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Spring Force question

  1. Oct 10, 2015 #1
    Hello,

    If we have a spring at rest and has a constant k=2 N/m (at its natural length with one end at origin (x=0) and the other end held stationary) is having a force applied on it. The force varies in same way as the the spring with function 2X. My question is how is it possible to start the spring moving if the spring is applying the same force as the force applied on it but in the opposite direction. In other words, booth forces are expressed by 2x how do they not cancel and how is it moving. Don't we need to apply a small force at first just to keep it moving.
     
  2. jcsd
  3. Oct 10, 2015 #2
    You must consider only forces acting on the spring. The forces do not cancel because they are acting on different objects. Your hand applies a force on the spring and the spring applies the same force on your hand
     
  4. Oct 10, 2015 #3
    But what if I apply a constant force of 10N would this continue for ever even if the spring stops my hand. Would the spring apply a force of 10N immediately after its contact with my hand at the origin( x=0)
     
  5. Oct 10, 2015 #4
    The spring would apply a force equal to the force your hand applies (Newton's 3rd Law).
     
  6. Oct 10, 2015 #5
    So the spring will apply 10 Newtons immediately, but what about f=kx. will this add to the 10 Newton when the end of the spring starts to change position.
     
  7. Oct 10, 2015 #6
    The "problem" is that you cannot do that. You cannot apply the 10 N force instantaneosly. The force will increase gradualy from 0 to 10 N as the spring expands.
     
  8. Oct 10, 2015 #7
    If x equals 0, in the formula, wouldn't F = 0? Since F=-kx and x=0, 2 x 0 = 0.
     
  9. Oct 11, 2015 #8
    x is actually the amount by which the spring is stretched (or compressed). If you apply a constant force on a spring, for example, by hanging the spring from the ceiling and hanging a fixe weight from the other end of the spring, then the spring will stretch until, kx is equal to the weight.
     
  10. Oct 11, 2015 #9
    I can see it more clearly now so the newton's third law force pairs of an object interacting with a spring(compressing or stretching ) always equal to XK. So in the vertically hung spring case, if the spring force equal the weight of the object that is attached to it will the object stop completely or will the spring stretch more to stop the object from going down?
     
  11. Oct 14, 2015 #10
    Actually can someone answer BlueberryPi's question? How can you apply any force at all on the spring if the spring can't apply any force on you when x=0? How can you provide the necessary force to compress or elongate the spring when the spring is at equilibrium?
     
  12. Oct 14, 2015 #11
    Well, you can then pose the same question about any object in equilibrium, not just the spring. The point is again, a careful look at Newton's third law. The question is not, can or can't the spring apply force. Whatever force is applied on the spring, irrespective of whether it is in equilibrium or not, the spring will apply an equal force right back. If a big man and a little child are holding hands and skating, the force by the man on the kid, is exactly, all the time, equal in magnitude and opposite in direction to the force by the kid on the man. The question, "can the kid exert that much force?" does not arise. Newton's third does not say how big or small the action reaction pair are. They are always equal, in magnitude.
     
  13. Oct 14, 2015 #12
    This is in reply to Ibraheem: The experiment is done like this.
    1. First you establish the equilibrium position. You hang the spring from the ceiling, attach the mass at the bottom, place your hand below the mass, and slowly lower it until you can remove your hand without the mass oscillating. In this position, the mass is not moving, and it is in equilibrium. So in this position, the upward force from the spring is exactly equal to the downward force of gravity, and both forces are acting on the mass.
    2. Second, you hold the mass and either lower it, or raise it just a little bit, vertically, thus either stretching or compressing the spring, and let go. The mass will now oscillate up and down. During oscillation, the mass is in equilibrium at the central position of the oscillation, which is the equilibrium position that you established in step 1. At that position, the the velocity of the mass is not zero, but the acceleration is zero because it is in equilibrium. At that position, every time the mass crosses that position, the spring force balances gravity. At other positions in the oscillation, the acceleration is not zero, the mass is not in equilibrium, and the spring force and gravity are not equal in magnitude.
     
  14. Oct 14, 2015 #13

    Right but if we look at Newton's 3rd Law from the reverse perspective. Then if the spring applies a force of 0 because its displacement is x=0, then you must also apply a force of 0 because the forces must be equal and opposite. So how can you ever get a nonzero displacement?
     
