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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
Ans.
Ans.
=-
70.2
6
0 Unstable
u=12.1°,
d2V
du2=0.5[-3(9.81) sin 12.1° -14(9.81) cos 12.1°]
d2V
du2=l(-Wsin u-2W
Dcos u)
u=12.1°
tan u=W
2W
D
=3(9.81)
14(9.81) =0.2143
dV
du
=l(Wcos u-2W
Dsin u)=0
=Wl sin u-W
Dl(3 -2 cos u)
V=2Wy1-W
Dy2
y2=l+2l(1 -cos u)=l(3 -2 cos u)
y1=l
l=500 mm
500 mm
500 mm
A
C
500 mm
uu
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
du
11–37. Continued
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
k
u
SOLUTION
du
11–38 Continued
2
V
2
V
11–39.
SOLUTION
However, for a smallangle ,.Thus,
The elastic potentialenergy of the torsional spring canbe computed using
,where .Thus,
The totalpotentialenergy of the systemis
Equilibrium Configuration: Taking the first derivative of ,we have
Equilibrium requires .Thus,
Stability: The second derivative of is
To have neutral equilibrium at ,.Thus,
Ans.
W=8k
3L
-3WL
2
+4k=0
d2V
du22u=0°
=0u=0°
d2V
du2=-
3WL
2+4k
V
u=0°
ua-3WL
2+4kb=0
dV
du
=0
dV
du
=-
3WL
2u+4ku=ua-3WL
2+4kb
V
V=V
g+V
e=3WL
2a1-u2
2b+2ku2
V
g=1
2k(2u)2=2ku2
b=2uV
e=1
2kb2
V
g=Wy =Wa3
2Lba1-u2
2b=3WL
2a1-u2
2b
cos u1-u2
2
u
A
k
L
2
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
*11–40.
SOLUTION
P
otential Function: First, we must determine the center of gravity of the cylinder.By
referring to
Fig. a,
W
ith 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,
T
hus,
Equilibrium Configuration:
Taking the first derivative of V,
Equilibrium requires
.Thus,
Stability:
The second derivative of Vis
To
have neutral equilibrium at ,.Thus,
Ans.
d=12h;2(-12h)2-4(3)(6h2)
2(3) =0.5858h=0.586h
6h2-12hd +3d2=0
-W
B
6h2-12hd +3d2
4(3h-d)
R
cos 0° =0
d2V
d2u2u=0°
=0u=0°
d2V
d2u
=-W
B
6h2-12hd +3d2
4(3h-d)
R
cos u
sin u=0u=0°
-W
B
6h2-12hd +3d2
4(3h-d)
R
sin u=0
dV
du
=0
dV
du
=-W
B
6h2-12hd +3d2
4(3h-d)
R
sin u
V=V
g=Wy =W
B
6h2-12hd +3d2
4(3h-d)
R
cos u
y
=(y-d)cos u=
B
6h2-d2
4(3h-d)-d
R
cos u=
B
6h2-12hd +3d2
4(3h-d)
R
cos u
h
rpr2h-1
3rpr2d
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
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
SOLUTION
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
G
3.5 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
ba
af
f
u
SOLUTION
where
is a constant and
Ans.W
2=W
1ab
absin u
cos f
dV
du
=W
1bsin u-W
2acos (90° -u-g)
=-W
1bcos u+W
2asin (90° -u-g)
V=-W
1y1+W
2y2
f=(90° -u-g)g
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,
Ans.
or,
Ans.u=33.0°
A
36 -16 sin2u
B
1
2=6.667 cos u
u=0°
sin u=0
Wc4-0.8333(36 -16 sin2u)-1
2(32 cos u)dsin udu=0
dU=0; Wdy=0
dy=4 sin udu+1.667a1
2b(36 -16 sin2u)-1
2(-32 sin ucos u)du
y=-4 cos u+1.667 236 -16 sin2u
x+4 cos u+y=2.667x
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.
B
C
D
Ans:
u=0°
u=33.0°
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°.
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
At
Ans.m=5.29 kg
dV
du
`
u=30°
=-
29.43mcos 30°
210 -6 sin 30°
+58.86 cos 30° =0
u=30°,
=-
29.43m cos u
210 -6 sin u
+58.86 cos u
dV
du
=-9.81mc-1
2110 -6 sin u2-1
21-6 cos u2d +58.86 cos u
dV
du
`
u=30°
=0.
=29.43m-9.81m1210 -210 -6 sin u2+58.86 sin u
=9.81m33-1210 -210 -6 sin u24 +117.7210.5 sin u2
V=V
g=W
1y1+W
2y2
y2=0.5 sin um.y1=3-l=33-1210 -210 -6 sin u24 m
l=lAB -lA¿B=210 -210 -6 sin u
lAB =21
2+32=210 m
lA¿B=21
m
C
B
3m
1172
11–47.
SOLUTION
Equilibrium Position: The system is in equilibrium if .
Stability:
Thus, the cylinder is in unstable equilibrium at (Q.E.D.)u=0°
d2V
du22u=0°
=-mgd cos 0° =-mgd 60
d2V
du2=-mgd cos u
sin u=0u=0°
dV
du
=-mgd sin u=0
dV
du
=0
V=V
g=Wy =mg(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
Take the positive root
Ans.L=8 in.
L=20 ;2(-20)2-4(5)(-160)
10
0=5L2-20L-160
0=
2a16
12 b(5) -aL
2-2ba2L
12 b(5)
a16
12 b(5) +a2L
12 b(5)
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
Ans:
L=8 in.
1174
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
Require, Ans.
or
Ans.u=cos-1d
4aba
-0.5774 d-a
sin 60° (-2 cos u)=0
u=0°sin u=0
dV
du
=Wc(-0.5774 d) sin u-a
sin 60° (-1.5 sin ucos u-0.5 sin ucos u)d=0
V
=Wy
=a
sin 60° (0.75 cos2u-0.25 sin2u)
AF =a
sin 60° (sin (60° +u)) sin (60° -u)
AD =a
sin 60° (sin (60° +u))
AD
sin a
=a
sin 60°
AF =AD sin f=AD sin (60° -u)
d
G60
60
u
Ans:
u=0°
u=cos-1
a
d
4ab
