1159
*11–36.
Determine the angle for equilibrium and investigate the
stability at this position. The bars each have a mass of 3 kg
and the suspended block Dhas a mass of 7 kg. Cord DC has
a total length of 1 m.
u
SOLUTION
l=500 mm
500 mm
500 mm
A
C
500 mm
uu
Ans:
1160
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
500 mm
AC
500 mm
k 1.5 kN/m
uu
SOLUTION
Potential Function. The Datum is established through point A, Fig. a. Since
the centers of gravity of the bars are above the datum, their potential
energies are positive. Here
y=0.25 sin
u and the spring stretches
x=
3
2(0.5 cos u)–0.5
4
–0.1 =(cos u–0.6) m. Thus
V=V
e
+V
g
dV
du
dV
du
Solving numerically,
Using the trigonometry identity
sin 2u=2 sin u cos u,
dV
du
=–750 sin 2u+900 sin u+49.05 cos u
du
du
du
du
2=0.
11–37. Continued
u=51.22°
Thus, the system is in stable equilibrium at
u=51.2°.
u
=4.713°
1162
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. 1.5 m
1.5 m
B
A
C
k
u
SOLUTION
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
Using this result,
=–9(156.96) sin u cos u+[3(156.96) +235.44] sin u
Using the trigonometry identity
sin 2
u
=2 sin
u
cos
u
,
11–38 Continued
du
du
2
2
2
du
2=0.
=706.32 cos 60°–1412.64 cos [2(60°)] =1059.16 70
11–39.
To have neutral equilibrium at ,.Thus,
d2V
du22u=0°
=0u=0°
u=0°
ua–3WL
2+4kb=0
du
A
L
L
2
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 uL1–u2>2
1165
*11–40.
SOLUTION
P
otential Function: First, we must determine the center of gravity of the cylinder.By
referring to
Fig. a,
(1)
y=
©yCm
©m=
h
2(rpr2h)–d
4a1
3rpr2db
rpr2h–1
3rpr2d
=6h2–d2
4(3h–d)
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 din order for it to remain in stable
equilibrium.
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 B2 m
k 500 N/m
A
B
2 m
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=2 cos
u
,
and the spring stretches
x=2 sin 60°–2 sin
u
=2(sin 60°–sin
u
).
Thus
V=V
e
+V
g
=
1
2
kx2+Wy
1
2
u=24.62°
u=24.6°
1167
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
3
2
uf
l
l
A
l
SOLUTION
u=9.18°
11–43.
The truck has a mass of 20 Mg and a mass center at G.
Determine the steepest grade along which it can park
without overturning and investigate the stability in this
position.
u
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,
.
Stability:
1.5 cos u–3.5 sin u=0
y=(1.5 sin u+3.5 cos u)m
G
u
3.5 m
1.5 m
1.5 m
1169
*11–44.
The small postal scale consists of a counterweight
connected to the members having negligible weight.
Determine the weight that is on the pan in terms of the
angles and and the dimensions shown. All members are
pin connected.
fu
W
2
W
1
,
W
2
W
1
ba
af
f
u
y2=asin f=asin (90° –u–g)
y1=bcos u
1170
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 Cand rod BD,
determine the angle for equilibrium. Neglect the weight of
the rods and the slider.
u
SOLUTION
Thus,
6
x
=16
(x+4 cos u)+y
x=2(6)2–(4 sin u)2=236 –16 sin2u
6 in.
θ
10 in.
4 in.
A
B
C
D
Ans:
11–46.
If the uniform rod OA has a mass of 12 kg, determine the
mass mthat will hold the rod in equilibrium when
Point Cis coincident with Bwhen OA is horizontal. Neglect
the size of the pulley at B.
u=30°.
SOLUTION
Geometry: Using the law of cosines,
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.
dna,ereH
Equilibrium Position: The system is in equilibrium if
dV
du
`
u=30°
=0.
y2=0.5 sin um.y1=3–l=33–1210 –210 –6 sin u24 m
lA¿B=21
2+32–2112132cos190° –u2=210 –6 sin u
m
1m
A
C
B
3m
Ans:
1172
11–47.
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,
.
Equilibrium Position: The system is in equilibrium if .
dV
du
=0
y=r+dcos u
The cylinder is made of two materials such that it has a mass
of mand a center of gravity at point G. Show that when G
lies above the centroid Cof the cylinder, the equilibrium is
unstable.
C
G
a
r
1173
*11–48.
SOLUTION
To remain in neutral equilibrium, the center of gravity must be located at A.
y=
©yW
©W
The bent rod has a weight of 5 lb ft. A pivot of negligible
size is attached at its center Aand the rod is balanced as
shown. Determine the length Lof its vertical segments so
that it remains in neutral equilibrium. Neglect the thickness
of the rod.
8 in. 8 in.
2 in.
LL
A
11–49.
T
h
e tr
i
angu
l
ar
bl
oc
k
of we
i
g
h
t Wrests on t
h
e smoot
h
corners which are a distance aapart. If the block has three
equal sides of length d, determine the angle for
equilibrium.
u
SOLUTION
dV
du
=Wc(–0.5774 d) sin u–a
sin 60° (–1.5 sin ucos u–0.5 sin ucos u)d=0
AD
sin a
=a
sin 60°
AF =AD sin f=AD sin (60° –u)
d
a
G60
60
u