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Springs / restoring force

  1. Jun 11, 2014 #1
    Let's say we have a spring with 10 coils and a spring constant of k=100 and compress from the right it a distance of 1 cm, we have to apply force F = k*x = 100*0.01 = 1 N. But if we then compress it from both left and right, and apply half the force from the first case to each side, shouldn't that give an equal displacement?

    I understand it doesn't because of the restoring force of the spring which in the second case will be in the same directions as the forces applied. But I don't understand exacly how the restoring force is related to the forces applied. The spring constant must be the same in each case, right? So only the force applied and the displacement must change. So then how can we find out how much the string is displaced from equilibrium in the second case?

    (Btw this is not a homework question, I just made this example up in order to understand the principle behind how springs work.)
     
  2. jcsd
  3. Jun 11, 2014 #2

    jbriggs444

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    When you apply this 1 N force, what holds the spring in place?
     
  4. Jun 11, 2014 #3
    Ah, I forgot to say that. I meant that in the first case the left end of the spring is attached to a wall.
     
  5. Jun 11, 2014 #4

    Nugatory

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    And what is the magnitude and direction of the force that the wall applies to the spring? Answer that question and you'll see the answer to the questions in your first post.
     
  6. Jun 11, 2014 #5

    Doc Al

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    The spring is compressed by forces at both sides in each case. In the first case, the wall provides one of the forces. To compress the spring by x, a force of k*x must be applied to each end.
     
  7. Jun 11, 2014 #6
    The wall would also make F = 1 N on the spring from the left? So actually then the spring will be exposed to 2 N in the first case?
     
  8. Jun 11, 2014 #7

    jbriggs444

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    The spring is under 1 N of compression throughout its length.

    The force that one end exerts on your hand is 1N. The force that your hand exerts on the end is 1N. If you were to break the spring in the middle (and keep it straight), the two pieces would be exerting 1N on each other. The force that the wall exerts on the spring is 1N. The force that the spring exerts on the wall is 1N.
     
  9. Jun 11, 2014 #8
    But if the spring is exposed to 1 N from each side, why isn't it exposed to 2 N?
     
  10. Jun 11, 2014 #9

    Doc Al

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    For one thing, those forces are equal and opposite. So they add to zero, not 2 N. (The net force on the spring is zero.)

    If a spring is compressed by a distance x, how much force must it exert? On which end does it exert that force?
     
  11. Jun 11, 2014 #10
    If the net force is zero then why is there any restoring force at all?

    Since it's being compressed from both ends it must exert a force towards both ends.

    But I don't understand, in the equation F=-k*x, F represents the force exerted by the spring. Does that equation refer to the force exerted in only one direction then?
     
  12. Jun 11, 2014 #11

    Doc Al

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    The net force is irrelevant. For example: If an elephant steps on a peanut, the net force on the peanut is zero. Does that mean the peanut will not be squashed? Of course not. The elephant and the floor exert forces that compress the peanut.

    To compress a spring with 10 N of force, thus creating 10 N of compression throughout the spring, you have to push each end with 10 N.

    Yes.

    It tells you the force exerted by the spring at each end.
     
  13. Jun 11, 2014 #12
    But let's say you hold each end of a spring with each hand. If you press only with your right hand with 1 N and hold the other hand fixed, there is 1 N being pressed on each side. Right? But what then if you press with both hands, 1 N from each hand? Then it's also 1 N from each side or? What's the difference?
     
  14. Jun 11, 2014 #13

    Nugatory

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    Forces only add when they're acting on the same thing.

    To compress a spring, you always have to push on both ends - otherwise, the spring won't compress, it will just accelerate as you push it away from you. So there are two ways of viewing a compressed spring, whether it's the wall or your left hand that applies the 1N force that counterbalances the 1N force you're applying with your right hand:

    1) 1N is applied to the spring from one side and 1N is applied to the spring from the otherside. They're both being applied to the same thing (the spring) and they sum to zero (opposite directions so we have ##-1+1=0##); the net force is zero so the spring as a whole doesn't move. However, various parts of the spring may move around until all the forces on each individual part also balance.

    2) 1N is applied to one end of the spring. This causes that end of the spring to accelerate, compressing the spring so that it starts to exert an opposing force on the end of the spring. When the spring compresses enough to generate 1N of resistance, the two forces on the end of the spring cancel and the acceleration stops....

    But wait a moment! If the rest of the spring is exerting a force of 1N on the end of the spring, then by the second law the end of the spring must also be exerting 1N on the rest of the spring. Why doesn't the rest of the spring accelerate under this force, moving away? The answer, of course, is that wall or your other hand is applying a 1N force in the other direction. When we reach equilibrium, every point in the spring is pushing against its neighbors with a force of 1N, and experiencing equal and opposite forces from its neighbors.
     
  15. Jun 11, 2014 #14
    That makes sense, although I still don't find it perfectly clear how these two situations are different:

    1. Pressing right hand with 1 N, holding left hand still.
    2. Pressing both hands with 1 N EACH.

    I understand both hands create pressure in both cases, but there must be some difference if one hand is only held still (like a wall) or if it ALSO presses. As I see it, in the second case the force being pressed on the string must be twice as strong as the first case. Am I right?
     
