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Friction on an inclined plane

  1. Oct 23, 2009 #1
    I am having problems with a simple problem with a mass (m) stationary on an inclined plane of angle θ.

    I have obtained the equations for the reaction/normal force and the force acting on the mass parallel to the plane:

    Reaction force / normal force (R) = mgcosθ
    Parallel force = mgsinθ

    Also, I know that the static friction force is given by Friction force = (mu static) x R.

    My question is this:

    For the mass to be stationary, obviously the parallel force must equal the static friction force in magnitude. But if the angle, θ of the inclined slope is increased, logically both forces must also increase in magnitude by the same amount to keep equilibrium.

    Since mu static is a constant, this means that as the parallel force increases with θ, R must increase in the (mu static x R) equation to provide a frictional force of equal magnitude to the parallel force. But R, and hence the frictional force, do not increase with θ as the cosine in the reaction force equation decreases with an increase in θ between 0 and 90 degrees, also decreasing the frictional force.

    So how can there be equilibrium (which there obviously is!)?

    As far as I can see, since the two forces involve sine or cosine, they cannot both increase at the same time between 0 and 90 degrees.

    Any help would be greatly appreciated.

    Many thanks

    Paul
     
    Last edited: Oct 23, 2009
  2. jcsd
  3. Oct 23, 2009 #2

    Dale

    Staff: Mentor

    The correct equation is [tex]F \leq \mu_s R[/tex]. The static friction force is whatever force is required for the object to not move up to a maximum given by [tex]\mu_s R[/tex]
     
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