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Swing question

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data

    A swing is made from a rope that will tolerate a maximum
    tension of 800N without breaking. Initially, the swing
    hangs vertically. The swing is then pulled back at an
    angle of 60.0 degree with respect to the vertical and
    released from rest. What is the mass of the heaviest
    person who can ride the swing?



    2. Relevant equations
    Fc = mv^2/r
    trig ratios


    3. The attempt at a solution

    I initially just thought of doing 800N / 9.81m/s^2 which give 81.5kg or so but i thought that was too easy and can't be right so I did some more complex stuff:

    using unit circle:
    r = sin60r = 0.5r


    Fc = mv^2/r
    I forgot how but i somehow ended up in the step below, where 0.5r cancels out and in the end and the end result goes back to m = 800/9.81
    800 = (9.81m(0.5r))/0.5r

    still pretty sure that is wrong and I am really lost, so help please o;
     
  2. jcsd
  3. Oct 13, 2011 #2
    The swing is basically, undergoing motion in a vertical circle. So for this to happen, at every instant the swing needs a radially inward component of the net force(centripetal) which is equal to mv^2/R.

    Now it is crucial in this problem that you understand, the tension in the string will be maximum, when it is at, the lowest point in it's motion, I.e when weight and tension are anti parallel.

    So at this instant,

    T - mg= mv^2/R


    Now you want to know the maximum mass of the person, so put the maximum value in for the tension.

    Use conservation of energy to find v^2 and solve the above eqn for m
     
  4. Oct 13, 2011 #3
    At what point after release will the tension in the rope be the greatest? Once you decide on that, then draw a free body diagram of the weight on the swing noting all forces. Relate those forces to the tension in the rope. From the sound of the problem statement, I assume this is a one rope swing.
     
  5. Oct 13, 2011 #4
    The problem can be worked from an energy standpoint or from the acceleration of a mass as it is affected by gravity. I don't know which subject you are studying.
     
  6. Oct 13, 2011 #5
    Ok so I made some progress from all the help you guys give and now I am at :

    T - mg = mv^2/r

    mgh = 1/2mv^2
    v^2 = 2gh

    and since i figured h was 0.5r so:

    v^2 = 9.81r

    replacing that in initial yields:
    800 - 9.81m = 9.81rm/r
    800 = 19.62m
    m = 40.8kg

    DID I DO IT RIGHT?
     
  7. Oct 13, 2011 #6
    Looks good to me.
     
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