CHAPTER 10
PROBLEM 10.1
Determine the vertical force P that must be applied at C to maintain
the equilibrium of the linkage.
SOLUTION
Assume δ y A
δ yC = 2δ yA
δ yD = δ yA
1
2
1
2
δ yE = δ yD = δ yA
1.25 m
1
2
δ yG = δ yE = δ yA
Also
δθ =
δ yA
a
Virtual Work: We apply both P and M to member ABC.
δ U = 100δ y A − Pδ yC − M δθ + 60δ yD + 50δ yE − 40δ yG = 0
�δ y �
�1
�
�1
�
100δ y A − P(2δ y A ) − M � A � + 60δ y A + 50 � δ y A � − 40 � δ y A � = 0
�2
�
�2
�
� a �
M
+ 60 + 25 − 20 = 0
100 − 2 P −
a
M
= 165 N
a
2P +
Now from Eq. (1) for M = 0
2 P = 165 N
(1)
P = 82.5 N ��
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1699 1701
PROBLEM 10.2
Determine the horizontal force P that must be applied at A to maintain
the equilibrium
um of the linkage.
SOLUTION
Assume δθ
δ x A = 0.25dq
δ yC = 0.1dq
d yD = d yC = 0.1dq
δφ =
δ yD
2
= δθ
0.15 3
d xG = 0.375df
= 0.375
2
δθ
3
= 0.25dq
Virtual Work: We shall assume that a force P and a couple M are applied to member ABC
C as shown.
δ U = − Pδ x A − M δθ + 30δ yC + 40δ y D + 45df + 80δ xG = 0
− (0.25dq)
−P
dq) − M dq + 30(0.1dq
dq
dq) + 40(0.1dq
dq) + 45
2
dq + 80(0.25dq)
dq) = 0
dq
3
−0.25P − M + 3 + 4 + 30 + 20 = 0
(0.25 m)P + M = 57 N ⋅ m
Making M = 0 in Eq. (1):
P = +228 N
(1)
P = 228 N
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1700
PROBLEM 10.3
Determine the couple M that must be applied to member ABC to
maintain the equilibrium of the linkage.
SOLUTION
Assume δ y A
δ yC = 2δ y A
δ yD = δ y A
1
2
1
2
δ yE = δ yD = δ y A
1
2
δ yG = δ yE = δ y A
Also
δθ =
δ yA
a
Virtual Work: We apply both P and M to member ABC.
δ U = 100δ y A − Pδ yC − M δθ + 60δ yD + 50δ yE − 40δ yG = 0
�δ y �
�1
�
�1
�
100δ y A − P(2δ y A ) − M � A � + 660
0δ y A + 550
0 � δ y A � − 440
0 � δ yA � = 0
�2
�
�2
�
� a �
M
+ 60
100 − 2 P −
60 + 2
25
5 − 20
20 = 0
a
2P +
Now from Eq. (1) for
M
=1
165 N
a
(1)
P = 0 and
and a = 0.3
M
= 165 N
0.3 m
M = 49.5 N ⋅ m
��
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for or by any means, without the prior written permission of the publisher, or used beyond the limited
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raw-Hill for their individual course preparation. If you are a student using this Manual,
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permission.
1701
PROBLEM 10.4
Determine the couple M that must be applied to member ABC to
maintain the equilibrium of the linkage.
SOLUTION
Assume δθ
d xA = 0.25dq
d yC = 0.1dq
d yD = d yC = 0.1dq
δφ =
δ yD
2
= δθ
0.15 3
d xG = 0.375df
= 0.375
2
δθ
3
= 0.25dq
Virtual Work: We shall assume that a force P and a couple M are applied to member ABC as shown.
δ U = − Pδ x A − M δθ + 30δ yC + 40δ yD + 180δφ + 80δ xG = 0
−P(0.25dq) − M dq + 30(0.1dq) + 40(0.1dq) + 45
2
dq + 80(0.25dq) = 0
3
−0.25P − M + 3 + 4 + 30 + 20 = 0
(0.25)P + M = 57 N ⋅ m
P=0
Now from Eq. (1) for
(1)
M = 57 N ⋅ m
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1704
1702
PROBLEM 10.5
K
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
om sketch
δ y A = 4δ yC
Thus
(a)
y A = 4 yC
Virtual Work:
δ U = 0: Pδ y A − Fδ yC = 0
P (4δ yC ) − F δ yC = 0
P=
F = 300 N: P =
1
F
4
1
(300 N) = 75 N
4
P = 75 N
(b)
Free body: Corkscrew
ΣFy = 0
R + P − F = 0;
R + 75 N − 300 N = 0
On corkscrew:
R = 225 N ,
On battle:
R = 225 N
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1703
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 δ y A :
d yC =
1
0.2 m
d yA; d yC = d yA
2
0.4 m
Since bar CD move in translation
δ yE = δ yF = δ yC
or
Virtual Work:
δ yE = δ y F =
1
yA
2
δ U = 0: − Pδ y A + (100 N)δ yE + (150 N)δ yF = 0
− Pδ y A + 100
1
1
δ y A + 150 δ y A = 0
2
2
P = 125 N
P = 125.0 N
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1706
1704
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
yG = 4 yC
yH = 4 yC
yF = 3 yC
yE = 2 yC
δ yH = 4δ yC
δ yF = 3δ yC
δ yE = 2δ yC
For spring:
∆ = yF − yC
Q = Force in spring (assumed in tension)
Q = + k ∆ = k ( yF − yC ) = k (3 yC − yC ) = 2kyC
(1)
C = 120 N, E = F = H = 0
(a)
Virtual Work:
δ U = 0: −(120 N)δ yC + Qδ yC − Qδ yF = 0
−120δ yC + Qδ yC − Q (3δ yC ) = 0
Q = −60 N
Eq. (1):
Q = 2kyC , − 60 N = 2(15 kN/m) yC ,
At Point G:
Q = 60.0 N C �
yC = −2 mm
y G = 8.00 mm �
yG = 4 yC = 4(−2 mm) = −8 mm
C = H = 120 N, E = F = 0
(b)
Virtual Work:
δ U = 0: − (120 N)δ yC − (120 N) yH + Qδ yC − Qδ yF = 0
−120δ yC − 120(4δ yC ) + Qδ yC − Q(3δ yC ) = 0
Q = −300 N
Eq. (1):
At Point G:
Q = 2kyC
Q = 300 N C �
−300 N = 2(15 kN/m)yC , yC = −10 mm
yG = 4 yC = 4(−10 mm) = − 40 mm
y G = 40.0 mm ��
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1705 1707
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
yG = 4 yC
yH = 4 yC
yF = 3 yC
yE = 2 yC
δ yH = 4δ yC
δ yF = 3δ yC
δ yE = 2δ yC
∆ = yF − yC
For spring:
Q = Force in spring (assumed in tension)
Q = + k ∆ = k ( yF − yC ) = k (3 yC − yC ) = 2kyC
(1)
E = 120 N, C = F = H = 0
(a)
Virtual Work:
δ U = 0: − (120 N)δ yE + Qδ yC − Qδ yF = 0
−120(2δ yC ) + Qδ yC − Q (3δ yC ) = 0
Q = −120 N
Q = 2kyC , −120 N = 2(15 kN/m) yC ,
Eq. (1):
At Point G:
Q = 120.0 N C �
yC = −4 mm
y G = 16.00 mm �
yG = 4 yC = 4(−4 mm) = 16 mm
E = F = 120 N, C = H = 0
(b)
Virtual Work:
δ U = 0: − (120 N)δ yE − (120 N)δ yF + Qδ yC − Qδ yF = 0
−120(2δ yC ) − 120(3δ yC ) + Qδ yC − Q (3δ yC ) = 0
Q = −300 N
Eq. (1):
At Point G:
Q = 2kyC , − 300 N = 2(15 kN/m)yC ,
Q = 300 N C �
yC = −10 mm
yG = 4 yC = 4(−10 mm) = − 40 mm
y G = 40.0 mm ��
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1708
1706
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 A = 2l cos θ ; δ y A = −2l sin θ δθ
θ
θ
CD = 2l sin ; δ (CD ) = l cos δθ
2
2
Virtual Work:
δ U = 0: − Pδ y A − Qδ (CD ) = 0
θ �
�
− P( −2l sin θ δθ ) − Q � l cos δθ � = 0
2 �
�
Q = 2P
sin θ
��
cos(θ /2)
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1707 1709
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 A = 2l cos θ ; δ x A = 2l cos θ δθ
θ
θ
CD = 2l sin ; δ (CD ) = l cos δθ
2
2
Virtual Work:
δ U = 0: Pδ x A − Qδ (CD) = 0
θ �
�
P(2l cos θ δθ ) − Q � l cos δθ � = 0
2 �
�
Q = 2P
cos θ
��
cos(θ /2)
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1710
1708
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 A = 2l sin θ
δ x A = 2l cos θδθ
and
yF = 3l cos θ
δ yF = −3l sin θδθ
Virtual Work:
δ U = 0: Qδ x A + Pδ yF = 0
Q(2l cos θδθ ) + P(−3l sin θδθ ) = 0
Q=
3
P tan θ ��
2
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1709 1711
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 ∆AA′C :
A′C = a tan θ
y A = −( A′C ) = − a tan θ
δ yA = −
a
δθ
cos 2 θ
For ∆CC ′B :
BC ′ = l sin θ − A′C
= l sin θ − a tan θ
yB = BC ′ = l sin θ − a tan θ
δ yB = l cos θδθ −
Virtual Work:
a
δθ
cos 2 θ
δ U = 0: Qδ y A − Pδ yB = 0
a �
a
�
�
− Q�−
δθ − P � l cos θ −
2 �
cos 2 θ
� cos θ �
�
a �
� a �
�
Q�
= P � l cos θ −
�
2 �
cos 2 θ �
� cos θ �
�
�
� δθ = 0
�
�l
�
Q = P � cos3 θ − 1� ��
�a
�
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1712
1710
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 ∆ AA′C :
A′C = a tan θ
y A = −( A′C ) = − a tan θ
δ yA = −
a
δθ
cos 2 θ
For ∆ BB′C :
B′C = l sin θ − A′C
= l sin θ − a tan θ
B′C l sin θ − a tan θ
BB′ =
=
tan θ
tan θ
xB = BB′ = l cos θ − a
δ xB = −l sin θδθ
Virtual Work:
δ U = 0: Pδ xB − Qδ y A = 0
a
�
�
δθ � = 0
P(−l sin θδθ ) − Q � −
2
cos
θ
�
�
or
Pl sin θ cos 2 θ = Qa
1
Q = P sin θ cos 2 θ ��
a
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1711 1713
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
xD = 2l cosθ
δ A = 2lδθ
δ B = lδθ
so that δ xD = −2l sin θδθ
Virtual Work:
δ U = 0: − Qδ xD − Pδ A − Pδ B = 0
−Q (−2l sin θδθ ) − P(2lδθ ) − P(lδθ ) = 0
2Ql sin θ − 3Pl = 0
Q=
3 P
��
2 sin θ
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1714
1712
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
yG = h cos θ =
1
2
1
l tan α cos θ
2
δ yG = − l tan α sin θδθ
Virtual Work:
δ U = Wδ yG + M δθ = 0
� 1
�
W � − l tan α sin θδθ � + M δθ = 0
� 2
�
1
M = Wl tan α sin θ
2
��
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1713 1715
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
xB = l cos θ
δ xB = −l sin θδθ
(1)
yC = l sin θ
δ yC = l cos θδθ
Now
1
2
δ xB = lδθ
Substituting from Equation (1)
−l sin θδθ =
or
Virtual Work:
1
lδφ
2
δφ = −2sin θδθ
δ U = 0:
M δϕ + Pδ yC = 0
M (−2sin θδθ ) + P(l cos θδθ ) = 0
M=
or
1 cos θ
Pl
2 sin θ
M=
Pl
�
2 tan θ
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1716
1714
PROBLEM 10.17
Derive an expression for the magnitude of the couple M required to
maintain the equilibrium of the linkage shown.
SOLUTION
yE = 3a sin θ
δ yE = 3a cos θδθ
yF = 4a sin θ δ yF = 4a cos θδθ
Virtual Work:
δ U = 0:
− M δθ + Pδ yE + Pδ yF = 0
− M δθ + P(3a cos θδθ ) + P(4a cos θδθ ) = 0
M = 7Pa cos θ ��
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1715 1717
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
xD = l cos θ
δ xD = −l sin θδθ
yD = 3l sin θ
δ yD = 3l cos θδθ
Virtual Work:
δ U = 0: M δθ − ( P cos β )δ xD − ( P sin β )δ yD = 0
M δθ − ( P cos β )(−l sin θδθ ) − ( P sin β )(3l cos θδθ ) = 0
M = Pl (3sin β cos θ − cos β sin θ )
(a)
For P directed along BCD, β = θ
M = Pl (3sin θ cos θ − cos θ sin θ )
Equation (1):
(b)
M = Pl (2sin θ cos θ )
M = Pl sin 2θ ��
M = Pl (3sin 90° cos θ − cos 90° sin θ )
M = 3Pl cos θ ��
For P directed , β = 90°
Equation (1):
(c)
(1)
For P directed
Equation (1):
, β = 180°
M = Pl (3sin180° cos θ − cos180° sin θ )
M = Pl sin θ ��
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1718
1716
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) θ = 30°,
(b) θ = 150°.
