let us consider a small element at a distance x from the floor of lenght dy

so  dm=Mdx/L

so the velocity with whcih the element will strike the floor is

v=(2gx)^1/2

Therefore the momentum transferred to the floor is

M=(dm)v=Mdx(2gx)^1/2 /L

(because the element comes to rest)

so the force exerted on the floor changes in momentum is given by,

F₁=dM/dt=Mdx(2gx)^1/2 /Ldt

because v=dx/dt=(2gx)^1/2 (for the chain element)

F₁=M(2gx)² /L

=2Mgx/L

Again the force exerted due to x length of the chain on the floor due to its own weight is given by,

W=Mxg/L

so the total force exerted is given by

F=F₁+W

=2Mgx/L + Mgx/L

=3Mgx/L

Rate if satisfied. Plz correct me if i am wrong