- #1

Nathanael

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## Homework Statement

A uniform chain of length L and [constant] mass per unit length λ is suspended at one A end by an inextensible light string. The other end of the chain B is held at rest at level of end A of the chain. [See image.] Now if the end B of the chain is released under gravity, then find the tension in the

*string*when the end B has fallen a distance y

## Homework Equations

F=dp/dt

## The Attempt at a Solution

I was thinking the tension would be equal to the weight of the chain [itex]gλ(\frac{L}{2}+y)[/itex] plus the change in momentum per time of the chain that is transitioning from the right to left side (in the image).

The momentum per time would be equal to the speed times the mass per time, and the mass per time would be the speed times the linear density, so I get:

[itex]T=gλ(\frac{L}{2}+y)+λv^2[/itex]

v can be written in terms of y as [itex]v=\sqrt{2gy}[/itex] so I get:

[itex]T=gλ(\frac{L}{2}+3y)[/itex]

But this is apparently incorrect