)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.

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
  47 N

(b) When W = 60 N,
 N = 77 - 60 = 17 N
  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.

a) If the weight hanging on the other side of the pulley weighs is 30N than the force that the table exerts on the box is, N = 47N.

b) If the weight hanging on the other side of the pulley weighs is 60N than the force that the table exerts on the box is, N = 17N.

c) If the weight hanging on the other side of the pulley weighs is 90N than the force that the table exerts on the box is, N = -13N.

Given :

Box Weight = 77N

Solution :

From the given figure (attached below) we have the following equations:

T = W  ---- (1)

T + N = 77  ---- (2)

From equation (1) and (2) we get:

N = 77 - W --- (3)

a) When, W = 30N

N = 77 - 30 = 47N    (from equation (3))

b) When, W = 60N

N = 77 - 60 = 17N     (from equation (3))

c) When, W = 90N

N = 77 - 90 = -13N    (from equation (3))

Negative sign indicates that 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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The answer is 12, for the points