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How the coefficient of static friction works

  1. Oct 13, 2009 #1
    Not exactly a homework problem, but I'm trying to make sure I understand how the coefficient of static friction works.

    Given an object on an inclined plane and a question which asks for the minimum angle at which the object will begin to slide, I know that the formula to use is:

    mg sin([tex]\theta[/tex]) = [tex]\mu[/tex][tex]_{s}[/tex] mg cos([tex]\theta[/tex])

    which becomes

    tan[tex]^{-1}[/tex]([tex]\mu[/tex][tex]_{s}[/tex]) = [tex]\theta[/tex]

    I understand that mg times the sin of the angle represents the force which is working against friction. Is mu mg times the cos of the angle equal to the normal force?
  2. jcsd
  3. Oct 13, 2009 #2


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    Hi mvantuyl! :smile:

    (have a mu: µ and a theta: θ :wink:)
    No, mgcosθ is the normal force.

    So µsmgcosθ is the friction force, and mgsinθ is the component of the gravitational force which (in your terminology) is working against friction (and that's why they're equal). :smile:
  4. Oct 13, 2009 #3
    Re: Friction

    Thank you! That clears it up for me.
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