Cap 10 by Cristian Manrique

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Chapter 10, Solution 1.

Link ABC: 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 =  Link DEFG:

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 © 2007 The McGraw-Hill Companies.

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

Link ABC:

Link BF:

at D

δ yF = δ yB

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

Link DEFG: 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

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

P = − 28.0 lb P = 28.0 lb W

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Chapter 10, Solution 3.

Link ABC: 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 =  Link DEFG:

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

M = 18.00 N ⋅ m

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Chapter 10, Solution 4.

FA = 20 lb at A

Have

at D

FD = 30 lb

Link ABC:

δ y A = (16 in.) δθ

Link BF:

δ yF = δ yB

δ yB = (10 in.) δθ Link DEFG:

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

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

M = − 336.0 lb ⋅ in.

5  δθ  = 0 3  M = 28.0 lb ⋅ ft

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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

P = 132.0 lb

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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

∆xH = 60.0 mm

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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

∆xH = 140.0 mm

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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

∆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

∆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

P = 125 lb W

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Chapter 10, Solution 9.

Have

y A = 2l cosθ ;

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

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

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

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

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:

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

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

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

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 =

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

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 + 

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 )

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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

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

M = Pl sin θ

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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 =

(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

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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

(b) When

θ = 150°: sin φ = M = ( 2.5 in.)

2.5 sin150°, 10

sin (150° + 7.181° ) cos 7.181°

or M = 126.9 lb ⋅ ft

or M = 81.4 lb ⋅ ft

φ = 7.181°

(1.0 kip ) = 0.97722 kip ⋅ in.

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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

P = 369 lb

P = 477 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

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Chapter 10, Solution 21.

Consider a virtual rotation δφ

of link AB.

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

30.0°

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Chapter 10, Solution 22.

Consider a virtual rotation δφ

of link AB.

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

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Chapter 10, Solution 23.

From geometry

y A = 2l cosθ ,

CD = 2l sin

θ 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   sin θ θ  cos   2

Q = 2P

With

P = 60 lb,

( 75 lb ) = 2 ( 60 lb )

Q = 75 lb sin θ θ  cos   2

sin θ = 0.625 θ  cos   2

θ  θ  2sin   cos   2  2  = 0.625 θ  cos   2

θ = 36.42° θ = 36.4° (Additional solutions discarded as not applicable are θ = ± 180°)

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Chapter 10, Solution 24.

From the solution to Problem 10.16  M = Pl sin θ + 

2cosθ (1 − sin θ )

  5 + 8sin θ − 4sin 2 θ 

Substituting  13.5 N ⋅ m = ( 60 N )( 0.25 m ) sin θ +  or sin θ +

2cosθ (1 − sin θ ) 5 + 8sin θ − 4sin 2 θ

2 cosθ (1 − sin θ )

  5 + 8sin θ − 4sin θ  2

= 0.90

Solving numerically

θ = 57.5°

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Chapter 10, Solution 25.

OC = r

Geometry

cosθ =

xB =

δ xB =

OC r = OB xB

r cosθ r sin θ δθ cos 2 θ

y A = l cosθ ;

δ y A = − l sin θ δθ

Virtual Work:

δ U = 0: P ( − δ y A ) − Qδ xB = 0 Pl sin θ δθ − Q cos 2 θ =

r sin θ δθ = 0 cos 2 θ

Qr Pl

(1)

Then, with l = 15 in., r = 4.5 in., P = 15 lb, and Q = 30 lb

cos 2 θ = or

( 30 lb )( 4.5 in.) (15 lb )(15 in.)

θ = 39.231°

= 0.6

θ = 39.2°

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Chapter 10, Solution 26.

OC = r

Geometry

cosθ =

xB =

δ xB = y A = l cosθ ;

OC r = OB xB

r cosθ r sin θ δθ cos 2 θ

δ y A = − l sin θ δθ

Virtual Work:

δ U = 0: P ( − δ y A ) − Qδ xB = 0 Pl sin θ δθ − Q cos 2 θ =

r sin θ δθ = 0 cos 2 θ

Qr Pl

(1)

Then, with l = 14 in., r = 5 in., P = 75 lb, and Q = 150 lb

cos 2 θ = or

(150 lb )( 5 in.) ( 75 lb )(14 in.)

θ = 32.3115°

= 0.7143

θ = 32.3°

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Chapter 10, Solution 27.

We have

x A = ( a + b ) cosθ

δ x A = − ( a + b ) sin θ δθ

yG = a sin θ

δ y A = 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 or We have

T =

a W cot θ a+b

sin θ =

( 42 in.) BD BD = = AB a + b ( 42 in.) + ( 28 in.)

sin θ = 0.600

θ = 36.87° Thus

T =

( 42 in.) (160 lb ) cot 36.87° 42 in. ( ) + ( 28 in.)

= 127.99 lb

T = 128.0 lb

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Chapter 10, Solution 28.

Note that

y A = − ( 0.15 m ) tan θ yB = − ( 0.15 m ) tan θ + ( 0.9 m ) sin θ Then

δ y A = − ( 0.15 m ) sec2 θ δθ δ yB = − ( 0.15 m ) sec2 θ δθ + ( 0.9 m ) cosθ δθ Virtual Work:

δ U = 0: Qδ y A + Pδ yB = 0 or (135 N )  − ( 0.15 m ) sec2 θ δθ 

+ ( 75 N )  − ( 0.15 m ) sec2 θ δθ + ( 0.9 m ) cosθ δθ  = 0 or − 20.25 sec2 θ − 11.25sec2 θ + 67.5cosθ = 0 or − 31.5 + 67.5cos3 θ = 0 cos3 θ = 0.4667 or

θ = 39.1°

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Chapter 10, Solution 29. First note

BC = BD ∴ BCD is isosceles Then ∠BCD = ∠BDC = θ and BE = l sin θ = 2l sin φ or sin φ =

1 sin θ 2

Then cos φ δφ = or δφ =

1 cosθ δθ 2

1 cosθ δθ 2 cos φ

Now

xC = 2l cos φ − l cosθ

xD = 2l cos φ + l cosθ

= l ( 2cos φ − cosθ )

= l ( 2cos φ + cosθ )

Then

δ xC = l ( − 2sin φ δφ + sin θ δθ )

δ xD = l ( − 2sin φ δφ − sin θ δθ )

= l ( sin θ − cosθ tan φ ) δθ

= − l ( sin θ + cosθ tan φ ) δθ

Also

FSP = kxSP = k ( 3l − xD ) = kl 3 − ( 2cos φ + cosθ )  Virtual Work:

δ U = 0: Pδ xC − FSP δ xD = 0

Then P l ( sin θ − cosθ tan φ ) δθ  − kl 3 − ( 2 cos φ + cosθ )  l ( sin θ + cosθ tan φ ) δθ = 0

sin θ + cosθ tan φ or P = kl 3 − ( 2cos φ + cosθ )  sin θ − cosθ tan φ tan θ + tan φ = kl 3 − ( 2cos φ + cosθ )  tan θ − tan φ Now sin φ =

1 1 sin θ = sin 25° 2 2

or φ = 12.1991°

N tan 25° + tan12.1991°  ∴ P = 1600  ( 0.150 m )( 3 − 2cos12.1991° − cos 25° ) m tan 25° − tan12.1991°   or P = 90.9 N

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Chapter 10, Solution 30.

yE =

x x x + = 3 6 2

∴ δ yE =

1 δx 2

Linear Spring: FSP = ks = ( 5000 N/m )( x − 0.30 m ) Virtual Work: δ U = 0

δ U = − FSPδ x + Pδ E = 0 1  − ( 5000 N/m )( x − 0.30 m ) δ x + ( 900 N )  δ x  = 0 2  − 5000 x + 1500 + 450 = 0 x = 0.390 m

or x = 390 mm

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Chapter 10, Solution 31.

yC =

x 6

Linear Spring:

FSP = ks = ( 5000 N/m )( x − 0.30 m )

Virtual Work:

δU = 0

∴ δ yC =

1 δx 6

δ U = − FSPδ x + Pδ yC = 0 1  − ( 5000 N/m )( x − 0.30 m ) δ x + ( 900 N )  δ x  = 0 6  − 5000 x + 1500 + 150 = 0 x = 0.330 m

or x = 330 mm W

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Chapter 10, Solution 32.

First note:

yD = ( 250 mm ) sin θ

δ yD = ( 250 mm ) cosθ δθ

x A = 2 ( 300 mm ) cosθ

δ x A = − ( 600 mm ) sin θ ( − δθ )

Also, the spring force

FSP = k  x A − ( x A )0  = ( 2.5 N/mm )( 600cosθ − 600 cos 45° )( mm ) = (1500 N )( cosθ − cos 45° )

Virtual Work:

δ U = 0:

( 250 N ) δ yD − FSP δ xA = 0 ( 250 N )( 250 mm ) cosθ δθ − (1500 N )( cosθ 5 − 72 tan θ ( cosθ − cos 45° ) = 0

− cos 45° )( 600 mm ) sin θ δθ = 0

Solving numerically

θ = 15.03° and 36.9° W

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Chapter 10, Solution 33.

From geometry:

xC = 2 (15 in.) cosθ

δ xC = − ( 30 in.) sin θ δθ

yB = (15 in.) sin θ

δ yB = (15 in.) cosθ (δθ )

s = ( 30 − 30cosθ ) in. = 30 (1 − cosθ ) in. FSP = ks = (12.5 lb/in.) 30 (1 − cosθ ) in.

Then

= ( 375 lb )(1 − cosθ ) Virtual Work:

δ U = 0:

Pδ yB + FSP δ xC = 0

(150 lb )(15 in.) cosθ δθ

2250cosθ − 11250 (1 − cosθ ) sin θ = 0

(1 − cosθ ) tan θ

Solving numerically,

+ ( 375 lb )(1 − cosθ )   − ( 30in.) sin θ δθ  = 0

= 0.200

θ = 40.22° θ = 40.2° W

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Chapter 10, Solution 34.

From geometry:

xC = 2 (15 in.) cosθ

δ xC = − ( 30 in.) sin θ δθ

yB = (15 in.) sin θ

δ yB = (15 in.) cosθ δθ

s = ( 30 − 30cosθ ) in. = 30 (1 − cosθ ) in. FSP = ks = (12.5 lb/in.) 30 (1 − cosθ ) in.

Then

= ( 375 lb )(1 − cosθ ) Virtual Work:

δ U = 0:

Pδ yB + FSP δ xC = 0

P (15 in.)( cos 25° ) δθ  + ( 375 lb )(1 − cos 25° )   − ( 30 in.)( sin 25° ) δθ  = 0

P = 32.8 lb W

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Chapter 10, Solution 35.

s = rθ

δ s = r δθ Spring is unstretched at θ = 0°

FSP = ks = k r θ xC = l sin θ

δ xC = l cosθ δθ

Virtual Work:

δ U = 0:

Pδ xC − FSPδ s = 0 P ( l cosθ δθ ) − k r θ ( r δθ ) = 0 or

θ Pl = 2 cosθ kr

( 40 lb )(12 in.) ( 9 lb/in.)( 5 in.)2

Thus or

θ cosθ

= 2.1333

θ = 1.054 rad = 60.39°

θ = 60.4° W

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Chapter 10, Solution 36.

y A = l sin θ

δ y A = l cosθδθ Spring:

v = CD

Unstretched when

θ =0

so that

v0 =

For θ :

 90° + θ  v = 2l sin   2  

 

δ v = l cos  45° +

θ

 δθ 2

Stretched length:

θ  s = v − v0 = 2l sin  45° +  − 2l 2    θ  F = ks = kl  2sin  45° +  − 2  2    

Then Virtual Work:

δ U = 0: Pδ y A − F δ v = 0   θ θ   Pl cosθ δθ − kl  2sin  45° +  − 2  l cos  45° +  δθ = 0 2 2     or

P 1 = kl cosθ =

1 cosθ

 θ  θ θ     2sin  45° +  cos  45° +  − 2 cos  45° +   2 2 2       θ  θ θ     2sin  45° +  cos  45° +  cosθ − 2 cos  45° +   2 2 2     

θ  cos  45° +  2  =1− 2 cosθ

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Now, with P = 150 lb, l = 30 in., and k = 40 lb/in.