SOLUTION
Analysis of the geometry:
Law of sines
sin φ sin θ
=
AB
BC
sin φ =
AB
sin θ
BC
(1)
Now
xC = AB cos θ + BC cos φ
δ xC = − AB sin θδθ − BC sin φδφ
Now, from Equation (1)
or
cos φδφ =
δφ =
(2)
AB
cos θδθ
BC
AB cos θ
δθ
BC cos φ
(3)
From Equation (2)
� AB cos θ
�
δθ �
� BC cos φ �
δ xC = − AB sin θδθ − BC sin φ �
or
δ xC = −
AB
(sin θ cos φ + sin φ cos θ )δθ
cos φ
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1717 1719
PROBLEM 10.19 (Continued)
AB sin(θ + φ )
δθ
cos φ
Then
δ xC = −
Virtual Work:
δ U = 0: − Pδ xC − M δθ = 0
� AB sin(θ + φ ) �
− P �−
δθ � − M δθ = 0
cos φ
�
�
M = AB
Thus,
sin(θ + φ )
P
cos φ
(4)
P = 4 kN, θ = 30°
(a)
Eq. (1):
Eq. (4):
sin φ =
50 mm
sin 30°
200 mm
M = (0.05 m)
φ = 7.181°
sin (30°+7.181°)
(4 kN)
cos 7.818°
M = 121.8 N ⋅ m
�
M = 78.2 N ⋅ m
��
P = 4 kN, θ = 150°
(b)
Eq. (1):
Eq. (4):
sin φ =
50 mm
sin160°
200 mm
M = (0.05 m)
φ = 7.181°
sin (150°+7.181°)
(4 kN)
cos 7.181°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1720
1718
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) θ = 60°, (b) θ = 120°.
SOLUTION
Analysis of the geometry:
Law of sines
sin φ sin θ
=
AB
BC
sin φ =
AB
sin θ
BC
(1)
Now
xC = AB cos θ + BC cos φ
δ xC = − AB sin θδθ − BC sin φδφ
Now, from Equation (1)
or
cos φδφ =
δφ =
(2)
AB
cos θδθ
BC
AB cos θ
δθ
BC cos φ
(3)
From Equation (2)
� AB cos θ
�
δθ �
δ xC = − AB sin θδθ − BC sin φ �
� BC cos φ �
or
δ xC = −
AB
(sin θ cos φ + sin φ cos θ )δθ
cos φ
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1719 1721
PROBLEM 10.20 (Continued)
AB sin(θ + φ )
δθ
cos φ
Then
δ xC = −
Virtual Work:
δ U = 0: − Pδ xC − M δθ = 0
� AB sin(θ + φ ) �
− P �−
δθ � − M δθ = 0
cos φ
�
�
M = AB
Thus,
sin(θ + φ )
P
cos φ
(4)
M = 100 N ⋅ m, θ = 60°
(a)
Eq. (1):
Eq. (4):
sin φ =
50 mm
sin 60°
200 mm
100 N ⋅ m = (0.05 m)
φ = 12.504°
sin (60° + 12.504°)
P
cos 12.504°
P = 2047 N
P = 2.05 kN
�
P = 2.65 kN
�
M = 100 N ⋅ m, θ = 120°
(b)
Eq. (1):
Eq. (4):
sin φ =
50 mm
sin120° φ = 12.504°
200 mm
100 N ⋅ m = (0.05 m)
sin (120° + 12.504°)
P
cos12.504°
P = 2649 N
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1722
1720
PROBLEM 10.21
For the linkage shown, determine the couple M required for
equilibrium when l = 1.8 m, Q = 200 N, and q = 65°.
SOLUTION
δC =
1
2
lδφ
cos θ
Virtual Work:
δU = 0 :
M δφ − Qδ C = 0
M δφ − Q
1 l
δφ = 0
2 cos θ
M=
Data:
M=
1 Ql
2 cos θ
1 (200 N)(1.8 m)
= 425.9 N ⋅ m
2
cos 65°
M = 426 N ⋅ m
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1721 1723
PROBLEM 10.22
For the linkage shown, determine the force Q required for equilibrium
when l = 1.8 m, M = 3 kN ⋅ m, and q = 70°.
SOLUTION
δC =
1 lδφ
2 cos θ
Virtual Work:
δ U = 0:
M δφ − Q
M δφ − Qδ C = 0
1 l
δφ = 0
2 cos θ
Q=
Data:
Q=
2 M cos θ
l
2(3000 N ⋅ m) cos70°
= 1140 N
1.8 m
Q = 1.14 kN
70.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1724
1722
PROBLEM 10.23
Determine the value of θ corresponding to the equilibrium position of
the mechanism of Problem 10.11 when P = 180 N and Q = 640 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 A = 2l sin θ
δ x A = 2l cos θ δθ
yF = 3l cosθ
δ yF = −3l sin θ δθ
δ U = 0:
Qδ x A + Pδ yF = 0
Q(2l cosθ δθ ) + P (−3l sin θ δθ ) = 0
Q=
Data:
3
P tan θ
2
P = 180 N Q = 640 N
(640 N) =
3
(180 N) tanq
2
θ = 67.1°
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1723
1725
PROBLEM 10.24
Determine the value of θ corresponding to the equilibrium position of
the mechanism of Problem 10.9 when P = 80 N and Q = 100 N.
SOLUTION
From geometry
y A = 2l cos θ , δ y A = −2l sin θ δθ
θ
θ
CD = 2l sin , δ (CD ) = l cos δθ
2
2
Virtual Work:
δ U = 0:
− Pδ y A − Qδ (CD) = 0
θ �
�
− P(−2l sin θ δθ ) − Q � l cos δθ � = 0
2 �
�
Q = 2P
or
sin θ
cos ( θ2 )
P = 80 N, Q = 100 N
With
(100 N) = 2(80 N)
sinθ
cos ( θ2 )
sin θ
= 0.625
cos ( θ2 )
or
2sin ( θ2 ) cos ( θ2 )
cos ( θ2 )
= 0.625
θ = 36.42°
θ = 36.4° ��
(Additional solutions discarded as not applicable are θ = ±180°)
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1726
1724
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 θ corresponding to equilibrium.
SOLUTION
Dimensions in mm.
y A = 400sin θ
δ y A = 400 cos θ δθ
xD = 100 cos θ
δ xD = −100sin θ δθ
Virtual Work:
δ U = −(160 N)δ yA − (800 N)δ xD = 0
−(160)(400 cosθ δθ ) − (800)(−100sin θ δθ ) = 0
sin θ
= 0.8; tan θ = 0.8
cos θ
θ = 38.7° �
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1725 1727
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 θ corresponding to
equilibrium.
SOLUTION
y A = (0.4 m) sin θ
δ y A = (0.4 m) cos θ δθ
Virtual Work:
δ U = −(160 N)δ y A + (24 N ⋅ m)δθ = 0
−(160 N)(0.4 m)cosθ δθ + (24 N ⋅ m)δθ = 0
cos θ = 0.375
θ = 68.0° �
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1728
1726
PROBLEM 10.27
Determine the value of θ corresponding to the equilibrium position of the rod
of Problem 10.12 when l = 600 mm, a = 100 mm, P = 100 N, and Q = 160 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 ∆ AA′C :
A′C = a tan θ
y A = −( A′C ) = − a tan θ
δ yA = −
a
δθ
cos 2 θ
For ∆CC ′B :
BC ′ = l sin θ − A′C
= l sin θ − a tan θ
yB = BC ′ = l sin θ − a tan θ
δ yB = l cos θ δθ −
a
δθ
cos 2 θ
Virtual Work:
δ U = 0: − Qδ y A − Pδ yB = 0
−Q −
Q
or
Q=P
a
a
δθ − P l cos θ −
δθ = 0
2
cos θ
cos 2 θ
a
a
= P l cos θ −
cos 2 θ
cos 2 θ
l
cos3 θ − 1
a
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1727
1729
PROBLEM 10.27 (Continued)
with
l = 600 mm, a = 100 mm, P = 100 N, and Q = 160 N
(160 N) = (100 N)
or
600 mm
cos3q − 1
100 mm
cos3 θ = 0.4333
θ = 40.8°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1730
1728
PROBLEM 10.28
Determine the values of θ 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 ∆ AA′C :
A′C = a tan θ
y A = −( A′C ) = − a tan θ
δ yA = −
a
δθ
cos 2 θ
For ∆BB′C :
B′C = l sin θ − A′C
= l sin θ − a tan θ
BB′ =
B′C l sin θ − a tan θ
=
tan θ
tan θ
xB = BB′ = l cos θ − a
δ xB = −l sin θ δθ
Virtual Work:
δ U = 0: Pδ xB − Qδ y A = 0
a
�
�
P(−l sin θ δθ ) − Q � −
δθ � = 0
2
� cos θ
�
Pl sin θ cos 2 θ = Qa
or
l
Q = P sin θ cos 2 θ
a
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1729 1731
PROBLEM 10.28 (Continued)
with
l = 600 mm, a = 100 mm, P = 50 N, and Q = 90 N
90 N = (50 N)
or
600 mm
sin θ cos 2 θ
100 mm
sin θ cos 2 θ = 0.300
θ = 19.81° and 51.9° �
Solving numerically
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1732
1730
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 θ = 30°. For the loading shown, derive an equation in P, θ, l,
and k that must be satisfied when the system is in equilibrium.
SOLUTION
yE = l cos θ
δ yE = −l sin θ δθ
Spring:
Unstretched length = 2l
x = 2(2l sin θ ) = 4l sin θ
δ x = 4l cos θ δθ
F = k ( x − 2l )
F = k ( 4l sin θ − 2l )
Virtual Work:
δ U = 0:
Pδ yE − F δ x = 0
P(−l sin θ δθ ) − k (4l sin θ − 2l )(4l cos θ δθ ) = 0
− P sin θ − 8kl (2sin θ − 1) cos θ = 0
or
P
cos θ
= (1 − 2sin θ )
8kl
sin θ
P 1 − 2sin θ
=
�
8kl
tan θ
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1731 1733
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 = 200 mm and neglecting the weight of
the rods, determine the value of θ corresponding to equilibrium
when P = 160 N.
SOLUTION
From the analysis of Problem 10.29
Then with
P 1 − 2sin θ
=
8kl
tan θ
P = 160 N, l = 0.2 m, and k = 300 N/m
160 N
1 − 2sin θ
=
tan θ
8(300 N/m)(0.2 m)
or
1 − 2sin θ 1
= = 0.3333
tan θ
3
θ = 25.0°
Solving numerically
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1734
1732
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 θ = 30°. Knowing that
l = 0.2 m and neglecting the weight of the rods, determine the
value of θ corresponding to equilibrium when P = 160 N.
SOLUTION
yC = l cos θ , δ yC = −l sin θδθ
Spring:
Unstretched length = 2l
x = 2(2l sin θ ) = 4l sin θ
δ x = 4l cos θδθ
F = k ( x − 2l )
F = k (4l sin θ − 2l )
Virtual Work:
δ U = 0: − Pδ yC − Fδ x
− P(−l sin θδθ ) − k (4l sin θ − 2l )(4l cos θδθ ) = 0
P sin θ − 8kl (2sin θ − 1) cos θ = 0
P
cos θ
= (2sin θ − 1)
8kl
sin θ
or
l = 200 mm, k = 300 N/m, and P = 160 N
with
(160 N)
cos θ
= (2sin θ − 1)
8(300 N/m)(0.2)
sin θ
or
(2sin θ − 1)
cos θ 1
=
sin θ 3
Solving numerically
θ = 39.65°
and
θ = 68.96°
θ = 39.7°
θ = 69.0°
and
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1733 1735
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 θ
corresponding to equilibrium.
SOLUTION
xC = (0.4 m)sin θ
δ xC = 0.4cos θδθ
y A = (0.2 m) cos θ
δ y A = −0.2sin θδθ
Spring:
Unstretched length = 0.2 m
F = k (0.2 m − y A ) = k (0.2 − 0.2 cos θ )
= (3000 N/m)(0.2)(1 − cos q )
F = 600(1 − cos θ )
Virtual Work:
δ U = 0: (150 N)δ xC + F δ y A = 0
150(0.4 cos θδθ ) + 600(1 − cos θ )(−0.2sin θδθ ) = 0
150(0.4) 1 1
= :
= (1 − cos θ ) tan θ
600(0.2) 2 2
θ = 57.2°
Solve by trial and error:
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1736
1734
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 θ corresponding to equilibrium.
