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If you apply the same force to both sides of a spring, how much does each side compress?

  1. Sep 23, 2014 #1
    If you apply a force of 5N to one side of a spring, and 5N to the other side, then the spring will compress according to F=kX. If K=1, then X=5m.

    This is similar to what happens when you apply a force of 5N to a spring attached to a wall, right? the wall applies an equal and opposite force back since the spring is not accelerating.

    But if you apply the force on your own to both sides, how do you know which side compresses? And does the same result occur? Whereby you only consider the compression of one side with the 5N and ignore the 5N applied to the other side (since its used to prevent translational motion)?
     
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  3. Sep 23, 2014 #2

    Simon Bridge

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    In this question, the center of the spring remains stationary - so treat it as two springs, half-length, attached to a wall.
    Do you know how the value of k changes with the (instretched) length of the spring?

    You can also do it by comparison with the "attached to a wall" situation - to get the end compression you need to keep track of the end position and the middle position.
     
  4. Sep 24, 2014 #3
    So if i treat it as 2 springs, of half length, then the spring constant is double since the spring is cut in half to form the two springs.
    F=KX

    k=2 (double the original).

    5=2X

    x=2.5meters

    So each spring compresses 2.5 meters.

    Total compression would be 5 meters (2.5m from each side). How does this compare to the wall situation? I understand that in the wall situation the spring compresses 5M total as well, but its only compressing from one side isnt it? In the case with the wall, the middle position of the spring moves as it gets compressed, while in the case where the spring is in the air and you push from both sides, the middle point stays constant.

    Force wise they're both the same but the mid point of the spring and compression (from which side it compresses) seem different.
     
  5. Sep 24, 2014 #4

    A.T.

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    A spring compresses along it's
    entire length.

    And that's all that matters according to Hooke's law.


    Why should the compression depend on where the spring is?
     
  6. Sep 24, 2014 #5
    I see. So when we say a spring compresses by X, it's the entire spring that compresses rather than it compressing from the end from where it's compressed (sorry if that's confusing).

    So when a spring compresses against a wall, is the same thing happening? When we say the spring compresses by 5 meters, it's compressing by 2.5 meters on each half, which is why the case is the same as if you apply 5 Newtons to both sides in the air (2.5 meters from each side and since it compresses along its entire length, the compression is the same as if you held the spring up against the wall).
     
  7. Sep 24, 2014 #6

    olivermsun

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    Well, the force applied is the same so the compression is the same.
    The only difference is that the wall doesn't move to meet you, so you have to move the "free" end twice as far if you want to keep applying that force.
     
  8. Sep 24, 2014 #7

    A.T.

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    Right, X is the change in length, not how much one of the ends moved. It doesn't matter if that change in length is realized by moving one end or both. Same length change means you have the same force.
     
  9. Sep 24, 2014 #8

    Simon Bridge

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    The "spring" in Hook's Law compresses uniformly along it's entire length no matter what.
    Real-life springs can be more complicated - having more than one kind of force acting on them.

    The spring cannot tell how it is being compressed, only that it is compressed.
     
  10. Sep 24, 2014 #9

    Nugatory

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    This may be redundant at this point, but here's an example:

    If I apply 3N to the left-hand end of a spring and 5N to the right-hand end.... The spring will compress exactly as if it were being forced against a wall by a force of 3N on one end, and it accelerate to the left according to ##F=ma## with a force of 2N (2N be the net force when you have 3N working on side against 5N on the other).
     
  11. Oct 2, 2014 #10
    Hello,

    This is true if the spring has mass .

    But if the spring is massless ,the forces on the two ends should be equal .So,what would happen if unequal forces 3N and 5N are applied on a massless spring ?
     
  12. Oct 2, 2014 #11

    Nugatory

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    Any time that you see a "massless" object in a physics problem, you have to assume that the forces on it are exactly balanced. Otherwise you'd get an infinite acceleration and other impossible results.

    Of course there's no such thing as a massless spring in the real world. However, pretending that there could be simplifies many problems.
     
  13. Oct 2, 2014 #12
    Right.

    But the question still remains - What would happen if unequal forces 3N and 5N are applied simultaneously to the two ends of an ideal massless spring ?

    Will it be 3N or 5N across the spring or some other condition holds?
     
  14. Oct 2, 2014 #13

    Nugatory

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    The spring would move in the direction of the 2N net force with infinite acceleration. :)

    OK, that's not a very sensible answer - so we conclude that this is not one of those problems that is made simpler by assuming an ideal massless spring.

    But suppose that we instead give the spring a very small but non-zero mass, something like .002 kg which is about the weight of a small paperclip. Now if we try pushing on one side of it with a force of 3N and other side with a force of 5N, we will find that a second later the spring is moving at a kilometer per second.

    Now where are you going to find devices that can apply a constant force of a few newtons to an object that is moving away from them at speeds of many kilometers per second? It's not easy to push on something that won't stick around to be pushed.

    So in practice what happens? You get an enormous acceleration for a moment and then something breaks, the spring goes flying away like a bullet, the forces are no longer acting on the spring, and a fraction of a second later we're solving a different physics problem.
     
  15. Oct 2, 2014 #14
    Thank you :) . Your explanation is very nice .

    Could you reflect more on this .
     
  16. Oct 2, 2014 #15

    Nugatory

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    Imagine a massless bead at each end of the spring. Look at the forces acting on those beads (and remember that the net force on the beads has to be zero because they're massless) in both cases, and you'll see it works.
     
  17. Oct 2, 2014 #16

    olivermsun

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    I'm a little confused myself by the massless bead example. Are you using them to demonstrate that the change in force has to be applied across the spring (since there can be no forces on the ends)?
     
  18. Oct 2, 2014 #17

    Nugatory

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    The question was why does applying a force of 5N to one and and 3N to other end produce the same compression as applying 3N to one end and 5N to other. Draw the force diagrams on these hypothetical beads for the two cases and it will (I hope) become more intuitively clear why this should be so.
     
  19. Oct 2, 2014 #18

    olivermsun

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    I'm not the OP; I'm just asking what motivated that example.
     
  20. Oct 2, 2014 #19
    Okay you lost me a bit in considering it as a spring with mass.

    If it's a massless spring and you apply 3 newtons from one side and 5 newtons from the other, then the compression and direction would be determined by the net force.

    Then if it's against a wall, is it the same thing as applying 10 newtons from one side and 5 newtons from the other side? The net force would be the same as if you applied 5 newtons to a spring against a wall wouldnt it?
     
  21. Oct 2, 2014 #20

    olivermsun

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    If the spring is assumed massless, then I don't really see how it makes sense to say 5 N is applied from one side and only 3 N from the other. As has been discussed above, the interpretation would be that the spring is accelerates to one side due to the 2 N net force… but the mass is zero so it makes no sense…

    If it's up against a wall, then the wall is going to exert whatever force is necessary to oppose the other force (since it can't move). If you put 5 N into the spring at the free end, the spring will compress until it's able to transmit the force to the wall, which will push back with 5 N.

    It's kind of like the springs on your car. The car pushes down with F = mg, the springs compress until F = -kx, and at that point the ground is pushing back up with the opposite and equal force F = -mg (since it is able to support your car).
     
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