Chain falling out of a horizontal tube onto a table

[haruspex]

No retardation condition is fulfilled when no motion is happening. So, yes, when the chain is not moving at all, then Tension is more than when it is moving. So, this statement is true.


I'm still trying to understand this.

Let's say that before the chain starts to move, the tension at the rightmost point of the purely horizontal portion of the chain is $T_{when\,static}$.
Now, let's say, the chain is in moving condition, then the tension at the same location is $T_{when\,moving}$.
Let $F_{on\,element}$ be the leftward force on the element which is just at the bend.
then $T_{when\,moving}$ < $T_{when\,static}$ as long as there is a horizontal chain to talk about.
When you say, that

It means that equation wise, it would be like this:
$T_{when\,moving}$ = $T_{when\,static}$ - $F_{on\,element}$ . ...........................................(i)
And from earlier posts, we have established that magnitude of $F_{on\,element}$ = $\lambda v^{2}$.

But, $T_{when\,static}$ = $\lambda hg$, and $T_{when\,moving}$ = $\lambda h(g - \frac {dv}{dt})$
So, $T_{when\,moving}$ = $T_{when\,static}$ - $\lambda h \frac {dv}{dt}$ ...........................(ii)

Clearly, equations (i) and (ii) are not same, the last term that is.

At this point I'm certain that I'm misunderstanding the critical point of view/ information about what you've written. But I've read your above post multiple times, and have spent much time thinking about it, but I still don't understand it.

The retardation force on the horizontal portion is not the only reason the tension reduces. Consider a massless string but with a mass on each end. The tension is reduced when accelerating even though there is no retardation force. This corresponds to the $-\lambda\frac{dv}{dt}$ term in the vertical.
The correct comparison is with the massless string with two point masses model, though the vertical equation may be complicated by table impact.

 

[haruspex]

Thinking about it some more, I have decided it is not helpful to describe it the way I did, with the $\lambda v^2$ being subtracted from the tension both sides. That would make sense if the quadrant contained a contraption that first brought each element of the horizontal stream to rest then kicked it out vertically at the same speed, but that’s not how it works.
We know that overall the quadrant must exert a force $\sqrt 2\lambda v^2$ down and left at 45°, but it does that by some pattern of normal forces around the quadrant. Since these are normal to the chain, they do not affect the tension.

Taking it in stages…
Scenario 1: vertical portion only, with each element hitting the table being magically returned to the top and retaining its velocity.
Clearly the portion is in free fall, $\dot v=g$.
Scenario 2: As 1, but the returned element starts as stationary.
This creates a retardation force $\lambda v^2$, so $\dot v=g-v^2/h$.
If we consider the tension immediately below the point where the element reappears, $T_o=\lambda v^2$.
Scenario 3: As 2, but the returned element is returned to the horizontal inlet to the quadrant instead of to the vertical outlet.
Now, on the outlet side, the element is already up to speed, so there is no retardation force for that. Instead, there is a tension $T_i=\lambda v^2$ on the inlet side, just to the right of the returned element. This is transferred around the quadrant to the outlet side, so we still have $T_o=\lambda v^2$.
Scenario 4: As 3, but the returned element is introduced with its prior speed, now horizontal.
That brings the tension back to zero.
Scenario 5: Adds the whole horizontal portion.
$T=\lambda (L-h-x)\dot v$.

 

[NTesla]

The retardation force on the horizontal portion is not the only reason the tension reduces. Consider a massless string but with a mass on each end. The tension is reduced when accelerating even though there is no retardation force. This corresponds to the −λdvdt term in the vertical.

Yes. I agree with this. However, I suppose you meant to write $\lambda h \frac {dv}{dt}$ instead of $\lambda \frac {dv}{dt}$. It was just a minor typo, I think.

overall the quadrant must exert a force 2λv2 down and left at 45°, but it does that by some pattern of normal forces around the quadrant. Since these are normal to the chain, they do not affect the tension.

Yes I agree.


For Scenario 2, you wrote:

This creates a retardation force λv2, so v˙=g−v2/h.
If we consider the tension immediately below the point where the element reappears, To=λv2.