(150 lb ) ( 40 lb/in.)( 30 in.) or Solving numerically,

θ  cos  45° +  2  =1− 2 cosθ θ  cos  45° +  2  = 0.61872 cosθ θ = 17.825°

θ = 17.83°

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Chapter 10, Solution 37.

y A = l sin θ

From geometry:

δ y A = l cosθ δθ xC = l cosθ + l sin θ = l ( cosθ + sin θ )

yC = l sin θ − l cosθ = l ( sin θ − cosθ ) lCD = l

( cosθ

+ sin θ ) + ( sin θ − cosθ ) − ( −1)  2

= l 3 + 2sin θ − 2 cosθ

δ lCD = l

cosθ + sin θ δθ 3 + 2sin θ − 2 cosθ

FSP = k ( lCD − l )

and

= kl

(

)

3 + 2sin θ − 2 cosθ − 1

Virtual Work:

δU = 0: δθ

P ( l cosθ δθ ) − kl

 1 − 

Pδ y A − FSPδ lCD = 0

(

  cosθ + sin θ 3 + 2sin θ − 2cosθ − 1 l δθ  = 0  3 + 2sin θ − 2cosθ 

)

 P  (1 + tan θ ) = kl 3 + 2sin θ − 2cosθ  1

600 N ( 4000 N/m )( 0.8m )

= 0.1875 Solving numerically

θ = 10.77° W

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Chapter 10, Solution 38.

yC = ( 375 mm ) tan θ

Have

δ yC = ( 375 mm ) sec2 θ δθ

δ S = ( 75 mm ) δθ FSP = kyC = ( 0.8 N/mm )( 375 mm ) tan θ = ( 300 N ) tan θ Virtual Work:

δ U = 0:

Pδ S − FSPδ yC = 0

( 480 N )( 75 mm ) δθ or

− ( 300 N ) tan θ  ( 375 mm ) sec 2 θ δθ = 0

3.125 tan θ sec2 θ = 1

Solving numerically,

θ = 16.41° W

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Chapter 10, Solution 39.

y A = l sin θ

Have

δ y A = l cosθ δθ

Virtual Work:

δ U = 0:

Pδ y A − M δθ = 0 Pl cosθ δθ − Kθ δθ = 0

or Thus

θ cosθ

Pl K

( 2000 N )( 0.25 m ) ( 225 N ⋅ m/rad )

= 2.2222 rad

Solving numerically

θ = 61.2° W

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Chapter 10, Solution 40.

y A = l sin θ

Have

δ y A = l cosθ δθ

Virtual Work:

δ U = 0:

Pδ y A − M δθ = 0 Pl cosθ δθ − Kθ δθ = 0

Then

cosθ

θ cosθ

Pl K

( 6300 N )( 0.25 m ) = 7 rad ( 225 N ⋅ m/rad )

θ = 7 cosθ

Plotting y = θ and y = 7 cosθ in the range 0 ≤ θ ≤

5π reveals three points of intersection, and thus 2

three roots:

Then, for each range

π  0 < θ < 90°  0 < θ <  :  2

θ = 1.37333 rad

θ = 78.7° W

 3π  270° < θ < 360°  < θ < 2π  :  2 

θ = 5.6522 rad

θ = 324° W

5π   360° < θ < 450°  2π < θ < :  2

θ = 6.6160 rad

θ = 379° W

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Chapter 10, Solution 41.

d AC = 212 + 7.22 = 22.2 in.

Have

tan φ =

7.2 21

φ = 18.9246°

α =θ +φ

Now

By Law of Cosines: 2 d AB = 22.22 + 102 − 2 ( 22.2 )(10 ) cos α

d AB = 592.84 − 444 cos α ( in.)

and

δ d AB =

222sin α δα ( in.) 592.84 − 444cos α

By Virtual Work:

δ U = 0:

Pδ yD − Fcyl δ d AB = 0 

222 sin α  592.84 − 444cos α

(120 lb )(8 in.) δα − Fcyl  Fcyl =

   in. δα = 0  

4.3243 592.84 − 444 cos α ( lb ) sin α

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Given data: Thus

θ = 60°, α = 60° + 18.9246° = 78.9246° Fcyl =

4.3243 592.84 − 444cos 78.9246° ( lb ) sin 78.9246°

= 99.270 lb d AB = 592.84 − 444cos 78.9246° = 22.529 in. By Law of Sines: 10 22.529 = sin β sin 78.9246°

β = 25.824° Fcyl = 99.3 lb

44.7°

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Chapter 10, Solution 42.

d AC = 212 + 7.22 = 22.2 in.

Have

tan φ =

7.2 21

φ = 18.9246°

α =θ +φ

Now

By Law of Cosines: 2 d AB = 22.22 + 102 − 2 ( 22.2 )(10 ) cos α

d AB = 592.84 − 444 cos α ( in.) and

δ d AB =

222sin α δα ( in.) 592.84 − 444cos α

By Virtual Work:

δ U = 0:

Pδ yD − Fcyl δ d AB = 0 

222 sin α  592.84 − 444cos α

(120 lb )(8 in.) δα − Fcyl  Fcyl =

   in. δα = 0  

4.3243 592.84 − 444 cos α ( lb ) sin α

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Thus

105 lb =

4.3243 592.84 − 444 cos α ( lb ) sin α

24.28125 sin α = 592.84 − 444cos α

589.5791sin 2 α = 592.84 − 444 cos α

(

)

589.5791 1 − cos 2 α = 592.84 − 444 cos α

cos 2 α − 0.75307 cos α + 0.0055309 = 0 Solving

α = 41.785° and α = 89.575°

θ = α − φ = α − 18.9246° Thus

θ = 22.9° and θ = 70.7°

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Chapter 10, Solution 43.

First note

θ + β = 90° α + β = 90°

∴α = θ

and

s = aθ

(length of cord unwound for rotation θ )

Now

yO = a (1 − cosθ ) (Distance O moves down for rotation θ )

and

yP = s + y

yP = a θ + a (1 − cosθ ) (Distance P moves down for rotation θ )

Then

δ yP = ( a + a sin θ ) δθ

By the Law of Cosines 2 lSP = ( 4a ) + ( 2a ) − 2 ( 4a )( 2a ) cosθ 2

lSP = 2a 5 − 4cosθ

and

δ lSP =

(

FSP = k lSP − ( lSP )0  = k 2a 5 − 4 cosθ − 2a = 2ka

(

4a sin θ δθ 5 − 4 cosθ

)

5 − 4 cosθ − 1

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By Virtual Work:

δ U = 0:

Pδ yP − FSPδ lSP = 0 P ( a + a sin θ ) δθ − 2ka

(

 4a sin θ 5 − 4cosθ − 1   5 − 4 cosθ 

)

  δθ = 0 

 P  (1 + sin θ ) − sin θ  5 − 4cosθ + sin θ = 0  ka 8  

Then

  12 lb (1 + sin θ ) − sin θ  5 − 4cosθ + sin θ = 0   8 (15 lb/in.)( 7.5 in.)  1   75 (1 + sin θ ) − sin θ  5 − 4cosθ + sin θ = 0  

Solving numerically,

θ = 15.27° W

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Chapter 10, Solution 44.

For Bar ABC:

δα =

δ yC

2a For Bar CD, using Law of Cosines

(where a = 15 in. )

a 2 = lC2 + lD2 − 2lC lD cos 55° Then with a = constant, we have 0 = 2lC δ lC + 2lDδ lD − 2 (δ lC ) lD cos 55° − 2lC (δ lD ) cos 55°

Since

δ lC = − δ yC :

( lC

− lD cos 55° ) δ yC = ( lD − lC cos55° ) δ lD

For the given position of member CD, ∆CDE is isosceles. Thus

lD = a

Then

( 2a cos 55° − a cos 55° ) δ yC

δ lD =

and

lC = 2a cos 55°

(

)

= a − 2a cos 2 55° δ yD

cos 55° δ yC 1 − 2cos 2 55°

By Virtual Work:

δ U = 0:

M δα − Pδ lD = 0  cos 55°  δ y  M  C  − P  δ yC = 0 2  2a   1 − 2 cos 55° 

M 1 − 2cos 2 55° 2a cos 55°

( 320 lb ⋅ in.) 2 (15 in.)

Thus for given data 1 − 2cos 2 55° = 6.3605 lb cos 55° P = 6.36 lb

35° W

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Chapter 10, Solution 45.

We have x A = (10 in.) sin θ + ( 24 in.) cosθ

δ x A = (10 cosθ − 24sin θ ) in. δθ

Law of Cosines:

( CD )2 = ( BC )2 + ( BD )2 − 2 ( BC )( BD ) cosθ = ( 20 in.) + ( 60 in.) − 2 ( 20 in.)( 60 in.) cosθ 2

(

) (

)

= 4000 in 2 − 2400 in 2 cos θ =  400 (10 − 6cos θ ) in 2  Differentiating:

(

)

2 ( CD ) δ ( CD ) = 2400 in 2 sin θ δθ

δ ( CD ) =

(1200 in ) 2

 20 10 − 6 cos θ in.  

sin θ δθ

 60 sin θ  δ ( CD ) =  in. δθ  10 − 6 cosθ 

or Virtual Work:

δ U = Pδ x A + FCD δ ( CD ) = 0

( 4000 lb ) (10 cos 60° − 24sin 60° ) in. δθ or

 60sin 60°  + FCD  in. δθ = 0 10 6 cos 60 − °  

FCD = 3214.9 lb

FCD = 3.21 kips

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Chapter 10, Solution 46.

Triangle ADE:

tan α =

( 2.7 ft ) (1.5 ft )

= 1.800

α = 60.945° AD =

yC = (15 ft ) sin θ

( 2.7 ft ) sin 60.945°

= 3.0887 ft

δ yC = (15 ft ) cosθ δθ

Law of Cosines: BD 2 = AB 2 + AD 2 − 2 ( AB )( AD ) cos (α + θ ) 2

= ( 7.2 ft ) + ( 3.0887 ft ) − 2 ( 7.2 ft )( 3.0887 ft ) cos (α + θ ) = 61.38 − 44.4773cos (α + θ )  ft 2

2 ( BD )(δ BD ) = 44.4773 sin (α + θ ) δθ  44.4773 sin (α + θ )

δ BD =  

2 ( BD )



δθ  ft 

Virtual Work:

δ U = 0: − P δ yC + FBD δ BD = 0  44.4773sin (α + θ )  − ( 500 lb )(15 ft ) cosθ δθ + FBD  δθ  ft = 0   2 ( BD )     cosθ FBD = 337.25 BD  lb sin (α + θ )   We have θ = 20°

BD 2 = 61.38 − 44.4773 cos ( 60.945° + 20° )

BD = 7.3743 ft Thus

FBD = 337.25

cos 20° ( 7.3743) lb sin ( 60.945° + 20° )

= 2366.5 lb

FBD = 2370

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Chapter 10, Solution 47.

