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 ?

**Click to See Answer : **

Sol: Since chain is uniform, the mass of ‘y’ part of chain will be ((M/L) y). When this part reaches table, its total force exerted must be equal to the weight of y part resting on table + force due to the momentum

imparted

imparted

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

$\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} $