SOLUTION
xC = 2l cos θ
δ xC = −2l sin θδθ
yB = l sin θ δ yB = l cos θδθ
F = ks = k (2l − xC ) = 2kl (1 − cos θ )
Virtual Work:
δ U = 0: F δ xC + W δ yB = 0
2kl (1 − cos θ )( −2l sin θδθ ) + W (l cos θδθ ) = 0
4kl 2 (1 − cos θ )sin θ = Wl cos θ
or
Given:
(1 − cos θ ) tan θ =
l = 0.3 m, W = 600 N, k = 2500 N/m
600 N
4(2500 N/m)(0.3 m)
Then
(1 − cos θ ) tan θ =
or
(1 − cos θ ) tan θ = 0.2
Solving numerically
W
4kl
θ = 40.22°
θ = 40.2° ��
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1735 1737
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 θ, W, l, and k that must be satisfied when the linkage
is in equilibrium.
SOLUTION
xC = 2l cos θ
δ xC = −2l sin θδθ
yB = l sin θ δ yB = l cos θδθ
F = ks = k (2l − xC ) = 2kl (1 − cos θ )
Virtual Work:
δ U = 0: F δ xC + W δ yB = 0
2kl (1 − cos θ )( −2l sin θδθ ) + W (l cos θδθ ) = 0
4kl 2 (1 − cos θ )sin θ = Wl cos θ
or
(1 − cos θ ) tan θ =
W
4kl
(1 − cos θ ) tan θ =
From above
W
��
4kl
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you are using it without permission.
1738
1736
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 θ
corresponding to equilibrium for the data indicated.
P = 300 N, l = 400 mm, k = 5 kN/m.
SOLUTION
y A = l sin θ
δ y A = l cos θδθ
Spring:
v = CD
Unstretched when
θ =0
so that
v0 = 2l
For θ :
� 90° + θ �
v = 2l sin �
�
� 2 �
�
�
δ v = l cos � 45° +
θ�
� δθ
2�
Stretched length:
θ�
�
s = v − v0 = 2l sin � 45° + � − 2l
2�
�
Then
�
�
θ�
�
F = ks = kl � 2sin � 45° + � − 2 �
2�
�
�
�
Virtual Work:
δ U = 0: Pδ y A − F δ v = 0
�
�
θ�
θ�
�
�
Pl cos θδθ − kl � 2sin � 45° + � − 2 � l cos � 45° + � δθ = 0
2�
2�
�
�
�
�
or
P
1 �
θ� �
θ�
θ ��
�
�
=
� 2sin � 45° + � cos � 45° + � − 2 cos � 45° + � �
kl cos θ �
2�
2�
2 ��
�
�
�
=
1
cos θ
�
θ� �
θ�
θ ��
�
�
� 2sin � 45° + � cos � 45° + � cos θ − 2 cos � 45° + � �
2�
2�
2 ��
�
�
�
�
=1− 2
cos ( 45° θ2 )
cos θ
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1737 1739
PROBLEM 10.35 (Continued)
Now, with
P = 300 N, l = 400 mm, and k = 5 kN/m
cos ( 45° +
(300 N)
=1− 2
(5000 N/m)(0.4 m)
cos θ
or
cos ( 45° + θ2 )
cos θ
θ
2
)
= 0.60104
θ = 22.6° ��
Solving numerically
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you are using it without permission.
1740
1738
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 θ
corresponding to equilibrium for the data indicated.
P = 300 N, l = 0.3 m, k = 4 kN/m
SOLUTION
From the analysis of Problem 10.35, we have
cos ( 45° + θ2 )
P
=1− 2
cos θ
kl
with
P = 300 N, l = 0.3 m and k = 4 kN/m
cos ( 45° + θ2 )
(300 N)
=1− 2
(4000 N/m)(0.3 m)
cosq
or
cos ( 45° + θ2 )
cosθ
= 0.53033
θ = 51.1°
Solving numerically
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you are using it without permission.
1739 1741
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 θ corresponding to
equilibrium. The constant of the spring is k = 5 kN/m, and the spring is
unstretched when θ = 0.
SOLUTION
s = rθ
δ s = rδθ
F = ks = krθ
yC = l sin θ
δ yC = l cosθδθ
Virtual Work:
δ U = − Fδ s + Wδ yC = 0
− krθ (rδθ ) + W (l cosθδθ ) = 0
θC
(360 N)(0.3 m)
Wl
= 2 =
= 1.500
cos θ kr
(5000 N/m)(0.12 m)2
Solving by trial and error:
θ = 0.91486 rad
θ = 52.4°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1742
1740
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 θ corresponding to equilibrium. The
constant of the spring is k = 4 kN/m, and the spring is unstretched
when θ = 90°.
SOLUTION
�π
�
s = r � −θ �
�2
�
δ s = − rδθ
�π
�
F = ks = kr � − θ �
�2
�
CD = 2l sin
θ
2
θ �1 �
δ (CD) = 2l cos � δθ �
2�2 �
θ
= l cos δθ
2
Virtual Work:
Since F tends to decrease s and P tends to decrease CD, we have
δ U = − F δ s − Pδ (CD ) = 0
θ �
�π
�
�
− kr � − θ � (− rδθ ) − P � l cos δθ � = 0
2
2 �
�
�
�
π
2
−θ
cos θ
2
Solving by trial and error:
=
pl
(240 N)(0.3 m)
= 1.25
=
2
kr
(4000 N/m)(0.12 m)2
θ = 0.33868 rad
θ = 19.40° ��
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1741 1743
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 θ corresponding to the position
of equilibrium when P = 100 N, l = 250 mm, and K = 12.5 N · m/rad.
SOLUTION
y A = l sin θ
We have
δ y A = l cos θδθ
Virtual Work:
δ U = 0: Pδ y A − M δθ = 0
Pl cos θδθ − Kθδθ = 0
or
with
θ
cos θ
Pl
K
(1)
P = 100 N, l = 250 mm and K = 12.5 N ⋅ m/rad
θ
cos θ
or
=
θ
cos θ
=
(100 N)(0.250 m)
12.5 N ⋅ m/rad
= 2.0000
θ = 59.0° ��
Solving numerically
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1744
1742
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 � θ � 90°, 270° � θ � 360°, 360° � θ � 450°.
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 θ 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 = 350 N, l = 250 mm and K = 12.5 N ⋅ m/rad
We have
or
θ
cos θ
θ
cos θ
=
(350 N)(0.250 m)
12.5 N ⋅ m/rad
= 7 or θ = 7 cos θ
(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)}, t = 0.5 * Pi/2);
Thus, we plot y = θ and y = 7 cos θ in the range
0 �θ �
5π
2
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1743 1745
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 θ .
�π �
0 � θ � 90° � � ;
�2�
θ = 1.37333 rad, θ = 78.69°
θ = 78.7° ��
�
� 3π
�
� θ � 2π � ; θ = 5.65222 rad, θ = 323.85° �
270 � θ � 360° �
2
�
�
θ = 324° ��
�
5π
�
360 � θ � 450° � 2π � θ �
2
�
θ = 379° ��
�
�;
�
θ = 6.61597 rad, θ = 379.07 �
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1746
1744
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 θ = 65°.
SOLUTION
We have
yA = (1.44 m)cosq dyA = −1.44 sinqdq
( BD )2 = ( BC )2 + (CD ) 2 − 2( BC )(CD ) cos(180° − θ )
= (0.9)2 + (0.48)2 + 2(0.9)(0.48)cosq
( BD )2 = 1.0404 + 0.864cosq
Differentiating:
(1)
2( BD )δ ( BD ) = −0.864 sinqdq
δ ( BD) = −
0.432
sinqdq
BD
(2)
Virtual Work: Noting that P tends to decrease yA and FBD tends to increase BD, write
δ U = − Pδ y A + FBDδ ( BD ) = 0
−P(−1.44 sinqdq) + FBD −
FBD =
or, since
0.432
sinqdq = 0
BD
1
( BD) P
0.3
P = 3 kN:
FBD =
3000
(BD) = (10000 N)(BD)
0.3
(3)
Making θ = 65° in Eq. (1), we have
(BD)2 = 1.0404 + 0.864 cos 65° = 1.4055
BD = 1.1855
Carrying into Eq. (3):
FBD = (10000 N)(1.1855) = 11855 N
FBD = 11.86 kN
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1745
1747
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 θ 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):
FBD = (10000 N)(BD)
(b)
For ( FBD )max = 12.5 kN, we have
12500 N = (10000 N)(BD)
BD = 1.25
Carrying this value into Eq. (1) of Problem 10.41, write
(BD)2 = 1.0404 + 0.864cosq
(1.25)2 = 1.0404 + 0.864cosq
cos θ = 0.60428
θ = 52.8°
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1748
1746
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 θ = 55°.
SOLUTION
y A = (0.8 m)sin θ
δ y A = 0.8cos θδθ
CD 2 = BC 2 + BD 2 − 2( BC )( BD ) cos(90° − θ )
CD 2 = 0.52 + 1.52 − 2(0.5)(1.5)sin θ
CD 2 = 2.5 − 1.5sin θ
2(CD )(δ CD ) = −1.5cos θδθ
Virtual Work:
(1)
δ CD = −
3cos θ
δθ
4CD
δ U = 0: − (10 kN)δ y A − FCDδ CD = 0
� 3cos θ
�
δθ � = 0
−10(0.8cos θδθ ) − FCD � −
� 4CD
�
FCD =
32
CD
3
(2)
For θ = 55°:
Eq. (1):
Eq. (2):
CD 2 = 2.5 − 1.5sin 55° = 1.2713; CD = 1.1275 m
FCD =
32
32
CD = (1.1275) = 12.027 kN
3
3
FCD = 12.03 kN
��
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1747 1749
PROBLEM 10.44
The position of member ABC is controlled by
the hydraulic cylinder CD. Determine the angle θ
knowing that the hydraulic cylinder exerts a 15-kN
force on pin C.
SOLUTION
y A = (0.8 m)sin θ
δ y A = 0.8cos θδθ
CD 2 = BC 2 + BD 2 − 2( BC )( BD ) cos(90° − θ )
CD 2 = 0.52 + 1.52 − 2(0.5)(1.5)sin θ
CD 2 = 2.5 − 1.5sin θ
2(CD )(δ CD ) = −1.5cos θδθ ;
Virtual Work:
(1)
δ CD = −
3cos θ
δθ
4CD
δ U = 0: − (10 kN)δ y A − FCDδ CD = 0
� 3cos θ
�
δθ � = 0
−10(0.8cos θδθ ) − FCD � −
� 4CD
�
FCD =
32
CD
3
(2)
For FCD = 15 kN:
Eq. (2):
Eq. (1):
15 kN =
32
CD :
3
CD =
(1.40625) 2 = 2.5 − 1.5sin θ :
45
= 1.40625 m
32
sin θ = 0.34831
θ = 20.38°
θ = 20.4° ��
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1750
1748
PROBLEM 10.45
The telescoping arm ABC is used to provide an elevated platform
for construction workers. The workers and the platform together
weigh 2 kN and their combined center of gravity is located
directly above C. For the position when θ = 20°, determine the
force exerted on pin B by the single hydraulic cylinder BD.
SOLUTION
In ∆ ADE :
From the geometry:
AE 0.81 m
=
DE 0.45 m
α = 60.945°
0.81 m
= 0.93 m
AD =
sin 60.945°
tan α =
yC = (4.5 m)sin q
δ yC = (4.5 m)cosqdq
Then, in triangle BAD:
Angle BAD = α + θ
Law of cosines:
BD 2 = AB 2 + AD 2 − 2( AB )( AD ) cos(α + θ )
or
BD 2 = (2.16 m)2 + (0.93 m)2 − 2(2.16 m)(0.93 m)cos(a + q)
BD 2 = 5.5305 m2 − (4.0176 cos(a + q ))m2
(1)
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1749
1751
PROBLEM 10.45 (Continued)
And then
2( BD )(δ BD) = (4.0176 sin(a + q))dq
4.0176 sin(a + q)
δ BD =
dq
2(BD)
Virtual Work:
δ U = 0: − Pδ yC + FBDδ BD = 0
−(2000N)(4.5 m) cosqdq + FBD
Substituting
or
FBD = 4480.3
(4.0176 m2) sin(a + q)
δθ = 0
2( BD )
cos θ
BD N/m
sin(α + θ )
(2)
Now, with θ = 20° and α = 60.945°
Equation (1):
BD 2 = 5.5305 − 4.0176cos(60.945° + 20°)
BD = 2.213 m
Equation (2):
FBD = 4480.3
or
cos 20°
(2.213) N/m
sin(60.945° + 20°)
FBD = 9282.9 N
FBD = 9.28 kN
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1752
1750
PROBLEM 10.46
Solve Problem 10.45 assuming that the workers are lowered to
a point near the ground so that θ = −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 2 kN and their
combined center of gravity is located directly above C. For the
position when θ = 20°, determine 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 θ = −20°, we have
Equation (1):
BD2 = 5.5305 − 4.0176cos(60.945° − 20°)
BD = 1.58 m
Equation (2):
FBD = 4408.3
cos(−20°)
(1.58)
sin(60.945° − 20°)
FBD = 9987.4 N
or FBD = 9.99 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1751
1753
PROBLEM 10.47
A block of weight W is pulled up a plane forming an angle α with the horizontal by a force P directed along
the plane. If µ 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 12 if the block is to
remain in place when the force P is removed.
