111
E
QUILIBRIUM AND
E
LASTICITY
11.1.
IDENTIFY:
Use Eq.(11.3) to calculate
cm
x
. The center of gravity of the bar is at its center and it can be treated as
a point mass at that point.
SET UP:
Use coordinates with the origin at the left end of the bar and the
x
+
axis along the bar.
1
2.40 kg,
m
=
2
1.10 kg,
m
=
3
2.20 kg.
m
=
EXECUTE:
11
2 2
33
cm
123
(2.40 kg)(0.250 m)
0
(2.20 kg)(0.500 m)
0.298 m
2.40 kg 1.10 kg
2.20 kg
mx
x
mmm
++
+
+
==
=
+
+
. The fulcrum
should be placed 29.8 cm to the right of the lefthand end.
EVALUATE:
The mass at the righthand end is greater than the mass at the lefthand end. So the center of gravity
is to the right of the center of the bar.
11.2.
IDENTIFY:
Use Eq.(11.3) to calculate
cm
x
of the composite object.
SET UP:
Use coordinates where the origin is at the original center of gravity of the object and
x
+
is to the right.
With the 1.50 g mass added,
cm
2.20 cm
x
=−
,
1
5.00 g
m
=
and
2
1.50 g
m
=
.
1
0
x
=
.
EXECUTE:
22
cm
12
x
mm
=
+
.
2c
m
2
5.00 g 1.50 g
(2
.
2
0
cm
)
9
.
5
3
1.50 g
xx
m
⎛⎞
−
=
−
⎜⎟
⎝⎠
.
The additional mass should be attached 9.53 cm to the left of the original center of gravity.
EVALUATE:
The new center of gravity is somewhere between the added mass and the original center of gravity.
11.3.
IDENTIFY:
The center of gravity of the combined object must be at the fulcrum. Use Eq.(11.3) to calculate
cm
x
SET UP:
The center of gravity of the sand is at the middle of the box. Use coordinates with the origin at the
fulcrum and
x
+
to the right. Let
1
25.0 kg
m
=
, so
1
0.500 m
x
=
. Let
2s
a
n
d
=
, so
2
0.625 m
x
.
cm
0
x
=
.
EXECUTE:
cm
0
x
+
+
and
1
21
2
0.500 m
(25.0 kg)
20.0 kg
0.625 m
x
x
=
−
.
EVALUATE:
The mass of sand required is less than the mass of the plank since the center of the box is farther
from the fulcrum than the center of gravity of the plank is.
11.4.
IDENTIFY:
Apply the first and second conditions for equilibrium to the trap door.
SET UP:
For
0
z
τ
=
∑
take the axis at the hinge. Then the torque due to the applied force must balance the
torque due to the weight of the door.
EXECUTE:
(a)
The force is applied at the center of gravity, so the applied force must have the same magnitude as
the weight of the door, or 300 N. In this case the hinge exerts no force.
(b)
With respect to the hinges, the moment arm of the applied force is twice the distance to the center of mass, so
the force has half the magnitude of the weight, or 150 N . The hinges supply an upward force of
300 N 150 N 150 N.
−=
EVALUATE:
Less force must be applied when it is applied farther from the hinges.
11.5.
IDENTIFY:
Apply
0
z
=
∑
to the ladder.
SET UP:
Take the axis to be at point
A
. The freebody diagram for the ladder is given in Figure 11.5. The torque
due to
F
must balance the torque due to the weight of the ladder.
EXECUTE:
(8.0 m)sin40
(2800 N)(10.0 m), so
5.45 kN
FF
°=
=
.
11
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Chapter 11
EVALUATE:
The force required is greater than the weight of the ladder, because the moment arm for
F
is less
than the moment arm for
w
.
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 Spring '11
 Shaefer
 Force, Friction, freebody diagram

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