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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–1.
Use the method of virtual work to determine the tensions in
cable AC. The lamp weighs 10 lb.
B
C
45°
SOLUTION
A
30°
Free Body Diagram: The tension in cable AC can be determined by releasing
cable AC. The system has only one degree of freedom defined by the independent
coordinate u. When u undergoes a positive displacement du, only FAC and the weight
of lamp (10 lb force) do work.
Virtual Displacement: Force FAC and 10 lb force are located from the fixed point B
using position coordinates yA and xA.
xA = l cos u
dxA = - l sin udu
(1)
yA = l sin u
dyA = l cos udu
(2)
Virtual–Work Equation: When yA and xA undergo positive virtual displacements
dyA and dxA, the 10 lb force and horizontal component of FAC, FAC cos 30°, do
positive work while the vertical component of FAC, FAC sin 30°, does negative work.
dU = 0;
10dyA - FAC sin 30°dyA + FAC cos 30°, dxA = 0
(3)
Substituting Eqs. (1) and (2) into (3) yields
(10 cos u - 0.5FAC cos u - 0.8660FAC sin u)ldu = 0
Since ldu Z 0, then
FAC =
10 cos u
0.5 cos u + 0.8660 sin u
At the equilibrium position u = 45°,
FAC =
10 cos 45°
= 7.32 lb
0.5 cos 45° + 0.8660 sin 45°
Ans.
Ans:
FAC = 7.32 lb
1123
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–2.
The scissors jack supports a load P. Determine the axial
force in the screw necessary for equilibrium when the jack
is in the position u. Each of the four links has a length L and
is pin-connected at its center. Points B and D can move
horizontally.
P
SOLUTION
x = L cos u,
dx = -L sin u du
y = 2L sin u,
dy = 2L cos u du
dU = 0;
- Pdy - Fdx = 0
C
D
A
B
u
-P(2L cos u du) - F(- L sin u du) = 0
Ans.
F = 2P cot u
Ans:
F = 2P cot u
1124
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–3.
If a force of P = 5 lb is applied to the handle of the
mechanism, determine the force the screw exerts on the cork
of the bottle. The screw is attached to the pin at A and passes
through the collar that is attached to the bottle neck at B.
P 5 lb
D
SOLUTION
A
Free - Body Diagram: When u undergoes a positive virtual angular displacement of
du, the dash line configuration shown in Fig. a is formed. We observe that only the
force in the screw Fs and force P do work when the virtual displacements take place.
u 30°
3 in.
B
Virtual Displacement: The position of the points of application for Fs and P are
specified by the position coordinates yA and yD, measured from the fixed point B,
respectively.
yA = 2(3 sin u)
dyA = 6 cos udu
(1)
yD = 6(3 sin u)
dyD = 18 cos udu
(2)
Virtual Work Equation: Since P acts towards the positive sense of its corresponding
virtual displacement, it does positive work. However, the work of Fs is negative
since it acts towards the negative sense of its corresponding virtual displacement.
Thus,
dU = 0;
PdyD +
A - FSdyA B = 0
(3)
Substituting P = 5 lb, Eqs. (1) and (2) into Eq. (3),
5(18 cos udu)FS (6 cos udu) = 0
cos udu(90 - 6FS) = 0
Since cos udu Z 0, then
90 - 6FS = 0
Ans.
FS = 15 lb
Ans:
FS = 15 lb
1125
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–4.
The disk has a weight of 10 lb and is subjected to a vertical
force P = 8 lb and a couple moment M = 8 lb # ft.
Determine the disk’s rotation u if the end of the spring
wraps around the periphery of the disk as the disk turns.
The spring is originally unstretched.
M 8 lb ft
1.5 ft
k 12 lb/ft
SOLUTION
P 8 lb
dyF = dyP = 1.5du
PdyP + Mdu - FdyF = 0
dU = 0;
8(1.5 du) + 8 du - F(1.5 du) = 0
u(20 - 1.5F) = 0
Since du Z 0
20 -1.5F = 0
F = 13.33 lb
F = kx
Where x = 1.5u
13.33 = 12(1.5u)
Ans.
u = 0.7407 rad = 42.4°
Ans:
u = 42.4°
1126
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–5.
The punch press consists of the ram R, connecting rod AB,
and a flywheel. If a torque of M = 75 N # m is applied to the
flywheel, determine the force F applied at the ram to hold
the rod in the position u = 60°.
B
200 mm
600 mm
R
u
F
M
A
Solution
Free Body Diagram. The system has only one degree of freedom, defined by the
independent coordinate u. When u undergoes a positive angular displacement du as
shown in Fig. a, only force F and couple moment M do work.
Virtual Displacement. The position of force F is measured from fixed point O by
position coordinate xA. Applying the law of cosines by referring to Fig. b,
0.62 = x2A + 0.22 - 2xA(0.2) cos u
(1)
Differentiating the above expression,
0 = 2xAdxA + 0.4xA sin u du - 0.4 cos u dxA
dxA =
0.4xA sin u
du
0.4 cos u - 2xA
(2)
Virtual–Work Equation. When point A undergoes a positive virtual displacement,
and the flywheel undergoes positive virtual angular displacement du, both F and M
do negative work.
(3)
- FdxA - Mdu = 0
dU = 0;
Substituting Eq. (2) into (3)
Since du ≠ 0, then
ca
0.4 xA sin u
bF + M d du = 0
0.4 cos u - 2xA
0.4 xA sin u
F + M = 0
0.4 cos u - 2xA
F = -a
0.4 cos u - 2xA
bM
0.4 xA sin u
(4)
At the equilibrium position u = 60°, Eq. (1) gives
0.62 = xA2 + 0.22 - 2xA(0.2) cos 60°
xA = 0.6745 m
Substitute M = 75 N # m, u = 60° and this result into Eq. (4)
F = -c
0.4 cos 60° - 2(0.6745)
0.4(0.6745) sin 60°
= 368.81 N = 369 N
d (75)
Ans.
Ans:
F = 369 N
1127
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–6.
The flywheel is subjected to a torque of M = 75 N # m.
Determine the horizontal compressive force F and plot the
result of F (ordinate) versus the equilibrium position u
(abscissa) for 0° … u … 180°.
B
200 mm
R
u
F
M
Solution
Free Body Diagram. The system has only one degree of freedom, defined by the
independent coordinate u. When u undergoes a positive angular displacement du as
shown in Fig. a, only force F and couple moment M do work.
Virtual Displacement. The position of force F is measured from fixed point O by
position coordinate xA. Applying the law of cosines by referring to Fig. b,
0.62 = xA2 + 0.22 - 2xA(0.2) cos u
(1)
Differentiating the above expression,
0 = 2xAdxA + 0.4xA sin u du - 0.4 cos u dxA
dxA =
0.4xA sin u
0.4 cos u - 2xA
(2)
Virtual–Work Equation. When point A undergoes a positive virtual displacement
and the flywheel undergoes positive virtual angular displacement du, both F and M
do negative work.
(3)
- FdxA - Mdu = 0
dU = 0;
Substituting Eq. (2) into (3)
ca
Since du ≠ 0, then
0.4 xA sin u
bF + M d du = 0
0.4 cos u - 2xA
a
Here, M = 75 N # m. Then
0.4 xA sin u
bF + M = 0
0.4 cos u - 2xA
F = -M a
0.4 cos u - 2xA
b
0.4 xA sin u
F = - 75 a
0.4 cos u - 2xA
b
0.4 xA sin u
(4)
Using Eq. (1) and (4), the following tabulation can be computed. Subsequently, the
graph of F vs u shown in Fig. c can be plotted
u(deg.)
0
xA(m)
0.80
F(N)
∞
30
60
90
120
150
0.7648 0.6745 0.5657 0.4745 0.4184
580
369
375
524
1060
600 mm
180
0.4
15
73.0
0.7909 0.6272
∞
1128
1095
356
A
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–7.
When u = 20°, the 50-lb uniform block compresses the
two vertical springs 4 in. If the uniform links AB and CD
each weigh 10 lb, determine the magnitude of the applied
couple moments M needed to maintain equilibrium when
u = 20°.
k
2 lb/in.
k
B
D
2 lb/ in.
