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Friction - what direction?

  1. Aug 13, 2007 #1
    Hey,

    Im making a snowboarding game. I'm using Coulomb's equation for friction:

    Ff = coef_of_fric * normal force

    This gives me the size of the frictional force. But how is the direction calculated? Is it the opposite of the direction of the velocity of the snowboarder? Or is it the opposite direction of the force applied on the snowboarder, gravity in this case. I thought the latter would have been the correct answer. However, when a snowboarder is moving along a flat surface, the force due to gravity is cancelled out by the normal. Though the snowboarder is still moving, so there is still some friction. I think I might be missing a force, or vector or something somewhere...

    Can someone help me please?f
     
  2. jcsd
  3. Aug 13, 2007 #2

    Doc Al

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    Friction opposes slipping between surfaces, in this case between the board and the snow. It will be directed opposite to the snowboard's velocity.
     
  4. Aug 14, 2007 #3
    Well the easy way(since its a game) would to multiply the velocity by a constant less than one.And when the velocity is really small just make it 0.
     
  5. Aug 14, 2007 #4

    Doc Al

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    Well, that would certainly be easier than modeling it correctly.
     
  6. Aug 14, 2007 #5
    Be carefull when using that formula. µ*Fn is the maximum friction force, so a better formular is

    where µ is coeff'o'friction and Fn is normal force. The magnitude of Fn is so that the resulting force is zero, i.e. constant velocity.
     
  7. Aug 14, 2007 #6

    robphy

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    The first is appropriate for kinetic or "sliding" friction. The µ here is the coefficient of kinetic-friction µk. The net force on the object need not be zero in this case.

    The second is appropriate for static friction, where it is Ff (not Fn) that is the frictional force that is needed to yield a zero net force, up to that maximum of µ*Fn. The µ here is the coefficient of static-friction µs.

    The magnitude of Fn is generally not determined by Ff... but by other forces in the problem. (Fn and Ff are perpendicular components ("the legs") of the total reaction force R.)
     
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