SOLUTION
Input work = Pδ x
Output work = (W sin α )δ x
Efficiency:
η=
W sin αδ x
Pδ x
or η =
W sin α
P
ΣFx = 0: P − F − W sin α = 0 or
ΣFy = 0: N − W cos α = 0 or
(1)
P = W sin α + F
(2)
N = W cos α
F = µ N = µW cos α
Equation (2):
P = W sin α + µW cos α = W (sin α + µ cos α )
Equation (1):
η=
W sin α
W (sin α + µ cos α )
or η =
1
1 + µ cot α
��
If block is to remain in place when P = 0, we know (see Chapter 8) that φs � α or, since
µ = tan φs,
µ cot α � tan α cot α = 1
Multiply by cot α :
Add 1 to each side:
µ � tan α
1 + µ cot α � 2
Recalling the expression for η , we find
η�
1
��
2
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1754
1752
PROBLEM 10.48
Denoting by µ s the coefficient of static friction between the block
attached to rod ACE and the horizontal surface, derive expressions
in terms of P, µ s, and θ for the largest and smallest magnitude of
the force Q for which equilibrium is maintained.
SOLUTION
For the linkage:
ΣM B = 0: − x A +
xA
P
P = 0 or A =
2
2
Then:
F = µ s A = µs
Now
x A = 2l sin θ
P 1
= µs P
2 2
δ x A = 2l cos θ δθ
and
yF = 3l cos θ
δ yF = −3l sin θ δθ
Virtual Work:
δ U = 0: (Qmax − F )δ x A + Pδ yF = 0
1
�
�
� Qmax − 2 µ s P � (2l cos θ δθ ) + P(−3l sin θ δθ ) = 0
�
�
or
Qmax =
3
1
P tan θ + µ s P
2
2
Qmax =
P
(3 tan θ + µ s ) �
2
For Qmin, motion of A impends to the right and F acts to the left. We change µ s to − µ s and find
Qmin =
P
(3tan θ − µ s ) ��
2
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1753 1755
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 θ = 30°, l = 0.2 m,
and P = 40 N.
SOLUTION
Using the results of Problem 10.48 with
θ = 30°
l = 0.2 m
P = 40 N, and µ s = 0.15
We have
P
(3 tan θ + µ s )
2
(40 N)
(3tan 30° + 0.15)
=
2
= 37.64 N
Qmax =
Qmax = 37.6 N �
and
P
(3 tan θ − µ s )
2
(40 N)
(3tan 30° − 0.15)
=
2
= 31.64 N
Qmin =
Qmin = 31.6 N ��
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1756
1754
PROBLEM 10.50
Denoting by µ s 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 µ s � tan θ .
SOLUTION
Member BC: We have
xB = l cos θ
δ xB = −l sin θδθ
(1)
and
yC = l sin θ
δ yC = l cos θδθ
(2)
Member AB: We have
1
2
δ xB = lδφ
Substituting from Equation (1),
−l sin θδθ =
1
lδφ
2
δφ = −2sin θδθ
or
(3)
Free body of rod BC
For M max, motion of collar C impends upward
ΣM B = 0: Nl sin θ − ( P + µ s N )(l cos θ ) = 0
N tan θ − µ s N = P
N=
�
P
tan θ − µ s
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1755 1757
PROBLEM 10.50 (Continued)
Virtual Work:
δ U = 0: M δφ + ( P + µs N )δ yC = 0
M (−2sin θδθ ) + ( P + µs N )l cos θδθ = 0
M max =
P + µ s tan θP− µ
( P + µs N )
s
l
l=
2 tan θ
2 tan θ
or M max =
Pl
�
2(tan θ − µs )
If µs = tan θ , M = ∞, system becomes self-locking�
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1758
1756
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 θ = 35°, l = 600 mm,
and P = 300 N.
SOLUTION
From the analysis of Problem 10.50, we have
M max =
With
Pl
2(tan θ + µ s )
θ = 35°, l = 0.6 m, P = 300 N
(300 N)(0.6 m)
2(tan 35° − 0.4)
= 299.80 N ⋅ m
M max =
M max = 300 N ⋅ m ��
For M min , motion of C impends downward and F acts upward. The equations of Problem 10.50 can still be
used if we replace µs by − µs . Then
M min =
Pl
2(tan θ + µ s )
Substituting,
(300 N)(0.6 m)
2(tan 35° + 0.4)
= 81.803 N ⋅ m
M min =
M min = 81.8 N ⋅ m ��
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1757 1759
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 12 .
SOLUTION
Recall Figure 8.9a. Draw force triangle
Q = W tan(θ + φs )
y = x tan θ so that δ y = δ x tan θ
Input work = Qδ x = W tan(θ + φs )δ x
Output work = W δ y = W (δ x) tan θ
Efficiency:
η=
W tan θδ x
;
W tan(θ + φs )δ x
η=
tan θ
��
tan(θ + φs )
From Page 432, we know the jack is self-locking if
φs � θ
Then
so that
θ + φs � 2θ
tan(θ + φs ) � tan 2θ �
From above
η=
tan θ
tan(θ + φs )
It then follows that
η�
tan θ
tan 2θ
But
Then
tan 2θ =
2 tan θ
1 − tan 2 θ
η�
tan θ (1 − tan 2 θ ) 1 − tan 2 θ
=
2 tan θ
2
η�
1
��
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1760
1758
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 displacement δ yE for Point E.
From similar triangles:
2.1
δ yE = 1.75δ yE
1.2
2.7
δ yC =
δ yE = 2.25δ yE
1.2
0.5
0.5
(2.25δ yE ) = 1.25δ yE
δ yB =
δ yC =
0.9
0.9
δ yD =
Virtual Work:
δ U = (2 kN)δ yB + (3 kN)δ yD − Eδ yE = 0
2(1.25δ yE ) + 3(1.75δ yE ) − Eδ yE = 0
E = +7.75 kN
E = 7.75 kN ��
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1759 1761
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 displacement δ yH to Point H, keeping member FH horizontal.
From the geometry of the diagram:
δ yF = δ yG = δ yH
0.9
0.9
δ yF =
δ yH = 0.75δ yH
1.2
1.2
1.5
1.5
(0.75δ yH ) = 1.25δ yH
δ yC =
δ yD =
0.9
0.9
0.5
0.5
(1.25δ yH ) = 0.69444δ yH
δ yB =
δ yC =
0.9
0.9
δ yD =
Virtual Work:
δ U = (2 kN)δ yB + (3 kN)δ yD − (5 kN)δ yG + H δ yH = 0
2(0.69444δ yH ) + 3(0.75δ yH ) − 5δ yH + H δ yH = 0
H = +1.3611 kN
H = 1.361 kN �
Couple at H. We rotate beam FH through δθ about Point H.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1762
1760
PROBLEM 10.54 (Continued)
From the geometry of the diagram:
δ yG = 1.2δθ
δ yF = 1.8δθ
0.9
0.9
(1.8δθ ) = 1.35δθ
δ yF =
1.2
1.2
1.5
1.5
(1.35δθ ) = 2.25δθ
δ yC =
δ yD =
0.9
0.9
0.5
0.5
(2.25δθ ) = 1.25δθ
δ yB =
δ yC =
0.9
0.9
δ yD =
Virtual Work:
δ U = (2 kN)δ yB + (3 kN)δ yD − (5 kN)δ yG + M H δθ = 0
2(1.25δθ ) + 3(1.35δθ ) − 5(1.2δθ ) + M H δθ = 0
MH = −0.550 kN ⋅ m
M H = 550 N ⋅ m
�
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1761 1763
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 θ = 55°.
SOLUTION
Virtual Work: Assume both δ yA and δ CD increase
δ U = 0: − (10 kN)δ y A − FCDδ CD = 0
Substitute:
δ y A = 15 mm and FCD = 12.03 kN
−(10 kN)(15 mm) − (12.03 kN)δ CD = 0
δ CD = −12.47 mm
δ CD = 12.47 mm shorter ��
The negative sign indicates that CD shortened
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1764
1762
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 2 kN and
their combined center of gravity is located directly above
C. For the position when θ = 20°, determine the force
exerted on pin B by the single hydraulic cylinder BD.
SOLUTION
Virtual Work: Assume both δ yC and δ BD increase
dU = 0: − (2000 N)d yC + FBDdBD = 0
Substitute:
d yC = 60 mm and
FBD = 9282.9 N
−(2000 N)(60 mm) + (9282.9 N)dBD = 0
dBD = 12.93 mm
The positive sign indicates that cylinder BD increases in length
dBD = 12.9 mm longer
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1763
1765
PROBLEM 10.57
Determine the vertical movement of joint D if the length of
member BF is increased by 38 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.
ΣMH = 0: −P(4 m) + E(12 m) = 0
E=
ΣFy = 0:
P
3
3
P
FBF − = 0
5
3
FBF =
5
P
9
Virtual Work:
We remove member BF and replace it with forces FBF and −FBF at pins F and B, respectively. Denoting the
virtual displacements of Points B and F as δ rB and δ rF , respectively, and noting that P and δ D have the
same direction, we have
Virtual Work:
δ U = 0:
Pδ D + FBF ⋅ δ rF + (−FBF ) ⋅ δ rB = 0
Pδ D + FBF δ rF cos θ F − FBF δ rB cos θ B = 0
Pδ D − FBF (δ rB cos θ B − δ rF cos θ F ) = 0
where (δ rB cos θ B − δ rF cos θ F ) = δ BF , which is the change in
length of member BF. Thus,
Pδ D − FBF δ BF = 0
PdD −
5
P (38 mm) = 0
9
δ D = 21.1 mm
δ D = 21.1 mm
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1766
1764
PROBLEM 10.58
Determine the horizontal movement of joint D if the
length of member BF is increased by 38 mm. (See the hint
for Problem 10.57.)
SOLUTION
Apply horizontal load P at D.
ΣMH = 0: P(3 m) − E y (12 m) = 0
Ey =
ΣFy = 0:
P
4
3
P
FBF − = 0
5
4
FBF =
5
P
12
We remove member BF and replace it with forces FBF and −FBF at pins F and B, respectively. Denoting the
virtual displacements of Points B and F as δ rB and δ rF , respectively, and noting that P and δ D have the
same direction, we have
Virtual Work:
δ U = 0:
Pδ D + FBF ⋅ δ rF + (−FBF ) ⋅ δ rB = 0
Pδ D + FBF δ rF cos θ F − FBF δ rB cos θ B = 0
Pδ D − FBF (δ rB cos θ B − δ rF cos θ F ) = 0
where (δ rB cos θ B − δ rF cos θ F ) = δ BF , which is the change in
length of member BF. Thus,
Pδ D − FBF δ BF = 0
Pd D −
5
P (38 mm) = 0
12
d D = 15.8 mm
d D = 15.8 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1765
1767
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 θ = 30°. For the loading shown, derive
an equation in P, θ, l, and k that must be satisfied when the system
is in equilibrium.
SOLUTION
AE = x = 2(2l sin θ ) = 4l sin θ
Spring:
x0 = 4l sin 30° = 2l
Unstretched length:
Deflection of spring
s = x − x0
s = 2l (2sin θ − 1)
1
V = ks 2 + PyE
2
1
2
V = k [ 2l (2sin θ − 1) ] + P(−l cos θ )
2
dV
= 4kl 2 (2sin θ − 1)2 cos θ + Pl sin θ = 0
dθ
(2sin θ − 1)
cos θ
P
+
=0
sin θ 8kl
P 1 − 2sin θ
=
�
8kl
tan θ
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1768
1766
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 = 200 mm
and neglecting the weight of the rods, determine the value of q
corresponding to equilibrium when P = 160 N.
SOLUTION
Using the result of Problem 10.59, with
P = 160 N
l = 200 mm and k = 300 N/m
P 1 − 2sin θ
=
8kl
tan θ
or
160 N
1 − 2sinq
=
tanq
8(300 N/m)(0.2 m)
1
=
3
θ = 25.0°
Solving numerically,
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1767
1769
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 θ
corresponding to equilibrium.
