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Friction on two sides

  1. Oct 2, 2008 #1
    A 55 kg rock climber is climbing a "chimney" between two rock slabs with her shoes on one side and her back against the other. The static coefficient of friction between her shoes and the rock is 1.1; between her back and the rock it is 0.80. She has reduced her push against the rock until her back and her shoes are on the verge of slipping. What is the magnitude of her push against the rock, and what fraction of her weight is supported by her shoes?

    Okay, so I think I should set it up like this (though I'm not sure if this is right):
    friction(shoes) + friction(back) = weight

    I plugged in the numbers and now I have:
    (1.1)*(N1) + (0.8)*(N2) = 539

    Where do i go from here (if this is right so far)?
     
  2. jcsd
  3. Oct 2, 2008 #2

    Doc Al

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    Staff: Mentor

    So far, so good. See if you can figure out the relationship between N1 and N2 by analyzing horizontal forces on the climber.
     
  4. Oct 2, 2008 #3
    does she have to exert a force proportional to the coefficients of friction?
     
  5. Oct 2, 2008 #4

    Doc Al

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    Staff: Mentor

    Analyze the horizontal forces.
     
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