1 ft
1 ft
SOLUTION
u
Free Body Diagram: The system has only one degree of freedom defined by the
independent coordinate u. When u undergoes a positive displacement du, only the
spring forces Fsp, the weight of the block (50 lb), the weights of the links (10 lb) and
the couple moment M do work.
u
M
A
1 ft
2 ft
M
C
4 ft
Virtual Displacements: The spring forces Fsp, the weight of the block (50 lb) and the
weight of the links (10 lb) are located from the fixed point C using position
coordinates y3, y2 and y1 respectively.
y3 = 1 + 4 cos u
y2 = 0.5 + 4 cos u
y1 = 2 cos u
(1)
dy3 = -4 sin udu
(2)
dy2 = -4 sin udu
(3)
dy1 = - 2 sin udu
Virtual–Work Equation: When y1, y2 and y3 undergo positive virtual displacements
dy1, dy2 and dy3, the spring forces Fsp, the weight of the block (50 lb) and the weights
of the links (10 lb) do negative work. The couple moment M does negative work
when the links undergo a positive virtual rotation du.
dU = 0;
- 2Fspdy3 - 50dy2 - 20dy1 - 2Mdu = 0
(4)
Substituting Eqs. (1), (2) and (3) into (4) yields
(8Fsp sin u + 240 sin u - 2M) du = 0
Since du Z 0, then
8Fsp sin u + 240 sin u - 2M = 0
M = sin u(4Fsp + 120)
At the equilibrium position u = 20°, Fsp = kx = 2(4) = 8 lb.
M = sin 20°[4(8) + 120] = 52.0 lb # ft
Ans.
Ans:
M = 52.0 lb # ft
1129
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–8.
The bar is supported by the spring and smooth
collar that allows the spring to be always perpendicular to
the bar for any angle u. If the unstretched length of the
spring is l0, determine the force P needed to hold the bar in
the equilibrium position u. Neglect the weight of the bar.
a
A
B
u
k
C
l
SOLUTION
s = a sin u,
ds = a cos u du
y = l sin u,
dy = l cos u du
Fs = k(a sin u - l0)
dU = 0;
P
Pdy - Fsds = 0
Pl cos du - k(a sin u - l0) a cos u du = 0
P =
ka (a sin u - l0)
l
Ans.
Ans:
P =
1130
ka (a sin u - l0)
l
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–9.
The 4-ft members of the mechanism are pin-connected at
their centers. If vertical forces P1 = P2 = 30 lb act at C and
E as shown, determine the angle u for equilibrium. The
spring is unstretched when u = 45°. Neglect the weight of
the members.
P1
P2
E
C
k = 200 lb/ft
2 ft
2 ft
SOLUTION
y = 4 sin u,
x = 4 cos u
dy = 4 cos u du,
dU = 0;
B
2 ft
dx = - 4 sin u du
A
- Fsdx - 30dy - 30dy = 0
θ
2 ft
D
3 - Fs1 -4 sin u2 - 6014 cos u24du = 0
Fs = 60 a
cos u
b
sin u
Since Fs = k14 cos u - 4 cos 45°2 = 20014 cos u - 4 cos 45°2
60 cos u = 8001cos u - cos 45°2sin u
sin u - 0.707 tan u - 0.075 = 0
Ans.
u = 16.6°
and
Ans.
u = 35.8°
Ans:
u = 16.6°
u = 35.8°
1131
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–10.
The thin rod of weight W rest against the smooth wall and
floor. Determine the magnitude of force P needed to hold it
in equilibrium for a given angle u.
B
SOLUTION
l
Free-Body Diagram: The system has only one degree of freedom defined by the
independent coordinate u. When u undergoes a positive displacement du, only the
weight of the rod W and force P do work.
Virtual Displacements: The weight of the rod W and force P are located from the
fixed points A and B using position coordinates yC and xA, respectively
yC =
1
sin u
2
dyC =
xA = l cos u
1
cos udu
2
dxA = - l sin udu
P
A
θ
(1)
(2)
Virtual-Work Equation: When points C and A undergo positive virtual displacements
dyC and dxA, the weight of the rod W and force F do negative work.
(3)
dU = 0; - WdyC - PdyA = 0
Substituting Eqs. (1) and (2) into (3) yields
a Pl sin u -
Wl
cos ub du = 0
2
Since du Z 0, then
Pl sin u P =
Wl
cos u = 0
2
W
cot u
2
Ans.
Ans:
P =
1132
W
cot u
2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–11.
If each of the three links of the mechanism have a mass of
4 kg, determine the angle u for equilibrium. The spring,
which always remains vertical, is unstretched when u = 0°.
A
M 30 N m
u
200 mm
k 3 kN/ m
D
200 mm
Solution
B
Free Body Diagram. The system has only one degree of freedom, defined by the
independent coordinate u. When u undergoes a positive angular displacement du
as shown in Fig. a, only the weights W1 = W2 = W3 = W, couple moment M, and
spring force Fsp do work.
Virtual Displacement. The positions of the weights W1, W2, W3 and spring force
Fsp are measured from fixed point A using position coordinates y1, y2, y3 and y4
respectively
y1 = 0.1 sin u
dy1 = 0.1 cos u du
(1)
y2 = 0.2 sin u + 0.1
dy2 = 0.2 cos u du
(2)
y3 = 0.1 sin u + 0.2
dy3 = 0.1 cos u du
(3)
y4 = 0.5 sin u
dy4 = 0.2 cos u du
(4)
Virtual Work Equation. When all the weights undergo positive virtual displacement,
all of them do positive work. However, Fsp does negative work when its undergoes
positive virtual displacement. Also, M does positive work when it undergoes positive
virtual angular displacement.
dU = 0;
W1dy1 + W2dy2 + W3dy3 - Fspdy4 + Mdu = 0
(5)
Substitute Eqs (1), (2) and (3) into (5), using W1 = W2 = W3 = W.
W(0.1 cos u du) + W(0.2 cos u du) + W(0.1 cos u du) - Fsp(0.2 cos u du) + Mdu = 0
(0.4 W cos u - 0.2 Fsp cos u + M)du = 0
Since d ≠ 0, then
0.4 W cos u - 0.2 Fsp cos u + M = 0
Here M = 30 N # m, W = 4(9.81)N = 39.24 N and Fsp = kx = 3000(0.2 sin u)
= 600 sin u. Substitute these results into this equation,
0.4(39.24) cos u - 0.2(600 sin u)cos u + 30 = 0
15.696 cos u - 120 sin u cos u + 30 = 0
1133
u
200 mm
C
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–11. Continued
Solve numerically:
Ans.
u = 23.8354° = 23.8°
or
Ans.
u = 72.2895° = 72.3°
Note: u = 23.8° is a stable equilibrium, while u = 72.3° is an unstable one.
Ans:
u = 23.8°
u = 72.3°
1134
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–12.
The disk is subjected to a couple moment M. Determine the
disk’s rotation u required for equilibrium. The end of the
spring wraps around the periphery of the disk as the disk
turns. The spring is originally unstretched.
k 4 kN/ m
0.5 m
M 300 N m
Solution
Free Body Diagram. The system has only one degree of freedom, defined by the
independent coordinate u. When u undergoes a positive angular displacement du as
shown in Fig. a, only the spring force Fsp and couple moment M do work.
Virtual Work Equation. When the disk undergoes a positive angular displacement
du, correspondingly point A undergoes a positive displacement of dxA. As a result
couple moment M does positive work whereas spring force Fsp does negative work.
dU = 0;
(1)
Mdu + ( - FspdxA) = 0
Here, Fsp = kxA = 4000(0.50) = 2000u and dxA = 0.5du. Substitute these results
into Eq. (1)
300du - 2000u(0.5du) = 0
(300 - 1000u)du = 0
Since du ≠ 0,
300 - 1000u = 0
u = 0.3 rad = 17.19° = 17.2° Ans.
Ans:
u = 17.2°
1135
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–13.
