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The coefficient of friction at a specific angle.

  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data

    A block of mass 'm' is placed on an inclined plane which has an adjustable angle with the horizontal. The angle is increased slowly from zero until the block just starts to slide. This angle is called the critical angle, θs.

    Express the coefficient of friction in terms of θs.

    2. Relevant equations

    F = m*a
    friction = μ*(normal force, N)
    μ = coefficient of friction

    3. The attempt at a solution

    Fx = m*ax
    m*ax = m*g*sinθs - μN
    m*ay = m*g*cosθs - N

    After I set both equations equal to the normal force, N, I got:

    μ = (m*g*sins-m*ax) / (m*g*cosθs-m*ay)

    If I am allowed to set the acceleration equal to zero, I would get μ = tanθs, which is the correct answer. But since the block "just starts to slide" at θs, I don't think the acceleration would equal zero... Help please?
     
  2. jcsd
  3. Feb 12, 2012 #2
    It is assumed that acceleration is equal to 0 as the block just starts to slide. Hence the forces in play are equal, otherwise the friction force would be less than ma and the equation would become an inequality.
     
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