Input work = P δ x Output work = (W sin α ) δ x Efficiency:

η=

W sin αδ x Pδ x

η=

ΣFx = 0: P − F − W sin α = 0 ΣFy = 0: N − W cos α = 0

W sin α P

(1)

P = W sin α + F

(2)

N = W cos α

F = µ N = µW cosα Equation (2): Equation (1):

P = W sin α + µW cosα = W ( sin α + µ cosα )

η=

W sin α W ( sin α + µ cos α )

η=

1 1 + µ cot α

If block is to remain in place when P = 0, we know (see page 416) that φ s ≥ α or, since

µ = tan φs , Multiply by cot α :

µ ≥ tan α µ cot α ≥ tan α cot α = 1

Add 1 to each side:

1 + µ cot α ≥ 2

Recalling the expression for η, we find

η≤

1 2

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Chapter 10, Solution 48.

xB = l cosθ

Link BC:

δ xB = − l sin θ δθ or δ xB = l sin θ δθ yC = l sin θ

δ yC = l cosθ δθ

δ xB =

Link AB:

1 l δφ 2

1 l δφ = l sin θ δθ 2 δφ = 2sin θ δθ

Thus

Virtual Work:

δ U = 0:

M maxδφ − ( P + µ s N ) δ yC = 0 M max ( 2sin θ δθ ) − ( P + µ s N ) ( l cosθ δθ ) = 0

P + µs N l 2 tan θ Link BC Free-Body Diagram: M max =

+ ΣM B = 0:

N ( l sin θ ) − ( P + µ s N ) l cosθ = 0 N tan θ − P − µ s N = 0

or N =

P tan θ − µ s

Substituting N into relationship for M max :

P+ M max =

µs P Pl ( tan θ − µ s + µ s ) tan θ − µ s l = 2 tan θ 2 tan θ ( tan θ − µ s ) M max =

For µ s ≥ tan θ , we have M max = ∞; the system becomes self-locking.

 2 ( tan θ − µ s ) 

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Chapter 10, Solution 49.

Largest value of M is obtained from the solution of Problem 10.48

M max =

 2 ( tan θ − µ s ) 

Thus M max =

( 400 N )( 0.500 m ) 2 ( tan 35° − 0.30 )

= 249.87 N ⋅ m M max = 250 N ⋅ m

Smallest value of M occurs when the friction force in Problem 10.48 is directed upward instead of downward. The equations obtained in Problem 10.48 may be used if we replace µ s by − µ s . Thus

M min =

P − µs N l 2 tan θ

N =

P tan θ + µ s

and

M min =

Thus

M min =

 2 ( tan θ + µ s ) 

( 400 N )( 0.500 m ) 2 ( tan 35° + 0.30 )

= 99.98 N ⋅ m M min = 100.0 N ⋅ m

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Chapter 10, Solution 50.

For the linkage:

ΣM B = 0: − x A +

xA P=0 2

F = µs A = µs

Then:

P 2

P 1 = µs P 2 2

x A = 2l sin θ

Now

δ x A = 2l cosθ δθ yF = 3l cosθ

and

δ yF = − 3l sin θ δθ Virtual Work:

δ U = 0:

( Qmax

− F ) δ x A + Pδ yF = 0

1    Qmax − µ s P  ( 2l cosθ δθ ) + P ( −3l sin θ δθ ) = 0 2   or

Qmax =

3 1 P tan θ + µ s P 2 2 Qmax =

P ( 3tan θ + µs ) 2

For Qmin , motion of A impends to the right and F acts to the left. We change µ s to − µ s and find

Qmin =

P ( 3tan θ − µ s ) 2

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Chapter 10, Solution 51.

Using the results of Problem 10.50 with

θ = 30°, l = 0.2 m, P = 40 N, and µ s = 0.15 We have

Qmax = =

P ( 3tan θ + µs ) 2

( 40 N ) 2

( 3tan 30° + 0.15)

= 37.64 N Qmax = 37.6 N and

Qmin = =

P ( 3tan θ − µ s ) 2

( 40 N ) 2

( 3tan 30° − 0.15)

= 31.64 N Qmin = 31.6 N

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Chapter 10, Solution 52. Recall Figure 8.9a. Draw force triangle

Q = W tan (θ + φs ) y = x tan θ so that δ y = δ x tan θ Input work = Qδ x = W tan (θ + φs ) δ x Output work = W δ y = W (δ x ) tan θ

η=

Efficiency:

W tan θδ x ; W tan (θ + φs ) δ x

η=

tan θ tan (θ + φs )

From page 432, we know the jack is self-locking if

φs ≥ θ θ + φ s ≥ 2θ

Then

tan (θ + φ s ) ≥ tan 2θ

so that

η=

From above

tan θ tan (θ + φ s )

η≤

It then follows that

tan 2θ =

But

Then

η≤

(

tan θ 1 − tan 2 θ 2 tan θ

tan θ tan 2θ

2 tan θ 1 − tan 2 θ

) = 1 − tan θ

∴ η ≤

1 2

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Chapter 10, Solution 53.

To determine Ay, consider a vertical displacement δ y A :

Note that δ y A = δ yB = δ yC

δ yC

δ yE

300 mm 240 mm

δ yE

750 mm

δ yG

360 mm

or δ yE = 2.5 δ y A or δ yG =

360 ( 2.5δ y A ) = 3.75δ y A 240

Virtual Work:

δ U = 0: Ay δ y A + ( 960 N ) δ yB − ( 240 N ) δ yG = 0 Ay δ y A + ( 960 N ) δ y A − ( 240 N )( 3.75 δ y A ) = 0

Ay = − 60 N or A y = 60 N To determine A x , consider a horizontal displacement δ x A : Virtual Work: δ U = 0:

Axδ x A = 0, or Ax = 0

A = 60.0 N

Thus the total force reaction is: To determine M A , consider a counterclockwise rotation δθ A :

Note that

δ yB = 600 δθ A

δ yC = 900 δθ A

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δ yC

δ yE

300 mm

240 mm

δ yE

750 mm

δ yG

360 mm

or δ yE =

750 ( 900 δθ A ) = 2250 δθ A 300

or δ yG =

360 ( 2250 δθ A ) = 3375δθ A 240

Virtual Work:

δ U = 0:

M A δθ A + ( 960 N ) δ yB − ( 240 N ) δ yG = 0 M A δθ A + ( 960 N )( 600 δθ A mm ) − ( 240 N )( 3375 δθ A mm ) = 0 M A = 234000 N ⋅ mm, or M = 234 N ⋅ m

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Chapter 10, Solution 54.

To determine Dy , consider a vertical displacement δ yD :

Note that

δ yD

δ yE

300 mm

240 mm

δ yE

δ yE = 3.5 δ yD

δ yG

δ yG =

1050 mm

360 mm

360 ( 3.5δ yD ) = 5.25δ yD 240

Virtual Work:

δ U = 0: Dy δ yD + ( 240 N ) δ yG = 0 Dy δ yD + ( 240 N )( 5.25 δ yD ) = 0

Dy = −1260 N or

D y = 1.260 kN

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Chapter 10, Solution 55.

From the solution of Problem 10.41

Fcyl = 99.270 lb By Virtual Work:

δ U = 0:

Fcyl ( − δ d AB ) − P δ yD = 0

( 99.270 lb )( − δ d AB ) − (120 lb )(1.2 in.) = 0 δ d AB = −1.45059 in. or δ d AB = 1.451 in. ( shorter )

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Chapter 10, Solution 56.

From Problem 10.46

FBD = 2370 lb as shown.

By Virtual Work: Assume both δ yC and δ BD increase.

δ U = 0:

− ( 500 lb ) δ yC + FBD δ BD = 0 − ( 500 lb )( 2.5 in.) + ( 2370 lb ) δ BD = 0

δ BD = 0.5274 in. Thus δ BD = 0.527 in. Longer

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Chapter 10, Solution 57.

Apply vertical load P at C: ΣM B = 0: (12 m ) J y − ( 3 m ) P = 0

Jy =

P 4

ΣFx = 0: J x = 0 ΣFy = 0:

P 3 − FFG = 0 4 5

FFG =

5 P (T) 12

Virtual Work: Remove member FG and replace it with forces FFG and − FFG at pins F and G, respectively. Denoting the virtual displacements of F and G as δ rF and δ rG , respectively, and noting that P and δ yC have the same direction, have by virtual work.

δ U = 0: Pδ yC + FFG ⋅δ rF + ( − FFG ) ⋅δ rG = 0 Pδ yC + FFG δ rF cosθ F − FFG δ rG cosθ G = 0 Pδ yC − FFG (δ rG cosθ G − δ rF cosθ F ) = 0 Where (δ rG cosθ G − δ rF cosθ F ) = δ FG , which is the change in length of member FG. Thus

Pδ yC − FFG δ FG = 0  5  Pδ yC −  P  ( 30 mm ) = 0  12 

δ yC = 12.50 mm or δ yC = 12.50 mm

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Chapter 10, Solution 58.

Apply a horizontal load P at C: ΣFx = 0: P − J x = 0 ∴ J x = P ΣM B = 0: (12 m ) J y − ( 2.25 m ) P = 0

Jy = ΣFy = 0:

3 P 16 3 3 P − FFG = 0 16 5

FFG =

5 P (T) 16

Virtual Work: Remove member FG and replace it with forces FFG and − FFG at pins F and G, respectively. Since P and δ xC have the same direction, and since FFG tends to decrease the length FG, have by virtual work.

δ U = 0: P δ xC − FFG δ FG = 0  5  P δ xC −  P  ( 30 mm ) = 0  16 

δ xC = 9.375 mm or δ xC = 9.38 mm

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Chapter 10, Solution 59.

Spring:

s = x − 0.3 m yE = −

x x x − =− 3 6 2

Potential Energy:

V =

1 2 ks + WyE 2

1 2 x k ( x − 0.3) − W   2 2

For equilibrium:

dV 1 = k ( x − 0.3) − W = 0 dx 2 1 2

( 5000 N/m )( x − 0.3) m − ( 900 N ) = 0 Solving

x = 0.390 m

x = 390 mm

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Chapter 10, Solution 60.

Given:

( xSP )0

= 300 mm

k = 5 kN/m From geometry:

1 x yC = −   3 2 

=−

x 6

sSP = x − ( xSP )0

and

= ( x − 0.3) m Potential Energy: V = VSP + VFC

1 2  1  k ( x − 0.3) + FC  − x  2  6 

For equilibrium:

dV = 0: dx

k ( x − 0.3) −

1 FC = 0 6 1 6

( 5000 N/m )( x − 0.3) m − ( 900 N ) = 0 or x = 0.330 m x = 330 mm

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Chapter 10, Solution 61.

k = 2.5 kN/m

Given:

( SSP )0 = 0

From geometry:

θ = 45°

yD = − ( 0.25 m ) sin θ

( xSP )0 = 2 ( 0.3 m ) cos 45° S SP = ( xSP )0 − x A = ( 0.6 m ) cos 45° − 2 ( 0.3 m ) cosθ = ( 0.6 m )( cos 45° − cosθ ) Potential Energy:

V = VSP + VFD =

For equilibrium:

1 2 2 k ( 0.6 m ) ( cos 45° − cosθ ) − ( 250 N )  − ( 0.25 m ) sin θ  2

dV = 0: dθ k ( 0.36 )( sin θ )( cos 45° − cosθ ) + 62.5cosθ = 0 62.5 N ⋅ m

tan θ ( cos 45° − cosθ ) = −

tan θ ( cos 45° − cosθ ) = − 0.06944

( 2500 N/m ) ( 0.36 m 2 )

Solving numerically and

θ = 15.03° θ = 36.9°

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Chapter 10, Solution 62.

k = 12.5 s=0

lb in.

P = 150 lb

when θ = 0

From geometry: yB = − (15 in.) sin θ s = 30 in. − 2 (15 in.) cosθ = ( 30 in.)(1 − cosθ ) Potential Energy:

V = VSP + VP =

1 2 ks + P yB 2

1 2 2 k ( 30 in.) (1 − cosθ ) + P  − (15 in.) sin θ  2

For equilibrium

dV = 0: dθ or

(

)

k 900 in 2 (1 − cos θ )( sin θ ) − (15 in.) cos θ P = 0

(

)

lb   2 12.5  900 in (1 − cos θ )( sin θ ) − (15 in.)(150 lb ) cos θ = 0 in.

11250 (1 − cosθ )( sin θ ) − 2250 cosθ = 0

(1 − cosθ ) tan θ

Solving numerically,

= 0.200

θ = 40.22°

θ = 40.2°

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Chapter 10, Solution 63.

k = 12.5 s=0

lb in.

