A 50-N crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50. A 20-N force is applied to the crate acting to the right. What is the resulting static friction force acting on the crate?

a. 20 N to the left.

b. 25 N to the left.

c. 25 N to the right.

d. None of the above; the crate starts to move.

e. 20 N to the right.

Question

Everett

Physics

7 December, 19:43

A 50-N crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50. A 20-N force is applied to the crate acting to the right. What is the resulting static friction force acting on the crate?

a. 20 N to the left.

b. 25 N to the left.

c. 25 N to the right.

d. None of the above; the crate starts to move.

e. 20 N to the right.

+2

Answers (1)

Know the Answer?

Emmanuel Powers · Answer 1

b. 25 N to the left.

Explanation:

Hi there!

The static friction force is calculated as follows:

Fr = N · μ

Where:

Fr = friction force.

N = normal force.

μ = coefficient of static friction.

If the object is not being accelerated in the vertical direction, the normal force is equal to the weight of crate (with opposite sign). Then:

Fr = 50 N · 0.50

Fr = 25 N

Since the 20-N force is applied to the right of the box, the friction force will be directed to the left because the friction force opposes the displacement. Then, the right answer is "b": 25 N to the left.