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Find tension in a rope of a mountain climber

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A 735N mountain climber is rappelling down the face of a vertical cliff as shown in the diagram. If the rope makes an angle of 12deg with the vertical face, what is the tension in the rope?


    2. Relevant equations
    Fnet=ma
    trig functions
    Ff=uFN


    3. The attempt at a solution
    I seperated force into x and y components

    X
    ---
    FN-Tx=0
    therefore FN=Tx

    Y
    ----
    Ff+Ty-mg=0
    Ff=uFN=uTx
    uTx+Ty-mg=0
    uTcos78+Tsin78-mg=0

    I need help from here...the answer in the book says 751N, so i don't know...maybe i setup my net force equation wrong?
     
  2. jcsd
  3. Nov 15, 2009 #2
    Re: Equilibrium

    The mountain climber is repelling down the mountain, not sliding down the mountain. Consider this when deriving your equations.
     
  4. Nov 15, 2009 #3
    Re: Equilibrium

    Oh, so does that mean no friction force then? Or does that mean I need to change a positive sign to a negative sign in one or more of my equations? Thank you.
     
  5. Nov 15, 2009 #4
    Re: Equilibrium

    [tex]\mu_s[/tex] isn't a given quantity, if you need it, where are you going to get it from?
     
  6. Nov 15, 2009 #5
    Re: Equilibrium

    there will be no frictional force
     
  7. Nov 15, 2009 #6
    Re: Equilibrium

    correct ^^
     
  8. Nov 15, 2009 #7
    Re: Equilibrium

    I'm lost...i think the angle to use with the trig functions is 78deg, but i'm not sure...i'm not sure how to proceed further....
     
  9. Nov 15, 2009 #8
    Re: Equilibrium

    ok, what do we know?

    we know that the tension force has two components and we know that its y-component must equal (mg) in order for the man not to accelerate downward.

    we know that the force of the mountain on the man must equal the x-component of the tension force in order for the man not to accelerate through the mountain.

    we have the adjacent side of our force vector and we have the angle between the adjacent and the hypotenuse. Can you take it from here?
     
  10. Nov 15, 2009 #9
    Re: Equilibrium

    I will try and get back...thank you for help.
     
  11. Nov 15, 2009 #10
    Re: Equilibrium

    my pleasure, keep in mind that cosine is defined adjacent over hypotenuse.
     
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