θ = 25°

when θ = 0°

From geometry:

yB = − (15 in.) sin θ s = 30 in. − 2 (15 in.) cosθ = ( 30 in.)(1 − cosθ )

Potential Energy:

V = VSP + VP =

1 2 ks + P yB 2

1 2 2 k ( 30 in.) (1 − cosθ ) + P  − (15 in.) sin θ  2

dV = 0: dθ

For equilibrium

(

)

k 900 in 2 (1 − cos θ )( sin θ ) − (15 in.) cos θ P = 0 or

(

)

lb   2 12.5  900 in (1 − cos 25°)( sin 25°) − (15 in.)( cos 25°) P = 0 in. or

P = 32.8 lb

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Chapter 10, Solution 64.

 90° + θ  v = 2l sin   2  

Spring

θ  v = 2l sin  45° +  2  Unstretched (θ = 0 )

v0 = 2l sin 45° =

θ  s = v − v0 = 2l sin  45° +  − 2l 2 

Deflection of spring

V =

  1 2 1 θ  ks + Py A = kl 2  2sin  45° +  − 2  + P ( −l sin θ ) 2 2 2   

  θ θ dV   = kl 2  2sin  45° +  − 2  cos  45° +  − Pl cosθ = 0 dθ 2 2      θ  θ θ  P    2sin  45° +  cos  45° +  − 2 cos  45° +   = cosθ 2 2 2   kl    

θ P  cosθ − 2 cos  45° +  = cosθ 2  kl  Divide each member by cosθ

θ  cos  45° +  2 P  = 1− 2 cosθ kl

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Then with P = 150 lb, l = 30 in. and k = 40 lb/in.

θ  cos  45° +  2 150 lb  1− 2 = cosθ 40 lb/in. ( )( 30 in.) = 0.125

θ  cos  45° +  2  = 0.618718 cosθ

Solving numerically,

θ = 17.83°

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Chapter 10, Solution 65.

y A = − l sin θ

From geometry:

( cosθ

lCD = l

+ sin θ ) + 1 − ( cosθ − sin θ )  2

= l 3 + 2sin θ − 2 cosθ

sSP = ( lCD − l ) =l

Potential Energy:

(

)

3 + 2sin θ − 2 cosθ − 1

V = VSP + VP 1 2 = ksSP + Py A 2 2 1 = kl 2 3 + 2sin θ − 2cosθ − 1 + P ( − l sin θ ) 2 dV = 0: dθ  cosθ + sin θ  3 + 2sin θ − 2cosθ − 1 − P l cosθ = 0 3 + 2sin θ − 2 cosθ   1 P  (1 + tan θ ) = kl 3 + 2sin θ − 2cosθ  600 N = ( 4000 N/m )( 0.8 m ) = 0.1875

(

For equilibrium:

 kl 2     1 − 

)

(

)

Solving numerically

θ = 10.77°

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Chapter 10, Solution 66.

From geometry yC = d AC tan θ

d AC = 375 mm

yP = r θ

r = 75 mm

Potential Energy:

V = VSP + VP = =

1 2 kyC − PyP 2

1 2 kd AC tan 2 θ − P r θ 2

For equilibrium:

dV = 0: dθ

2 tan θ sec 2 θ − P r = 0 kd AC

( 0.8 N/mm )( 375 mm )2 tan θ sec2 θ − ( 480 N )( 75 mm ) = 0 3.125 tan θ sec2 θ = 1 Solving numerically

θ = 16.41°

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Chapter 10, Solution 67.

125θ = 375φ ,

Since

xC = xD ,

Also

 250  xE = ( 250 mm ) φ =  θ  mm  3 

φ=

θ 3

Potential Energy:

V = − M θ + Q xE − PyG

= − Mθ +

250 100 Qθ − Pθ 3 3

For Equilibrium:

dV =θ: dθ

−M +

250 100 Q− P=0 3 3

Thus at equilibrium, V is constant and the equilibrium is neutral.

Q.E.D.

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Chapter 10, Solution 68.

First note

y B = yF

For small θ , φ :

yB = (10 in.)θ yF = ( 6 in.) φ 10θ = 6 φ

Thus

5 3

φ= θ y A = (16 in.)θ

Also

yG = (12 in.) φ = ( 20 in.)θ h + xD ( h = constant ) = h + ( 8 in.)θ xD = ( 4.8 in.) φ = ( 8 in.)θ Potential Energy: V = VFA + VP + VW = − FA y A − PyG + Wyw = − ( 20 lb )(16 in.)θ − P ( 20 in.)θ + ( 30 lb )  − ( h + 8θ ) in. = ( − 320 − 20 P − 240 )θ − 30 h  ( lb ⋅ in.)

For equilibrium:

dV =0 dθ

− 320 − 20 P − 240 = 0 V is a constant therefore equilibrium is neutral. Q.E.D.

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Chapter 10, Solution 69.

Potential Energy: l  l  V = − W  cosθ  − W  sin θ  2 2     =−

Wl ( cosθ + sin θ ) 2

dV Wl = ( sin θ − cosθ ) dθ 2

For equilibrium: dV = 0: dθ

sin θ − cosθ = 0 tan θ = 1

θ = 45° and θ = −135° d 2V Wl = ( cosθ + sin θ ) 2 dθ 2

Stability:

θ = 45°: θ = −135°:

d 2V Wl = ( 0.707 + 0.707 ) > 0 2 dθ 2 d 2V Wl = ( − 0.707 − 0.707 ) < 0 2 dθ 2

Stable Unstable

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Chapter 10, Solution 70.

Potential Energy: l  l  V = − WCD  cosθ  − WAB  sin θ  2  2  = − WCD

 l WAB sin θ   cosθ + 2 WCD 

WAB m AB g 300 = = = 0.6 WCD mCD g 500

But

V = − WCD

Thus

l ( cosθ + 0.6sin θ ) 2

dV l = WCD ( sin θ − 0.6cosθ ) dθ 2 For Equilibrium: dV = 0: dθ

sin θ − 0.6cosθ = 0 tan θ = 0.6

θ = 31.0° and θ = −149.0° Stability:

d 2V l = WCD ( cosθ + 0.6sin θ ) 2 2 dθ

θ = 31.0° :

d 2V l = WCD ( cos 31.0° + 0.6sin 31.0° ) > 0 2 2 dθ Stable

θ = −149.0°: = WCD

d V l = WCD cos ( −149.0° ) + 0.6sin ( −149.0° )  2 2 dθ

l  − 0.8572 + 0.6 ( − 0.5150 )  < 0 2 Unstable

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Chapter 10, Solution 71.

Let each rod be of length L and weight W. Then the potential energy V is

L  L  V = W  sin θ  + W  cos 2θ  2  2  Then

dV W = L cosθ − WL sin 2θ dθ 2 For equilibrium

dV W = 0: L cosθ − WL sin 2θ = 0 dθ 2 or

cosθ − 2sin 2θ = 0

Solving numerically or using a computer algebra system, such as Maple, gives four solutions:

θ = 1.570796327 rad = 90.0° θ = −1.570796327 rad = 270° θ = 0.2526802551 rad = 14.4775° θ = 2.888912399 rad = 165.522° Now

1 d 2V = − WL sin θ − 2WL cos 2θ 2 2 dθ

1  = −WL  sin θ + 2cos 2θ  2  continued

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At θ = 14.4775°

d 2V 1  = −WL  sin14.4775° + 2cos  2 (14.4775° )   2 2 dθ   = −1.875WL ( < 0 )

∴ θ = 14.48°, Unstable

At θ = 90°

d 2V 1  = −WL  sin 90° + 2 cos180° 2 dθ 2  = 1.5WL ( > 0 )

∴ θ = 90°, Stable

At θ = 165.522°

d 2V 1  = −WL  sin165.522° + 2cos ( 2 × 165.522° )  2 dθ 2  = −1.875WL ( < 0 )

∴ θ = 165.5°, Unstable

At θ = 270°

d 2V 1  = −WL  sin 270° + 2cos 540°  2 dθ 2  = 2.5WL ( > 0 )

∴ θ = 270°, Stable

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Chapter 10, Solution 72.

Potential energy

l  l  V = W  cos1.5θ  + W  cosθ  2  2 

W = mg

dV Wl Wl = ( −1.5sin1.5θ ) + ( − sin θ ) dθ 2 2 =−

Wl (1.5sin1.5θ + sin θ ) 2

d 2V Wl = − ( 2.25cos1.5θ + cosθ ) 2 2 dθ For equilibrium

dV = 0: 1.5sin1.5θ + sin θ = 0 dθ

Solutions: One solution, by inspection, is θ = 0, and a second angle less than 180° can be found numerically:

θ = 2.4042 rad = 137.8° Now

d 2V Wl = − ( 2.25cos1.5θ + cosθ ) 2 dθ 2

At θ = 0:

d 2V Wl = − ( 2.25cos 0° + cos 0° ) 2 dθ 2

=− At θ = 137.8°:

Wl ( 3.25) ( < 0 ) 2

∴ θ = 0, Unstable

d 2V Wl  2.25cos (1.5 × 137.8° ) + cos137.8° =− 2 2  dθ

Wl ( 2.75) ( > 0 ) 2

∴ θ = 137.8°, Stable

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Chapter 10, Solution 73.

Potential Energy

V =

1 Kθ 2 − Pl sin θ 2

dV = Kθ − Pl cosθ dθ d 2V = K + Pl sin θ dθ 2

dV K = 0: cosθ = θ dθ Pl

Equilibrium: For

P = 2 kN,

l = 250 mm, cosθ =

K = 225 N ⋅ m/rad

225 N ⋅ m/rad θ ( 2000 N )( 0.25 m )

= 0.450θ Solving numerically, we obtain

θ = 1.06896 rad = 61.247° θ = 61.2° Stability d 2V = ( 225 N ⋅ m/rad ) + ( 2000 N )( 0.25 m ) sin 61.2° > 0 dθ 2 ∴ Stable

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Chapter 10, Solution 74.

Potential Energy

V =

1 Kθ 2 − Pl sin θ 2

dV = Kθ − Pl cosθ dθ d 2V = K + Pl sin θ dθ 2 Equilibrium

dV K = 0: cosθ = θ dθ Pl

P = 6.3 kN, l = 250 mm, and K = 225 N ⋅ m/rad

For

cosθ =

225 N ⋅ m/rad θ ( 6300 N )( 0.25 m ) cosθ =

or Solving numerically,

θ 7

θ = 1.37333 rad, 5.652 rad, and 6.616 rad θ = 78.7°, 323.8°, 379.1°

Stability At θ = 78.7°:

d 2V = ( 225 N ⋅ m/rad ) + ( 6300 N )( 0.25 m ) sin 78.7° dθ 2

= 1769.5 N ⋅ m > 0 At θ = 323.8°:

d 2V = ( 225 N ⋅ m/rad ) + ( 6300 N )( 0.25 m ) sin 323.8° dθ 2

= − 705.2 N ⋅ m < 0 At θ = 379.1°:

∴ θ = 78.7°, Stable

∴ θ = 324°, Unstable

d 2V = ( 225 N ⋅ m/rad ) + ( 6300 N )( 0.25 m ) sin 379.1° dθ 2

= 740.37 N ⋅ m > 0

∴ θ = 379°, Stable

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Chapter 10, Solution 75.

ySP = r (θ − θ 0 ) ,

Have

yB = l AB cosθ ,

r = 4 in.,

θ 0 = 20° =

π 9

rad

l AB = 18 in.

Potential Energy: V = =

1 2 kySP + WyB 2 1 2 2 kr (θ − θ 0 ) + Wl AB cosθ 2

dV = kr 2 (θ − θ 0 ) − Wl AB sin θ dθ d 2V = kr 2 − Wl AB cosθ dθ 2

For equilibrium: dV = 0: dθ

( 4.5 lb/in.)( 4 in.)2 

4

−

π

 − W (18 in.) sin = 0 9 4

W = 2.4683 lb

W = 2.47 lb

Stability: d 2V π 2 = ( 4.5 lb/in.)( 4 in.) − ( 2.4683 lb )(18 in.) cos 2 4 dθ = 40.6 lb ⋅ in. > 0

∴ Stable

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Chapter 10, Solution 76.