A 5-kg uniform serving table is supported on each side by
pairs of two identical links, AB and CD, and springs CE. If
the bowl has a mass of 1 kg, determine the angle u where the
table is in equilibrium. The springs each have a stiffness of
k = 200 N>m and are unstretched when u = 90°. Neglect
the mass of the links.
250 mm
150 mm
A
E
k
C
250 mm
u
u
SOLUTION
Free -Body Diagram: When u undergoes a positive virtual angular displacement of
du, the dash line configuration shown in Fig. a is formed. We observe that only the
spring force Fsp, the weight Wt of the table, and the weight Wb of the bowl do work
when the virtual displacement takes place. The magnitude of Fsp can be computed
using the spring force formula, Fsp = kx = 200 A 0.25 cos u B = 50 cos u N.
D
B
150 mm
Virtual Displacement: The position of points of application of Wb, Wt, and Fsp are
specified by the position coordinates yGb, yGt, and xC, respectively. Here, yGb and yGt
are measured from the fixed point B while xC is measured from the fixed point D.
yGb = 0.25 sin u + b
dyGb = 0.25 cos udu
(1)
yGt = 0.25 sin u + a
dyGt = 0.25 cos udu
(2)
xC = 0.25 cos u
dxC = -0.25 sin udu
(3)
Virtual Work Equation: Since Wb, Wt, and Fsp act towards the negative sense of
their corresponding virtual displacement, their work is negative. Thus,
dU = 0;
-WbdyGb +
A - WtdyGt B + A - FspdxC B = 0
(4)
1
5
Substituting Wb = a b(9.81) = 4.905 N, Wt = a b(9.81) = 24.525 N,
2
2
Fsp = 50 cos u N, Eqs. (1), (2), and (3) into Eq. (4), we have
- 4.905(0.25 cos udu) - 24.525(0.25 cos udu) - 50 cos u( -0.25 sin udu) = 0
du A - 7.3575 cos u + 12.5 sin u cos u B = 0
Since du Z 0, then
- 7.3575 cos u + 12.5 sin u cos u = 0
cos u( -7.3575 + 12.5 sin u) = 0
Solving the above equation,
cos u = 0
Ans.
u = 90°
- 7.3575 + 12.5 sin u = 0
Ans.
u = 36.1°
Ans:
u = 90°
u = 36.1°
1136
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–14.
A 5-kg uniform serving table is supported on each side by
two pairs of identical links, AB and CD, and springs CE. If
the bowl has a mass of 1 kg and is in equilibrium when
u = 45°, determine the stiffness k of each spring. The springs
are unstretched when u = 90°. Neglect the mass of the links.
250 mm
150 mm
A
E
k
C
250 mm
u
u
SOLUTION
Free -Body Diagram: When u undergoes a positive virtual angular displacement of
du, the dash line configuration shown in Fig. a is formed. We observe that only the
spring force Fsp, the weight Wt of the table, and the weight Wb of the bowl do work
when the virtual displacement takes place. The magnitude of Fsp can be computed
using the spring force formula, Fsp = kx = k A 0.25 cos u B = 0.25 k cos u.
D
B
150 mm
Virtual Displacement: The position of points of application of Wb, Wt, and Fsp are
specified by the position coordinates yGb, yGt, and xC, respectively. Here, yGb and yGt
are measured from the fixed point B while xC is measured from the fixed point D.
yGb = 0.25 sin u + b
dyGb = 0.25 cos udu
(1)
yGt = 0.25 sin u + a
dyGt = 0.25 cos udu
(2)
xC = 0.25 cos u
dxC = -0.25 sin udu
(3)
Virtual Work Equation: Since Wb, Wt, and Fsp act towards the negative sense of
their corresponding virtual displacement, their work is negative. Thus,
dU = 0;
-WbdyGb +
A - WtdyGt B + A - FspdxC B = 0
(4)
1
5
Substituting Wb = a b(9.81) = 4.905 N, Wt = a b(9.81) = 24.525 N,
2
2
Fsp = 0.25k cos u N, Eqs. (1), (2), and (3) into Eq. (4), we have
- 4.905(0.25 cos udu) - 24.525(0.25 cos udu) - 0.25k cos u(-0.25 sin udu) = 0
du A -7.3575 cos u + 0.0625k sin u cos u B = 0
Since du Z 0, then
- 7.3575 cos u + 0.0625k sin u cos u = 0
117.72
k =
sin u
When u = 45°, then
117.72
k =
= 166 N>m
sin 45°
Ans.
Ans:
k = 166 N>m
1137
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–15.
The service window at a fast-food restaurant consists of glass
doors that open and close automatically using a motor which
supplies a torque M to each door. The far ends, A and B,
move along the horizontal guides. If a food tray becomes
stuck between the doors as shown, determine the horizontal
force the doors exert on the tray at the position u.
a
a
a
u
C
M
a
u
A
M
B
D
SOLUTION
x = 2a cos u,
dU = 0;
dx = - 2a sin u du
-M du - F dx = 0
-M du + F (2a sin u)du = 0
F =
M
2a sin u
Ans.
Ans:
F =
1138
M
2a sin u
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–16.
The members of the mechanism are pin connected. If a
vertical force of 800 N acts at A, determine the angle u for
equilibrium. The spring is unstretched when u = 0°. Neglect
the mass of the links.
B
k 6 kN/m
1m
1m
1m
Solution
D
u
A
Free Body Diagram. The system has only one degree of freedom, defined by the
independent coordinate u. When u undergoes a positive angular displacement du as
shown in Fig. a, only spring force Fsp and force P do work.
800 N
Virtual Displacement. The positions of Fsp and P are measured from fixed point B
using position coordinates yc and yA respectively.
yc = 1 sin u dyc = cos u du
(1)
yA = 3(1 sin u) dyA = 3 cos u du
(2)
Virtual Work Equation. When Fsp and P undergo their respective positive virtual
displacement, P does positive work whereas Fsp does negative work.
(3)
- Fspdyc + PdyA = 0
dU = 0;
Substitute Eqs. (1) and (2) into (3)
- Fsp(cos udu) + P(3 cos udu) = 0
( -Fsp cos u + 3P cos u)du = 0
Since du ≠ 0, and assuming u 6 90°, then
- Fsp cos u + 3P cos u = 0
Fsp = 3P
Here Fsp = kx = 6000(1 sin u) = 6000 sin u and P = 800 N, Then
6000 sin u = 3(800)
sin u = 0.4
Ans.
u = 23.58° = 23.6°
Ans:
u = 23.6°
1139
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–17.
When u = 30°, the 25-kg uniform block compresses the two
horizontal springs 100 mm. Determine the magnitude of the
applied couple moments M needed to maintain equilibrium.
Take k = 3 kN>m and neglect the mass of the links.
50 mm
k
D
300 mm
100 mm
200 mm
M
C
100 mm
Solution
A
u
k
u
M
B
Free Body Diagram. The system has only one degree of freedom, defined by the
independent coordinate u. When u undergoes a positive angular displacement du as
shown in Fig. a, only spring force Fsp, the weight of the block W, and couple moment
M do work.
Virtual Displacement. The positions of Fsp and W are measured from fixed point B
using position coordinates x and y respectively.
x = 0.3 cos u + 0.05 dx = -0.3 sin u du
(1)
y = 0.3 sin u + 0.1 dy = 0.3 cos udu
(2)
Virtual–Work Equation. When Fsp, W and M undergo their respective positive
virtual displacement, all of them do negative work. Thus
dU = 0;
- 2Fsp dx - Wdy - 2Mdu = 0
(3)
Substitute Eqs. (1) and (2) into (3),
- 2Fsp( -0.3 sin udu) - W(0.3 cos udu) - 2Mdu = 0
(0.6Fsp sin u - 0.3 W cos u - 2M)du = 0
Since du ≠ 0, then
0.6 Fsp sin u - 0.3 W cos u - 2M = 0
(4)
M = 0.3 Fsp sin u - 0.15 W cos u
When u = 30°, Fsp = kx = 3000(0.1) = 300 N. Also W = 25(9.81) = 245.25 N.
Substitute these values into Eq. 4.
M = 0.3(300) sin 30° - 0.15(245.25) cos 30° = 13.14 N # m = 13.1 N # m Ans.
