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Tyrolean Traverse/Static Equilibrium Problem

  1. Dec 2, 2008 #1
    1. The problem statement, all variables and given/known data

    In 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 75-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?

    I attatched the picture from masteringphysics.

    2. Relevant equations

    [tex]\sum{F}=0[/tex]

    3. The attempt at a solution

    T=2800 (this is 28 kN/10 for the safety factor)
    m=75 kg

    [tex]\sum{F}=2Tsin\vartheta-mg=0[/tex]

    [tex]\vartheta=sin^{-1}\frac{mg}{2T}=7.54 degrees[/tex]

    [tex]cos\vartheta=\frac{x}{12.5}[/tex]

    x=12.39 m

    Obviously, this answer is too large, and I've verified that masteringphysics wont accept it, but i'm still pretty confused as to where I went wrong. Thanks in advance for the help!
     

    Attached Files:

  2. jcsd
  3. Dec 2, 2008 #2

    rl.bhat

    User Avatar
    Homework Helper

    x/12.5 is tan(theta), not cos (theta)
     
  4. Dec 2, 2008 #3
    doh! I knew it would be something simple that was throwing me off. Well thanks for the help...I got 1.7 m now, and masteringphysics approves, so all is right with the world.
     
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