# Static Friction of box on a ramp

1. Jun 15, 2010

### elbonymcbride

1. The problem statement, all variables and given/known data
A 100kg box is placed on a ramp. As part of the ramp is raised, the box begins to move downward just as the angle of inclination reaches 15 degrees. What is the coefficient of static friction between the box and the ramp?

2. Relevant equations
Fn = mg-Fsin15

3. The attempt at a solution
Fcos15 = Mk(mg-Fsin15)
F(cos15+sin15) = Mkmg
F=Mkmg/cos15+Mksin15
= 0.2(100kg)(9.8)\cos15+0.2sin15
I continued to solve this problem, however the answer that I got was not one of my answer choices. I believe that I used the wrong equation; but even with the right equation, I never know where to start. Can anybody help me?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 15, 2010

### Staff: Mentor

Hint: At the point where the box just barely begins to slide, the force acting down the ramp (the component of gravity) is just overcoming the static friction.

What's the component of gravity down the ramp?
What's the maximum value of static friction?

These both depend on the angle. Set up an equation and [STRIKE]solve for the angle[/STRIKE]. Edit: You are given the angle; I meant set up an equation and solve for the coefficient of friction.

Last edited: Jun 15, 2010
3. Jun 15, 2010

### elbonymcbride

would the maximum value of static friction be zero?
How would I find the component of gravity with the information given?

4. Jun 15, 2010

### Staff: Mentor

No. Hint: How does friction force relate to the normal force? How does the normal force depend on the angle?
You are given the angle of the ramp. Use a bit of trig to find the component parallel to the ramp.

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