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Kinetic Friction!

  1. Jul 24, 2009 #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, as shown in the figure below. Fortunately, he's roped to a 980 kg rock located 51 m from the edge of the cliff. Assume that the coefficient of kinetic friction between rock and ice is 5.5×10−2. What is his acceleration, and how much time does he have before the rock goes over the edge? Neglect the rope's mass. What is his acceleration? How much time does he have before the rock goes over the edge?




    2. Relevant equations
    F=ma


    3. The attempt at a solution

    equation of the climber:
    Ft-mg=may

    equation of the rock:
    Ft=Fr

    I found the tension of the ropes by this equation: Ft=ukN=(.055)(980)(9.8) = 528.22N. I plugged this back into the climber's equation and got -2.76 m/s^2 but this answer is wrong. Help please!
     
  2. jcsd
  3. Jul 24, 2009 #2
    The equation of the rock is wrong
    The tension is not the same as friction because the rock moves with acceleration.
    Use [tex]\sum F[/tex] = ma to get the equation for the rock
     
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