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Why is the tension of a rope not doubled with one weight on each side?

  1. Jul 19, 2011 #1
    Suppose you put two weights at each end of a string or rope and then throw it away. As we all know the weights will circulate around the center of mass.

    Now the problem is what is the tension of the string.

    I am confused about this since there seem to be two contributions to the tension. First the first weight's centripetal force around the center of mass and secondly the second weight's centripetal force around the very same center of mass.

    But it turns out the two are equal and that you should only count one of them, not add both.

    My question is, what physical law says the forces should not be added? After all, at the center of mass, there will be two forces in each direction.
  2. jcsd
  3. Jul 19, 2011 #2


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    Staff: Mentor

    Let's say you weigh 150lb. You are pushing on the ground with 150lbthe of force and it is pushing back with 150lbthe of force. Now insert a scale in between your feet and the ground. Does it read 150 or 300?
  4. Jul 19, 2011 #3
    Ok, so basically you are referring to Newtons third law. That is the law I was asking for.

    If I pull a rope with a force 600 N and my buddy is pulling 600 N at the other end, he is not contributing to some extra tension in the rope, but if she is pulling with a force 700 N, then I need to apply extra strength or I will accelerate into the ground. And the dynamometer (or whatever it is called) will read 700 N, assuming I have a firm grip when falling.

    In my problem case there always is a firm grip between the two weights.

    I guess sometimes, or even very often, you have to get back to basics! Good thing there is a physics forum because in scool, university or before university, you don't have time to discuss.
  5. Jul 21, 2011 #4
    I would have answered your question with an example more along the lines of yours...you know, a rope spinning with weights...

    for example, let's say that you have a rope with a 10 lb weight on one end and the other end of the rope is tied to a post....and then, you throw the weight around...what is the tension in the rope? is it easier to visualize?...maybe?

    here is another example to help (confuse) you further...

    take the same rope from above with a 10lb weight on one end, remove it from the post and now put a 20 lb weight on that end and throw the rope and the two weight like you were talking originally...what is the tension on the rope? is it the tension of the centripetal force of the 10lb? the 20? the summation of the two? what is it?
  6. Jul 22, 2011 #5
    It is difficult to visualize, but easier to do the calculation. This is the very same problem I originally stated, so the answer is something I have already calculated. It is proportional to the reduced mass of the system.

    So if you throw the 10 lb weight (m1) with a speed v, and the rope is l units long, then I think the centripetal force will be (I am not at home right now and cannot check) [tex]\frac{v^{2}m_{1}m_{1}}{l(m_{1}+m_{2})}[/tex] This obviously the same for both the first and the second weight. But, and this is where it gets difficult, you only count one of them to get the tension of the rope. So you either take the tension of the centripetal force of the 10 lb weight or ditto of the 20 lb weight. This is true even if the second weight goes to infinity, which is the usual situation.

    All this is straightforward, but at the same time confusing. For example, if the second weight goes to infinity, then, if you stand in the middle of the rope, isnt it true that you only feel a pull from the first weight?
  7. Jul 22, 2011 #6


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    It's the definition of tension, if there is 100 Newtons of tension in a (non-accelerating) rope, then that rope exerts 100 Newtons of inwards force at each end of the rope. Tension is the result of net outward forces on an object, such as a rope. If the forces are equal and opposing, then there is no acceleration, and the tension is the same throughout the rope. If the forces are not equal and opposing then there is acceleration and the tension at any point on the rope will depend on the remaining mass aft (relative to the direction of acceleration) of that point, times the rate of acceleration.

    Compression works in the same way. If there is 100 Newtons of compression on a (non-accelerating) spring, then there is 100 Newtons of inwards force on each end of the spring.
  8. Jul 22, 2011 #7

    we are at the top of a building
    we put two identical weights at the ends of a rope and
    very skillfully, we give the entire system a good spin right around the center of mass
    and we let the system start falling to the ground

    as it is, the rope-weight system does not care that is falling, and for analysis purposes, we could consider that the system is stationary right in front of us and nicely spinning around the center of mass

    you follow so far?

    for as long as the center of mass is stationary...
    what if we magically insert a post and a bearing right in the middle of the rope?
    what if we cut one side of the rope and get rid of one weight?
    Do you think the remaining weight noticed what we did?
    What is the tension along the rope?
  9. Jul 23, 2011 #8


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    The tension in the rope, with the two masses spinning around, is just enough to keep each mass rotating in a circle (radius of twice the length of the rope). If you swung just one mass around a fixed point, attached with half the length of rope and at the same rate, then the tension would have to be exactly the same. The mass doesn't care what's providing the force - fixed point or another mass - it just keeps on the circular path with the same tension.

    Oh, I see I have said more or less the same thing. There you go.
  10. Jul 25, 2011 #9
    I assume you mean, by a post and bearing, a mass of infinite weight. Here are then my answers:

    The remaining weight did not notice what happened and the tension of the rope is [tex]T=mr\omega^{2}[/tex] where r is half the length of the rope. Right?
  11. Jul 25, 2011 #10
    that's correct.
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