Chain falling out of a horizontal tube onto a table

[Steve4Physics]

@NTesla, maybe this is one source of your difficulty...

Now, in the $\hat{i}$ direction, the force is due to the weight of the hanging part of the chain,

That's not entirely correct. The horizontal section of the chain is accelerated by tension,T, acting at its right-most end:

We have to treat the falling vertical chain section as if there are only two forces acting on it*. The forces are its weight, $\lambda h g$ downwards and $T$ acting upwards at its top end:

The acceleration, $a$, is the same for both sections, so you can (using N2L) construct 2 equations and eliminate T. This gives an equation relating $x$ and $a$.

Hint: When you get this equation that remember $a = v \frac {dv}{dx}$.

*The absence of a force from the ground, which affects the motion of the vertical section of the chain, has been discussed by @jbriggs444 and @haruspex.

 

[jbriggs444]

"The portion of the chain that is sliding across the table will not be accelerating." Why can't it have the same |acceleration| as the rest of the chain?

It could if, for instance, we let the chain ride in a frictionless tube on the table. Then the section of chain in the tube would be accelerating (we assume without evidence that the links can sustain a compressive load as long as they are forced to glide in a straight line). But then the compressive load from this would feed back into the falling chain. So then we would have to extend the guide tube upward from the table. With this in place the whole chain would count in our equation for acceleration.

Arguably, that would make the problem both too easy to solve and too messy to pose. We would regain mechanical energy conservation. The solution proposed in the OP would apply.

 

[NTesla]

Equations that you wrote are not legible. Kindly see the attached pic.

 

[NTesla]

@NTesla, maybe this is one source of your difficulty...


That's not entirely correct. The horizontal section of the chain is accelerated by tension,T, acting at its right-most end:
View attachment 367233
We have to treat the falling vertical chain section as if there are only two forces acting on it*. The forces are its weight, $\lambda h g$ downwards and $T$ acting upwards at its top end:
View attachment 367231
The acceleration, $a$, is the same for both sections, so you can (using N2L) construct 2 equations and eliminate T. This gives an equation relating $x$ and $a$.

Hint: When you get this equation that remember $a = v \frac {dv}{dx}$.

*The absence of a force from the ground, which affects the motion of the vertical section of the chain, has been discussed by @jbriggs444 and @haruspex.

Yes, and I've already mentioned that in post#22.

But there's another problem that I've mentioned in post#22, that is the cause of much problem.

 

[Philip Wood]

I know this is the wrong place to ask, but I need help with this site and don't know where to find it... (a) I can't find the 'edit' button on my longish post above with red error messages. (b) I can't understand the error messages (c) I can't find the latex guide, the one that uses a quadratic equation as an example.

 

[haruspex]

But there's another problem that I've mentioned in post#22, that is the cause of much problem.

Please show the details of your differentiation of the horizontal momentum. Looks like you missed a term.

 

[haruspex]

I know this is the wrong place to ask, but I need help with this site and don't know where to find it... (a) I can't find the 'edit' button on my longish post above with red error messages. (b) I can't understand the error messages (c) I can't find the latex guide, the one that uses a quadratic equation as an example.

Are you writing multiple lines of LaTeX inside one pair of $ $ brackets? Try not doing that.
And please use the preview button (magnifying glass at right hand end of controls) before posting.

 

[NTesla]

Please show the details of your differentiation of the horizontal momentum. Looks like you missed a term.

Here's all the steps:
$\Delta\vec{P} = \vec{P_{f}} - \vec{P_{i}}$
$\vec{P_{i}} = \lambda (l - x - h)v\hat{i} - \lambda hv\hat{j}$
$\vec{P_{f}} = \lambda (l - (x + \Delta{x}) - h)(v + \Delta{v}) \hat{i} - \lambda h(v + \Delta{v}) \hat{j}$

In the $\hat{i}$ direction: $\Delta P\hat{i} = \lambda(l - (x + \Delta x) - h)(v + \Delta v)\hat i - \lambda (l - x - h)v\hat i$
= $\lambda((l - x - h)\Delta v - \Delta x v)\hat i$ (ignoring $\Delta x \Delta v$).

