# Overcoming Friction Problem

1. Oct 11, 2009

### PhysicsIsRuff

1. The problem statement, all variables and given/known data
A 75kg box slides down a 25.0 degree ramp with an acceleration of 3.60m/s$$^{2}$$
a.) Find $$\mu$$k between the box and the ramp.
b.) What acceleration would a 175kg box have on this ramp?

2. Relevant equations
Fapplied,y=(?N)(sin[STRIKE]0[/STRIKE])
Fapplied,x=(?N)(cos[STRIKE]0[/STRIKE])
Fg=mg
$$\sum$$Fy=Fn+Fapplied,y-Fg-0
Fk=$$\mu$$kFn
$$\sum$$Fx=Fapplied,x-Fk=ma$$_{2}$$

3. The attempt at a solution
Given: ax=3.60m/s
a.) Fg=(75kg)(9.81m/s$$^{2}$$)=736N
b.) m=175kg

I do not understand how to get Fk or Fn without first finding $$\mu$$, so I am stuck for now...

Last edited: Oct 12, 2009
2. Oct 11, 2009

### Delphi51

You can't find the normal force or solve for the coefficient without separating the mg into components parallel and perpendicular to the ramp.

3. Oct 11, 2009

### rl.bhat

The acceleration of the sliding box is given by
ma = mgsinθ - μmgcosθ
Solve for μ.

4. Oct 11, 2009

### PhysicsIsRuff

So in that case, I should change it into
Fapplied,y=736cos25 down
Fapplied,x=736sin25 right
and continue from there? Or are those symbols for something else?

Is this how it should look in graph form?

@rl.bhat ah, so it is that simple... is the acceleration constant or 9.81m/s^2

Last edited: Oct 11, 2009
5. Oct 12, 2009

### rl.bhat

a is given in the problem and it is constant. g is 9.8 m/s^2

6. Oct 12, 2009

### PhysicsIsRuff

So 0 = mgsinθ - μmgcosθ
Ok, thank you both for the help, now to lock this thread...

7. Oct 12, 2009

### rl.bhat

It is not correct.
3.6 m/s^2 = mgsinθ - μmgcosθ