Question

A chain hangs on a thread and touches a surface of a table by its lower end. Show that after the thread has been burnt the force exerted on the table by the falling part of the chain at any moment is twice the weight of the part already resting on the table.

Answer 1

The descending part of the chain is in free fall, it has speed $\nu =\sqrt{2gh}$ at the instant, all its points have descended a distance $y$ The length of the chain which lands on the floor during the differential time interval $dt$ following this instant is $vdt$
For the incoming chain element on the floor:
from $d{p}_y=F_{\nu }dt$ (where $y-$ axis is directed down)
$0-\left(\lambda vdt\right)nu=F_{y}dt$
or $F_y=-\lambda {\nu }^2=-2\lambda gy$
Hence the force exerted on the falling chain equals $\lambda {\nu }^2$ and is directed upwards. therefore from third law the force centred by the falling chain on the table at the same instant of time becomes $\lambda {\nu }^2$ and is directed upwards. therefore from third law the force exerted by the falling chain on the table at the same instant of time becomes $\lambda {\nu }^2$ and is directed downwards
Since a length of chain of weight $\left(\lambda yg\right)$ already lies on the table the total force on the floor is $\left(2\lambda yg\right)+\left(\lambda yg\right)=\left(3\lambda yg\right)$ or the weight of a length $3y$ of chain.