This is the way I'm seeing this: When scenario 2 starts with Scenario 1, this means the vertical portion is in free fall. Now the element that is about to be attached to the top of this freely falling chain, is also falling downward with the velocity of the chain. But just before it has to attach itself to the freely falling chain, it has to suddenly lose all it's velocity. So this requires an upward impulse on this element whose magnitude is $\lambda v^{2}$. This impulse brings the velocity of that element to zero. But you've written:"This creates... ". I don't understand what is creating this upward force.

For scenario 3 you've written:
"Now, on the outlet side, the element is already up to speed, so there is no retardation force for that. Instead, there is a tension $T_{i} =λv^{2}$ on the inlet side, just to the right of the returned element. This is transferred around the quadrant to the outlet side, so we still have $T_{o} =λv^{2}$."

For the element on the outlet, you've mentioned that "there is no retardation force for that". And then in the next sentence you've written: "this is transferred around the quadrant, so we still have $T_{o}$". I don't understand this at all.

And consequently I couldn't understand scenarios 4 and 5.

Although I appreciate your kind help very much, but could you kindly explain in some other way. I found this different scenarios situation extremely difficult to grasp.

 

[haruspex]

Yes. I agree with this. However, I suppose you meant to write $\lambda h \frac {dv}{dt}$ instead of $\lambda \frac {dv}{dt}$. It was just a minor typo, I think.

Yes I agree.


For Scenario 2, you wrote:

This is the way I'm seeing this: When scenario 2 starts with Scenario 1, this means the vertical portion is in free fall. Now the element that is about to be attached to the top of this freely falling chain, is also falling downward with the velocity of the chain. But just before it has to attach itself to the freely falling chain, it has to suddenly lose all it's velocity. So this requires an upward impulse on this element whose magnitude is $\lambda v^{2}$. This impulse brings the velocity of that element to zero. But you've written:"This creates... ". I don't understand what is creating this upward force.

For scenario 3 you've written:
"Now, on the outlet side, the element is already up to speed, so there is no retardation force for that. Instead, there is a tension $T_{i} =λv^{2}$ on the inlet side, just to the right of the returned element. This is transferred around the quadrant to the outlet side, so we still have $T_{o} =λv^{2}$."

For the element on the outlet, you've mentioned that "there is no retardation force for that". And then in the next sentence you've written: "this is transferred around the quadrant, so we still have $T_{o}$". I don't understand this at all.

And consequently I couldn't understand scenarios 4 and 5.

Although I appreciate your kind help very much, but could you kindly explain in some other way. I found this different scenarios situation extremely difficult to grasp.

I am sorry, but I am rather snowed now. I'll try to get back to this in a day or two.

 

[jbriggs444]

This is the way I'm seeing this: When scenario 2 starts with Scenario 1, this means the vertical portion is in free fall. Now the element that is about to be attached to the top of this freely falling chain, is also falling downward with the velocity of the chain. But just before it has to attach itself to the freely falling chain, it has to suddenly lose all it's velocity. So this requires an upward impulse on this element whose magnitude is $\lambda v^{2}$. This impulse brings the velocity of that element to zero. But you've written:"This creates... ". I don't understand what is creating this upward force.

The element that was returned to the top never left the system. The impulse that brought it to momentary rest was an external force.

The description of the scenario requires this external force. If @haruspex wishes to think of the scenario modification as having "created" the required external force, that is fine.

For scenario 3 you've written:
"Now, on the outlet side, the element is already up to speed, so there is no retardation force for that. Instead, there is a tension $T_{i} =λv^{2}$ on the inlet side, just to the right of the returned element. This is transferred around the quadrant to the outlet side, so we still have $T_{o} =λv^{2}$."

For the element on the outlet, you've mentioned that "there is no retardation force for that". And then in the next sentence you've written: "this is transferred around the quadrant, so we still have $T_{o}$". I don't understand this at all.

In scenario 3 (returned links return to the inlet side and are brought to rest by some external force) do you agree that we have tension $T_i$ at the inlet? This tension is responsible for the internal force that brings the returned links up to horizontal velocity $v$.

So we have tension $T_i = \lambda v^2$ at the inlet. The mass of the chain within the curve is infinitesimal. Do you agree that it follows that the tension $T_o$ at the outlet must be the same?

