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Rope tension problem

  1. Mar 8, 2004 #1
    http://abs-5.me.washington.edu/me230/hw_solut.pdf

    Could someone please look at this website. Specifically problem 14-54. Why is a force on the box 3F? I don't understand why it is 3F. I only see two ropes going from the box.
     
    Last edited: Mar 8, 2004
  2. jcsd
  3. Mar 8, 2004 #2

    NateTG

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    There's a couple of ways to look at this:

    You assume that bot of the pieces of rope have the same tension, when they do not.

    For example, the rope that goes from the crate to the center of a pulley can be considered to be a part of the crate. (The length of that rope never changes.) That will give you that there are three connections.

    Another thing you can use is 'mechanical advantage' which is based on work. If you check, you will see that for every three feet of rope that the motor pulls, the box moves one foot. This indicates that there is a mechanical advantage of three, so the force is three times as great.
     
  4. Mar 8, 2004 #3
    Thanks for your reply:)

    I get the three connections part, but the second part I don't get.

    "If you check, you will see that for every three feet of rope that the motor pulls, the box moves one foot."

    How would I check this?
     
  5. Mar 8, 2004 #4

    NateTG

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    You build a model?

    I had some stuff about movable pulleys in Jr. High school. It essentially comes down to three ropes (again) otherwise, I'm not sure.
     
  6. Mar 9, 2004 #5

    HallsofIvy

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    The easy way is to look at the picture. You will see that, at one point, a vertical line will cross the cable hauling the box three times: three ropes, mechanical advantage of 3.

    More detail: Let "a" be the distance between the two pulleys. The distance from the pulley fixed to the wall and the wall is irrelevant as is the distance from the pulley fixed to the crate and the crate. You can think of the those pulleys as attached directly to the appropriate surface. In fact, we can think of the motor as being right up to the wall so that the total length of the cable (the important part of it anyway!) is 3a. In order to move the crate a distance, d, a must be shortened by d also so that the rope must be shortened by 3d. The work done is the force necessary to move the crate times d: Fcd. The work done by the motor pulling the cable is the force of the motor times 3d: Fm(3d). Of course, those are equal: 3dFm= Fcd so
    Fm= Fc/3.
     
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