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Question how to find acceleration with kinetic friction involved.

  1. Jan 2, 2013 #1
    1. The problem statement, all variables and given/known data

    [URL=http://imageshack.us/photo/my-images/132/81100769.png/][PLAIN]http://img132.imageshack.us/img132/2433/81100769.png[/URL] Uploaded with ImageShack.us[/PLAIN]

    I have a question on part of this problem.

    Paul accidentally falls off the edge of a glacier as shown in Fig 4-21 (p 103). He
    is tied by a long rope to Steve, who has a climbing ax. Before Steve sets his ax
    to stop them, he slides without friction along the ice, attached by the rope to
    Paul. Assume no friction between the rope and the glacier. Find the
    acceleration of each person and the tension in the rope. (figured this part out already)

    Questions: The two climbers each have masses of 110kg each, the coefficient of kinetic friction between steve and the rock is .5 and theta is 15 degrees. If steve has 3.2 meters to reach the edge of the cliff how long till he reaches the cliffs edge. I think I use Δx=1/2at^2?

    question. after steve goes over the cliffs edge what is the tension in the rope (have no idea how to do this part)



    2. Relevant equations
    these are the solutions for frictionless.
    Steve:
    ƩS: Fx : T + msg sin = msax
    ƩFy : Fn – msg cos = ms · 0

    Paul:
    ƩP: Fy : T – mpg = may

    g(mp + ms sin)
    __________________=ax
    (ms + mp)

    3. The attempt at a solution

    I know how to figure out acceleration when its frictionless , but I dont know how with friction. I am thinking I add the kinetic friction vector opposite direction of tension (to pic above) and when I sum the forces I am subtracting it off (ƩFx: T + mgsintheta - fk) for Steve in the y direction. So this:
    g(mp + ms sin)
    __________________=ax
    (ms + mp)

    would be this:
    g(mp + ms sin)
    __________________-0.5 (kinetic friction)=ax
    (ms + mp)

    The line is a divison line....
     
    Last edited: Jan 2, 2013
  2. jcsd
  3. Jan 2, 2013 #2

    haruspex

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    0.5 is the coefficient of friction. So what is the frictional force opposing motion?

    After going over the edge, suppose the tension in the rope is T. What then would be the acceleration of each climber?
     
  4. Jan 3, 2013 #3
     
  5. Jan 3, 2013 #4

    haruspex

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    You don't seem to know what is meant by a 'coefficient of friction'. It's the ratio between the magnitude of the frictional force and that of the normal force between the surfaces.
    What is the magnitude of the normal force between Steve and the ground?
     
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