SOLUTION
P = 600 N
l = 0.3 m, and k = 2500 N ⋅ m
We have
600 N
4(2500 N/m)(0.3 m)
= 0.2
(l − cos θ )tan θ =
θ = 40.2° �
Solving numerically
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1770
1768
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 θ, W, l, and k that must be satisfied
when the linkage is in equilibrium.
SOLUTION
1 2
ks + PyB
2
1
V = k (2i − xC )2 + PyB
2
xC = 2l cos θ and yB = −l sin θ
V=
Thus
1
k (2l − 2l cos θ ) 2 − Pl sin θ
2
= 2kl 2 (1 − cos θ ) 2 − Pl sin θ
V=
dV
= 2kl 2 2(1 − cos θ ) sin θ − Pl cos θ = 0
dθ
(1 − cos θ )tan θ =
P
�
4kl
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1769 1771
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 θ
corresponding to equilibrium for the data indicated.
P = 300 N, l = 400 mm, k = 5 kN/m.
SOLUTION
Spring
� 90° + θ �
v = 2l sin �
�
� 2 �
θ�
�
v = 2l sin � 45° + �
2�
�
Unstretched (θ = 0)
Deflection of spring
v0 = 2l sin 45° = 2l
θ�
�
s = v − v0 = 2l sin � 45° + � − 2l
2�
�
2
�
�
θ�
1
1
�
V = ks 2 + PyA = kl 2 � 2sin � 45° + � − 2 � + P(−l sin θ )
2
2
2�
�
�
�
�
�
θ�
θ�
dV
�
�
= kl 2 � 2sin � 45° + � − 2 � cos � 45° + � − Pl cos θ = 0
dθ
2�
2�
�
�
�
�
�
θ� �
θ�
θ ��
�
�
� 2sin � 45° + � cos � 45° + � − 2 cos � 45° + � � =
2�
2�
2 ��
�
�
�
�
θ�
�
cos θ − 2 cos � 45° + � =
2�
�
Divide each member by cos θ
1− 2
cos ( 45° + θ2 )
cos θ
=
P
cos θ
kl
P
cos θ
kl
P
kl
Then with P = 300 N, l = 0.4 m and k = 5000 N/m
1− 2
or
cos ( 45° + θ2 )
cos θ
cos ( 45° + θ2 )
cos θ
300 N
(5000 N/m)(0.4 m)
= 0.15
=
= 0.60104
θ = 22.6° �
Solving numerically
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1772
1770
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 θ
corresponding to equilibrium for the data indicated.
P = 300 N,
l = 0.3 m,
k = 4 kN/m
SOLUTION
Using the results of Problem 10.63 with P = 300 N, l = 0.3 m and k = 4 kN/m, we have
1− 2
cos ( 45° + θ2 )
cos θ
=
P
kl
300 N
(4000 N/m)(0.3 m)
= 0.25
=
or
Solving numerically
cos ( 45° + θ2 )
cos θ
= 0.53033
θ = 51.058°
θ = 51.1°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1771
1773
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
AE = x = 2(2l sin θ ) = 4l sin θ
Spring:
Unstretched length:
Deflection of spring
x0 = 4l sin 30° = 2l
s = x − x0
s = 2l (2sin θ − 1)
1
V = ks 2 + PyC
2
1
= k[2l (2sin θ − 1)]2 + P(l cos θ )
2
V
dV
dθ
P
cos θ
(1 − 2sin θ )
+
sin θ 8kl
P
8kl
with
We have
or
Solving numerically
= 2kl 2 (2sin θ − 1) 2 + Pl cos θ
= 4kl 2 (2sin θ − 1)2 cos θ − Pl sin θ = 0
=0
=
2sin θ − 1
tan θ
P = 160 N, l = 0.2 m and k = 300 N/m
160 N
2sinq − 1
=
8(300 N/m)(0.2 m)
tan θ
2sin − 1 1
=
tan θ
3
θ = 39.65° and 68.96°
θ = 39.7°
θ = 69.0°
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1774
1772
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 θ corresponding to
equilibrium. The constant of the spring is k = 4 kN/m, and the spring
is unstretched when θ = 90°.
SOLUTION
�π
�
s = r � −θ �
�2
�
For spring:
Ve =
For force P:
1 2 1 2 �π
�
ks = kr � − θ �
2
2
2
�
�
x = 2l sin
2
θ
2
V p = Px = 2 Pl sin
θ
2
2
1
θ
�π
�
V = Ve + V p = kr 2 � − θ � + 2 Pl sin
2
2
�2
�
dV
θ
�π
�
= −kr 2 � − θ � + Pl cos
2
dθ
�2
�
Then
θ
�π
�
= −(4000 N/m)(0.12 m) 2 � − θ � + (240 N)(0.3 m) cos = 0
2
2
�
�
π
2
Solving by trial and error:
− θ = 1.25cos
θ
2
θ = 0.33868 rad
θ = 19.40° �
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1773 1775
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
yA = b − u
yC = b + 2u
y D = −u
1
yE = − u
2
1
yG = + u
2
V = (100 N)y A + PyC + (60 N)yD + (50 N)yE + (40 N)yG
� 1 �
�1 �
= 100(b − u ) + P(b + 2u ) + 60(−u ) + 50 � − u � + 40 � u �
� 2 �
�2 �
dV
= −100 + 2 P − 60 − 25 + 20 = 0
du
P = 82.5 N
Substituting P = 82.5 N in the expression obtained for V :
V = (100 + 82.5)b + (−100 + 165 − 60 − 25 + 20)u
V = 182.5 b = constant
Since V is constant, the equilibrium is neutral ��
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1776
1774
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 A = 2u
y E = −u
y F = −u
V = PyA + (100 N)yE + (150 N)yF
V = P (2u ) + (100 N)(−u ) + (150 N)( −u )
dV
= 2 P − 100 − 150 = 0
du
P = 125 N
Now, substitute P = 125 N in expression for V, making V = 0. Thus, V is constant and
equilibrium is neutral.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1775 1777
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
�l
�
�l
�
V = W � cos 2θ � − W � cos θ �
2
2
�
�
�
�
dV
l
= W ( −2sin 2 + sin θ )
dθ
2
2
1
d V
= W ( −4 cos 2θ − cos θ )
2
2
dθ
Equilibrium:
or
Solving
dV
Wl
= 0:
(−2sin 2θ + sin θ ) = 0
dθ
2
sin θ (−4cos θ + 1) = 0
θ = 0, 75.5°, 180°, and 284°
Stability:
d 2V
l
= W (−4 cos 2θ − cos θ )
2
2
dθ
At θ = 0:
l
d 2V
= W ( −4 − 1) � 0
2
2
dθ
At θ = 75.5°:
d 2V
l
= W (−4(−.874) − .25) � 0
2
2
dθ
At θ = 180°:
d 2V
l
= W ( −4 + 1) � 0
2
dθ 2
At θ = 284°:
d 2V
l
= W (−4(−.874) − .25) � 0
2
2
dθ
θ = 0, Unstable �
θ = 75.5°, Stable �
θ = 180.0°, Unstable �
θ = 284°, Stable ��
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1778
1776
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 3 N and that rod CD
weighs 2 N, determine the positions of equilibrium of the system
and state in each case whether the equilibrium is stable, unstable,
or neutral.
SOLUTION
V = −WAB
l
l
sin θ − WCD
cos 2θ
2
2
1
dV
= − WAB l cos θ + WCD l sin 2θ
dθ
2
dV
=0
dθ
Equilibrium:
1
− WAB l cos θ + WCD l sin 2θ = 0
2
−WAB cos θ + 4WCD sin θ cos θ = 0
−WAB cos θ 1 −
WCD
sin θ = 0
WAB
cos θ = 0 and sin θ =
WAB
4WCD
θ = 90°, θ = 270°, θ = sin −1
WAB
4WCD
(1)
WAB = 3 N and WCD = 2 N, We have
For
3
8
θ = 90°, θ = 270°, θ = 22.0°, θ = 158.0°
θ = 90°, θ = 270°, θ = sin −1
1
d 2V
= WAB sin θ + 2WCD cos 2θ ) l
2
2
dθ
Stability:
(2)
θ = 270° :
d 2V
1
= (3) sin(270°) + 2(2) cos(540°) l = −5.5l
2
2
dθ
θ = 22.0°:
1
d 2V
= (3) sin 220° + 2(2) cos 44.0° l
2
2
dθ
0, unstable
0 Stable
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you are using it without permission.
1777
1779
PROBLEM 10.70 (Continued)
θ = 90°:
1
d 2V
= (3)sin 90° + 2(2) cos180° l = 2.5l
2
2
dθ
θ = 158.0° :
0, Unstable
d 2V
1
= (3) sin158° + 2(2) cos 316° l = 3.44l
2
2
dθ
0, Stable
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1780
1778
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.
SOLUTION
Potential energy
� l
�
�l
�
V = W � − sin θ � + W � cos θ � W = mg
� 2
�
�2
�
l
= W (cosθ − sin θ )
2
dV Wl
=
( − sin θ − cos θ )
dθ
2
d 2V Wl
=
(sin θ − cos θ )
2
dθ 2
For equilibrium:
or
Thus
dV
= 0: sin θ = − cos θ
dθ
tan θ = −1
θ = −45.0° and θ = 135.0°
Stability:
At θ = −45.0°:
d 2V Wl
=
[sin( −45°) − cos 45°]
2
dθ 2
Wl �
2
2�
−
=
�� −
��0
2 � 2
2 ��
θ = −45.0°, Unstable �
At θ = 135.0°:
d 2V Wl
=
(sin135° − cos135°)
2
dθ 2
Wl � 2
2�
=
+
��
��0
2 � 2
2 ��
θ = 135.0°, Stable �
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1779 1781
PROBLEM 10.72
Two uniform rods, AB and CD, are attached to gears of equal radii
as shown. Knowing that MAB = 3.5 kg and MCD = 1.75 kg, determine
the positions of equilibrium of the system and state in each case
whether the equilibrium is stable, unstable, or neutral.
SOLUTION
Potential energy
l
l
V = (3.5 kg × 9.81 m/s 2 ) − sin θ + (1.75 kg × 9.81 m/s 2 ) cos θ
2
2
= (8.5838 N)l (−2sin θ + cos θ )
dV
= (8.5838 N)l ( −2 cos θ − sin θ )
dθ
d 2V
= (8.5838 N)l (2sin θ − cos θ )
dθ 2
Equilibrium:
or
Thus
dV
= 0: − 2 cos θ − sin θ = 0
dθ
tan θ = −2
θ = −63.4° and 116.6°
Stability
At θ = −63.4°:
d 2V
= (8.5838 N)l[2sin(−63.4°) − cos( −63.4°)]
dθ 2
= (8.5838 N)l (−1.788 − 0.448) 0
θ = −63.4°, Unstable
At θ = 116.6°:
d 2V
= (8.5838 N)l[2sin(116.6°) − cos(116.6°)]
dθ 2
= (8.5838 N)l (1.788 + 0.447) 0
θ = 116.6°, Stable
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1782
1780
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 Kθ 2 , where K is the torsional spring constant and θ 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 θ
corresponding to the position of equilibrium when P = 100 N, l = 250 mm,
and K = 12.5 N ⋅ m/rad.
SOLUTION
Potential energy
V=
1
Kθ 2 − Pl sin θ
2
dV
= Kθ − Pl cos θ
dθ
d 2V
= K + Pl sin θ
dθ 2
Equilibrium:
For
dV
K
= 0: cos θ = θ
dθ
Pl
P = 100 N, l = 0.25 m., K = 12.5 N ⋅ m/rad
12.5 N ⋅ m/rad
θ
(100)(0.25 m)
= 0.500θ
cos θ =
Solving numerically, we obtain
θ = 1.02967 rad = 59.000°
θ = 59.0° ��
Stability
d 2V
= (12.5 N ⋅ m/rad) + (100 N)(0.25 m)sin 59.0° � 0
dθ 2
Stable �
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1781 1783
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 �θ � 90°, 270°�θ � 360°, 360° �θ � 450°.
SOLUTION
Potential energy
V=
1
Kθ 2 − Pl sin θ
2
dV
= Kθ − Pl cos θ
dθ
d 2V
= K + Pl sin θ
dθ 2
Equilibrium
dV
K
= 0: cos θ = θ
dθ
Pl
P = 350 N, l = 0.250 m and
For
cos θ =
cos θ =
or
K = 12.5 N ⋅ m/rad
12.5 N ⋅ m/rad
θ
(350 N)(0.250 m)
θ
7
Solving numerically
θ = 1.37333 rad, 5.652 rad, and 6.616 rad
or
θ = 78.7°, 323.8°, 379.1° �
Stability at θ = 78.7°:
d 2V
= (12.5 N ⋅ m/rad) + (350 N)(0.250 m)sin 78.7°
dθ 2
θ = 78.7°, Stable ��
= 98.304 � 0
At θ = 323.8°:
d 2V
= (12.5 N ⋅ m/rad) + (350 N)(0.250 m)sin 323.8°
dθ 2
= −39.178 N ⋅ m � 0
At At θ = 379.1°:
θ = 324°, Unstable ��
d 2V
= (12.5 N ⋅ m/rad) + (350 N)(0.250 m)sin 379.1°
dθ 2
θ = 379°, Stable �
= 44.132 N ⋅ m � 0
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1784
1782
PROBLEM 10.75
A load W of magnitude 100 N is applied to the mechanism at C.
