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Forces in Equilibrium acting on wall climber

  1. May 1, 2015 #1
    < Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

    Hello and thank you in advance for any help. I am currently studying physics and having difficulty working out a forces at equilibrium problem. With limited given variables, I am unable to see the path to a solution.

    The problem is as such:

    A 650N climber is resting mid-climb. If the co-efficient of friction between her feet and the vertical wall is 0.65 and the angle between the rope and the cliff is 8°, what is the tension in the rope? Find normal force (net force "X" direction)

    Fg=650N (Ff÷Fn)=0.65 T = ? Fn = ?
    Sum Fx = 0N Sum Fy = 0N
    (Tx-Fn) = 0N (Ty+Ff) - Fg = 0N

    And this is where I can't seem to find the relation...

    Any help is greatly appreciated :)
     
    Last edited by a moderator: May 1, 2015
  2. jcsd
  3. May 1, 2015 #2
    Is there a diagram that goes with this?

    Seems like the answer would depend on where the rope was tied on the body and where the center of mass of the climber is on the body.
     
  4. May 1, 2015 #3
    This is all that is given for the problem. The black is original. The green was me.
    Snapshot.jpg
     
  5. May 1, 2015 #4
    Typically that would be the case. But since the frictional force depends upon the normal force, which in turn depends upon the horizontal component of the tension, some substitutions can be made. Of course, we do have to assume that the climber is a rigid body.

    That looks correct to me. What is the equation for the frictional force?

    Also, be careful with the angles. It is a tad bit tricky, since you are given the angle between the rope and the wall instead of the angle between the rope and the climber.
     
    Last edited: May 1, 2015
  6. May 1, 2015 #5
    The only equation to calculate force of friction that I know is Ff = µ x Fn

    It may be the substitutions you mentioned that I am not understanding. Could you please elaborate?
     
  7. May 1, 2015 #6
    Okay, good. Do you see where you can substitute that?

    Next you have to express the horizontal and vertical tension in the rope in terms of the angle between the rope and the cliff.
     
  8. May 1, 2015 #7

    I can see now that you can substitute (Ty + Ff) - Fg for ((Ty + (µ x Tx))-Fg. So that would mean (T(0.990) + µT(0.139)) - 650 N = 0 N But how do I carry on from there?
    Tx = T sin 8 or T(0.139) = Fn
    Ty = T cos 8 or T(0.990)
     
  9. May 1, 2015 #8
    Good job.
    You are given the coefficient of static friction between the climber and the cliff. You only have one unknown in your equation: T. You can solve for T, right? After you have T it is a simple calculation to find the horizontal component of the tension.
     
  10. May 1, 2015 #9
    I see now. Perfect! Thank you very much.
     
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