solution_manual_vme_9e_chapter_10 - · pdf file3. proprietary material. © 2010 the...

CHAPTER 3CHAPTER 10CHAPTER 10

3

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

PROBLEM 10.1

Determine the vertical force P that must be applied at C to maintain the equilibrium of the linkage.

SOLUTION

Assume d yA Ø

d dy yC A= 2 ≠

d dy yD A= Ø

d d dy y yE D A= =1

2

1

2Ø

d d dy y yG E A= = 1

2≠

Also dqd

=y

aA

Virtual Work: We apply both P and M to member ABC.

d d d dq d d d

d dd

U y P y M y y y

y P y My

a

A C D E G

A AA

= - - + + - =

- -

100 60 50 40 0

100 2( )ÊÊËÁ

ˆ¯̃

+ + ÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃

=

- - + +

60 501

240

1

20

100 2 60 25

d d dy y y

PM

a

A A A

-- =20 0

2 165PM

a+ = N (1)

Now from Eq. (1) for M = 0 2 165P = N P = 82 5. N Ø

4


PROBLEM 10.2

Determine the horizontal force P that must be applied at A to maintain the equilibrium of the linkage.

SOLUTION

Assume dq

d dqxA = 0 25. ¨

d dqyC = 0 1. Ø

d d dqy yD C= = 0 1. Ø

dd

dqf = =yD

0 15

2

3.

d d

dq

xG =

= ÊËÁ

ˆ¯̃

0 375

0 3752

3

.

.

f

= 0 25. dq Æ

Virtual Work: We shall assume that a force P and a couple M are applied to member ABC as shown.

d d dq d d d dU P x M y y xA C D G= + + + + =- - 150 200 22 5 400 0. f

- -P M( . ) ( . ) ( . ) . ( .0 25 150 0 1 200 0 1 22 52

3400 0dq dq dq dq dq+ + + Ê

ËÁˆ¯̃

+ 225 0dq ) =

- -0 25 15 20 15 100 0. P M + + + + =

( . )0 25 150m N mP M+ = ◊ (1)

Making M = 0 in Eq. (1): P = +600 N P = 600 N Æ

5


PROBLEM 10.3

Determine the couple M that must be applied to member ABC to maintain the equilibrium of the linkage.

SOLUTION

Assume d yA Ø

d dy yC A= 2 ≠

d dy yD A= Ø

d d dy y yE D A= =1

2

1

2Ø

d d dy y yG E A= = 1

2≠

Also dqd

=y

aA

Virtual Work: We apply both P and M to member ABC.

d d d dq d d d

d dd

U y P y M y y y

y P y My

a

A C D E G

A AA

= - - + + - =

- -

100 60 50 40 0

100 2( )ÊÊËÁ

ˆ¯̃

+ + ÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃

=

- - + +

60 501

240

1

20

100 2 60 25

d d dy y y

PM

a

A A A

-- =20 0

2 165PM

a+ = N (1)

Now from Eq. (1) for P a= =0 0 3and .

M

0 3165

. mN= M = 49 5. N m◊

6


SOLUTION

Assume dq

d dqxA = 0 25. ¨

d dqyC = 0 1. Ø

d d dqy yD C= = 0 1. Ø

dd

dqf = =yD

0 15

2

3.

d d

dq

xG =

= ÊËÁ

ˆ¯̃

0 375

0 3752

3

.

.

f

= 0 25. dq Æ

Virtual Work: We shall assume that a force P and a couple M are applied to member ABC as shown.

d d dq d d d dU P x M y y xA C D G= + + + + =- - 150 200 22 5 400 0. f

- -P M( . ) ( . ) ( . ) . ( .0 25 150 0 1 200 0 1 22 52

3400 0dq dq dq dq dq+ + + Ê

ËÁˆ¯̃

+ 225 0dq ) =

- -0 25 15 20 15 100 0. P M + + + + =

( . )0 25 150m N mP M+ = ◊ (1)

Now from Eq. (1) for P = 0 M = 150 N m◊

PROBLEM 10.4

Determine the couple M that must be applied to member ABC to maintain the equilibrium of the linkage.

7


PROBLEM 10.5

Knowing that the maximum friction force exerted by the bottle on the cork is 300 N, determine (a) the force P that must be applied to the corkscrew to open the bottle, (b) the maximum force exerted by the base of the corkscrew on the top of the bottle.

SOLUTION

From sketch y yA C= 4

Thus d dy yA C= 4

(a) Virtual Work: d d dU P y F yA C= - =0 0:

P y F yC C( )4 0d d- =

P F= 1

4

F P= = =3001

4300 7N: N) 5N(

P = 75 0. N ≠

(b) Free body: Corkscrew

+ S - -F R P F Ry = + = + =0 0 75 300; N N 0

On corkscrew: R = 225N ≠, On battle: R = 225N Ø

8


PROBLEM 10.6

The two-bar linkage shown is supported by a pin and bracket at B and a collar at D that slides freely on a vertical rod. Determine the force P required to maintain the equilibrium of the linkage.

SOLUTION

Assume d yA ≠:

d d d dy y y yC A C A= =200 1

2

mm

400 mm; Ø

Since bar CD move in translation

d d dy y yE F C= =

or d dy y yE F A= = 1

2Ø

Virtual Work: d d d dU P y y yA E F= + + =0 500 750 0: ( ) ( )- N N

-P y y yA A Ad d d+ ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

=5001

2750

1

20

P = 625N

P = 625N Ø

9


PROBLEM 10.7

A spring of constant 15 kN/m connects Points C and F of the linkage shown. Neglecting the weight of the spring and linkage, determine the force in the spring and the vertical motion of Point G when a vertical downward 120-N force is applied (a) at Point C, (b) at Points C and H.

SOLUTION

y y

y y y y

y y y y

y y y y

G C

H C H C

F C F C

E C E C

== == == =

4

4 4

3 3

2 2

d dd dd d

For spring: D = -y yF C

Q = Force in spring (assumed in tension)

Q k k y y k y y kyF C C C C= + = - = - =D ( ) ( )3 2 (1)

(a) C E F H= = = =120 0N,

Virtual Work:

d d d dU y Q y Q yC C F= - + - =0 120 0: ( )N

- + - =120 3 0d d dy Q y Q yC C C( )

Q = -60 N Q = 60 0. N C

Eq. (1): Q ky y yC C C= - = = -2 60 2 15 2, ( ) ,N kN/m mm

At Point G: y yG C= = - = -4 4 2 8( )mm mm yG = 8 00. mm Ø

(b) C H E F= = = =120 0N,

Virtual Work:

d d d dU y y Q y Q yC H C F= - - + - =0 120 120 0: ( ) ( )N N

- - + - =120 120 4 3 0d d d dy y Q y Q yC C C C( ) ( )

Q = -300 N Q = 300 N C

Eq. (1): Q ky y yC C C= - = = -2 300 10N 2(15 kN/m) mm,


10


PROBLEM 10.8

A spring of constant 15 kN/m connects Points C and F of the linkage shown. Neglecting the weight of the spring and linkage, determine the force in the spring and the vertical motion of Point G when a vertical downward 120-N force is applied (a) at Point E, (b) at Points E and F.

SOLUTION

y y

y y y y

y y y y

y y y y

G C

H C H C

F C F C

E C E C

== == == =

4

4 4

3 3

2 2

d dd dd d

For spring: D = -y yF C

Q = Force in spring (assumed in tension)

Q = + = - = - =k k y y k y y kyF C C C CD ( ) ( )3 2 (1)

(a) E C F H= = = =120 0N,

Virtual Work: d d d dU y Q y Q yE C F= - + - =0 120 0: ( )N

- + - =120 2 3 0( ) ( )d d dy Q y Q yC C C

Q = -120 N Q = 120 0. N C

Eq. (1): Q ky y yC C C= - = = -2 120 2 15 4, ( ) ,N kN/m mm

At Point G: y yG C= = - =4 4 4 16( )mm mm yG = 16 00. mm Ø

(b) E F C H= = = =120 0N,

Virtual Work:

d d d d dU y y Q y Q yE F C F= - - + - =0 120 120 0: ( ) ( )N N

- - + - =120 2 120 3 3 0( ) ( ) ( )d d d dy y Q y Q yC C C C

Q = -300 N Q = 300 N C

Eq. (1): Q ky y yC C C= - = = -2 300 10, ,N 2(15 kN/m) mm


11


PROBLEM 10.9

Knowing that the line of action of the force Q passes through Point C, derive an expression for the magnitude of Q required to maintain equilibrium.

SOLUTION

We have y l y lA A= = -2 2cos ; sinq d q dq

CD l CD l= =22 2

sin ; ( ) cosq d q dq

Virtual Work:

d d dU P y Q CDA= - - =0 0: ( )

- - - ÊËÁ

ˆ¯̃

=P l Q l( sin ) cos22

0q dq q dq Q P= 22

sin

cos( )

qq /

12


PROBLEM 10.10

Solve Problem 10.9 assuming that the force P applied at Point A acts horizontally to the left.

PROBLEM 10.9 Knowing that the line of action of the force Q passes through Point C, derive an expression for the magnitude of Q required to maintain equilibrium.

SOLUTION

We have x l x lA A= =2 2cos ; cosq d q dq

CD l CD l= =22 2

sin ; ( ) cosq d q dq

Virtual Work:

d d dU P x Q CDA= - =0 0: ( )

P l Q l( cos ) cos22

0q dq q dq- ÊËÁ

ˆ¯̃

= Q P= 22

cos

cos( )

qq /

13


PROBLEM 10.11

The mechanism shown is acted upon by the force P derive an expression for the magnitude of the force Q required to maintain equilibrium.

SOLUTION

Virtual Work:

We have x l

x lA

A

==

2

2

sin

cos

qd qdq

and y lF = 3 cosq

d qdqy lF = -3 sin

Virtual Work: d d dU Q x P yA F= + =0 0:

Q l P l( cos ) ( sin )2 3 0qdq qdq+ - = Q P= 3

2tanq

14


PROBLEM 10.12

The slender rod AB is attached to a collar A and rests on a small wheel at C. Neglecting the radius of the wheel and the effect of friction, derive an expression for the magnitude of the force Q required to maintain the equilibrium of the rod.

SOLUTION

For DAA C¢ :

¢ == - ¢ = -

= -

A C a

y A C a

ya

A

A

tan

( ) tan

cos

qq

dq

dq2

For DCC B¢ :

BC l A C

l a

y BC l a

y la

B

B

¢ = - ¢= -= ¢ = -

= -

sin

sin tan

sin tan

cosco

qq q

q q

d qdqss2 q

dq

Virtual Work: d d d

qdq q

qdq

U Q y P y

Qa

P la

Q

A B= - =

- -ÊËÁ

ˆ¯̃

- -ÊËÁ

ˆ¯̃

=

0 0

02 2

:

coscos

cos

aaP l

a

coscos

cos2 2qq

qÊËÁ

ˆ¯̃

= -ÊËÁ

ˆ¯̃

Q Pl

a= -Ê

ËÁˆ¯̃

cos3 1q

15


PROBLEM 10.13

The slender rod AB is attached to a collar A and rests on a small wheel at C. Neglecting the radius of the wheel and the effect of friction, derive an expression for the magnitude of the force Q required to maintain the equilibrium of the rod.

SOLUTION

For D AA C¢ :¢ =

= - ¢ = -

= -

A C a

y A C a

ya

A

A

tan

( ) tan

cos

qq

dq

dq2

For DBB C¢ :

¢ = - ¢

= -

¢ = ¢ = -

= ¢

B C l A C

l a

BBB C l a

x BBB

sin

sin tan

tan

sin tan

tan

q

q q

qq q

q

== -

= -

l a

x lB

cos

sin

q

d qdq

Virtual Work:

d d dU P x Q yB A= - =0 0:

P l Qa

( sin )cos

- - -ÊËÁ

ˆ¯̃

=qdqq

dq2

0

or Pl Qasin cosq q2 = Q Pa

= 1 2sin cosq q

16


PROBLEM 10.14

Derive an expression for the magnitude of the force Q required to maintain the equilibrium of the mechanism shown.

SOLUTION

We have x l x l

A l

B l

D D= = -==

2 2

2

cos sinq d qdqd dqd dq

so that

Virtual Work:

d d d dU Q x P A P BD= - - - =0 0:

- - - - =- =

Q l P l P l

Ql Pl

( sin ) ( ) ( )

sin

2 2 0

2 3 0

qdq dq dqq Q

P= 3

2 sinq

17


PROBLEM 10.15

A uniform rod AB of length l and weight W is suspended from two cords AC and BC of equal length. Derive an expression for the magnitude of the couple M required to maintain equilibrium of the rod in the position shown.

SOLUTION

y h l

y l

G

G

= =

= -

cos tan cos

tan sin

q a q

d a qdq

1

21

2

Virtual Work:

d d dqU W y MG= + = 0

W l M

M Wl

-ÊËÁ

ˆ¯̃

+ =

=

1

20

1

2

tan sin

tan sin

a qdq dq

a q

18


PROBLEM 10.16

Derive an expression for the magnitude of the couple M required to maintain the equilibrium of the linkage shown.

SOLUTION

We have x lB = cosq

d qdqx lB = - sin (1)

y lC = sinq

d qdqy lC = cos

Now d dqx lB = 1

2

Substituting from Equation (1)

- =l lsinqdq d1

2f

or d qdqf = -2sin

Virtual Work: dU = 0: M P yCd df + = 0

M P l( sin ) ( cos )- + =2 0qdq qdq

or M Pl= 1

2

cos

sin

qq

MPl=

2tanq

19


PROBLEM 10.17

Derive an expression for the magnitude of the couple M required to maintain the equilibrium of the linkage shown.

SOLUTION

y a y a

y a y aE E

F F

= == =

3 3

4 4

sin cos

sin cos

q d qdqq d qdq

Virtual Work:

dU = 0: - + + =M P y P yE Fdq d d 0

- + + =M P a P adq qdq qdq( cos ) ( cos )3 4 0 M Pa= 7 cosq

20


PROBLEM 10.18

The pin at C is attached to member BCD and can slide along a slot cut in the fixed plate shown. Neglecting the effect of friction, derive an expression for the magnitude of the couple M required to maintain equilibrium when the force P that acts at D is directed (a) as shown, (b) vertically downward, (c) horizontally to the right.

