The Falling Chain

Relevant wiki: Newton's Second Law

Let $x$ is the length of the chain that has fallen to the ground, thus the chain still in the air will be $L-x$ . The mass $m$ of the chain in the air will be $m = \frac{M(L-x)}{L}.$ Only the link that collides with the ground experience the force and hence the remaining chain in the air can be treated in the free fall. Therefore, the speed $v$ of the chain in the air will be $v = \frac{dx}{dt} = \sqrt{2gx}.$ Thus, the momentum $p$ of the chain in the air will be $\vec{p} = \frac{M(L-x)}{L} \sqrt{2gx} (-\hat{j}).$ Now, the net force on the chain $F_\text{net}$ will be $\vec F_\text{net} = F_{\text{table}} \hat{j} - mg \hat{j}.$ Also, from the Newton's Second law $\vec F_\text{net} = \frac{d \vec p}{dt}.$ Hence, $\begin{aligned} F_{\text{table}} \hat{j} - mg \hat{j} &= \frac{dp}{dt} \left(\frac{M(L-x)}{L} \sqrt{2gx} (-\hat{j}) \right). \\ &= \frac{M}{L}(-2gx + (L-x)g)(-\hat{j}). \\ F_{\text{table}} \hat{j} &= \frac{3Mgx}{L} (\hat{j}). \\ F_{\text{table}} &= K \times \frac{Mgx}{L} = \frac{3Mgx}{L}. \\ \text{Hence,} \\ K &= \boxed{3}. \\ \end{aligned}$

Assuming $\rho$ to be constant, $\frac {M}{L} = \frac {dm}{dx}$ . By Newton's third law we can say that the normal reaction $(N)$ is equal to the force we are trying to find. Also we will have to account for the change in momentum of the chain just before it hits the ground. This can be written as

$\frac {dp}{dt} = \frac {dm}{dt}V = \frac {M}{L} \frac {dx}{dt} V = \frac {M}{L}V^{2} = \frac {M}{L}(2xg)$

$\therefore N = \frac {Mxg}{L} + \frac {2Mxg}{L} = \boxed {\frac {3Mxg}{L}}$