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Homework Help: Finding tension in tarzan swing

  1. Oct 20, 2006 #1
    Tarzan (mass m) is trying to swing across a river on a long vine (length L) Assume that he starts from rest. Determine the tension in the rope at the bottom of the arc as a function of the angle theta, the arc angle between his starting point and ending point and whatever other variables necessary. Hints: Draw Free body diagrams for Tarzan when the rope is vertical.

    I drew the free body diagram. It has tension pointing up and gravity pointing down. However, I don't know what to do.
  2. jcsd
  3. Oct 20, 2006 #2

    Doc Al

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    So far, so good. What does Newton's 2nd law tell you? Hint: What's the acceleration?
  4. Oct 20, 2006 #3


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    You mean, tension pointing in the direction of the rope, right?
    Use the fact that energy is conserved.

    Btw, I'm sure I saw a few more 'Tarzan' problems in the last few days, so using the search option might be useful. :smile:
  5. Oct 20, 2006 #4
    Using the law of conservation of energy

    K_i + U_i + W_other = K_f + U_f

    (1/2)m(v_i)^2 + mgLcos(theta) = (1/2)m(v_f)^2 + mgL

    I let the height equals zero at the top

    v_i is zero since it starts from rest

    mgLcos(theta) = (1/2)m(v_f)^2 + mgL

    the masses cancel

    gLcos(theta) = (1/2)(v_f)^2 + gL
    gLcos(theta) - gL = (1/2)(v_f)^2
    2(gLcos(theta) - gL) = (v_f)^2

    I know the the motion is circular so I know

    F = m(v^2)/r

    In this case, v is v_f and r is L

    F = m(v^2)/L

    I want to solve for the force of tension which is in the same direction as the centripetal force in this case


    F = m(2gLcos(theta) - gL) / L
    F = mL(2gcos(theta) - g)/L
    F = m (2gcos(theta) - g)
    F = mg (2cos(theta) - 1)

    T = mg(2cos(theta) - 1)

    Is this correct?
  6. Oct 20, 2006 #5

    Doc Al

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    Just a few corrections:
    If you let the height = 0 at the top (top of the vine, I presume), then the initial and final heights will be negative, since they are below the top:
    (1/2)m(v_i)^2 - mgLcos(theta) = (1/2)m(v_f)^2 - mgL

    The main thing is that Tarzan drops: his final height is lower than the initial.

    Realize that the left hand side of this is negative--the correction above will fix that.

    Right approach, but realize that the tension is not the only force acting on Tarzan. (Use the net force in the above equations.)
  7. Oct 20, 2006 #6
    how about making tarzan's weight equal and opposite to the tension in the rope?
  8. Oct 20, 2006 #7


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    That will not do. When Tarzan swings on the rope he is travelling on a circular path, which means he is accelerating. There must be a net force acting upwards when he is at the bottom of the swing. The tension must be greater than his weight.

    Tarzan will feel like his weight is the tension in the rope, because he has to pull that hard to hold on. But what he feels is not his true weight.
    Last edited: Oct 20, 2006
  9. Oct 20, 2006 #8

    Doc Al

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    Huh? You're supposed to solve for the tension, not just assume it to be something. (The tension doesn't equal Tarzan's weight--if it did, he'd be in equilibrium.)
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