SOLUTION

We have x lD = cosq

d qdqx lD = - sin

y lD = 3 sinq

d qdqy lD = 3 cos

Virtual Work: d dq b d b dU M P x P yD D= - - =0 0: ( cos ) ( sin )

M P l P ldq b qdq b qdq- - - =( cos )( sin ) ( sin )( cos )3 0

M Pl= -( sin cos cos sin )3 b q b q (1)

(a) For P directed along BCD, b q=

Equation (1): M Pl= -( sin cos cos sin )3 q q q q

M Pl= ( sin cos )2 q q M Pl= sin2q

(b) For P directed Ø, b = ∞90

Equation (1): M Pl= ∞ - ∞( sin cos cos sin )3 90 90q q M Pl= 3 cosq

(c) For P directedÆ, b = ∞180

Equation (1): M Pl= ∞ - ∞( sin cos cos sin )3 180 180q q M Pl= sinq

21


PROBLEM 10.19

A 4-kN force P is applied as shown to the piston of the engine system. Knowing that AB = 50 mm and BC = 200 mm, determine the couple M required to maintain the equilibrium of the system when (a) q = 30°, (b) q = ∞150 .

SOLUTION

Analysis of the geometry:

Law of sinessin sinfAB BC

= q

sin sinf = AB

BCq (1)

Now x AB BCC = +cos cosq f

d qdq dx AB BCC = - -sin sinf f (2)

Now, from Equation (1) cos cosf fd qdq= AB

BC

or d q dqff

= AB

BC

cos

cos (3)

From Equation (2)

d qdq q dqx AB BCAB

BCC = - -ÊËÁ

ˆ¯̃

sin sincos

cosf

f

or d q q dqxAB

C = - +cos

(sin cos sin cos )f

f f

22


PROBLEM 10.19 (Continued)

Then d q dqxAB

C = - +sin( )

cos

ff

Virtual Work: d d dqU P x MC= - - =0 0:

- - +È

ÎÍ

˘

˚˙ - =P

ABM

sin( )

cos

q dq dqff

0

Thus, M AB P= +sin( )

cos

q ff

(4)

(a) P = =4 kN, 30q ∞

Eq. (1): sin sinf = ∞5030

mm

200 mm f = ∞7 181.

Eq. (4): M =∞

( .cos .

(0 057 818

4 m)sin (30 + 7.181 )

kN)∞ ∞ M = ◊121 8. N m

(b) P = =4 kN, 150q ∞

Eq. (1): sin sin .f f= ∞ = ∞50160 7 181

mm

200 mm

Eq. (4): M = ( . (0 05 4 m)sin (150 +7.181 )

cos 7.181 kN)

∞ ∞∞

M = ◊78 2. N m

23


PROBLEM 10.20

A couple M of magnitude 100 N · m is applied as shown to the crank of the engine system. Knowing that AB = 50 mm and BC = 200 mm, determine the force P required to maintain the equilibrium of the system when (a) q = ∞60 , (b) q = ∞120 .

SOLUTION

Analysis of the geometry:

Law of sinessin sinfAB BC

= q

sin sinf = AB

BCq (1)

Now

x AB BCC = +cos cosq f

d qdq dx AB BCC = - -sin sinf f (2)

Now, from Equation (1) cos cosf fd qdq= AB

BC

or d q dqff

= AB

BC

cos

cos (3)

From Equation (2)

d qdq q dqx AB BCAB

BCC = - -ÊËÁ

ˆ¯̃

sin sincos

cosf

f

or d q q dqxAB

C = - +cos

(sin cos sin cos )f

f f

24



Then d q dqxAB

C = - +sin( )

cos

ff

Virtual Work: d d dqU P x MC= - - =0 0:

- - +È

ÎÍ

˘

˚˙ - =P

ABM

sin( )

cos

q dq dqff

0

Thus, M AB P= +sin( )

cos

q ff

(4)

(a) M = = ∞100 60 N m,◊ q

Eq. (1): sin sin .f f= ∞ = ∞5060 12 504

mm

200 mm

Eq. (4): 100 0 0560 12 504

12 504 N m m◊ =

∞ + ∞∞

( . )sin ( . )

cos .P

P = 2047 N P = 2 05. kN Æ

(b) M = = ∞100 120 N m,◊ q

Eq. (1): sin sin .f f= ∞ = ∞50120 12 504

mm

200 mm

Eq. (4): 100 0 0512 504

12 504 N m m)

sin (120◊ ∞ ∞∞

= +( .

. )

cos .P

P = 2649 N P = 2 65. kN Æ

25


PROBLEM 10.21

For the linkage shown, determine the couple M required for equilibrium when l = 0 5. m, Q = 200 N, and q = ∞65 .

SOLUTION

ddq

Cl

=12 fcos

Virtual Work:

dU = 0 : M Q Cd df - = 0

M Qldq

df f- ÊËÁ

ˆ¯̃

=1

20

cos

MQl= 1

2 cosq

Data: M = = ◊1

2

200118 31

(

cos.

N)(0.5m)

65N m

∞ M = ◊118 3. N m

26


PROBLEM 10.22

For the linkage shown, determine the force Q required for equilibrium when l = 400 mm, M = ◊75 N m, and q = ∞70 .

SOLUTION

d dq

Cl= 1

2

fcos

Virtual Work:

dU = 0: M Q Cd df - = 0

M Qldq

df f- ÊËÁ

ˆ¯̃

=1

20

cos

QM

l= 2 cosq

Data: l M Q= = ◊ = =

=

0 4 752 75

0 4128 258

70

. , ,( )

( ..m N m

N m cos70

m)N

◊ ∞

∞q

Q = 128.3N 70 0. ∞

27


PROBLEM 10.23

Determine the value of q corresponding to the equilibrium position of the mechanism of Problem 10.11 when P = 225N and Q = 800 N.

PROBLEM 10.11 The mechanism shown is acted upon by the force P derive an expression for the magnitude of the force Q required to maintain equilibrium.

SOLUTION

Virtual Work:

x lA = 2 sinq

d q dqx lA = 2 cos

y lF = 3 cosq

d q dqy lF = -3 sin

dU = 0: Q x P yA Fd d+ = 0

Q l P l( cos ) ( sin )2 3 0q dq q dq+ - =

Q P= 3

2tanq

Data: P Q= =225N 800 N

( ( )8003

2225N) N tan= q q = ∞67 1.

28


PROBLEM 10.24

Determine the value of q corresponding to the equilibrium position of the mechanism of Problem 10.9 when P = 80 N and Q = 100 N.

SOLUTION

From geometry

y l y l

CD l CD l

A A= = -

= =

2 2

22 2

cos , sin

sin , ( ) cos

q d q dqq d q dq

Virtual Work:

dU = 0: - - =P y Q CDAd d ( ) 0

- - - ÊËÁ

ˆ¯̃

=P l Q l( sin ) cos22

0q dq q dq

or Q P= ( )22

sin

cos

qq

With P Q= =80 100 N, N

(1002

N) 2(80 N)sin

cos= ( )

qq

sin

cos.

qq2

0 625( ) =

or 2

0 6252 2

2

sin cos

cos.

q q

q

( ) ( )( ) =

q = ∞36 42. q = ∞36 4.

(Additional solutions discarded as not applicable are q = ± ∞180 )

29


PROBLEM 10.25

Rod AB is attached to a block at A that can slide freely in the vertical slot shown. Neglecting the effect of friction and the weights of the rods, determine the value of q corresponding to equilibrium.

SOLUTION

Dimensions in mm

y

yA

A

==

400

400

sin

cos

qd q dq

x

xD

D

== -

100

100

cos

sin

qd q dq

Virtual Work: d d dU y xA D= - - =( (160 800 0 N) N)

- - - =( )( ( )( sin )160 400 800 100 0 cos )q dq q dq

sin

cos. ; tan .

qq

q= =0 8 0 8 q = ∞38 7.

30


PROBLEM 10.26

Solve Problem 10.25 assuming that the 800-N force is replaced by a 24-N m◊ clockwise couple applied at D.

PROBLEM 10.25 Rod AB is attached to a block at A that can slide freely in the vertical slot shown. Neglecting the effect of friction and the weights of the rods, determine the value of q corresponding to equilibrium.

SOLUTION

yA = ( . )sin0 4 m q

d q dqyA = ( . )cos0 4 m

Virtual Work: d d dqU yA= - + =( (160 24 N) N m) 0◊

- + =( (160 24 N)(0.4 m)cos N m) 0q dq dq◊

cos .q = 0 375 q = ∞68 0.

31


PROBLEM 10.27

Determine the value of q corresponding to the equilibrium position of the rod of Problem 10.12 when l = 750 mm, a = 125 mm, P = 125 N, and Q = 200 N.

PROBLEM 10.12 The slender rod AB is attached to a collar A and rests on a small wheel at C. Neglecting the radius of the wheel and the effect of friction, derive an expression for the magnitude of the force Q required to maintain the equilibrium of the rod.

SOLUTION

For D ¢AA C : ¢ =A C a tanq

y A C aA = - ¢ = -( ) tanq

dq

dqya

A = -cos2

For D ¢CC B:

BC l A C¢ = - ¢sinq

= -l asin tanq q

y BC l aB = ¢ = -sin tanq q

d q dqq

dqy la

B = -coscos2

Virtual Work:

d d dU Q y P yA B= - - =0 0:

- -ÊËÁ

ˆ¯̃

- -ÊËÁ

ˆ¯̃

=Qa

P la

coscos

cos2 20

qdq q

qdq

Qa

P la

coscos

cos2 2qq

qÊËÁ

ˆ¯̃

= -ÊËÁ

ˆ¯̃

or Q Pl

a= -Ê

ËÁˆ¯̃

cos3 1q

32



with l a P Q= = = =750 125 125 200mm, mm, N, and N

( ( cos200 125750

13N) N)mm

125mm=

ÊËÁ

ˆ¯̃

q -

or cos .3 0 4333q = q = ∞40 8.

33


PROBLEM 10.28

Determine the values of q corresponding to the equilibrium position of the rod of Problem 10.13 when l = 600 mm, a = 100 mm, P = 50 N, and Q = 90 N.

PROBLEM 10.13 The slender rod AB is attached to a collar A and rests on a small wheel at C. Neglecting the radius of the wheel and the effect of friction, derive an expression for the magnitude of the force Q required to maintain the equilibrium of the rod.

SOLUTION

For D ¢AA C : ¢ =A C a tanq

y A C aA = ¢ =- -( ) tanq

dq

dqya

A = -cos2

For D ¢BB C :

¢ = ¢

=B C l A C

l a

sin

sin tan

qq q

--

BBB C l a

¢ = ¢ =tan

sin tan

tanqq q

q-

x BB l aB = ¢ = cosq -

d q dqx lB = - sin

Virtual Work:

d d dU P x Q yB A= =0 0: -

P l Qa

( sin )cos

- - -q dqq

dq2

0ÊËÁ

ˆ¯̃

=

Pl Qasin cosq q2 =

or Q Pl

a= sin cosq q2

34



with l a P Q= = = =600 100 50 90 mm, mm, N, and N

90 50600 2 N N)

mm

100 mm= ( sin cosq q

or sin cos .q q2 0 300=

Solving numerically q = ∞19 81. and 51 9. ∞

35


PROBLEM 10.29

Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is k, and the spring is unstretched when q = ∞30 . For the loading shown, derive an equation in P, q, l, and k that must be satisfied when the system is in equilibrium.

SOLUTION

y lE = cosq

d q dqy lE = - sin

Spring:

Unstretched length = 2l

x l l= =2 2 4( sin ) sinq q

d q dqx l= 4 cos

F k x l= -( )2

F k l l= -( )4 2sinq

Virtual Work:

dU = 0: P y F xEd d- = 0

P l k l l l( sin ) ( sin )( cos )- - - =q dq q q dq4 2 4 0

- - - =P klsin ( sin )cosq q q8 2 1 0

or P

kl81 2= -( sin )

cos

sinq q

q

P

kl8

1 2= - sin

tan

qq

36


PROBLEM 10.30

Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is 300 N/m, and the spring is unstretched when q = ∞30 . Knowing that l = 250 mm and neglecting the weight of the rods, determine the value of q corresponding to equilibrium when P = 200 N.

SOLUTION

From the analysis of Problem 10.29 P

kl8

1 2= - sin

tan

qq

Then with P l k= = =200 0 25 300 N, m, and N/m.

200

8 300 0 25

1 2 N

N/m)( m)( .

sin

tan= - q

q

or 1 2 1

30 3333

- sin

tan.

qq

= =

Solving numerically q = ∞25 0.

37


PROBLEM 10.31

Solve Problem 10.30 assuming that force P is moved to C and acts vertically downward.

PROBLEM 10.30 Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is 300 N/m, and the spring is unstretched when q = ∞30 . Knowing that 250 mm and neglecting the weight of the rods, determine the value of q corresponding to equilibrium when P = 200 N.

SOLUTION

y l y lC C= =cos , sinq d qdq-

Spring:

Unstretched length == ===

2

2 2 4

4

2

l

x l l

x l

F k x l

( sin ) sin

cos

(

q qd qdq

- ))

( sin )F k l l= 4 2q -Virtual Work:

d d dU P y F xC= 0: - -

- - - -P l k l l l( sin ) ( sin )( cos )qdq q qdq4 2 4 0=

P klsin ( sin )cosq q q- -8 2 1 0=

or P

kl82 1= ( sin )

cos

sinq q

q-

with l k P= = =250 300 200 mm, N/m, and N

(

( . )( sin )

cos

sin

200

8 300 0 252 1

N)

N/m)( m= q q

q-

or ( sin )cos

sin2 1

1

3q q

q- =


and q = ∞68 96. q = ∞39 7.

and q = ∞69 0.

38


PROBLEM 10.32

Rod ABC is attached to blocks A and B that can move freely in the guides shown. The constant of the spring attached at A is k = 3 kN/m, and the spring is unstretched when the rod is vertical. For the loading shown, determine the value of q corresponding to equilibrium.

SOLUTION

x

xC

C

==

( . )sin

. cos

0 4

0 4

m qd qdq

y

yA

A

==

( . )cos

. sin

0 2

0 2

m qd qdq-

Spring:

Unstretched length = 0 2. m

F k y k

F

A= ===

( . ) ( . . cos )

( )( . )( cos )

( c

0 2 0 2 0 2

300 0 2 1

600 1

m

N/m

- --

-

qq

oos )q

Virtual Work: d d dU x F yC A= + =0 150 0: ( )N

150 0 4 600 1 0 2 0( . cos ) ( cos )( . sin )qdq q qdq+ =- -

150 0 4

600 0 2

1

2

1

21

( . )

( . ): ( cos ) tan= = - q q

Solve by trial and error: q = ∞57 2.

