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CHAPTER 6
Problem
3. A 3700-kg barge is being pulled along a canal by two mules, as shown in Fig. 6-59. The tension in each tow rope is 1100
N, and the ropes make 25
°
angles with the forward direction. What force does the water exert on the barge
(a) if it moves with constant velocity and (b) if it accelerates forward at
0.16m/s
2
?
Solution
The horizontal forces on the barge are the two tensions and the resistance of the water, as shown on Figure 6-59. The net
force is in the
x
direction, so
2
T
cos25
°
!
F
res
=
ma
x
,
since
T
1
=
T
2
=
T
.
(a) If
a
x
=
0,
F
res
=
2(1100 N)cos25
° =
1.99 kN.
(b) If
a
x
=
0.16 m/s
2
,
F
res
=
1.99 kN
!
(3700 kg)(0.16 m/s
2
)
=
1.40 kN.
FIGURE 6-59 Problem 3 Solution.
Problem
7. A block is launched up a frictionless ramp that makes an angle of 35º to the horizontal. If the block’s initial speed is
2.2 m/s,
how far up the ramp does it slide?
Solution
The acceleration up the ramp is
!
g
sin35
°
,
so the block goes a distance in this direction calculated from the equation
v
0
x
2
!
2
g
sin
35
°
(
x
!
x
0
)
=
0
.
Thus,
x
!
x
0
=
(
2
.
2 m
/
s
)
2
=
2
(
9
.
8 m
/
s
2
)sin
35
° =
43
.
1 cm
.

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*Sign up*CHAPTER 6
123
Problem
9. A 15-kg monkey hangs from the middle of a massless rope as shown in Fig. 6-60. What is the tension in the rope?
Compare with the monkey’s weight.
Solution
The sum of the forces at the center of the rope (shown on Fig. 6-60) is zero (if the monkey is at rest),
T
1
+
T
2
+
W
=
0
.
The
x
component of this equation requires that the tension is the same on both sides:
T
1
cos8
° +
T
2
cos172
° =
0,
or
T
1
=
T
2
.
The
y
component gives
2
T
sin8
° =
W
,
or
T
=
W
=
2
sin
8
° =
3
.
59
W
=
3
.
59
(
15 kg
)(
9
.
8 m
/
s
2
)
=
528 N
.
FIGURE 6-60 Problem 9 Solution.
Problem
11. A 10-kg mass is suspended at rest by two strings attached to walls, as shown in Fig. 6-62. Find the tension forces in the
two strings.
Solution
The force diagram is superimposed on Fig. 6-62. Since the mass is at rest, the sum of the forces is zero,
T
1
+
T
2
+
W
=
0
,
which is true for the
x
and
y
components separately,
T
1
cos 45
°
!
T
2
=
and
T
1
sin45
°
!
W
=
0.
Solving for the
magnitudes of the tensions, and substituting 98 N for the weight, we find
T
1
=
2
98 N
=
139 N
, and
T
2
=
T
1
=
=
98 N
.
FIGURE 6-62 Problem 11 Solution.