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HW 2 Solutions
Test Your Understanding 11.1: Conditions for Equilibrium
A drum major marching in a parade tosses his baton in the air. As the baton rises and falls, it
rotates at a uniform rate.
Which condition(s) for equilibrium does this motion satisfy?
the first condition for equilibrium
neither the first condition nor the second condition for equilibrium
the zero condition for equilibrium
the second condition for equilibrium
both the first condition and the second condition for equilibrium
The baton has a downward linear acceleration due to gravity, so the first condition for
equilibrium (that the net force on the object is zero) is not satisfied. The baton rotates at a
constant rate and so has zero
angular
acceleration, so the second condition for equilibrium (that
the net torque on the object is zero) is satisfied.
Torques on a Ruler.
A centimeter ruler, balanced at its center point, has two coins placed on it, as shown in the figure.
One coin, of mass
M
1
= 10 g, is placed at the zero mark; the other, of unknown mass
M
2
, is placed
at the 4.7 cm mark. The center of the ruler is at the 3.0 cm mark. The ruler is in equilibrium; it is
perfectly balanced.
Does the pivot point (i.e., the triangle in the diagram upon
which the ruler balances) exert a force on the ruler? Does
it exert a nonzero torque about the pivot?
Yes/No
Note that the weight of the ruler itself also does not exert
a torque with respect to the pivot point since the rule is
uniform and is pivoted at its midpoint.
Although the pivot exerts a force on the ruler it does not exert a torque with respect to the pivot
point. Why not?
Because the distance from the exerted force to the pivot is zero.
Find
the
mass
M
2
.
Second
condition
for
equilibrium:
0
i
i
τ
=
∑
11
2 2
2
2
30
10
3
1.7
1.7
Mgl Mgl
g cm M
cm M
g
⋅=
⋅
⋅
= ⋅
=
Exercise 11.6
Two people are carrying a uniform wooden board that is 3.0 m long and weighs 160 N.
If one
person applies an upward force equal to 60 N at one end, at what point does the other person lift?
Begin with a freebody diagram of the board.
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DENTIFY:
Apply the first and second conditions of equilibrium to the board.
SET UP:
The freebody diagram for the board is given in Figure 11.6. Since the board is uniform its center of
gravity is 1.50 m from each end. Apply
0
y
F
=
∑
, with
y
+
upward. Apply
0
τ
=
∑
with the axis at the end
where the first person applies a force and with counterclockwise torques positive.
EXECUTE:
0
y
F
=
∑
gives
12
0
FFw
+−=
and
21
160 N
60 N 100 N
FwF
=− =
−
=
.
0
=
∑
gives
2
(1.50 m)
0
Fx w
−=
and
2
160 N
2.40 m
100 N
w
x
F
⎛⎞
==
=
⎜⎟
⎝⎠
. The other person lifts with a force of
100 N at a point 2.40 m from the end where the other person lifts.
EVALUATE:
By considering the axis at the center of gravity we can see that a larger force is applied by the
person who pushes closer to the center of gravity.
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 Spring '11
 EIns

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