60 N ) + c a 0.180 m 0.240 m.

(6)

Finally substituting into (1) and (2) gives: Βx = −80.0 N

By = − Ay − C y = 12.50 N − 35.0 N = −22.5 N Therefore:

A = − (12.50 N ) j + ( 60.0 N ) k

B = − ( 80.0 N ) i − ( 22.5 N ) j

C = ( 80.0 N ) i + ( 35.0 N ) j

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 138.

Free-Body Diagram:

ΣFx = 0:

Bx = 0

ΣM D( x - axis) = 0:

(80 N )( 2.6 m ) − Bz ( 2 m ) = 0 or B = (104.0 N ) k

Bz = 104.000 N

ΣM D ( z - axis) = 0:

C y ( 4 m ) − 144 N ⋅ m = 0

C y = 36.000 N

ΣM D( y - axis) = 0:

− C z ( 4 m ) − (104 N )( 6 m ) + ( 80 N )( 6 m ) = 0 C z = − 36.000 N and C = ( 36.0 N ) j − ( 36.0 N ) k

ΣFy = 0:

Dy + 36 = 0, or Dy = −36.000 N

ΣFz = 0:

Dz − 36 + 104 − 80 = 0, or Dz = 12.000 N

Therefore:

D = − ( 36.0 N ) j + (12.00 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 139.

Free-Body Diagram:

ΣFx = 0:

Bx = 0

ΣM D ( x - axis) = 0:

(80 N )( 2.6 m ) − Bz ( 2 m ) = 0 or B = (104.0 N ) k

Bz = 104.000 N

ΣM D ( z - axis) = 0:

Cy (4 m ) = 0

Cy = 0

ΣM D ( y - axis) = 0:

− C z ( 4 m ) − (104 N )( 6 m ) + ( 80 N )( 6 m ) = 0 C z = − 36.000 N and C = − ( 36.0 N ) k

ΣFy = 0: ΣFz = 0:

Dy = 0

Dz − 36 + 104 − 80 = 0, or Dz = 12.000 N

Therefore:

D = (12.00 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 140.

Free-Body Diagram:

Express the forces in terms of rectangular components:

(

)

W = − ( mg ) j = − ( 3 kg ) 9.81 m/s 2 j = − ( 29.43 N ) j

N B = N B ( 0.8j + 0.6k )

( xB )2 + ( 325 + 75)2 + (100 )2

LAB = 525 mm =

xB = 325 mm

Then,

TBC = TBC

uuur BC = TBC BC

325i + 400 j − 100k

( 325)

2

2

+ ( 400 ) + ( −100 )

2

=

13 16 4 TBC i + TBC j − TBC k 21 21 21 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Equilibrium:

ΣM A = 0:

rG/ A × W + rB/ A × N B + rC/ A × TBC = 0

i j k i j k i j k T or 162.5 − 200 50 + 325 − 400 100 N B + 650 0 0 BC = 0 21 − 29.43 0 0 0 0.8 0.6 13 16 − 4 Equating the coefficients of the unit vectors to zero:

i:

1471.5 − 320 N B = 0 N B = 4.5984 N

or

N B = ( 3.6787 N ) j + ( 2.7590 N ) k

j:

− 195 N B +

2600 TBC = 0 21 TBC = 7.2425 N 13 ( 7.2425 N ) = 0 21 Ax = −4.4835 N

ΣFx = 0:

Ax +

ΣFy = 0:

Ay − 29.43 N + 3.6787 N +

16 ( 7.2425 N ) = 0 21

Ay = 20.233 N

ΣFy = 0:

Az + 2.7590 N −

4 ( 7.2425 N ) = 0 21

Az = −1.37948 N Therefore: (a)

TBC = 7.24 N

(b)

A = − ( 4.48 N ) i + ( 20.2 N ) j − (1.379 Ν ) k N B = ( 3.68 N ) j + ( 2.76 N ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 141.

Free-Body Diagram:

(a) The force acting at E on the free-body diagram of rod AB is perpendicular to AB and CD. Letting λ E = direction cosines for force E, λE =

rB/ A × k rB/ A × k

 − ( 32 in.) i + ( 24 in.) j − ( 40 in.) k  × k = 2 2 ( 32 ) + ( 24 ) in.