  16. Jun 11, 2014 #15

    Doc Al

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    While only one hand might move, both are pressing with 1 N.

    The only difference is that you held one hand still. Both are still pressing with 1 N of force. The spring compresses the same amount either way.
     
  17. Jun 11, 2014 #16

    Doc Al

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    You are wrong. In both cases each hand presses with 1 N of force once the spring is compressed.
     
  18. Jun 11, 2014 #17
    Then it is the same as pressing with one hand and the static wall in the other end. Les us call this: case A.

    Explained later.

    Let us call this: case B.


    The difference between case A and case B is "the timing of the forces".

    In case A, at first ( let say, at t=t_0 ) the force you apply with your right hand is greater than the force the wall apply at the other end, that is why the center of mass of the spring starts to move to the left. As soon as the spring starts deforming (compressing), the force it exerts on the wall (and so the force the wall exerts on the left end of the spring) is increasing. If at a given moment in time it gets to be equal to the force your right hand is applying, then at this very moment the spring stops compressing and its center of mass stops moving.

    If then you increase the force your right hand was applying to the right end of the spring, then this process starts again, because the force at the right end will be once again greater than the force at the left end of the spring, so the spring starts compressing again, the force the left end of the spring exerts on the wall (and so the force the wall exerts on the left end of the spring) is increasing again, until it reaches again a moment in time when the force the left end of the spring exerts on the wall (and so the force the wall exerts on the left end of the spring) gets to be equal to the force your right hand is applying, then again at this very moment the spring stops compressing and its center of mass stops moving.

    This is why in this case A, the center of mass of the spring is moving to the left.


    In case B, we suppose both hands are applying exactly the same force at the same moments of time. That is the difference (and that is why I said "the timing of the forces"). In this case, the center of mass of the spring does not move, because at any given moment of time, the total force (at the right end + at the left end, of the spring) adds to zero (their are equal in magnitude but of opposite sense).

    That was not true in case A. In case A, at every moment of time when the spring is compressing (and its center of mass is moving to the left), the force at the right end of the spring is greater than the force at the left end of the spring (that is why its center of mass is moving to the left). The force at the left end of the spring (the one the wall exerts on the left end of the spring) is increasing, "trying" to be equal to the force you are applying at the right end of the spring, but it is always "catching up", it takes some time for it to be equal to the force you are applying at the right end of the spring, (and "during this time of catching up" it is when the spring is compressing and its center of mass moving to the left). When it really gets to be equal (to the force you are applying at the right end of the spring), then the spring stops compressing (and its center of mass stops moving).

    That is the main difference between case A and case B.
     
  19. Jun 11, 2014 #18
    matt that makes sense also but I still can't determine from this whether the force that is exerted on the spring is different in each case.

    This is how I think of it:
    image.png

    In case 1 force is applied only from one hand, while in case two from two hands. It's easy to see that in case 2 the displacement of the spring is twice as large, and so the force must be twice as large also, since it's the same spring in both cases, it has the same spring constant. What is wrong about this picture?
     
  20. Jun 11, 2014 #19

    Doc Al

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    This seems different that what you were describing. (I don't know why you have a box and two springs. Can we assume you are still thinking about a single spring?)

    In case 2 you have a total compression of 2x, so of course the force you must exert (on both ends!) must be twice as much. (In your previous posts, you claimed that the same force was exerted in case 2, which is not what you are describing here.)
     
    Last edited: Jun 11, 2014
  21. Jun 11, 2014 #20
    No.

    If in case 1 your right hand apply 1 N, and in case 2 both hands apply 1 N, then in your picture of case 2 the distance at each end will be x/2.

    Why? Well, I tried to explain it in my previous post.

    In both cases, the process of "compressing" (and center of mass moving to the left) in case 1, and "compressing" (and center of mass not moving) in case 2, end exactly the moment in time when the net force on any particle of the spring is zero. And this situation is reached exactly when the distance between left and right ends of the spring has varied a certain amount (you called it x ). And that is because the system is conservative.

    Read again my last post and try to visualize what is really happening to each particle of the spring (you can even visualize it better if you imagine the spring is made up of only three particles, 1 2 and 3, separated some distance, and the force between 1 and 2, and the force between 2 and 3, depends only on the distance between 1 and 2, and the distance between 2 and 3 ) both in case 1 and in case 2.

    I would gladly put some pictures (with a spring made up of only three particle) with the forces and sure you would understand it much better, but I don't know how to put or make pictures yet.

    The thing is that, both in case 1 and case 2, it all ends when (imagine just three particles making up the spring) the force particle 1 exerts on particle 2 (and 2 on 1, and 2 on 3, and 3 on 2 ) is exactly 1 N, and THIS happens (regardless of how it got there, which is different in case 1 and case 2, but it does not matter) exactly when the distance between particle 1 and 2, and the distance between particle 2 and 3, has changed a certain amount (which will be the same in case 1 and case 2 ).
     
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