What is the Force Between Links in a Falling Chain?

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The discussion centers on the dynamics of a falling chain, specifically the forces between its links and the time it takes for the top link to reach the table. The time to fall is calculated as sqrt(2s/a), based on kinematic equations. The force between the links is debated, with considerations of whether it is solely the weight of the chain below or if tension plays a role, leading to the equation Mg-T=Ma. The complexity of falling chain problems is highlighted, noting that they remain an area of ongoing research. Understanding the tension's impact on the vertical acceleration of both the top and bottom links is essential for a complete analysis.
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Homework Statement


A uniform chain of length l and mass M contains many links. It is held above a table so that one end is just touching the table top. The chain is released freely. What is the force between the links? What is the time for the topmost link to fall to the table?

Homework Equations



v=u+at, v^2 = u^2 + 2as

The Attempt at a Solution


I think that the time to reach the table is sqrt(2s/a), because v=at, so (at)^2=2as, so rearranging that would give me that answer. I'm trying to visualise the part about the force though; is it that the only force exerted on each chain the weight of the chain below it? Or is it tension, in which case Mg-T=Ma, so T=Mg-Ma?
 
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Falling chain problems are still somewhat of an open research area because the correct model is not entirely clear. Here is a paper on the topic:

"Falling chains as variable mass systems: theoretical model and experimental analysis," by C.A. de Sousa, P.M. Gordo, and P. Costa, Physics Education, 28 May, 2012.

The particular problem stated by the OP is not addressed in this paper, but some of the relevant modeling questions are.
 
PhysicsKid99 said:

Homework Statement


A uniform chain of length l and mass M contains many links. It is held above a table so that one end is just touching the table top. The chain is released freely. What is the force between the links? What is the time for the topmost link to fall to the table?

Homework Equations



v=u+at, v^2 = u^2 + 2as

The Attempt at a Solution


I think that the time to reach the table is sqrt(2s/a), because v=at, so (at)^2=2as, so rearranging that would give me that answer. I'm trying to visualise the part about the force though; is it that the only force exerted on each chain the weight of the chain below it? Or is it tension, in which case Mg-T=Ma, so T=Mg-Ma?
Suppose there is some tension in the chain as it falls. What does that tell you about the vertical acceleration of the top link? What does it tell you about the vertical acceleration of the bottom link that's not yet in contact with the table?
 

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