Knowing that the spring is unstretched when θ = 15°, determine that
value of θ corresponding to equilibrium and check that the equilibrium
is stable.
SOLUTION
We have
yC = l cos θ
1
π
k[ r (θ − θ0 )]2 + WyC θ0 = 15° = rad
2
12
1
= kr 2 (θ − θ0 )2 + Wl cos θ
2
V=
dV
= kr 2 (θ − θ0 ) − Wl sin θ
dθ
Equilibrium
dV
= 0: kr 2 (θ − θ0 ) − wl sin θ = 0
dθ
with
W = 100 N, R = 2500 N/m, l = 0.4 m, and r = 0.1 m
(2500 N/m)(0.01 m2) q −
or
Solving numerically
π
12
(1)
− (100 N)(0.4 m)sin q = 0
0.625θ − sin θ = 0.16362
θ = 1.8145 rad = 103.97°
θ = 104.0°
Stability
d 2V
= kr 2 − Wl cos θ
dθ 2
or
= 25 − 40cosq
For θ = 104.0°:
= 34.7 m ⋅ N > 0
(2)
Stable
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1783 1785
PROBLEM 10.76
A load W of magnitude 100 N is applied to the mechanism at C.
Knowing that the spring is unstretched when θ = 30°, determine that
value of θ corresponding to equilibrium and check that the equilibrium
is stable.
SOLUTION
Using the solution of Problem 10.75, particularly Equations (1), with 15° replace by 30°
For equilibrium
kr 2 θ −
π
6
π
6
rad :
− Wl sin θ = 0
k = 2500 N/m, W = 100 N, r = 0.1 m, and l = 0.4 m
With
(2500 N/m)(0.01 m2) q −
π
6
− (100 N)(0.4 m)sin q = 0
25q − 13.09 − 40sin q = 0
or
θ = 1.9870 rad = 113.8°
Solving numerically,
θ = 113.8°
Stability: Equation (2), Problem 10.75:
d 2V
= kr 2 − Wl cos θ
2
dθ
or
= 25 − 40cosq
For θ = 113.8°:
= 41.1 m ⋅ N > 0
Stable
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1786
1784
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 θ, W, l, and k that must be
satisfied when the rod is in equilibrium.
SOLUTION
Elongation of spring:
s = l sin θ + l cos θ − l
s = l (sin θ + cos θ − 1)
Potential energy:
1 2
1
ks − W sin θ W = mg
2
2
1 2
l
= kl (sin θ + cos θ − 1)2 − mg sin θ
2
2
V=
dV
1
= kl 2 (sin θ + cos θ − 1)(cos θ − sin θ ) − mgl cos θ
dθ
2
Equilibrium:
(1)
dV
mg
= 0: (sin θ + cos θ − 1)(cos θ − sin θ ) −
cos θ = 0
2kl
dθ
mg �
�
=0 �
or cos θ �(sin θ + cos θ − 1)(1 − tan θ ) −
2kl ��
�
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1785 1787
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 θ corresponding to
equilibrium when W = 300 N, l = 400 mm, and k = 1500 N/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 θ (sin θ + cos θ − 1)(1 − tan θ ) −
mg
=0
2kl
Now, with W = 300 N, l = 0.4 m, and k = 1500 N/m
300 N
W
=
= 0.25
2kl 2(0.4 m)(1500 N/m)
cos θ [ (sin θ + cos θ − 1)(1 − tan θ ) − 0.25] = 0
Thus:
cos θ = 0 and (sinθ + cos θ − 1)(1 − tan θ ) = 0.25
First equation yields θ = 90°. Solving the second equation by trial, we find θ = 9.39° and 34.16°
Values of θ for equilibrium are
θ = 9.39°, 34.2°, and 90.0°
Stability we differentiate Eq. (1) of Problem 10.77.
d 2V
1
= kl 2 [(cosθ − sin θ )(cos θ − sin θ ) + (sin θ + cos θ − 1)(− sin θ − cos θ )] + Wl sinq
2
2
dq
= kl 2 cos 2 θ + sin 2 θ − 2 cos θ sin θ − sin 2 θ − cos 2 θ − 2 cos θ sin θ + sin θ + cos θ +
= kl 2
1+
W
sin θ
2kl
W
sin θ + cos θ − 2sin 2θ
2kl
d 2V
= kl 2 (1.25sin θ + cos θ − 2sin 2θ )
dθ 2
θ = 9.39° :
d 2V
= kl 2 (1.25sin 9.4 + cos9.4 − 2sin18.8)
dθ 2
= kl 2 (+0.55)
Stable
0
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1788
1786
PROBLEM 10.78 (Continued)
θ = 34.2° :
d 2v
= kl 2 (1.25sin 34.2° + cos 34.3° − 2sin 68.4°)
2
dθ
= kl 2 (−0.33) � 0
θ = 90.0° :
Unstable ��
d 2V
= kl 2 (1.25sin 90° + cos 90° − 2sin180°)
2
dθ
= kl 2 (1.25) � 0
Stable �
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1787 1789
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
s = l2 + y2 − l
Deflection of spring = s, where
ds
y
=
2
dy
l − y2
1 2
y
ks − W
2
2
dV
ds 1
= ks
− W
dy
dy 2
dV
= k l2 + y2 − l
dy
V=
Potential energy:
(
)
y
1
− W
l 2 + y2 2
�
�
l
� y − 1W
= k �1 −
2
2 �
�
2
l +y �
�
�
�
dV
l
�y= 1W
= 0: �1 −
2
2
�
dy
2 k
l + y ��
�
Equilibrium
W = 80 N, l = 0.500 m, and k = 600 N/m
Now
Then
or
Solving numerically,
�
0.500 m
�1 −
�
(0.500) 2 + y 2
�
�
�1 − 0.500
�
0.25 + y 2
�
�
� y = 1 (80 N)
�
2 (600 N/m)
�
�
� y = 0.066667
�
�
y = 0.357 m
y = 357 mm �
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1790
1788
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 = 550 mm, and k = 600 N/m.
SOLUTION
Spring deflections
S AD = l − l 2 − y 2
S BC = l 2 + y 2 − l
1 2
1 2
y
−W
kS AD + kS BC
2
2
2
2
1
1
V = k l − l 2 − y2 + k
2
2
�
dV
y
= k l − l 2 − y2 �
2
� l − y2
dy
�
V=
)
(
(
(
)
) − W 2y
�
�
� + k ( l + y − l )�
�
�
�
�
l 2 + y2 − l
2
2
��
� �
dV
l
l
= 0: ��
− 1� + � 1 −
2
2
2
��
� �
dy
l + y2
� �
�� l − y
2
� W
�−
l 2 + y 2 �� 2
y
��
�� y = W
��
2k
��
Data: W = 80 N, l = 0.5 m, k = 600 N/m
�
0.5
0.5
�
−
2
2
� (0.5) − y
(0.5) 2 + y 2
�
�
80
�y=
= 0.066667
2(1200)
�
�
y = 0.252 m
Solve by trial and error:
y = 252 mm �
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1789 1791
PROBLEM 10.81
A spring AB of constant k is attached to two identical gears as
shown. Knowing that the spring is undeformed when θ = 0,
determine two values of the angle θ corresponding to equilibrium
when P = 150 N, a = 100 mm, b = 75 mm, r = 150 mm, and
k = 1 kN/m. State in each case whether the equilibrium is stable,
unstable, or neutral.
SOLUTION
Elongation of spring
s = 2(a sin θ ) = 2a sin θ
1 2
ks − Pbθ
2
1
= k (2a sin θ ) 2 − Pbθ
2
V=
dV
= 4ka 2 sin θ cosθ − Pb
dθ
= 2ka 2 sin 2θ − Pb
Equilibrium
(1)
dV
Pb
= 0: sin 2θ =
dθ
2ka 2
sin 2q =
(150 N)(0.075 m)
; sin2q = 0.5625
2(1000 N/m)(0.1 m)2
2θ = 34.229° and 145.771°
θ = 17.11° and 72.9°
Stability: We differentiate Eq. (1)
d 2V
= 4ka 2 cos 2θ
dθ 2
θ = 17.11° :
d 2v
= 4ka 2 cos 34.2° = 4ka 2 (0.83)
2
dθ
θ = 72.9°:
d 2v
= 4ka 2 cos145.8° = 4ka 2 (−0.83)
dθ 2
0
Stable
0
Unstable
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1792
1790
PROBLEM 10.82
A spring AB of constant k is attached to two identical gears as
shown. Knowing that the spring is undeformed when θ = 0, and
given that a = 60 mm, b = 45 mm, r = 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 θ 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 = 2(a sin θ ) = 2a sin θ
Potential energy
1 2
1
ks − Pbθ = k (2a sin θ )2 − Pbθ
2
2
2
2
V = 2ka sin θ − Pbθ
V=
Equilibrium
dV
dV
= 0:
= 4ka 2 sin θ cos θ − Pb
dθ
dθ
= 2ka 2 sin 2θ − Pb = 0
sin 2θ =
(b)
P b
Pb
; For Pmax ; max2 = 1
2
2ka
2ka
Pmax (0.045 m)
(a)
2(6000 N/m)(0.06 m) 2
For P =
1
Pmax ,
2
(1)
=1
Pmax = 960 N �
1
sin 2θ = ; 2θ = 30° and 150°
2
θ = 15.00° and 75.0° ��
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1791 1793
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 β = 30° and P = Q = 400 N,
determine the value of the angle θ corresponding to equilibrium.
SOLUTION
Law of Sines
yA
L
=
sin(90° + β − θ ) sin(90 − β )
yA
L
=
cos(θ − β ) cos β
or
yA = L
cos(θ − β )
cos β
From the figure:
yB = L
cos(θ − β )
− L cos θ
cos β
Potential Energy:
� cos(θ − β )
�
cos(θ − β )
V = − PyB − Qy A = − P � L
− L cos θ � − QL
cos β
cos β
�
�
� sin(θ − β )
�
sin(θ − β )
dV
= − PL � −
+ sin θ � + QL
dθ
cos β
cos β
�
�
= L( P + Q )
sin(θ − β )
− PL sin θ
cos β
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1794
1792
PROBLEM 10.83 (Continued)
dV
sin(θ − β )
= 0: L( P + Q )
− PL sin θ = 0
cos β
dθ
Equilibrium
or
( P + Q) sin(θ − β ) = P sin θ cos β
( P + Q)(sin θ cos β − cos θ sin β ) = P sin θ cos β
or
−( P + Q ) cos θ sin β + Q sin θ cos β = 0
−
P + Q sin β sin θ
+
=0
Q cos β cos θ
tan θ =
With
P+Q
tan β
Q
(2)
P = Q = 400 N, β = 30°
tan θ =
800 N
tan 30° = 1.1547
400 N
θ = 49.1° ��
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1793 1795
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 β = 30°, P = 100 N,
and Q = 25 N, determine the value of the angle θ corresponding to
equilibrium.
SOLUTION
Using Equation (2) of Problem 10.83, with P = 100 N, Q = 25 N, and β = 30°, we have
(100 N)(25 N)
tan 30°
(25 N)
= 57.735
θ = 89.007°
tan θ =
θ = 89.0° ��
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1796
1794
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 θ corresponding to
equilibrium when W = 200 N, r = 180 mm, and k = 3 kN/m
SOLUTION
Stretch of Spring
s = AB − r
s = 2(r cos θ ) − r
s = r (2cos θ − 1)
Potential Energy:
Equilibrium
1 2
ks − Wr sin 2θ
W = mg
2
1
V = kr 2 (2 cos θ − 1) 2 − Wr sin 2θ
2
dV
= − kr 2 (2 cos θ − 1)2sin θ − 2Wr cos 2θ
dθ
V=
dV
= 0: − kr 2 (2 cos θ − 1) sin θ − Wr cos 2θ = 0
dθ
(2cos θ − 1) sin θ
W
=−
cos 2θ
kr
Now
Then
Solving numerically,
W
(200 N)
=
= 0.37037
kr (3000 N/m)(0.18 m)
(2cos θ − 1)sin θ
= −0.37037
cos 2θ
θ = 0.95637 rad = 54.8°
θ = 54.8°
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1795
1797
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 θ corresponding to
equilibrium when W = 200 N, r = 180 mm, and k = 3 kN/m
SOLUTION
Stretch of spring
s = AB − r = 2( r cos θ ) − r
s = r (2cos θ − 1)
1
V = ks 2 − Wr cos 2θ
2
1 2
= kr (2 cos θ − 1) 2 − Wr cos 2θ
2
dV
= −kr 2 (2 cos θ − 1)2sin θ + 2Wr sin 2θ
dθ
Equilibrium
dV
= 0: − kr 2 (2 cos θ − 1) sin θ + Wr sin 2θ = 0
dθ
− kr 2 (2cos θ − 1) sin θ + Wr (2sin θ cos θ ) = 0
(2cos θ − 1) sin θ W
=
2 cos θ
kr
or
Now
Then
W
(200 N)
=
= 0.37037
kr (3000 N/m)(0.18 m)
2cos θ − 1
= 0.37037
2 cos θ
θ = 37.4°
Solving
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1798
1796
PROBLEM 10.87
Cart B, which weighs 75 kN, rolls along a sloping track that forms
an angle β 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 β indicated.
