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Question about Kinetic Friction

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Run a trial with an angle of 28 degrees so the block is in translation equilibrium. Show how to calculate the mu_k value.
    http://www.lon-capa.org/~mmp/applist/si/plane.htm

    2. Relevant equations

    F_fk = mu_k * F_n

    3. The attempt at a solution

    I have solved for Normal force and the frictional force and plugged them into the equation to get the coefficient of friction, I get roughly 0.53.
    However, on the site given above, to get it in rough equilibrium, the mu_k is 0.44 while the mu_s is roughly 0.53

    Just wondering, if an object is in equilibrium, are both coefficients the same?
     
  2. jcsd
  3. Sep 28, 2008 #2
    No. The coefficient of static friction will pretty much always be larger, the reason being that it's harder to get something moving than to keep it moving.

    If it's in equilibrium and you're measuring the coefficient of friction by when it begins to move, that's the static coefficient. If it's moving and you're trying to measure when it's moving at a constant velocity, that's going to be the kinetic coefficient. Does that help?
     
  4. Sep 28, 2008 #3
    Yes that does help, Thanks

    But any ideas on how to solve for it?
     
  5. Sep 28, 2008 #4
    Solve for which? Static or kinetic coefficient?
     
  6. Sep 28, 2008 #5
    Kinetic
     
  7. Sep 28, 2008 #6
    Okay, well think about it. The simplest way to find that is to set the frictional force equal to the force down the incline. What would have to be the case for you to be able to do that if the object is already moving?
     
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