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
100 mm
400 mm
F
M
u
Ans:
1144
11–21.
SOLUTION
From Eq. (1)
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
Ans:
11–22.
The spring is unstretched when If
determine the angle for equilibrium. Due to the roller
guide, the spring always remains vertical. Neglect the weight
of the links.
u
P=8 lb,u=0°.
SOLUTION
2ft
2ft
k50 lb/ft
Ans:
11–23.
4in. 4in. x
A
B
CG
E
D
2in.
F
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
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
Substituting and Eqs.(1) and (2) into Eq. (3),
W
G=20 lb
dyG
Ans:
1147
*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
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,
Since ,then
du Z0
dyG
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
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
s=2a2+c2–2accos (u+90°)
θ
bd
G
a
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
Thus,the equilibrium configuration at is stable.Ans.
x=0.8333 ft
V=(20x3–10x2–25x–10) ft#lb,
Ans:
1150
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
Ans:
1151
*11–28.
SOLUTION
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:
1152
11–29.
SOLUTION
For equilibrium:
Stability:
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:
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:
Ans:
1154
11–31.
k100 N/m
400 mm
D
C
A
u
SOLUTION
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
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
=0
y=2Lcos ux=2Lsin u
k
L
W
LL
L
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,
Equilibrium Position. The bar is in equilibrium of
dV
du
=0
dV
du
dV
du
du
du
du
du
u=90°
Thus, the bar is in unstable equilibrium at
u=90°
1157
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.
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
Equilibrium Position. The bar is in equilibrium if
dV
du
=0.
dV
du
u=0°
u=72.88°=72.9°
Using the trigonometry identity
sin 2
u
=2 sin
u
cos
u
,
dV
du
du
Stability. The equilibrium conguration is stable if d
2
V
du
du
2
du
270, unstable if d
2
V
260
Thus, the bar is in unstable equilibrium at
u=0°.
1158
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
2 ft
B
AC
θ
2 ft
k
Ans: