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

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(m)/(s^2)\\F_(max)=134.06N`

Answer 2

value of fs,max under the circumstances is

f = 42.56 N

Step-by-step 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