PHY 2053
Recitation Week # 12
Problem #
Ch. 9 #
12,25,37,44,49,57,60
12.
A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m
of the plank hanging over the right support (see the drawing). To what distance
x
can a person who
weighs 450 N walk on the overhanging part of the plank before it just begins to tip?
REASONING
At every instant before the plank begins to
tip, it
is in
equilibrium, and the net torque on it is zero:
0
τ
Σ
=
(Equation 9.2). When the person reaches
the maximum distance
x
along the overhanging
part, the plank is just about to rotate about the right
support. At that instant, the plank loses contact
with the left support, which consequently exerts no force on it. This leaves only three vertical forces
acting on the plank: the weight
W
of the plank, the force
F
R
due to the right support, and the force
P
due
to the person (see the freebody diagram of the plank). The force
F
R
acts at the right support, which we
take as the axis, so its lever arm is zero. The lever arm for the force
P
is the distance
x
. Since
counterclockwise is the positive direction, Equation 9.2 gives
W
W
0
or
W
W
Px
x
P
τ
Σ
=

=
=
l
l
(1)
SOLUTION
The weight
W
= 225 N of the plank is known, and the force
P
due to the person is equal to
the person’s weight:
P
= 450 N. This is because the plank supports the person against the pull of gravity,
and Newton’s third law tells us that the person and the plank exert forces of equal magnitude on each
other. The plank’s weight
W
acts at the center of the uniform plank, so we have (see the drawing)
1
1
W
W
2
2
or
d
L
L
d
+
=
=

l
l
(2)
where
d
= 1.1 m is the length of the overhanging part of the
plank, and
L
= 5.0 m is the length of the entire
plank.Substituting Equation (2) into Equation (1), we obtain
(
29
(
29
(
29
1
1
2
2
225 N
5.0 m
1.1m
0.70 m
450 N
W
L
d
x
P


=
=
=
*
25.
A 1220N uniform beam is attached to a vertical wall at one
end and is supported by a cable at the other end. A 1960N crate
hangs from the far end of the beam. Using the data shown in the
x
L/2
F
R
W
P
d
W
l
Freebody diagram of the plank
Axis
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drawing, find
(a)
the magnitude of the tension in the wire and
(b)
the magnitude of the horizontal and
vertical components of the force that the wall exerts on the left end of the beam.
REASONING
The drawing shows the beam and the five forces that act on it: the horizontal and vertical
components
S
x
and
S
y
that the wall exerts on the left end of the beam, the weight
W
b
of the beam, the
force due to the weight
W
c
of the crate, and the tension
T
in the cable. The beam is uniform, so its center
of gravity is at the center of the beam, which is where its weight can be assumed to act. Since the beam is
in equilibrium, the sum of the torques about any axis of rotation must be zero
(
29
0
τ
Σ
=
, and the sum of
the forces in the horizontal and vertical directions must be zero
(
29
0,
0
x
y
F
F
Σ
=
Σ
=
. These three
conditions allow us to determine the magnitudes of
S
x
,
S
y
, and
T
.
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 Spring '09
 Staff
 Angular Momentum, Moment Of Inertia, στ

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