Chain falling out of a horizontal tube onto a table

[jbriggs444]

I alluded to that in post #44, but I think it is ok because we can instead view it in terms a tension transmitted around the curve and write the separate equations for the accelerations of the two portions.
I believe this leads to the same equation.

I agree completely. The chain traverses a one dimensional fixed path. There is no net force nor freedom to move other than along the path. The chain moves along this path in lock step, all pieces keeping pace with all others. Accordingly, we are free to view the chain as if it were a single entity moving along a straight line. Though, obviously, only one part is subject to gravity. And we still have the issue with links dribbling off the one end.

Similarly, in pulley problems with two masses dangling from a single ideal pulley, it is valid to add the masses together to get a single effective mass for the combined system.

 

[NTesla]

I believe this leads to the same equation.

Which equation are you referring to ?

If we happen to write separate equations for accelerations for the two parts of the chain (horizontal and vertical), then we do reach the correct differential equation. I've been through that route and I understand how to solve this question using that route.

However, in my current attempt to solve this problem, I've taken the route of finding the change in momentum and equating it to the net external force, as may be deduced from my post #38 and 49.

I'm asking for help in writing the correct equations for $\vec {P_{i}}$ or $\vec {P_{f}}$ or the force applicable on the system. Kindly go through the equations that I've written in those aforementioned posts, and let me know which term is wrong or if any other term needs to be included in those expressions. After @haruspex's post in post#48, I tried writing the equation for force = rate of change of momentum of the system and this is the equation i've landed with:
$$ \vec {P_{f}} - \vec {P_{i}} = \lambda ((l - (x + \Delta x) - h)(v + \Delta v)- \lambda (l - x - h)v)\hat i + (h (v+\Delta v)-\Delta xv - hv)(-\hat {j})$$

$$\vec F_{ext} = \lambda hg(-\hat j) + \lambda \frac {\Delta x}{\Delta t}v \hat j$$

Now, by equating these two equations, i'm getting: $$((l - (x + \Delta x) - h)(v + \Delta v)- (l - x - h)v)\hat i + (h (v+\Delta v)-\Delta xv - hv)(-\hat {j}) = hg\Delta t(-\hat j) + {\Delta x}v \hat j $$

I don't know where to proceed from here.

 

[jbriggs444]

$$\vec F_{ext} = \lambda hg(-\hat j) + \lambda \frac {\Delta x}{\Delta t}v \hat j$$

The claim that $\vec F_\text{ext}$ is given by $\lambda hg(-\hat j) + \lambda \frac {\Delta x}{\Delta t}v \hat j$ is clearly incorrect.

We have the downward force from gravity given by $\lambda h g (-\hat j)$. That much is correct.

We have the upward momentum flow as downward moving chain links leave the system. That is indeed given by $\lambda \frac {\Delta x}{\Delta t}v$.

But we also have the force from the mouth of the tube which delivers a downward momentum flow due to the chain links emerging downward. The mouth is accelerating these links downward. It also delivers a leftward momentum flow due to the chain links entering the mouth from the left. It is bringing these links to a stop in the horizontal direction. It also delivers an upward force equal to the tension in the chain at the mouth. Finally it delivers a rightward force equal to the tension in the chain at the mouth.

It is way easier to make an argument and reduce the problem to one dimension.

 

[haruspex]

I don't know where to proceed from here.

Well, obviously, the usual steps of cancelling out the non-infinitesimal terms and discarding the second order infinitesimals. What does that give you?