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I Understanding the force on an item placed on a spring

  1. Aug 1, 2017 #1
    I hope you enjoy the image I made to facilitate my query. I wanted to add Thor as my force input but don't want any copyright infringement..

    If I have a compression spring with a spring rate of K, on a fixed surface with an apple on top and a hammer.... If hammer applies force N, what does the apple see prior to the max deflection of the spring?

    What if instead of an apple I had put a solid steel cube or something that won't easily yield? Does the cube only get some of the force after the spring has bottomed out?

    GDHuSiF.png
     
  2. jcsd
  3. Aug 1, 2017 #2

    phinds

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    Given the size of the spring in your illustration, what the apple sees is SQUISH !!!
     
  4. Aug 1, 2017 #3
    Yes, I regret the apple.... For the sake of the exercise lets say it's a titanium apple expertly painted....
     
  5. Aug 1, 2017 #4

    phinds

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    Ah, well in that case you're getting into actual physics, and I'm no good at that. Ask @Drakkith. He knows everything. Well, he says he does :smile:
     
  6. Aug 1, 2017 #5

    Drakkith

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    If we assume a perfectly rigid object (apple, cube, or whatever), the force on the object from the hammer is the same as the force on the spring from the object. The force applied by the hammer will generally increase as the spring compresses, and I believe it reaches a maximum at the point of maximum deflection. That appears to follow from the spring equation since the force increases as the distance the spring is compressed increases.
     
  7. Aug 1, 2017 #6

    Drakkith

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    Nonsense. I don't know where you get your information from, sir. (there's a joke in there somewhere):-p
     
  8. Aug 1, 2017 #7
    Maybe my question should have been asked in terms of energy, I'm actually so bad at physics I can't even ask a straight question.

    As phinds pointed out, the apple will squish if it wasn't perfectly rigid. So in that case the apple is damping the force of the hammer on the spring? I guess the squish-point would be where the spring has depressed to the point where kx is now greater than the yield point of the apple?

    I'll try and ask a better question:

    Pertaining again to the picture, if a certain amount of energy is spend applying the force, how is that energy distributed and in what order?
     
  9. Aug 1, 2017 #8

    Drakkith

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    In a simplified sense, yes.

    For a rigid object on the spring, all of the energy is spent compressing the spring. For non-rigid objects, some energy is spent deforming the object. There's no order in which these occur. They occur at the same time.
     
  10. Aug 1, 2017 #9
    Aha, now we have arrived at my question!

    In super slow-mo...wouldn't we see the hammer hit the apple, the hammer-and-spring depress until kx > yield of apple, then apple explode?
     
  11. Aug 1, 2017 #10

    phinds

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    Not exactly. At the very least, you have to take the inertia of the apple into account. If you are using the hammer to just press on the apple instead of striking it, then you can ignore the inertia of the apple and you would be right, more or less. For the hammer hitting the apple, the apple would start to deform before it transmitted any energy to the spring. The apple would "explode" only if you hit it very rapidly in which case that would happen pretty much before ANY energy had been transmitted to the spring.
     
  12. Aug 2, 2017 #11

    jbriggs444

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    Nonsense. The force on the object from the hammer is the same as the force on the hammer from the object. That is all that can be said.

    "perfectly rigid" turns out to be a very poor way to specify the relevant behavior of the apple. It leads to infinities and does not predict the outcome of the collision. If you add "perfectly elastic" to the description of the collision then we can predict the outcome in the two-body case.

    [Multi-body simultaneous collisions of rigid objects (e.g. Newton's cradle) turn out to still be ambiguous]

    Now then.... If the collision is elastic and the apple is rigid then we can predict the motion of the apple if given the mass and velocity of the hammer and the mass of the apple. One can proceed to see what happens to the spring.
    Again, this is nonsense.
     
  13. Aug 2, 2017 #12

    Drakkith

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    I'm not sure I understand your objection, jbriggs. What about my explanation contradicts this?

    Can you elaborate? I can't see anywhere else that the energy of the hammer is spent. Just saying that my answer is nonsense does not help me understand why.
     
  14. Aug 2, 2017 #13

    jbriggs444

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    You had written:
    That is utterly incorrect. It would apply to a massless object, not to a rigid object.
    It is nonsense for two reasons.

    First, there is a difference between "rigid" and "elastic". Rigid means that it retains its shape. Elastic means that collisions preserve energy. The two are not equivalent. In particular, in a collision between two rigid objects one has an infinite force over a zero displacement. Mathematics ceases to be predictive in this realm. One can have a collision between rigid objects that fails to preserve kinetic energy in spite of the fact that no deformation has taken place. [Think irresistable force and immovable object]

    Second, there is a difference between "rigid" and "massless". A rigid object has momentum. The net force on a massive object that accelerates is not zero.
     
    Last edited: Aug 2, 2017
  15. Aug 2, 2017 #14
    Maybe I should slow things down to further hone my question.

    Let's please consider the the force a slow compression with a finite amount of energy (now I regret the hammer shoulda used a linear actuator). Then lets say it takes 10lbf of force against the apple for it to deform to the point of yielding (kind of an ambiguous hypothetical for an apple but I hope you get my drift) Further, let's say it takes 10lbs of force to depress the spring to max displacement.

    Let's say I have enough battery to apply 10 lbf. Now we have apple and spring assembly compressed with 10lbf.

    Intuitively I see a spring partially depressed and an intact apple. Assuming I don't have a fool's inituition (big assumption), my question remains the same, what determines how energy has been distributed. The energy in my battery has done some work to move the actuator, the apple has absorbed some energy in it's deformation, the spring has converted some to potential. Of course, tons of loss everywhere.
     
  16. Aug 2, 2017 #15
    Are you giving the apple a high speed rap with the hammer, or are you applying the loading very gradually?
     
  17. Aug 2, 2017 #16

    phinds

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    His most recent post says "slow compression"
     
  18. Aug 2, 2017 #17
    Ash. That makes things a lot easier.
     
  19. Aug 2, 2017 #18

    jbriggs444

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    Yes, for slow compression I can rescind my criticism of the characterization by @Drakkith and agree that a [nearly] rigid object that is [very] slowly accelerated will pick up negligible energy from both deformation and motion so that we can ignore it entirely.
     
  20. Aug 2, 2017 #19

    Drakkith

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    Okay. My mistake then.
     
  21. Aug 3, 2017 #20
    Can anyone input on the extension of this question, back to a non-rigid apple...

     
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