*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.
For
Ans.
Ans.F=512 N
(–50 +0.09769F)du =0
dx=-0.09769 du
u=60°,
x=0.4405 m
dU=0; –50du –Fdx=0
0=0+2xdx+0.2xsin udu–0.2 cos udx
100 mm
400 mm
F
M
u
1144
11–21.
SOLUTION
From Eq. (1)
Ans.F=50020.04 cos2u+0.6
(0.2 cos u+20.04 cos2u+0.6) sin u
x=0.2 cos u+20.04 cos2u+0.6
2
x=0.2 cos u;20.04 cos2u+0.6
2, since20.04 cos2u+0.6 70.2 cos u
x2–0.2xcos u–0.15 =0
F=50(2x–0.2 cos u)
0.2xsin u
–50du –Fa0.2xsin u
0.2 cos u–2xbdu =0, du Z0
dU=0; –50du –Fdx=0
dx=a0.2xsin u
0.2 cos u–2xbdu
0=0+2xdx+0.2xsin udu–0.2 cos udx
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.
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
Assume , so .
Ans.u=9.21°
200 sin u=32
cos ®Z0®690°
–100 sin u(2 cos udu)+8(4 cos udu)=0
dU=0; –F
sdy1+Pdy2=0
F
s=50(2 sin u)=100 sin u
y2=4 sin u+4, dy2=4 cos udu
y1=2 sin u,dy1=2 cos udu
4ft
4ft
2ft
2ft
k50 lb/ft
P
u
11–23.
4in. 4in. x
A
B
CG
E
D
2in.
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
*11–24.
4in. 4in. x
A
B
CG
E
D
2in.
SOLUTION
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
(1)
(2)
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,
(3)
Substituting ,,Eqs.(1) and (2) into Eq. (3),
Since ,then
Ans.x=16 in.
2(4+x)–40 =0
du Z0
du
C
2(4+x)–40
D
=0
2(4+x)du –20(2du)=0
W
G=2lbW
F=20 lb
dU=0;W
GdyG+
A
–W
FdyF
B
=0
dyF=2du
dyG=(4+x)du
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
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
ac
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 requires .Thus,
and Ans.
Stability:The second derivative of Vis
At ,
Thus,the equilibrium configuration at is stable.Ans.
At ,
Thus,the equilibrium configuration at is unstable.Ans.x
=-
0.5 ft
d2V
dx22x=-0.5 ft
=120(–0.5)–20 =-80 60
x=-0.5ft
x=0.8333 ft
d2V
dx22x=0.8333 ft
=120(0.8333)–20 =80 70
x=0.8333 ft
d2V
dx2=120x–20
–0.5 ftx=0.833 ft
x=
20 ;3(–20)2–4(60)(–25)
2(60)
60x2–20x–25 =0
dV
dx =0
V=(20x3–10x2–25x–10) ft#lb,
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
48 sin2u+15 sin u–24 =0
2411–2 sin2u2–15 sin u=0
24 cos 2u–15 sin u=0
dV
du
=0;
V=12 sin 2u+15 cos u
1151
*11–28.
Ans.
Ans.
Ans.
d2V
dx2=470 Stablex=0.167 m,
d2V
dx2=-460 Unstablex=0,
d2V
dx2=48x–4
x=0 and x=0.167 m
124x–42x=0
dV
dx
=24x2–4x=0
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,
dx
dx
2=470
11–29.
SOLUTION
For equilibrium:
Ans.
and
Ans.
Stability:
Ans.
Ans.
Ans.
d2V
du2=15 70, Stableu=90°,
d2V
du2=-24.4 60, Unstableu=141°,
d2V
du2=-24.4 60, Unstableu=38.7°,
d2V
du2=-40 cos 2u–25 sin u
u=cos–10=90°
u=sin–1a25
40 b=38.7° and 141°
1–40 sin u+252cos u=0
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,
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
V=4x3–x2–3x+10
11–31.
k100 N/m
400 mm
400 mm
D
C
B
A
u
SOLUTION
(1)
Ans.
stable Ans.
d2V
du2=5.886 sin u+(16) cos u=17.0 70
u=20.2°
dV
du
=-(5.886) cos u+16(sin u)=0
=-(0.2)(sin u)3(9.81) +1
2(100)
C
(0.4)2(2)(1 –cos u)
D
V=V
g+V
e
=(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
*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
V=V
e+V
g
k
L
W
L
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
B
SOLUTION
u=90°
u=90°
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.
u=72.88°=72.9°
0.5 m
k 2 kN/m
C
B
A
u
SOLUTION
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