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
×
=
°
=
θ
=
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
θ
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
=
=
°
°
5.13:
a) In the absence of friction, the force that the rope between the blocks exerts on block
B
will be
the component of the weight along the direction of the incline,
α
w
T
sin
=
. b) The tension in the upper
rope will be the sum of the tension in the lower rope and the component of block
A
’s weight along the
incline,
.
sin
2
sin
sin
α
=
α
+
α
w
w
w
c) In each case, the normal force is
.
cos
α
w
d) When
,
,
0
w
n
=
=
α
when
.
0
,
90
=
°
=
α
n
5.15:
a)
The tension is related to the masses and accelerations by
.
2
2
2
1
1
1
a
m
g
m
T
a
m
g
m
T
=

=

b) For the bricks accelerating upward, let
a
a
a
=

=
2
1
(the counterweight will accelerate down).
Then, subtracting the two equations to eliminate the tension gives
.
s
m
96
.
2
kg
0
.
15
kg
0
.
28
kg
0
.
15
kg
0
.
28
s
m
80
.
9
or
,
)
(
)
(
2
2
1
2
1
2
2
1
1
2
=
+

=
+

=
+
=

m
m
m
m
g
a
a
m
m
g
m
m
c) The result of part (b) may be substituted into either of the above expressions to find the tension
N.
191
=
T
As an alternative, the expressions may be manipulated to eliminate