Angle β = 30°.
SOLUTION
x = (4 m) tan θ
(1)
yB = x sin β = 4 tan θ sin β
AC = (4 m) cos θ
For x = 0,
Stretch of spring.
( AC )0 = 4 m
s = AC − ( AC )0 =
V=
=
4
� 1
�
− 4 = 4�
− 1�
cos θ
� cos θ
�
1 2
ks − (75 kN) yB
2
2
1
� 1
�
(5 kN/m) 16 �
− 1� − (75 kN)4 tan θ sin β
2
� cos θ
�
sin β
dV
� 1
� sin θ
− 300
= 80 �
− 1�
2
dθ
cos 2 θ
� cos θ
� cos θ
Equilibrium
Given:
Eq. (2):
Solve by trial and error:
Eq. (1):
dV
� 1
�
= 0: �
− 1� sin θ = 3.75sin β
dθ
� cos θ
�
(2)
β = 30°, sin θ = 0.5
� 1
�
� cos θ − 1� sin θ = 3.75(0.5) = 1.875
�
�
θ = 70.46°
x = (4 m) tan 70.46°
x = 11.27 m ��
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1797 1799
PROBLEM 10.88
Cart B, which weighs 75 kN, rolls along a sloping track that forms
an angle β 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 β indicated.
Angle β = 60°.
SOLUTION
x = (4 m) tan θ
(1)
yB = x sin β = 4 tan θ sin β
AC = (4 m) cos θ
For x = 0,
Stretch of spring.
( AC )0 = 4 m
s = AC − ( AC )0 =
V=
=
4
� 1
�
− 4 = 4�
− 1�
cos θ
� cos θ
�
1 2
ks − (75 kN) yB
2
2
1
� 1
�
(5 kN/m) 16 �
− 1� − (75 kN)4 tan θ sin β
2
� cos θ
�
sin β
dV
� 1
� sin θ
− 300
= 80 �
− 1�
2
dθ
cos 2 θ
� cos θ
� cos θ
Equilibrium
Given:
Eq. (2):
dV
� 1
�
= 0: �
− 1� sin θ = 3.75sin β
dθ
� cos θ
�
(2)
β = 60°, sin θ = 0.86603
� 1
�
� cos θ − 1� sin θ = 3.75(0.86603) = 3.2476
�
�
Solve by trial and error:
θ = 76.67°
Eq. (1):
x = (4 m) tan 26.67°
x = 16.88 m ��
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1800
1798
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 = 2CD.
SOLUTION
For both (a) and (b): Since P and − P are vertical, they form a couple of moment
M P = + Pl sin θ
The forces F and −F exerted by springs must, therefore, also form a couple, with moment
M F = − Fa cos θ
We have
dU = M P dθ + M F dθ
= ( Pl sin θ − Fa cos θ )dθ
but
Thus,
�1
�
F = ks = k � a sin θ �
2
�
�
1
�
�
dU = � Pl sin θ − ka 2 sin θ cos θ � dθ
2
�
�
From Equation (10.19), page 580, we have
dV = − dU = − Pl sin θ dθ +
or
1 2
ka sin 2θ dθ
4
1
dV
= − Pl sin θ + ka 2 sin 2θ
dθ
4
and
1
d 2V
= − Pl cos θ + ka 2 cos 2θ
2
2
dθ
For θ = 0:
1
d 2V
= − Pl + ka 2
2
2
dθ
(1)
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1799 1801
PROBLEM 10.89 (Continued)
For Stability:
d 2V
1
� 0, − Pl + ka 2 � 0
2
2
dθ
or (for Parts a and b)
P�
Note: To check that equilibrium is unstable for P =
ka 2
2l
, we differentiate (1) twice:
d 3V
= + Pl sin θ − ka 2 sin 2θ = 0, for
dθ 3
d 4V
= Pl cos θ − 2ka 2 cos 2θ
dθ 4
For θ = 0
Thus, equilibrium is unstable when
ka 2
��
2l
θ = 0,
d 4V
ka 2
2
Pl
ka
2
=
−
=
− 2ka 2 � 0
2
dθ 4
P=
ka 2
�
2l
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1802
1800
PROBLEM 10.90
Rod AB is attached to a hinge at A and to two springs, each of
constant k. If h = 500 mm, d = 240 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
xC = d sin θ
Potential Energy:
V =2
yB = h cos θ
1 2
kxC + WyB
2
= kd 2 sin 2 θ + Wh cos θ
dV
= 2kd 2 sin θ cos θ − Wh sin θ
dθ
= kd 2 sin 2θ − Wh sin θ
Then
d 2V
= 2kd 2 cos 2θ − Wh cos θ
2
dθ
and
(1)
For equilibrium position θ = 0 to be stable, we must have
d 2V
= 2kd 2 − Wh
2
dθ
kd 2
or
Note: For kd 2 = 12 Wh, we have
d 2V
dθ 2
0
1
Wh
2
(2)
= 0, so that we must determine which is the first derivative that is not
equal to zero. Differentiating Equation (1), we write
d 3V
= −4kd 2 sin 2θ + Wh sin θ = 0
3
dθ
d 4V
= −8kd 2 cos 2θ + Wh cos θ
dθ 2
For θ = 0:
for θ = 0
d 4V
= −8kd 2 + Wh
dθ 4
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1801 1803
PROBLEM 10.90 (Continued)
Since kd 2 = 12 Wh,
the
= −4Wh + Wh
0, we conclude that the equilibrium is unstable for kd 2 = 12 Wh and
sign in Equation (2) is correct.
With
Equation (2) gives
or
d 4V
dθ 4
W = 400 N,
k(0.24 m)2
k
h = 0.5 m, and
d = 0.24 m
1
(400 N)(0.5 m)
2
k
1736 N/m
1.74 kN/m
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1804
1802
PROBLEM 10.91
Rod AB is attached to a hinge at A and to two springs, each of
constant k. If h = 0.9 m, k = 1.5 kN/m, and W = 300 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.
SOLUTION
Using Equation (2) of Problem 10.90 with
h = 0.9 m, k = 1.5 kN/m, and W = 300 N
(1500 N/m)d 2
or
d2
d
1
(300 N)(0.9 m)
2
0.09 m 2
0.3 m
smallest d = 300 mm
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1803 1805
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
s=
L
2L
sin φ =
sin θ
3
3
For small values of φ and θ
φ = 2θ
2L
�L
� 1
cos θ � + ks 2
V = P � cos φ +
3
�3
� 2
2
1 � 2L
PL
�
(cos 2θ + 2cos θ ) + k �
sin θ �
3
2 � 3
�
2 2
PL
(−2sin 2θ − 2sin θ ) + kL sin θ cos θ
=
3
9
2
PL
(2sin 2θ + 2sin θ ) + kL2 sin 2θ
=−
3
9
4
PL
(4 cos 2θ + 2 cos θ ) + kL2 cos 2θ
=−
3
9
V=
dV
dθ
d 2V
dθ 2
when θ = 0:
d 2V
6 PL 4 2
=−
+ kL
2
3
9
dθ
For stability:
4
d 2V
� 0, − 2 PL + kL2 � 0
2
9
dθ
2
P � kL ��
9
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1806
1804
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.
SOLUTION
a=
2L
L
sin θ = sin φ
3
3
For small values of φ and θ
φ = 2θ
s=
L
sin θ
3
L
� 2L
� 1
cos θ + cos φ � + ks 2
V = P�
3
� 3
� 2
=
1 �L
PL
�
(2 cos θ + cos 2θ ) + k � sin θ �
3
2 �3
�
2
dV PL
kL2
=
(−2sin θ − 2sin 2θ ) +
sin θ cos θ
3
9
dθ
2 PL
dV
kL2
=−
(sin θ + sin 2θ ) +
sin 2θ
3
18
dθ
2 PL
d 2V
kL2
θ
θ
=
−
+
+
(cos
2
cos
2
)
cos 2θ
3
9
dθ 2
when θ = 0:
dV 2
kL2
=
−
2
PL
+
9
dθ 2
For stability:
d 2V
kL2
PL
�
�0
0,
−
2
+
9
dθ 2
P�
1
kL ��
18
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1805 1807
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) sin θ
V = P(2l cos θ ) +
= 2 Pl cos θ +
1 2
ks
2
1
k (l − a) 2 sin 2 θ
2
dV
= −2 Pl sin θ + k (l − a) 2 sin θ cos θ
dθ
1
= −2 Pl sin θ + k (l − a )2 sin 2θ
2
2
d V
= −2 Pl cos θ + k (l − a) 2 cos 2θ
dθ 2
when
Stability:
θ = 0:
(1)
d 2V
= −2 Pl + k (l − a) 2
2
dθ
d 2V
� 0: − 2 Pl + k (l − a) 2 > 0
dθ 2
To check whether equilibrium is unstable for P =
k (l − a )2
2l
P�
k (l − a) 2
�
2l
, we differentiate
Eq. (1) twice:
d 3V
= 2 Pl sin θ − 2k (l − a) 2 sin 2θ = 0, For θ = 0
3
dθ
d 4V
= 2 Pl cos θ − 4k (l − a) 2 cos 2θ
dθ 4
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1808
1806
PROBLEM 10.94 (Continued)
For G = 0 and
P=
k (l − a) 2
2l
d 4V
= 2 Pl − 4k (l − a )2
dθ 4
= k (l − a)3 − 4k (l − a) 2 � 0
Thus equil is unstable for
P=
k (l − a) 2
�
2l
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1807 1809
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 = 480 mm, b = 400 mm, and P = 600 N.
SOLUTION
First note
For small values of θ and φ :
or
A = a sin θ = b sin φ
aθ = bφ
a
b
φ= θ
V = P( a + b) cos φ − 2Q (a + b) cos θ
= ( a + b) P cos
a
θ − 2Q cos θ
b
dV
a
a
= ( a + b) − P sin θ + 2Q sin θ
b
dθ
b
d 2V
a2
a
=
+
−
(
)
a
b
P cos θ + 2Q cos θ
2
2
b
dθ
b
when θ = 0:
a2
d 2V
a
b
P + 2Q
=
(
+
)
−
dθ 2
b2
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1810
1808
PROBLEM 10.95 (Continued)
d 2V
dθ 2
Stability:
0: −
or
with
P = 600 N,
Equation (1):
Q
a2
P + 2Q
b2
0
P
2
b2
Q
a2
(1)
Q
a2
P
2b 2
(2)
a = 0.48 m,
and b = 0.4 m
(0.48 m)2
(600 N) = 432 N
2(0.4 m)2
Q > 432 N
For stability
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1809 1811
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.
SOLUTION
Using Equation (2) of Problem 10.95 with
Q = 45 N, a = 150 mm, and b = 200 mm
Equation (2)
(200 mm) 2
(45 N)
(150 mm) 2
= 160.000 N
P�2
P � 160.0 N �
For stability
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1812
1810
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 θ1 = θ2 = 0. Determine the range of values of P for
which the equilibrium position is stable.
SOLUTION
We have
xB = l sin θ
xC = l sin θ1 + l sin θ 2
yC = l cos θ1 + l cos θ 2
V = PyC +
or
V = Pl (cos θ1 + cos θ 2 ) +
1 2 1 2
kxB + kxC
2
2
1 2
kl ��sin 2 θ1 + (sin θ1 + sin θ 2 ) 2 ��
2
For small values of θ1 and θ 2 :
1
1
sin θ1 ≈ θ1 , sin θ 2 ≈ θ 2 , cosθ1 ≈ 1 − θ12 , cosθ 2 ≈ 1 − θ 22
2
2
Then
and
� θ2
θ2
V = Pl �1 − 1 + 1 − 2
�
2
2
�
� 1 2 2
2
�� + kl �θ1 + (θ1 + θ 2 ) �
�
�
� 2
∂V
= − Plθ1 + kl 2 [θ1 + (θ1 + θ 2 )]
∂ θ1
∂V
= − Plθ 2 + kl 2 (θ1 + θ 2 )
∂ θ2
∂ 2V
= − Pl + 2kl 2
2
∂ θ1
∂ 2V
= − Pl + kl 2
2
∂ θ2
∂ 2V
= kl 2
∂ θ1∂ θ 2
Stability: Conditions for stability (see Page 583).