Have

ySP = r (θ − θ 0 ) ,

r = 4 in.,

yB = l AB cosθ ,

l AB = 18 in.

θ 0 = 20° =

π 9

rad

Potential Energy: 1 2 kySP + WyB 2 1 2 = kr 2 (θ − θ 0 ) + Wl AB cosθ 2

V =

dV = kr 2 (θ − θ 0 ) − Wl AB sin θ dθ d 2V = kr 2 − Wl AB cosθ dθ 2 For equilibrium: dV = 0: dθ

Solving numerically:

( 4.5 lb/in.)( 4 in.)2 θ 

−

π

 − ( 6.6 lb )(18 in.) sin θ = 0 9

π  θ −  − 1.65sin θ = 0 9  θ = 1.90680 rad = 109.252°

θ = 109.3° W

Stability: d 2V 2 = ( 4.5 lb/in.)( 4 in.) − ( 6.6 lb )(18 in.) cos109.252° dθ 2 = 111.171 lb ⋅ in. > 0

∴ Stable W

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Chapter 10, Solution 77.

Note:  y =  20 − 

(11)2 − ( 20 − x )2  in. 

)

(

= 20 − − x 2 + 40 x − 279 in.

Potential Energy: V =

1 1 2 2 k ( x − 7.5 ) + k ( y − 7.5) + WA ( 20 ) + WB y 2 2

(

1 1 2 k ( x − 7.5 ) + k 12.5 − − x 2 + 40 x − 279 2 2 Equilibrium Condition: =

dV = 0: dx

)

(

+ 20WA + WB 20 − − x 2 + 40 x − 279

  40 − 2 x  k ( x − 7.5 ) + k 12.5 − − x 2 + 40 x − 279  −  2 − x 2 + 40 x − 279   

)

(

− WB

40 − 2 x 2

2 − x + 40 x − 279

Simplifying, 12.5 k ( x − 20 ) + 12.5k − x 2 + 40 x − 279 + WB ( x − 20 ) = 0 Substituting

k = 1 lb/in.,

WB = 10 lb:

12.5 (1 lb/in.) ( x − 20 ) in. + 12.5 (1 lb/in.)

(

)

− x 2 + 40 x − 279 in. + (10 lb ) ( x − 20 ) in. = 0

)

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or or

( x − 20 ) +

− x 2 + 40 x − 279 + 0.8 ( x − 20 ) = 0

− x 2 + 40 x − 279 = 1.8 ( 20 − x )

− x 2 + 40 x − 279 = ( 36 − 1.8 x )

4.24 x 2 − 169.6 x + 1575 = 0 169.6 ±

( −169.6 )2 − 4 ( 4.24 )(1575) 2 ( 4.24 )

Then

x = 14.6579 in.

Since

x ≤ 20 in.

and

x = 25.342 in.

x = 14.66 in. W

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Chapter 10, Solution 78.

Deflection of spring = s, where

l 2 + y2 − l

s =

ds = dy V =

Potential Energy:

y 2

l − y2 1 2 y ks − W 2 2

dV ds 1 = ks − W dy dy 2

dV =k dy

(

l 2 + y2 − l

 = k 1 −   Equilibrium Now Then

or Solving numerically,

)

l + y

−

1 W 2

  y − 1W 2 2  2 l + y  l

 dV = 0: 1 −  dy 

(

y 2

 y = 1W 2 2  2 k l + y  l

)

W = mg = (12 kg ) 9.81 m/s 2 = 117.72 N, l = 0.75 m, and k = 900 N/m

 1 −   

0.75 m

( 0.75 m )2  1 −  

  y = 1 (117.72 N )  2 ( 900 N/m ) + y 2 

  y = 0.0654 0.5625 + y 2  0.75

y = 0.45342 m y = 453 mm

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Chapter 10, Solution 79.

(a)

We note that in U ABC, 1 ∠A = ∠B = 180° − ( 90° + θ )  2 = 45° −

θ 2

θ  Thus AB = 2a cos  45° −  2 

θ θ  = 2a  cos 45° cos + sin 45° sin  2 2  = 2a

= Elongation of Spring:

2 θ θ  cos + sin  2  2 2

θ θ  2 a  cos + sin  2 2 

s = AB − 2 a

θ θ   = 2 a  cos + sin − 1 2 2   Potential Energy:

V =

1 2 ks − W ( l sin θ ) 2

V =

1 θ θ   k 2a 2  cos + sin − 1 − Wl sin θ 2 2 2  

( )

θ θ θ θ θ θ  = ka 2  cos 2 + sin 2 + 1 + 2cos sin − 2cos − 2sin  − Wl sin θ 2 2 2 2 2 2  θ θ  = ka 2 1 + 1 + sin θ − 2cos − 2sin  − Wl sin θ 2 2  θ θ Wl    = 2ka 2 1 − cos − sin  + ka 2 1 − 2  sin θ 2 2 ka    continued

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dV θ 1 θ Wl  1  = 2ka 2  sin − cos  + ka 2 1 − 2  cosθ 2 2 2 dθ ka  2     θ θ  Wl  = ka 2  −  cos − sin  + 1 − 2  cosθ  2 2  ka     For Equilibrium:

dV = 0: dθ (b)

cos

θ 2

− sin

Wl   = 1 − 2  cosθ 2  ka 

(1)

Given data:

k = 75 lb/in., a = 10 in., l = 15 in., and W =100 lb cos

Using equation (1):

cos

θ 2

− sin

 (100 lb )(15 in.)  cosθ = 1 − 2  ( 75 lb/in.)(10 in.)2  

Letting

cosθ = cos 2

Then

cos

θ 2

− sin

= 0.8 cosθ

− sin 2

θ 2

θ θ  θ θ  = 0.8  cos − sin  cos + sin  2 2 2 2  

Which yields cos

θ 2

− sin

θ 2

and

The first equation yields In the second equation let x = cos

θ 2

cos

+ sin

θ 2

= 1.25

and the equation becomes

x 2 − 1.25 x + 0.28125 = 0

Solving

x = 0.95572 and x = 0.29428 2

= cos −1 0.95572 and

θ 2

θ = 90.0° W

= 45°,

x + 1 − x 2 = 1.25

= cos −1 0.29428

θ = 34.2° and θ = 145.8° W Stability:  d 2V 1 2  θ θ Wl   = ka sin + cos − 2 1 − 2  sin θ  2 2 2 dθ ka    2  =

1 2 θ θ  ka  sin + cos −1.6sin θ  2 2 2  

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θ = 34.2°: θ = 90.0°: θ = 145.8°:

d 2V 1 2 = ka ( 0.2940 + 0.9558 − 0.8993) > 0 dθ 2 2 d 2V 1 2 = ka ( 0.707 + 0.707 − 1.6 ) < 0 dθ 2 2 d 2V 1 2 = ka ( 0.9558 + 0.2940 − 0.8993) > 0 dθ 2 2

Stable W Unstable W Stable W

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Chapter 10, Solution 80.

Note:

xSPA = rA sin θ xSPB = xSPA xSP = xSPA + xSPB = 2rA sin θ ,

yBLOCK = rθ , V =

Potential Energy:

rA = 150 mm

r = 200 mm

1 2 kxSP − mgyBLOCK 2 1 2 k ( 2rA sin θ ) − mgr θ 2

= 2krA2 sin 2 θ − mgr θ

dV = 2krA2 ( 2sin θ cosθ ) − mgr = 2krA2 sin 2θ − mgr dθ d 2V = 4krA2 cos 2θ 2 dθ

(1)

Equilibrium Condition:

dV = 0: dθ

2krA2 sin 2θ − mgr = 0

(

)

Thus

2 ( 2000 N/m )( 0.15 m ) sin 2θ − m 9.81 m/s 2 ( 0.2 m ) = 0

m = 45.872 sin 2θ ( kg )

(a)

From Eq. (2),

with m ≥ 0:

(b)

For Stable equilibrium:

d 2V >0 dθ 2

Then from Eq. (1)

cos 2θ > 0

(2) 0 ≤ m ≤ 45.9 kg W

0 ≤ θ ≤ 45.0° W

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Chapter 10, Solution 81.

Note:

xSPA = rA sin θ xSPB = xSPA xSP = xSPA + xSPB = 2rA sin θ ,

rA = 150 mm

yBLOCK = r θ , V =

Potential Energy:

r = 200 mm

1 2 kxSP − mgyBLOCK 2 1 2 k ( 2rA sin θ ) − mgr θ 2

= 2krA2 sin 2 θ − mgr θ

dV = 2krA2 ( 2sin θ cosθ ) − mgr = 2krA2 sin 2θ − mgr dθ d 2V = 4krA2 cos 2θ dθ 2

(1)

Equilibrium Condition:

dV = 0: dθ

2krA2 sin 2θ − mgr = 0

Thus

2 ( 2000 N/m )( 0.15 m ) sin 2θ − m 9.81 m/s 2 ( 0.2 m ) = 0

m = 45.872 sin 2θ ( kg )

(

)

(2)

20 = 45.872 sin 2θ Solving

θ = 12.9243°

and

θ = 77.076°

Stability: Using Eq. (1)

θ = 12.9243°:

d 2V = 4krA2 cos ( 2 × 12.9243° ) > 0 dθ 2

θ = 77.076°:

d 2V = 4krA2 cos ( 2 × 77.076° ) < 0 dθ 2

∴ θ = 12.92°, Stable W ∴ θ = 77.1°, Unstable W

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Chapter 10, Solution 82.

(a)

sSP = l cosθ 0 − l cosθ = l ( cosθ 0 − cosθ ) Potential Energy:

V = =

1 2 ksSP − mg ( 2l sin θ ) 2 1 2 2 kl ( cosθ 0 − cosθ ) − 2mgl sin θ 2

dV = kl 2 ( cosθ 0 − cosθ ) sin θ − 2mgl cosθ dθ Equilibrium: dV = 0: dθ

(1)

kl 2 ( cosθ 0 − cosθ ) sin θ − 2mgl cosθ = 0

Since cosθ = 0 is not a solution of the equation, we can divide all terms by kl 2 cosθ and write

( cosθ 0 − cosθ ) tan θ

2mg kl

(2)

The spring is unstretched for θ = 0 thus θ 0 = 0 in Eq. (2) and we have

(1 − cosθ ) tan θ

2mg W kl

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(b)

For the given data

(1 − cosθ ) tan θ Solving by trial and error:

(

2 ( 5 kg ) 9.81 m/s 2

)

(800 N/m )( 0.250 m )

= 0.4905

θ = 51.96°,

θ = 52.0° W

Stability: Differentiating Eq. (1):

d 2V = kl 2 sin 2 θ + cosθ 0 cosθ − cos 2 θ + 2mgl sin θ 2 dθ

(

)

2mg   sin θ  = kl 2  cosθ 0 cosθ − cos 2θ + kl  

(3)

For θ 0 = 0, θ = 51.96°, and the given data

d 2V = kl 2 ( cos 51.96° − cos103.92° + 0.4905sin 51.96° ) dθ 2 = 1.2431 kl 2 > 0

θ = 52.0°, Stable W

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Chapter 10, Solution 83.

(1)

Law of Sines:

yA l = sin ( 90° + β − θ ) sin ( 90° − β ) yA l = cos (θ − β ) cos β yA = l

From Eq. (1):

yB = l

cos (θ − β ) cos β

cos (θ − β ) − l cosθ cos β

Potential Energy:

 l cos (θ − β )  cos (θ − β ) − l cosθ  − Ql V = − PyB − Qy A = − P  cos β cos β    sin (θ − β )  sin (θ − β ) dV = − Pl  − + l sin θ  + Ql dθ cos β cos β  

Equilibrium:

dV = 0: dθ

( P + Q )( sin θ cos β

( P + Q ) sin (θ

− β ) = P sin θ cos β

− cosθ sin β ) = P sin θ cos β

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Substracting P ( sin θ cos β ) from each member yields

− ( P + Q ) cosθ sin β + Q sin θ cos β = 0 or Given data:

tan θ =

P+Q tan β Q

β = 30°, P = Q = 400 N tan θ =

( 400 N ) + ( 400 N ) ( 400 N )

tan 30°

tan θ = 2 ( 0.57735 ) = 1.1547

θ = 49.1°

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Chapter 10, Solution 84.

