A woman exerts a horizontal force of 113 N on a crate with a mass of 31.2 kg. a. If the crate doesn't move, what's the magnitude of the static friction force (in N)?
b. What is the minimum possible value of the coefficient of static friction between the crate and the floor?

a) 113N

b) 0.37

Explanation:

a) Using the Newton's second law:

\sum Fx =ma

Since the crate doesn't move (static), acceleration will be zero. The equation will become:

\sum Fx = 0

\sumFx = Fm - Ff = 0.

Fm is the applied force

Ff is the frictional force

Since Fm - Ff = 0

Fm = Ff

This means that the applied force is equal to the force of friction if the crate is static.

Since applied force is 113N, hence the magnitude of the static friction force will also be 113N

b) Using the formula

Ff = nR

n is the coefficient of friction

R is the reaction = mg

R = 31.2 × 9.8

R = 305.76N

From the formula

n = Ff/R

n = 113/305.76

n = 0.37

Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37

wind moving at higher speeds will transfer more energy to the water, resulting in stronger waves.