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Homework Help: Length of a spring under its own weight

  1. Jul 20, 2007 #1
    1. The problem statement, all variables and given/known data
    I was thinking over lunch today about this old problem from intro physics but I can't remember whether or not this is the correct solution.

    What is the final length L of a spring of mass M and spring constant k, initially of length L0 after it is left to hang under its own weight?

    2. Relevant equations

    3. The attempt at a solution
    Divide the unstretched spring into infinitesimal segments dl. Each of these segments will stretch an infinitesimal distance dx under the weight of an infinitesimal mass dm by the relation g dm = k dx. Let [tex]\lambda[/tex] be the linear density of the spring, so that [tex]dm = \lambda \,dl[/tex]. Then,

    [tex]g \lambda \,dl = k \,dx[/tex], and
    [tex]dx = \frac{g}{k}\lambda dl[/tex].

    Integrating LHS from 0 to X and RHS from 0 to L0, we find that [tex]X = \frac{g}{k}\lambda L_0[/tex], so the final length of the spring is L = L0 + X or

    [tex]L = L_0(1 + \frac{g}{k}\lambda) = \frac{g}{k}M + L_0[/tex].

    Is that right?
    Last edited: Jul 20, 2007
  2. jcsd
  3. Jul 20, 2007 #2


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    I'm no student of the calculus, but your solution is identical to the case where the spring is massless and its weight is concentrated at the far end. That doesn't sound right. I believe your extension X should be 1/2 of your calculated value, i.e. X =Mg/2k. Comments welcome.
  4. Jul 20, 2007 #3
    The com is at L/2, but only half of the spring is effectively pulling that mass up, so 2kx=mg, which gives you x=mg/2k.
  5. Jul 20, 2007 #4
    Yeah, now that I think about it, that should be completely wrong, since the segments at the top are under greater load than the ones at the bottom. I'll think about it again in a bit.
  6. Jul 20, 2007 #5
    I think youre forgetting that each element has more than just dm*g acting on it (the masses of the elements below it are acting on each element too).
  7. Jul 20, 2007 #6
    Right, that's what I mean. Okay, so including the fact that ALL the masses below a certain point contribute to the stretching of the segment at that point, we have

    [tex]k \,dx = \bigg( \frac{l}{L_0} \bigg) g \, dm[/tex]

    [tex]dx = \frac{gl}{L_0 k} \frac{M}{L_0} \, dl[/tex]

    [tex]dx = \frac{Mgl}{{L_0}^2 k} \, dl[/tex]

    [tex]x = \frac{Mg}{{L_0}^2 k} \int_0^{L_0} l \, dl[/tex]

    [tex]x = \frac{Mg}{{L_0}^2 k} \frac{{L_0}^2}{2}[/tex]

    [tex]x = \frac{Mg}{2k}[/tex]

    Ahh, thanks guys :smile:
    Last edited: Jul 20, 2007
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