= 0.6i + 0.8 j Also,

W = − (10 lb ) j B = Bk

E = E ( 0.6i + 0.8 j) From free-body diagram of rod AB

ΣM A = 0: rG/ A × W + rE/ A × E + rB/ A × B = 0 i j k i j k i j k ∴ −16 12 −20 (10 lb ) + −24 18 −30 E + −32 24 −40 B = 0 0 −1 0 0.6 0.8 0 0 0 1

( −20i + 16k )(10 lb ) + ( 24i − 18 j − 30k ) E + ( 24i + 32 j) B = 0 From k-coefficient

160 − 30 E = 0 ∴ E = 5.3333 lb

and

E = 5.3333 lb ( 0.6i + 0.8 j) E = ( 3.20 lb ) i + ( 4.27 lb ) j

or (b) From j-coefficient

−18 ( 5.3333 lb ) + 32 B = 0 ∴ B = 3.00 lb

or

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

B = ( 3.00 lb ) k

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From free-body diagram of rod AB

ΣF = 0: A + W + E + B = 0 Ax i + Ay j + Az k − (10 lb ) j + ( 3.20 lb ) i + ( 4.27 lb ) j + ( 3.00 lb ) k = 0 From i-coefficient

Ax + 3.20 lb = 0

∴ Ax = −3.20 lb j-coefficient

Ay − 10 lb + 4.27 lb = 0

∴ Ay = 5.73 lb k-coefficient

Az + 3.00 lb = 0

∴ Az = −3.00 lb Therefore

A = − ( 3.20 lb ) i + ( 5.73 lb ) j − ( 3.00 lb ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 142.

Free-Body Diagram:

There is only one unknown of interest and, therefore only one equation is needed:

ΣM AB = 0 Geometry:

 1.05 m   = 16.2602°  3.5 m 

θ = tan −1 

xG = (1.25 m ) cos16.2602° = 1.2 m yG = 1.95 m − (1.25 m ) sin16.2602° = 1.6 m

λ BA

uuur BA = = BA

−3.6i + 1.05j

( −3.6 )2 + (1.05 )2

=−

24 7 i+ j 25 25 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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rK/A = (1.8 m ) i − ( 0.525 m ) j + ( 0.225 m ) k rG/A = (1.2 m ) i − (1.95 m − 1.6 m ) j + ( 0.45 m ) k

= (1.2 m ) i − ( 0.35 m ) j + ( 0.45 m ) k

(

)

W = − ( mg ) j = − ( 25 kg ) 9.81 m/s 2 j = − ( 245.25 kg ) j

PHG = PHG

uuuur HG = PHG HG

−0.05i + 1.6 j − 0.4 k

( −0.05)

2

2

+ (1.6 ) + ( −0.4 )

2

32 8   1 = PHG  − i + j− k 33 33   33

Now,

ΣM BA = 0:

(

)

(

)

λ BA ⋅ rK / A × W + λ BA ⋅ rG/ A × PHG = 0

− 24 7 0 − 24 7 0 1 PHG + 1.2 − 0.35 0.45 =0 1.8 − 0.525 0.225 25 25 )( 33) ( −1 −8 0 − 245.25 0 32

−

1324.35 342.45 + P =0 25 25 ( )( 35) HG

Therefore: PHG = 127.620 N or PHG = 127.6 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 143.

Free-Body Diagram:

There is only one unknown of interest and, therefore only one equation is needed: ΣM AB = 0

Geometry:

 1.05 m   = 16.2602°  3.5 m 

θ = tan −1 

xI = ( 2.50 m ) cos16.2602° = 2.4 m yI = 1.95 m − ( 2.50 m ) sin16.2602° = 1.25 m

λ BA =

uuur BA = BA

−3.6i + 1.05j

( −3.6 )

2

+ (1.05 )

2

=−

24 7 i+ j 25 25 continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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rK/A = (1.8 m ) i − ( 0.525 m ) j + ( 0.225 m ) k rI/A = ( 2.4 m ) i − ( 0.7 m ) j + ( 0.45 m ) k

(

)

W = − ( mg ) j = − ( 25 kg ) 9.81 m/s 2 j = − ( 245.25 kg ) j

PJI

uur JI = PJI = PJI JI

−0.025i + 1.25j − 0.25k

( −0.025)2 + (1.25)2 + ( −0.25)2

50 10   1 = PJI  − i + j + k 51 51   51

Now,

ΣM BA = 0:

(

)

(

)

λ BA ⋅ rK / A × W + λ BA ⋅ rI / A × PJI = 0

− 24 7 0 − 24 7 0 1 PJI + 12.4 − 0.7 0.45 =0 1.8 − 0.525 0.225 25 25 51) ( )( −1 50 −10 0 − 245.25 0 −

1324.35 536.85 + P =0 25 ( 25)( 51) JI

Therefore: PJI = 125.811 N or PJI = 125.8 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 144.

