A uniform chain of mass M and length L is suspended vertically such...

Let $\lambda = \dfrac{M}{L}$ be the linear mass density of the chain.

When a length $x$ of the chain has reached the floor, the chain has fallen through a distance $x$. The velocity of the falling portion of the chain is $v = \sqrt{2gx}$.

The force exerted by the chain on the floor consists of two components.

The weight of the portion of the chain lying on the floor is given by:
$W = (\lambda x)g = \dfrac{Mgx}{L}$

The thrust force due to the continuous impact of the falling chain is equal to the rate of change of momentum. The rate at which mass strikes the floor is $\dfrac{dm}{dt} = \lambda v$. Since the chain comes to rest immediately upon striking the floor, the thrust force is:
$F_{thrust} = v \dfrac{dm}{dt} = v(\lambda v) = \lambda v^2$

Substituting $v^2 = 2gx$, we get:
$F_{thrust} = \lambda (2gx) = \dfrac{2Mgx}{L}$

The total force exerted on the floor is the sum of these two forces:
$F_{total} = W + F_{thrust} = \dfrac{Mgx}{L} + \dfrac{2Mgx}{L} = \dfrac{3Mgx}{L}$