Weight, Normal Force, and a Box

A friend has given you a special gift, a box of mass 10.0 kg with a mystery surprise inside. The box is resting on the smooth (frictionless) horizontal surface of a table.

(a) Determine the weight of the box and the normal force exerted on it by the table.

(b) Now your friend pushes down on the box with a force of 40.0 N , as in fig (b) .
Again determine the normal force exerted on the box by the table.

(c) If your friend pulls upward on the box with a force of 40.0 N (fig. (c)), what now is the normal force exerted on the box by the table.

Approach : The box is at rest on the table, so the net force on the box in each case is zero. (Newton's Second Law).
The weight of the box equals mg in all three cases.

Solution:

(a) The weight of the box is mg = (10.0 kg) (8.80 m/s^2) = 98.0 N

This force acts downward.

The normal force exerted upward on it by the table. Fig.(a)

The upward direction as the positive y direction;

The net force along y direction on the box = F (y) = F (N) - mg .

The box is at rest, so the net force on it = 0.

F(y) = F (N) - mg = 0

F (N) = mg

The normal force on the box, exerted by the table , is 98.0 N upward. It has magnitude equal to the box's weight.


(b) Your friend is pushing down on the box with a force of 40.0 N .

Now there are three forces acting on th ebox. Fig. (b)

The weight of the box = mg = 98.0 N

The net force = F (y) = F (N) - mg - 40.0 N = 0 ( a = 0)

For Normal Force ,

F(N) = mg + 40.0 N = 98.0 N + 40.0 N = 138.0 N


(c) The box's weight is still 98.0 N and acts downward.

The force exerted by your friend and the normal force
both act upward ( positive direction). Fig. (c)

The box does not move, since your friend's upward force is less than the weight.
The net force = 0 (a =0)

F (y) = F (N) - mg + 40.0 N = 0

F (N) = mg - 40.0 N = 98.0 N - 40.0 N = 58.0 N.

The table does not push against the full weight of the box because of the upeard pull exerted by your friend.