Lagrangian of System

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1. Dec 3, 2016

Molar

1. The problem statement, all variables and given/known data
Uniform chain of length L is kept of a horizontal table in such a way that l of its length keeps hanging from the table. If the whole system is in equilibrium, find the Lagrangian of the system.

2. Relevant equations
Lagrangian of the system = Kinetic energy (T) - Potential energy (V)
T = (mv^2)/2
V = mgh

3. The attempt at a solution
In this case, V = 0 - mgl = -mgl
My question is what will be the kinetic energy of the system ?

I think as the system is in equilibrium, there is no motion in any portion of the chain. So, K.E. = 0
Am I wrong ?

2. Dec 3, 2016

Dilemma

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3. Dec 3, 2016

Molar

What was that @Dilemma Can you please fill in the blanks ?

4. Dec 4, 2016

Dilemma

I tried to solve your problem calculating the forces acting on the system but then I noticed that you were using the energy changes of the system. However, if I were you, I would introduce a friction force of μ to the system. There is no way this system can stay like that.

5. Dec 4, 2016

hilbert2

There's no way for that system to be in equilibrium unless there's static friction or if the chain is being pulled from the left side end. If you want to use the formula $V = mgh$ for the potential energy, you must interpret $h$ as the vertical (z) coordinate of the center of mass of the chain. If we set $z=0$ on the top of the table, what is the $z$ coordinate of the CM?

6. Dec 5, 2016

Molar

In that case, the position of the CM will be -l/2 k hat ,right ?

Yeah friction force is holding the chain above the table in its position, and that is F = μm1g, where m1 is the mass of the chain above the table
And the corresponding energy will be E = μm1g(L-l), right ?