A box with a weight of 50 N rests on a horizontal surface. A person pulls horizontally on it with a force of 15 N and it does not move. To start it moving, a second person pulls vertically upward on the box. If the coefficient of static friction is 0.4, what is the smallest vertical force for which the box moves?
A) 5 N
B) 12 N
C) 20 N
D) 25 N
E) 35 N

Question

A box with a weight of 50 N rests on a horizontal surface. A person pulls horizontally on it with a force of 15 N and it does not move. To start it moving, a second person pulls vertically upward on the box. If the coefficient of static friction is 0.4, what is the smallest vertical force for which the box moves? 
A) 5 N 
B) 12 N 
C) 20 N 
D) 25 N 
E) 35 N

Quizplus · Accepted Answer

The force of friction keeping the box from moving horizontally is 15 N according to the person pulling with 15 N, but the maximum force of friction is 0.4(50 N) = 20 N. This means the box could begin to move horizontally if a force greater than 15 N but less than or equal to 20 N were applied horizontally.

To start moving the box vertically, the force of the second person pulling upward must be greater than or equal to the force of gravity on the box. Thus, the smallest vertical force required to move the box is the weight of the box which is 50 N.

Therefore, the total force acting on the box is the horizontal force plus the vertical force:

$\sqrt{(15 	ext{ N})^2 + (50 	ext{ N})^2} = 52.1 	ext{ N}$

This force must be greater than the force of friction:

$52.1 	ext{ N} > 20 	ext{ N}$

This inequality is satisfied, so the box will begin to move.

A Box with a Weight of 50 N Rests on a Horizontal