39


PROBLEM 10.33

A load W of magnitude 600 N is applied to the linkage at B. The constant of the spring is k = 2.5 kN/m, and the spring is unstretched when AB and BC are horizontal. Neglecting the weight of the linkage and knowing that l = 300 mm, determine the value of q corresponding to equilibrium.

SOLUTION

x l x l

y l y l

F ks k l x kl

C C

B B

C

= =

= =

= = =

2 2

2 2

cos sin

sin cos

( )

q d qdq

q d qdq

-

- (( cos )1- q

Virtual Work: d d dU F x W yC B= + =0 0:

2 1 2 0kl l W l( cos )( sin ) ( cos )- -q qdq qdq+ =

4 12kl Wl( cos )sin cos- q q q=

or ( cos ) tan14

- q q = W

kl

Given: l W k= = =0 3 600 2500. , m N, N/m

Then ( cos ) tan.

1600

0 3- q q = N

4(2500 N/m)( m)

or ( cos ) tan .1 0 2- q q =

Solving numerically q = ∞40 22. q = ∞40 2.

40


PROBLEM 10.34

A vertical load W is applied to the linkage at B. The constant of the spring is k, and the spring is unstretched when AB and BC are horizontal. Neglecting the weight of the linkage, derive an equation in q, W, l, and k that must be satisfied when the linkage is in equilibrium.

SOLUTION

x l x l

y l y l

F ks k l x kl

C C

B B

C

= =

= =

= = =

2 2

2 2

cos sin

sin cos

( )

q d qdq

q d qdq

-

- (( cos )1- q

Virtual Work: d d dU F x W yC B= + =0 0:

2 1 2 0kl l W l( cos )( sin ) ( cos )- -q qdq qdq+ =

4 12kl Wl( cos )sin cos- q q q=

or ( cos ) tan14

- q q = W

kl

From above ( cos ) tan14

- q q = W

kl

41


PROBLEM 10.35

Knowing that the constant of spring CD is k and that the spring is unstretched when rod ABC is horizontal, determine the value of q corresponding to equilibrium for the data indicated.

P l k= = =300 N, 400 mm, 5 kN/m

SOLUTION

y lA = sinq

d qdqy lA = cos

Spring: v CD=

Unstretched when q = 0

so that v l0 2=

For q : v l

v l

= ∞ +ÊËÁ

ˆ¯̃

= ∞ +ÊËÁ

ˆ¯̃

290

2

452

sin

cos

q

d q dq

Stretched length: s v v l l= = ∞ +ÊËÁ

ˆ¯̃

- -0 2 452

2sinq

Then F ks kl= = ∞ +ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚̇

2 452

2sinq -

Virtual Work:

d d dU P y F vA= =0 0: -

Pl kl lcos sin cosqdq q q dq- -2 452

2 452

0∞ +ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚̇

∞ +ÊËÁ

ˆ¯̃

=

or P

kl= ∞ +Ê

ËÁˆ¯̃

∞ +ÊËÁ

ˆ¯̃

∞ +ÊËÁ

ˆ¯̃

È12 45

245

22 45

2cossin cos cos

qq q q-

ÎÎÍ˘˚̇

= ∞ +ÊËÁ

ˆ¯̃

∞ +ÊËÁ

ˆ¯̃

∞ +ÊËÁ

ˆ¯̃

12 45

245

22 45

2cossin cos cos cos

qq q q q-ÈÈ

ÎÍ˘˚̇

=∞ +( )

1 245 2-

cos

cos

q

q

42



Now, with P l k= = =300 400 5 N, mm, and kN/m

( )

( )( . )

cos

cos

300

5000 0 41 2

45 2 N

N/m m=

∞ +( )-

q

q

or cos

cos.

450 601042∞ +( )

=q

qSolving numerically q = ∞22 6.

43


PROBLEM 10.36

Knowing that the constant of spring CD is k and that the spring is unstretched when rod ABC is horizontal, determine the value of q corresponding to equilibrium for the data indicated.

P l k= = =500 N, 600 mm, 4 kN/m

SOLUTION

From the analysis of Problem 10.35, we have

P

kl=

∞ +( )1 2

45 2-cos

cos

q

qwith P l k= = =500 0 6 4000N, m and N/m.

(

( )( . )

cos

cos

500

4000 0 61 2

45 2N)

N/m m=

∞ +( )-

q

q

or cos

cos.

450 559792∞ +( )

=q


44


PROBLEM 10.37

A load W of magnitude 360 N is applied to the mechanism at C. Neglecting the weight of the mechanism, determine the value of q corresponding to equilibrium. The constant of the spring is k = 4 kN/m, and the spring is unstretched when q = 0.

SOLUTION

s r

s r

F ks kr

y l

y l

C

C

=

=

= =

=

=

q

d dq

q

q

d qdq

sin

cos

Virtual Work: d d dU F s W yC= + =- 0

-kr r W lq dq qdq( ) ( cos )+ = 0

q

qC Wl

krcos

( )( . )

( )( . ).= = =

2 2

360 0 375

4000 0 151 500

N m

N/m m

Solving by trial and error: q = 0 91486. rad q = ∞52 4.

45


PROBLEM 10.38

A force P of magnitude 240 N is applied to end E of cable CDE, which passes under pulley D and is attached to the mechanism at C. Neglecting the weight of the mechanism and the radius of the pulley, determine the value of q corresponding to equilibrium. The constant of the spring is k = 4 kN/m, and the spring is unstretched when q = ∞90 .

SOLUTION

s r

s r

F ks kr

CD l

= ÊËÁ

ˆ¯̃

=

= = ÊËÁ

ˆ¯̃

=

p q

d dqp q

q

2

2

22

-

-

-

sin

d q dq

q dq

( ) cos

cos

CD l

l

= ÊËÁ

ˆ¯̃

=

22

1

2

2

Virtual Work:

Since F tends to decrease s and P tends to decrease CD, we have

d d dU F s P CD= =- - ( ) 0

- - - -kr r P lp q dq q dq2 2

0ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

=( ) cos

p

q

q2

22 2

240 0 3

4000 0 121 25

-cos

( )( . )

( )( . ).= = =pl

kr

N m

N/m m


46


PROBLEM 10.39

The lever AB is attached to the horizontal shaft BC that passes through a bearing and is welded to a fixed support at C. The torsional spring constant of the shaft BC is K; that is, a couple of magnitude K is required to rotate end B through 1 rad. Knowing that the shaft is untwisted when AB is horizontal, determine the value of q corresponding to the position of equilibrium when P l= =100 N, 250 mm, and K = 12.5 N m/rad.◊

SOLUTION

We have y lA = sinq

d qdqy lA = cos

Virtual Work:

d d dqU P y MA= =0 0: -

Pl Kcosqdq qdq- = 0

or q

qcos= Pl

K (1)

with P l K= = =100 250 12 5 N, mm and N m/rad. ◊

q

qcos

( )( . )

.=

100 0 250

12 5

N m

N m/rad◊

or q

qcos.= 2 0000


47


PROBLEM 10.40

Solve Problem 10.39 assuming that P = 350 N, l = 250 mm, and K = 12.5 N m/rad.◊ Obtain answers in each of the following quadrants: 0 90 , 270 360 , 360 450 . q q q∞ ∞ ∞ ∞ ∞

PROBLEM 10.39 The lever AB is attached to the horizontal shaft BC that passes through a bearing and is welded to a fixed support at C. The torsional spring constant of the shaft BC is K; that is, a couple of magnitude K is required to rotate end B through 1 rad. Knowing that the shaft is untwisted when AB is horizontal, determine the value of q corresponding to the position of equilibrium when P = 100 N, l = 250 mm, and K = 12.5 N m/rad.◊

SOLUTION

Using Equation (1) of Problem 10.39 and

P l K= = =350 250 12 5 N, mm and N m/rad. ◊

We have q

qcos

( )( . )=

350 0 250 N m

12.5 N m/rad◊

or q

qq q

coscos= =7 7or (1)

The solutions to this equation can be shown graphically using any appropriate graphing tool, such as Maple, with the command: plot ({theta, 7 cos(theta)} Pi/2 ;* , . * )t = 0 5

Thus, we plot y y= =q q and in the range7cos

05

2 q p

48



We observe that there are three points of intersection, which implies that Equation (1) has three roots in the specified range of q.

0 902

1 37333 78 69 q p q q∞ÊËÁ

ˆ¯̃

= = ∞; . . rad, q = ∞78 7.

270 3603

22 5 65222 323 85 q p q p q q∞Ê

ËÁˆ¯̃

= = ∞; . . rad, q = ∞324

360 450 25

26 61597 379 07 q p q p q q∞Ê

ËÁˆ¯̃

= =; . . rad, q = ∞379

49


PROBLEM 10.41

The position of boom ABC is controlled by the hydraulic cylinder BD. For the loading shown, determine the force exerted by the hydraulic cylinder on pin B when q = ∞65 .

SOLUTION

We have

y yA A= =( )cos sin1200 1200mm q d qdq-

( ) ( ) ( ) ( )( )cos( )

( ) ( ) (

BD BC CD BC CD2 2 2

2 2

2 180

750 400 2 750

= + ∞

= + +

- - q

))( )cos400 q

( ) , , cosBD 2 722 500 600 000= + q (1)

Differentiating: 2 600 000( ) ( ) , sinBD BDd qdq= -

d qdq( ),

sinBDBD

= -300 000

(2)

Virtual Work: Noting that P tends to decrease yA and FBD tends to increase BD, write

d d dU P y F BDA BD= + =- ( ) 0

- - -P FBDBD( sin ),

sin1200300 000

0qdq qdq+ ÊËÁ

ˆ¯̃

=

F BD PBD = 1

250( )

or, since P F BD BD BDBD= = =30003000

25012N: in mm, in N( ) ( )( )( )P (3)

Making q = ∞65 in Eq. (1), we have

( ) , , cos

.

BD

BD

2 722 500 600 000 65

987 963

= + ∞ == mm

Carrying into Eq. (3): FBD = =( )( . ) , .12 987 963 11 855 6 N FBD = 11 86. kN

50


PROBLEM 10.42

The position of boom ABC is controlled by the hydraulic cylinder BD. For the loading shown, (a) express the force exerted by the hydraulic cylinder on pin B as a function of the length BD, (b) determine the smallest possible value of the angle q if the maximum force that the cylinder can exert on pin B is 12.5 kN.

SOLUTION

(a) See solution of Problem 10.41 for the derivation of Eq. (3):

F BDBD = ( )( )12

(b) For ( ) . ,maxFBD = =12 5 12 500kN N, we have (BD in mm, FBD

in N)

12500 12

1041 67

==

( )( )

.

BD

BD mm

Carrying this value into Eq. (1) of Problem 10.41, write

( ) , , cos

( . ) , , cos

cos

BD 2

2

722 500 600 000

1041 67 722 500 600 000

= +

= +

q

qqq = 0 60428. q = ∞52 8.

51


PROBLEM 10.43

The position of member ABC is controlled by the hydraulic cylinder CD. For the loading shown, determine the force exerted by the hydraulic cylinder on pin C when q = ∞55 .

SOLUTION

y

y

CD BC BD BC BD

C

A

A

==

= + - ∞ -

( . )sin

. cos

( )( )cos( )

0 8

0 8

2 902 2 2

m qd qdq

q

DD

CD

2 2 2

2

0 5 1 5 2 0 5 1 5

2 5 1 5

= + -

= -

. . ( . )( . )sin

. . sin

q

q (1)

2 1 53

4( )( ) . cos

cosCD

CDCD CDd qdq d q dq= - = -

Virtual Work: d d dU y FA CD CD= - - =0 10 0: ( )kN

- - -ÊËÁ

ˆ¯̃

=10 0 83

40( . cos )

cosqdq q dqFCDCD

F CDCD = 32

3 (2)

For q = ∞55 :

Eq. (1): CD CD2 2 5 1 5 55 1 2713 1 1275= - ∞ = =. . sin . ; . m

Eq. (2): F CDCD = = =32

3

32

31 1275 12 027( . ) . kN FCD = 12 03. kN

52


PROBLEM 10.44

The position of member ABC is controlled by the hydraulic cylinder CD. Determine the angle q knowing that the hydraulic cylinder exerts a 15-kN force on pin C.

SOLUTION

y

y

CD BC BD BC BD

C

A

A

==

= + ∞

( . )sin

. cos

( )( )cos( )

0 8

0 8

2 902 2 2

m qd qdq

q- -

DD

CD

2 2 2

2

0 5 1 5 2 0 5 1 5

2 5 1 5

= +

=

. . ( . )( . )sin

. . sin

-

-

q

q (1)

2 1 53

4( )( ) . cos ;

cosCD

CDCD CDd qdq d q dq= =- -

Virtual Work: d d dU y FA CD CD= - =0 10 0: ( )- kN

- -10 0 83

40( . cos )

cosqdq q dqFCDCD -Ê

ËÁˆ¯̃

=

F CDCD = 32

3 (2)

For FCD = 15 kN:

Eq. (2): 1532

3kN = CD : CD = =45

321 40625. m

Eq. (1): ( . ) . . sin : sin .1 40625 2 5 1 5 0 348312 = - =q q

q = ∞20 38. q = ∞20 4.

53


SOLUTION

In D ADE : tan.

.

.

sin ..

a

a

= =

= ∞

=∞

=

AE

DE

AD

0.9m

m

mm

0 5

60 945

0 9

60 9451 02956

From the geometry: y

yC

C

==

( )sin

( )cos

5

5

m

m

qd qdq

Then, in triangle BAD: Angle BAD = +a q

Law of cosines:

BD AB AD AB AD2 2 2 2= + - +( )( )cos( )a q

or BD2 2 22 4 1 02956 2 2 4 1 02956= + - +( . ) ( . ) ( . )( . )cos( )m m m m a q

BD2 6 81999 4 94189= - +. ( . cos( ))m m2 2a q (1)

PROBLEM 10.45

The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together weigh 2500 N and their combined center of gravity is located directly above C. For the position when q = ∞20 , determine the force exerted on pin B by the single hydraulic cylinder BD.

54



And then

2 4 94189

4 94189

2

( )( ) ( . sin( ))

. sin( )

( )

BD BD

BDBD

d a q dq

d a q dq

= +

= +

Virtual Work: d d dU P y F BDC BD= - + =0 0:

Substituting - ++È

ÎÍ

˘

˚˙ =( )( )cos

( . )sin( )

( )2500 5

4 94189

20

2

N mm

qdqa q

dqFBDBD

or F BDBD =+

È

ÎÍ

˘

˚˙5058 79.

cos

sin( )

qa q

N/m (2)

Now, with q = ∞20 and a = ∞60 945.

