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Hangmans Drop?

  1. Dec 16, 2007 #1
    OK here goes....

    I have a 20kg weight attached to a 10 meter long wire rope.
    The rope is fixed vertically to a cross beam.
    The rope has an elastic stretch limit of 47.5 kN/mm^2.
    If I lift the 20kg weight up 1 meter and then let it go what would the maximum force exerted by weight be on the wire rope?

    It's been a very very long time since I studied Newton so forgive me if I'm barking loudly up the wrong tree!

    I started of with Potential Energy = Mass X Gravity X Height but this didn't seem to cut the mustard so I then went for:

    Elastic Potential Energy = 1/2 x (k Spring Constant) x (L Spring Length)^2

    But I'm not sure if this is the correct equation to use....

    Impulse = (Mass x Velocity Initial) - (Mass x Velocity Final)

    Which again doesn't seem right but maybe I'm getting confused...

    Tension = Mass x Gravity x SQRT(2xHeight/Length Increase)

    So I have a selection of formulas that I think I should be using but I don't think I have the correct data to complete the formulas.

    So this is what I came up with but I don't really know if I'm going along the right lines or not.

    Velocity = sqrt 2 x9.81x1 = 4.429446918 m/s
    Potential Energy = 20x9.81x1 = 196.2 Joules
    Momentum = 20 x 4.429446918 = 88.58893836kg m /s
    Force = Change in Momentum / Time = 88.58893836 / 0.1 Second = 885.8893836

    So just how far wide of the mark am I?

    Attached Files:

  2. jcsd
  3. Dec 16, 2007 #2


    Staff: Mentor

    Use energy conservation, no need for momentum considerations here. The gravitational potential energy (h = 1m + x) at the top goes to elastic potential energy at the bottom (the rope stretches by x). Solve for x then put x into the elastic stretch formula to get the force.
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