Question

A worker pushes horizontally on a 36.0 kg crate with a force of magnitude 112 N. The coefficient of static friction between the crate and the floor is 0.380. (a) What is the value of fs,max under the circumstances

Answer 1

Answer:

value of fs,max under the circumstances is

f = 42.56 N

Explanation:

The formula is µ = f / N, where µ is the coefficient of friction, f is the amount of force that resists motion, and N is the normal force. Normal force is the force at which one surface is being pushed into another.

So f = µN

f = 0.380 * 112

f = 42.56 N

Answer 2

The maximum value for the force of friction is given by:

$F_{max}=\mu_sN$

Where $\mu_s$ is the static friction between the body and the surface and N is the normal force on the body. Using the free body diagram, we know that:

$N=mg$

Replacing this and solving:

$F_{max}=\mu_s mg\\F_{max}=0.38*36kg*9.8\frac{m}{s^2}\\F_{max}=134.06N$