Tension in the chain from a distance

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    Chain Tension
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The discussion revolves around calculating the tension in a vertically suspended chain at a distance y from the support. The initial approach uses Newton's second law, concluding that the net force is zero, leading to the equation T = Mg. However, the correct formula for tension at distance y is given as (Mg(L-y))/L. Participants are encouraged to consider the weight distribution along the chain, particularly at midpoint analysis, to understand the tension variation. The conversation highlights the importance of recognizing how weight is distributed in a hanging chain.
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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}##
 
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How about considering a point half way ? Is the full weight hanging from that point ?
 
I think the approach can be implemented in the problem.
 
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