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Length of a massive elastic string hung by it's end

  • Thread starter Heirot
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  • #1
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Homework Statement



Given a homogenous string of mass M, length L and elasticity k, find the length L' when it's hung by one end in a constant gravity field.

Homework Equations



Hooke's law.

The Attempt at a Solution



I don't know how to apply Hooke's law in this situation. Please help.
Thanks
 

Answers and Replies

  • #2
tiny-tim
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Hi Heirot! :smile:

It will be in equilibrium when its total potential energy (gravitational plus spring) is a minimum. :wink:

(can you see why?)
 
  • #3
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Yes, I can see that :) I just don't know what to minimize mathematically because the streching won't be uniform.
 
  • #4
tiny-tim
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… the streching won't be uniform.
Yes it will …

consider a small section of spring from rest-length x to x + dx …

the external forces on it are T(x + dx) and -T(x) and Mgdx/L, so … ? :smile:
 
  • #5
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So dT = - Mg/L dx -> T(x) = Mg/L (L-x) if x is measured form the top. But this we already knew. How does this help in calculating the streaching?
 
  • #6
tiny-tim
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Now use Hooke's law
 
  • #7
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But spring constant k is a global property of the string, not local. Recall that if we cut the string in half, the new constant is 2k, not k. So, what would be the string constant for an infinitesimal part of the string?
 
  • #8
tiny-tim
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Do it for the (L-x) part of the spring
 
  • #9
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Well, all below x, we can replace by an external force Mg (L-x)/L. If the upper part of the string were massless, we would have Mg(L-x)/L= k L/x s, where s is the displacement. I can't figure out how to include the mass of the string.
 
  • #10
tiny-tim
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I'm confused … where does Hooke's law (which is only concerned with internal tension) come into that? …

and what does the mass have to do with it? :confused:
 
  • #11
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Can you please show me your solution so that I can precisely tell you where I think the problem lies?
 
  • #12
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After having thought about it for a week, still not being able to solve it, I was thinking about your approach. Would you say that L' = L + Mg/k?
 
Last edited:
  • #13
tiny-tim
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Yup! … gravitational force = elastic force (so net force = 0) … Mg = (L' - L)k. :smile:
 
  • #14
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It just occured to me now, if we divide the unstreched length L into equal parts of length dx then the spring constant for that part is k'=kL/dx. On the other hand, a change in the length dl(x) of dx at some point x is dl(x) = F(x)/k'. Integration gives L' = L + Mg/2k as is pretty much expected.

Thanks for your help, tiny-tim :)
 

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