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5.10.
Model:
We assume that the box is a point particle that is acted on only by the tension in the rope and the
pull of gravity. Both the forces act along the same vertical line.
Visualize:
Solve: (a)
Since the box is at rest,
a
y
=
0 m/s
2
and the net force on it must be zero:
( )( )
2
net
0 N
50 kg
9.8 m/s
490 N
FTw
Twm
g
=−=
⇒== =
=
(b)
Since the box is rising at a constant speed, again
a
y
=
0 m/s
2
,
=
0 N, and
net
F
490 N
Tw
=
=
.
(c)
The velocity of the box is irrelevant, since only a
change
in velocity requires a nonzero net force. Since
a
y
=
5.0
,
2
m/s
( )( )
2
net
50 kg
5.0 m/s
250 N
250 N
250 N
490 N
740 N
y
FTwm
a
=
=
⇒=
+=
+
=
(d)
The situation is the same as in part (c), except that the rising box is slowing down. Thus
a
y
=
–5.0
and we
have instead
2
m/s
( )( )
2
net
50 kg
5.0 m/s
250 N
250 N
250 N
490 N
240 N
y
a
=
−
=
−
−
−
+
=
Assess:
For parts (a) and (b) the zero accelerations immediately imply that the box’s weight must be exactly
balanced by the upward tension in the rope. For part (c) the tension not only has to support the box’s weight but
must also accelerate it upward, hence,
T
must be greater than
w
. When the box accelerates downward, the rope need
not support the entire weight, hence,
T
is less than
w
.
5.14.
Model:
We assume that the passenger is a particle acted on by only two vertical forces: the downward
pull of gravity and the upward force of the elevator floor.
Visualize:
Please refer to Figure Ex5.14. The graph has three segments corresponding to different conditions: (1)
increasing velocity, meaning an upward acceleration, (2) a period of constant upward velocity, and (3) decreasing
velocity, indicating a period of deceleration (negative acceleration).
Solve:
Given the assumptions of our model, we can calculate the acceleration for each segment of the graph and
then apply Equation 5.10. The acceleration for the first segment is
2
10
8 m/s
0 m/s
4 m/s
2 s
0 s
y
vv
a
tt
−
−
==
=
−−
()
2
2
app
2
4 m/s
4
1
1
75 kg
9.8 m/s
1
1035 N
9.8 m/s
9.8
y
a
ww
m
g
g
⇒=+=
+
=
+ =
For the second segment,
a
y
=
0 m/s
2
and the apparent weight is
2
2
app
1
75 kg
9.8 m/s
740 N
m
g
g
=+
=
For the third segment,
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View Full DocumentDynamics I: Motion Along a Line
52
()
2
32
2
2
app
2
0 m/s
8 m/s
2 m/s
10 s
6 s
2 m/s
1
75 kg
9.8 m/s
1
0.2
590 N
9.8 m/s
y
vv
a
tt
ww
−
−
==
=
−
−−
−
⇒=+
=
−=
Assess:
As expected, the apparent weight is greater than normal when the elevator is accelerating upward and
lower than normal when the acceleration is downward. When there is no acceleration the weight is normal. In all
three cases the magnitudes are reasonable, given the mass of the passenger and the accelerations of the elevator.
5.15.
Model:
We assume that the safe is a particle moving only in the
x
direction. Since it is sliding during the
entire problem, we can use the model of kinetic friction.
Visualize:
Solve:
The safe is in equilibrium, since it’s not accelerating. Thus we can apply Newton’s first law in the vertical
and horizontal directions:
net
B
C
k
k
B
C
0 N
350 N
385 N
735 N
x
x
F
FFF f
fFF
=Σ
=
+
−
=
⇒
=
+
=
+
=
( )( )
2
net
0 N
300 kg
9.8 m/s
2940 N
y
y
FF
n
w
n
w
m
g
=
−
=
⇒
=
=
=
=
Then, for kinetic friction:
k
kk
k
735 N
0.25
2940 N
f
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 Spring '06
 staff
 Force, Gravity

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