Tension in moving chain Suppose I have a chain in equilibrium position (a catenary). The only force acting on it is gravity. Now, say that I move this chain in such a way that every point traces this catenary. Basically, the chain "flows through its shape". What would be the tension now? I do not want to ignore gravity. I've read several times that the tension in a moving chain is $\lambda v^2$ ($\lambda$ is the mass/unit length, $v$ is the speed) , but these models ignore gravity. Is the tension in this case the tension in the static case + $\lambda v^2$? If so, how could I prove it?
 A: The tension will not remain the same.
We can simplify as follows: take a catenary curve that is very narrow.
(For instance, we can take a catenary curve such that the height is several times the horizontal distance between the suspension points.)
With a very narrow catenary curve the sides are close to vertical.
Take the part of the chain that is descending. In the case of a static chain gravity is tensioning the chain.
Let us assume that chain is fed to the catenary-shaped section along a horizontal track. The chain has to go from horizontal velocity to (close to) vertical velocity. One way to implement that would be to run the chain over a wheel.
Now we go to the state where the chain has a velocity.
As the chain goes over the wheel, changing the direction of motion from horizontal to (close to) vertical: some of the gravity will go to accelerating the chain, rather than contributing to tension.
This demonstrates that the tension distribution will not remain the same. The higher the velocity, the larger the deviation from the static case.
If you want the shape of the chain to remain the same then you have to constrain the chain to a channel that is in the shape of the catenary curve. Conversely, without that guiding channnel the shape of the chain would change away from the catenary curve.
