Solution manual vector mechanics engineers dynamics 8th beer chapter 10

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Chapter 10, Solution 1.

Assume δθ clockwise
Then, for point C

δ xC = (125 mm ) δθ
and for point D

δ xD = δ xC = (125 mm ) δθ
and for point E

 250 mm 
2
 δ xD = δ xD
3
 375 mm 

δ xE = 

δ xD = ( 375 mm ) δφ

Thus

(125 mm ) δθ = ( 375 mm ) δφ
1
3

δφ = δθ

(

)

 100

δ G = 100 2 mm δφ = 
 3

Then

2 mm  δθ

 100

 100

2 mm  δθ cos 45° = 
mm  δθ
 3

 3

δ yG = δ G cos 45° = 
Virtual Work:

δ U = 0:

Assume P acts downward at G

( 9000 N ⋅ mm ) δθ − (180 N )(δ xE

mm ) + P (δ yG mm ) = 0

2

 100 
δθ  = 0
9000 δθ − 180  × 125 δθ  + P 
3

 3

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

P = 180.0 N

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Chapter 10, Solution 2.

FA = 20 lb at A

Have

FD = 30 lb

δ y A = (16 in.) δθ

at D

δ yF = δ yB

δ yB = (10 in.) δθ
δ yF = ( 6 in.) δφ = (10 in.) δθ

or

5
3

δφ = δθ
δ yG = (12 in.) δφ = ( 20 in.) δθ
d ED =

( 5.5 in.)2 + ( 4.8 in.)2
5

= 7.3 in.

δ D = ( 7.3 in.) δφ =  × 7.3 in.  δθ
3

δ xD =
Virtual Work:

4.8
4.8  5

δD =
 × 7.3 in.  δθ = ( 8 in.) δθ
7.3
7.3  3

Assume P acts upward at G

δ U = 0:

FAδ y A + FDδ xD + Pδ yG = 0

or

( 20 lb ) (16 in.) δθ  + ( 30 lb ) (8 in.) δθ  + P ( 20 in.) δθ  = 0

or

P = − 28.0 lb
P = 28.0 lb W

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

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Chapter 10, Solution 3.

Assume δθ clockwise
Then, for point C

δ xC = (125 mm ) δθ
and for point D

δ xD = δ xC = (125 mm ) δθ
and for point E

 250 mm 
2
 δ xD = δ xD
3
 375 mm 

δ xE = 

δ xD = ( 375 mm ) δφ

Thus

(125 mm ) δθ = ( 375 mm ) δφ
or

Virtual Work:

δ U = 0:

1
3

δφ = δθ
Assume M acts clockwise on link DEFG

( 9000 N ⋅ mm ) δθ − (180 N )(δ xE

mm ) + M δφ = 0

2

1 
9000 δθ − 180  ⋅125 δθ  + M  δθ  = 0
3

3 
or

M = 18000 N ⋅ mm

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

M = 18.00 N ⋅ m

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Chapter 10, Solution 4.

FA = 20 lb at A

Have

at D

FD = 30 lb

δ y A = (16 in.) δθ

δ yF = δ yB

δ yB = (10 in.) δθ

δ yF = ( 6 in.) δφ = (10 in.) δθ
or

d ED =

5
3

δφ = δθ

( 5.5 in.)2 + ( 4.8 in.)2
5
3

= 7.3 in.

δ D = ( 7.3 in.) δφ =  × 7.3 in.  δθ
δ xD =
Virtual Work:

4.8
4.8  5

δD =
 × 7.3 in.  δθ = ( 8 in.) δθ
7.3
7.3  3

Assume M acts

on DEFG

δ U = 0:

FAδ y A + FDδ xD + M δφ = 0

or

( 20 lb ) (16 in.) δθ  + ( 30 lb ) (8 in.) δθ  + M 

or

M = − 336.0 lb ⋅ in.

5 
δθ  = 0
3 
M = 28.0 lb ⋅ ft

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

W

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Chapter 10, Solution 5.

Assume δθ

δ x A = 10 δθ in.

δ yC = 4 δθ in.
δ yD = δ yC = 4 δθ in.

δφ =

δ yD
6

=

2
δθ
3
2

δ xG = 15 δφ = 15  δθ  = 10 δθ in.
3
Virtual Work:

δ U = 0:

Assume that force P is applied at A.

δ U = − Pδ x A + 30 δ yC + 60 δ yD + 240 δφ + 80 δ xG = 0
2 
− P (10 δθ in.) + ( 30 lb )( 4 δθ in.) + ( 60 lb )( 4 δθ ) + ( 240 lb ⋅ in.)  δθ 
3 
+ ( 80 lb )(10 δθ in.) = 0
−10P + 120 + 240 + 160 + 800 = 0
10P = 1320

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

P = 132.0 lb

W

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Chapter 10, Solution 6.