Equation (1):

BD

BD

BD

2

2

6 8199 4 94189 60 945 20

6 04213

2 4581

= - ∞ + ∞

==

. . cos( . )

.

. m

Equation (2):

FBD = ∞∞ + ∞

È

ÎÍ

˘

˚˙5058 79

20

60 945 202 4581.

cos

sin( . )( . )m N/m

or FBD = 11 832 55, . N FBD = 11 83. kN

55


PROBLEM 10.46

Solve Problem 10.45 assuming that the workers are lowered to a point near the ground so that q = - ∞20 .

PROBLEM 10.45 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together weigh 2500 N and their combined center of grav-ity is located directly above C. For the position when q = ∞20 , deter-mine the force exerted on pin B by the single hydraulic cylinder BD.

SOLUTION

Using the figure and analysis of Problem 10.45, including Equations (1) and (2), and with q = - ∞20 , we have

Equation (1): BD

BD

BD

2

2

6 8199 4 94189 60 945 20

3 0871

1 7570

= - ∞ - ∞

==

. . cos( . )

.

. m

Equation (2): FBD = - ∞∞ - ∞

5058 7920

60 945 201 7570.

cos( )

sin( . )( . )

FBD = 12 745 04, . N or FBD = 12 75. kN

56


PROBLEM 10.47

A block of weight W is pulled up a plane forming an angle a with the horizontal by a force P directed along the plane. If m is the coefficient of friction between the block and the plane, derive an expression for the mechanical efficiency of the system. Show that the mechanical efficiency cannot exceed 1

2 if the block is to remain in place when the force P is removed.

SOLUTION

Input work

Output work

==

P x

W x

da d( sin )

Efficiency:

h add

h a= =W x

P x

W

P

sin sinor (1)

+ SF P F W P W Fx = - - = = +0 0: sin sina aor (2)

+ SF N W N Wy = - = =0 0: cos cosa aor

F N W= =m m acos

Equation (2): P W W W= + = +sin cos (sin cos )a m a a m a

Equation (1): h aa m a

hm a

=+

=+

W

W

sin

(sin cos ) cotor

1

1

If block is to remain in place when P = 0, we know (see Chapter 8) that fs a or, since

m m a= tan , tanfs

Multiply by cot :a m a a acot tan cot = 1

Add 1 to each side: 1 2+ m acot

Recalling the expression for h, we find h 1

2

57


SOLUTION

For the linkage:

+ SM xx

PP

B AA= - + = =02

02

: or A ≠

Then: F AP

Ps s s= = =m m m2

1

2

Now x l

x lA

A

==

2

2

sin

cos

qd q dq

and y l

y lF

F

== -

3

3

cos

sin

qd q dq

Virtual Work: d d dU Q F x P yA F= - + =0 0: ( )max

Q P l P lsmax ( cos ) ( sin )-ÊËÁ

ˆ¯̃

+ - =1

22 3 0m q dq q dq

or Q P Psmax tan= +3

2

1

2q m

QP

smax ( tan )= +2

3 q m

For Qmin, motion of A impends to the right and F acts to the left. We change ms to -ms and find

QP

smin ( tan )= -2

3 q m

PROBLEM 10.48

Denoting by ms the coefficient of static friction between the block attached to rod ACE and the horizontal surface, derive expressions in terms of P, ms, and q for the largest and smallest magnitude of the force Q for which equilibrium is maintained.

58


PROBLEM 10.49

Knowing that the coefficient of static friction between the block attached to rod ACE and the horizontal surface is 0.15, determine the magnitude of the largest and smallest force Q for which equilibrium is maintained when q = ∞ =30 , 0.2 m,l and P = 40 N.

SOLUTION

Using the results of Problem 10.48 with

q

m

= ∞== =

30

0 2

40 0 15

l

P s

.

.

m

N, and

We have QP

smax ( tan )

( )( tan . )

.

= +

= ∞ +

=

23

40

23 30 0 15

37 64

q m

N

N

Qmax .= 37 6 N

and QP

smin ( tan )

( )( tan . )

.

= -

= ∞ -

=

23

40

23 30 0 15

31 64

q m

N

N

Qmin .= 31 6 N

59


PROBLEM 10.50

Denoting by ms the coefficient of static friction between collar C and the vertical rod, derive an expression for the magnitude of the largest couple M for which equilibrium is maintained in the position shown. Explain what happens if m qs tan .

SOLUTION

Member BC: We have

x lB = cosq

d qdqx lB = - sin (1)

and

y lC = sinq

d dy lC = cosq q (2)

Member AB: We have

d dx lB = 1

2f

Substituting from Equation (1),

- =l lsinqdq d1

2f

or d qdqf = -2sin (3)

Free body of rod BC

For Mmax, motion of collar C impends upward

+ SM Nl P N lB s= - + =0 0: sin ( )( cos )q m q

N N P

NP

s

s

tan

tan

q m

q m

- =

=-

60



Virtual Work:

d df m dU M P N yC= + + =0 0; ( )s

M P N l

MP N

lPs

s sP

( sin ) ( ) cos

( )

tanmaxtan

- + + =

=+

=+ -

2 0

2

qdq m qdq

mq

m q mss l2tanq

or MPl

smax (tan )

=-2 q m

If m qs M= = •tan , , system becomes self-locking

61


PROBLEM 10.51

Knowing that the coefficient of static friction between collar C and the vertical rod is 0.40, determine the magnitude of the largest and smallest couple M for which equilibrium is maintained in the position shown, when q = 35°, l = 600 mm, andP = 300 N.

SOLUTION

From the analysis of Problem 10.50, we have

MPl

smax (tan )

=+2 q m

With q = ∞ = =35 0 6 300, .l Pm, N

Mmax( )( . )

(tan . )

.

=∞ -

= ◊

300 0 6

2 35 0 4

299 80

N m

N m Mmax = 300 N m◊

For Mmin, motion of C impends downward and F acts upward. The equations of Problem 10.50 can still be used if we replace ms by -ms . Then

MPl

smin (tan )

=+2 q m

Substituting,

Mmin( )( . )

(tan . )

.

=∞ +

= ◊

300 0 6

2 35 0 4

81 803

N m

N m Mmin .= 81 8 N m◊

62


PROBLEM 10.52

Derive an expression for the mechanical efficiency of the jack discussed in Section 8.6. Show that if the jack is to be self-locking, the mechanical efficiency cannot exceed 1

2.

SOLUTION

Recall Figure 8.9a. Draw force triangle

Q W

y x y x

Q x W

s= +

= =

= =

tan( )

tan tan

tan(

q

q d d q

d q

f

so that

Input work ++

= =

fs x

W y W x

)

( ) tan

d

d d qOutput work

Efficiency: h qdq d

=+

W x

W xs

tan

tan( );

f h q

q=

+tan

tan( )fs

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

Then

f

fs

q

q q+ s 2

so that tan( ) tanq q+ fs 2

From above h qq

=+

tan

tan( )fs

It then follows that h qq

tan

tan2

But tantan

tan2

2

1 2q q

q=

-

Then h q qq

q

tan ( tan )

tan

tan1

2

1

2

2 2- = - h £ 1

2

63


PROBLEM 10.53

Using the method of virtual work, determine the reaction at E.

SOLUTION

We release the support at E and assume a virtual displacementd yE for Point E.

From similar triangles:

d d d

d d d

d d

y y y

y y y

y y

D E E

C E E

B C

= =

= =

= =

2 1

1 21 75

2 7

1 22 25

0 5

0 9

0

.

..

.

..

.

.

.55

0 92 25 1 25

.( . ) .d dy yE E=

Virtual Work:

d d d dU y y E yB D E= + - =( ) ( )2 3 0kN kN

2 1 25 3 1 75 0( . ) ( . )d d dy y E yE E E+ - =

E = +7 75. kN E = 7 75. kN ≠

64


PROBLEM 10.54

Using the method of virtual work, determine separately the force and couple representing the reaction at H.

SOLUTION

Force at H. We give a vertical virtual displacementd yH to Point H, keeping member FH horizontal.

From the geometry of the diagram:

d d d

d d d d

d d

y y y

y y y y

y y

F G H

D F H H

C D

= =

= = =

= =

0 9

1 2

0 9

1 20 75

1 5

0 9

1

.

.

.

..

.

.

.55

0 90 75 1 25

0 5

0 9

0 5

0 91 25 0 6944

.( . ) .

.

.

.

.( . ) .

d d

d d d

y y

y y y

H H

B C H

=

= = = 44d yH

Virtual Work: d d d d d

dU y y y H y

yB D G H

H

= + - + =+

( ( (

( . ) ( .

2 3 5 0

2 0 69444 3 0 7

kN) kN) kN)

55 5 0d d dy y H yH H H) - + =

H = +1 3611. kN

H = 1 361. kN ≠

Couple at H. We rotate beam FH throughdq about Point H.

65



From the geometry of the diagram:

d dq d dq

d d dq dq

d

y y

y y

y

G F

D F

C

= =

= = =

=

1 2 1 8

0 9

1 2

0 9

1 21 8 1 35

1

. .

.

.

.

.( . ) .

.55

0 9

1 5

0 91 35 2 25

0 5

0 9

0 5

0 92 25

.

.

.( . ) .

.

.

.

.( . )

d dq dq

d d dq

y

y y

D

B C

= =

= = == 1 25. dq

Virtual Work:

d d d d dqdq dq

U y y y MB D G H= + - + =+

( ( (

( . ) ( . )

2 3 5 0

2 1 25 3 1 35

kN) kN) kN)

-- + =5 1 2 0( . )dq dqMH

MH = - ◊0 550. kN m MH = ◊550 N m

66


PROBLEM 10.55

Referring to Problem 10.43 and using the value found for the force exerted by the hydraulic cylinder CD, determine the change in the length of CD required to raise the 10-kN load by 15 mm.

PROBLEM 10.43 The position of member ABC is controlled by the hydraulic cylinder CD. For the loading shown, determine the force exerted by the hydraulic cylinder on pin C when u = 55.

SOLUTION

Virtual Work: Assume bothd yA anddCD increase

d d dU y FA CD CD= - - =0 10 0: ( kN)

Substitute: d y FA CD= =15 12 03 mm and kN.

- - =( ( .10 12 03 0 kN)(15 mm) kN)dCD

dCD = -12 47. mm

The negative sign indicates that CD shortened dCD = 12 47. mm shorter

67


PROBLEM 10.56

Referring to Problem 10.45 and using the value found for the force exerted by the hydraulic cylinder BD, determine the change in the length of BD required to raise the platform attached at C by 60 mm.

PROBLEM 10.45 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together weigh 2500-N and their combined center of gravity is located directly above C. For the position when q = ∞20 , determine the force exerted on pin B by the single hydraulic cylinder BD.

SOLUTION

Virtual Work: Assume bothd yC anddBD increase

d d dU y FC BD BD= - + =0 2500 0: ( N)

Substitute: d y FC BD= =60 11 830mm and N,

- + =( ( ,2500 11 830 0 N)(60 mm) N)dBD

dBD = +12 6796. mm

The positive sign indicates that cylinder BD increases in length dBD = 12 68. mm longer

68


PROBLEM 10.57

Determine the vertical movement of joint D if the length of member BF is increased by 40 mm (Hint: Apply a vertical load at joint D, and, using the methods of Chapter 6, compute the force exerted by member BF on joints B and F. Then apply the method of virtual work for a virtual displacement resulting in the specified increase in length of member BF. This method should be used only for small changes in the lengths of members.)

SOLUTION

Apply vertical load P at D. + SM P EH = - + =0 10 30 0: m) m)( (

E = P

3Ø

+

SF FP

y BF= - =03

5 30:

F PBF = 5

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 of Points B and F as drB and drF, respectively, and noting that P and d D

u ruu have the same

direction, we have

Virtual Work:

d d d dU P D BF F BF B= + + - =0 0: F r F r◊ ◊( )

P D F r F rBF F F BF B Bd d q d q+ - =cos cos 0

P D F r rBF B B F Fd d q d q- - =( cos cos ) 0

where ( cos cos ) ,d q d q dr rB B F F BF- = which is the change inlength of member BF. Thus,

P D F

P D P

D

BF BFd d

d

d

- =

- ÊËÁ

ˆ¯̃

=

=

0

5

940 0

22 222

(

.

mm)

mm d D = 22 2. mm Ø

69


PROBLEM 10.58

Determine the horizontal movement of joint D if the length of member BF is increased by 40 mm (See the hint forProblem 10.57.)

SOLUTION

Apply horizontal load P at D.

+

SM P EH y= - =0 7 5 30 0: m) m)( . (

EyP=4

Ø

+

SF FP

y BF= - =03

5 40:

F PBF = 5

12

We remove member BF and replace it with forces FBF and -FBF at pins F and B, respectively. Denoting the virtual displacements of Points B and F as drB and drF, respectively, and noting that P and d D

u ruu have the same

direction, we have

Virtual Work:

d d d dU P D BF F BF B= + + - =0 0: F r F r◊ ◊( )

P D F r F rBF F F BF B Bd d q d q+ - =cos cos 0

P D F r rBF B B F Fd d q d q- - =( cos cos ) 0

where ( cos cos ) ,d q d q dr rB B F F BF- = which is the change inlength of member BF. Thus,

P D F

P D P

D

BF BFd d

d

d

- =

- ÊËÁ

ˆ¯̃

=

=

0

5

1240 0

16 667

(

.

mm)

mm d D = 16.67mm

+

70


PROBLEM 10.59

Using the method of Section 10.8, solve Problem 10.29.

PROBLEM 10.29 Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is k, and the spring is unstretched when q = ∞30 . For the loading shown, derive an equation in P, q, l, and k that must be satisfied when the system is in equilibrium.

SOLUTION

Spring: AE x l l= = =2 2 4( sin ) sinq q

Unstretched length: x l l0 4 30 2= ∞ =sin

Deflection of spring

s x x

s l

V ks Py

V k l P l

E

= -

= -

= +

= -[ ] + -

0

2

2

2 2 1

1

2

1

22 2 1

( sin )

( sin ) ( cos

q

q q ))

( sin ) cos sindV

dkl Pl

qq q q= - + =4 2 1 2 02

( sin )cos

sin2 1

80q q

q- + =P

kl

P

kl8

1 2= - sin

tan

qq

71


PROBLEM 10.60


PROBLEM 10.30 Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is 300 N/m, and the spring is unstretched when q = ∞30 . Knowing that l = 250 mm and neglecting the weight of the rods, determine the value of q corresponding to equilibrium when P = 200 N .

