Tension in the chain from a distance

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    Chain Tension
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SUMMARY

The discussion focuses on calculating the tension in a vertically suspended chain of mass M and length L. The tension at a distance y from the rigid support is derived using Newton's second law, resulting in the formula T = Mg(L-y)/L. The participants explore the implications of this formula, particularly when considering points along the chain, such as halfway down, and whether the full weight is supported at that point. The conversation emphasizes the importance of understanding how tension varies along the length of the chain.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Basic principles of mechanics related to tension in strings and chains
  • Concept of gravitational force (g) acting on mass
  • Knowledge of how to apply formulas to physical scenarios
NEXT STEPS
  • Study the derivation of tension in different configurations of chains and ropes
  • Learn about static equilibrium and its applications in mechanics
  • Explore the concept of distributed loads in structural engineering
  • Investigate the effects of varying mass distributions on tension calculations
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Students of physics, mechanical engineers, and anyone interested in understanding the mechanics of tension in chains and ropes.

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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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