Question

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

Answer 1

Answer:

The resulting static friction force acting on the crate is 10 N.

Explanation:

Given that,

Coefficient of static friction = 0.50

Normal force = 20 N

We need to calculate the resulting static friction force acting on the crate

Using formula of static friction force

$f_{s}=\mu_{s}N$

Where, N = normal force

$\mu_{s}$ = Coefficient of static friction

Put the value into the formula

$f_{s}=0.50\times20$

$f_{s}=10\ N$

Hence, The resulting static friction force acting on the crate is 10 N.