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Spring with Coulomb Friction

  1. Oct 14, 2014 #1
    1. The problem statement, all variables and given/known data
    A mass m is attached to a spring with spring constant k. There is a coefficient of static friction,
    us
    The coefficient of kinetic friction is uk
    Suppose you pull the mass to the right and release it from rest.
    You find there is a limiting value of x = A0 > 0 below which the
    mass just sticks and does not move. For x > A0 , it starts sliding
    when you release it from rest. Find A0 .

    2. Relevant equations
    ## x''(t) + \omega x(t) - ( \mu mg)/k =0 ##

    3. The attempt at a solution
    I solved the ODE for Simple Harmonic Motion, and I get that ##x(t)=B sin ( \omega t) + C ( \omega t) -umg/k##, but I'm not sure where to go from there. The derivative at x = A0 must be zero, but how does that help me find A0 itself?
     
    Last edited: Oct 14, 2014
  2. jcsd
  3. Oct 14, 2014 #2
    The place to go is back to Newton's 2nd law. Your DE is wrong, and your thinking about static friction is in error. Please try again.
     
  4. Oct 14, 2014 #3
    In light of your post, I figured this:
    At x=A0, the block is at rest, so the forces acting on it must be balanced. Thus, -kx=umg, and at A0, -kA0=umg, so solving for A0 gives: A0=-umg/k

    Am I on the right track now?
     
  5. Oct 15, 2014 #4
    Wouldn't A0 be positive since the question indicates A0 > 0 ?
     
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