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

## 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.

Hooke's law.

## The Attempt at a Solution

Thanks

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tiny-tim
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Hi Heirot!

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

(can you see why?)

Yes, I can see that :) I just don't know what to minimize mathematically because the streching won't be uniform.

tiny-tim
Homework Helper
… 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 … ?

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?

tiny-tim
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Now use Hooke's law

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?

tiny-tim
Homework Helper
Do it for the (L-x) part of the spring

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.

tiny-tim
Homework Helper
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?

Can you please show me your solution so that I can precisely tell you where I think the problem lies?

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?

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tiny-tim