Question

A 40 N crate rests on a rough horizontal floor. A 12 N horizontal force is then applied to it. If the coecients of friction are µs = 0.5 and µk = 0.4, the magnitude of the frictional force on the crate is:

Answer 1

Answer:

Frictional force, f = 20 N

Explanation:

It is given that,

Weight of the crate, W = 40 N

Horizontal force acting on the crate, F = 12 N

The coefficient of static friction, $\mu_s=0.5$

The coefficient of kinetic friction, $\mu_s=0.4$

Let f is the frictional force acting on the crate. Friction is an opposing force. It is equal to the product of coefficient friction and the normal force. It is given by :

$f=\mu_s\times N$

$f=0.5\times 40$

f = 20 N

So, the frictional force on the crate is 20 N. Hence, this is the required solution.