# Tension in the chain from a distance

1. Dec 10, 2014

### Ashes Panigrahi

1. The problem statement, all variables and given/known data
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

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

3. 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.
$\frac{Mg(L-y)}{L}$

2. Dec 10, 2014

### BvU

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

3. Dec 10, 2014

### Ashes Panigrahi

I think the approach can be implemented in the problem.