Free-Body Diagram:

Express forces in terms of their rectangular components: uuur BG − 40 i + 74 j − 32 k 37 16   20 TBG = TBG j− k = TBG = TBG  − i + 2 2 2 BG 45 45 45   − + + − 40 74 32 ( ) ( ) ( )

TBH = TBH

uuuur BH = TBH BH

30 i + 60 j − 60 k

( 30 )2 + ( 60 )2 + ( − 60 )2

2 2  1 = TBH  i + j − k  3 3  3

P = − ( 75 lb ) j

λ AD

uuur AD = = AD

80 i − 60 j 2

(80 ) + ( − 60 )

2

= 0.8 i − 0.6 j

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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rB/A = ( 40 in.) i rC/A = ( 80 in.) i Now,

ΣM AD = 0:

(

)

(

)

(

)

λ AD ⋅ rB/ A × TBG + λ AD ⋅ rB/ A × TBH + λ AD ⋅ rC/ A × P = 0

0.8 0 − 0.6 0.8 0 − 0.6 0.8 0 − 0.6 TBG TBH + 40 0 0 + 80 0 40 0 0 0 =0 45 3 − 20 37 −16 1 2 −2 0 − 75 0

−

888 48 TBG − TBH + 3600 = 0 45 3

Noting that TBG = TBH = T and solving:

T = 100.746 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or T = 100.7 lb

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Chapter 4, Solution 145.

Free-Body Diagram:

Express forces in terms of their rectangular components: uuur BG − 40 i + 74 j − 32 k 37 16   20 = TBG = TBG  − i + TBG = TBG j− k 2 2 2 BG 45 45 45   − + + − 40 74 32 ( ) ( ) ( )

P = − ( 75 lb ) j

λ AD =

uuur AD = AD

80 i − 60 j

(80 )

2

+ ( − 60 )

2

= 0.8 i − 0.6 j

rB/ A = ( 40 in.) i rC/ A = ( 80 in.) i Now,

ΣM AD = 0:

(

)

(

)

λ AD ⋅ rB/ A × TBG + λ AD ⋅ rC/ A × P = 0

0.8 0 − 0.6 0.8 0 − 0.6 TBG + 80 0 40 0 0 0 =0 45 − 20 37 −16 0 − 75 0

−

888 TBG + 3600 = 0 45

Solving for TBG :

TBG = 182.432 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or TBG = 182.4 lb

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Chapter 4, Solution 146.

Free-Body Diagram:

Express forces in terms of their rectangular components:

W1 = W2 = − ( 30 lb ) j uuuur ( x − 6) i + y j − 6 k BH T=T =T BH ( x − 6 )2 + y 2 + ( − 6 )2

λ AF =

uuur AF = AF

6i − 3 j − 6k

( 6)

2

2

+ ( − 3) + ( − 6 )

2

=

2 1 2 i− j− k 3 3 3

rG/ A = ( 3 ft ) i rB/ A = ( 6 ft ) i rI / A = ( 6 ft ) i − ( 3 ft ) k continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Now,

(

ΣM AF = 0:

)

(

)

(

)

λ AF ⋅ rG/ A × W1 + λ AF ⋅ rB/ A × T + λ AF ⋅ rI / A × W2 = 0

−1 − 2 2 −1 − 2 1 0 0  + 6 0 0  3 0 − 30 0 ( x − 6) y − 6 2 3

60 + ( −36 − 12 y )

T 3

( x − 6 )2 +

T 3

( x − 6)

2

2

+ y + 36

y 2 + 36

−1 − 2 1 0 −3   = 0 3 0 − 30 0

2 + 6

+ 60 = 0

Solving for T : T =

30 30 + y

( x − 6 )2 +

y 2 + 36

It is thus clear that for a given y, T will have its minimum value when x = 6 ft. Denoting this minimum by Tm:

Tm =

30 3+ y

y 2 + 36

Now to find the minimum of Tm, differentiate Tm with respect to y and equate the derivative to zero.

dTm dy

 1    ( 3 + y ) 36 + y 2  2  = 

(

)

−

1 2

(

( 2 y ) − 36 + y 2

(3 + y )

2

)

1 2



(1) 30 

=0

Setting the numerator equal to zero and simplifying:

( 3 + y ) y − y 2 − 36 = 0 y = 12 ft x = 6 ft, y = 12 ft

(a) Minimum occurs at: (b) Using the expression for T:

Tmin =

30 3 + 12

( 6 − 6 )2 + (12 )2 + 36

= 26.833 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or Tmin = 26.8 lb

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Chapter 4, Solution 147.

