solution_manual_vme_9e_chapter_10 - · pdf file3. proprietary material. © 2010 the...
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CHAPTER 3CHAPTER 10CHAPTER 10
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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
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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 10.2
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
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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 10.3
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◊
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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.
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.
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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 10.5
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
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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 10.6
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
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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 10.7
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,
At Point G: y yG C= = - = -4 4 10 40( )mm mm yG = 40 0. mm Ø
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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 10.8
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
At Point G: y yG C= = - = -4 4 10 40( )mm mm yG = 40 0. mm Ø
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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 10.9
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 /
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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 10.10
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 /
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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 10.11
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
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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 10.12
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
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
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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 10.13
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
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
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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 10.14
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
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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 10.15
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
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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 10.16
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
MPl=
2tanq
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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 10.17
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
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20
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 10.18
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
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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 10.19
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
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22
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 10.19 (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
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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 10.20
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
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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 10.20 (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) 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
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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 10.21
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
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26
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 10.22
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. ∞
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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 10.23
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.
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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 10.24
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
With P Q= =80 100 N, N
(1002
N) 2(80 N)sin
cos= ( )
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 )
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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 10.25
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 .
q= =0 8 0 8 q = ∞38 7.
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30
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 10.26
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.
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31
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 10.27
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
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32
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 10.27 (Continued)
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.
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33
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 10.28
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
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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 10.28 (Continued)
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. ∞
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35
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 10.29
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
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36
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 10.30
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
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.
= =
Solving numerically q = ∞25 0.
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37
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 10.31
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- =
Solving numerically q = ∞39 65.
and q = ∞68 96. q = ∞39 7.
and q = ∞69 0.
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38
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 10.32
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
- --
-
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.
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39
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 10.33
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.
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40
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 10.34
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
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41
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 10.35
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-
ÎÎ͢˚̇
= ∞ +ÊËÁ
ˆ¯̃
∞ +ÊËÁ
ˆ¯̃
∞ +ÊËÁ
ˆ¯̃
12 45
245
22 45
2cossin cos cos cos
qq q q q-ÈÈ
Î͢˚̇
=∞ +( )
1 245 2-
cos
cos
q
q
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42
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 10.35 (Continued)
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.
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43
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 10.36
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
qSolving numerically q = ∞36 6.
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44
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 10.37
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.
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45
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 10.38
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
Solving by trial and error: q = 0 33868. rad q = ∞19 40.
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46
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 10.39
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
Solving numerically q = ∞59 0.
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47
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 10.40
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
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48
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 10.40 (Continued)
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
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49
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 10.41
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
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50
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 10.42
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.
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51
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 10.43
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
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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 10.44
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.
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53
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.
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.
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54
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 10.45 (Continued)
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
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55
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 10.46
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
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56
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 10.47
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
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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.
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.
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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 10.49
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
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59
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 10.50
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
- =
=-
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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 10.50 (Continued)
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
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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 10.51
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◊
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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 10.52
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
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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 10.53
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 ≠
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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 10.54
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.
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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 10.54 (Continued)
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
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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 10.55
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
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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 10.56
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
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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 10.57
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
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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 10.58
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
+
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70
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 10.59
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
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71
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 10.60
Using the method of Section 10.8, solve Problem 10.30.
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
or 1 2 200
8 300
1
3
- =
=
sin
tan (
N
N/m)(0.25 m)
Solving numerically, q = ∞25 0.
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72
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 10.61
Using the method of Section 10.8, solve Problem 10.33.
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)
Solving numerically q = ∞40 2.
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73
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 10.62
Using the method of Section 10.8, solve Problem 10.34.
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
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74
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 10.63
Using the method of Section 10.8, solve Problem 10.35.
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
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
qSolving numerically q = ∞22 6.
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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 10.64
Using the method of Section 10.8, solve Problem 10.36.
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.
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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 10.65
Using the method of Section 10.8, solve Problem 10.31.
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
+ =
= -
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.
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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 10.66
Using the method of Section 10.8, solve Problem 10.38.
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
Solving by trial and error: q = 0 33868. rad q = ∞19 40.
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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 10.67
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
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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 10.68
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.
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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 10.69
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
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
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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 10.70
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)
= ∞ = ∞ + ∞ÈÎÍ
˘˚̇
= -2701
215 270 2 10 540 27 5
2
2: ( )sin( ) ( )cos( ) .
d V
dl l 0, unstable
= ∞ = ∞ + ∞ÈÎÍ
˘˚̇
22 01
215 22 2 10 44 0 0
2
2. : ( )sin ( )cos .
d V
dl Stable
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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 10.70 (Continued)
= ∞ = ∞ + ∞ÈÎÍ
˘˚̇
= -901
215 90 2 10 180 12 5 0
2
2: ( )sin ( )cos . ,
d V
dl l Unsttable
= ∞ = ∞ + ∞ÈÎÍ
˘˚̇
=158 01
215 158 2 10 316 17 196
2
2. : ( )sin ( )cos .
d V
dl l 00, Stable
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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.
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.
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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 10.72
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
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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 10.73
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
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
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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 10.74
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
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
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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
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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 10.76
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
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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 10.77
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
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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 10.78
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
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PROBLEM 10.78 (Continued)
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
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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
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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
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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
= ∞ = ∞ =17 11 4 34 2 4 0 83 02
22 2. : cos . ( . )
d v
dka ka Stable
= ∞ = ∞ = -72 9 4 145 8 4 0 83 02
22 2. : cos . ( . )
d v
dka ka Unstable
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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 10.82
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
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96
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.
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.
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97
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 10.83 (Continued)
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
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.
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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 10.84
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.
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99
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 10.85
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 )
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.
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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 10.86
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
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.
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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.
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¯̃
= =
Solve by trial and error: q = ∞70 46.
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∞.
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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.
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¯̃
= =
Solve by trial and error: q = ∞76 67.
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 .∞
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103
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 10.89
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= - +
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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 10.89 (Continued)
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
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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 10.90
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= - +
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106
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 10.90 (Continued)
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
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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.
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108
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 10.92
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
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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
( 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.
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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
= - - = =
= - -
sin ( ) sin ,
cos (
For 0
)) cos2 2q
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PROBLEM 10.94 (Continued)
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
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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.
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PROBLEM 10.95 (Continued)
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
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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.
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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
΢˚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).
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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.
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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.
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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
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
q
) sin
cos
cos
wl
vkr pl
vkr wl
v
∂∂
∂∂
∂∂ ∂∂q2
2= - kr
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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
¯˜
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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.
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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.
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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.
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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.
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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.
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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
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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 10.106
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.
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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
= ===
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.
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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.
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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
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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
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PROBLEM 10.110 (Continued)
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
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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
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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