The argument is that any difference in tension is given by the acceleration of the entire chain multiplied by the mass of just the section that exists within the 90 degree curve (the quadrant). We can make the quadrant as small as we please. So the tension difference can be made as small as we please.

It is the tension that @haruspex speaks of as having been "transferred around the quadrant".

 

[NTesla]

do you agree that we have tension Ti at the inlet?

Yes. I agree. Reason: There's no point on the chain that doesn't have tension, except the ends of the chain. So, yes, there is tension at the bend. However, I'm not sure whether the tension at the inlet would be $\lambda v^{2}$ or not. This is because, $\lambda v^{2}$ is the leftward force on the element which is on the bend. Just before this element, tension is = $\lambda(l - h - x)$. But right after the element, I don't know what the tension will be.

Do you agree that it follows that the tension To at the outlet must be the same?

I agree with the calculation done by @haruspex in post#87 that $T_v$ = $T_h$.

So the tension difference can be made as small as we please.

Yes, @haruspex has done the calculation which shows that $T_v$ = $T_h$ in post#87

However, as I'm mentioned in post#93, I'm finding not just difficult to understand the explanation given in post#92 and 95, but near impossible to understand as to how the explanation given in these 2 posts is helping me understand the question that I had raised in post#84.

 

[jbriggs444]

Yes. I agree. Reason: There's no point on the chain that doesn't have tension, except the ends of the chain. So, yes, there is tension at the bend. However, I'm not sure whether the tension at the inlet would be $\lambda v^{2}$ or not. This is because, $\lambda v^{2}$ is the leftward force on the element which is on the bend. Just before this element, tension is = $\lambda(l - h - x)$. But right after the element, I don't know what the tension will be.

Which scenario are you contemplating here? I made it quite clear that I was writing about scenario 3.

In this scenario there is nothing to the left of the element that is entering the inlet to the quadrant. The only source of tension in this scenario is the impulsive acceleration of the incremental added mass, bringing that incremental mass up to speed to match the rest of the chain.

The formula $\lambda(l - h - x)$ does not apply in scenario 3. The tension "just before this element" is undefined because there is absolutely nothing before this element.

 

[NTesla]

As I've mentioned in post#93 and 96, I'm finding it not just difficult to understand the explanation given in post#92 and 95, but near impossible to understand as to how the explanation given in these 2 posts would help me understand the question that I had raised in post#84. And post#97 is in the same line of explanation.

I'm just not able to understand how all these explanations tie up altogether in answering the question that I posed in post#84.

In your previous post you've written:

In this scenario there is nothing to the left of the element that is entering the inlet to the quadrant.

This i'm failing to find useful. Whatever element is at the bend, there is chain to the left of that element and also the right of that element. And I understand that you specifically wrote this about scenario 3. But this is where I'm not able to understand any further. This different scenario situation is not at all helpful to me. I'm not able to put together how the scenario wise explanation is useful.

I request you to kindly explain the answer to the question in post#84, in some other way.

 

[jbriggs444]

As I've mentioned in post#93 and 96, I'm finding it not just difficult to understand the explanation given in post#92 and 95, but near impossible to understand as to how the explanation given in these 2 posts would help me understand the question that I had raised in post#84. And post#97 is in the same line of explanation.

If you have something to say, say it. Do not say that you already said it.

I am not going to go back to prior posts to try to tease out some unseen meaning. If it was not clear then, it will not be clear now.

I'm just not able to understand how all these explanations tie up altogether in answering the question that I posed in post#84.

Again, this is unhelpful.

In your previous post you've written:

"In this scenario there is nothing to the left of the element that is entering the inlet to the quadrant."

This i'm failing to find useful. Whatever element is at the bend, there is chain to the left of that element and also the right of that element.

What does "at the bend" mean?

In the scenario 3 that I am considering, the bend has a small size. It has an inlet. It has an outlet. It has a curved region in between. If we consider an element, it is much smaller than the bend. The element has a position within the bend. Simply saying "at the bend" is not sufficiently precise.

In scenario 3 there is nothing to the left of the bend. Chain materializes exactly at the entrance to the bend. There is nothing to the left of the chain segments that are materializing there.

