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Rope tension?

  1. Apr 18, 2009 #1
    I'm a bit confused here with determining the tension of a rope. A very simple example will do in this case. Lets consider a weight of G hanging from a weightless rope which is tied to a ring hanging from the ceiling. The weight pulls down on the rope with the force of G and the ring pulls up on the rope cancelling the effect of the weight i.e. supporting the rope-weight system.

    Now to me it would seem logical that as a result of these TWO forces, the overall force experienced by the rope is 2xG. But is that actually correct? Thanks for your help:)
     
  2. jcsd
  3. Apr 18, 2009 #2

    Doc Al

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    Actually, the resultant of those two forces on the rope is ZERO. The net force on the rope is zero.

    Of course, what you are really asking is: Shouldn't the rope tension be 2xG, no just G? No. The only way to create a tension in the rope is to pull both ends with the same force. The tension is the force that the rope is exerting. It exerts the same force G (not 2xG) at each end.
     
  4. Apr 18, 2009 #3
    Still don't get it:( I drew a FBD of the rope. If it¨s correct, then howcome one force is neglected in the actual calculation?
     

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  5. Apr 18, 2009 #4

    Doc Al

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    I don't understand. What force is neglected in what calculation? Your FBD shows two forces: N (up) and G (down). They better add to zero.

    If you want to consider rope tension, draw a FBD of the hanging mass. Two forces act on it: The rope tension T (up) and its weight G (down). The mass is in equilibrium, so the net force on it must be zero, thus the rope tension must equal the weight (not twice the weight).
     
  6. Apr 18, 2009 #5
    The part I don't understand is: Why does the pull of the ceiling/ring not contribute to the tension of the rope?
     
  7. Apr 18, 2009 #6

    Doc Al

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    The pull from the ceiling/ring contributes to the rope tension in exactly the same way that the pull from the weight does. There must be an equal pull at both ends to create tension in the rope.

    If you cut the ring so that the ceiling no longer exerts a force on the top end of the rope, then there will be no tension in the rope. It just falls.

    Here's a related example that often gives folks trouble. Compare these two scenarios:
    (1) Attach a rope to a wall and pull the free end with force F. What's the tension in the rope?
    (2) Two people hold opposite ends of a rope and both pull with force F. What's the tension in the rope?
     
  8. Apr 18, 2009 #7
    Thanks for trying!:) I've already considered both of the examples that you mentioned but I just think the result should again be two "pulls" because their are two "pulls" acting on the rope. But hey, maybe some rainy day I'll finally get it...
     
  9. Apr 18, 2009 #8

    Doc Al

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    Did you draw yourself a FBD of the hanging mass, as I suggested in post #4?
     
  10. Apr 18, 2009 #9
    Yep, and I understand the bottom part of the rope. The problem is the top...
     
  11. Apr 18, 2009 #10

    Hurkyl

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    Have you considered going back to the definition of tension? Maybe the idea you have in your head is the wrong one?
     
  12. Apr 18, 2009 #11

    Doc Al

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    Well, there's only one tension throughout the rope--the same everywhere.

    To analyze the forces at the top, consider the "rope + hanging mass" as a single system. What forces act on it?

    By the way, Hurkyl is hinting at an important point. The "tension" in the rope is a condition of the rope (putting all parts of it under stress) when it is is subjected to a pull at each end.
     
  13. Apr 18, 2009 #12
    They've promissed rain for next week. i think I'll just wait for that;)
     
  14. Apr 18, 2009 #13

    Doc Al

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    Up to you, but the "analysis" needed to understand forces at the top of the rope should take you about 1 second of work. :wink:
     
  15. Apr 18, 2009 #14
    Here's an idea that comes from the Light And Matter text by Crowell...

    Suppose you tie a goat to a stake in the ground and he pulls as hard as he can. The tension in that rope is one "goat" of force.

    Now replace the stake in the ground with another goat identical to the first. We're inclined to think that the tension must be two "goats" of force. But that's not right, because the rope doesn't know what's the origin of the force holding it there. It's still just one goat's worth of tension. It could be the stake in the ground, or the second goat, or a big magnet. It doesn't matter to the rope.

    Crowell has a nice discussion of this if you google for "light and matter."
     
  16. Apr 18, 2009 #15

    Doc Al

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    That example is equivalent to the one I tried to give in post #6. (But I'm happy to plug Crowell's books--I recommend them. :smile:)
     
  17. Apr 18, 2009 #16
    Your question pertains to the subject of "Statics"; the study of structures in static (non-moving) equilibrium. By the nature of your problem, no component of the assembly is moving. Good - so let's study it using the principles of statics.

    One can take a slice (called a section) anywhere on the structure (say, the rope), and consider the forces on each side of the section. Because the structure is static, the forces will ALWAYS be equal and opposite. (If they weren't equal and opposite then there would be a net differential. Such a differential would propel that section into movement, as in F=ma.

    What you had done in your opening statement was to sum the forces around both sides of the section; therein you got zero. Good, as it should be. When the sum of the forces at a section equal zero, then you have proven there is no net force remaining - thus the section is in static equilibrium. So, there must be no movement.

    The above could be merely an exercise in accounting. If a lender borrows $100 from a bank. Then the lender has +$100 while the bank has -$100; together this financial system has $0 money overall. It would be wrong to take the absolute values of both moneys and say the system has $200 in it.

    Note: if there is movement, then one has moved into the realm of "dynamics" - the study of structures in movement. In that case, there would be a net force once you have summed around a section.

    One could ask what is the point of determining the forces at a section for statics, if you know that the sum of the forces equal zero. Well, what is of interest is the magnitude of the force on one side of the section. That level of force is often what is of interest, such as when designing the necessary strength of the rope in supporting the intended weight below.
     
    Last edited: Apr 18, 2009
  18. Apr 18, 2009 #17
    You're really putting time into this guys! And I think I finally might be starting to accept the truth as it is:) The goats and the money really helped!
     
  19. Apr 18, 2009 #18
    Don't know if this will help, but here goes. Have you ever been fishing & weighed a fish with one of those spring-scales? You hold the top ring of the scale, and hang the fish from the bottom ring. The spring stretches & a pointer tells you how much the fish weighs.

    Now (for the ideal massless scale) whats the force on your hand pulling up? its the weight of the fish, right? Whats the force on the bottom ring of the scale? Its "one fish, pulling down" right? What your original question amounts to, is "why does the scale read '1 fish' instead of 2."
     
    Last edited: Apr 18, 2009
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