# Homework hellp. coefficent of friction.

## Homework Statement

An 18.0 kg box is released on a 37.0 degree incline and accelerates down the incline at 0.270 m/s/s. Find the friction force impeding it's motion. How large is the coefficient of friction?

Ff=uFn

## The Attempt at a Solution

I have no idea how to do this.

## The Attempt at a Solution

Can you do the problem if there were no friction, that is just asking for acceleration down an inclined plane?

ohh, maybe how can i do it without friction though?

Well you can't but the principles used apply:

In other words the acceleration from gravity along the surface of the ramp is sin(theta) * g. It is not 9.8. So if there were no friction a=9.8 sin(37). What is this come out to be?

5.9?

OK. Now we are told that the acceleration is not 5.9 but a measly 0.27.

So lets write an eqn using Newtons laws.

ma=sum of all forces ? is the frictional force

18.0*0.27=18*5.9-? Lets not worry about the components that make up ? yet. Can you compute ?

4.86=106.2-? so 101.34? or is there more than one other force?, Thank you for helping I'm sorry I'm really bad at physics.

Should be 18(5.9-0.27)

Ok now the tricky part. This frictional force is equal to the normal force * u where u is the coefficient of friction.

On level ground normal force is just opposite the weight. But on a plane it is
weight* cos(37)

So normal force is 18*9.8 cos (37). Lets call this N.

Then we have N*u=the same ? as you calculated before. can you get u?

176.4 cos(37)=140.9
N*u=101.34
140.9u=101.34
so u=.719

By george I think you got it. It makes more sense with a free body diagram.

Thank you so much!