Note:

(a)

xE = 2 x D

δ xE = 2 δ xD

xG = 3xD

δ xG = 3δ xD

x H = 4 xD

δ xH = 4 δ xD

xI = 5 x D

δ x I = 5 δ xD

Virtual Work:

δ U = 0: FG δ xG − FSPδ xI = 0

( 90 N )( 3δ xD ) − FSP ( 5δ xD ) = 0
or
Now

FSP = 54.0 N W

FSP = k ∆xI
54 N = ( 720 N/m ) ∆xI
∆xI = 0.075 m

and

1
4

1
5

δ xD = δ x H = δ x I
∆xH =

4
4
∆xI = ( 0.075 m ) = 0.06 m
5
5
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

∆xH = 60.0 mm

W

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(b)

Virtual Work:

δ U = 0: FG δ xG + FH δ xH − FSP (δ xI ) = 0

( 90 N )( 3 δ xD ) + ( 90 N )( 4 δ xD ) − FSP ( 5δ xD ) = 0
or
Now

FSP = 126.0 N W

FSP = k ∆xI

126.0 N = ( 720 N/m ) ∆xI
∆xI = 0.175 m
From Part (a)

∆xH =

4
4
∆xI = ( 0.175 m ) = 0.140 m
5
5
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

∆xH = 140.0 mm

W

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Chapter 10, Solution 7.

Note:

(a)

xE = 2 x D

δ xE = 2 δ xD

xG = 3xD

δ xG = 3δ xD

xH = 4 xD

δ xH = 4 δ xD

xI = 5 xD

δ xI = 5 δ x D

Virtual Work:

δ U = 0: FE δ xE − FSP δ xI = 0

( 90 N )( 2 δ xD ) − FSP ( 5δ xD ) = 0
or
Now

FSP = 36.0 N

FSP = k ∆xI

36 N = ( 720 N/m ) ∆xI
∆xI = 0.050 m
and

1
4

1
5

δ xD = δ xH = δ xI
∆xH =

4
4
∆xI = ( 0.050 m ) = 0.04 m
5
5
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

∆xH = 40.0 mm

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(b)

Virtual Work:

δ U = 0: FD δ xD + FEδ xE − FSPδ xI = 0

( 90 N ) δ xD + ( 90 N )( 2 δ xD ) − FSP ( 5 δ xD ) = 0
or
Now

FSP = 54.0 N

FSP = k ∆xI

54 N = ( 720 N/m ) ∆xI
∆xI = 0.075 m

From Part (a)

∆xH =

4
4
∆xI = ( 0.075 ) = 0.06 m
5
5
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

∆xH = 60.0 mm

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Chapter 10, Solution 8.

Assume δ y A

δ yA
16 in.

=

δ yC
8 in.

1
2

δ yC = δ y A

;

Bar CFDE moves in translation
1
2

δ yE = δ yF = δ yC = δ y A
Virtual Work:

δ U = 0: − P (δ y A in.) + (100 lb )(δ yE in.) + (150 lb )(δ yF in.) = 0
1

1

− P δ y A + 100  δ y A  + 150  δ y A  = 0
2

2

P = 125 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

P = 125 lb W

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Chapter 10, Solution 9.

Have

y A = 2l cosθ ;

θ

CD = 2l sin ;
2

δ y A = −2l sin θ δθ
θ
δ ( CD ) = l cos δθ
2

Virtual Work:

δ U = 0: − Pδ y A − Qδ ( CD ) = 0

θ 

− P ( −2l sin θ δθ ) − Q  l cos δθ  = 0
2 

Q = 2P

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

sin θ
W
θ 
cos  
2

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Chapter 10, Solution 10.

Virtual Work:
Have

x A = 2l sin θ

δ x A = 2l cosθ δθ
and

yF = 3l cosθ

δ yF = −3l sin θ δθ
Virtual Work:

δ U = 0: Qδ x A + Pδ yF = 0
Q ( 2l cosθ δθ ) + P ( −3l sin θ δθ ) = 0

Q=

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

3
P tan θ
2

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Chapter 10, Solution 11.

Virtual Work:
We note that the virtual work of Ax , Ay and C is zero, since A is fixed and C is ⊥ to δ xC .

