91Ó°ÊÓ

The chain length is L

The mass is M

The total force exerted by the chain is calculatedby adding the gravitationalforce and impulse force acting on the chain.

The force exerted by the chain,

$F=F_1=F_2.........\left(1\right)$

Where,

$$\begin{gathered} F_1=\frac{dp}{dt} \\ =\frac{d\left(mv\right)}{dt} \\ =m\frac{dv}{dt}+v\frac{dm}{dt}.....\left(2\right) \end{gathered}$$

The first term is 0 because only the rate of change of the amount of chain hitting the table per unit is considered. So a differential mass element of the chain will be,

$dm=\frac{M}{L}dx.....\left(3\right)$

Substitute Equation (3) in Equation (2),

$$\begin{gathered} F_1=v\frac{Mdx}{Ldt} \\ =\frac{M}{L}{v}^2 \end{gathered}$$

F₁ = impulse Force acting on the chain $⇒\frac{M}{L}{v}^2$

After period of all , the velocity of the chain $v^2=2gx$

So F₁ = $2mg\frac{x}{L}$

x = The distance con ered

$$\begin{gathered} F_1=TheGravitationalforceisgivensimplybytheweightonthechain \\ {F}_2=mg=\frac{x}{L} \end{gathered}$$

The total force exerted by the chain is addition of impulse force and the gravitational force,

From Equation (1),

$$\begin{gathered} F=2mg\frac{x}{L}+mg\frac{x}{L} \\ =3mg\frac{x}{L}....\left(4\right) \end{gathered}$$

If we consider the instantaneous rate of change momentum of the chain as the last link hits the table, x will be L

So, put x = L in Equation (4),

F = 3mg

Hence, the force exerted by the table on the chain is