Free-Body Diagram:

Express forces in terms of their rectangular components:

W1 = W2 = − ( 30 lb ) j uuuur BH − 6i + y j − 6k =T T=T BH ( −6 )2 + y 2 + ( − 6 )2 uuur AF 6i − 3 j − 6k 2 1 2 = = i− j− k λ AF = AF ( 6 ) 2 + ( − 3)2 + ( − 6 ) 2 3 3 3 rG/ A = ( 3 ft ) i rB/ A = ( 6 ft ) i rI / A = ( 6 ft ) i − ( 3 ft ) k continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Now,

(

ΣM AF = 0:

)

(

)

(

)

λ AF ⋅ rG/ A × W1 + λ AF ⋅ rB/ A × T + λ AF ⋅ rI / A × W2 = 0

2 −1 − 2 2 −1 − 2 1 3 0 0  + 6 0 0 3 −6 y −6 0 − 30 0

60 + ( − 36 − 12 y )

( −6 )2

T 3

( − 6 )2 +

2 −1 − 2 1 + 6 0 −3   = 0  3 + y 2 + 36 0 − 30 0 T 3

y 2 + 36

+ 60 = 0

Solving for T :

T =

30 30 + y

T =

30 3+ y

( −6 )2 +

y 2 + 36

y 2 + 72

Now to find the minimum of T, differentiate T with respect to y and equate the derivative to zero.

dTm dy

 1    ( 3 + y ) 72 + y 2  2  = 

(

)

−

1 2

( 2 y ) − ( 72 +

(3 + y )

2

1 2 2 y

)



(1) 30 

=0

Setting the numerator equal to zero and simplifying:

( 3 + y ) y − y 2 − 72 = 0, or

y = 24 ft x = 0, y = 24 ft

(a) Minimum occurs at: (b) Using the expression for T:

Tmin =

30 3 + 24

( 24 )2 + 72

= 28.284 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or

Tmin = 28.3 lb

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Chapter 4, Solution 148.

Free-Body Diagram:

Express forces in terms of their rectangular components:

(

)

(

)

(

)

WAB = − (1.25 kg/m ) 9.81 m/s 2 ( 0.9 m ) j = − (11.0363 Ν ) j WBC = − (1.25 kg/m ) 9.81 m/s 2 ( 0.3 m ) j = − ( 3.6788 Ν ) j

WCD = − (1.25 kg/m ) 9.81 m/s 2 (1.35 m ) j = − (16.5544 Ν ) j

uuur FE T=T =T FE

−0.6i + 0.9 j − 1.35k

( − 0.6 )

2

2

+ ( 0.9 ) + ( −1.35 )

2

=

T ( −2i + 3j − 4.5k ) 33.25

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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λ AD

uuur AD = = AD

0.9i − 0.3j − 1.35k

( 0.9 )

2

2

+ ( −0.3) + ( −1.35 )

2

=

6 2 9 i− j− k 11 11 11

rG/ A = ( 0.45 m ) i

rF / A = ( 0.6 m ) i rB/ A = ( 0.9 m ) i

rH/A = ( 0.9 m ) i − ( 0.3 m ) j − ( 0.675 m ) k Now,

ΣM AD = 0:

(

)

(

)

(

)

(

)

λ AD ⋅ rG/ A × WAB + λ AD ⋅ rF / A × T + λ AD ⋅ rB/ A × WBC + λ AD ⋅ rH / A × WCD = 0

6 −2 −9 6 − 2 −9 6 −2 −9 T 1 1 0.45 0 0   + 0.6 0 0 0 0   + 0.9 11 11 33.25    11  0 −11.0363 0 0 − 3.6788 0 − 2 3 − 4.5 6 −2 −9 1 + 0.9 − 0.3 − 0.675   = 0  11  0 −16.5544 0

4.0634 − 0.34054T + 2.7089 + 6.0950 = 0

Solving for T:

TBG = 37.785 N

or TBG = 37.8 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 149.

Free-Body Diagram:

Express forces in terms of their rectangular components:

(

)

(

)

(

)

WAB = − (1.25 kg/m ) 9.81 m/s 2 ( 0.9 m ) j = − (11.0363 Ν ) j WBC = − (1.25 kg/m ) 9.81 m/s 2 ( 0.3 m ) j = − ( 3.6788 Ν ) j

WCD = − (1.25 kg/m ) 9.81 m/s 2 (1.35 m ) j = − (16.5544 Ν ) j

uuur CE T=T =T CE

−0.9 i + 1.2 j − 1.35 k

( − 0.9 )

2

2

+ (1.2 ) + ( −1.35 )

2

=

T ( −3 i + 4 j − 4.5 k ) 45.25

continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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λ AD

uuur AD = = AD

0.9 i − 0.3 j − 1.35 k

( 0.9 )

2

2

+ ( − 0.3) + ( −1.35 )

