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Posted
2 minutes ago, zetetic56 said:

Yep, I got that part wrong.

I'm guessing you are right and the "infinitesimal steps" of the different movements and shifts would need to be analyzed to resolve the overall process.

Obviously, beyond anything I can do.

Thank you for considering my question and thank you for all of your insights and thank you (especially) for your patience with me. : )

Take care and 'till another time. : ) !!!

This was fun. Same to you.

Posted (edited)
On 6/12/2023 at 7:55 AM, Genady said:

This was ...

I hope you still have some interest in this.  I have a different question but it's related to what we've been talking about here.  I was going to just move on from this changing pressure stuff but I've had this other related question that seems makes sense to ask about now and here.

It's the same kinda thing but it's about a possibly vertically moving boat rather than a possibly horizontally moving wall.

apaddlewheelinaboat01.jpg.ba47a9ecb7d70cf93c9059436c708eae.jpg

There is a cup of water and a boat.  The boat is also filled with water.  Submerged in the water in the boat is a paddle wheel.  The paddle wheel is fixed in place within the boat but it can rotate around its central axis.

The paddle wheel and the boat are the same density as the fluid (except for the part of the boat above the water line which needs to be hypothetically massless so that the water level inside the boat is the same as the water level outside the boat).

Everything is at rest.

apaddlewheelinaboat02.jpg.6fc8552bc43f2bb0925c67bc4240de24.jpg

The paddle wheel is then set into motion.  And so the paddle wheel moves the fluid around inside the boat.  And so the fluid inside the boat moves relative to the inside boat walls while the fluid outside the boat remains at rest.

apaddlewheelinaboat03.jpg.6ec3ae7ff46677c88c937a47ec137201.jpg

Will the boat rise (while the fluid outside the boat falls)?

My guess is "yes".

[ But, as always, I always seem to ask these questions that seem to me at first to be super simple but then turn out to be very complicated. ]

What do ya' think?

Thank you. : )

[ Edit: If there is an atmosphere fluid pushing down on the top of the water fluid then that might make this more complicated.  It, perhaps, might be simpler to think of this thought experiment as happening in a vacuum (?)  or maybe not (?) ]

 

Edited by zetetic56
Posted
1 hour ago, zetetic56 said:

I hope you still have some interest in this.  I have a different question but it's related to what we've been talking about here.  I was going to just move on from this changing pressure stuff but I've had this other related question that seems makes sense to ask about now and here.

It's the same kinda thing but it's about a possibly vertically moving boat rather than a possibly horizontally moving wall.

apaddlewheelinaboat01.jpg.ba47a9ecb7d70cf93c9059436c708eae.jpg

There is a cup of water and a boat.  The boat is also filled with water.  Submerged in the water in the boat is a paddle wheel.  The paddle wheel is fixed in place within the boat but it can rotate around its central axis.

The paddle wheel and the boat are the same density as the fluid (except for the part of the boat above the water line which needs to be hypothetically massless so that the water level inside the boat is the same as the water level outside the boat).

Everything is at rest.

apaddlewheelinaboat02.jpg.6fc8552bc43f2bb0925c67bc4240de24.jpg

The paddle wheel is then set into motion.  And so the paddle wheel moves the fluid around inside the boat.  And so the fluid inside the boat moves relative to the inside boat walls while the fluid outside the boat remains at rest.

apaddlewheelinaboat03.jpg.6ec3ae7ff46677c88c937a47ec137201.jpg

Will the boat rise (while the fluid outside the boat falls)?

My guess is "yes".

[ But, as always, I always seem to ask these questions that seem to me at first to be super simple but then turn out to be very complicated. ]

What do ya' think?

Thank you. : )

[ Edit: If there is an atmosphere fluid pushing down on the top of the water fluid then that might make this more complicated.  It, perhaps, might be simpler to think of this thought experiment as happening in a vacuum (?)  or maybe not (?) ]

 

I say, no. Regardless of what happens inside the boat, the Archimede's principle holds.

Posted
1 hour ago, Genady said:

I say, no. Regardless of what happens inside the boat, the Archimede's principle holds.

Okay, yeah, I wasn't thinking in terms of Archimedes Principle.

apaddlewheelinaboat05.jpg.294ed1f12cc88d14b82bea4d0a619ca3.jpg

So when the paddle wheel is at rest, and given Archimedes Principle, and given that the density of the boat and the paddle wheel and the fluid in the boat are all the same as the density of the displaced fluid, then there would be no net force in any direction.

apaddlewheelinaboat04.jpg.5a019f9abc589bcc8d469df065f54b55.jpg

Also going on here is there is pressure from the fluid inside the boat pushing down on the boat and there is pressure from the fluid outside the boat pushing up on the boat.

apaddlewheelinaboat07.jpg.614390e444227566a7c0f3377f72b2b1.jpg

And so my question is: is it possible for the downward pressure on the inside of the boat from the fluid to decrease, and for this to have no effect or no impact on the overall net force in this system?