(1)

Law of Sines:

yA l = sin ( 90° + β − θ ) sin ( 90° − β ) yA l = cos (θ − β ) cos β yA = l

From Eq. (1):

yB = l

cos (θ − β ) cos β

cos (θ − β ) − l cosθ cos β

Potential Energy:

 cos (θ − β )  cos (θ − β ) − l cosθ  − Ql V = − PyB − Qy A = − P l cos β cos β    sin (θ − β )  sin (θ − β ) dV = − Pl  − + l sin θ  + Ql dθ cos β cos β  

Equilibrium:

dV = 0: dθ

( P + Q )( sin θ cos β

( P + Q ) sin (θ

− β ) = P sin θ cos β

− cosθ sin β ) = P sin θ cos β

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Substracting P ( sin θ cos β ) from each member yields

− ( P + Q ) cosθ sin β + Q sin θ cos β = 0 or Given data:

tan θ =

P+Q tan β Q

P = 100 N, Q = 25 N, β = 30° tan θ =

(100 N ) + ( 25 N ) ( 25 N )

tan 30°

= 5 ( 0.57735 ) = 2.8868

θ = 70.9°

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Chapter 10, Solution 85.

First note, by Law of Cosines: 2

θ θ θ 2   d 2 = (16 ) + 16 sin  − 2 (16 )  16 sin  cos 2 2 2   d = 16 1 + sin 2

θ 2

− sin θ in.

xSP = rA θ

Also note Potential Energy:

V =

1 2 kxSP + WD yD 2

1 2 k ( rAθ ) + WD ( yD )0 − (16 − d ) sin 60° 2

  1 2 2 θ krAθ + WD ( yD )0 − 16 − 16 1 + sin 2 − sin θ  2 2  

   sin 60°  

Equilibrium condition:

dV = 0: dθ

krA2θ

+ 16WD sin 60°

krA2θ + 4WD sin 60°

1 θ θ   sin cos − cosθ  2 2 2  1 + sin 2

− sin θ

sin θ − 2cosθ 1 + sin

2θ

− sin θ

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Substituting,

( 2.5 lb/in.)( 2 in.)2 θ or

θ 1 + sin 2

θ 2

Solving numerically,

1 + sin 2

θ 2

− sin θ + 4 ( 25 lb ) sin 60° ( sin θ − 2cosθ ) = 0

− sin θ + 8.6603 ( sin θ − 2 cosθ ) = 0

θ = 1.08572 rad or

θ = 62.2°

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Chapter 10, Solution 86.

First note that cable tension is uniform throughout, hence

FSP1 = FSP2 k1x1 = k2 x2

x2 =

k1 6 lb/in. x1 = x1 k2 3 lb/in.

x2 = 2 x1

Now, with C midway between the pulleys,

2d = 16 in. + x1 + x2

d =8+ Then

1 ( x1 + x2 ) 2

y 2 = d 2 − 82 2

1   = 8 + ( x1 + x2 )  − 82 2   1 2 = 8 ( x1 + x2 ) + ( x1 + x2 ) 4 1 2 = 8 ( x1 + 2 x1 ) + ( x1 + 2 x1 ) 4 9 = 24 x1 + x12 in 2 4

( )

1 96 x1 + 9 x12 2 Potential Energy: y=

V =

1 2 1 k1x1 + k2 x22 − Wy 2 2

1 2 1 2 1  k1x1 + k2 ( 2 x1 ) − W  96 x1 + 9 x12  2 2 2  

1 1 ( k1 + 4k2 ) x12 − W  96 x1 + 9 x12  2 2 

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Equilibrium condition:



 =0  4 96 x + 9 x 2  1 1  

dV = 0: dx1

( k1 + 4k2 ) x1 − W 

(

6 + 4 ( 3)  lb/in. × ( x1 ) in.  

18x1 96 x1 + 9 x12 − 600 − 112.5x1 = 0

x1 = 2.7677 in.

Then

)

1 96 x1 + 9 x12 in. − ( 25 lb ) ( 96 + 18 x1 ) in. = 0  4

Solving,

96 + 18x1

1 2 96 ( 2.7677 ) + 9 ( 2.7677 ) 2

= 9.1466 in.

y = 9.15 in.

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Chapter 10, Solution 87.

Stretch of Spring s = AB − r

s = 2 ( r cosθ ) − r s = r ( 2cosθ − 1) Potential Energy:

V =

1 2 ks − Wr sin 2θ 2

V =

1 2 2 kr ( 2 cos θ − 1) − Wr sin 2θ 2

W = mg

dV = −kr 2 ( 2 cosθ − 1) 2sin θ − 2Wr cos 2θ dθ Equilibrium

dV = 0: dθ

− kr 2 ( 2cosθ − 1) sin θ − Wr cos 2θ = 0

( 2cosθ

− 1) sin θ

cos 2θ

Now

Then Solving numerically,

=−

(

W kr

)

( 20 kg ) 9.81 m/s2 W = = 0.36333 kr ( 3000 N/m )( 0.180 m ) ( 2cosθ

− 1) sin θ

cos 2θ

= −0.36333

θ = 0.9580 rad = 54.9°

θ = 54.9°

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Chapter 10, Solution 88.

Have Then

l AD = 2r sin θ

y A = − LAD sin ( 90° − θ ) − 45°

= − 2r sin θ sin ( 45° − θ ) Also for spring

s = l AB − r

= 2r cosθ − r = r ( 2 cosθ − 1) V = VSP + Vm

Potential Energy:

1 2 ks + mgy A 2

1 2 2 kr ( 2 cosθ − 1) − 2mgr sin θ sin ( 45° − θ ) 2

For Equilibrium:

dV = 0: dθ

kr 2 ( − 2sin θ )( 2cosθ − 1) − 2mgr cosθ sin ( 45° − θ ) − sin θ cos ( 45° − θ )  = 0 or or

− kr sin θ ( 2 cosθ − 1) − mg sin ( 45° − 2θ ) = 0 sin θ ( 2cosθ − 1) sin ( 45° − 2θ )

(

)

( 20 kg ) 9.81 m/s2 mg =− =− kr ( 3000 N/m )( 0.18 m ) = − 0.36333

Solving numerically

θ = 46.6°

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Chapter 10, Solution 89.

xC = d sin θ

Have

yB = h cosθ

1  V = 2  kxC2  + WyB 2 

Potential Energy:

= kd 2 sin 2 θ + Wh cosθ dV = 2kd 2 sin θ cosθ − Wh sin θ dθ

Then

= kd 2 sin 2θ − Wh sin θ d 2V = 2kd 2 cos 2θ − Wh cosθ dθ 2

and

(1)

For equilibrium position θ = 0 to be stable, we must have d 2V = 2kd 2 − Wh > 0 dθ 2

kd 2 >

1 Wh 2

(2)

1 d 2V = 0, so that we must determine which is the first derivative that is not Wh, we have 2 dθ 2 equal to zero. Differentiating Equation (1), we write

Note: For kd 2 =

d 3V = −4kd 2 sin 2θ + Wh sin θ = 0 dθ 3

for θ = 0

d 4V = −8kd 2 cos 2θ + Wh cosθ dθ 4 For θ = 0:

d 4V = −8kd 2 + Wh dθ 4

1 d 4V 1 Wh, = −4Wh + Wh < 0, we conclude that the equilibrium is unstable for kd 2 = Wh 4 2 2 dθ and the > sign in Equation (2) is correct. Since kd 2 =

With Equation (2) gives

W = 160 lb, h = 50 in., and d = 24 in. 2

k ( 24 in.) > or

1 (160 lb )( 50 in.) 2

k > 6.944 lb/in.

k > 6.94 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 10, Solution 90.

Using Equation (2) of problem 10.89 with

h = 30 in.,

k = 4 lb/in.,

kd 2 > or

( 4 lb/in.) d 2

and

W = 40 lb

1 Wh 2 1 ( 40 lb )( 30 in.) 2

d 2 > 150 in 2 or

d > 12.247 in. Smallest d = 12.25 in.

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Chapter 10, Solution 91.

Consider a small clockwise rotation θ of the plate about its center. Then

V = 2VP + 4VSP

where

a  VP = P  cosθ  2  = VSP =

and

1 ( Pa cosθ ) 2 1 2 kySP 2 2

d =

Now

= and

a 2   +a 2 a 5 2

 θ   α = 180° − φ +  90° −   2   

θ  = 90° −  φ −  2  Then

 a   5 θ  sin α ySP =     2 =

 a θ   θ 5 sin 90° −  φ −   2 2   

a θ  θ 5 cos  φ −  2 2  continued

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VSP =

and

1 a θ   k θ 5 cos  φ −   2  2 2  

5 2 2 θ  ka θ cos 2  φ −  8 2 

∴ V = Pa cosθ +

5 2 2 θ  ka θ cos 2  φ −  2 2 

Then  5 θ dV  = − Pa sin θ + ka 2  2θ cos 2  φ −  8 2 dθ  

θ  θ   1  + θ 2  −  cos  φ −  sin  φ −  2  2   2  = − Pa sin θ +

 5 2 θ 1  ka  2θ cos 2  φ −  + θ 2 sin ( 2φ − θ )  2 2 2   

 5 θ d 2V  = − Pa cosθ + ka 2  2 cos 2  φ −  2 2 2 dθ  

θ  θ  1  − 2θ  −  cos  φ −  sin  φ −  + θ sin ( 2φ − θ ) 2  2  2  −

1 2  θ cos ( 2φ − θ )  2 

= − Pa cosθ + −

5 2 θ 3  ka  2cos 2  φ −  + θ sin ( 2φ − θ ) 2 2 2  

1 2  θ cos ( 2φ − θ )  2 

  1 5 θ  θ 3 d 3V  = Pa sin θ + ka 2  4  −  cos  φ −  sin  φ −  + sin ( 2φ − θ ) 3 2 2  2 2 dθ    2 −

3 1  θ cos ( 2φ − θ ) − θ cos ( 2φ − θ ) + θ 2 sin ( 2φ − θ )  2 2 

= Pa sin θ + +

5 2 1 5 ka  sin ( 2φ − θ ) − θ cos ( 2φ − θ ) 2 2 2 

1 2  θ sin ( 2φ − θ )  2 

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When θ = 0,

dV = 0 for all values of P. dθ

For stable equilibrium when θ = 0, require d 2V 5 > 0: − Pa + ka 2 2cos 2 φ > 0 2 2 dθ

(

cos φ =

Now, when θ = 0,

)

a 2

a 5 2

1 5

1 ∴ − Pa + 5ka 2   > 0 5 P < ka

or When P = ka ( for θ = 0 ) :

dV =0 dθ d 2V =0 dθ 2 d 3V 5 = ka 2 sin 2φ > 0 ⇒ unstable 3 4 dθ

∴ Stable equilibrium for 0 ≤ P < ka W

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Chapter 10, Solution 92.

Spring:

L 2L sin φ = sin θ 3 3

For small values of φ and θ :

φ = 2θ

2L L  1 V = P  cos φ + cosθ  + ks 2 3 3   2 =

PL 1 2L ( cos 2θ + 2cosθ ) + k  sin θ  3 2  3 

dV PL 2 = ( −2sin 2θ − 2sin θ ) + kL2 sin θ cosθ dθ 3 9 =−

PL 2 ( 2sin 2θ + 2sin θ ) + kL2 sin 2θ 3 9

d 2V PL 4 =− ( 4 cos 2θ + 2cosθ ) + kL2 cos 2θ 2 3 9 dθ When

For stability:

θ = 0:

d 2V 6 PL 4 2 =− + kL 2 3 9 dθ

d 2V 4 > 0: − 2PL + kL2 > 0 2 9 dθ 0≤ P<

2 kL W 9

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Chapter 10, Solution 93.