Dividing both sides by $\Delta t$, we get:
$\frac{\Delta P}{\Delta t} = \lambda((l - x - h)\frac{\Delta v}{\Delta t} - \frac{\Delta x}{\Delta t} v)$

Now, in the limit, $\Delta t\rightarrow 0$ the equation becomes:
in the $\hat i$ direction: $$\frac {dP}{dt} = \lambda((l - x - h)\frac {dv}{dt} - \frac {dx}{dt}v)$$

$\Rightarrow$ For the horizontal part of the chain: $\frac {dP}{dt} = \lambda((l - x - h)\frac{dv}{dt} - v^{2})$

Now, by comparing the force applicable on the horizontal and vertical part of the chains, for same acceleration throughout, we get: Force applicable on the horizontal part of the chain, $F_{ext}$ = $\lambda h(g - \frac{\mathrm{d} v}{\mathrm{d} t})$.

$\therefore$ $\frac {dP}{dt} = \lambda((l - x - h)\frac{dv}{dt} - v^{2})$ = $\lambda h(g - \frac{\mathrm{d} v}{\mathrm{d} t})$.

But, this is not the right differential equation which leads to the correct answer. As I had mentioned in post#22, for reaching the correct differential equation, the $v^{2}$ term needs to vanish. For that to happen the term $( - \Delta x v\hat i)$ terms needs to vanish. That can only happen if there's an additional term $+ \Delta x v\hat i$ in the expression for $\vec {P_{f}}$. However, I do not see how or why there could be an additional term $( + \Delta x v\hat i)$ in the expression for $\vec {P_{f}}$. This is where I'm facing the problem.

 

[haruspex]

Here's all the steps:
$\Delta\vec{P} = \vec{P_{f}} - \vec{P_{i}}$
$\vec{P_{i}} = \lambda (l - x - h)v\hat{i} - \lambda hv\hat{j}$
$\vec{P_{f}} = \lambda (l - (x + \Delta{x}) - h)(v + \Delta{v}) \hat{i} - \lambda h(v + \Delta{v}) \hat{j}$

In the $\hat{i}$ direction: $\Delta P\hat{i} = \lambda(l - (x + \Delta x) - h)(v + \Delta v)\hat i - \lambda (l - x - h)v\hat i$
= $\lambda((l - x - h)\Delta v - \Delta x v)\hat i$ (ignoring $\Delta x \Delta v$).

Dividing both sides by $\Delta t$, we get:
$\frac{\Delta P}{\Delta t} = \lambda((l - x - h)\frac{\Delta v}{\Delta t} - \frac{\Delta x}{\Delta t} v)$

Now, in the limit, $\Delta t\rightarrow 0$ the equation becomes:
in the $\hat i$ direction: $$\frac {dP}{dt} = \lambda((l - x - h)\frac {dv}{dt} - \frac {dx}{dt}v)$$

$\Rightarrow$ For the horizontal part of the chain: $\frac {dP}{dt} = \lambda((l - x - h)\frac{dv}{dt} - v^{2})$

Now, by comparing the force applicable on the horizontal and vertical part of the chains, for same acceleration throughout, we get: Force applicable on the horizontal part of the chain, $F_{ext}$ = $\lambda h(g - \frac{\mathrm{d} v}{\mathrm{d} t})$.

$\therefore$ $\frac {dP}{dt} = \lambda((l - x - h)\frac{dv}{dt} - v^{2})$ = $\lambda h(g - \frac{\mathrm{d} v}{\mathrm{d} t})$.

But, this is not the right differential equation which leads to the correct answer. As I had mentioned in post#22, for reaching the correct differential equation, the $v^{2}$ term needs to vanish. For that to happen the term $( - \Delta x v\hat i)$ terms needs to vanish. That can only happen if there's an additional term $+ \Delta x v\hat i$ in the expression for $\vec {P_{f}}$. However, I do not see how or why there could be an additional term $( + \Delta x v\hat i)$ in the expression for $\vec {P_{f}}$. This is where I'm facing the problem.

Ok, I understand now. You have fallen into the classic trap of applying $F=\frac{d(mv)}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}$ to a variable mass system.
Consider a cart rolling on a level surface and leaking water. If we apply that equation to the cart plus contents we conclude that the cart will accelerate! The error is that the first part of that equation only applies in a closed system, but in closed systems m does not vary.

 

[Philip Wood]

Equations that you wrote are not legible. Kindly see the attached pic.