Ans:
M = 13.1 N # m
1140
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–18.
The “Nuremberg scissors” is subjected to a horizontal force
of P = 600 N. Determine the angle u for equilibrium. The
spring has a stiffness of k = 15 kN>m and is unstretched
when u = 15°.
E
u
200 mm
A
k
D
SOLUTION
B
200 mm
C
Free - Body Diagram: When u undergoes a positive virtual angular displacement of
du, the dash line configuration shown in Fig. a is a formed. We observe that only the
spring force Fsp acting at points A and B and the force P do work when the virtual
displacements take place. The magnitude of Fsp can be computed using the spring
force formula,
Fsp = kx = 15(103) C 2(0.2 sin u) - 2(0.2 sin 15°) D = 6000(sin u - 0.2588) N
Virtual Displacement: The position of points A and B at which Fsp acts and point C
at which force P acts are specified by the position coordinates yA, yB, and yC,
measured from the fixed point E, respectively.
yA = 0.2 sin u
dyA = 0.2 cos udu
(1)
yB = 3(0.2 sin u)
dyB = 0.6 cos udu
(2)
yC = 8(0.2 sin u)
dyB = 1.6 cos udu
(3)
Virtual Work Equation: Since Fsp at point A and force P acts towards the positive
sense of its corresponding virtual displacement, their work is positive. The work of
Fsp at point B is negative since it acts towards the negative sense of its
corresponding virtual displacement. Thus,
Fsp dyA + A - FspdyB B + PdyC = 0
dU = 0;
(4)
Substituting Fsp = 6000(sin u - 0.2588), P = 600 N, Eqs. (1), (2), and (3) into Eq. (4),
6000(sin u - 0.2588)(0.2 cos udu - 0.6 cos udu) + 600(1.6 cos udu) = 0
cos udu C -2400(sin u - 0.2588) + 960 D = 0
Since cos udu Z 0, then
- 2400(sin u - 0.2588) + 960 = 0
Ans.
u = 41.2°
Ans:
u = 41.2°
1141
P
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11–19.
The “Nuremberg scissors” is subjected to a horizontal force
of P = 600 N. Determine the stiffness k of the spring for
equilibrium when u = 60°. The spring is unstretched when
u = 15°.
E
u
200 mm
A
k
D
SOLUTION
B
200 mm
C
P
Free - Body Diagram: When u undergoes a positive virtual angular displacement of
du, the dash line configuration shown in Fig. a is formed. We observe that only the
spring force Fsp acting at points A and B and the force P do work when the virtual
displacements take place. The magnitude of Fsp can be computed using the spring
force formula.
Fsp = kx = k C 2(0.2 sin u) - 2(0.2 sin 15°) D = (0.4)k(sin u - 0.2588) N
Virtual Displacement: The position of points A and B at which Fsp acts and point C
at which force P acts are specified by the position coordinates yA, yB, and yC,
measured from the fixed point E, respectively.
yA = 0.2 sin u
dyA = 0.2 cos udu
(1)
yB = 3(0.2 sin u)
dyB = 0.6 cos udu
(2)
yC = 8(0.2 sin u)
dyB = 0.6 cos udu
(3)
Virtual Work Equation: Since Fsp at point A and force P acts towards the positive
sense of its corresponding virtual displacement, their work is positive. The work of
Fsp at point B is negative since it acts towards the negative sense of its
corresponding virtual displacement. Thus,
dU = 0;
Fsp dyA +
A - FspdyB B + PdyC = 0
(4)
Substituting Fsp = k(sin u - 0.2588), P = 600 N, Eqs. (1), (2), and (3) into Eq. (4),
(0.4)k(sin u - 0.2588)(0.2 cos udu - 0.6 cos udu) + 600(1.6 cos udu) = 0
cos udu C - 0.16k(sin u - 0.2588) + 960 D = 0
Since cos udu Z 0, then
- 0.16k(sin u - 0.2588) + 960 = 0
k =
6000
sin u - 0.2588
When u = 60°,
k =
6000
= 9881N>m = 9.88 kN>m
sin 60° - 0.2588
Ans.
Ans:
k = 9.88 kN>m
1142
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–20.
The crankshaft is subjected to a torque of M = 50 N # m.
Determine the horizontal compressive force F applied to
the piston for equilibrium when u = 60°.
400 mm
100 mm
u
F
M
SOLUTION
(0.4)2 = (0.1)2 + x2 - 2(0.1)(x)(cos u)
0 = 0 + 2x dx + 0.2x sin u du - 0.2 cos u dx
dU = 0;
For u = 60°,
- 50du - Fdx = 0
x = 0.4405 m
Ans.
dx = - 0.09769 du
( -50 + 0.09769F) du = 0
Ans.
F = 512 N
Ans:
dx = - 0.09769 du
F = 512 N
1143
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–21.
The crankshaft is subjected to a torque of M = 50 N # m.
Determine the horizontal compressive force F and plot
the result of F (ordinate) versus u (abscissa) for
0° … u … 90°.
400 mm
100 mm
u
F
M
SOLUTION
(0.4)2 = (0.1)2 + x 2 - 2(0.1)(x)(cos u)
(1)
0 = 0 + 2x dx + 0.2x sin u du - 0.2 cos u dx
dx = a
dU = 0;
- 50du - F a
F =
0.2x sin u
b du
0.2 cos u - 2x
-50du - Fdx = 0
0.2x sin u
b du = 0,
0.2 cos u - 2x
du Z 0
50(2x - 0.2 cos u)
0.2x sin u
From Eq. (1)
x2 - 0.2x cos u - 0.15 = 0
x =
0.2 cos u ; 20.04 cos2 u + 0.6
,
2
x =
0.2 cos u + 20.04 cos2 u + 0.6
2
F =
since20.04 cos2 u + 0.6 7 0.2 cos u
500 20.04 cos2 u + 0.6
Ans.
(0.2 cos u + 20.04 cos2 u + 0.6) sin u
Ans:
F =
1144
50020.04 cos2 u + 0.6
(0.2 cos u + 20.04 cos2 u + 0.6) sin u
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–22.
The spring is unstretched when u = 0°. If P = 8 lb,
determine the angle u for equilibrium. Due to the roller
guide, the spring always remains vertical. Neglect the weight
of the links.
k 50 lb/ft
2 ft
2 ft
SOLUTION
y1 = 2 sin u,
dy1 = 2 cos u du
y2 = 4 sin u + 4,
dy2 = 4 cos u du
Fs = 50(2 sin u) = 100 sin u
dU = 0;
4 ft
u
- Fs dy1 + Pdy2 = 0
4 ft
- 100 sin u(2 cos u du) + 8(4 cos u du) = 0
Assume ® 6 90°, so cos ® Z 0.
P
200 sin u = 32
Ans.
u = 9.21°
Ans:
u = 9.21°
1145
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–23.
Determine the weight of block G required to balance the
differential lever when the 20-lb load F is placed on the pan.
The lever is in balance when the load and block are not on
the lever. Take x = 12 in.
4 in.
x
4 in.
C
A
G
B
E
D
2 in.
SOLUTION
Free-Body Diagram: When the lever undergoes a virtual angular displacement of
du about point B, the dash line configuration shown in Fig. a is formed. We observe
that only the weight WG of block G and the weight WF of load F do work when the
virtual displacements take place.
F
Virtual Displacement: Since dyG is very small, the vertical virtual displacement of
block G and load F can be approximated as
dyG = (12 + 4)du = 16du
(1)
dyF = 2du
(2)
Virtual Work Equation: Since WG acts towards the positive sense of its
corresponding virtual displacement, its work is positive. However, force WF does
negative work since it acts towards the negative sense of its corresponding virtual
displacement. Thus,
dU = 0;
WGdyG +
A - WFdyF B = 0
(3)
Substituting WG = 20 lb and Eqs. (1) and (2) into Eq. (3),
WG A 16du B - 20(2du) = 0
du(16WG - 40) = 0
Since du Z 0, then
16WG - 40 = 0
Ans.
WG = 2.5 lb
Ans:
WG = 2.5 lb
1146
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–24.
