# A very flexible chain of length L and mass M is vertically suspended with its lower end just touching the table. If it is released…

Q: A very flexible chain of length L and mass M is vertically suspended with its lower end just touching the table. If it is released so that each link strikes the table and comes to rest. What force the chain will exert on the table at the moment ‘y’ part of length falls on the table ?

$\large F = \frac{M}{L} y g + \frac{\frac{M}{L}dy \sqrt{2 g y}}{dt}$
$\large F = \frac{M}{L} y g + \frac{M}{L}.v \sqrt{2 g y}$
$\large F = \frac{M}{L} y g + \frac{M}{L}.\sqrt{2 g y} .\sqrt{2 g y}$
$\large F = 3\frac{M g y}{L}$