  15. Oct 14, 2015 #14
    By applying a force that increases from zero to some finite value.
     
  16. Oct 14, 2015 #15
    But for the spring to compress it must receive a nonzero force before it is compressed. But before it is compressed it can only receive a force of 0 because kx=0.
     
  17. Oct 14, 2015 #16
    The force and compression are simultaneous. There is no before and after.

    Same as to move you need to change your position. But if your original speed was zero you cannot change position so you cannot move. Are you saying that nothing can start moving? :)
     
  18. Oct 14, 2015 #17
    no..but the spring force case is different. If I compress the spring a distance x, then I put 1/2kx^2 J into the system. So I need to apply an average force of 1/2kx to do that. But how can I ever apply that force if I have to compress the spring by a infintesimal amount x before any force can be applied at all? I can only apply a force when the compression amount is nonzero. But for there to be a nonzero compression a force must be applied.
     
  19. Oct 14, 2015 #18

    sophiecentaur

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    Isn't the resulting motion SHM, impressed on a net mean acceleration? That would give a phase difference between the instantaneous velocity and the spring compression
    That would be a form of Zeno's paradox, I think.
     
  20. Oct 14, 2015 #19

    jbriggs444

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    For a massless spring, you do not need to apply any force to get it to move. But no real spring is massless so... For a spring with mass the force you apply to the spring is not always given by f=kx and the force that the spring applies to you is not always given by f=kx. The delta is given by f=ma.
     
  21. Oct 14, 2015 #20
    That example was not about the spring but just a simple motion.
    I was thinking about one of Zeno's paradoxes, yes.

    For the spring, the question was about force and displacement. They are in phase. Velocity has indeed a phase difference.
     
  22. Oct 14, 2015 #21
    So look at the process carefully, as an experiment that you are performing. You stretched the spring from equilibrium and let it go. Nothing prevents you from exerting a non-zero force to stretch the spring. The spring will just keep exerting an equal force on you. Newton's third law does not say that if an object is NOT exerting a force on you, you CANNOT exert a force on it. It only says that IF you exert a force on the object, THEN the object will exert an equal force on you. Certainly, if the OBJECT exerts a force on you, YOU will exert an equal force back on it.
     
  23. Oct 15, 2015 #22

    sophiecentaur

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    This question is the core of the problem with this thread. You cannot instantly apply a 10N force to the spring unless you displace it instantaneously by10k. You have to start with an infinitessimal reaction force. The intuitive response to the OP just takes you into seeing the problem as a paradox, when it isn't. The problem becomes the same as all proportional relationships (y=ax) and no one has a problem with what happens at the origin. (Even if they do, a course of Mathematical Analysis can put them straight.)
     
  24. Oct 15, 2015 #23

    sophiecentaur

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    Yes. People seem to be obsessed with Newton 3 at the moment and they are trying to apply it where it just doesn't apply,
     
  25. Oct 15, 2015 #24
    Is it that the spring force is only equal to kx when there is no net force acting on it? Because in order for me to compress or stretch the spring from its equilibrium position I must apply a nonzero force when it is at equilibrium. That means that the spring must also apply an equal nonzero force. So in that case the springs force is not kx, because it applies a force but x=0
     
  26. Oct 15, 2015 #25

    sophiecentaur

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    Whatever position the spring is in, if it is in equilibrium (i.e it could be loaded or not) then the force to extend it (or compress it) from that position is kx when the extension is measured from the start (equilibrium) position. That assumes the spring follows Hooke's Law.
    If you plot F against the length then you get a straight diagonal line with slope k. You can start where you like (pre-loaded) and you still move up and down the same diagonal line. So the answer to your question is no.
    Why don't you just read about this - for instance, this link? All will be made clear if you follow logically what the link says. It is not always made clear that there is nothing special about starting with an unloaded spring - but, if you think about it, any real spring has mass and will be compressing or stretching under its own weight. No one says you need to take that into account - you just plot the load / extension to find k.
    Why do you find this a problem? Any two objects interacting will have equal and opposite forces between them. This (as has been mentioned previously) is nothing to do with the k of a spring.
     
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