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Mountain climber 2d forces question.

  1. Feb 27, 2013 #1
    1. a 52 Kg mountain climber is suspended from a cliff by a rope.

    Given a few bits of information the angle formed by the rope to the climber is 31 degrees

    The angle of the climbers legs is 15 degrees north of horizontal

    (a) Find the tension in the rope and the force that the mountain climber must exert
    with her feet on the vertical rock face to remain stationary. Assume that
    the force is exerted parallel to her legs. Also, assume negligible force
    exerted by her arms.

    (b) What is the minimum coefficient of friction
    between her shoes and the cliff?

    note: part of the weight of the climber is held up my the rope and part by her legs.




    any help here would be great. I can account for the forces in a free body diagram and can account for a weight force of 509.6 N, but I am a bit lost in how to set up the rest of the system to account for the different angles.

    full problem at http://openstaxcollege.org/files/textbook_version/hi_res_pdf/9/Physics_col11406_CP-1-002-DW-preflight.pdf [Broken]

    ch 5 question 17
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 27, 2013 #2
    Start with the free body diagram of the climber, and ask yourself what the conditions are for her to be stationary.
     
  4. Feb 27, 2013 #3
    ok here is the first part i am stuck at.

    If i isolate the tension force then i can see that there is a 31 degree angle east of south...
    the components of y would be Ft cos (31) and x would be Ft sin (31)

    Where i get confused is the next step shown by my instructor is this equation.

    y = Ft cos 31 + Fc sin 15 -Fg = 0

    where Fc sin 15 is the force of static friction.

    It seems like there are a lot of steps missed here.
    I can usually fill in the gaps my teacher leaves but this seems like a BIG gap.
     
  5. Feb 27, 2013 #4
    to answer your question the forces would have to cancel each other out.
     
  6. Feb 27, 2013 #5
    Yes, you canceled the forces out in the y component, now do the same with the x component.
     
  7. Feb 27, 2013 #6
    so here is where i am at...

    y: Ft cos 31 + Fc sin 15 - Fg = 0
    x: ft sin 31 - fc cos 15 = 0

    using substitution i can solve for Fc

    cos 31 x (Fc cos 15/sin 31) + Fc sin 15 - Fg = 0

    when i try to solve for elements of cos 15 and sin I am getting different square roots divided by 4.

    i dont know how to handle that information in the equation or if there is a more simple way to resolve he equation.
     
  8. Feb 27, 2013 #7
    I don't see a problem with solving for Fc here...
     
  9. Feb 27, 2013 #8
    I adjusted the equation to solve for tension as that is what I actually want to know so:

    Ft x cos 31 + (Ft sin 31/cos 15) x sin 15 - Fg = O

    When I enter the known values i have this:

    Ft x .857 + (Ft x .515 / square root of 6 + square root of 2 / 4) x (square root of 6 - square root of 2 / 4) - 509.6 = 0

    I am confused why the sin and cos of 15 are not coming up with something more concrete than square + square/ 4
     
  10. Feb 27, 2013 #9
    I am not sure why you are trying to solve an exact value for sin 15 and cos 15. They are not special angles, so you can leave them in decimal form.
     
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