Balloons on a string - Lagrangians? The other day, at the convocation of university, there were some balloons tied to a long piece of string, that floated in the air, and as the speeches droned on I wondered what curve they formed. (yellow and blue balloons here:here )
So I thought that I would 'formalize' the problem like this. Suppose that there are two posts, separated horizontally by a distance of $D$, with a string of length $L$ (where $L > D$) connecting the two posts, and $n$ identical balloons evenly spaced along the string of length $L$. 
In this, the blue dots represent the balloons, the red represents the string of length $L$. The way I tried to tackle the problem was through Lagrangian mechanics, with the degrees of freedom being the angles from one balloon to the next balloon, with a couple of constraints on the angles (such that the string length total is $L$, etc...) but I don't really have any experience doing Lagrangians, so I got stuck. 
 A: The answer will depend on the volume of the balloons, temperature of air etc. But you can think of it like this. In equilibrium, each of the balloon is fixed in space. So imagine the situation to be a rope tethered at 2 pegs on a vertical wall on (the end points) with n pegs in between these 2, over which the rope passes. Remember , all of this is hanging in a vertical wal, just supported at equal imtervals by pegs. But you already know that the shape between any two pegs will be an unsymetrical catenary. And this will be true for any 2 consecutive pairs of pegs. You can calculate the positions of the balloons in equilibrium using tension in the string, and free body diagrams(but you will need ballon volume and temperature of gases inside and outside). This is just an idea. Try it out. It should work.
A: In the limit of an infinite number of infinitesimally spaced balloons, the balloons are effectively applying a constant buoyant force per unit arc length of the curve.  This would therefore be exactly the same as the problem of a heavy rope/chain hanging under its own weight, but with the effective direction of $\vec{g}$ reversed.  In other words, the curve should be an inverted catenary.
If you're curious as to how to solve this with calculus of variations (not Lagrangians, strictly speaking), you can find any number of derivations that use various techniques to derive this result.  Here's a derivation of the catenary using calculus of variations, for example.