And I understand that you specifically wrote this about scenario 3. But this is where I'm not able to understand any further. This different scenario situation is not at all helpful to me. I'm not able to put together how the scenario wise explanation is useful.

If I understand you correctly, you find scenario 3 unhelpful. So you have discarded it and are commenting on the original scenario instead.

It is unhelpful to comment on one scenario while pretending to comment on another.

I request you to kindly explain the answer to the question in post#84, in some other way.

I decline to read post #84 in an attempt to figure out the question you cannot be bothered to ask here.

 

[NTesla]

I decline to read post #84 in an attempt to figure out the question you cannot be bothered to ask here.

Who says I cannot be bothered to ask the question here. Did you ask ? Even once ? The answer is NO. Not even once. If you didn't understand the question itself, you should have asked question, or atleast let me know that you are not able to understand the question. I would have tried to frame the question differently. But no. Helping someone was not really your intention, but only pretending to help.

Your entire previous post is indicative of you being unreasonably resentful and irritated even when there's no cause for that. If you bother to go through this post of mine, maybe you'll start to see reason as to why I referred to post#84.

As long as I've seen this forum, it is a standard practice in this forum to refer to any previous post in order to refer to the calculations/ paragraph therein. I've seen it done many times earlier. When I did it, it became a problem for you. !! I have done the same on many earlier occasions, and no one had any problem with it. You are the 1st person who is having problem with it. Everyone else seems to understand that referring to a previous post is an efficient way for written conversation. If that's a problem for you, the problem lies with you, not me.

IF you would have read my post#84, you would have come to understand that the question that I raised in that post is not a one liner. It took me quite some equations and words to express the question. Had it been a one liner question or if you had asked to repeat the question again, even once, I would have definitely repeated as often as possible/required. But you have clearly declined to even click a button to go back and go through the post. This behaviour of your is manifestly unreasonable. Would it have hurt your ego to click a button to go back to read that post ? If it does, then i'll post it here for your convenience, so that you don't have to bear the heavy burden of moving the mouse to click the button for going to page 3. Here is the screenshot of that post:

I am not going to go back to prior posts to try to tease out some unseen meaning. If it was not clear then, it will not be clear now.

Had you attempted to read that post of mine, which you've clearly refused to do, you would have found that it is quite self explanatory. There's no "unseen meaning". It's as simple and plain as it could be. And, if my question wasn't clear to you, did you ask me anything to clear up that question. You didn't, not even once !!

You didn't even understand the question that I had asked. So, why were you answering, in your previous posts if you were not even aware as to what I had asked. Twice already I have mentioned that this explanation of scenarios is not helpful to me and had requested to explain it differently. But you chose to ignore that and continued on. Instead of only pretending to help, if you had really wanted to help someone, you would have atleast tried to read the question.

Again, this is unhelpful.

Of course it's unhelpful to you.

What does "at the bend" mean?

It would have been clear to you had you bothered to read that post of mine. You asked this question only because of your reluctance to even read questions before answering. In order to actually help someone, one must atleast take effort to read the question. But if one is going to only pretend to be helpful, then there's no recourse.

It is unhelpful to comment on one scenario while pretending to comment on another.

You have not even bothered to read the question that has been raised, and are only pretending to help others by doing the minimum and the funny thing is that you are blaming others that they are pretending.

Few years ago also, I had once interacted with you and seeing your foul behavior back then also, I had blocked you, and had forgotten all about blocking you. But when you posted your answer for this question, I unblocked you, shouldn't have done so, but I did, to see what you might have written. But you are such an egoistic person that conversing with you is a painful waste of time, everytime. It's such a disappointing experience talking to you.

Now, don't bother responding as I'm going to block you again. And I'm quite sure you will be reporting this post to the forum as offensive. But that won't change the truth that it's your own egoistic immature self, that is at the root of your problem. So, go ahead, do report.

 

[jbriggs444]

Who says I cannot be bothered to ask the question here. Did you ask ? Even once ? The answer is NO. Not even once. If you didn't understand the question itself, you should have asked question, or atleast let me know that you are not able to understand the question. I would have tried to frame the question differently. But no. Helping someone was not really your intention, but only pretending to help.

Since this is your attitude, I am out. Live long and prosper.