What a chain! 4

First of all lets find out the Tension as a function of distance falled(y) or $T(y)$ .

So We can see that $\dfrac{L+y}{2} \lambda =m$ , where m=mass of the left portion.

$\dfrac{dm}{dt} = \dfrac{\lambda}{2} \dfrac{dy}{dt}$

$\implies F_{thrust} = \dfrac{\lambda v}{2}$ and $v=\sqrt{2gy}$ .

Now the force due to the mass of the rope,

$F_{weight}= T- mg$ .

Also $F_{weight}-F_{thrust}=ma$ ....

Hence $F_{weight}-F_{thrust}= m \times 0$ as there is no accleration of the left chain.

$F_{weight} = F_{thrust} \implies F_{weight}=\dfrac{\lambda v}{2} = \lambda gy$

So putting in the equation with Tension we get

$T(y)\quad =\quad \dfrac { \lambda g }{ 2 } (L+ 3y)$

Now we know $v^2=2gy$ and $v=gt$

So $y=\dfrac{2g}{t}$

Again $T(t)\quad =\quad \dfrac { \lambda g }{ 2 } (L+ 3\dfrac{2g}{t}) \implies \dfrac{Mg}{2}+\dfrac{3Mg^2t^2}{4L}$

So $a+b+c=2+3+4=9$

The tension rises as a function of time as mass of the hanging chain ( effective weight ) increases