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
\[F=\frac{M}{L}xg+\frac{M}{L}2gy=\frac{M}{L}g\left( x+2y
\right)\].
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