If the load F weighs 20 lb and the block G weighs
2 lb, determine its position x for equilibrium of the
differential lever. The lever is in balance when the load and
block are not on the lever.
4 in.
x
4 in.
C
A
G
B
E
D
2 in.
SOLUTION
Free -Body Diagram: When the lever undergoes a virtual angular displacement of
du about point B, the dash line configuration shown in Fig. a is formed. We observe
that only the weight WG of block G and the weight WF of load F do work when the
virtual displacements take place.
F
Virtual Displacement: Since dyG is very small, the vertical virtual displacement of
block G and load F can be approximated as
dyG = (4 + x)du
(1)
dyF = 2du
(2)
Virtual Work Equation: Since WG acts towards the positive sense of its
corresponding virtual displacement, its work is positive. However, force WF does
negative work since it acts towards the negative sense of its corresponding virtual
displacement. Thus,
dU = 0;
WGdyG +
A -WFdyF B = 0
(3)
Substituting WF = 20 lb, WG = 2 lb, Eqs.(1) and (2) into Eq. (3),
2(4 + x)du - 20(2du) = 0
du C 2(4 + x) - 40 D = 0
Since du Z 0, then
2(4 + x) - 40 = 0
Ans.
x = 16 in.
Ans:
x = 16 in.
1147
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–25.
The dumpster has a weight W and a center of gravity at G.
Determine the force in the hydraulic cylinder needed to
hold it in the general position u.
b
a
SOLUTION
G
d
θ
s = 2a2 + c2 - 2a c cos (u + 90°)
= 2a2 + c2 + 2a c sin u
c
1
ds = (a2 + c2 + 2a c sin u)- 2 ac cos u du
y = (a + b) sin u + d cos u
dy = (a + b) cos u du - d sin u du
dU = 0;
Fds - Wdy = 0
1
F(a2 + c2 + 2a c sin u)- 2 ac cos u du - W(a + b) cos u du + Wd sin u du = 0
F = a
W(a + b - d tan u)
b 2a2 + c2 + 2a c sin u
ac
Ans.
Ans:
F =
1148
W(a + b - d tan u)
ac
2a2 + c 2 + 2ac sin u
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–26.
The potential energy of a one-degree-of-freedom system is
defined by V = (20x3 - 10x2 - 25x - 10) ft # lb, where x is
in ft. Determine the equilibrium positions and investigate the
stability for each position.
SOLUTION
Equilibrium Configuration: Taking the first derivative of V, we have
dV
= 60x2 - 20x - 25
dx
Equilibrium requires
dV
= 0. Thus,
dx
60x 2 - 20x - 25 = 0
x =
20 ; 3( -20)2 - 4(60)(- 25)
2(60)
Ans.
x = 0.833 ft and - 0.5 ft
Stability: The second derivative of V is
d2V
= 120x - 20
dx2
At x = 0.8333 ft,
d2V 2
= 120(0.8333) - 20 = 80 7 0
dx2 x = 0.8333 ft
Ans.
Thus, the equilibrium configuration at x = 0.8333 ft is stable.
At x = -0.5ft,
d 2V 2
= 120( - 0.5) - 20 = - 80 6 0
dx2 x = -0.5 ft
Ans.
Thus, the equilibrium configuration at x = -0.5 ft is unstable.
Ans:
x = - 0.5 ft unstable
x = 0.833 ft stable
1149
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–27.
If the potential function for a conservative one-degree-offreedom system is V = (12 sin 2u + 15 cos u) J,where
0° 6 u 6 180°, determine the positions for equilibrium and
investigate the stability at each of these positions.
SOLUTION
V = 12 sin 2u + 15 cos u
dV
= 0;
du
24 cos 2u - 15 sin u = 0
2411 - 2 sin2 u2 - 15 sin u = 0
48 sin2 u + 15 sin u - 24 = 0
Choosing the angle 0° 6 u 6 180°
u = 34.6°
Ans.
u = 145°
Ans.
and
d2V
= - 48 sin 2u - 15 cos u
du2
u = 34.6°,
d 2V
= - 57.2 6 0
du2
Unstable
Ans.
u = 145°,
d2V
= 57.2 7 0
du2
Stable
Ans.
Ans:
Unstable at u = 34.6°
stable at u = 145°
1150
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–28.
If the potential function for a conservative one-degree-offreedom system is V = 18x3 - 2x 2 - 102 J, where x is
given in meters, determine the positions for equilibrium and
investigate the stability at each of these positions.
SOLUTION
V = 8x3 - 2x2 - 10
dV
= 24x2 - 4x = 0
dx
124x - 42x = 0
x = 0
Ans.
and x = 0.167 m
d2V
= 48x - 4
dx2
x = 0,
d 2V
= -4 6 0
dx2
x = 0.167 m,
Ans.
Unstable
d2V
= 4 7 0
dx2
Ans.
Stable
Ans:
x = 0.167 m
d 2V
= -4 6 0 Unstable
dx2
d 2V
= 4 7 0 Stable
dx2
1151
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–29.
If the potential function for a conservative one-degree-offreedom system is V = 110 cos 2u + 25 sin u2 J, where
0° 6 u 6 180°, determine the positions for equilibrium and
investigate the stability at each of these positions.
SOLUTION
V = 10 cos 2u + 25 sin u
For equilibrium:
dV
= - 20 sin 2u + 25 cos u = 0
du
1- 40 sin u + 252 cos u = 0
u = sin-1 a
25
b = 38.7° and 141°
40
Ans.
and
u = cos-1 0 = 90°
Ans.
Stability:
d2V
= - 40 cos 2u - 25 sin u
du2
u = 38.7°,
d2V
= - 24.4 6 0,
du2
Unstable
Ans.
u = 141°,
d2V
= - 24.4 6 0,
du2
Unstable
Ans.
Stable
Ans.
u = 90°,
d 2V
= 15 7 0,
du2
Ans:
u = 38.7° unstable
u = 90° stable
u = 141° unstable
1152
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–30.
If the potential energy for a conservative one-degree-offreedom system is expressed by the relation
V = (4x3 - x 2 - 3x + 10) ft # lb, where x is given in feet,
determine the equilibrium positions and investigate the
stability at each position.
SOLUTION
V = 4x3 - x2 - 3x + 10
Equilibrium Position:
dV
= 12x 2 - 2x - 3 = 0
dx
x =
2 ; 2(- 2)2 - 4(12)(- 3)
24
x = 0.590 ft
and
Ans.
- 0.424 ft
Stability:
d 2V
= 24x - 2
dx 2
At x = 0.590 ft
d2V
= 24(0.590) - 2 = 12.2 7 0
dx 2
Stable
Ans.
At x = -0.424 ft
d 2V
= 24( - 0.424) - 2 = -12.2 6 0
dx 2
Unstable
Ans.
Ans:
x = - 0.424 ft unstable
x = 0.590 ft stable
1153
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–31.
The uniform link AB has a mass of 3 kg and is pin connected
at both of its ends. The rod BD, having negligible weight,
passes through a swivel block at C. If the spring has a
stiffness of k = 100 N>m and is unstretched when u = 0°,
determine the angle u for equilibrium and investigate the
stability at the equilibrium position. Neglect the size of the
swivel block.
400 mm
D
k 100 N/m
A
C
u
SOLUTION
400 mm
s = 2(0.4)2 + (0.4)2 - 2(0.4)2 cos u
B
= (0.4)22(1 - cos u)
V = Vg + Ve
= -(0.2)(sin u)3(9.81) +
1
(100) C (0.4)2(2)(1 - cos u) D
2
dV
= - (5.886) cos u + 16(sin u) = 0
du
(1)
Ans.
u = 20.2°
d2V
= 5.886 sin u + (16) cos u = 17.0 7 0
du2
stable
Ans.
Ans:
u = 20.2°
stable
1154
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–32.
The spring of the scale has an unstretched length
of a. Determine the angle u for equilibrium when a weight W
is supported on the platform. Neglect the weight of the
members. What value W would be required to keep
the scale in neutral equilibrium when u = 0°?
