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Finding the angle at which a box starts to slide down a ramp

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data
    A box of mass   4.0 kg is placed at rest on a ramp. The coefficients of static and kinetic
    friction are $&  0.35 and $%  0.22, respectively. As the ramp is slowly raised from a
    horizontal orientation, find the critical angle above which the box starts to slide.


    2. Relevant equations
    ƩF=ma


    3. The attempt at a solution
    I break ƩF=ma down to

    F(gravity) x sin∅ - μ(kinetic) x F(normal) = ma

    This is as far as I get before I start to get confused. In order to solve for ∅ I need to find accerlation, which i cant think to do without knowing the angle, and the normal force. I can see that the normal force is equal to mgcos∅.


    Thanks for the help
     
  2. jcsd
  3. Oct 14, 2012 #2

    TSny

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    Gold Member

    Hello, joe426
    Note that you are looking for the maximum angle before the block begins to slide. So, what would the acceleration be? Should the friction force in your equation be static or kinetic?
     
  4. Oct 15, 2012 #3
    The acceleration would be 0.

    The equation would then be

    ƩFx = ma = 0
    mgsinθ - μmg = 0
    θ = sin^-1 (.35)
    θ = 20 degrees

    Thank you!
     
  5. Oct 15, 2012 #4

    TSny

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    Yes, a = 0. But, earlier you stated that the normal force would be mgcos∅. Now you have it as just mg.
    Which is correct?
     
  6. Oct 15, 2012 #5
    I think mgcosθ is correct but now I dont know how I would solve for theta.

    mgsinθ - μsmgcosθ = 0
     
  7. Oct 15, 2012 #6
    mg( sinθ - μcosθ) = 0
    sinθ - μcosθ = 0
    μ = sinθ/cosθ = tanθ

    θ = arctan μ
     
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