Thus:

δ U = 0:

Pδ xD + Qδ yD = 0

xD = 3l cosθ

δ xD = − 3l sin θ δθ

yD = l sin θ

δ yD = l cosθ δθ

P ( − 3l sin θ δθ ) + Q ( l cosθ δθ ) = 0

− 3Pl sin θ + Ql cosθ = 0
Q=

3P sin θ
= 3P tan θ
cosθ

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

Q = 3P tan θ W

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Chapter 10, Solution 12.

x A = ( a + b ) cosθ

δ x A = − ( a + b ) sin θ δθ

yG = a sin θ

δ yG = a cosθ δθ

Virtual Work:
The reactions at A and B are perpendicular to the displacements of A and B hence do no work.

δ U = 0:

T δ x A + W δ yG = 0
T  − ( a + b ) sin θ δθ  + W ( a cosθ δθ ) = 0

− T ( a + b ) sin θ + Wa cosθ = 0
T =

a
W cot θ
a+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

T =

a
W cot θ W
a+b

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Chapter 10, Solution 13.

yH = 2l sin θ

Note:

l = 600 mm ( length of a link )

Where
Then

δ yH = 2l cosθ δθ

Also

1
1
1
W = mg = ( 450 kg ) 9.81 m/s 2
2
2
2

(

)

= 2207.3 N
2

d AF

3

5

=  l cosθ  +  l sin θ 
4

4

=

δ d AF =

l
9 + 16sin 2 θ
4
l 16sin θ cosθ
δθ
4 9 + 16sin 2 θ

= 4l

Virtual Work:

2

sin θ cosθ
9 + 16sin 2 θ

δθ

1 
Fcyl δ d AF −  W  δ yH = 0
2 

δ U = 0:

sin θ cosθ
Fcyl  4l
δθ  − ( 2.2073 kN )( 2l cosθ δθ ) = 0

9 + 16sin 2 θ

Fcyl
For θ = 30°

sin θ
9 + 16sin 2 θ

Fcyl

= 1.10365 kN

sin 30°
9 + 16sin 2 30°

= 1.10365 kN
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

Fcyl = 7.96 kN W

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Chapter 10, Solution 14.

From solution of Problem 10.13:
Fcyl

sin θ
9 + 16sin 2 θ

= 1.10365 kN

Fcyl = 35 kN

Then for

( 35 kN )

sin θ
9 + 16sin 2 θ

= 1.10365 kN

( 31.713 sin θ )2 = 9 + 16sin 2 θ
sin 2 θ =

9
989.71
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

θ = 5.47° W

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Chapter 10, Solution 15.

yB = a sin θ ⇒ δ yB = a cosθδθ

ABC:

yC = 2a sin θ ⇒ δ yC = 2a cosθδθ
CDE: Note that as ABC rotates counterclockwise, CDE rotates clockwise
while it moves to the left.

δ yC = aδφ

Then

2a cosθδθ = aδφ

or

δφ = 2 cosθδθ

or
Virtual Work:

δ U = 0: − Pδ yB − Pδ yC + M δφ = 0
− P ( a cosθδθ ) − P ( 2a cosθδθ ) + M ( 2cosθδθ ) = 0
or 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

3
Pa W
2

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Chapter 10, Solution 16.

First note l sin θ +

3
l sin φ = l
2

sin φ =

2
(1 − sin θ )
3

or
Then

2
cos φ δφ = − cosθ δθ
3

or

δφ = −

2 cosθ
δθ
3 cos φ
=

Now

xC = − l cosθ +

Then

δ xC = l sin θ δθ −

5 + 8sin θ − 4sin 2 θ

3
l cos φ
2
3
l sin φ δφ
2

 − 2 cosθ  
3
= l sin θ − sin φ 
  δθ
2
 3 cos φ  

= l ( sin θ + cosθ tan φ ) δθ

= l sin θ +


2cosθ (1 − sin θ )

 δθ
5 + 8sin θ − 4sin θ 
2

Virtual Work:

δ U = 0: M δθ − Pδ xC = 0

M δθ − Pl sin θ +


2 cosθ (1 − sin θ )

 δθ = 0
5 + 8sin θ − 4sin 2 θ 

or M = Pl sin θ +


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

2cosθ (1 − sin θ )

5 + 8sin θ − 4sin 2 θ 

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Chapter 10, Solution 17.

