# Find the static coefficient of friction

Havenater23

## Homework Statement

You pull on a crate with a force of 50N at an angle of 25 degrees. The box doesn't move.?
What is the coefficient of friction and what is the frictional force?

I understand the forces are balanced, but are they balanced in the x and y direction. I've done the problem , but I get a coefficient of 2.14 , which I know isn't right. Can

## Answers and Replies

Homework Helper
I can't find the coefficient without knowing the mass. It would be useful to see your work.
Do we know the box will move if the force is slightly larger? If not, we can only find a minimum value for the coefficient.

Homework Helper
Hi Havenater23! Show us your full calculations, and then we'll see what went wrong, and we'll know how to help! Havenater23
Okay, well basically this how I did it.

This is a rope pulling on the crate above the horizon at 25 degrees.
The components :
X= cos(25) 50N
Y= sin(25) 50N

Then I start calculating Net Force of Y
Fy = Fn - mg - sin(25) 50 N = ma
but A=o , because it's not moving .
So I guess it means Fn and Fg = sin(25) 50 N

Then calculate Net force of X
Fx = cos(25) 50 N - Ff = ma
Well Ff = sin25 50 N * u
Ff = Fn * u

Since a=o in x direction as well
Ff=cos(25) 50 N
Cos(25) 50 N = sin25 50 N * u

but my answer comes out too big.
I get like 2.14
Help ?

Havenater23
Can anyone help me? I know it's just a small error, I just don't know what.

Homework Helper
Fy = Fn - mg - sin(25) 50 N = ma
but A=o , because it's not moving .
So I guess it means Fn and Fg = sin(25) 50 N
Something wrong here. The mg is down, the sin(25)*50 is up. They should have opposite signs. The normal force is the total force pressing the crate down:
Fn = mg - 50*sin(25)
(This counteracted by the floor pushing up, but we aren't interested in that!)

Your horizontal part looked good, but you had the wrong Fn so it will change quite a bit when you redo it.