question_answer

A very flexible uniform chain of mass M and length L is suspended vertically so that its lower end just touches the surface of a table. When the upper end of the chain is released, it falls with each link coming to rest the instant it strikes the table. Find the force exerted by the chain on the table at the moment when x part of chain has already rested on the table.

Answer:

                In this problem, force exerted by the chain on the table \[F={{F}_{1}}+{{F}_{2}}\] where \[{{F}_{1}}\] = weight of chain already on the table = \[\frac{M}{L}x.g\]                                        ...(ii) and \[{{F}_{2}}\] = rate of change of momentum of chain at the instant it strikes the table. To calculate\[{{F}_{2}}\] , let us consider a small element dy of the chain at a height y above the table, Fig. 3(HT).6. mass of the element\[=\frac{M}{L}dy\] velocity of the element on striking the table,         \[\upsilon =\sqrt{2gy}\] \[\therefore \]  \[dp=\left( \frac{M}{L}dy \right)\sqrt{2gy}\]                       ??? (iii) \[{{F}_{2}}=\frac{dP}{dt}=\frac{M}{L}\frac{dy}{dt}\sqrt{2gy}\] As \[\frac{dy}{dt}=\upsilon =\sqrt{2gy}\]                              \[\therefore \]  \[{{F}_{2}}=\frac{M}{L}(2gy)\] Putting in (i), we get \[F=\frac{M}{L}xg+\frac{M}{L}2gy=\frac{M}{L}g\left( x+2y \right)\].