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Force vectors - leverage?

  1. Sep 12, 2008 #1
    Consider the following scenario.

    There is a train engine on a cliff above a plain. Imagine the engine is directly on the cliff edge. On the plain below, there is a carriage. Both tracks are flat and horizontal.

    The engine is connected directly to the carriage by a cable. This cable when taut obviously is oriented at some angle theta below horizontal.

    The engine now tries to pull the carriage. Its pulling force tensions the cable. Obviously the carriage will move. The engine applies force H.

    The engine is applying only a horizontal force. The carriage can only move horizontally. I appreciate that, in the cable, there will be this horizontal force, plus a vertical element necessary to make up the 'hypotenuse of the triangle'. So there will be a resultant lift force on the carriage, and an opposite downward force on the engine (which the cliff plateau opposes), given by H*tan(theta). Correct?

    The cable must therefore have a higher total tension than is actually being applied by the engine.

    Question: how do we refer do this force multiplication? Is it a form of leverage?

    P.S. The best real world example I can think of is pushing horizontally on the pole of a tent. It is relatively easy to pull out the peg/anchor. Just trying to figure out the language to cover this.
  2. jcsd
  3. Sep 13, 2008 #2


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    Well, the tent example is definitely leverage. Leverage has to do with angle, but distance is also important. Leverage has to do with the ability to turn something. So in the tent example, the angle at which you push is important - but so is how far from the peg you push. The further from the peg, the greater your ability to turn the peg with a small amount of force.

    I'm not sure how the train example is related to the tent example.
  4. Sep 13, 2008 #3
    It is precisely the same, think about it :)

    To transliterate the train example: the train engine is you pushing horizontally on the tent pole. The carriage on the plain below is the peg. The cable connecting the two is the peg rope. Where the vertical tent pole offers an opposing force upward, the cliff plateau does the same to the train engine. Got it?
  5. Sep 13, 2008 #4


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    In the tent example, the further from the peg you push, the easier it is for you to turn the peg out of the ground. But it doesn't seem like the carriage becomes easier to pull the further the train is from it.:confused:
  6. Sep 13, 2008 #5
    Consider the initial state only then, you are right that they only compare at that point. This is what I'm interested in, the train example seemed like the best and most easily visualized model.
  7. Sep 17, 2008 #6
    Too hard for all you smart chimps huh?
  8. Sep 17, 2008 #7
    Hmm. I disagree. Think of the extremes:

    If the rope is very long such that the cable is nearly horizontal, it's almost as if the train and carriage were on the same plane. Disregard the weight of the cable of course (^:

    On the other hand, if the rope is only an inch longer than the height of the cliff, most of the force is applied to tension (pulling the train down/pulling the carriage up), and very little goes into horizontal motion.

    Or am I missing something?
  9. Sep 18, 2008 #8
    I think you are missing something. The horizontal force in the line has to always equal the force applied by the engine, with the tension in the cable equal to that plus the vertical force resulting from the 'leverage' for lack of a better term.

    I will quote myself:

    Imagine a different scenario, whereby the line from the engine is attached to an immovable point on the track below. The engine can now no longer move; nonetheless it tries. It applies a horizontal force by way of its wheels on its track. On the plain below, the point on the track is subjected to a force from the angled cable.

    Clearly, this point must provide an equal and opposite force. 1) Horizontally - it has to be equal to that of the engine, otherwise it will move as the carriage would. 2) Vertically, clearly it must resist the force given by H*tan(theta), else it will pop out and fly upward!
  10. Sep 18, 2008 #9
    This is the same scenario by the way, I'm just trying to make it easier to visualize.

    Here's a picture, they always help!

    Attached Files:

  11. Sep 18, 2008 #10
    So the red force is that applied by the engine, and the blue force comes from the opposing support force of the plateau. Summed they give a tension in the cable greater than the red H alone.

    Is this leverage? What do we call it? Am I wrong completely?
  12. Sep 25, 2008 #11
    Aw now c'mon. I refuse to let this die.
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