) 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:
38.47 m
Explanation:
To find the height of the building, we will use the following equation
Where yf is the final height, yi is the initial height, viy is the initial vertical velocity, t is the time, and a is the acceleration due to gravity.
If the brick is in flight for 3.1 s, we can say that when t = 3.1s, yf = 0 m. So, replacing
viy = (16 m/s)sin(10) = 2.78 m/s
a = -9.8 m/s²
we get
Solving for yi
Therefore, the height of the building is 38.48 m