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

Answer 1

Answer:

12.5N

Explanation:

Data;

W = 50N

Fh = 15N

μ = 0.4

Fv= ?

The second attachment is showing the resolution of the vectors.

The normal force acting on the box (N) = W - Fv

The frictional force ( ⃗f) = Fh

Frictional force ⃗f = μN = μ(W - Fv)

μW - μFv = Fh

Fv = W - (Fh/μ)

Fv = 50 - (15/0.4) = 50 - 37.5

Fv = 12.5N

The smallest vertical force which moves the box is 12.5N