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Coefficient of static friction on an inclined plane

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data

    A crate, mass = 275 kg, sits at rest on a surface that is inclined at 20.0° above the horizontal. A horizontal force (parallel to the ground), F = 583 N is required to just start the crate moving down the incline.


    2. The attempt at a solution
    U = (((mg sin (∅)) + (583 cos (70)) / (mg cos (∅))
    I get .443, but the answer is .630. What am I doing wrong?
     
  2. jcsd
  3. Oct 3, 2012 #2

    Doc Al

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    Staff: Mentor

    You are assuming that the normal force remains mgcosθ, but the presence of the horizontal force changes that. Also, what's the component of that horizontal force parallel to the surface?
     
  4. Oct 3, 2012 #3
    Just figured it out on my own. But to answer your question it ends up being

    μs = (w sin ∅) + (583 cos ∅) / (w cos ∅) - (583 sin ∅)

    Thanks for the help!
     
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