Chapter 12 – Student Solutions Manual
5. Three forces act on the sphere: the tension force
G
T
of the rope
(acting along the rope), the force of the wall
N
F
G
(acting h
away from the wall), and the force of gravity
mg
orizontally
G
(acting
downward). Since the sphere is in equilibrium they sum to zero. Le
θ
be the angle between the rope and the vertical. Then Newton’s
second law gives
t
vertical component :
T
cos
–
mg
= 0
horizontal component:
F
N
–
T
sin
= 0.
(a) We solve the first equation for the tension:
T
=
mg
/ cos
. We
substitute
cos
to obtain
=+
LL
r
/
22
222
(0.85 kg)(9.8 m/s ) (0.080 m)
(0.042 m)
9.4 N
0.080 m
mg L
r
T
L
+
+
==
=
.
(b) We solve the second equation for the normal force:
sin
N
FT
=
.
Using
sin
rL
r
/
, we obtain
2
(0.85 kg)(9.8 m/s )(0.042 m)
4.4 N.
(0.080 m)
N
Tr
mg L
r
r
mgr
F
Lr
+
=
=
=
++
7. We take the force of the left pedestal to be
F
1
at
x
= 0, where the
x
axis is along the
diving board. We take the force of the right pedestal to be
F
2
and denote its position as
x
=
d
.
W
is the weight of the diver, located at
x
=
L
. The following two equations result
from setting the sum of forces equal to zero (with upwards positive), and the sum of
torques (about
x
2
) equal to zero:
12
1
0
()
FFW
Fd W L d
0
+
−=
+
(a) The second equation gives
1
3.0m
(580 N)= 1160 N
1.5m
Ld
FW
d
⎛⎞
−
=−
−
⎜⎟
⎝⎠
which should be rounded off to
. Thus,

3
1
1.2 10 N
F
×
3
1
1
.
2
1
0
N
.
F
=×
(b) Since
F
1
is negative, indicating that this force is downward.
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View Full Document(c) The first equation gives
21
580 N+1160 N=1740 N
FWF
=−=
which should be rounded off to
. Thus,
3
2
1.7 10 N
F
=×
3
2

1
.
7
1
0
N
F
.
(d) The result is positive, indicating that this force is upward.
(e) The force of the diving board on the left pedestal is upward (opposite to the force of
the pedestal on the diving board), so this pedestal is being stretched.
(f) The force of the diving board on the right pedestal is downward, so this pedestal is
being compressed.
11. The
x
axis is along the meter stick, with the origin at the
zero position on the scale. The forces acting on it are shown
on the diagram below. The nickels are at
x
=
x
1
= 0.120 m,
and
m
is their total mass. The knife edge is at
x
=
x
2
= 0.455 m
and exerts force
G
. The mass of the meter stick is
M
, and the
force of gravity acts at the center of the stick,
x
=
x
F
3
= 0.500
m. Since the meter stick is in equilibrium, the sum of the
torques about
x
2
must vanish:
Mg
(
x
3
–
x
2
) –
mg
(
x
2
–
x
1
) = 0.
Thus,
32
0.455m 0.120m
(10.0g)=74.4 g.
0.500m 0.455m
xx
Mm
⎛⎞
−−
==
⎜⎟
⎝⎠
21. We consider the wheel as it leaves the lower floor. The floor no longer exerts a force
on the wheel, and the only forces acting are the force
F
applied horizontally at the axle,
the force of gravity
mg
acting vertically at the center of the wheel, and the force of the
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 Winter '08
 Naik
 Physics, Force, Friction, Gravity, Angle of inclination, Fgx

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