Have
xB = l sin θ

δ xB = l cosθδθ
y A = l cosθ

δ y A = − l sin θδθ
Virtual Work:

δ U = 0: M δθ − Pδ xB + Pδ y A = 0
M δθ − P ( l cosθδθ ) + P ( − l sin θδθ ) = 0
M = Pl ( sin θ + cosθ )

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

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Chapter 10, Solution 18.

xD = l cosθ

Have

δ xD = − l sin θδθ
yD = 3l sin θ

δ yD = 3l cosθδθ
Virtual Work:

δ U = 0: M δθ − ( P cos β ) δ xD − ( P sin β ) δ yD = 0

M δθ − ( P cos β )( − l sin θδθ ) − ( P sin β )( 3l cosθδθ ) = 0
M = Pl ( 3sin β cosθ − cos β sin θ )

(1)

(a) For P directed along BCD, β = θ
Equation (1):

M = Pl ( 3sin θ cosθ − cosθ sin θ )
M = Pl ( 2sin θ cosθ )

M = Pl sin 2θ

(b) For P directed , β = 90°
Equation (1):

M = Pl ( 3sin 90° cosθ − cos 90° sin θ )
M = 3Pl cosθ

(c) For P directed
Equation (1):

, β = 180°
M = Pl ( 3sin180° cosθ − cos180° sin θ )

M = Pl sin θ

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

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Chapter 10, Solution 19.

Analysis of the geometry:

Law of Sines

sin φ
sin θ
=
AB
BC
sin φ =

AB
sin θ
BC

(1)

Now

xC = AB cosθ + BC cos φ

δ xC = − AB sin θδθ − BC sin φδφ
cos φδφ =

Now, from Equation (1)

δφ =

or

(2)

AB
cosθδθ
BC

AB cosθ
δθ
BC cos φ

(3)

From Equation (2)

 AB cosθ

δθ 
 BC cos φ

δ xC = − AB sin θδθ − BC sin φ 
or

Then

δ xC = −

AB
( sin θ cos φ + sin φ cosθ ) δθ
cos φ

δ xC = −

AB sin (θ + φ )
δθ
cos φ
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

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δ U = 0: − Pδ xC − M δθ = 0

Virtual Work:

 AB sin (θ + φ ) 
δθ  − M δθ = 0
−P −
cos φ

M = AB

Thus,

sin (θ + φ )
P
cos φ

(4)

For the given conditions: P = 1.0 kip = 1000 lb, AB = 2.5 in., and BC = 10 in.:
(a) When

θ = 30°: sin φ =
M = ( 2.5 in.)

2.5
sin 30°,
10

sin ( 30° + 7.181° )
cos 7.181°

φ = 7.181°

(1.0 kip ) = 1.5228 kip ⋅ in.
= 0.1269 kip ⋅ ft
or M = 126.9 lb ⋅ ft

(b) When

θ = 150°: sin φ =
M = ( 2.5 in.)

2.5
sin150°,
10

sin (150° + 7.181° )
cos 7.181°

φ = 7.181°

(1.0 kip ) = 0.97722 kip ⋅ in.
or M = 81.4 lb ⋅ ft

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

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Chapter 10, Solution 20.

M = AB

From the analysis of Problem 10.19,

sin (θ + φ )
P
cos φ

Now, with M = 75 lb ⋅ ft = 900 lb ⋅ in.
(a) For θ = 60°

sin φ =

2.5
sin 60°,
10

( 900 lb ⋅ in.) = ( 2.5 in.)

φ = 12.504°

sin ( 60° + 12.504° )
cos12.504°

( P)

P = 368.5 lb

or

P = 369 lb

(b) For θ = 120°

sin φ =

2.5
sin120°,
10

( 900 lb ⋅ in.) = ( 2.5 in.)
or

φ = 12.504°

sin (120° + 12.504° )
cos12.504°

( P)

P = 476.7 lb
P = 477 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

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Chapter 10, Solution 21.

Consider a virtual rotation δφ

Then δ B = aδφ
Note that δ yB = δ B cosθ = a cosθδφ
Disregarding the second-order rotation of link BC,

δ yC = δ yB = a cosθδφ
Then δ C =

Virtual Work: δ U = 0:

δ yC
a cosθδφ
a
=
=
δφ
sin θ
sin θ
tan θ

M δφ − Pδ C = 0

 a

δφ  = 0
M δφ − P 
 tan θ

or M tan θ = Pa
Thus ( 27 N ⋅ m ) tan30° = P ( 0.45 m )
P = 34.6 N
P = 34.6 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

30.0°

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Chapter 10, Solution 22.

Consider a virtual rotation δφ

Then δ B = a δφ
Note that δ yB = δ B cosθ = a cosθ δφ
Disregarding the second-order rotation of link BC,

δ yC = δ yB = a cosθ δφ
Then δ C =

Virtual Work: δ U = 0:

δ yC
a cosθ δφ
a
=
=
δφ
sin θ
sin θ
tan θ

M δφ − Pδ C = 0
 a

δφ  = 0
M δφ − P 
 tan θ

or M tan θ = Pa
Thus M tan 40° = (135 N )( 0.60 m )
M = 96.53 N ⋅ m
M = 96.5 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