I know, but I can't find an 'edit' button, so I can't do anything about them.

 

[berkeman]

I can't find the latex guide

The link to the "LaTeX Guide" is below the Edit window on the left.

 

[NTesla]

Ok, I understand now. You have fallen into the classic trap of applying $F=\frac{d(mv)}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}$ to a variable mass system.
Consider a cart rolling on a level surface and leaking water. If we apply that equation to the cart plus contents we conclude that the cart will accelerate! The error is that the first part of that equation only applies in a closed system, but in closed systems m does not vary.

By first part, did you mean this equation: F = d(mv)/dt.?

If you meant this equation, then starting from momentum calculation, what would be the correct equation to write ?

I've studied how the equation for velocity in rocket propulsion is calculated. Therein also, change in momentum of the (rocket + ejected gas) is equated to $F_{ext}dt$. This calculation leads to the correct form of Newton's 2nd law:
$m\frac {d\vec{v}}{dt} = \vec{F}_{ext} + \vec{v}_{rel}\frac {dm}{dt}$.

I've also tried to do the same for the Question that I've posted. I don't understand why it's wrong to use the same concept here.

In the scenario that you gave, of a leaking cart: We begin our calculation by considering the cart + the water within the cart + the leaking water drop as our system and then finding change in momentum at some time $t$ and some other time $t + \Delta t$, and then since external force in horizontal direction is presumably zero in this cart scenario, we equate that change in momentum of the system to zero and then we will get our differential equation.

 

[NTesla]

I know, but I can't find an 'edit' button, so I can't do anything about them.

If you are accessing this forum using your phone, then see this to see where the edit button is:

If you are working on desktop, then also, you will find the edit button. I'm on my phone, so can't send the desktop screenshot right now. But it's there.

 

[haruspex]

By first part, did you mean this equation: F = d(mv)/dt.?

Yes.

I've studied how the equation for velocity in rocket propulsion is calculated. Therein also, change in momentum of the (rocket + ejected gas) is equated to $F_{ext}dt$.

Right, but that's rocket+fuel+exhaust, which is a closed system.
There are two issues here.
First, somehow an element leaves the horizontal segment to become part of the vertical segment, which must involve forces that do not appear in your equation. Now, there’s probably an argument that gets around that because the tension must be the same both sides of the bend.
More seriously, you have overlooked the "exhaust" consisting of the element has hit the table. To include that, you must add in the normal force from the table that brings that element to rest.

 

[NTesla]

Yes.


Right, but that's rocket+fuel+exhaust, which is a closed system.
There are two issues here.
First, somehow an element leaves the horizontal segment to become part of the vertical segment, which must involve forces that do not appear in your equation. Now, there’s probably an argument that gets around that because the tension must be the same both sides of the bend.
More seriously, you have overlooked the "exhaust" consisting of the element has hit the table. To include that, you must add in the normal force from the table that brings that element to rest.

Even if we include the force exerted by the table on the element which has brought the element to rest, it will all be happening in the $\hat j$ direction. That force won't impact the momentum change in $\hat i$ direction.

Starting from momentum calculation, what would be the correct expression for $\vec{P_{i}}$ and $\vec{P_{f}}$ according to you ?

 

[haruspex]

Even if we include the force exerted by the table on the element which has brought the element to rest, it will all be happening in the $\hat j$ direction. That force won't impact the momentum change in $\hat i$ direction.

Right, but it does affect this:

Now, by comparing the force applicable on the horizontal and vertical part of the chains

 

[NTesla]

Right, but it does affect this:

I don't think it does. The force exerted by the table on the element which comes to rest, is not impacting any other element in the entire chain. That force is not going to affect the acceleration of the hanging part of the chain. That force's entire role is limited to that element only which is touching the table and coming to rest.

 

[haruspex]

I don't think it does. The force exerted by the table on the element which comes to rest, is not impacting any other element in the entire chain. That force is not going to affect the acceleration of the hanging part of the chain. That force's entire role is limited to that element only which is touching the table and coming to rest.

But that element is the "exhaust". The loss of it is removing momentum from the system. It is the water leaking from the cart.

 

[NTesla]

But that element is the "exhaust". The loss of it is removing momentum from the system. It is the water leaking from the cart.

I can see why it would seem like the water leaking from the cart scenario.

