Question

A chain hangs on a thread and touches the surface of a table by its lower end. After the thread has been burned through, the force exerted on the table by the falling part of the chain is given as $x\lambda ygN$. Find $x$. ($y$ is the descent and $\lambda$ is the mass per unit length)

Answer 1

The descending part of the chain is in free fall, it has speed $v=\sqrt{2gh}$ 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 $d{p}_y=F_{v}dt$ (where y - axis is directed down)
$0-\left(\lambda vdt\right)v=F_{y}dt$
or $F_y=-\lambda v^2=-2\lambda gy$
Hence, the force exerted on the falling chain equals $\lambda v^2$ and is directed upward. Therefore from third law the force exerted by the falling chain on the table at the same instant of time becomes $\lambda v^2$ and is directed downward. Since a length of chain of weight ($\lambda yg$) 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.