"Doubling up" effect when pulling a chain on a table

[Father_Ing]

Homework Statement

A chain with length L lies straight on a frictionless horizontal surface. You grab one end and pull it back along itself in a parallel manner (see Fig. 5.27). Assume that you pull it at constant velocity v.

Relevant Equations

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In the book, it is stated that if your hand move a distance x, then x/2 is the length of the moving part of the chain because the chain gets “doubled up.” as in the image below.
I don't get the meaning of this. For example, if our hand move L metres from initial position, shouldn't the moving part is not half of it, but instead, it is still L (the same as how rope and pulley system works)?

Can anyone please explain to me about this matter?

 

[Orodruin]

Homework Statement:: A chain with length L lies straight on a frictionless horizontal surface. You grab one end and pull it back along itself in a parallel manner (see Fig. 5.27). Assume that you pull it at constant velocity v.
Relevant Equations:: -

if our hand move L/2 metres from initial position as in the image below, the moving part is not half of it, but instead, it is still L/2

No it is not. When your hand has moved L/2, the total length of chain that your hand has passed is L/2. Only half of this length is moving, the rest is still lying still on the table.

 

[Father_Ing]

Ah.. I should have said the hand moved at L distance in the example. My bad. (I already edited it)

Isn't there any tension between each part? If the rest of the chain (such as the other end of the chain) does not move at all, what force that cancel the leftward tension on this part?

 

[Orodruin]

The force that is continually supplying acceleration to the parts of the chain that starts moving.

 

[Father_Ing]

You mean the force that the hand gives?
But, that force is not applied to the end that is standing still.

 

[Orodruin]

The force acts on the moving end. The tension in the moving part is constant and equal to this force. At the bend of the chain, more and more chain starts moving. This is where the tension accelerates the chain to start moving and therefore the tension in the chain is maintained.

 

[Lnewqban]

... In the book, it is stated that if your hand move a distance x, then x/2 is the length of the moving part of the chain because the chain gets “doubled up.” as in the image below.

Could you post the exact wording shown in that book?
"... the length of the moving part of the chain" sounds very confusing to me.

 

[Father_Ing]

Could you post the exact wording shown in that book?
"... the length of the moving part of the chain" sounds very confusing to me.

"Let x be the distance your hand has moved.Then, x/2 is the length of the moving part of the chain, because the chain gets "doubled up""

 

[Father_Ing]

The force acts on the moving end. The tension in the moving part is constant and equal to this force. At the bend of the chain, more and more chain starts moving. This is where the tension accelerates the chain to start moving and therefore the tension in the chain is maintained.

Why the other end(the one that is standing still) does not accelerate? How can we know that the tension in this end is zero?

 

[Lnewqban]

"Let x be the distance your hand has moved.Then, x/2 is the length of the moving part of the chain, because the chain gets "doubled up""

Thank you.
This is how I understand it:
Your hand grabs the left end of the chain and relocates it all the way to the right by a distance equaling two lengths of the chain.
The original right end of the chain remains at the same spatial point, only that at the end of the 2L hand movement, it becomes the left side.

In summary, the chain has flipped over about its free end.
We can say that such a movement is equivalent to a translation of the whole chain by a L distance, at constant velocity (as stated in the problem).

The input work or energy must be equal to the out put work.
Therefore, if output work is FxL, the hand average work must have been 0.5Fx2L.
Note that the hand force needed to move each link of the chain increases with time, as more links adquire that constant velocity (as less links remain in repose on that table).

 

[Orodruin]

Why the other end(the one that is standing still) does not accelerate? How can we know that the tension in this end is zero?

There is no force present to accelerate that end. The chain only propagates the force along the chain direction and only when the chain is taut.

In summary, the chain has flipped over about its free end.
We can say that such a movement is equivalent to a translation of the whole chain by a L distance, at constant velocity (as stated in the problem).

The input work or energy must be equal to the out put work.
Therefore, if output work is FxL, the hand average work must have been 0.5Fx2L.
Note that the hand force needed to move each link of the chain increases with time, as more links adquire that constant velocity (as less links remain in repose on that table).

No, this is wrong. The force needed does not increase with time as the moving part of the chain is linearly gaining both mass and momentum (the force supplied by the hand is the added momentum per time). Once the full chain is in motion, no force is required to keep it moving at constant velocity.

You also need to be very careful with respect to work done in this problem. Pulling the chain at speed v means adding momentum mv^2/2L per time (mass of the moving part increases by mv/2L per time and it has velocity v after starting to move). This is therefore the force required to pull the chain until it is all in motion. The total work done over the time taken to move the hand the full distance of 2L before the entire chain is in motion is therefore mv^2, but the total kinetic energy of the chain is half of this. This is because the entire process may be seen as a series of inelastic collisions between the moving and non-moving chain links.

 

[Mister T]

(the same as how rope and pulley system works)?

No, because there is no fixed pulley. The place where the the chain makes its 180 degree bend moves along the table top, it doesn't stay in one place as it would if you had a fixed pulley.