If there is no effect on the overall net force from this change in pressure, then the boat won't rise.  But I have a hard time seeing how this could not have an impact on this system.

Or, perhaps, it does have an impact on this system, but there is also some other physical dynamic than comes into play when the paddle wheel moves the fluid, and this other thing then offsets the impact from the downward decrease in pressure on the inside of the boat, and so that's how the boat does not rise?

Or, something else?

(?)   : )

 

 

 

 

Posted
4 minutes ago, zetetic56 said:

Okay, yeah, I wasn't thinking in terms of Archimedes Principle.

apaddlewheelinaboat05.jpg.294ed1f12cc88d14b82bea4d0a619ca3.jpg

So when the paddle wheel is at rest, and given Archimedes Principle, and given that the density of the boat and the paddle wheel and the fluid in the boat are all the same as the density of the displaced fluid, then there would be no net force in any direction.

apaddlewheelinaboat04.jpg.5a019f9abc589bcc8d469df065f54b55.jpg

Also going on here is there is pressure from the fluid inside the boat pushing down on the boat and there is pressure from the fluid outside the boat pushing up on the boat.

apaddlewheelinaboat07.jpg.614390e444227566a7c0f3377f72b2b1.jpg

And so my question is: is it possible for the downward pressure on the inside of the boat from the fluid to decrease, and for this to have no effect or no impact on the overall net force in this system?

If there is no effect on the overall net force from this change in pressure, then the boat won't rise.  But I have a hard time seeing how this could not have an impact on this system.

Or, perhaps, it does have an impact on this system, but there is also some other physical dynamic than comes into play when the paddle wheel moves the fluid, and this other thing then offsets the impact from the downward decrease in pressure on the inside of the boat, and so that's how the boat does not rise?

Or, something else?

(?)   : )

 

 

 

 

1. What does happen on the other side of the wheel?

2. Consider the following situation. Assume that the fluid inside the boat is sealed. Heat it up. The pressure inside the boat will rise. Will the boat sink?

Posted
2 hours ago, Genady said:

1. What does happen on the other side of the wheel?

apaddlewheelinaboat13.jpg.65614f57afd70a676fd318a19f8250fb.jpg

The downward pressure from the fluid decreases when the fluid is in motion and the upward pressure from the fluid decreases when the fluid is in motion (or so I would think).

apaddlewheelinaboat09.jpg.1ac9e5ca7d2e4f51e3a433e8d7eeeac3.jpg

If instead of having a paddle wheel in a boat filled with fluid, we had a paddle wheel in a round submarine filled with fluid, then when the paddle wheel is at rest there would be upward and downward pressures on the walls of the submarine from the fluid inside and from the fluid outside.

apaddlewheelinaboat08.jpg.871d4ef8730841d7baeef4b73e9dccf2.jpg

And when the paddle wheel moves the fluid around there would be a decrease in downward pressure on the inside of the sub and there would be a decrease in upward pressure on the inside of the sub.

And so, if the sub was neutrally buoyant when the paddle wheel was at rest, I would suspect that there would be equal decreases in upward and downward pressures and so I would suspect that the sub would also remain neutrally buoyant when the paddle wheel moves.

apaddlewheelinaboat12.jpg.eb07d8e3c5de1721e1ba4df89d766aaf.jpg

With a boat filled with fluid and an open top and nothing above, if there is a decrease in downward pressure on the boat as the fluid moves around within it, I don't see a decrease in upward pressure on the boat as well (as there was with the water filled sub).

And so I end up with the suspicion that the boat will rise.

2 hours ago, Genady said:

2. Consider the following situation. Assume that the fluid inside the boat is sealed. Heat it up. The pressure inside the boat will rise. Will the boat sink?

apaddlewheelinaboat14.jpg.8b5efb0a01afca301a4bcf50eef33a73.jpg

If there is a deck on the boat then there will be downward and upward pressures on the boat from the fluid inside it.

apaddlewheelinaboat15.jpg.9d2994b3e2dea30d2620d411ceebd5ff.jpg

And if that fluid is heated up inside the boat then there will be an increase in downward and upward pressures on the boat from the fluid inside.

And my guess is that like when the fluid moved in the sub and there was a decrease in downward and upward pressures from the fluid inside, that here where there is an increase in downward and upward pressures from the fluid inside that if the boat and its contents were neutrally buoyant when it was cool then it will be neutrally buoyant when it is hot.

---

apaddlewheelinaboat16.jpg.16ddfedc345f75bd7a1175a5d50685cc.jpg

If

1. The decrease in downward pressure on the bottom of the boat from the fluid inside means a greater overall upward force on the bottom of the boat

While if

2. The decrease in upward pressure from the fluid inside the boat is not on the boat, but is just on the open nothingness above the boat, and so if this does not create a corresponding greater overall downward force on the boat

Then

3. I would expect the boat to rise (and the fluid outside the boat to fall).

And this is why my guess (as of now) is "yes, the boat will rise".

But I'm sure there are lots of things I'm still missing here.  And so maybe it won't rise.

