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CHAPTER 14 Solutions
Problem
2. A body is subject to three forces:
F
1
=
2
î
+
2
ˆ
j
N
,
applied at the point
x
=
2 m,
y
=
0 m;
F
2
=
!
2
î
!
3
ˆ
j
N
,
applied
at
x
=
!
1 m,
y
=
0;
and
F
3
=
1
ˆ
j
N
,
applied at
x
=
!
7 m,
y
=
1 m.
(a) Show explicitly that the net force on the body
is zero. (b) Show explicitly that the net torque about the origin is zero. (c) To confirm the assertion following
Equation 14-4 that the net torque must be zero about any other point, evaluate the net torque about the point (3 m, 2 m),
the point (
−
7 m, 1 m), and about any other point of your choosing.
Solution
(a)
!
F
i
=
(
2
î
+
2
ˆ
j
"
2
î
"
3
ˆ
j
+
ˆ
j
)
N
=
0
.
(b)
(
! "
i
)
0
= [
2
î
#
(
2
î
+
2
ˆ
j
)
+
(
$
î
)
#
(
$
2
î
$
3
ˆ
j
)
+ (
$
7
î
+
ˆ
j
)
#
ˆ
j
]
N
%
m
=
(
4
+
3
!
7
)
ˆ
k
N
"
m
=
0
.
(c) For any point
r
0
=
(
x
0
î
+
y
0
ˆ
j
)
m
,
!
(
r
i
"
r
0
)
#
F
i
=
{[(
2
"
x
0
)
î
"
y
0
ˆ
j
]
#
(
2
î
+
2
ˆ
j
)
+
[(
!
1
!
x
0
)
î
!
y
0
ˆ
j
]
"
(
!
2
î
!
3
ˆ
j
)
+
!
7
!
x
0
)
î
+
(
1
!
y
0
)
ˆ
j
]
"
ˆ
j
}
N
#
m
=
4
+
3
!
7
)
+
x
0
(
!
2
+
3
!
1
)
+
y
0
(
2
!
2
+
0
)]
ˆ
k
N
"
m
=
0
.
Problem
3. Suppose the force
F
3
in the preceding problem is doubled so the forces no longer balance and the body is therefore
accelerating. Show that (a) the torque about the point
(
!
7 m, 1 m)
is still zero, but that (b) the torque about the origin is
no longer zero. What is the torque about the origin?
Solution
(a) Since
r
3
=
(
!
7
î
+
ˆ
j
)
m
is the point of application of
F
3
, the total torque about
r
3
is just due to
F
1
and
F
2
:
(
!
"
i
)
3
=
(
r
1
!
r
3
)
"
F
1
+
(
r
2
!
r
3
)
"
F
2
=
[(
2
î
+
7
î
!
ˆ
j
)(
2
î
+
2
ˆ
j
)
+
(
!
î
+
7
î
!
ˆ
j
!
2
î
!
3
ˆ
j
)]
N
#
m
=
9
"
2
)
ˆ
k
!
(
1
"
2
)(
!
ˆ
k
)
+
6
!
(
"
3
)
ˆ
k
+
(
1
!
2
)(
"
ˆ
k
)]
N
#
m
=
0
.
(b)
(
!
i
)
0
=
!
(
r
i
#
F
i
)
=
[
2
î
#
(
2
î
+
2
ˆ
j
)
+
(
$
î
)
#
(
2
î
$
3
ˆ
j
)
+
(
$
7
î
+
ˆ
j
)
#
(
2
ˆ
j
N
!
m
=
[
4
ˆ
k
+
3
ˆ
k
"
14
ˆ
k
]
N
!
m
=
"
7
ˆ
k
N
!
m
.
Problem
9. Figure 14-29
a
shows a thin, uniform square plate of mass
m
and side
l
.
The plate is in a vertical plane. Find the
magnitude of the gravitational torque on the plate about each of the three points shown.
Solution
The center of gravity is at the center of a uniform plate. In calculating the gravitational torque, one may consider the entire
weight as acting at the center of gravity. (a)
r
A
=
2
l
=
2
at 135
°
from the weight of the plate, so
!
A
=
l
=
2)
mg
sin135
° =
1
2
mg
l
.
(b)
r
B
is colinear with the weight, so
B
=
0.
(c)
C
=
1
2
l
mg
sin90
° =
1
2
mg
l
(but note that
C
"!
A
).
(We also
assumed that
B
and
C
are at the centers of their respective sides. Alternatively, the torques can be found from the lever arms
shown.)
FIGURE 14-29(
a
) Problem 9 Solution.

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