SOLUTION

Using the result of Problem 10.59, with

P

l k

P

kl

=

= =

= -

200

0 25 300

8

1 2

N

m and N/m.

sin

tan

qq

or 1 2 200

8 300

1

3

- =

=

sin

tan (

qq

N

N/m)(0.25 m)

Solving numerically, q = ∞25 0.

72


PROBLEM 10.61


PROBLEM 10.33 A load W of magnitude 600 N is applied to the linkage at B. The constant of the spring is k = 2 5. kN/m, and the spring is unstretched when AB and BC are horizontal. Neglecting the weight of the linkage and knowing that l = 300 mm, determine the value of q corresponding to equilibrium.

SOLUTION

P

l k

== =

600

0 3 2500

N

m, and N m. ◊

We have ( cos )tan

.

l - =

=

q q 600

0 2

N

4(2500 N/m)(0.3 m)


73


PROBLEM 10.62


PROBLEM 10.34 A vertical load W is applied to the linkage at B. The constant of the spring is k, and the spring is unstretched when AB and BC are horizontal. Neglecting the weight of the linkage, derive an equation in q, W, l, and k that must be satisfied when the linkage is in equilibrium.

SOLUTION

V ks Py

V k i x Py

x l y l

B

C B

C B

= +

= - +

= = -

1

21

22

2

2

2( )

cos sinq qand

Thus V k l l Pl

kl Pl

= - -

= - -

1

22 2

2 1

2

2 2

( cos ) sin

( cos ) sin

q q

q q

dV

dkl Pl

qq q q= - - =2 2 1 02 ( cos )sin cos ( cos )tan1

4- =q q P

kl

74


PROBLEM 10.63


PROBLEM 10.35 Knowing that the constant of spring CD is k and that the spring is unstretched when rod ABC is horizontal, determine the value of q corresponding to equilibrium for the data indicated.

P l k= = =300 N, 400 mm, 5 kN/m

SOLUTION

Spring v l

v l

= ∞ +ÊËÁ

ˆ¯̃

= ∞ +ÊËÁ

ˆ¯̃

290

2

2 452

sin

sin

q

q

Unstretched ( )q = 0 v l l0 2 45 2= ∞ =sin

Deflection of spring s v v l l

V ks Py klA

= - = ∞ +ÊËÁ

ˆ¯̃

-

= + = ∞ +ÊËÁ

ˆ

0

2 2

2 452

2

1

2

1

22 45

2

sin

sin

q

q¯̃̄

-ÈÎÍ

˘˚̇

+ -

= ∞ +ÊËÁ

ˆ¯̃

-ÈÎÍ

˘˚̇

2

2 452

2 4

2

2

P l

dV

dkl

( sin )

sin cos

q

qq

552

0∞ +ÊËÁ

ˆ¯̃

- =q qPl cos

2 452

452

2 452

sin cos cos∞ +ÊËÁ

ˆ¯̃

∞ +ÊËÁ

ˆ¯̃

- ∞ +ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚̇

=q q q P

klccos

cos cos cos

q

q q q- ∞ +ÊËÁ

ˆ¯̃

=2 452

P

kl

Divide each member by cos q 1 245 2-

∞ +( )=

cos

cos

q

qP

klThen with P l= =300 0 4 N, m. and k = 5000 N/m

1 245 300

0 15

2-∞ +( )

=

=

cos

cos

.

q

q N

(5000 N/m)(0.4 m)

or cos

cos.

450 601042∞ +( )

=q


75


PROBLEM 10.64


PROBLEM 10.36 Knowing that the constant of spring CD is k and that the spring is unstretched when rod ABC is horizontal, determine the value of q corresponding to equilibrium for the data indicated.

P l k= = =500 N, 600 mm, 40 kN/m

SOLUTION

Using the results of Problem 10.63 with P l k= = =500 0 6 4000N, m and N/m, we have.

1 245

500

5

24

2-∞ +( )

=

=

=

cos

cos

q

qP

klN

(4000 N/m)(0.6 m)

or cos

cos.

450 559792∞ +( )

=q

q

Solving numerically q = ∞36 55. q = ∞36 6.

76


PROBLEM 10.65


PROBLEM 10.31 Solve Problem 10.30 assuming that force P is moved to C and acts vertically downward.

SOLUTION

Spring: AE x l l= = =2 2 4( sin ) sinq q

Unstretched length: x l l0 4 30 2= ∞ =sin

Deflection of spring s x x

s l

V ks Py

k l P l

C

= -= -

= +

= - +

0

2

2

2 2 1

1

21

22 2 1

( sin )

[ ( sin )] ( cos )

q

q q

V kl Pl

dV

dkl Pl

= - +

= - - =

-

2 2 1

4 2 1 2 0

1

2 2

2

( sin ) cos

( sin ) cos sin

(

q q

qq q q

228

0

8

2 1

sin )cos

sinsin

tan

q qq

qq

+ =

= -

P

klP

kl

with P l k= = =200 250 300N, mm and N/m

We have ( sin

tan

200 2 1N)

8(300 N/m)(250 mm)= -q

q

or 2 1 1

3

sin

tan

- =q

Solving numerically q = ∞ ∞39 65. and 68.96 q = ∞39 7.

q = ∞69 0.

77


PROBLEM 10.66


PROBLEM 10.38 A force P of magnitude 240 N is applied to end E of cable CDE, which passes under pulley D and is attached to the mechanism at C. Neglecting the weight of the mechanism and the radius of the pulley, determine the value of q corresponding to equilibrium. The constant of the spring is k = 4 kN/m, and the spring is unstretched when q = ∞90 .

SOLUTION

For spring: s r

V ks kre

= -ÊËÁ

ˆ¯̃

= = -ÊËÁ

ˆ¯̃

p q

p q

2

1

2

1

2 22 2

2

For force P: x l

V Px Plp

=

= =

22

22

sin

sin

q

q

Then V V V kr Pl

dV

dkr Pl

e p= + = -ÊËÁ

ˆ¯̃

+

= - -ÊËÁ

ˆ¯̃

+

1

2 22

2

2

22

2

p q q

qp q

sin

cosqq

p q q

p

2

40002

2402

0

2

= - -ÊËÁ

ˆ¯̃

+ =( ( )cos N/m)(0.12 m) N)(0.3 m2

-- =q q1 25

2. cos


78


PROBLEM 10.67

Show that equilibrium is neutral in Problem 10.1.

PROBLEM 10.1 Determine the vertical force P that must be applied at C to maintain the equilibrium of the linkage.

SOLUTION

We have y b u

y b u

y u

y u

y u

A

C

D

E

G

= -= += -

= -

= +

2

1

21

2

V y Py y y y

b u P b u

A C D E G= + + + +

= - + +

( ( ( (

( ) (

100 60 50 40

100 2

N) N) N) N)

)) ( )+ - + -ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

= - + - - + =

60 501

240

1

2

100 2 60 25 20 0

u u u

dV

duP

P == 82 5. N

Substituting P = 82 5. N in the expression obtained for V :

V b u

V b

= + + - + - - += =

( . ) ( )

.

100 82 5 100 165 60 25 20

182 5 constant

Since V is constant, the equilibrium is neutral

79


PROBLEM 10.68

Show that equilibrium is neutral in Problem 10.6.

PROBLEM 10.6 The two-bar linkage shown is supported by a pin and bracket at B and a collar at D that slides freely on a vertical rod. Determine the force P required to maintain the equilibrium of the linkage.

SOLUTION

y u

y u

y u

V Py y y

V P u u

A

E

F

A E F

== -= -= + += + - +

2

500 750

2 500

( (

( ) ( ( ) (

N) N)

N) 7750

2 500 750 0

625

N)

N

( )-

= - - =

=

u

dV

duP

P

Now, substitute P = 625N in expression for V, making V = 0. Thus, V is constant and equilibrium is neutral.

80


PROBLEM 10.69

Two uniform rods, each of mass m and length l, are attached to drums that are connected by a belt as shown. Assuming that no slipping occurs between the belt and the drums, determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION

W mg

V Wl

Wl

dV

dW

l

d V

=

= ÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃

= - +

22

2

22 2

2

cos cos

( sin sin )

q q

qq

ddW

qq q

2

1

24 2= - -( cos cos )

Equilibrium: dV

d

Wl

qq q= - + =0

22 2 0: ( sin sin )

or sin ( cos )q q- + =4 1 0

Solving q = 0, 75.5 , 180 , and 284∞ ∞ ∞

Stability: d V

dW

l2

2 24 2

qq q= - -( cos cos )

At q = 0: d V

dW

l2

2 24 1 0

q= - -( ) q = 0, Unstable

At q = ∞75 5. : d V

dW

l2

2 24 874 25 0

q= - - -( ( . ) . ) q = ∞75 5. , Stable

At q = ∞180 : d V

dW

l2

2 24 1 0

q= - +( ) q = ∞180 0. , Unstable

At q = ∞284 : d V

dW

l2

2 24 874 25 0

q= - - -( ( . ) . ) q = ∞284 , Stable

81


PROBLEM 10.70

Two uniform rods AB and CD, of the same length l, are attached to gears as shown. Knowing that rod AB weighs 15 N and that rod CD weighs 10 N, determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION

V W

lW

l

dV

dW l W l

AB CD

AB CD

= - ÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃

= - +

2 22

1

2

sin cos

cos si

q q

qq nn2q

Equilibrium: dV

d

W l W l

W W

W

AB CD

AB CD

AB

q

q q

q q q

=

- + =

- + =

-

0

1

22 0

4 0

cos sin

cos sin cos

coss sinq q1 0-ÊËÁ

ˆ¯̃

=W

WCD

AB

cos sin

, , sin

q q

q q q

= =

= ∞ = ∞ = -

04

90 2704

1

andW

W

W

W

AB

CD

AB

CD

(1)

For W WAB CD= =15 10N and N we have,

q q q

q q q q

= ∞ = ∞ =

= ∞ = ∞ = ∞ = ∞

-90 2703

890

1, , sin

, 270 , 22.0 , 158.0

Stability: d V

dW W lAB CD

2

2

1

22 2

qq q= +Ê

ËÁˆ¯̃

sin cos ) (2)

qq

= ∞ = ∞ + ∞ÈÎÍ

˘˚̇

= -2701

215 270 2 10 540 27 5

2

2: ( )sin( ) ( )cos( ) .

d V

dl l 0, unstable

qq

= ∞ = ∞ + ∞ÈÎÍ

˘˚̇

22 01

215 22 2 10 44 0 0

2

2. : ( )sin ( )cos .

d V

dl Stable

82



qq

= ∞ = ∞ + ∞ÈÎÍ

˘˚̇

= -901

215 90 2 10 180 12 5 0

2

2: ( )sin ( )cos . ,

d V

dl l Unsttable

qq

= ∞ = ∞ + ∞ÈÎÍ

˘˚̇

=158 01

215 158 2 10 316 17 196

2

2. : ( )sin ( )cos .

d V

dl l 00, Stable

83


SOLUTION

Potential energy

V Wl

Wl

W mg

Wl

dV

d

Wl

= -ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

=

= -

=

2 2

2

2

sin cos

(cos sin )

(

q q

q q

q-- -

= -

sin cos )

(sin cos )

q q

qq qd V

d

Wl2

2 2

For equilibrium: dV

dqq q= = -0: sin cos

or tanq = -1

Thus q q= - ∞ = ∞45 0. and 135.0

Stability:

At q = - ∞45 0. : d V

d

Wl

Wl

2

2 245 45

2

2

2

2

20

q= - ∞ - ∞

= - -Ê

ËÁˆ

¯̃

[sin( ) cos ]

q = - ∞45 0. , Unstable

At q = ∞135 0. : d V

d

Wl

Wl

2

2 2135 135

2

2

2

2

20

q= ∞ - ∞

= +Ê

ËÁˆ

¯̃

(sin cos )

q = ∞135 0. , Stable

PROBLEM 10.71

Two uniform rods, each of mass m, are attached to gears of equal radii as shown. Determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

84


PROBLEM 10.72

Two uniform rods, AB and CD, are attached to gears of equal radii as shown. Knowing that WAB = 40 N and WCD = 20 N, determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION

Potential energy

Vl l

l

= -ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

= - +

( ) sin ( ) cos

( ( sin cos )

402

202

10 2

N N

N)

q q

q q

dV

dl

d V

dl

qq q

qq q

= - -

= -

( ( cos sin )

( ( sin cos )

10 2

10 22

2

N)

N)

Equilibrium: dV

dqq q= - - =0 2 0: cos sin

or tanq = -2

Thus q = - ∞ ∞63 4 116 6. .and

Stability

At q = - ∞63 4. : d V

dl

l

2

210 2 63 4 63 4

10 1 788 0 44q

= - ∞ - - ∞

= - -

( [ sin( . ) cos( . )]

( ) ( . .

N)

N 88 0)

q = - ∞63 4. , Unstable

At q = ∞116 6. : d V

dl

l

2

210 2 116 6 116 6

10 1 788 0 447q

= ∞ - ∞

= +

( [ sin( . ) cos( . )]

( ) ( . .

N)

N )) 0

q = ∞116 6. , Stable

85


PROBLEM 10.73

Using the method of Section 10.8, solve Problem 10.39. Determine whether the equilibrium is stable, unstable, or neutral. (Hint: The potential energy corresponding to the couple exerted by a torsion

spring is 12

2Kq , where K is the torsional spring constant and q is the

angle of twist.)

PROBLEM 10.39 The lever AB is attached to the horizontal shaft BC that passes through a bearing and is welded to a fixed support at C. The torsional spring constant of the shaft BC is K; that is, a couple of magnitude K is required to rotate end B through 1 rad. Knowing that the shaft is untwisted when AB is horizontal, determine the value of q corresponding to the position of equilibrium when P = 100 N, l = 250 mm, and K = 12 5. N m/rad.◊

SOLUTION

Potential energy V K Pl

dV

dK Pl

d V

dK Pl

= -

= -

= +

1

22

2

2

q q

qq q

qq

sin

cos

sin

Equilibrium: dV

d

K

Plqq q= =0: cos

For P l K= = =100 0 25 12 5 N m N m/rad, . , . ◊

cos( )( .

.

q q

q

=

=

12.5 N m/rad

m)

◊100 0 25

0 500

Solving numerically, we obtain

q = = ∞1 02967 59 000. .rad q = ∞59 0.

Stability

d V

d

2

212 5 100 59 0 0

q= + ∞( . ( sin . N m/rad) N)(0.25 m)◊ Stable

86


PROBLEM 10.74

In Problem 10.40, determine whether each of the positions of equilibrium is stable, unstable, or neutral. (See hint for Problem 10.73.)

