Tension in the chain from a distance

[ubergewehr273]

Homework Statement

A chain of mass $M$ and length $L$ held vertically by fixing its upper end to a rigid support. The tension in the chain at a distance $y$ from the rigid support is

Homework Equations

$F=ma$ (Newton's 2nd law)

The Attempt at a Solution

Since net acceleration, $a=0$,
Therefore, $F=0$
Hence, $T=Mg$,
Then onward I don't seem to get my head around.
The answer given is :
$\frac{Mg(L-y)}{L}$

 

[BvU]

How about considering a point half way ? Is the full weight hanging from that point ?

 

[ubergewehr273]

I think the approach can be implemented in the problem.