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Forces/Tension, please check my work

  1. Sep 24, 2006 #1
    Hi, I have tried to solve the following problem, and I have my work for it.

    So first off, I calculated the force of the tightrope walker:

    Then, since the walker is in the middle of the rope, I assumed that the tension of each side of the rope would be 2000N. Plus, the triangle that she forms with the rope because she's in the center is isosceles. Therefore, the angle on both sides formed with the horizontal will be equal.

    Next, I solved for the y-component of the diagram. The sum of the forces must equal 0 since there is no acceleration.

    2000*sin(x) + 2000*sin(x) + 666.4N*sin(270) = 0
    2000*sin(x) + 2000*sin(x) = 666.4
    4000*sin(x) = 666.4
    sin(x) = .1666
    x = 9.59 degrees

    Where am I making a mistake? Did I assume something incorrectly?
  2. jcsd
  3. Sep 24, 2006 #2
    Nevermind, I figured out my problem.
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