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Climber hanging from a cliff and acceleration

  1. Feb 18, 2007 #1
    1. The problem statement, all variables and given/known data
    A 75 kg climber finds himself dangling over the edge of an ice cliff. Fortunately, he's roped to a 920 kg rock located 51 m from the edge of the cliff. Assume that the coefficient of kinetic friction between rock and ice is 5.3×10^−2 . What is his acceleration? Neglect the rope's mass.


    2. Relevant equations
    Tension(rock)=tension(climber)=T
    acceleration(rock)=acceleration(climber)=a

    Climber(y):T-m(c)*g=m(c)*a
    Rock(x):T+f(k)=m(r)*a
    Rock(y):N-m(r)*g=0


    3. The attempt at a solution
    T=m(c)*a+m(c)*g
    f(k)=mu(k)*N
    N=m(r)*g

    m(c)*a+m(c)*g+mu(k)*m(r)*g=m(r)*a
    a=[m(c)*g+mu(k)*m(r)*g]/[m(r)-m(c)]
    a=1.44 m/s^2

    The computer is telling me that I have the wrong answer. Do I need to take the 51 m into account somehow? Any help would be greatly appreciated!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 18, 2007 #2
    friction opposes relative motion. the rock wants to move in the direction of the tension of the rope, but the friction opposes it, which means its direction is opposite.
     
  4. Feb 18, 2007 #3
    Thanks! That gave me the right answer :)
     
  5. Feb 18, 2007 #4
    Write out the individual equations of motion for the rock and the man, add them to eliminate T and solve for a. You do not need to take the length of the rope into account.
     
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