From geometry:

xC = − a sin θ = − 2a sin φ For small values of θ , φ

θ = 2φ or φ =

1 θ 2

y A = a cosθ + 3 a cos φ

θ  = a  cosθ + 3cos  2  For spring:

s = xC = − a sin θ Potential Energy:

V = VSP + VP

1 θ 2  k ( − a sin θ ) + Pa  cosθ + 3cos  2 2 

dV 3 θ  = ka 2 cosθ sin θ − Pa  sin θ + sin  2 2 dθ  d 2V 3 θ  = ka 2 − sin 2 θ + cos 2 θ − Pa  cosθ + cos  4 2 dθ 2 

(

For stable equilibrium:

)

d 2V >0 dθ 2

Then, with θ = 0

3  ka 2 − Pa 1 +  > 0 4  or

4 ka 7

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Chapter 10, Solution 94.

Consider a small disturbance of the system defined by the angle θ . Have xC = 2a sin θ = a sin φ For small θ :

2θ = φ

Now, the Potential Energy is V =

1 2 kxB + PyE 2

xB = a sin θ

where

yE = yC + yE/C

and

= 2a cosθ + 2a cos φ = 2a ( cosθ + cos 2θ ) Then and

V =

1 2 2 ka sin θ + 2Pa ( cosθ + cos 2θ ) 2

dV 1 = ka 2 ( 2sin θ cosθ ) − 2Pa ( sin θ + 2sin 2θ ) dθ 2 =

1 2 ka sin 2θ − 2Pa ( sin θ + 2sin 2θ ) 2

d 2V = ka 2 cos 2θ − 2Pa ( cosθ + 4 cos 2θ ) dθ 2 For θ = 0 and for stable equilibrium: d 2V >0 dθ 2 or

ka 2 − 2 Pa (1 + 4 ) > 0

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1 ka 10 ∴ 0≤ P<

Check stability for

ka 10

d 3V = −2ka 2 sin 2θ + 2Pa ( sin θ + 8sin 2θ ) dθ 3 d 4V = −4ka 2 cos 2θ + 2Pa ( cosθ + 16cos 2θ ) dθ 4 Then, with

θ =0

and

ka 10

dV =0 dθ d 2V =0 dθ 2 d 3V =0 dθ 3 d 4V  1  = −4ka 2 + 2  ka  ( a )(1 + 16 ) 4 dθ  10 

= −0.6ka 2 < 0 ⇒ Unstable

1 ka W 10

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Chapter 10, Solution 95.

Displacements:

xG = xC = a sin θ

tan φ =

xG a + c − a cosθ

a sin θ a + c − a cosθ

sin θ c 1 + − cosθ a

Differentiating both sides with respect to θ :

 1  dφ =  2   cos φ  dθ

For θ = φ = 0:

c   cosθ 1 + − cosθ  − sin θ ( sin θ ) a   2 c   + − 1 cos θ   a  

c dφ a = a 2 = dθ  c  c   a

Potential Energy:

V = W1 y1 + W2 y2 = m1gb ( cos φ − 1) + m2 ga (1 − cosθ ) dV dφ = − m1gb sin φ + m2 ga sin θ dθ dθ 2  d 2φ  d 2V  dφ  = − − m gb cos φ m gb sin φ  2  + m2 ga cosθ 1 1   dθ 2  dθ   dθ 

(1)

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For θ = φ = 0, and recalling Eq. (1), 2

d 2V a = − m1gb   − 0 + m2 ga 2 dθ c For stability we need

d 2V >0 dθ 2

Thus

m1 < m2

m1g

ba 2 < m2 ga c2

c2 ab

The smallest value of m1 for stable equilibrium is thus

m1 = m2

c2 ab

Note: To determine whether the equilibrium is stable when m1 has the exact value we found, we should d 3V d 4V m c2 determine the values of the derivatives and for m1 = 2 . In practice, however we shall want to 4 3 dθ dθ ( ab ) keep m1 below this value.

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Chapter 10, Solution 96.

A = a sin θ = b sin φ

First note For small values of θ and φ :

aθ = bφ a b

φ= θ V = P ( a + b ) cos φ − 2Q ( a + b ) cosθ   a  = ( a + b )  P cos  θ  − 2Q cos θ    b    a  dV a  = ( a + b )  − P sin  θ  + 2Q sin θ  dθ b b      a2  d 2V a  a b = + ( )  − 2 P cos  θ  + 2Q cosθ  2 dθ b   b  When θ = 0:

Stability:

 a2  d 2V a b = + ( )  − 2 P + 2Q  2 dθ  b 

d 2V a2 > 0: − P + 2Q > 0 dθ 2 b2 P<2

or With

b2 Q a2

(1)

a2 P 2b 2 P = 600 N, a = 480 mm and b = 400 mm Q>

(2)

1 ( 480 mm ) Q> ( 600 N ) = 432 N 2 ( 400 mm )2 Q > 432 N

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Chapter 10, Solution 97.

xB = l sin θ

Have

xC = l sin θ1 + l sin θ 2

yC = l cosθ1 + l cosθ 2

V = PyC + or

1 2 1 2 kxB + kxC 2 2

V = Pl ( cosθ1 + cosθ 2 ) +

1 2 2 2 kl sin θ1 + ( sin θ1 + sin θ 2 )    2

For small values of θ1 and θ 2 :

sin θ1 ≈ θ1, Then

and

sin θ 2 ≈ θ 2 ,

1 cosθ1 ≈ 1 − θ12 , 2

cos θ 2 ≈ 1 −

1 2 θ2 2

 θ2 θ2  1 2 V = Pl 1 − 1 + 1 − 2  + kl 2 θ12 + (θ1 + θ 2 )    2 2 2   ∂V = − Plθ1 + kl 2 θ1 + (θ1 + θ 2 )  ∂θ1 ∂V = − Plθ 2 + kl 2 (θ1 + θ 2 ) ∂θ 2

∂ 2V = − Pl + 2kl 2 ∂θ12

∂ 2V = − Pl + kl 2 ∂θ 22

∂ 2V = kl 2 ∂θ1∂θ 2 continued

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Stability For

Conditions for stability (see page 583).

∂V ∂V = =0 ∂θ1 ∂θ 2

θ1 = θ 2 = 0:

( condition satisfied )

 ∂ 2V  ∂ 2V ∂ 2V <0   − ∂θ12 ∂θ 22  ∂θ1∂θ 2  Substituting,

( kl ) − ( −Pl + 2kl ) ( −Pl + kl ) < 0 2

k 2l 4 − P 2l 2 + 3Pkl 3 − 2k 2l 4 < 0 P 2 − 3klP + k 2l 2 > 0 Solving, or

3− 5 kl 2

P >

P < 0.382kl

P > 2.62kl

3+ 5 kl 2

∂ 2V > 0: − Pl + 2kl 2 > 0 ∂θ12 or

1 kl 2

∂ 2V > 0: − Pl + kl 2 > 0 ∂θ 22 or

P < kl

Therefore, all conditions for stable equilibrium are satisfied when 0 ≤ P < 0.382kl

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Chapter 10, Solution 98.

From the analysis of Problem 10.97 with l = 400 mm

and

k = 1.25 kN/m

P < 0.382kl = 0.382 (1250 N/m )( 0.4 m ) = 191 N 0 â&#x2030;¤ P < 191.0 N

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Chapter 10, Solution 99.

V =

Have

1 1 2 2 k ( aθ 2 ) + k ( a sin θ1 + a sin θ 2 ) + P ( 2a cosθ1 + a cosθ 2 ) 2 2 ∂V = ka 2 ( sin θ1 + sin θ 2 ) cosθ1 − 2Pa sin θ1 ∂θ1

Then

1  = ka 2  sin 2θ1 + cosθ1 sin θ 2  − 2 Pa sin θ1 2  and

∂ 2V = ka 2 ( cos 2θ1 − sin θ1 sin θ 2 ) − 2Pa cosθ1 ∂θ12 ∂ 2V = ka 2 cosθ1 cosθ 2 ∂θ1∂θ 2

Also

∂V = ka 2θ 2 + ka 2 ( sin θ1 + sin θ 2 ) cosθ 2 − Pa sin θ 2 ∂θ 2

1   = ka 2θ 2 + ka 2  sin θ1 cosθ 2 + sin 2θ 2  − Pa sin θ 2 2   and

∂ 2V = ka 2 + ka 2 ( − sin θ1 sin θ 2 + cos 2θ 2 ) − Pa cosθ 2 ∂θ 22

θ1 = θ 2 = 0

When

∂V =0 ∂θ1 ∂ 2V = ka 2 − 2Pa ∂θ12

∂ 2V = ka 2 ∂θ1∂θ 2

∂V =0 ∂θ 2

∂ 2V = ka 2 + ka 2 − Pa = 2ka 2 − Pa ∂θ 22

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∂V = 0: ∂θ1

Apply Equations 10.24

condition satisfied

∂V = 0: ∂θ 2 2

 ∂ 2V  ∂ 2V ∂ 2V < 0:   − ∂θ12 ∂θ 22  ∂θ1∂θ 2 

)(

)

− 2Pa 2ka 2 − Pa < 0

2P 2 − 5kaP + k 2a 2 > 0

Also

k 2a 2 − 2k 2a 2 + 5kaP − 2 P 2 < 0

Expanding

k 2a 2 − ( ka − 2P )( 2ka − P ) < 0

( ka ) − ( ka

condition satisfied

5 − 17 ka 4

and

P < 0.21922ka

and

P >

5 + 17 ka 4

P > 2.2808ka

δ 2V > 0: ka 2 − 2 Pa > 0 δθ12

δ 2V > 0: 2ka 2 − Pa > 0 δθ 22

1 ka 2

P < 2ka

∴ For stable equilibrium when θ1 = θ 2 = 0: 0 ≤ P < 0.219ka

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Chapter 10, Solution 100.

Potential Energy:

V = = Then

1 2 1 2 kx1 + kx2 + Py 2 2 1 1 2 2 k ( a sin θ1 + a sin θ 2 ) + k ( a θ 2 ) + P ( 2a cosθ1 + a cosθ 2 ) 2 2

∂V = ka 2 ( sin θ1 + sin θ 2 ) cosθ1 − 2P a sin θ1 ∂θ1

1  = ka 2  sin 2θ1 + cosθ1 sin θ 2  − 2Pa sin θ1 2  ∂ 2V = ka 2 ( cos 2θ1 − sin θ1 sin θ 2 ) − 2Pa cosθ1 ∂θ12 ∂ 2V = ka 2 cosθ1 cosθ 2 ∂θ1∂θ 2

∂V = ka 2 ( sin θ1 + sin θ 2 ) cosθ 2 + ka 2θ 2 − Pa sin θ 2 ∂θ 2

1   = ka 2  sin θ1 cosθ 2 + sin 2θ 2  + ka 2θ 2 − Pa sin θ 2 2   ∂ 2V = ka 2 ( − sin θ1 sin θ 2 + cos 2θ 2 ) + ka 2 − Pa cosθ 2 ∂θ 22

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θ1 = θ 2 = 0

When

∂ 2V = ka 2 ∂θ1∂θ 2

∂V =0 ∂θ1

∂V =0 ∂θ 2

∂ 2V = ka 2 − 2 Pa 2 ∂θ1 ∂ 2V = ka 2 + ka 2 − Pa = 2ka 2 − Pa 2 ∂θ 2 Apply Eq. 10.24:

∂V = 0: ∂θ1

Condition satisfied

∂V = 0: ∂θ 2

Condition satisfied

 ∂ 2V  ∂ 2V ∂ 2V < 0:   − ∂θ12 ∂θ 22  ∂θ1∂θ 2 

( ka ) − ( ka 2

)(

)

− 2Pa 2ka 2 − Pa < 0

k 2a 2 − ( ka − 2P )( 2ka − P ) < 0

k 2a 2 − 2k 2a 2 + 5Pka − 2P 2 < 0 2P 2 − 5P ka + k 2a 2 > 0 5 − 17 ka 4

and

P < 0.21922 ka

and

P > 2.2808 ka

∂ 2V >0 ∂θ12

5 + 17 ka 4

ka 2 − 2Pa > 0 or

1 ka 2

∂ 2V < 0: ∂θ 22

2ka 2 − Pa > 0 or

P < 2ka

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, for stable equilibrium when θ1 = θ 2 = 0:

0 ≤ P < 0.21922 ka with k = 2 kN/m and a = 350 mm

0 ≤ P < 0.21922 ( 2000 N/m )( 0.35 m ) or

0 ≤ P < 153.5 N

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Chapter 10, Solution 101.

