Net Vertical Force on a Slipping Chain on a Table

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

The initial shape of chain is like an inverted "L" with end B just touching the floor . Height of table is "h" .

My problem is in identifying the net force on the chain in vertical direction .At any instant of time there are three parts of chain , first on table , second hanging part of length "h" and third is lying on the floor in an unordered fashion .

The normal force on the first part is balanced by the weight of that part .

I think the normal force on the third part is not equal to the weight of the chain lying on the floor . What will be the vertical force on the second and third parts ?

 

[haruspex]

The initial shape of chain is like an inverted "L" with end B just touching the table . Height of table is "h" .

I can't make sense of the description. If end B is hanging h below the table how can it be touching the table? And it does not say the height of the table is h.
Should the text say that end B just touches the floor?

 

[Jahnavi]

Should the text say that end B just touches the floor?

Yes .

Typing mistake .I will edit it .

 

[haruspex]

Yes .

Typing mistake .I will edit it .

Ok.
There is a second difficulty with the question. (I am assuming the chain lies straight on the table, not in a heap.)
In reality, the chain leaving the table will have acquired horizontal velocity, so will not descend vertically. This makes it extremely hard to analyse. To avoid that, the question ought to include some device to deflect the chain downwards at the edge of the table.
Glossing over that, what other approach can you think of instead of using forces and accelerations?

 

[Jahnavi]

what other approach can you think of instead of using forces and accelerations?

I don't think energy is conserved . Is it ?

 

[haruspex]

I don't think energy is conserved . Is it ?

Ah, right, forget that.
Consider the tension, T, at the edge of the table. What forces are the on the horizontal part? What acceleration results?
What about the vertical part?

 

[Jahnavi]

Consider the tension, T, at the edge of the table. What forces are the on the horizontal part? What acceleration results?

Just the tension T . T = (Mx/L)a . x is length on the table .

What about the vertical part

This is what I am unsure about .Please read my first post .

The acceleration will be same as horizontal part .

What is the force on the lowest part of chain (just when it stops ) ?

 

[haruspex]

What is the force on the lowest part of chain

Do you mean the force from the part of the chain that has already landed? Why would there be any?

 

[Jahnavi]

Do you mean the force from the part of the chain that has already landed? Why would there be any?

There is a change in momentum of the lowermost tip . Wouldn't there be an upward force on the hanging part from the heap ? I don't know .I am not sure .

 

[haruspex]

There is a change in momentum of the lowermost tip . Wouldn't there be an upward force on the hanging part from the heap ? I don't know .I am not sure .

When an object falls to the floor, there is a force that brings it to rest, but there is no force the instant before it hits the floor. Treat the vertical part of the chain the same way, only include the section that has not quite reached the floor.

 

[Jahnavi]

In that case , (Mhg/L) -T = (Mhg/L)a

 

[haruspex]

In that case , (Mhg/L) -T = (Mhg/L)a

Yes.

 

[Jahnavi]

Thanks !