# Question involving Newton's Laws and Friction

## Homework Statement

A worker pulls horizontally on a rope that is attached to a 10 kg crate that is resting on a rough floor. The coefficients of static and kinetic friction are 0.5 and 0.3 respectively. The worker pulls with a force of 45N. What is the frictional force exerted by the surface?

We did this in class and the answer was 45N but I do not understand why that is. Can someone explain?

A worker pulls horizontally on a rope that is attached to a 10 kg crate that is resting on a rough floor. The coefficients of static and kinetic friction are 0.5 and 0.3 respectively. The worker pulls with a force of 45N. What is the frictional force exerted by the surface?

We did this in class and the answer was 45N but I do not understand why that is. Can someone explain?

So vertical direction of the pull does not need to be taken into account, therefore the normal force = weight.

The weight = mg or (10kg)(9.8) = 98 N

so the normal force = 98N

since f = (coefficient) * N

f = .5 * 98 = 49N, so the worker would need to pull with a force GREATER than 49 N in order to start accelerating the box, but since he pulls with only 45N, the force of friction will be equal to the pulling force..

so therefore f = force of pull = 45 N

so f = 45N