A woman exerts a horizontal force of 113 N on a crate with a mass of 31.2 kg. Required:
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 is not moving then its acceleration will be zero. The equation will become:

\sum Fx = 0

\sumFx = 0

Fm - Ff = 0.

Fm is the moving force

Ff is the frictional force

Fm = Ff

This means that the moving 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

m is the mass of the crate = 31.2kg

g is the acceleration due to gravity = 9.8m/s²

R = 31.2 × 9.8

R = 305.76N

Recall that;

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

i would think it's food chain

d.7.5cm is your answer