PROBLEM 10.40 Solve Problem 10.39 assuming that P = 350 N, l = 250 mm, and K = 12 5. N m/rad.◊ Obtain answers in each of the following quadrants: 0 90 , 270 360 , 360 450 . q q q∞ ∞ ∞ ∞ ∞

SOLUTION

Potential energy V K Pl

dV

dK Pl

d V

dK Pl

= -

= -

= +

1

22

2

2

q q

qq q

qq

sin

cos

sin

Equilibrium dV

d

K

Plqq q= =0: cos

For P l K= = =

=

350 0 250 12 5

350

N m and N m/rad

12.5 N m/rad

N)(0

, . .

cos(

◊◊q..250 m)

q

or cosq q=7

Solving numerically q = 1 37333 6 616. . rad, 5.652 rad, and rad

or q = ∞ ∞ ∞78 7 323 8 379 1. , . , .

Stability at q = ∞78 7. : N m/rad) N)(0.250 m)d V

d

2

212 5 350 78 7

q= + ∞( . ( sin .◊

= 98 304 0. q = ∞78 7. , Stable

At q = ∞323 8. : d V

d

2

212 5 350 323 8

q= + ∞( . ( sin . N m/rad) N)(0.250 m)◊

= -39 178 0. N m◊ q = ∞324 , Unstable

At At q = ∞379 1. : d V

d

2

212 5 350 379 1

q= + ∞( . ( sin . N m/rad) N)(0.250 m)◊

= 44 132. N m 0◊ q = ∞379 , Stable

87


PROBLEM 10.75

A load W of magnitude 500 N is applied to the mechanism at C. Knowing that the spring is unstretched when q = ∞15 , determine that value of q corresponding to equilibrium and check that the equilibrium is stable.

SOLUTION

We have y l

V k r Wy

kr Wl

C

C

=

= - + = ∞ =

= - +

cos

[ ( )]

( ) co

q

q q q p

q q

1

215

12

2

02

0

20

2

rad

1ss

( ) sin

q

qq q qdV

dkr Wl= - -2

0

Equilibrium dV

dkr wl

qq q q= - - =0 02

0: ( ) sin (1)

with W k l r= = = =500 10 000 0 5 0 125N, N/m, m, and m, . .

( , ) (10 00012

500N/m)(0.125 m N)(0.5 m)sin 02 q p q-ÊËÁ

ˆ¯̃

- =

or 0 625 0 16362. sin .q q- =

Solving numerically q = =1 8145. rad 103.97∞

q = ∞104 0.

Stability d V

dkr Wl

2

22

qq= - cos (2)

or = -156 25 250. cosq

For q = ∞104 0. : = 216 73. N m 0◊ Stable

88


PROBLEM 10.76

A load W of magnitude 500 N is applied to the mechanism at C. Knowing that the spring is unstretched when q = ∞30 , determine that value of q corresponding to equilibrium and check that the equilibrium is stable.

SOLUTION

Using the solution of Problem 10.111, particularly Equations (1), with 15° replace by 306

∞ÊËÁ

ˆ¯̃

prad :

For equilibrium kr Wl2

60q p q-Ê

ËÁˆ¯̃

- =sin

With k W r l= = = =10 000 500 0 5 0 125, . .N/m, N, m, and m

( , ) ( )sin10 0006

500 0N/m)(0.125 m N)(0.5 m2 q p q-ÊËÁ

ˆ¯̃

- =

or 156 25 81 8123 250 0. . sinq q- - =

Solving numerically, q = =1 9870. rad 113.8∞

q = ∞113 8.

Stability: Equation (2), Problem 111:

d V

dkr Wl

2

22

qq= - cos

or = -156 25 250. cosq

For q = ∞113 8. : = 257 136. N m 0◊ Stable

89


PROBLEM 10.77

A slender rod AB, of weight W, is attached to two blocks A and B that can move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when AB is horizontal. Neglecting the weight of the blocks, derive an equation in q, W, l, and k that must be satisfied when the rod is in equilibrium.

SOLUTION

Elongation of spring: s l l l

s l

= + -= + -

sin cos

(sin cos )

q qq q 1

Potential energy: V ks W W mg

kl mgl

= - =

= + - -

1

2

1

21

21

2

2

2 2

sin

(sin cos ) sin

q

q q q

dV

dkl mgl

qq q q q q= + - - -2 1

1

2(sin cos )(cos sin ) cos (1)

Equilibrium: dV

d

mg

klqq q q q q= + - - - =0 1

20: (sin cos )(cos sin ) cos

or cos (sin cos )( tan )q q q q+ - - -ÈÎÍ

˘˚̇

=1 12

0mg

kl

90


PROBLEM 10.78

A slender rod AB, of weight W, is attached to two blocks A and B that can move freely in the guides shown. Knowing that the spring is unstretched when AB is horizontal, determine three values of q corresponding to equilibrium when W = 1500 N, l = 400 mm, and k = 15 kN/m . State in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION

Using the results of Problem 10.77, particularly the condition of equilibrium

cos (sin cos )( tan )q q q q+ - - -ÈÎÍ

˘˚̇

=1 12

0mg

kl

Now, with W l k= = =1500 400 15N, mm, and kN/m

W

kl2

1500 1

80 125=

ÊËÁ

ˆ¯̃

= =N

2 15000Nm

(0.4 m).

Thus: cos (sin cos )( tan ) .q q q q+ - - -[ ] =1 1 0 125 0

cos cos )( tan ) .q q q q= + - - =0 1 1 0 125and (sin

First equation yields q = ∞90 . Solving the second equation by trial, we find q = ∞ ∞9 39. and 34.16Values of q for equilibrium are

q = ∞ ∞ ∞9 39. , 34.2 , and 90.0

Stability we differentiate Eq. (1).

d y

dskl

2

22 1= - - + + - - -[(cos sin )(cos sin ) (sin cos )( sin cosq q q q q q q q ))] sin

cos sin cos sin sin cos cos sin

+

= + - - - -

1

2

2 22 2 2 2 2

wl

kl

q

q q q q q q q qq q q q

q q

+ + +ÈÎÍ

˘˚̇

= +ÊËÁ

ˆ¯̃

+ -

sin cos sin

sin cos sin

W

kl

klW

kl

2

12

2 22 qq

qq q q

ÈÎÍ

˘˚̇

= + -d V

dkl

2

22 1 125 2 2( . sin cos sin )

q = ∞9 39. : d V

dkl

2

22 1 125 9 4 9 4 2 18 8

q= + -( . sin . cos . sin . )

= + >kl2 0 53 0( . ) Stable

91



q = ∞34 2. : d v

dkl

2

22 1 125 34 2 34 2 2 68 4

q= ∞ + ∞ - ∞( . sin . cos . sin . )

= -kl2 0 400 0( . ) Unstable

q = ∞90 0. : d V

dkl

2

22 1 125 90 90 2 180

q= ∞ + ∞ - ∞( . sin cos sin )

= kl2 1 125 0( . ) Stable

92


PROBLEM 10.79

A slender rod AB, of weight W, is attached to two blocks A and B that can move freely in the guides shown. Knowing that the spring is unstretched when y = 0, determine the value of y corresponding to equilibrium when W = 80 N, l = 500 mm, and k = 600 N/m.

SOLUTION

Deflection of spring = s, where s l y l

ds

dy

y

l y

= + -

=-

2 2

2 2

Potential energy: V ks Wy

dV

dyks

ds

dyW

dV

dyk l y l

y

l yW

kl

l

= -

= -

= + -( )+

-

= -

1

2 2

1

2

1

2

1

2 2

2 2

2

2

++

Ê

ËÁÁ

ˆ

¯˜˜

-y

y W2

1

2

Equilibrium dV

dy

l

l yy

W

k= -

+

Ê

ËÁÁ

ˆ

¯˜˜

=0 11

22 2:

Now W l k= = =80 0 500 600 N, m, and N/m.

Then 10 500

0 500

1

2

80

6002 2-

+

Ê

ËÁÁ

ˆ

¯˜˜

=.

( . )

(

(

m N)

N/m)yy

or 10 500

0 250 066667

2-

+

Ê

ËÁÁ

ˆ

¯˜˜

=.

..

yy

Solving numerically, y = 0 357. m y = 357 mm

93


PROBLEM 10.80

Knowing that both springs are unstretched when y = 0, determine the value of y corresponding to equilibrium when W = 80 N,l = 500 mm, and k = 600 N/m.

SOLUTION

Spring deflections

S l l y

S l y l

AD

BC

= - -

= + -

2 2

2 2

V kS kS W

y

V k l l y k l y l Wy

dV

dy

AD BC= + -

= - -( ) + + -( ) -

1

2

1

2 2

1

2

1

2 2

2 2

2 22

2 22

== - -( )-

Ê

ËÁÁ

ˆ

¯˜˜

+ + -( )+

Ê

ËÁÁ

ˆ

¯˜˜

-k l l yy

l yk l y l

y

l y

W2 2

2 2

2 2

2 2 2

dV

dy

l

l y

l

l yy

W

k=

--

Ê

ËÁÁ

ˆ

¯˜˜

+ -+

Ê

ËÁÁ

ˆ

¯˜˜

È

ÎÍÍ

˘

˚˙˙

=0 1 122 2 2 2

:

Data: W l k= = =80 600 N, 0.5 m, N/m

0 5

0 5

0 5

0 5

80

2 12000 066667

2 2 2 2

.

( . )

.

( . ) ( ).

--

+

È

ÎÍÍ

˘

˚˙˙

= =y y

y

Solve by trial and error: y = 0 252. m y = 252 mm

94


PROBLEM 10.81

A spring AB of constant k is attached to two identical gears as shown. Knowing that the spring is undeformed when q = 0, determine two values of the angle q corresponding to equilibrium when P = 150 N, a = 100 m, b = 75 m, r = 150 m, and k = 1 kN/m. State in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION

Elongation of spring

s a a

V ks Pb

k a Pb

dV

dka

= =

= -

= -

=

2 2

1

21

22

4

2

2

2

( sin ) sin

( sin )

sin

q q

q

q q

qq ccos

sin

q

q

-

= -

Pb

ka Pb2 22 (1)

Equilibrium dV

d

Pb

kaqq= =0 2

2 2: sin

sin( )( . )

( )( . ); sin .2

150 0 075

2 1000 0 12 0 5625

2q q= =

N m

N/m m

2 34 229 145 771q = ∞ ∞. .and

q = ∞ ∞17 11 72 9. .and

Stability: We differentiate Eq. (1)

d V

dka

2

224 2

qq= cos

qq

= ∞ = ∞ =17 11 4 34 2 4 0 83 02

22 2. : cos . ( . )

d v

dka ka Stable

qq

= ∞ = ∞ = -72 9 4 145 8 4 0 83 02

22 2. : cos . ( . )

d v

dka ka Unstable

95


PROBLEM 10.82

A spring AB of constant k is attached to two identical gears as shown. Knowing that the spring is undeformed when q = 0, and given that a b r= = =60 mm, 45 mm, 90 mm, and k = 6 kN/m, determine (a) the range of values of P for which a position of equilibrium exists, (b) two values of q corresponding to equilibrium if the value of P is equal to half the upper limit of the range found in Part a.

SOLUTION

Elongation of spring

s a a= =2 2( sin ) sinq q

Potential energy

V ks Pb k a Pb

V ka Pb

= - = -

= -

1

2

1

22

2

2 2

2 2

q q q

q q

( sin )

sin

Equilibrium

dV

d

dV

dka Pb

ka Pb

q qq q

q

= = -

= - =

0 4

2 2 0

2

2

: sin cos

sin (1)

sin ; ;maxmax2

2 21

2 2q = =Pb

kaP

P b

kaFor

(a) Pmax ( . )

( )( . )

0 045

2 6000 0 061

2

m

N/m m= Pmax = 960 N

(b) For P P= 1

2 max, sin ;21

22 30 150q q= = ∞ ∞and

q = ∞ ∞15 00 75 0. .and

96


SOLUTION

Law of Sines y L

y L

A

A

sin( ) sin( )

cos( ) cos

90 90∞ + -=

-

-=

b q b

q b b

or y LA = -cos( )

cos

q bb

From the figure: y L LB = - -cos( )

coscos

q bb

q

Potential Energy: V Py Qy P L L QLB A= - - = - - -È

ÎÍ

˘

˚˙ - -cos( )

coscos

cos( )

cos

q bb

q q bb

dV

dPL QL

L P Q

qq b

bq q b

b= - - - +

È

ÎÍ

˘

˚˙ + -

= +

sin( )

cossin

sin( )

cos

( )sin(qq b

bq- -)

cossinPL

PROBLEM 10.83

A slender rod AB is attached to two collars A and B that can move freely along the guide rods shown. Knowing that b = ∞30 and P = Q = 400 N, determine the value of the angle q corresponding to equilibrium.

97



Equilibrium dV

dL P Q PL

qq b

bq= + - - =0 0: ( )

sin( )

cossin

or ( )sin( ) sin cosP Q P+ - =q b q b

( )(sin cos cos sin ) sin cosP Q P+ - =q b q b q b

or - + + =( )cos sin sin cosP Q Qq b q b 0

- + + =P Q

Q

sin

cos

sin

cos

bb

qq

0

tan tanq b= +P Q

Q (2)

With P Q= = = ∞400 30 N, b

tan tan .q = ∞ =80030 1 1547

N

400 N q = ∞49 1.

98


PROBLEM 10.84

A slender rod AB is attached to two collars A and B that can move freely along the guide rods shown. Knowing that b = ∞ =30 , 100 N,P and Q = 25 N, determine the value of the angle q corresponding to equilibrium.

SOLUTION

Using Equation (2) of Problem 10.83, with P Q= = = ∞100 25 30 N, N, and b , we have

tan( )( )

( )tan

.

.

q

q

= ∞

== ∞

100 25

2530

57 735

89 007

N N

N

q = ∞89 0.

99


PROBLEM 10.85

Collar A can slide freely on the semicircular rod shown. Knowing that the constant of the spring is k and that the unstretched length of the spring is equal to the radius r, determine the value of q corresponding to equilibrium when W r= =250 N, 225mm, and k = 3kN/m.

SOLUTION

Stretch of Spring s AB r

s r r

s r

= -= -= -

2

2 1

( cos )

( cos )

qq

Potential Energy: V ks Wr W mg

V kr Wr

dV

dkr

= - =

= - -

= -

1

22

1

22 1 2

2

2 2

2

2 sin

( cos ) sin

( cos

q

q q

qqq q q- -1 2 2 2) sin cosWr

Equilibrium dV

dkr Wr

qq q q= - - - =0 2 1 2 02: ( cos )sin cos

( cos )sin

cos

2 1

2

q qq

- = -W

kr

Now W

kr= =

( )

( )( . ).