2

=

6 2 9 i− j− k 11 11 11

rG/ A = ( 0.45 m ) i

rC/ A = ( 0.9 m ) i − ( 0.3 m ) j rB/ A = ( 0.9 m ) i

rH/A = ( 0.9 m ) i − ( 0.3 m ) j − ( 0.675 m ) k Now,

ΣM AD = 0:

(

)

(

)

(

)

(

)

λ AD ⋅ rG/ A × WAB + λ AD ⋅ rC/ A × T + λ AD ⋅ rB/ A × WBC + λ AD ⋅ rH / A × WCD = 0

6 −2 −9 6 − 2 −9 6 −2 −9 T 1 1 0.45 0 0   + 0.9 − 0.3 0 0 0   + 0.9 11 11 45.25    11  0 −11.0363 0 0 − 3.6788 0 − 3 4 − 4.5 6 −2 −9 1 + 0.9 − 0.3 − 0.675   = 0  11  0 −16.5544 0

4.0634 − 0.32840T + 2.7089 + 6.0950 = 0

Solving for T:

T = 39.182 N

or T = 39.2 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 150.

Free-Body Diagram:

Express forces in terms of their rectangular components: uuur FG 18 i − 6 j − 9 k T TFG = TFG = TFG = FG ( 6 i − 2 j − 3 k ) 2 2 2 FG 7 (18) + ( − 6 ) + ( − 9 )

λ AB

uuur AB = = AB

ΣM AB = 0:

13.5 i + 9 j − 27 k

(13.5)2 + ( 9 )2 + ( − 27 )2

=

3 2 6 i+ j− k 7 7 7

λ AB ⋅ ( rAE × FE ) + λ AB ⋅ ( rBG × TFG ) = 0 3 2 −6 3 2 −6 T 1 1 −1.5 3 − 9  18 + 13.5 −13 0 FG   = 0 7 7 7 0 1 0 6 − 2 −3

18 ( 9 + 27 ) +

TFG (117 + 162 − 468 + 81) = 0 7

Solving for TFG: TFG = 42.000 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or

TFG = 42.0 lb

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Chapter 4, Solution 151.

Free-Body Diagram:

(a) The location of D follows from the geometry of the problem. Since the steel plate is rectangular rD/ A is perpendicular to rB/ A and therefore:

rD/ A ⋅ rB/ A = 0 Denoting the coordinates of D by (0, y, z): rD/ A = − ( 0.1 m ) i + yj + ( z − 0.7 m ) k

rB/ A = ( 0.3 m ) i − ( 0.4 m ) k

and

Thus, rD/ A ⋅ rB/ A = − 0.03 − 0.4 z + 0.28 = 0 or

z = 0.625 m.

rD/ A =

( − 0.1 m )2 +

2

y 2 + ( 0.625 m − 0.7 m ) = 0.75 m continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Solving for y:

y = 0.73951 m x = 0, y = 0.740 m, z = 0.625 m

Location of D is therefore: (b) Consider moment equilibrium about axis AB:

λ AB

uuur AB = = AB

0.3i − 0.4k

( 0.3)2 + ( − 0.4 )2

= 0.6i − 0.8k

rD/ A = − ( 0.1 m ) i + ( 0.73951 m ) j − ( 0.075 m ) k

rD/B = − ( 0.4 m ) i + ( 0.73951 m ) j + ( 0.625 m − 0.3 m ) k N D = N Di

(

)

W = − ( mg ) j = − ( 40 kg ) 9.81 m/s 2 j = − ( 392.4 N ) j Then,

ΣM AB = 0:

(

)

(

)

λ AB ⋅ rD/ A × N D + λ AB ⋅ rG/B × W = 0

0.6 0 − 0.8 0.6 0 −0.8 − 0.1 0.73951 − 0.075 + − 0.2 0.36976 0.1625 = 0 ND − 392.4 0 0 0 0 0.59161 N D − 24.525 = 0 N D = 41.455 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

or N D = ( 41.455 N ) i

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Chapter 4, Solution 152.

Free-Body Diagram: From free-body diagram of beam

ΣFx = 0: Bx = 0

so that

B = By

ΣFy = 0: A + B − (100 + 200 + 300 ) N = 0 A + B = 600 N

or

Therefore, if either A or B has a magnitude of the maximum of 360 N,

the other support reaction will be < 360 N ( 600 N − 360 N = 240 N ) .

(100 N )( d ) − ( 200 N )( 0.9 − d ) − ( 300 N )(1.8 − d )

ΣM A = 0:

+ B (1.8 − d ) = 0 d =

or

720 − 1.8B 600 − B

Since B ≤ 360 N,

d =

720 − 1.8 ( 360 ) 600 − 360

ΣM B = 0:

= 0.300 m

or

d ≥ 300 mm

(100 N )(1.8) − A (1.8 − d ) + ( 200 N )( 0.9 ) = 0 d =

or

1.8 A − 360 A

Since A ≤ 360 N,

d =

1.8 ( 360 ) − 360 360

= 0.800 m

or

d ≤ 800 mm

or 300 mm ≤ d ≤ 800 mm

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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Chapter 4, Solution 153.

