(adsbygoogle = window.adsbygoogle || []).push({}); 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

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

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