The upper end of the uniform chain of linear mass density $\mu$ is lowered with a speed $v$ as shown in the figure. The reaction force offered by the ground after the chain falls through a distance $y.$
 

 

The each falling link collides with fallen links with a relative velocity, ${\overrightarrow{v}}_{rel}=-v\widehat{j}$

where $v=$velocity of the falling chin, we have the formula for the variable mass system “m”  given as: ${\overrightarrow{F}}_{ext}+{\overrightarrow{F}}_{thrust}={\overrightarrow{F}}_{net},$where ${\overrightarrow{F}}_{thrust}={\overrightarrow{v}}_{rel}\frac{d{m}^{\text{'}}}{dt}$

where ${\overrightarrow{v}}_{rel}=-v\widehat{j}$ and $\frac{d{m}^{\text{'}}}{dt}=\mu v$ (positive, as mass $m^{\text{'}}$ is increasing)

This gives ${\overrightarrow{F}}_{thrust}=-\mu v^2\widehat{j}$

Force equation on the heap: $N-m^{\text{'}}g+F_{thrust}=0$

where $m^{\text{'}}=\mu y$ and $F_{thrust}=\mu v^2$ or $N=\mu (v^2+gy)$