Free-Body Diagram: Cmax = 1000 N

Have

C 2 = C x2 + C y2

Now

∴ Cy =

(1000 )2 − Cx2

(1)

From free-body diagram of pedal ΣFx = 0: C x − Tmax = 0 ∴ C x = Tmax

(2)

ΣM D = 0: C y ( 0.4 m ) − Tmax ( 0.18 m ) sin 60° = 0 ∴ C y = 0.38971Tmax

(3)

Equating the expressions for C y in Equations (1) and (3), with C x = Tmax from Equation (2) 2 (1000 )2 − Tmax

= 0.389711Tmax

2 ∴ Tmax = 868,150

and

Tmax = 931.75 N or Tmax = 932 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 154.

Free-Body Diagram: From free-body diagram of inverted T-member

ΣM C = 0: T ( 25 in.) − T (10 in.) − ( 30 lb )(10 in.) = 0 ∴ T = 20 lb

or T = 20.0 lb ΣFx = 0: C x − 20 lb = 0

∴ C x = 20 lb C x = 20.0 lb

or

ΣFy = 0: C y + 20 lb − 30 lb = 0 ∴ C y = 10 lb C y = 10.00 lb

or Then and

C =

C x2 + C y2 =

( 20 )2 + (10 )2

= 22.361 lb

 Cy  −1  10   = tan   = 26.565°  20   Cx 

θ = tan −1 

or

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

C = 22.4 lb

26.6°

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Chapter 4, Solution 155.

Free-Body Diagram:

From free-body diagram of frame with T = 300 lb

 5 ΣFx = 0: C x − 100 lb +   300 lb = 0  13  ∴ C x = −15.3846 lb

or

C x = 15.3846 lb

 12  ΣFy = 0: C y − 180 lb −   300 lb = 0  13 

∴ C y = 456.92 lb Then and

C =

or

C y = 456.92 lb

(15.3846 )2 + ( 456.92 )2

C x2 + C y2 =

= 457.18 lb

 Cy  −1  456.92   = tan   = −88.072°  −15.3846   Cx 

θ = tan −1 

or C = 457 lb

88.1°

 12   ΣM C = 0: M C + (180 lb )( 20 in.) + (100 lb )(16 in.) −   300 lb  (16 in.) = 0  13  

∴ M C = −769.23 lb ⋅ in. or M C = 769 lb ⋅ in.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 156.

Free-Body Diagram: (a) From free-body diagram of rod AB

ΣM C = 0: P ( l cosθ ) + P ( l sin θ ) − M = 0 or sin θ + cosθ = (b) For

M

Pl

M = 150 lb ⋅ in., P = 20 lb, and l = 6 in. sin θ + cosθ =

150 lb ⋅ in.

5 = = 1.25 20 lb 6 in. ( )( ) 4 sin 2 θ + cos 2 θ = 1

Using identity

(

sin θ + 1 − sin 2 θ

(1 − sin θ ) 2

1 2

)

1 2

= 1.25

= 1.25 − sin θ

1 − sin 2 θ = 1.5625 − 2.5sin θ + sin 2 θ 2sin 2 θ − 2.5sin θ + 0.5625 = 0

Using quadratic formula

sin θ = = or

− ( −2.5) ±

( 6.25) − 4 ( 2 )( 0.5625) 2 ( 2)

2.5 ± 1.75 4

sin θ = 0.95572 ∴ θ = 72.886°

and

sin θ = 0.29428

θ = 17.1144°

and

or θ = 17.11° and θ = 72.9°

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 157.

From geometry of forces

 yBE    1.5 ft 

β = tan −1  where

yBE = 2.0 − yDE = 2.0 − 1.5 tan 35° = 0.94969 ft

 0.94969  ∴ β = tan −1   = 32.339°  1.5  and

α = 90° − β = 90° − 32.339° = 57.661° θ = β + 35° = 32.339° + 35° = 67.339°

Applying the law of sines to the force triangle,

200 lb T B = = sin θ sin α sin 55° or

(a)

( 200 lb ) sin 67.339°

T =

=

T B = sin 57.661° sin 55°

( 200 lb )( sin 57.661° ) sin 67.339°

= 183.116 lb or T = 183.1 lb

(b)

B=

( 200 lb )( sin 55° ) sin 67.339°

= 177.536 lb or B = 177.5 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

32.3°

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Chapter 4, Solution 158.

