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

Heirot

## 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

Homework Helper
Hi Heirot! It will be in equilibrium when its total potential energy (gravitational plus spring) is a minimum. (can you see why?)

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

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 … ? Heirot
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?

Homework Helper
Now use Hooke's law

Heirot
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?

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

Heirot
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.

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

Heirot
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:
Yup! … gravitational force = elastic force (so net force = 0) … Mg = (L' - L)k. 