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Rock climber tension in rope

  1. Sep 24, 2015 #1
    1. The problem statement, all variables and given/known data
    A 65kg rock climber is abseiling down a rock, consider the situation when he's in equilibrium and standing at an angle on the surface of the rock, hanging from a rope.

    The rope is at an angle of 40 degrees to the horizontal, calculate the tension in the rope

    2. Relevant equations
    W = mg.
    Knowledge about components of forces.

    3. The attempt at a solution
    I drew a free body diagram of the climber and used Newton's Third Law to form a right triangle with the tension as the hypotenuse which led me to sin 40 = W/T so T = W/sin 40. but the solution states that T = W sin 40 I don't understand why?
     
  2. jcsd
  3. Sep 24, 2015 #2
    Would it be possible for you to provide a diagram? It's a little hard to picture the system you are describing.

    Chet
     
  4. Sep 24, 2015 #3
    Foto am 25.09.15 um 1.46 vorm..jpg
     
  5. Sep 24, 2015 #4
    If angle of the rock face is unspecified, there doesn't seem to be enough information provided to solve the problem.

    Chet
     
  6. Sep 24, 2015 #5
    The question does not contain enough information to yield a unique solution, however given the solution T = W sin 40° you can infer that the rock face is vertical and that the climber forms a right angle with the rope. T = W / sin 40° implies that the climber forms a right angle with the wall. If the wall were not vertical (as you have drawn) it gets more complicated.

    Bad question.
     
  7. Sep 25, 2015 #6
    As you say '' you can infer that the rock face is vertical and that the climber forms a right angle with the rope" lets assume all the conditions you mentioned were a given, then how would you go about solving the problem i.e. would there be enough information to solve it?
     
  8. Sep 25, 2015 #7
    Yes, provided there is no friction between the climber's foot and the vertical rock.

    Chet
     
  9. Sep 25, 2015 #8
    And how would you solve it?
     
  10. Sep 25, 2015 #9
    The way that you did it originally, and, as you and MrAnchovy showed, the correct answer should be with the sine in the denominator (not the book answer). There is no way that the tension is less the w.

    Chet
     
  11. Sep 25, 2015 #10
    Thank you
     
  12. Sep 25, 2015 #11
    You can get to the book answer if there is sufficient friction between boot and rock to maintain a contact 50 degrees from horizontal so the climber forms a right angle with the rope. Of course neither this nor the other solution are real world examples of abseiling.
     
  13. Sep 25, 2015 #12
    Can you explain how ( if that were the case)? Even with this condition I still don't understand how the (book) solution came to be.
     
  14. Sep 25, 2015 #13

    haruspex

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    I read the problem a bit differently. I suggest the rope is parallel to the rock face and that the climber is standing perpendicularly to the rock face. If you assume the rope is tied to the climber at mass centre then you can show friction does not enter into it.
     
  15. Sep 25, 2015 #14
    Yeah but the rope has to make a 40 degree angle with the horizontal.
     
  16. Sep 25, 2015 #15
    Draw it, with the (vertical) rock face as the hypotenuse.
     
  17. Sep 25, 2015 #16
    In my judgement, we're all spending too much of our valuable time on this crazy problem statement. I'm outta here.

    Chet
     
  18. Sep 26, 2015 #17

    haruspex

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    So make the rock face do the same.
     
  19. Sep 26, 2015 #18
    @MrAnchovy: How do I get the weight component for T = w x sin40? Is this diagram correct? IMG_1225.JPG
     
  20. Sep 28, 2015 #19
    It's not incorrect, but it's not complete. Extend the rope marked T with a line. Add a line perpendicular to this extension to the tip of the arrow marked W. This triangle shows the components of W along the rope (W sin 40°) and along the body of the abseiler (W cos 40°).
     
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