1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook