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1143
*11–20.
The crankshaft is subjected to a torque of
Determine the horizontal compressive force Fapplied to
the piston for equilibrium when u=60°.
M=50 N #m.
SOLUTION
(0.4)2=(0.1)2+x2-2(0.1)(x)(cos u)
100 mm
400 mm
F
M
u
Ans:
11–21.
SOLUTION
(1)
From Eq. (1)
x=0.2 cos u;20.04 cos2u+0.6
x2-0.2xcos u-0.15 =0
(0.4)2=(0.1)2+x2-2(0.1)(x)(cos u)
Thecrankshaft is subjected to atorque of
Determine the horizontal compressive force Fand plot
the result of F(ordinate) versus (abscissa) for
0° …u…90°.
u
M=50 N
#
m.
100 mm
400 mm
F
M
u
11–22.
Ans:
11–23.
4in. 4in. x
A
B
CG
E
D
2in.
F
Determine the weight of block required to balance the
differential lever when the 20-lb load Fisplaced on the pan.
The lever is in balance when the load and block are not on
the lever. Take .x=12 in
G
SOLUTION
Free - Body Diagram: When the lever undergoes a virtual angular displacement of
about point B,the dash line configuration shown in Fig.aisformed. We observe
that only the weight WGofblock Gand the weight WFofload Fdowork when the
virtual displacements take place.
du
Ans:
1147
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
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*11–24.
4in. 4in. x
A
B
CG
E
D
2in.
F
SOLUTION
Free - Body Diagram: When the lever undergoes a virtual angular displacement of
about point B,the dash line configuration shown in Fig.aisformed. We observe
that only the weight WGofblock Gand the weight WFofload Fdowork when the
virtual displacements take place.
Virtual Displacement: Since is very small,the vertical virtual displacement of
block Gand load Fcan be approximated as
Virtual Work Equation: Since WGacts towards the positive sense of its
corresponding virtual displacement, its work is positive.However,force WFdoes
negative work since it acts towards the negative sense of its corresponding virtual
displacement.Thus,
dyG
du
If the load weighs 20 lb and the block weighs
2lb,determine its position for equilibrium of the
differential lever. The lever is in balance when the load and
block are not on the lever.
x
GF
Ans:
1148
11–25.
The dumpster has a weight Wand a center of gravity at G.
Determine the force in the hydraulic cylinder needed to
hold it in the general position .u
SOLUTION
=2a2+c2+2acsin u
s=2a2+c2-2accos (u+90°)
θ
bd
G
a
Ans:
1149
11–26.
The potential energy of a one-degree-of-freedom system is
defined by where xis
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
dx =60x2-20x-25
V=(20x3-10x2-25x-10) ft#lb,
Ans:
11–27.
If the potential function for a conservative one-degree-of-
fe
rehw si metsys modeer
determine the positions for equilibrium and
investigate the stability at each of these positions.
0° 6u6180°,
(12 sin 2u+15 cos u)J,V=
SOLUTION
24 cos 2u-15 sin u=0
dV
du
=0;
V=12 sin 2u+15 cos u
*11–28.
SOLUTION
dV
dx
=24x2-4x=0
V=8x3-2x2-10
If the potential function for a conservative one-degree-of-
f erehw si metsys modeer xis
given in meters, determine the positions for equilibrium and
investigate the stability at each of these positions.
V=18x3-2x2-102J,
Ans:
x=0.167 m
2
11–29.
SOLUTION
For equilibrium:
Stability:
d2V
du2=-40 cos 2u-25 sin u
dV
du
=-20 sin 2u+25 cos u=0
V=10 cos 2u+25 sin u
If the potential function for a conservative one-degree-of-
fe
rehw si metsys modeer
determine the positions for equilibrium and
investigate the stability at each of these positions.
0° 6u6180°,
V=110 cos 2u+25 sin u2J,
Ans:
u=38.7° unstable
1153
11–30.
If t
h
e potent
i
a
l
energy for a conservat
i
ve one-
d
egree-of-
freedom system is expressed by the relation
, where xis given in feet,
determine the equilibrium positions and investigate the
stability at each position.
V=(4x3-x2-3x+10) ft #lb
SOLUTION
Equilibrium Position:
Stability:
d2V
dx2=24x-2
V=4x3-x2-3x+10
Ans:
11–31.
k100 N/m
400 mm
400 mm
D
C
B
A
u
SOLUTION
=(0.4)22(1 -cos u)
s=2(0.4)2+(0.4)2-2(0.4)2cos u
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 and is unstretched when ,
determine the angle for equilibrium and investigate the
stability at the equilibrium position. Neglect the size of the
swivel block.
u
u=0°k=100 N>m
Ans:
1155
*11–32.
The spring of the scale has an unstretched length
of a.Determine the angle for equilibrium when a weight W
is supported on the platform. Neglect the weight of the
members.What value Wwould be required to keep
the scale in neutral equilibrium when u=0°?
u
SOLUTION
Potential Function: The datum is established at point A. Since the weight Wis
above the datum, its potential energy is positive.From the geometry, the spring
stretches and .
Equilibrium Position: The system is in equilibrium if .
dV
du
=4kL2sin ucos u-2WL sin u=0
dV
du
=0
=1
2kx2+Wy
V=V
e+V
g
y=2Lcos ux=2Lsin u
k
L
W
LL
L
a
u
u
1156
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°.
4 m
k 400 N/m
A
B
u
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,
V=V
e
+V
g
=
1
2
kx2+Wy
Equilibrium Position. The bar is in equilibrium of
dV
du
=0
dV
du
=6400 sin u cos u-4830.4 cos u=0
Stability. The equilibrium conguration is stable if d
2
V
du
270, unstable if d
2
V
du
260
and neutral if d
2
V
du
2=0.
At
u=90°
,
Ans:
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.
D
1 m
0.5 m
k 2 kN/m
C
B
A
u
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,
V=V
e
+V
g
Equilibrium Position. The bar is in equilibrium if
dV
du
=0.
=-500 sin u cos u+147.15 sin u=0
Using the trigonometry identity
sin 2
u
=2 sin
u
cos
u
,
dV
du
=-250 sin 2u+147.15 sin u
Stability. The equilibrium conguration is stable if d
2
V
du
270, unstable if d
2
V
du
260
and neutral if d
2
V
du
2=0.
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.
SOLUTION
V=2(8)(1 sin u)+1
2
k(4 cos u-1)2
dV
du
=16 cos u+k(4 cos u-1)(-4 sin u)
2 ft
B
AC
θ
2 ft
k
Ans:
