What is the reading on a scale when a chain falls onto it?

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Homework Help Overview

The problem involves a chain of mass M and length L that is released from a vertical position and falls onto a scale. Participants are discussing the reading on the scale as a function of the length of the chain that has fallen, denoted as x. The context includes concepts from mechanics, particularly forces acting on the chain and the scale during the fall.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the forces acting on the chain and the scale, including tension, normal force, and weight. There is a focus on whether tension in the chain disappears instantaneously and how that affects the reading on the scale. Questions are raised about the forces involved when the chain is falling versus when it has come to rest on the scale.

Discussion Status

The discussion is active, with participants exploring different interpretations of the forces at play. Some have suggested that the scale exerts a force on the chain as it comes to rest, while others are questioning the assumptions about tension and inertia. There is no explicit consensus, but various productive lines of reasoning are being examined.

Contextual Notes

Participants are considering the implications of neglecting the size of individual links in the chain and the assumptions about the instantaneous loss of tension. The discussion reflects a range of interpretations regarding the dynamics of the falling chain and its interaction with the scale.

honlin
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Homework Statement


A chain of mass M and length L is suspended vertically with its lowest end touching a scale. The chain is released and falls onto the scale. What is the reading of the scale when a length of chain, x , has fallen? (Neglect the size of individual links.)

Homework Equations


Weight = mg, Resultant force = ma, Newton third law.

The Attempt at a Solution


When the chain is suspended, by drawing the free body diagram of the chain, it should experienced three forces, upward tension, normal reaction by the scale, and its own weight.
When the chain falls onto the scale, it experience 3 forces too, upward tension, normal reaction and the weight of the the fallen chain.The weight of the fallen chain, W1 = (x/L)Mg. The upward tension has a reaction pair too, which is pointing downwards and has a magnitude of the weight of the chain which is still suspended. Therefore N2 = W2 = (L-x/L)Mg.
There are a total of 2 forces acting downward on the scale, the reaction of the tension, and the weight of the fallen chain. Therefore, the scale should read N2+W1=W1+W2 = W, the weight of the chain itself.
I do not have the answer for this question. Can someone please help me to check if my solution is correct because I am not confident with this answer. Thank you so much !
 
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I suspect that you are supposed to assume that the tension in the chain disappears almost instantaneously and the entire chain immediately goes into free fall.

If you do not assume this, then you would need to some additional assumptions about how quickly the tension is lost. If the chain were a "slinky", for example, this could happen:

 
Yeah, there shouldn't be any tension left when it is released, the string stays on top due to inertia. Therefore I should be calculating the downward resultant force, which is the chain that is still falling and the weight of the fallen chain. Is that right?
 
honlin said:
Yeah, there shouldn't be any tension left when it is released, the string stays on top due to inertia. Therefore I should be calculating the downward resultant force, which is the chain that is still falling and the weight of the fallen chain. Is that right?

The scale is doing two things:

1) Supporting the weight of the chain that has already fallen. And:

2) What else?
 
PeroK said:
The scale is doing two things:

1) Supporting the weight of the chain that has already fallen. And:

2) What else?

Supporting the weight of the chain that is still falling? Since the chain that is still falling is in contact with the scale, there must be a normal contact force right?
 
honlin said:
Supporting the weight of the chain that is still falling? Since the chain that is still falling is in contact with the scale, there must be a normal contact force right?

Not quite, the chain is falling and then it isn't falling any more. Something must have applied a force to slow it down!
 
PeroK said:
Something must have applied a force to slow it down!
Is it the tension of the chain that stop it from falling down?
 
honlin said:
Is it the tension of the chain that stop it from falling down?

If someone dropped a chain on your head, would it be the tension in the chain or your head that stopped it falling? Why would your head hurt?
 
So the scale is the one who exerted a force on the chain?
 
  • #10
honlin said:
So the scale is the one who exerted a force on the chain?

Exactly. The scale is not only supporting the weight of the chain has has fallen and come to rest, but is exerting a force on the chain that is currently hitting it and coming to rest.

To give you a hint: force equals rate of change of momentum. That will be key in solving this.
 

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