In case we consider the momentum lost because of that piece of element which touches the table and comes to stop, we need to re-write the equation for $\vec {P_{f}}$.

$\vec {P_{f}} = \lambda (l - (x + \Delta x) - h)(v + \Delta v)\hat i + \lambda h(v + \Delta v)(-\hat j) - \lambda \Delta x v(-\hat j)$

For the element which is touching the table and has come to rest: $\lambda \Delta xv = \vec {F}_{table}\Delta t$

$\Rightarrow F_{table} = \frac {\lambda \Delta xv}{\Delta t}(\hat j)$

Therefore, net external force on the entire chain which has an impact on the momentum of the chain is given by: $\vec {F}_{ext} = \lambda hg(-\hat j) + \frac {\lambda \Delta xv}{\Delta t}(\hat j)$

But now I'm thinking that will the force exerted by the tube on the horizontal part of the chain have any effect on the momentum of the entire chain. ? Should we also take that force in the equation $\vec {F}_ {ext} = \frac {d\vec P}{dt}$ ?

 

[haruspex]

But now I'm thinking that will the force exerted by the tube on the horizontal part of the chain have any effect on the momentum of the entire chain. ?

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.

 

[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?

 

[wrobel]

Most funny thing is that the system is Hamiltonian (see #18). Even the "energy" integral exists:
$$\frac{1}{2}\dot x^2+gh\ln(x+h)=\mathrm{const}$$

 

[NTesla]

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

Did you really thought that you were being helpful when you wrote that ?

When I wrote the last equation in post#52, it was obvious to me that even if I cancelled the 2nd order differentials and cancelled out the non-infinitesimals, the situation hasn't improved at all, in the direction of solving the question. But apparently, it wasn't obvious to you. I suppose it would have been obvious to you too, had you bothered to read the equation, instead of just glancing at the post. Atleast @jbriggs444 made that effort and came up with some reasoning why the last equation in post #52 wouldn't work, and I appreciate him for that.

In an attempt to make the situation plain and simple to understand, and to take the discussion forward, I'll write down the equation after having cancelled out the non-infinitesimal terms and discarding the second order infinitesimals.:

In the $\hat i$ direction, we have: $l\Delta v - x\Delta v - \Delta x v - h\Delta v = 0$
and in the $\hat j$ direction, we have: $h\Delta v -\Delta x v = hg - \frac {\Delta x}{\Delta t}v$

From these 2 equations, we have: $l\Delta v -x\Delta v = hg - \frac {\Delta x}{\Delta t}v + 2 \Delta x v$


Any idea of what to do next after the above equations ? Or are you going to say "Ok, I'll give you that one" and then abandon the whole thing, again ?

 

[NTesla]

Most funny thing is that the system is Hamiltonian (see #18). Even the "energy" integral exists:
$$\frac{1}{2}\dot x^2+gh\ln(x+h)=\mathrm{const}$$

Could you kindly write for Newtonian mechanics. I have only studied Newtonian mechanics. And what is funny here, I don't understand.

 

[NTesla]

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.

I would argue that instead of a tube, if the chain was kept on a table, and everything else remained the same, even then the chain would bend and move down at the edge of the table. I don't see how mouth of the tube has anything to do with why the chain would go down.

 

[jbriggs444]

I would argue that instead of a tube, if the chain was kept on a table, and everything else remained the same, even then the chain would bend and move down at the edge of the table. I don't see how mouth of the tube has anything to do with why the chain would go down.

The drawing shows the chain descending vertically. Not in some sort of [evolving!] parabola. This implies that the chain is constrained not just by the bottom of the tube but also by its top and right where the tube turns downward at its mouth.

If the chain fell in a curve then the length of the falling portion would not be $h$.

 

[NTesla]

The drawing shows the chain descending vertically. Not in some sort of [evolving!] parabola. This implies that the chain is constrained not just by the bottom of the tube but also by its top and right where the tube turns downward at its mouth.

If the chain fell in a curve then the length of the falling portion would not be $h$.

Yes, that makes sense.

So, the rightmost part of the mouth of the tube would exert a leftward force on the piece of the chain which is right at the bend, and the magnitude of that force would be exactly equal to cancel the horizontal momentum carried by that piece. So, that piece would only have downward momentum from that point on.