Free-Body Diagram: yED = xED = a,

Since Slope of ED is

45°

∴ slope of HC is

45°

DE =

Also and

2a

a 1 DH = HE =   DE = 2 2

For triangles DHC and EHC

sin β =

a = 25 mm and sin β =

R = 125 mm

25 mm = 0.141421 2 (125 mm )

∴ β = 8.1301° and

a 2R

c = R sin ( 45° − β )

Now For

a/ 2 = R

or β = 8.13°

c = (125 in.) sin ( 45° − 8.1301° ) = 75.00 in. or c = 75.0 in.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 159.

Free-Body Diagram:

First note

(

)

W = mg = (17 kg ) 9.81 m/s 2 = 166.77 N

h=

(1.2 )2 − (1.125)2

= 0.41758 m

From free-body diagram of plywood sheet  (1.125 m )  ΣM z = 0: C ( h ) − W   =0 2  

C ( 0.41758 m ) − (166.77 N ) ( 0.5625 m ) = 0 ∴ C = 224.65 N

or

C = − ( 225 N ) i

ΣM B( y -axis ) = 0: − ( 224.65 N ) ( 0.6 m ) + Ax (1.2 m ) = 0

or

∴ Ax = 112.324 N

A x = (112.3 N ) i

ΣM B( x-axis ) = 0: (166.77 N ) ( 0.3 m ) − Ay (1.2 m ) = 0 ∴ Ay = 41.693 N

or

A y = ( 41.7 N ) j

ΣM A( y -axis ) = 0: ( 224.65 N ) ( 0.6 m ) − Bx (1.2 m ) = 0 ∴ Bx = 112.325 N

or

B x = (112.3 N ) i

ΣM A( x-axis ) = 0: B y (1.2 m ) − (166.77 N ) ( 0.9 m ) = 0 ∴ By = 125.078 N

or

B y = (125.1 N ) j

∴ A = (112.3 N ) i + ( 41.7 N ) j

B = (112.3 N ) i + (125.1 N ) j C = − ( 225 N ) i

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 160.

Free-Body Diagram:

First note

(

)

W = mg = ( 30 kg ) 9.81 m/s 2 = 294.3 N

FEC = λ EC FEC = ( sin15° ) i + ( cos15° ) j FEC From free-body diagram of cover (a)

ΣM z = 0:

( FEC cos15° ) (1.0 m ) − W ( 0.5 m ) = 0 or FEC cos15° (1.0 m ) − ( 294.3 N )( 0.5 m ) = 0 ∴ FEC = 152.341 N

(b)

or FEC = 152.3 N

ΣM x = 0: W ( 0.4 m ) − Ay ( 0.8 m ) − ( FEC cos15° ) ( 0.8 m ) = 0 or ( 294.3 N )( 0.4 m ) − Ay ( 0.8 m ) − (152.341 N ) cos15° ( 0.8 m ) = 0

∴ Ay = 0 ΣM y = 0: Ax ( 0.8 m ) + ( FEC sin15° ) ( 0.8 m ) = 0

or Ax ( 0.8 m ) + (152.341 N ) sin15° ( 0.8 m ) = 0

∴ Ax = −39.429 N ΣFx = 0: Ax + Bx + FEC sin15° = 0 −39.429 N + Bx + (152.341 N ) sin15° = 0 ∴ Bx = 0 ΣFy = 0: FEC cos15° − W + By = 0 or (152.341 N ) cos15° − 294.3 N + B y = 0

∴ By = 147.180 N

or A = − ( 39.4 N ) i B = (147.2 N ) j

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 161.

Free-Body Diagram:

λCD =

First note

=

λCE = =

− ( 23 in.) i + ( 22.5 in.) j − (15 in.) k 35.5 in.

1 ( −23i + 22.5j − 15k ) 35.5

( 9 in.) i + ( 22.5 in.) j − (15 in.) k 28.5 in.

1 ( 9i + 22.5 j − 15k ) 28.5

W = − ( 285 lb ) j From free-body diagram of plate (a)

ΣM x = 0:

 22.5    22.5    T  (15 in.) −   T  (15 in.) = 0  35.5    28.5  

( 285 lb )( 7.5 in.) − 

∴ T = 100.121 lb

or T = 100.1 lb continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

 23   9  ΣFx = 0: Ax − T   +T =0 35.5    28.5 

(b)

 23  (  9  Ax − (100.121 lb )   + 100.121 lb )  =0 35.5    28.5  ∴ Ax = 33.250 lb

  22.5   ΣM B( z -axis ) = 0: − Ay ( 26 in.) + W (13 in.) − T    ( 6 in.) −   35.5   or

