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Really simple question about spring compression

  1. Mar 24, 2012 #1
    Assuming idealized springs under Hooke's Law:

    A spring 1m in length, mounted on the ground, compresses 10cm under its own weight.

    Next to it is a spring identical in material, construction, etc., but 2m long with twice the mass.

    If I understand correctly, doubled weight alone would suggest compression of 20cm. BUT, at twice the length, the spring constant (k) is halved. Thus, the expected compression is actually 40cm.

    Is this correct?
  2. jcsd
  3. Mar 24, 2012 #2


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    Staff: Mentor

    That's a good question, superalias!

    Instead of the 2m spring, I think we could substitute two springs identical to the first, carefully stacked one on top of the other (perhaps with a weightless board between them for the top spring to sit on).

    The top spring compresses 10cm under its own weight, as before. Now what's going on with the lower spring?
  4. Mar 24, 2012 #3
    Hello, and thank you! Yes, I tried to think of it that way too; they should be equivalent cases.

    So: With two springs, the top one compresses 10cm under its own weight. The bottom one compresses 10 cm under its own weight. The bottom one also compresses some under the top one's weight… but how much?

    I don't know the relationship between how much the spring compresses under its own weight vs under a weight on top, but I'd hope it's simple as this: The compression caused by its own weight is half the compression caused by the same weight riding on top, as it's the average of the latter compression and 0.

    If so, the bottom spring's self-compression of 10cm suggests it would compress 20cm by having that same weight on top instead.

    So: 10cm + 10cm + 20cm = 40cm.

    I suspect this is correct. :smile:

    But doesn't it suggest some odd results? Say the 1m spring instead compressed a big 60cm under its own weight (weak spring, high gravity?). The 2m spring would have to end up compressing 240cm – more than its length! (Similar results pop up just by assuming a stronger but long spring; say, lengthen the original spring from 1m to 12m, and its original 10cm compression now becomes something incalculable.)

    What explains this? Is it simply a matter of Hooke's Law being necessarily unworkable in those more extreme examples?
  5. Mar 24, 2012 #4
    According to my little knowledge about deformation, when a deformable bar is compressed, its diameter increases. For small deformations the increase in diameter may be ignored but in you example, the increase becomes important. As the bar becomes shorter, it becomes more stiff then , so the length doesn't become zero.
  6. Mar 24, 2012 #5


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    Homework Helper

    Hooke's law only applies when the material is deformed by a small amount, so that if the load is removed it will return to its original shape. What is "small" depends on the material.

    You can't physically compress a coil spring beyond the point where all the coils of the spring are touching each other - or at least, to do that would require a much larger force to squash the shape of the wire from being circular, compared with changing the spacing between the turns of the spring.

    If you stretch a coil spring a large amount (maybe to double its original length) you might permanently deform the spring so it doesn't return to its original length. Don't try this experiment unless it's OK for you to ruin the spring, of course!
  7. Mar 24, 2012 #6
    Yes, it makes sense that Hooke's Law will only apply over a limited distance, and strange calculation results would imply that it's no longer valid for the situation.

    Thanks for the replies!
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