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Tyrolean Traverse/Static Equilibrium Problem

  1. Sep 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Please read the WHOLE post before replying, because I dont want the answer, but how to obtain one of the equations.

    A mountain-climbing technique called the "Tyrolean traverse," a rope is anchored on both ends (to rocks or strong trees) across a deep chasm, and then a climber traverses the rope while attached by a sling as in the figure (Intro 1 figure) . This technique generates tremendous forces in the rope and anchors, so a basic understanding of physics is crucial for safety. A typical climbing rope can undergo a tension force of perhaps 28 kN before breaking, and a "safety factor" of 10 is usually recommended. The length of rope used in the Tyrolean traverse must allow for some "sag" to remain in the recommended safety range.

    Consider a 72-kg climber at the center of a Tyrolean traverse, spanning a 25-m chasm. To be within its recommended safety range, what minimum distance x must the rope sag?




    2. Relevant equations

    [tex]
    2Tsin\vartheta-mg = \Sigma
    [/tex]F

    3. The attempt at a solution
    I dont have a problem with the question itself, I have a problem with understanding one of the equations, 2TSin(Theta) = mg... Why is the TSin(Theta) multiplied by 2? I can't conceptually comprehend that... if the line can take only 2900 newtons, then multiplying it by two means we are putting a load of 4800 newtons on the whole line?...

    The answer is 1.5 meters
     
  2. jcsd
  3. Sep 11, 2010 #2

    rock.freak667

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    Homework Helper

    The tensions in the two sides would be the same if the weight is at the center.
     
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