  22.5     ( 6 in.) = 0 T    28.5  

  22.5   − Ay ( 26 in.) + ( 285 lb )(13 in.) − (100.121 lb )    ( 6 in.)  35.5      22.5   − (100.121 lb )    ( 6 in.) = 0  28.5   

∴ Ay = 109.615 lb   15   ΣM B( y -axis ) = 0: Az ( 26 in.) − T    ( 6 in.) −   35.5  

  23     (15 in.) T    35.5  

  15     9  − T    ( 6 in.) + T    (15 in.) = 0   28.5     28.5   or

1  −1 Az ( 26 in.) +  ( 90 + 345) − ( 90 − 135) (100.121 lb ) = 0 28.5  35.5  ∴ Az = 41.106 lb

or A = ( 33.3 lb ) i + (109.6 lb ) j + ( 41.1 lb ) k  22.5   22.5  ΣFy = 0: By − W + T   +T  + Ay = 0  35.5   28.5   22.5 22.5  B y − 285 lb + (100.121 lb )  +  + 109.615 lb = 0  35.5 28.5  ∴ By = 32.885 lb

 15   15  ΣFz = 0: Bz + Az − T   −T =0  35.5   28.5  15   15 Bz + 41.106 lb − (100.121 lb )  + =0 35.5 28.5   ∴ Bz = 53.894 lb

or B = ( 32.9 lb ) j + ( 53.9 lb ) k

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 4, Solution 162.

First note

Free-Body Diagram:

TBG = λBGTBG =

− (18 in.) i + (13.5 in.) k 2

2

(18) + (13.5) in.

TBG

= TBG ( −0.8i + 0.6k )

TDH = λDH TDH =

− (18 in.) i + ( 24 in.) j

(18)2 + ( 24 )2 in.

TDH

= TDH ( −0.6i + 0.8j) Since λFJ = λDH ,

TFJ = TFJ ( −0.6i + 0.8j) From free-body diagram of member ABF

ΣM A( x-axis ) = 0:

( 0.8TFJ ) ( 48 in.) + ( 0.8TDH )( 24 in.) − (120 lb )( 36 in.) − (120 lb )(12 in.) = 0 ∴ 3.2TFJ + 1.6TDH = 480

ΣM A( z -axis ) = 0:

(1)

( 0.8TFJ ) (18 in.) + ( 0.8TDH )(18 in.) − (120 lb )(18 in.) − (120 lb )(18 in.) = 0 ∴

− 3.2TFJ − 3.2TDH = −960

(2)

Equation (1) + Equation (2) Substituting in Equation (1)

ΣM A( y -axis ) = 0:

TDH = 300 lb

TFJ = 0

( 0.6TFJ ) ( 48 in.) + 0.6 ( 300 lb ) ( 24 in.) − ( 0.6TBG ) (18 in.) = 0 ∴ TBG = 400 lb continued

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

ΣFx = 0: − 0.6TFJ − 0.6TDH − 0.8TBG + Ax = 0

−0.6 ( 300 lb ) − 0.8 ( 400 lb ) + Ax = 0 ∴ Ax = 500 lb ΣFy = 0: 0.8TFJ + 0.8TDH − 240 lb + Ay = 0 0.8 ( 300 lb ) − 240 + Ay = 0

∴ Ay = 0 ΣFz = 0: 0.6TBG + Az = 0

0.6 ( 400 lb ) + Az = 0 ∴ Az = −240 lb Therefore,

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

A = ( 500 lb ) i − ( 240 lb ) k

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Chapter 4, Solution 163.

Free-Body Diagram: First note

λ AE =

− ( 70 mm ) i + ( 240 mm ) k

( 70 )

2

=

2

+ ( 240 ) mm

1 ( −7i + 24k ) 25

rC/ A = ( 90 mm ) i + (100 mm ) k FC = − ( 600 N ) j rD/ A = ( 90 mm ) i + ( 240 mm ) k T = λDF T =

=

− (160 mm ) i + (110 mm ) j − ( 80 mm ) k

(160 )2 + (110 )2 + (80 )2

mm

T

T ( −16 i + 11j − 8k ) 21

From the free-body diagram of the bend rod

(

)

(

)

ΣM AE = 0: λ AE ⋅ rC/ A × FC + λ AE ⋅ rD/ A × T = 0 ∴

−7 0 24 −7 0 24  600   T  90 0 100  =0  + 90 0 240  25 ( 21)   25   0 −1 0 −16 11 −8 600   T   + (18 480 + 23 760 )   =0  25   25 ( 21) 

( −700 − 2160 ) 

∴ T = 853.13 N

or T = 853 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.