W
L
L
k
SOLUTION
Potential Function: The datum is established at point A. Since the weight W is
above the datum, its potential energy is positive. From the geometry, the spring
stretches x = 2L sin u and y = 2L cos u.
V = Ve + Vg
L
L
u
u
a
=
1 2
kx + Wy
2
=
1
(k)(2 L sin u)2 + W(2L cos u)
2
= 2kL2 sin 2 u + 2WL cos u
Equilibrium Position: The system is in equilibrium if
dV
= 0.
du
dV
= 4kL2 sin u cos u - 2WL sin u = 0
du
dV
= 2kL2 sin 2u - 2WLsin u = 0
du
Solving,
u = 0°
or
u = cos-1 a
Stability: To have neutral equilibrium at u = 0°,
W
b
2kL
Ans.
d2V 2
= 0.
du2 u - 0°
d 2V
= 4kL2 cos 2 u - 2WL cos u
du2
d 2V
2
= 4kL2 cos 0° - 2WLcos 0° = 0
du2 u - 0°
Ans.
W = 2kL
Ans:
u = cos-1a
W = 2kL
1155
W
b
2kL
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–33.
The uniform bar has a mass of 80 kg. Determine the angle u
for equilibrium and investigate the stability of the bar when
it is in this position. The spring has an unstretched length
when u = 90°.
B
4m
k 400 N/ m
Solution
Potential Function. The Datum is established through point A, Fig. a. Since the
center of gravity of the bar is above the datum, its potential energy is positive. Here,
y = 2 sin u and the spring stretches x = 4(1 - sin u) m. Thus,
A
u
V = Ve + Vg
1 2
kx + Wy
2
1
= (400) 3 4(1 - sin u) 4 2 + 80(9.81)(2 sin u)
2
=
= 3200 sin2 u - 4830.4 sin u + 3200
Equilibrium Position. The bar is in equilibrium of
dV
= 0
du
dV
= 6400 sin u cos u - 4830.4 cos u = 0
du
cos u(6400 sin u - 4830.4) = 0
Solving,
Ans.
u = 90° or u = 49.00° = 49.0°
Using the trigonometry identify sin 2u = 2 sin u cos u,
dV
= 3200 sin 2u - 4830.4 cos u
du
d 2V
= 6400 cos 2u + 4830.4 sin u
du 2
d 2V
d 2V
Stability. The equilibrium configuration is stable if
7
0,
unstable
if
6 0
du 2
du 2
d 2V
and neutral if
= 0.
du 2
At u = 90°,
d 2V
2
= 6400 cos [2(90°)] + 4830.4 sin 90° = -1569.6 6 0
du 2 u = 90°
Thus, the bar is in unstable equilibrium at u = 90°
At u = 49.00°,
d 2V
= 6400 cos [2(49.00°)] + 4830.4 sin 49.00° = 2754.26 7 0
du2
Thus, the bar is in stable equilibrium at u = 49.0°.
Ans:
Unstable equilibrium at u = 90°
Stable equilibrium at u = 49.0°
1156
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11–34.
The uniform bar AD has a mass of 20 kg. If the attached
spring is unstretched when u = 90°, determine the angle u
for equilibrium. Note that the spring always remains in the
vertical position due to the roller guide. Investigate the
stability of the bar when it is in the equilibrium position.
C
k 2 kN/m
A
u
B
0.5 m
Solution
Potential Function. The Datum is established through point A, Fig. a. Since the
center of gravity of the bar is below the datum, its potential energy is negative. Here,
y = 0.75 cos u and the spring stretches x = 0.5 cos u. Thus,
1m
D
V = Ve + Vg
1 2
kx + ΣWy
2
1
= (2000)(0.5 cos u)2 +
2
=
3 - 20(9.81)(0.75 cos u) 4
= 250 cos2 u - 147.15 cos u
Equilibrium Position. The bar is in equilibrium if
dV
= 0.
du
dV
= - 500 sin u cos u + 147.15 sin u = 0
du
sin u(147.15 - 500 cos u) = 0
Solving,
u = 0°
Ans.
u = 72.88° = 72.9°
Using the trigonometry identity sin 2u = 2 sin u cos u,
dV
= - 250 sin 2u + 147.15 sin u
du
d 2V
= - 500 cos 2u + 147.15 cos u
du 2
d 2V
d 2V
Stability. The equilibrium configuration is stable if
7 0, unstable if
6 0
2
du
du 2
d 2V
and neutral if
= 0.
du 2
d 2V 2
= - 500 cos 0° + 147.15 cos 0° = -352.85 6 0
du 2 u = 0°
Thus, the bar is in unstable equilibrium at u = 0°.
At u = 72.88°,
d 2V 2
= - 500 cos [2(72.88°)] + 147.15 cos 72.88° = 456.69 7 0
du2 u = 72.88°
Thus, the bar is in stable equilibrium at u = 72.9°.
Ans:
Unstable equilibrium at u = 0°
Stable equilibrium at u = 72.9°
1157
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–35.
The two bars each have a weight of 8 lb. Determine the
required stiffness k of the spring so that the two bars are in
equilibrium when u = 30°. The spring has an unstretched
length of 1 ft.
B
2 ft
A
SOLUTION
V = 2(8)(1 sin u) +
θ
2 ft
k
C
1
k(4 cos u - 1)2
2
dV
= 16 cos u + k(4 cos u - 1)( - 4 sin u)
du
dV
= 16 cos u - 4k(4 cos u - 1) sin u
du
u = 30°,
dV
= 0
du
16 cos 30° - 4k(4 cos 30° - 1) sin 30° = 0
Ans.
k = 2.81 lb>ft
Ans:
k = 2.81 lb>ft
1158
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–36.
Determine the angle u for equilibrium and investigate the
stability at this position. The bars each have a mass of 3 kg
and the suspended block D has a mass of 7 kg. Cord DC has
a total length of 1 m.
A
SOLUTION
500 mm
500 mm
u
u
C
500 mm
l = 500 mm
y1 =
l
sin u
2
D
y2 = l + 2l(1 - cos u) = l(3 - 2 cos u)
V = 2Wy1 - WDy 2
= Wl sin u - WD l(3 - 2 cos u)
dV
= l(W cos u - 2WD sin u) = 0
du
tan u =
3(9.81)
W
= 0.2143
=
2WD
14(9.81)
Ans.
u = 12.1°
d2V
= l( -W sin u - 2WD cos u)
du2
u = 12.1°,
d 2V
= 0.5[- 3(9.81) sin 12.1° - 14(9.81) cos 12.1°]
du2
= -70.2 6 0
Ans.
Unstable
Ans:
u = 12.1°
Unstable
1159
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–37.
Determine the angle u for equilibrium and investigate the
stability at this position. The bars each have a mass of 10 kg
and the spring has an unstretched length of 100 mm.
500 mm
k 1.5 kN/ m u
u
A
500 mm
Solution
Potential Function. The Datum
the centers of gravity of the
energies are positive. Here
x = 3 2(0.5 cos u) - 0.5 4 - 0.1 =
is established through point A, Fig. a. Since
bars are above the datum, their potential
y = 0.25 sin u and the spring stretches
(cos u - 0.6) m. Thus
V = Ve + Vg
1 2
kx + ΣWy
2
1
= (1500)(cos u - 0.6)2 + 2[10(9.81)](0.25 sin u)
2
=
= 750 cos2 u - 900 cos u + 49.05 sin u + 270
Equilibrium Position. The system is in equilibrium if
dV
= 0.
du
dV
= - 1500 sin u cos u + 900 sin u + 49.05 cos u = 0
du
Solving numerically,
u = 4.713° = 4.71° or u = 51.22° = 51.2°
Ans.
Using the trigonometry identity sin 2u = 2 sin u cos u,
dV
= - 750 sin 2u + 900 sin u + 49.05 cos u
du
d 2V
= 900 cos u - 1500 cos 2u - 49.05 sin u
du 2
d 2V
d 2V
Stability. The equilibrium configurate is stable if
7
0,
unstable
if
6 0 and
du 2
du 2
d 2V
neutral if
= 0.
du 2
1160
500 mm
C
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–37. Continued
At u = 51.22°,
d 2V
2
= 900 cos 51.22° - 49.05 sin 51.22° - 1500 cos [2(51.22°)] = 848.77 7 0
du 2 u = 51.22°
Thus, the system is in stable equilibrium at u = 51.2°.