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1811 1813
PROBLEM 10.97* (Continued)
For
θ1 = θ 2 = 0:
∂V ∂V
=
= 0 (condition satisfied)
∂ θ1 ∂ θ 2
� ∂ 2V
��
� ∂ θ1∂ θ 2
Substituting
2
� ∂ 2V ∂ 2V
�0
�� −
2
2
� ∂ θ1 ∂ θ 2
(kl 2 ) 2 − (− Pl + 2kl 2 )(− Pl + kl ) � 0
k 2 l 4 − P 2 l 2 + 3Pkl 3 − 2k 2 l 4 � 0
P 2 − 3klP + k 2 l 2 � 0
3− 5
kl or
2
Solving
P�
or
P � 0.382kl or
P�
3+ 5
kl
2
P � 2.62kl
∂ 2V
� 0: − Pl + 2kl 2 � 0
∂ θ12
or
1
P � kl
2
∂ 2V
� 0: − Pl + kl 2 � 0
∂ θ 22
or
P � kl
Therefore, all conditions for stable equilibrium are satisfied when
0 � P � 0.382kl �
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1814
1812
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 θ1 = θ 2 = 0. Determine the range of values of P for which
the equilibrium position is stable.
SOLUTION
From the analysis of Problem 10.97 with
l = 800 mm and k = 2.5 kN/m
P � 0.382kl = 0.382(2500 N/m)(0.8 m) = 764 N
P � 764 N �
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1813 1815
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 θ1 = θ2 = 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 = rθ1 − rθ 2 = r (θ1 − θ 2 )
1 2
ks + pl cos θ1 − wl cos θ 2
2
1
= kr 2 (θ1 − θ 2 ) 2 + pl cos θ1 − wl cos θ 2
2
V=
∂v
= kr 2 (θ1 − θ 2 ) − pl sin θ1
∂ θ1
∂v
= − kr 2 (θ1 − θ 2 ) + wl sin θ 2
∂ θ2
∂ 2v
= kr 2 − pl cos θ1
∂ θ12
∂ 2v
= kr 2 + wl cos θ 2
2
∂ θ2
∂ 2v
= − kr 2
∂ θ1∂ θ 2
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1816
1814
PROBLEM 10.99* (Continued)
For
θ1 = θ 2 = 0:
∂ 2v
∂ 2v
r 2v
= kr 2 − pl ,
= + kr 2 + wl ,
= − kr 2
2
2
,
∂
θ
∂
θ
∂ θ1
∂ θ2
2
Stability conditions for stability (see Page 583)
� ∂ 2v
��
� ∂ θ1∂ θ 2
2
�
∂ 2v ∂ 2v
�0
⋅
�� −
2
2
� ∂ θ1 ∂ θ 2
(kr 2 ) 2 − (kr 2 − pl )(kr 2 + wl ) � 0
pl (kr 2 + wl ) − kr 2 wl � 0
P�
wkr 2
kr 2 �
�
; P�
2
l ��
kr + wl
�
�
+ w ��
W
kr
l
2
�
kr 2
∂ 2v
2
�
:
−
�
0;
P
�
kr
pl
l
∂ θ12
�
P�
We choose:
kr 2
l
�
�
�
�
�
� �
+ W ��
W
kr
l
2
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1815 1817
PROBLEM 10.100*
Solve Problem 10.99 knowing that k = 1600 N/m, r = 150 mm, l = 300
mm, and (a) W = 60 N, (b) W = 240 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
θ1 = θ 2 = 0 is stable.
SOLUTION
k = 1600 N/m
r = 0.15 m
l = 0.3 m
kr 2 (1600 N/m)(0.15 m)2
=
= 120 N
l
0.3 m
P < (120 N)
(a)
W = 60 N:
(b)
W = 240 N: P < (120 N)
60 N
(120 N) + (60 N)
240 N
(120 N) + (240 N)
P < 40 N
P < 80 N
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1818
1816
PROBLEM 10.101
Determine the vertical force P that must be applied at G to maintain
the equilibrium of the linkage.
SOLUTION
Assuming δ y A
it follows
d yC =
0.24
d yA = 1.5d yA
0.16
δ yE = δ yC = 1.5δ y A
d yD =
0.36
d yA = 3(1.5d yA) = 4.5d yA
0.12
d yG =
0.2
0.2
(1.5d yA) = 2.5d yA
d yA =
0.12
0.12
Then, by virtual work
δ U = 0: (300 N)δ y A − (100 N)δ yD + Pδ yG = 0
300δ y A − 100(4.5δ y A ) + P(2.5δ y A ) = 0
300 − 450 + 2.5 P = 0
P = 60.0 N
P = 60.0 N
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1817 1819
PROBLEM 10.102
Determine the couple M that must be applied to member DEFG to
maintain the equilibrium of the linkage.
SOLUTION
Assume δ y A :
d yC =
0.24
d yA = 1.5d yA , δ yE = δ yC = 1.5δ y A
0.16
d yD =
0.36
d yE = 3(1.5d yA) = 4.5d yA
0.12
df =
d yE
0.12
=
1
1.5d yA
=
d yA
0.08
0.12
Virtual Work:
δ U = 0:
(300 N)δ y A − (100 N)δ yD + M δφ = 0
1
d yA = 0
0.08
1
300 − 450 +
M=0
0.08
300d yA − 100(1.5d yA) + M
M = +12 N ⋅ m
M = 12 N ⋅ m
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1820
1818
B
A
q
M
q
E
P
l
l
F
– dxD
y
A
B
A
D
C
dq
P
q
M
Ex
D
l
C
E
F
Ey
x
xD
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1819 1821
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.
SOLUTION
Since AB = 440 mm, we have
(440 mm) 2 = (240 mm) 2 + b2 + c 2
b 2 = (440) 2 − (240) 2 − c 2 = 136 × 103 − c 2
2b δ b = −2c δ c
Differentiate:
c
b
δb = − δc
δb = −
cδ c
136 × 103 − c 2
Virtual Work:
δ U = 0: Pδ c + W δ b = 0
Pδ c = W
For W = 90 N:
(a)
136 × 103 − c 2
c
136 × 103 − c 2
(1)
When c = 80 mm:
Eq. (1):
(b)
P = 90
cδ c
P = 90
80
136 × 103 − (80) 2
P = 20.0 N �
When c = 280 mm:
Eq. (1)
P = 90
280
136 × 103 − (280) 2
P = 105.0 N �
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1822
1820
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
R
tan(90° − θ )
− Rδθ
δ yB =
cos 2 (90° − θ )
− Rδθ
δ yB = 2
sin θ
yB =
Virtual Work:
δ U = 0:
δ U = − M δθ − Pδ yB = 0
− M δθ + PR
1
δθ = 0
sin 2 θ
M=
PR
sin 2 θ
M = PR csc2 θ �
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1821 1823
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 θ 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
r
OC
=
cos θ =
OB xB
r
cos θ
r sin θ
δ xB =
δθ
cos 2 θ
y A = l cos θ
xB =
δ y A = −l sin θδθ
Virtual Work:
δ U = 0: P( −δ y A ) − Qδ xB = 0
Pl sin θδθ − Q
r sin θ
δθ = 0
cos 2 θ
cos 2 θ =
Qr
Pl
(1)
Then, with l = 200 mm, r = 60 mm, P = 40 N, and Q = 80 N
cos 2 θ =
or
(80 N)(60 mm)
= 0.6
(40 N)(200 mm)
θ = 39.231°
θ = 39.2° �
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1824
1822
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 = 2250 N/m, and the spring is
unstretched when θ = 0. Neglecting the weight of the mechanism,
determine the value of θ corresponding to equilibrium.
SOLUTION
s = rθ
δ s = rδθ
Spring is unstretched at θ = 0°
FSP = ks = krθ
xC = l sin θ
δ xC = l cosθδθ
Virtual Work:
δ U = 0: Pδ xC − FSPδ s = 0
P(l cos θδθ ) − krθ (rδθ ) = 0
Pl
θ
=
2
cos θ
kr
or
Thus
or
θ
(200 N)(0.24 m)
=
2
cos θ
(2250 N/m)(0.1 m)
θ
cos θ
= 2.1333
θ = 1.054 rad = 60.39°
θ = 60.4°
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1823
1825
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 80 mm long when
unstretched and that the constant of the spring is 1600 N/m,
determine the distance x corresponding to equilibrium when a
120-N load is applied at E as shown.
SOLUTION
Deformation of spring
s = EC − 0.08 m =
2x
− 0.08
3
2
V=
2x
1 2
x 1
− 0.08 − 20x
ks − (120 N) = (1600 N/m)
3
2
6 2
2x
2
dV
= 1600
− 0.08
− 20
dx
3
3
Equilibrium:
dV
=0
dx
3200 2x
− 0.08 − 20 = 0
3
3
3
2x
− 0.08 = 20
3
3200
x = 0.148 m
x = 148 mm
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1826
1824
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 80 mm long when unstretched and that the constant of
the spring is 1600 N/m, determine the distance x corresponding
to equilibrium when a 120-N load is applied at E as shown.
SOLUTION
Deformation of spring
s = EC− 0.08 m =
2x
− 0.08
3
1
5x 1
2x
− 0.08
V = ks2 − (120 N) = (1600 N/m)
2
6 2
3
2x
dV
= 1600
− 0.08
3
dx
Equilibrium:
dV
=0
dx
2
− 100x
2
− 100
3
3200 2x
− 0.08 − 100 = 0
3
3
100(3)
2x
− 0.08 =
3
3200
x = 0.2606 m
x = 261 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1825
1827
PROBLEM 10.110
Two uniform rods, each of mass m and length l, are attached to
gears as shown. For the range 0 � θ � 180°, determine the positions
of equilibrium of the system and state in each case whether the
equilibrium is stable, unstable, or neutral.
SOLUTION
Potential energy
�l
�
�l
�
V = W � cos1.5θ � + W � cos θ � W = mg
2
2
�
�
�
�
dV Wl
Wl
=
( −1.5sin1.5θ ) +
(− sin θ )
dθ
2
2
Wl
= − (1.5sin1.5θ + sin θ )
2
2
d V
Wl
= − (2.25cos1.5θ + cos θ )
2
2
dθ
For equilibrium
dV
= 0: 1.5sin1.5θ + sin θ = 0
dθ
Solutions: One solution, by inspection, is θ = 0, and a second angle less than 180° can be found numerically:
θ = 2.4042 rad = 137.8°
Now
d 2V
Wl
= − (2.25cos1.5θ + cos θ )
2
2
dθ
At θ = 0:
d 2V
Wl
= − (2.25cos 0° + cos 0°)
2
2
dθ
=−
Wl
(3.25)(� 0)
2
θ = 0, Unstable ��
�
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1828
1826
PROBLEM 10.110 (Continued)
At θ = 137.8°:
d 2V
Wl
= − [2.25cos(1.5 × 137.8°) + cos137.8°]
2
2
dθ
=
Wl
(2.75)(� 0)
2
θ = 137.8°, Stable �
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1827 1829
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 θ corresponding to
equilibrium when β = 10°.
SOLUTION
Detail
3
CG = r
8
V = W (− rθ sin β − (CG ) cos θ )
3
dV
�
�
= W � − r sin β + r sin θ �
dθ
8
�
�
Equilibrium:
3
dV
= 0, − sin β + sin θ = 0
dθ
8
3
sin β = sin θ
8
For β = 10°
3
sin10° = sin θ
8
(1)
sin θ = 0.46306, θ = 27.6° �
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1830
1828
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 β for
which a position of equilibrium exists, (b) the angle θ corresponding to
equilibrium when the angle β is equal to half the value found in Part a.
SOLUTION
Detail
3
CG = r
8
V = W (− rθ sin β − (CG ) cos θ )
3
dV
�
�
= W � − r sin β + r sin θ �
dθ
8
�
�
3
dV
= 0, − sin β + sin θ = 0
dθ
8
Equilibrium:
3
sin β = sin θ
8
(a)
For β max , θ = 90°
Eq. (1)
(b)
(1)
3
3
sin β max = sin 90°, sin β max = = 22.02°
8
8
β max = 22.0° �
When β = 12 β max = 11.01°
Eq. (1)
3
sin11.01° = sin θ ; sin θ = 0.5093
8
θ = 30.6° ��
Note: We can also use ∆CGD and law of sines to derive Eq. (1).
sin β sin θ
3
CG
=
; sin β =
sin θ ; sin β = sin θ
CG
CD
CD
8
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1829 1831