Solution manual vector mechanics engineers dynamics 8th beer chapter 04

COSMOS: Complete Online Solutions Manual Organization System

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

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.

or B y = 570 lb


COSMOS: Complete Online Solutions Manual Organization System

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


COSMOS: Complete Online Solutions Manual Organization System

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


COSMOS: Complete Online Solutions Manual Organization System

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


COSMOS: Complete Online Solutions Manual Organization System

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.


COSMOS: Complete Online Solutions Manual Organization System

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

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

or B = 15.35 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


COSMOS: Complete Online Solutions Manual Organization System

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

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.

or B = 14.72 kN


COSMOS: Complete Online Solutions Manual Organization System

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


COSMOS: Complete Online Solutions Manual Organization System

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


COSMOS: Complete Online Solutions Manual Organization System

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


COSMOS: Complete Online Solutions Manual Organization System

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


COSMOS: Complete Online Solutions Manual Organization System

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


COSMOS: Complete Online Solutions Manual Organization System

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.


COSMOS: Complete Online Solutions Manual Organization System

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°


COSMOS: Complete Online Solutions Manual Organization System

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°


COSMOS: Complete Online Solutions Manual Organization System

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°


COSMOS: Complete Online Solutions Manual Organization System

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
(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
∴ A = 275 N

, 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°


COSMOS: Complete Online Solutions Manual Organization System

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°


COSMOS: Complete Online Solutions Manual Organization System

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.


COSMOS: Complete Online Solutions Manual Organization System

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°


COSMOS: Complete Online Solutions Manual Organization System

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°


COSMOS: Complete Online Solutions Manual Organization System

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°


COSMOS: Complete Online Solutions Manual Organization System

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°