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Opposing forces

  1. Sep 5, 2006 #1
    I have a simple question to resolve a dispute at work.
    If I have two opposing forces connected by a rope, applying 50 lbs of force in opposite directions, do i have 50 lbs of force on the rope or do i have 100 lbs of force? (of course assuming no friction and what not)
     
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  3. Sep 5, 2006 #2

    berkeman

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    Hint -- If you tie the rope in a tree and hang a 50 pound weight on the bottom, what is the tension in the rope?
     
  4. Sep 5, 2006 #3
    i have two moving and opposing forces such as air cylinders...if i apply 50 lbs in both directions...what is the tension on the rope...isnt that the same as applying a 100 lb weight to one end of the rope in the tree?
     
  5. Sep 5, 2006 #4

    berkeman

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    In your examples, nothing is moving. You draw a force diagram at each end of the rope. There is an equal and opposite force from the rope in response to the applied force at the end. The tension in the rope is equal and opposite to the applied force.

    Think about your initial example. What is the difference from the rope's perspective whether there is a person on one end pulling against another person, versus just attaching the rope to a stationary wall?
     
  6. Sep 5, 2006 #5
    ok let me rephrase the question to make it more like my exact scenario...
    i have a cable with a connector that i need to try to pull apart. it takes 500 newtons to pull it apart. if i use two cylinders at either end of the cable to pull it. do they each have to apply 500 newtons or can i pull it apart with 250 newtons on each side?
     
  7. Sep 5, 2006 #6

    berkeman

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    You need to use a 500N cylinder at one end of the cable, and affix the other end to a stable support.

    Going back to the vertical analogy again -- Hang the cable from a stationary ceiling, and hang a 51kg mass at the bottom. This generates an F= ma = 500N force pulling down on the cable, which the cable opposes with 500N of tension. The tension is the same at the top of the cable, where the ceiling is "pulling" up with a force of 500N. The weight at the bottom is not accelerating, so the tension in the cable has to be the same as the 500N force of the weight being pulled down by gravity.
     
  8. Sep 5, 2006 #7
    in this situation both ends have to be attached to moving devices. i just was wondering if i needed to use cylinders that are each capable of applying 500N or if they would combine on the connector?
     
  9. Sep 5, 2006 #8
    well, now thinking about it, im not sure what is "the force working on the cable"...
    the tention is determined by the force the body trasfers from one side to another, so the answer would be 50 lb.

    btw, even if there was a force on only one side, i think that the tenstion would stay 50lb, since it would need to accelerate all the mass behind where the force is working.
     
  10. Sep 5, 2006 #9

    HallsofIvy

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    If you have 50 pounds force pulling one way and 50 pounds force pulling the other, that's 50 pounds tension in the rope.

    You should see that the scenario mentioned earlier, a 50 pound weight hanging off a cliff, supported by a rope tied to a tree, is exactly of that situation. The weight is pulling down with 50 pounds force, the tree is pulling up with 50 pounds force, and the tension in the rope is 50 pounds.
     
    Last edited: Sep 5, 2006
  11. Sep 5, 2006 #10
    Imagine cutting the rope. What force is required to keep the rope from accelerating? The same force!
     
  12. Sep 5, 2006 #11
    If you had 50lbs force pulling one end of the rope ,and 0lbs force pulling the other, you'd have zero tension in the rope.
     
  13. Sep 5, 2006 #12

    berkeman

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    ....actually a slight tension distribution -- because the whole rope then accelerates, it takes some force distribution (tension distribution) along the rope to accelerate it.

    Quiz question -- which end of the rope has the highest tension? The end being pulled or the open end?
     
  14. Sep 5, 2006 #13
    the end being pulled i belive.
    the body has the property of keeping the same acceleration.
    illustration:
    ----------X--------->
    ">" is where the force is working
    "X" is a point along the rope.
    at point X a force must pull all mass left to it.
    so the closer X is to the pulling point, the more mass it needs to accelerate, and by the 2nd law, it will exert more force, therefor more tension.
     
  15. Sep 5, 2006 #14

    berkeman

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    Very good anser TuviaDaCat. It's all about the F=ma mass that's being accelerated.
     
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