# Finding a Differential equation for falling chain

This is a chain wound loosely to a peg at the edge of a high platform. It has a total length of $L$ in feet and is hanging initially $x_{0}$ from the platform plus it's weight is $w\:\mathrm{lb/ft}$ and if it slides down smoothly and all resistive forces are neglected, what is the differential equation for the system? Numbers were give : $L=8$, $w=2$,$x_{0}=3$.

Ans: $$xv\frac{dv}{dx}+v^{2}=32x$$ Question and trial: I am having trouble finding this equation. The pulling force increases as the chain falls, this pulling force is $\alpha x$ where $\alpha$ is the weight per unit length. The acceleration thus becomes $$\alpha x = \frac{\alpha L}{g}a$$ $$xg =Lv\frac{dv}{dx}$$ So, what am I missing?, where did $v^{2}$ come from?

From Newton's law: $dp/dt=mg=vdm/dt+mdv/dt$. In our case m is not a constant but increase as the chain goes down as: $dm=vdt$. Replacing in the previous equation you get $mg=v^2 dt/dt+mdv/dt=v^2+mdv/dt=v^2+mxvdv/dx$, replacing $m=xw$ and changing the units to mks you get the equation given as a solution