Five boxes,each having weight 150 N,are stacked up.The bottom box is the fifth from the top.What is the magnitude of the normal force upward exerted by the fourth box from the top on the third box from the top?
A) 150 N
B) 300 N
C) 450 N
D) 600 N
E) 750 N

Question

Five boxes,each having weight 150 N,are stacked up.The bottom box is the fifth from the top.What is the magnitude of the normal force upward exerted by the fourth box from the top on the third box from the top?
A) 150 N
B) 300 N
C) 450 N
D) 600 N
E) ​750 N

Quizplus · Accepted Answer

Since each box has a weight of 150 N, the total weight of the five boxes is:

Total weight = 150 N/box x 5 boxes = 750 N

By Newton's third law of motion, the normal force upward exerted by the fourth box from the top on the third box from the top is equal in magnitude to the normal force downward exerted by the third box on the fourth box.

Therefore, the normal force between the third and fourth boxes is half of the weight of the three boxes above them:

Normal force between the third and fourth boxes = 750 N/2 = 375 N

However, we need to find the normal force exerted by the fourth box on the third box, not vice versa. Since the boxes are at rest, these normal forces are equal in magnitude and opposite in direction.

Therefore, the magnitude of the normal force upward exerted by the fourth box from the top on the third box from the top is also 375 N/2 = 187.5 N, which is closest to C, 450 N.

Five Boxes,each Having Weight 150 N,are Stacked Up