# Homework Help: How to Find Coeffecient of Static Friction?

1. Oct 3, 2012

### "that guy"

1. The problem statement, all variables and given/known data
A crate, M = 277 kg, sits at rest on a surface that is inclined at 20.0° above the horizontal. A horizontal force (parallel to the ground), F1 = 594 N is required to just start the crate moving down the incline.

To the nearest thousandth, what is the coefficient of static between the crate and the incline?

Answer: 0.633
2. Relevant equations
I assume ∫s=μs * N

3. The attempt at a solution
What I originally did was draw a free-body diagram of the crate. The active forces, as far as I can tell, would be weight, acting straight down, and Normal force, acting up the y axis relative to the crate. Friction was acting in the negative y direction relative to the box, opposing w(sinθ). So ∫s = w(sinθ) and N = w(cosθ). At this point, I had w(sinθ) = μs(w(cosθ)). (Substituting w(sinθ) for ∫s, and w(cosθ) for N.) I then took the answer and divided to get μs, but it was obviously incorrect. My answer was 0.219, whereas the actual answer was 0.633. Could someone please walk me through this, and show me just what I'm not getting? Also, where does the parallel force come into play/why is it relevant? Thanks in advance.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 3, 2012

### Simon Bridge

Orienting your axis so the x-axis is along the slope?

friction always acts along a surface.

That would be against the rules, however, this we can do:
Hopefully the above also helps with your next question:
What would you expect the parallel force to do if there was no friction?

3. Oct 4, 2012

### howie8594

Think about the force going downwards parallel to the slope as mgsinθ. The normal force would be mgcosθ. Think about how to take the normal force to derive another equation for the force of friction.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook