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Finding Tension in a Rope

  1. Jan 6, 2008 #1
    Tarzan, who weights 688 N, swings from a cliff at the end of a convenient vine that is 18 m long. From the top of the cliff to the bottom of the swing, he descends 3.2 m.

    How much work is done by the rope during Tarzan's swing?
    You would probably use the equation U = mgh and k = 1/2mv^2 so maybe W = u + k.

    What's the tension in the rope when he reaches the bottom of his swing assuming he started from rest. Answer Ft = 932.62 N.

    Can someone get me started, I have no clue what to do?
     
    Last edited: Jan 6, 2008
  2. jcsd
  3. Jan 6, 2008 #2
    For your second question, consider his motion as part of circular motion aka he is experiencing centripetal acceleration at the lowest point.
     
  4. Jan 6, 2008 #3

    Dick

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    For your first question, the force exerted by the rope is always perpendicular to the direction of Tarzan's motion. What does this tell you about the amount of work it does on him?
     
  5. Jan 6, 2008 #4
    trying...
     
    Last edited: Jan 6, 2008
  6. Jan 6, 2008 #5

    Dick

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    The tension in the rope has balance Tarzan's weight AND provide the force to produce the centripetal acceleration. BTW that formula is wrong, mgh isn't a force.
     
    Last edited: Jan 6, 2008
  7. Jan 6, 2008 #6
    Would that make it the tension force then? Also it asked for the work done by rope, is that the same as work done onto him?

    Thanks for the help because I managed to figure our the second question.
     
  8. Jan 6, 2008 #7

    Dick

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    A force does work on a body only if it has a component that points along the direction of motion of a body, right?
     
  9. Jan 6, 2008 #8
    So I'm guessing it's just Tarzan's weight then since it can't be the force to produce the centripetal acceleration?
     
    Last edited: Jan 6, 2008
  10. Jan 6, 2008 #9

    Dick

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    That's confusing. There are two forces in the problem, Tarzan's weight and the tension in the rope. Only one does work on Tarzan. Which one?
     
  11. Jan 6, 2008 #10
    I think it's the tension then.
     
  12. Jan 6, 2008 #11

    Dick

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    Wrong. Get your book out and look up how to find out how much work a force does on an object. This is often written as F.ds (dot product) or F*ds*cos(theta) where theta is the angle between the direction of displacement of the object and direction of the force. theta in this case is 90 degrees for the rope. The tension does no work.
     
  13. Jan 6, 2008 #12
    Ok then, I would use the equation W = Fxcos(theta).

    F = 688 N, x = 3.2 m

    W = 2201.6 N

    Is that right now?
     
  14. Jan 6, 2008 #13

    Dick

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    If you mean the work done by gravity, yes. Which is NOT the work done by the rope. Except the units of work are joules.
     
  15. Jan 6, 2008 #14
    I'm confused.... Do I calculate using the centripetal acceleration instead of gravity then?
     
  16. Jan 6, 2008 #15

    Dick

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    I'm confused as well. Is this still the first question? If so, I've been trying to hint (not very subtly) that the work done by the rope on Tarzan is ZERO. Big fat zero. The force it exerts on him is perpendicular to his direction of motion. In the formula W=F*ds*cos(theta), theta=90 degrees. cos(90 degrees)=0.
     
  17. Jan 6, 2008 #16
    What is it you want to calculate? The work done by the rope is 0. The work done by gravity is what you calculated. The work done by centripetal force is also 0 because it's always perpendicular the displacement of Tarzan.
     
  18. Jan 6, 2008 #17
    Work done by the rope. So that means that earth's gravity doesn't do any work on a satellite?
     
  19. Jan 6, 2008 #18

    Dick

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    If the satellite is in a circular orbit, no. Because it's kinetic energy doesn't change. If the orbit is not circular, then gravity does work on it while it approaches the earth (increasing kinetic energy) and negative work on it while it recedes (decreasing kinetic energy).
     
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