250

3000 0 2250 37037

N

N/m m

Then ( cos )sin

cos.

2 1

20 37037

q qq

- = -

Solving numerically, q = = ∞0 95637 54 8. . rad q = ∞54 8.

100


PROBLEM 10.86

Collar A can slide freely on the semicircular rod shown. Knowing that the constant of the spring is k and that the unstretched length of the spring is equal to the radius r, determine the value of q corresponding to equilibrium when W r= =250 N, 225 mm, and k = 3 kN/m.

SOLUTION

Stretch of spring

s AB r r r

s r

V ks Wr

kr

= - = -= -

= -

= -

2

2 1

1

22

1

22 1

2

2

( cos )

( cos )

cos

( cos

qq

q

q )) cos

( cos ) sin sin

2

2

2

2 1 2 2 2

-

= - - +

Wr

dV

dkr Wr

q

qq q q

Equilibrium

dV

dkr Wr

qq q q= - - + =0 2 1 2 02: ( cos )sin sin

- - + =kr Wr2 2 1 2 0( cos )sin ( sin cos )q q q q

or ( cos )sin

cos

2 1

2

q qq

- = W

kr

Now W

kr= =250

3000 0 2250 37037

N

N/m m( )( . ).

Then 2 1

20 37037

cos

cos.

qq- =

Solving q = ∞37 4.

101


SOLUTION

x = ( ) tan4 m q (1)

y x

ACB = =

=sin tan sin

( )cos

b q bq4

4 m

For x = 0, ( )AC 0 4= m

Stretch of spring. s AC AC

V ks yB

= - = - = -ÊËÁ

ˆ¯̃

= -

=

( )cos cos

( )

(

0

2

44 4

11

1

275

1

25

q q

kN

kN/m))cos

( ) tan sin161

1 75 42

qq b-Ê

ËÁˆ¯̃

- kN

dV

dq qqq

bq

= -ÊËÁ

ˆ¯̃

-801

1 3002 2cos

sin

cos

sin

cos

Equilibrium dV

dq qq b= -Ê

ËÁˆ¯̃

=01

1 3 75:cos

sin . sin (2)

Given: b q= ∞ =30 0 5, sin .

Eq. (2): 1

1 3 75 0 5 1 875cos

sin . ( . ) .q

q-ÊËÁ

ˆ¯̃

= =


Eq. (1): x = ∞( ) tan .4 70 46m x = 11 27. m

PROBLEM 10.87

Cart B, which weighs 75 kN, rolls along a sloping track that forms an angle b with the horizontal. The spring constant is 5 kN/m, and the spring is unstretched when x = 0. Determine the distance x corresponding to equilibrium for the angle b indicated.

Angle b = 30∞.

102


SOLUTION

x = ( ) tan4 m q (1)

y x

ACB = =

=sin tan sin

( )cos

b q bq4

4 m

For x = 0, ( )AC 0 4= m

Stretch of spring. s AC AC

V ks yB

= - = - = -ÊËÁ

ˆ¯̃

= -

=

( )cos cos

( )

(

0

2

44 4

11

1

275

1

25

q q

kN

kN/m))cos

( ) tan sin161

1 75 42

qq b-Ê

ËÁˆ¯̃

- kN

dV

dq qqq

bq

= -ÊËÁ

ˆ¯̃

-801

1 3002 2cos

sin

cos

sin

cos

Equilibrium dV

dq qq b= -Ê

ËÁˆ¯̃

=01

1 3 75:cos

sin . sin (2)

Given: b q= ∞ =60 0 86603, sin .

Eq. (2): 1

1 3 75 0 86603 3 2476cos

sin . ( . ) .q

q-ÊËÁ

ˆ¯̃

= =


Eq. (1): x = ∞( ) tan .4 26 67m x = 16 88. m

PROBLEM 10.88

Cart B, which weighs 75 kN, rolls along a sloping track that forms an angle b with the horizontal. The spring constant is 5 kN/m, and the spring is unstretched when x = 0. Determine the distance x corresponding to equilibrium for the angle b indicated.

Angle b = 60 .∞

103


PROBLEM 10.89

A vertical bar AD is attached to two springs of constant k and is in equilibrium in the position shown. Determine the range of values of the magnitude P of two equal and opposite vertical forces P and -P for which the equilibrium position is stable if (a) AB CD= , (b) AB CD= 2 .

SOLUTION

For both (a) and (b): Since P and -P are vertical, they form a couple of moment

M PlP = + sinq

The forces F and -F exerted by springs must, therefore, also form a couple, with moment

M FaF = - cosq

We have dU M d M d

Pl Fa d

P F= +

= -

q q

q q q( sin cos )

but F ks k a= = ÊËÁ

ˆ¯̃

1

2sinq

Thus, dU Pl ka d= -ÊËÁ

ˆ¯̃

sin sin cosq q q q1

22

From Equation (10.19), page 580, we have

dV dU Pl d ka d= - = - +sin sinq q q q1

422

or dV

dPl ka

qq q= - +sin sin

1

422

and d V

dPl ka

2

221

22

qq q= - +cos cos (1)

For q = 0: d V

dPl ka

2

221

2q= - +

104



For Stability: d V

dPl ka

2

220

1

20

q , - +

or (for Parts a and b) Pka

l

2

2

Note: To check that equilibrium is unstable for P kal=2

2 , we differentiate (1) twice:

d V

dPl ka

d V

dPl ka

3

32

4

42

2 0 0

2 2

qq q q

qq q

= + - = =

= -

sin sin , ,

cos cos

for

Forq = 0 d V

dPl ka

kaka

4

42

222

22 0

q= - = -

Thus, equilibrium is unstable when Pka

l=

2

2

105


PROBLEM 10.90

Rod AB is attached to a hinge at A and to two springs, each of constant k. If h d= =625mm, 300 mm, and W = 400 N, determine the range of values of k for which the equilibrium of the rod is stable in the position shown. Each spring can act in either tension or compression.

SOLUTION

We have x d y hC B= =sin cosq q

Potential Energy: V kx Wy

kd Wh

C B= +ÊËÁ

ˆ¯̃

= +

21

22

2 2sin cosq q

Then dV

dkd Wh

kd Wh

qq q q

q q

= -

= -

2

2

2

2

sin cos sin

sin sin

and d V

dkd Wh

2

222 2

qq q= -cos cos (1)

For equilibrium position q = 0 to be stable, we must have

d V

dkd Wh

2

222 0

q= -

or kd Wh2 1

2 (2)

Note: For kd Wh2 12= , we have d V

d

2

2 0q

= , so that we must determine which is the first derivative that is not equal

to zero. Differentiating Equation (1), we write

d V

dkd Wh

d V

dkd

3

32

4

22

4 2 0 0

8 2

qq q q

q

= - + = =

= -

sin sin

cos

for

qq q+Whcos

For q = 0: d V

dkd Wh

4

428

q= - +

106



Since kd Wh Wh Whd Vd

2 12 4 4 0= = - +, ,

4

q we conclude that the equilibrium is unstable for kd Wh2 1

2= and

the > sign in Equation (2) is correct.

With W h d= = =400 625 300N, mm, and mm

Equation (2) gives k( . ) ( )( . )0 31

2400 0 6252m N m

or k 1388 89. N/m k 1389 N/m

107


SOLUTION

Using Equation (2) of Problem 10.90 with

h k W= = =1 240m, l.2kN/m, and N

( ) ( )( )12001

2240 12N/m N md

or d2 0 1 . m2

d 0 3163. m smallest d = 0 316. m

PROBLEM 10.91

Rod AB is attached to a hinge at A and to two springs, each of constant k. If h k= =1m, l.2 kN/m, and W = 240 N, determine the smallest distance d for which the equilibrium of the rod is stable in the position shown. Each spring can act in either tension or compression.

108


PROBLEM 10.92

Two bars are attached to a single spring of constant k that is unstretched when the bars are vertical. Determine the range of values of P for which the equilibrium of the system is stable in the position shown.

SOLUTION

sL L= =3

2

3sin sinf q

For small values of f and qf q

f q

q q

=

= +ÊËÁ

ˆ¯̃

+

= + +

2

3

2

3

1

2

32 2

1

2

2

3

2V PL L

ks

VPL

kL

cos cos

(cos cos ) ssin

( sin sin ) sin cos

( si

q

qq q q q

ÊËÁ

ˆ¯̃

= - - +

= -

2

2

32 2 2

2

9

32

dV

d

PLkL

PLnn sin ) sin

( cos cos ) cos

2 22

92

34 2 2

4

92

2

2

22

q q q

qq q q

+ +

= - + +

kL

d V

d

PLkL

when q = 0: d V

d

PLkL

2

226

3

4

9q= - +

For stability: d V

dPL kL

2

220 2

4

90

q , - + P kL

2

9

109


SOLUTION

aL L= =2

3 3sin sinq f

For small values of f and q

f q= 2

sL=3

sinq

V PL L

ks

PLk

L

= +ÊËÁ

ˆ¯̃

+

= + + ÊË

2

3 3

1

2

32 2

1

2 3

2cos cos

( cos cos ) sin

q f

q q qÁÁˆ¯̃

= - - +

= - +

2

2

32 2 2

9

2

3

dV

d

PL kL

dV

d

PL

qq q q q

qq

( sin sin ) sin cos

(sin ssin ) sin

(cos cos ) cos

218

2

2

32 2

92

2

2

2

2

q q

qq q q

+

= - + +

kL

d V

d

PL kL

when q = 0: dV

dPL

kL2

2

2

29q

= - +

For stability: d V

dPL

kL2

2

2

0 29

0q

, - + P kL1

18

PROBLEM 10.93

Two bars are attached to a single spring of constant k that is unstretched when the bars are vertical. Determine the range of values of P for which the equilibrium of the system is stable in the position shown.

110


PROBLEM 10.94

Two bars AB and BC are attached to a single spring of constant k that is unstretched when the bars are vertical. Determine the range of values of P for which the equilibrium of the system is stable in the position shown.

SOLUTION

s l a

V P l ks

Pl k l a

dV

d

= -

= +

= + -

= -

( )sin

( cos )

cos ( ) sin

q

q

q q

q

21

2

21

2

2

2

2 2

PPl k l a

Pl k l a

d V

dPl

sin ( ) sin cos

sin ( ) sin

q q q

q q

q

+ -

= - + -

= -

2

2

2

2

21

22

2 ccos ( ) cosq q+ -k l a 2 2 (1)

when qq

= = - + -0 22

22: ( )

d V

dPl k l a

Stability: d V

dPl k l a

2

220 2 0

q : ( )- + - > P

k l a

l

( )- 2

2

To check whether equilibrium is unstable for P k l al= -( ) ,

2

2 we differentiate

Eq. (1) twice:

d V

dPl k l a

d V

dPl k l a

3

32

4

4

2 2 2 0

2 4

qq q q

qq

= - - = =

= - -

sin ( ) sin ,

cos (

For 0

)) cos2 2q

111



For G = 0 and Pk l a

l

d V

dPl k l a

k l a k l a

= -

= - -

= - - -

( )

( )

( ) ( )

2

4

42

3 2

2

2 4

4 0

q

Thus equilibrium is unstable for Pk l a

l= -( )2

2

112


SOLUTION

First note A a b= =sin sinq f

For small values of q f and : a bq f=

or f q= a

b

V P a b Q a b

a b Pa

bQ

dV

= + - +

= + ÊËÁ

ˆ¯̃

-ÈÎÍ

˘˚̇

( )cos ( )cos

( ) cos cos

f q

q q

2

2

dda b

a

bP

a

bQ

d V

da b

a

bP

qq q

q

= + - ÊËÁ

ˆ¯̃

+ÈÎÍ

˘˚̇

= + -

( ) sin sin

( ) co

2

2

2

2

2ss cos

a

bQq qÊ

ËÁˆ¯̃

+È

ÎÍ

˘

˚˙2

when q = 0: d V

da b

a

bP Q

2

2

2

22

q= + - +

Ê

ËÁˆ

¯̃( )

PROBLEM 10.95

The horizontal bar BEH is connected to three vertical bars. The collar at E can slide freely on bar DF. Determine the range of values of Q for which the equilibrium of the system is stable in the position shown when a = 600 mm, b = 500 mm, and P = 750 N.

113



Stability: d V

d

a

bP Q

2

2

2

20 2 0

q : - +

Pb

aQ 2

2

2 (1)

or Qa

bP

2

22 (2)

with P a b= = =750 0 6 0 5N, m, and m. .

Equation (1): Q ( . )

( . )

0 6

2 0 5750

2

2

m

m N

For stability Q 540 N

114


SOLUTION

Using Equation (2) of Problem 10.95 with

Q a b= = =45 150 200N, mm, and mm

Equation (2) P 2200

15045

160 000

2

2

( )

( )( )

.

mm

mmN

N=

For stability P 160 0. N

PROBLEM 10.96

The horizontal bar BEH is connected to three vertical bars. The collar at E can slide freely on bar DF. Determine the range of values of P for which the equilibrium of the system is stable in the position shown when a = 150 mm, b = 200 mm, and Q = 45 N.

115


PROBLEM 10.97*

Bars AB and BC, each of length l and of negligible weight, are attached to two springs, each of constant k. The springs are undeformed, and the system is in equilibrium when q

1 = q

2= 0. Determine the range of values of P for which the

equilibrium position is stable.

SOLUTION

We have x l

x l l

y l l

V Py kx kx

B

C

C

C B C

=

= +

= +

= + +

sin

sin sin

cos cos

q

q q

q q1 2

1 2

21

2

1

222

or V Pl kl= + + + +ÈÎ ˘̊(cos cos ) sin (sin sin )q q q q q1 22 2

1 1 221

2

For small values of q1and q2:

sin , sin , cos , cosq q q q q q q q1 1 2 2 1 12

2 221

1

21

1

2» » » »- -

Then V Pl kl= - + -Ê

ËÁˆ

¯̃+ + +( )È

Î˘˚1

21

2

1

212

22

212

1 22q q

q q q

and ∂∂

∂∂

∂∂

VPl kl

VPl kl

V

qq q q q

qq q q

q

11

21 1 2

22

21 2

2

12

= - + + +

= - + +

= -

[ ( )]

( )

PPl klV

Pl kl

Vkl

+ = - +

=

2 22

22

2

2

1 2

2

∂∂

∂∂ ∂

q

q q

Stability: Conditions for stability (see Page 583).

