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A aerialist on a high platform holds on to a trapeze attached to a sup

  1. Apr 19, 2013 #1
    1. The problem statement, all variables and given/known data

    A aerialist on a high platform holds on to a trapeze attached to a support by a 8.2-m cord. Just before he jumps off the platform, the cord makes an angle α of 39.8deg with the vertical. He jumps, swings down, then back up, releasing the trapeze at the instant it is 0.64 m below its initial height. Calculate the angle θ that the trapeze cord makes with the vertical at this instant.

    2. Relevant equations

    cos(theta) = A/H

    3. The attempt at a solution

    So I found the solution online.
    8.2 cos 39.8 = 6.3
    6.3 + 0.64 =6.94
    cos ^-1 (6.94 / 8.2) = 32.2 deg

    However I am not understanding it at all. I dont understand where is the initial position, when the rope is vertical or when the guy is holding on to it?
    I have attached the picture , can someone label on the picture or explain it please?

    My understanding is, the middle is 8.2m , the right rope is 6.3m, and the left rope is 6.94m
    but then the calculation doesnt really make sense.
    since if the middle is 8.2, to find l, u should do cos 39.8 = 8.2 / l, and l will be 10.67 instead.
     

    Attached Files:

  2. jcsd
  3. Apr 19, 2013 #2

    TSny

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    Homework Helper
    Gold Member

    The length of the rope is ##l## = 8.2 m. The length of the rope does not change.

    Examine some of the triangles shown in the attached figure.
     

    Attached Files:

    Last edited: Apr 19, 2013
  4. Apr 19, 2013 #3
    very helpful, thank you very much.
     
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