From sketch

y A = 4 yC

δ y A = 4δ yC

Thus, (a) Virtual Work:

δ U = 0: P=

Pδ y A − F δ yC = 0

1 F 4

F = 300 N:

1 ( 300 N ) = 75 N 4 P = 75.0 N

(b) Free body: Corkscrew

ΣFy = 0:

R+P−F =0

R + 75 N − 300 N = 0 R = 225 N

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Chapter 10, Solution 102.

First note, by the Law of Cosines 2

DB 2 = ( 3 ft ) + ( 2 ft ) − 2 ( 3 ft )( 2 ft ) cosθ

( )

= [13 − 12cosθ ] ft 2

DB = 13 − 12 cosθ Then

1 (12 )( sin θ ) δθ 2 13 − 12 cosθ

δ B = δ DB =

6sin θ δθ 13 − 12cosθ

δB =

Also

y A = 4.5cosθ

δ y A = −4.5sin θδθ

Then Virtual Work

δ U = 0: − ( 8 kips ) δ y A − FDBδ B = 0 Then

 6sin θ −8 ( −4.5sin θ ) δθ − FDB   13 − 12 cosθ

FDB = or For

(8)( 4.5sin θ ) 6sin θ

  δθ = 0 

13 − 12cosθ

FDB = 6 13 − 12cosθ

θ = 70° FDB = 17.895 kips

FDB = 17.90 kips

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Chapter 10, Solution 103.

Given:

l AB = 3.6 in. lBC = 1.6 in. lCD = 1.2 in. lDE = 1.6 in.

lEF = 1.6 in. lFG = 4.8 in. Assume δ y A :

δ yC =

(1.6 in.) δ y = 4 δ y ( 3.6 in.) A 9 A 4 9

δ yD = δ yC = δ y A δ yG =

δφ = Virtual Work:

( 4.8 in.) δ y = ( 4.8 in.)  4 δ y  = 2 δ y ( 3.2 in.) D ( 3.2 in.)  9 A  3 A δ yD

( 3.2 in.)

4 δ yA 9 ( 3.2 in.)

δ U = 0:

( 20 lb ) δ y A + Pδ yG + (180 lb ⋅ in.) δφ = 0   2 4  δ y A  + (180 lb ⋅ in.)  δ yA  = 0 3   9 ( 3.2 in.) 

( 20 lb ) δ y A + P  Solving

P = − 67.5 lb P = 67.5 lb

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Chapter 10, Solution 104.

Given:

l AB = 3.6 in. lBC = 1.6 in. lCD = 1.2 in. lDE = 1.6 in. lEF = 1.6 in. lFG = 4.8 in. Assume δ y A :

δ yC =

(1.6 in.) δ y = 4 δ y ( 3.6 in.) A 9 A

δ yD = δ yC = δφ = Virtual Work:

δ yD

4 δ yA 9

( 3.2 in.)

4 5 δ yA = δy 9 ( 3.2 in.) ( 36 in.) A

δ U = 0:

( 20 lb ) δ y A + (180 lb ⋅ in.) δφ + M δ φ = 0 or

 5   5  20 δ y A + 180  δ y A  + M  δ y A  = 0 36 36    

Solving

M = − 324.0 lb ⋅ in. or

27.0 lb ⋅ ft

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Chapter 10, Solution 105.

Have

xB = l cosθ

δ xB = −l sin θδθ

(1)

yC = l sin θ

δ yC = l cosθδθ δ xB =

Now

1 lδφ 2

Substituting from Equation (1) −l sin θδθ = or

1 lδφ 2

δφ = −2sin θδθ

Virtual Work:

δ U = 0: M δϕ + Pδ yC = 0 M ( −2sin θδθ ) + P ( l cosθδθ ) = 0 or

M =

1 cosθ Pl 2 sin θ

M =

Pl W 2 tan θ

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Chapter 10, Solution 106.

yE = l cosθ

δ yE = − l sin θ δθ Spring:

Unstretched length = 2 ( 2 l sin 30° ) = 2 l x = 2 ( 2l sin θ ) = 4 l sin θ

δ x = 4 l cos θ δ θ FSP = k ( x − 2l ) = k ( 4 l sin θ − 2 l )

Virtual Work:

δ U = 0:

P δ yE − FSP δ x = 0

P ( − l sin θ δθ ) − k ( 4 l sin θ − 2l )( 4l cosθ δθ ) = 0

− P sin θ − 8kl ( 2sin θ − 1) cosθ = 0

P 1 − 2sin θ = 8kl tan θ

We have

P = 40 lb, l = 10 in., and k = 1.5 lb/in.

Thus

( 40 lb ) 1 − 2sin θ = 8 (1.5 lb/in.)(10 in.) tan θ

Solving

θ = 24.98° or

θ = 25.0° W

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Chapter 10, Solution 107.

δ xB =

1 l δφ 2

δC =

δ Cx δ xB l δφ = = cosθ cosθ 2cosθ

Virtual Work:

δ U = 0:

M δφ − Q δ C = 0

 l M δφ − Q   2cosθ

  δφ = 0 

M =

Ql 2cosθ

Thus

M =

1 ( 40 lb )(1.8 ft ) = 85.18 lb ⋅ ft 2 cos 65°

M = 85.2 lb ⋅ ft

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Chapter 10, Solution 108.

Assuming

δ yA it follows

δ yC =

120 δ y A = 1.5δ y A 80

δ yE = δ yC = 1.5δ y A δ yD = δ yG =

180 δ yE = 3 (1.5δ y A ) = 4.5δ y A 60

100 100 δ yE = (1.5δ y A ) = 2.5δ y A 60 60

Then, by Virtual Work

δ U = 0:

( 300 N ) δ y A − (100 N ) δ yD + Pδ yG

300δ y A − 100 ( 4.5δ y A ) + P ( 2.5δ y A ) = 0 300 − 450 + 2.5P = 0 P = +60 N

P = 60 N

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Chapter 10, Solution 109.

Apply vertical load P at D.

ΣM H = 0: − P (12 m ) + E ( 36 m ) = 0 E= ΣFy = 0:

P 3

3 P FBF − =0 5 3 FBF =

5 P 9

Virtual Work: We remove member BF and replace it with forces FBF and −FBF at pins F and B, respectively. Denoting the virtual displacements ofuuu points B and r F as δ rB and δ rF , respectively, and noting that P and δ D have the same direction, we have Virtual Work:

δ U = 0: Pδ D + FBF ⋅δ rF + ( −FBF ) ⋅δ rB = 0 Pδ D + FBF δ rF cosθ F − FBF δ rB cosθ B = 0

Pδ D − FBF (δ rB cosθ B − δ rF cosθ F ) = 0 where (δ rB cosθ B − δ rF cosθ F ) = δ BF , which is the change in length of member BF. Thus, Pδ D − FBF δ BF = 0

5  Pδ D −  P  ( 75 mm ) = 0 9 

δ D = +41.67 mm δ D = 41.7 mm

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Chapter 10, Solution 110.

Apply horizontal load P at D.

ΣM H = 0: P ( 9 m ) − E y ( 36 m ) = 0

Ey = ΣFy = 0:

P 4

3 P FBF − =0 5 4

FBF =

5 P 12

We remove member BF and replace it with forces FBF and −FBF at pins F and B, respectively. Denoting the virtual displacements ofuuu points B and r F as δ rB and δ rF , respectively, and noting that P and δ D have the same direction, we have Virtual Work:

δ U = 0: Pδ D + FBF ⋅δ rF + ( −FBF ) ⋅δ rB = 0 Pδ D + FBF δ rF cosθ F − FBF δ rB cosθ B = 0

Pδ D − FBF (δ rB cosθ B − δ rF cosθ F ) = 0 where (δ rB cosθ B − δ rF cosθ F ) = δ BF , which is the change in length of member BF. Thus,

Pδ D − FBF δ BF = 0

 5  Pδ D −  P  ( 75 mm ) = 0  12 

δ D = 31.25 mm

δ D = 31.3 mm

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Chapter 10, Solution 111.

Potential Energy

 l  l  V = 3.5 kg × 9.81 m/s 2  − sin θ  + 1.75 kg × 9.81 m/s2  cosθ   2  2 

(

)

(

)

= ( 8.5838 N ) l ( −2sin θ + cosθ ) dV = ( 8.5838 N ) l ( −2cosθ − sin θ ) dθ d 2V = ( 8.5838 N ) l ( 2sin θ − cosθ ) dθ 2 Equilibrium: or Thus

dV = 0: − 2cosθ − sin θ = 0 dθ tan θ = −2

θ = −63.4°

and

116.6°

Stability At θ = −63.4°:

d 2V = ( 8.5838 N ) l  2sin ( −63.4° ) − cos ( −63.4° )  dθ 2 = ( 8.5838 N ) l ( −1.788 − 0.448 ) < 0 ∴ θ = −63.4°, Unstable

At θ = 116.6°:

d 2V = ( 8.5838 N ) l  2sin (116.6° ) − cos (116.6° )  dθ 2 = ( 8.5838 N ) l (1.788 + 0.447 ) > 0 ∴ θ = 116.6°, Stable

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Chapter 10, Solution 112.

s = l sin θ + l cosθ − l

Elongation of Spring:

= l ( sin θ + cosθ − 1) Potential Energy:

V = =

1 2 l ks − W sin θ 2 2 1 2 l 2 kl ( sin θ + cosθ − 1) − mg sin θ 2 2

dV 1 = kl 2 ( sin θ + cosθ − 1)( cosθ − sinθ ) − mgl cosθ dθ 2 Equilibrium:

dV = 0: dθ

mg   =0 cosθ ( sin θ + cosθ − 1)(1 − tan θ ) − 2kl  

Now with

W = mg = (125 kg ) 9.81 m/s 2 = 1226.25 N

( sin θ

+ cosθ − 1)( cosθ − sin θ ) −

(

)

l = 320 mm, and k = 15 kN/m,   1226.25 N cosθ ( sin θ + cosθ − 1)(1 − tan θ ) − =0 2 (15000 N/m )( 0.32 m )   or

cosθ ( sin θ + cosθ − 1)(1 − tan θ ) − 0.12773 = 0

mg cosθ = 0 2kl

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By inspection, one solution is Solving numerically:

cosθ = 0

θ = 90.0°

θ = 0.38338 rad = 9.6883° and θ = 0.59053 rad = 33.8351°

Stability d 2V 1 = kl 2 ( cosθ − sin θ )( cosθ − sin θ ) + ( sin θ + cosθ − 1)( − sin θ − cosθ )  + mgl sin θ 2 2 dθ

  mg  = kl 2 1 +  sin θ + cosθ − 2sin 2θ  2kl      (1226.25 N ) 2  sin θ + cosθ − 2sin 2θ  = (15000 N/m )( 0.32 m ) 1 +  2 (15000 N/m )( 0.32 m )   

= (1536 N ⋅ m ) [1.12773 sin θ + cosθ − 2sin 2θ ] Thus

At θ = 90°:

d 2V = 1732.2 > 0 dθ 2

∴ θ = 90.0°, Stable

At θ = 9.6883°:

d 2V = 786.4 > 0 dθ 2

∴ θ = 9.69°, Stable

At θ = 33.8351°:

d 2V = − 600.6 < 0 dθ 2

∴ θ = 33.8°, Unstable