116


PROBLEM 10.97* (Continued)

For q qq q1 2

1 2

0 0= = = =: condition satisfied∂∂

∂∂

V V( )

∂

∂ ∂∂∂

∂∂

2

1 2

2 2

12

2

22

0V V V

q q q q

Ê

ËÁˆ

¯̃-

Substituting ( ) ( )( )kl Pl kl Pl kl

k l P l Pkl k l

P klP

2 2 2

2 4 2 2 3 2 4

2

2 0

3 2 0

3

- + - +

- + -

- +

- �

�

kk l2 2 0�

Solving P kl P kl� �3 5

2

3 5

2

- +or

or P kl P kl� �0 382 2 62. .or

∂∂

2

12

20 2 0V

Pl klq

: - +

or P kl1

2

∂∂

2

22

20 0V

Pl klq

: - +

or P kl

Therefore, all conditions for stable equilibrium are satisfied when 0 0 382� �P kl.

117


SOLUTION

From the analysis of Problem 10.97 with

l k

P kl

= == =

800 2 5

0 382 0 382 2500 0 8 764

mm and kN/m

N/m m N

.

. . ( )( . ) P 764 N

PROBLEM 10.98*

Solve Problem 10.97 knowing that l = 800 mm and k = 2.5 kN/m.

PROBLEM 10.97* Bars AB and BC, each of length l and of negligible weight, are attached to two springs, each of constant k. The springs are undeformed, and the system is in equilibrium when q

1 = q

2 = 0. Determine the range of values of P for

which the equilibrium position is stable.

118


PROBLEM 10.99*

Two rods of negligible weight are attached to drums of radius r that are connected by a belt and spring of constant k. Knowing that the spring is undeformed when the rods are vertical, determine the range of values of P for which the equilibrium position q

1 = q

2 = 0 is stable.

SOLUTION

Left end of spring moves from a to b. Right end of spring moves from ¢a to ¢b . Elongation of spring

s a b ab r r r

V ks pl wl

kr

= ¢ ¢ - = - = -

= + -

=

q q q q

q q

q

1 2 1 2

21 2

21

1

21

2

( )

cos cos

( -- + -

= - -

= -

q q q

qq q q

qq

22

1 2

1

21 2 1

2

2

) cos cos

( ) sin

(

pl wl

vkr pl

vkr

∂∂∂∂ 11 2 2

2

12

21

2

22

22

2

1

- +

= -

= +

q q

qq

qq

q

) sin

cos

cos

wl

vkr pl

vkr wl

v

∂∂

∂∂

∂∂ ∂∂q2

2= - kr

119


PROBLEM 10.99* (Continued)

For q qq q q q1 2

2

12

22

22

22

2

20= = = - = + + = -: , ,,

∂∂

∂∂ ∂ ∂

vkr pl

vkr wl

r vkr

Stability conditions for stability (see Page 583)

∂∂ ∂

∂ ∂∂

∂ ∂∂

-

2

1 2

2 2

12

2

22

2 2 2 2

0v v v

kr kr pl kr wl

q q q q

Ê

ËÁˆ

¯̃

- +

-

( ) ( )( )

0

02 2pl kr wl kr wl( )+ -

Pwkr

kr wlP

kr

l

W

wkrl

2

2

2

2+ +

Ê

ËÁ

ˆ

¯˜;

∂∂

2

12

22

0v

kr pl Pkr

lq� � �: ;-

We choose: Pkr

l

W

Wkrl

2

2+

Ê

ËÁ

ˆ

¯˜

120


SOLUTION

k

r

l

kr

l

===

=

400

75

150

4000 0 0752 2

0 N/m

mm=0.075m

mm=0.15m

N/m m( )( . )

(( .0 15150

m)N=

(a) W P=+

75 15075

75N: N

N

(150 N) N ( )

( ) P 50 00. N

(b) W P=+

( : ( )( )

300 150300

300N) N

N

(150 N) N P 100 0. N

PROBLEM 10.100*

Solve Problem 10.99 knowing that k = 4 kN/m, r = 75 mm, l = 150 mm, and (a) W = 75 N, (b) W = 300 N.

PROBLEM 10.99* Two rods of negligible weight are attached to drums of radius r that are connected by a belt and spring of constant k. Knowing that the spring is undeformed when the rods are vertical, determine the range of values of P for which the equilibrium position q

1 = q

2 = 0 is stable.

121


SOLUTION

Assumingd yA ¯

it follows

d d dy y yC A A= =300

2001 5. ≠

d d dy y yE C A= = 1 5. ≠

d d d dy y y yD E A A= = =450

1503 1 5 4 5( . ) . ≠

d d d dy y y yG E A A= = =250

150

10

61 5 2 5( . ) . ¯

Then, by Virtual Work

d d d dU y y P yA D G= - + =0 1500 500 0: ( ) ( )N N

1500 500 4 5 2 5 0

1500 2250 2 5 0

d d dy y P y

P

A A A- + =

- + =

( . ) ( . )

.

P = 300 N P = 300 N

PROBLEM 10.101

Determine the vertical force P that must be applied at G to maintain the equilibrium of the linkage.

122


SOLUTION

Assume d yA ¯: d d dy y yC A A= =300

2001 5. ≠, d d dy y yE C A= = 1 5. ≠

d d d dy y y yD E A A= = =450

1503 1 5 4 5( . ) . ≠

dfd d

d= = =y y

yE AA150

1 5

1500 01

..

Virtual Work:

dU = 0: ( ) ( )

( . ) .

1500 500 0

1500 500 4 5 0 01 0

N Nd d dfd d d

y y M

y y M yA D

A A A

- + =- + ( ) =

11500 2250 0 01 0- + =. M

M = + = ◊75000 75N mm N m◊ M = 75.0 N m◊

PROBLEM 10.102

Determine the couple M that must be applied to member DEFG to maintain the equilibrium of the linkage.

123


SOLUTION

Choosing a coordinate system with origin at E,

x l x lD D= = -3 3cos sinq d q dq

Principle of Virtual Work: Noting that P acts in the positive x direction and M acts in the positive q direction, we write

d dq dU M P xD= + + =0 0:

+ + - =M P ldq qdq( sin )3 0

M P= 3 sinq

PROBLEM 10.103

Determine the magnitude of the couple M required to maintain the equilibrium of the mechanism shown.

124


SOLUTION

Since AB = 440 mm, we have

( ) ( )

( ) ( )

440 240

440 240 136 10

2 2 2 2

2 2 2 2 3 2

mm mm= + +

= - - = ¥ -

b c

b c c

Differentiate: 2 2

136 103 2

b b c c

bc

bc

bc c

c

d d

d d

dd

= -

= -

= -¥ -

Virtual Work:

d d dU P c W b= + =0 0:

P c Wc c

cd d=

¥ -136 103 2

For W = 90 N: Pc

c=

¥ -90

136 103 2 (1)

(a) When c = 80 mm:

Eq. (1): P =¥ -

9080

136 10 803 2( ) P = 20 0. N

(b) When c = 280 mm:

Eq. (1) P =¥ -

90280

136 10 2803 2( ) P = 105 0. N

PROBLEM 10.104

Collars A and B are connected by the wire AB and can slide freely on the rods shown. Knowing that the length of the wire is 440 mm and that the weight W of collar A is 90 N, determine the magnitude of the force P required to maintain equilibrium of the system when (a) c = 80 mm, (b) c = 280 mm.

125


PROBLEM 10.105

Collar B can slide along rod AC and is attached by a pin to a block that can slide in the vertical slot shown. Derive an expression for the magnitude of the couple M required to maintain equilibrium.

SOLUTION

yR

yR

yR

B

B

B

=∞ -

= -∞ -

= -

tan( )

cos ( )

sin

90

902

2

q

d dqq

d dqq

Virtual Work: d d dq dU U M P yB= = - - =0 0:

- + =M PRdqq

dq10

2sin

MPR=

sin2 q M PR= csc2 q

126


PROBLEM 10.106

A slender rod of length l is attached to a collar at B and rests on a portion of a circular cylinder of radius r. Neglecting the effect of friction, determine the value of q corresponding to the equilibrium position of the mechanism when l = 200 mm,r = 60 mm, P = 40 N, and Q = 80 N.

SOLUTION

Geometry OC r

OC

OB

r

xB

=

= =cosq

xr

xr

y l

y l

B

B

A

A

=

=

== -

cossin

coscos

sin

q

d qq

dq

qd qdq

2

Virtual Work:

d d dU P y Q xA B= - - =0 0: ( )

Pl Qr

sinsin

cosqdq q

qdq- =

20

cos2 q = Qr

Pl (1)

Then, with l r P Q= = = =200 60 40 80 mm, mm, N, and N

cos(

(.2 80

400 6q = = N)(60 mm)

N)(200 mm)

or q = ∞39 231. q = ∞39 2.

127


PROBLEM 10.107

A horizontal force P of magnitude 200 N is applied to the mechanism at C. The constant of the spring is k = 1 8. kN/m, and the spring is unstretched when q = 0. Neglecting the weight of the mechanism, determine the value of q corresponding to equilibrium.

SOLUTION

s r

s r

==

qd dq

Spring is unstretched at q = ∞0

F ks kr

x l

x l

SP

C

C

= ===

qq

d qdqsin

cos

Virtual Work:

d d dU P x F sC SP= - =0 0:

P l kr r( cos ) ( )qdq q dq- = 0

or Pl

kr2= q

qcos

Thus (

( cos

200

1800

N)(0.3 m)

N/m)(0.125)2= q

q

or q

qcos.= 2 1333

q = =1 054. rad 60.39∞ q = ∞60 4.

128


SOLUTION

Deformation of spring

s ECx

V ksx x

= - = -

= - = -ÊËÁ

ˆ¯̃

0 12

30 1

1

2120

6

1

21600 0 12

. .

( ( .

m

N) N/m)2

3

22

20

16002

30 1

2

320

-

= -ÊËÁ

ˆ¯̃

-

x

dV

dx

x.

Equilibrium: dV

dx

x= -ÊËÁ

ˆ¯̃

- =03200

3

2

30 1 20 0.

2

30 1 20

3

3200

2

3

3

160

1

10

19

1600 178125

x

x

x

-ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

= + =

=

.

.fi m

x = 178 1. mm

PROBLEM 10.108

Two identical rods ABC and DBE are connected by a pin at B and by a spring CE. Knowing that the spring is 100 mm. long when unstretched and that the constant of the spring is 1.6 kN/m, determine the distance x corresponding to equilibrium when a 120 N load is applied at E as shown.

129


PROBLEM 10.109

Solve Problem 10.108 assuming that the 120-N load is applied at C instead of E.

PROBLEM 10.108 Two identical rods ABC and DBE are connected by a pin at B and by a spring CE. Knowing that the spring is 100 mm long when unstretched and that the constant of the spring is 1.6 kN/m, determine the distance x corresponding to equilibrium when a 120-N load is applied at E as shown.

SOLUTION

Deformation of spring

s ECx

V ksx x

= - = -

= - = -ÊËÁ

ˆ¯̃

-

0 12

30 1

1

2120

5

6

1

21600 0 12

2

. .

( ( .

m

N) )2

31100

16002

30 1

2

3100

x

dV

dx

x= -ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

-.

Equilibrium: dV

dx

x= -ÊËÁ

ˆ¯̃

- =03200

3

2

30 1 100 0.

2

30 1 100

3

3200

2

3

300

32000 1

0 290625

x

x

x

-ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

= +

fi =

.

.

. m x = 291mm

130


PROBLEM 10.110

Two uniform rods, each of mass m and length l, are attached to gears as shown. For the range 0 180 q ∞, determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION

Potential energy V Wl

Wl

W mg= ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

=2

1 52

cos . cosq q

dV

d

Wl Wl

Wl

d V

d

qq q

q q

= - + -

= - +

21 5 1 5

2

21 5 1 5

2

( . sin . ) ( sin )

( . sin . sin )

qqq q

2 22 25 1 5= - +Wl

( . cos . cos )

For equilibrium dV

dqq q= + =0 1 5 1 5 0: . sin . sin

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

q = = ∞2 4042 137 8. . rad

Now d V

d

Wl2

2 22 25 1 5

qq q= - +( . cos . cos )

At q = 0: d V

d

Wl2

2 22 25 0 0

q= - ∞ + ∞( . cos cos )

= -Wl

23 25 0( . )( ) q = 0, Unstable

131



At q = ∞137 8. : d V

d

Wl2

2 22 25 1 5 137 8 137 8

q= - ¥ ∞ + ∞[ . cos( . . ) cos . ]

= Wl

22 75 0( . )( ) q = ∞137 8. , Stable

132


PROBLEM 10.111

A homogeneous hemisphere of radius r is placed on an incline as shown. Assuming that friction is sufficient to prevent slipping between the hemisphere and the incline, determine the angle q corresponding to equilibrium when b = ∞10 .

SOLUTION

CG r

V W r CG

dV

dW r r

=

= - -

= - +ÊËÁ

ˆ¯̃

3

8

3

8

( sin ( )cos )

sin sin

q b q

qb q

Equilibrium: dV

dqb q= - + =0

3

80, sin sin

sin sinb q= 3

8 (1)

For b = ∞10 sin sin103

8∞ = q sin . , .q q= = ∞0 46306 27 6

133


PROBLEM 10.112

A homogeneous hemisphere of radius r is placed on an incline as shown. Assuming that friction is sufficient to prevent slipping between the hemisphere and the incline, determine (a) the largest angle b for which a position of equilibrium exists, (b) the angle q corresponding to equilibrium when the angle b is equal to half the value found in Part a.

SOLUTION

Detail

CG r

V W r CG

dV

dW r r

=

= - -

= - +ÊËÁ

ˆ¯̃

3

8

3

8

( sin ( )cos )

sin sin

q b q

qb q

Equilibrium: dV

dqb q= - + =0

3

80, sin sin

sin sinb q= 3

8 (1)

(a) For b qmax, 90= ∞

Eq. (1) sin sin , sin .max maxb b= ∞ = = ∞3

890

3

822 02 bmax .= ∞22 0

(b) When b b= = ∞12 11 01max .

Eq. (1) sin . sin ; sin .11 013

80 5093∞ = =q q q = 30.6°

Note: We can also use DCGD and law of sines to derive Eq. (1).

sin sin; sin sin ; sin sin

b q b q b qCG CD

CG

CD= = = 3

8

solution_manual_vme_9e_chapter_10 - · pdf file3. proprietary material. © 2010 the...

Documents