At u = 4.713°,
d 2V 2
= 900 cos 4.713° - 49.05 sin 4.713° - 1500 cos [2(4.713°)] = - 586.82 6 0
du2 u = 4.713°
Thus, the system is in unstable equilibrium at u = 4.71°.
Ans:
Stable equilibrium at u = 51.2°
Unstable equilibrium at u = 4.71°
1161
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–38.
The two bars each have a mass of 8 kg. Determine the
required stiffness k of the spring so that the two bars are in
equilibrium when u = 60°. The spring has an unstretched
length of 1 m. Investigate the stability of the system at the
equilibrium position.
A
u
1.5 m
B
k
Solution
1.5 m
Potential Function. The Datum is established through point A, Fig. a. Since the
centers of gravity of the bars are below the datum, their potential energies are
negative. Here, y1 = 0.75 cos u, y2 = 1.5 cos u + 0.75 cos u = 2.25 cos u and the
spring stretches x = 2(1.5 cos u) - 1 = (3 cos u - 1) m. Thus,
V = Ve + Vg
=
=
1 2
kx + ΣWy
2
1
k(3 cos u - 1)2 + [ - 8(9.81)(0.75 cos u)] + [ - 8(9.81)(2.25 cos u)]
2
= 4.5 k cos2 u - 3 k cos u + 0.5 k - 235.44 cos u
= 4.5 k cos2 u - (3 k + 235.44) cos u + 0.5 k
Equilibrium Position. The system is in equilibrium if
dV
= 0
du
dV
= - 9 k sin u cos u + (3 k + 235.44) sin u = 0
du
sin u( -9 k cos u + 3 k + 235.44) = 0
Since sin u ≠ 0, then
- 9 k cos u + 3 k + 235.44 = 0
k =
When u = 60°,
k =
235.44
9 cos u - 3
235.44
= 156.96 N>m = 157 N>m
9 cos 60° - 3
Using this result,
dV
= -9(156.96) sin u cos u + [3(156.96) + 235.44] sin u
du
= - 1412.64 sin u cos u + 706.32 sin u
Using the trigonometry identity sin 2u = 2 sin u cos u,
dV
= 706.32 sin u - 706.32 sin 2u
du
d 2V
= 706.32 cos u - 1412.64 cos 2u
du 2
1162
Ans.
C
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–38
Continued
d 2V
d 2V
Stability. The equilibrium configuration is stable if
7 0, unstable if
6 0,
2
du
du 2
d 2V
and neutral if
= 0.
du 2
At u = 60°,
d 2V
2
= 706.32 cos 60° - 1412.64 cos [2(60°)] = 1059.16 7 0
du 2 u = 60°
Thus, the system is in stable equilibrium at u = 60°.
Ans:
k = 157 N>m
Stable equilibrium at u = 60°
1163
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–39.
A spring with a torsional stiffness k is attached to the pin
at B. It is unstretched when the rod assembly is in the
vertical position. Determine the weight W of the block that
results in neutral equilibrium. Hint: Establish the potential
energy function for a small angle u. i.e., approximate sin u L 0,
and cos u L 1 - u2>2.
L
2
A
L
2
SOLUTION
Potential Function: With reference to the datum, Fig. a, the gravitational potential
B
k
energy of the block is positive since its center of gravity is located above the datum. Here,
the rods are tilted with a small angle u. Thus, y =
3
L
cos u + L cos u = L cos u.
2
2
u2
.Thus,
2
u2
3WL
u2
3
b =
a1 - b
Vg = Wy = W a L b a1 2
2
2
2
L
2
C
1 -
However, for a small angle u, cos u
The elastic potential energy of the torsional spring can be computed using
1
Ve = kb 2, where b = 2u. Thus,
2
Vg =
1
k(2u)2 = 2ku2
2
The total potential energy of the system is
V = Vg + Ve =
u2
3WL
a1 - b + 2ku2
2
2
Equilibrium Configuration: Taking the first derivative of V, we have
3WL
3WL
dV
= u + 4ku = ua + 4k b
du
2
2
dV
= 0. Thus,
du
3WL
+ 4k b = 0
ua 2
Equilibrium requires
u = 0°
Stability: The second derivative of V is
3WL
d2V
+ 4k
= 2
du2
To have neutral equilibrium at u = 0°,
3WL
+ 4k = 0
2
8k
W =
3L
d2V
2
= 0. Thus,
du2 u = 0°
-
Ans.
Note: The equilibrium configuration of the system at u = 0° is stable if
8k d2V
8k d 2V
and
is
unstable
if
W
>
W<
>
0
< 0 .
3L du2
3L du2
1164
Ans:
W =
8k
3L
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–40.
A conical hole is drilled into the bottom of the cylinder, and
it is then supported on the fulcrum at A. Determine the
minimum distance d in order for it to remain in stable
equilibrium.
SOLUTION
Potential Function: First, we must determine the center of gravity of the cylinder. By
referring to Fig. a,
d 1
h
(rpr2h) - a rpr2d b
©yCm
2
4 3
6h2 - d 2
=
y =
=
1
©m
4(3h - d)
rpr 2h - rpr2d
3
(1)
With reference to the datum, Fig. a, the gravitational potential energy of the cylinder
is positive since its center of gravity is located above the datum. Here,
y = (y - d)cos u = B
6h2 - d2
6h2 - 12hd + 3d2
- d R cos u = B
R cos u
4(3h - d)
4(3h - d)
Thus,
V = Vg = Wy = W B
6h2 - 12hd + 3d2
R cos u
4(3h - d)
Equilibrium Configuration: Taking the first derivative of V,
dV
6h2 - 12hd + 3d2
= -W B
R sin u
du
4(3h - d)
Equilibrium requires
dV
= 0. Thus,
du
-WB
6h2 - 12hd + 3d2
R sin u = 0
4(3h - d)
sin u = 0
u = 0°
Stability: The second derivative of V is
6h2 - 12hd + 3d2
d 2V
= -W B
R cos u
2
4(3h - d)
du
To have neutral equilibrium at u = 0°,
-WB
d 2V 2
= 0. Thus,
d2u u = 0°
6h2 - 12hd + 3d2
R cos 0° = 0
4(3h - d)
6h2 - 12hd + 3d2 = 0
d =
12h ; 2( - 12h)2 - 4(3)(6h2)
= 0.5858h = 0.586h
2(3)
Ans.
Note: If we substitute d = 0.5858h into Eq. (1), we notice that the fulcrum must be
at the center of gravity for neutral equilibrium.
1165
Ans:
d = 0.586h
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–41.
The uniform rod has a mass of 100 kg. If the spring is
unstretched when u = 60°, determine the angle u for
equilibrium and investigate the stability at the equilibrium
position. The spring is always in the horizontal position due
to the roller guide at B
A
u
2m
k 500 N/m
B
2m
Solution
Potential Function. The Datum is established through point A, Fig. a. Since the
center of gravity of the bar is below the datum, its potential energy is negative. Here
y = 2 cos u, and the spring stretches x = 2 sin 60° - 2 sin u = 2(sin 60° - sin u).
Thus
V = Ve + Vg
=
1 2
kx + Wy
2
=
1
(500)[2(sin 60° - sin u)]2 + [ - 100(9.81)(2 cos u)]
2
= 1000 sin2 u - 100013 sin u - 1962 cos u + 750
dV
= 0.
Equilibrium Position. The bar is in equilibrium if
du
dV
= 2000 sin u cos u - 100013 cos u + 1962 sin u
du
Using the trigonometry identity sin 2u = 2 sin u cos u,
dV
= 1000 sin 2u - 100013 cos u + 1962 sin u = 0
du
Solved numerically,
Ans.
u = 24.62° = 24.6°
d 2V
= 2000 cos 2u + 100013 sin u + 1962 cos u
du 2
d 2V
d 2V
7 0, unstable if
6 0
Stability. The equilibrium configuration is stable if
2
du
du 2
d 2V
and neutral if
= 0.
du 2
At u = 24.62°,
d 2V
= 2000 cos[2(24.62°)] + 100013 sin 24.62° + 1962 cos 24.62° = 3811.12 7 0
du 2
Thus, the bar is in stable equilibrium at u = 24.6°.
