Tension in the string holding a falling chain

[TSny]

So that is how thrust force on left part is coming out to be λgy ?

yes

 

[TSny]

Which part of the string is in free fall ?Is left part in free fall ?

No parts of the left side are in free fall. All parts of the left side are at rest. The left side is just getting longer as more links are added to the left side.

Are you suggesting that only the right part is in free fall ?

Yes

Why do we get correct answer by assuming left part to be in free fall ?

Can you show how you get the answer assuming the left side is in free fall?

 

[Vibhor]

Can you show how you get the answer assuming the left side is in free fall?

Thrust force is λv² . If we assume tip of left string to be in free fall ,it falls by a distance y/2 . Hence the speed is √(2g(y/2)). This is how I got the thrust force λgy .

 

[TSny]

Thrust force is λv² . If we assume tip of left string to be in free fall ,it falls by a distance y/2 . Hence the speed is √(2g(y/2)). This is how I got the thrust force λgy .

I'm afraid I don't follow your argument.

haruspex outlined how to get the force in #27.

The rate, dm/dt, at which mass on the right is coming to rest is the rate at which mass on the right is reaching the (moving) blue line. This is λ times the velocity of the right side relative to the blue line: dm/dt = λ(v_{red dot} - v_{blue line}) =λ(v - v/2) = λv/2, where v is the velocity of free fall of the right side: v = √(2gy). The magnitude of the force required to bring elements of the right side to rest is equal to the magnitude of the rate of change of momentum of these elements:
F = (dm/dt)⋅v = (λv/2)⋅v = λv²/2 = λgy.

 

[Vibhor]

The rate, dm/dt, at which mass on the right is coming to rest is the rate at which mass on the right is reaching the (moving) blue line. This is λ times the velocity of the right side relative to the blue line: dm/dt = λ(v_{red dot} - v_{blue line}) =λ(v - v/2) = λv/2, where v is the velocity of free fall of the right side: v = √(2gy). The magnitude of the force required to bring elements of the right side to rest is equal to the magnitude of the rate of change of momentum of these elements:
F = (dm/dt)⋅v = (λv/2)⋅v = λv²/2 = λgy.

Thanks .This clears up the confusion.

 

[Vibhor]

I'm afraid I don't follow your argument.

Now ,even I don't follow my argument

The remarkable thing is that ,despite getting the dynamics of the problem wrong in every step ,I got the correct answer in first try .

 

[TSny]

The remarkable thing is that ,despite getting the dynamics of the problem wrong in every step ,I got the correct answer in first try .

That's happened to me before, too.

 

[Vibhor]

That's happened to me before, too.

Thanks a lot Sir .

Thank you very much @haruspex .