) a box weighing 77.0 n rests on a table. a rope tied to the box runs vertically upward over a pulley and a weight is hung from
the other end (fig. 4â45). determine the force that the table exerts on the box if the weight hanging on the other side of the pulley weighs (a) 30.0 n, (b) 60.0 n, and (c) 90.0 n.
1 answer:
Refer to the diagram shown below.
T = the tension in the rope
N = the the normal reaction (the force that the table exerts on the box)
W = the hanging weight
Assume that the pulley is frictionless.
For equilibrium,
T = W
and
T + N = 77
Therefore
N = 77 - W
(a) When W = 30 N,
N = 77 - 30 = 47 n
Answer: 47 N
(b) When W = 60 N,
N = 77 - 60 = 17 N
Answer: 17 N
(c) When W = 90 N
N = 77 - 90 = - 13 N
There is no normal reaction, and the system is no longer in equilibrium.
Instead, the box will be lifted by a force of 13 N, and the box will accelerate upward.
T = the tension in the rope
N = the the normal reaction (the force that the table exerts on the box)
W = the hanging weight
Assume that the pulley is frictionless.
For equilibrium,
T = W
and
T + N = 77
Therefore
N = 77 - W
(a) When W = 30 N,
N = 77 - 30 = 47 n
Answer: 47 N
(b) When W = 60 N,
N = 77 - 60 = 17 N
Answer: 17 N
(c) When W = 90 N
N = 77 - 90 = - 13 N
There is no normal reaction, and the system is no longer in equilibrium.
Instead, the box will be lifted by a force of 13 N, and the box will accelerate upward.
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Answer:
9) a = 25 [m/s^2], t = 4 [s]
10) a = 0.0875 [m/s^2], t = 34.3 [s]
11) t = 32 [s]
Explanation:
To solve this problem we must use kinematics equations. In this way we have:
9)
a)
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = acceleration [m/s^2]
x = distance = 200 [m]
Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.
0 = (100)^2 - (2*a*200)
a = 25 [m/s^2]
b)
Now using the following equation:
0 = 100 - (25*t)
t = 4 [s]
10)
a)
To solve this problem we must use kinematics equations. In this way we have:
Note: The positive sign of the equation means that the car increases his speed.
5^2 = 2^2 + 2*a*(125 - 5)
25 - 4 = 2*a* (120)
a = 0.0875 [m/s^2]
b)
Now using the following equation:
5 = 2 + 0.0875*t
3 = 0.0875*t
t = 34.3 [s]
11)
To solve this problem we must use kinematics equations. In this way we have:
10^2 = 2^2 + 2*a*(200 - 10)
100 - 4 = 2*a* (190)
a = 0.25 [m/s^2]
Now using the following equation:
10 = 2 + 0.25*t
8 = 0.25*t
t = 32 [s]