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Force exerted by spring

  1. Feb 19, 2015 #1
    So if I have a block of mass connected to a spring with spring constant k. If I pull the spring with force F, what is the force that is pulling the block? Is it F or the restoring force of the spring? I think it should be the restoring force of the spring, but if it is F, why is it so?
     
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  3. Feb 19, 2015 #2

    Stephen Tashi

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    Are you using the symbol "F" to represent a scalar? Or is "F" a vector?
     
  4. Feb 19, 2015 #3
    F is a vector. It is the force exerted.
     
  5. Feb 19, 2015 #4

    Stephen Tashi

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    The end of the spring attached to the block exerts a force on the block. If you're pulling on the other end of the spring, you're exerting a force on the spring, not on the block. Whether you call the force exerted by the spring on the block the "restoring force" of the spring depends on what you mean by a "restoring force". It might be that the force exerted by the spring on the block is equal to F as a vector. That depends on theoretical considerations, such as whether the spring is "massless".
     
  6. Feb 19, 2015 #5
    Consider 2 ends of a massless spring, where end A is attached to a block, and end B is being pulled by a force with magnitude F. The spring is streched X m.The magnitude of thd restoring force of the spring at end B, F(re)=/=F.What is the magnitude of the force the spring exerts on the block, at end A? Is it F, or F(re)?
     
  7. Feb 19, 2015 #6

    Stephen Tashi

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    You didn't define F(re).

    If you're imagining a steady state situation where the block and spring have constant acceleration A then the force of one the end of the spring on the block is equal, as a vector, to the force you are exerting on the other end of the spring. ( Considering the spring as a free body, there is zero net force on it. Since it is massless, if there were a net force on it, it would have infinite acceleration.)

    Perhaps you mean to ask if Hooke's law, F = k delta_x, applies to the spring. If this is a steady state situation, I'd say the spring is stretched to a length delta_x that makes Hooke's law hold.
     
  8. Feb 22, 2015 #7
    I've thought of this question myself before, and I think I know what's going on in your head. The answer I came up with may surprise you: You cannot pull on a massless spring at rest length. If a spring is at rest, then F= -kx= 0. So the spring exerts no force. Imagine a massless spring hanging vertically from the ceiling. Since the spring itself is massless, there must be some mass attached to its free end for you to be able to lengthen it from rest. If there is no such mass, you can never exerct "action" on your spring to lengthen it, since by newton's third law, your action force would equal to the zero reaction force of the spring at rest.

    As you can see, the massless spring is only an idealization, a model. It only makes sense to talk about a masless spring attached to some massive object. Now in this case, even if the spring is at rest length, you can still lengthen it by applying a force directly to the OBJECT. This way, initially the only force acting on the object is that exerted by your hand (or gravity if you let it hang), and the force in the spring will be initially zero since you are starting from rest length. So you have some non zero net force acting on your massive object which, according to Newton's second law provides acceleration to the object. Therefore it will start moving, and as soon as it displaces, the spring will start exerting force equal to -kx on the other end of the block, and also on the ceiling or wall to which it is attached.

    So one way to resolve this seemingly perpetual paradox was to add a massive object to the free end and pull on it. Another way is to be realistic about the spring, and assume it has mass. So if there's no block, then the spring has some mass, and now you can lengthen it from rest, since you can hold its massive tip in your fingers and pull on that mass ;)
     
  9. Feb 22, 2015 #8
    If the spring is already lengthened, then assuming it is mass-less, the forces on the two sides of the spring will be equal, regardless of where it is attached (wall or ceiling or block or your hand, doesn't matter). So the forces on the two ends of the spring would each equal -kx, where x is the change in length of the spring from rest.

    If you conduct the hook experiment to find the force of the spring, say by letting it hang from the ceiling and then connecting various blocks to it, You will find that F= - kx, where F is the force exerted by one side of the spring, doesn't matter which side.
     
  10. Feb 22, 2015 #9
    Thx a lot! I understand now :)
     
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