Ans:
Stable equilibrium at u = 24.6°
1166
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–42.
Each bar has a mass per length of m0. Determine the angles u
and f at which they are suspended in equilibrium. The
contact at A is smooth, and both are pin connected at B.
B
l
u
3l
2
Solution
A
l
2
Require G for system to be at its lowest point.
0 x =
f
l l
3
a b - (0.75 l sin 26.565°)a lb
4 2
2
= -0.20937 l
l
3
l + + l
2
2
lx
1
3
- a b(l) - l a b - (0.75 l cos 26.565°)a lb
2
2
2
y =
= -0.66874 l
l
3
l + + l
2
2
f = tan-1a
0.20937 l
b = 17.38° = 17.4°
0.66874 l
Ans.
Ans.
u = 26.565° - 17.38° = 9.18°
Ans:
f = 17.4°
u = 9.18°
1167
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–43.
The truck has a mass of 20 Mg and a mass center at G.
Determine the steepest grade u along which it can park
without overturning and investigate the stability in this
position.
G
3.5 m
SOLUTION
Potential Function: The datum is established at point A. Since the center of gravity
for the truck is above the datum, its potential energy is positive. Here,
y = (1.5 sin u + 3.5 cos u) m.
1.5 m
u
1.5 m
V = Vg = Wy = W(1.5 sin u + 3.5 cos u)
Equilibrium Position: The system is in equilibrium if
dV
= 0
du
dV
= W(1.5 cos u - 3.5 sin u) = 0
du
Since W Z 0,
1.5 cos u - 3.5 sin u = 0
Ans.
u = 23.20° = 23.2°
Stability:
d2V
= W(-1.5 sin u - 3.5 cos u)
du2
d2V
du2
u = 23.20°
= W( - 1.5 sin 23.20° - 3.5 cos 23.20°) = -3.81W 6 0
Thus, the truck is in unstable equilibrium at u = 23.2°
Ans.
Ans:
Unstable equilibrium at u = 23.2°
1168
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–44.
The small postal scale consists of a counterweight W1,
connected to the members having negligible weight.
Determine the weight W2 that is on the pan in terms of the
angles u and f and the dimensions shown. All members are
pin connected.
W2
a
u
SOLUTION
b
y1 = b cos u
a
f
f
W1
y2 = a sin f = a sin (90° - u - g)
where g is a constant and f = (90° - u - g)
V = -W1y1 + W2y2
= -W1 b cos u + W2 a sin (90° - u - g)
dV
= W1 b sin u - W2 a cos (90° - u - g)
du
b sin u
W2 = W1 a b
a cos f
Ans.
Ans:
b sin u
W2 = W1a b
a cos f
1169
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–45.
A 3-lb weight is attached to the end of rod ABC. If the rod
is supported by a smooth slider block at C and rod BD,
determine the angle u for equilibrium. Neglect the weight of
the rods and the slider.
C
6 in.
B
θ
D
4 in.
10 in.
SOLUTION
x = 2(6)2 - (4 sin u)2 = 236 - 16 sin2 u
A
16
6
=
x
(x + 4 cos u) + y
x + 4 cos u + y = 2.667x
y = - 4 cos u + 1.667 236 - 16 sin2 u
1
1
dy = 4 sin u du + 1.667a b (36 - 16 sin2 u)- 2 ( -32 sin u cos u)du
2
dU = 0;
W dy = 0
W c 4 - 0.8333(36 - 16 sin2 u)- 2 (32 cos u) d sin u du = 0
1
Thus,
sin u = 0
Ans.
u = 0°
or,
A 36 - 16 sin2u B 2 = 6.667 cos u
1
Ans.
u = 33.0°
Ans:
u = 0°
u = 33.0°
1170
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–46.
If the uniform rod OA has a mass of 12 kg, determine the
mass m that will hold the rod in equilibrium when u = 30°.
Point C is coincident with B when OA is horizontal. Neglect
the size of the pulley at B.
B
C
m
3m
SOLUTION
Geometry: Using the law of cosines,
2
A
1m
2
lA¿B = 21 + 3 - 2112132 cos190° - u2 = 210 - 6 sin u
e
O
lAB = 212 + 32 = 210 m
l = lAB - lA¿B = 210 - 210 - 6 sin u
Potential Function: The datum is established at point O. Since the center of gravity
of the rod and the block are above the datum, their potential energy is positive.
Here, y1 = 3 - l = 33 - 1210 - 210 - 6 sin u24 m and y2 = 0.5 sin u m.
V = Vg = W1y1 + W2y2
= 9.81m33 - 1210 - 210 - 6 sin u24 + 117.7210.5 sin u2
= 29.43m - 9.81m1210 - 210 - 6 sin u2 + 58.86 sin u
Equilibrium Position: The system is in equilibrium if
dV
= 0.
`
du u = 30°
1
1
dV
= - 9.81m c - 110 - 6 sin u2- 21- 6 cos u2 d + 58.86 cos u
du
2
= -
29.43m cos u
210 - 6 sin u
+ 58.86 cos u
At u = 30°,
dV
29.43m cos 30°
= + 58.86 cos 30° = 0
`
du u = 30°
210 - 6 sin 30°
Ans.
m = 5.29 kg
Ans:
m = 5.29 kg
1171
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–47.
The cylinder is made of two materials such that it has a mass
of m and a center of gravity at point G. Show that when G
lies above the centroid C of the cylinder, the equilibrium is
unstable.
G
C
r
SOLUTION
Potential Function: The datum is established at point A. Since the center of gravity
of the cylinder is above the datum, its potential energy is positive. Here,
y = r + d cos u.
V = Vg = Wy = mg(r + d cos u)
Equilibrium Position: The system is in equilibrium if
dV
= 0.
du
dV
= - mgd sin u = 0
du
sin u = 0
u = 0°
Stability:
d 2V
= - mgd cos u
du2
d 2V 2
= - mgd cos 0° = - mgd 6 0
du2 u = 0°
Thus, the cylinder is in unstable equilibrium at u = 0°
(Q.E.D.)
1172
a
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*11–48.
The bent rod has a weight of 5 lb ft. A pivot of negligible
size is attached at its center A and the rod is balanced as
shown. Determine the length L of its vertical segments so
that it remains in neutral equilibrium. Neglect the thickness
of the rod.
8 in.
2 in.
L
8 in.
A
L
SOLUTION
To remain in neutral equilibrium, the center of gravity must be located at A.
y =
0 =
©yW
©W
2a
L
2L
16
b (5) - a - 2 b a
b (5)
12
2
12
a
2L
16
b (5) + a
b (5)
12
12
0 = 5L2 - 20L - 160
L =
20 ; 2( - 20)2 - 4(5)(- 160)
10
Take the positive root
Ans.
L = 8 in.
Ans:
L = 8 in.
1173
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11–49.
The triangular block of weight W rests on the smooth
corners which are a distance a apart. If the block has three
equal sides of length d, determine the angle u for
equilibrium.
d
60
G 60
u
SOLUTION
a
AF = AD sin f = AD sin (60° - u)
a
AD
=
sin a
sin 60°
AD =
a
(sin (60° + u))
sin 60°
AF =
a
(sin (60° + u)) sin (60° - u)
sin 60°
=
y=
a
(0.75 cos2 u - 0.25 sin2 u)
sin 60°
d
23
cos u - AF
V = Wy
dV
a
( -1.5 sin u cos u - 0.5 sin u cos u)d = 0
= Wc( -0.5774 d) sin u du
sin 60°
Require, sin u = 0
or
- 0.5774 d -
Ans.
u = 0°
a
( -2 cos u) = 0
sin 60°
u = cos-1 a
d
b
4a
Ans.
Ans:
u = 0°
u = cos-1a
1174
d
b
4a