5.1:
a) The tension in the rope must be equal to each suspended weight, 25.0 N. b) If the
mass of the light pulley may be neglected, the net force on the pulley is the vector sum of
the tension in the chain and the tensions in the two parts of the rope; for the pulley to be
in equilibrium, the tension in the chain is twice the tension in the rope, or 50.0 N.
5.2:
In all cases, each string is supporting a weight
w
against gravity, and the tension in
each string is
w
. Two forces act on each mass
:
w
down and
)
(
w
T
up.
5.3:
a) The two sides of the rope each exert a force with vertical component
T
θ
sin
, and
the sum of these components is the hero’s weight. Solving for the tension
T
,
.
N
10
54
.
2
0.0
1
sin
2
)
s
m
(9.80
kg)
0
.
90
(
sin
2
3
2
w
T
b) When the tension is at its maximum value, solving the above equation for the angle
θ
gives
.
01
.
1
N)
10
50
.
2
(
2
s
m
(9.80
kg)
(90.0
arcsin
2
arcsin
4
2
T
w
5.4:
The vertical component of the force due to the tension in each wire must be half of
the weight, and this in turn is the tension multiplied by the cosine of the angle each wire
makes with the vertical, so if the weight is
.
48
arccos
and
cos
,
3
2
4
3
2
θ
θ
w
w
w
5.5:
With the positive
y
direction up and the positive
x
direction to the right, the free
body diagram of Fig. 5.4(b) will have the forces labeled
n
and
T
resolved into
x
 and
y

components, and setting the net force equal to zero,
.
0
sin
cos
0
sin
cos
w
T
n
F
n
T
F
y
x
Solving the first for
α
T
n
cot
and substituting into the second gives
w
T
T
T
T
sin
sin
sin
sin
cos
sin
sin
cos
2
2
2
and so
,
cos
cot
sin
cot
w
w
T
n
as in Example 5.4.
5.6:
N.
10
4.10
17.5
sin
)
s
m
(9.80
kg)
1390
(
sin
sin
3
2
α
mg
α
w
5.7:
a)
N.
10
23
.
5
cos
or
,
cos
4
40
cos
)
s
m
kg)(9.8
4090
(
2
θ
W
T
W
θ
T
B
B
b)
N.
10
36
.
3
40
sin
N)
10
23
.
5
(
sin
4
4
θ
T
T
B
A
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5.8:
a)
.
0
45
cos
30
cos
and
,
45
sin
30
sin
,
B
A
C
B
A
C
T
T
w
T
T
T
w
T
Since
,
45
cos
45
sin
adding the last two equations gives
,
)
30
sin
30
(cos
w
T
A
and so
.
732
.
0
366
.
1
w
T
w
A
Then,
.
897
.
0
45
cos
30
cos
w
T
T
A
B
b) Similar to part (a),
,
45
sin
60
cos
,
w
T
T
w
T
B
A
C
and
.
0
45
cos
60
sin
B
A
T
T
Again adding the last two,
,
73
.
2
)
60
cos
60
(sin
w
T
w
A
and
.
35
.
3
45
cos
60
sin
w
T
T
B
B
5.9:
The resistive force is
N.
523
m)
6000
m
200
)(
s
m
kg)(9.80
1600
(
sin
2
w
.
5.10:
The magnitude of the force must be equal to the component of the weight along the
incline, or
N.
337
0
.
11
sin
)
s
m
kg)(9.80
180
(
sin
2
θ
W
5.11:
a)
,
sin
N,
60
W
θ
T
W
so
,
sin45
N)
60
(
T
or
N.
85
T
b)
N.
60
45
cos
N
85
,
cos
2
1
2
1
F
F
θ
T
F
F
5.12:
If the rope makes an angle
with the vertical, then
0 110
1 51
sin
0 073
(the
denominator is the sum of the length of the rope and the radius of the ball). The weight is
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