Thank you so much for all your insights and help. : ) !!!

 

 

 

 

 

 

Posted (edited)
31 minutes ago, zetetic56 said:

apaddlewheelinaboat13.jpg.65614f57afd70a676fd318a19f8250fb.jpg

The downward pressure from the fluid decreases when the fluid is in motion and the upward pressure from the fluid decreases when the fluid is in motion (or so I would think).

apaddlewheelinaboat09.jpg.1ac9e5ca7d2e4f51e3a433e8d7eeeac3.jpg

If instead of having a paddle wheel in a boat filled with fluid, we had a paddle wheel in a round submarine filled with fluid, then when the paddle wheel is at rest there would be upward and downward pressures on the walls of the submarine from the fluid inside and from the fluid outside.

apaddlewheelinaboat08.jpg.871d4ef8730841d7baeef4b73e9dccf2.jpg

And when the paddle wheel moves the fluid around there would be a decrease in downward pressure on the inside of the sub and there would be a decrease in upward pressure on the inside of the sub.

And so, if the sub was neutrally buoyant when the paddle wheel was at rest, I would suspect that there would be equal decreases in upward and downward pressures and so I would suspect that the sub would also remain neutrally buoyant when the paddle wheel moves.

apaddlewheelinaboat12.jpg.eb07d8e3c5de1721e1ba4df89d766aaf.jpg

With a boat filled with fluid and an open top and nothing above, if there is a decrease in downward pressure on the boat as the fluid moves around within it, I don't see a decrease in upward pressure on the boat as well (as there was with the water filled sub).

And so I end up with the suspicion that the boat will rise.

apaddlewheelinaboat14.jpg.8b5efb0a01afca301a4bcf50eef33a73.jpg

If there is a deck on the boat then there will be downward and upward pressures on the boat from the fluid inside it.

apaddlewheelinaboat15.jpg.9d2994b3e2dea30d2620d411ceebd5ff.jpg

And if that fluid is heated up inside the boat then there will be an increase in downward and upward pressures on the boat from the fluid inside.

And my guess is that like when the fluid moved in the sub and there was a decrease in downward and upward pressures from the fluid inside, that here where there is an increase in downward and upward pressures from the fluid inside that if the boat and its contents were neutrally buoyant when it was cool then it will be neutrally buoyant when it is hot.

---

apaddlewheelinaboat16.jpg.16ddfedc345f75bd7a1175a5d50685cc.jpg

If

1. The decrease in downward pressure on the bottom of the boat from the fluid inside means a greater overall upward force on the bottom of the boat

While if

2. The decrease in upward pressure from the fluid inside the boat is not on the boat, but is just on the open nothingness above the boat, and so if this does not create a corresponding greater overall downward force on the boat

Then

3. I would expect the boat to rise (and the fluid outside the boat to fall).

And this is why my guess (as of now) is "yes, the boat will rise".

But I'm sure there are lots of things I'm still missing here.  And so maybe it won't rise.

Thank you so much for all your insights and help. : ) !!!

 

 

 

 

 

 

I am sure that the boat will not rise. Because if it rises, its weight is not balanced by the decreased buoyancy, and it goes down.

I cannot tell how the internal forces will redistribute to balance each other, but they will.

BTW, you might be interested in this discussion, which is somewhat related: https://www.scienceforums.net/topic/128670-buoyant-force/?do=findComment&comment=1227608

Feel free to skip unrelated comments in that thread.

 

Edited by Genady
Posted (edited)
3 hours ago, Genady said:

BTW ...

I think the pressure change stuff I’ve been asking about here is totally different from the proposed perpetual motion machine in the other thread on a mechanical level.  But I have been butting up against conservation of energy questions here and all pmms do the same and so I can see the similarities in that aspect.  I know most people hate perpetual motion machines, but I love analyzing them.  As you know I am self-educated in Physics and so take all that follows with a truck load of salt.  And if I’m repeating anything that was already said in the other thread, I apologize.  : )

---

pmm01.jpg.cea8466b16afb813d108497a249fef2b.jpg

Fascinating perpetual motion machine.  I haven’t seen this one before.

pmm02.jpg.532c37c23cfd3e2778a49a61e9a9b707.jpg

The first thing I noticed about this has to do with the basic idea of “pressure increases with depth”.

The containers either have a vacuum inside or a compressible/decompressible fluid inside.  And so after the collapsed containers have reached the bottom and are then on the other side where they expand, when the weight falls and they expand this either creates a larger vacuum or decompresses the fluid inside.  It takes energy to either one of these.  And this energy comes from the falling weight and the loss of gravitational potential energy.

The point is that the deeper this occurs in the column of fluid the greater the pressure the falling weight has to push back against to create a larger vacuum or more decompressed fluid.

And so, this is a limiting factor.  At some depth the pressure will be too great and the force of the weight will not be able to overcome the force from the pressure of the surrounding fluid.  And so, sadly if we want this pmm to succeed, this system cannot be made taller and taller and taller.  There is limiting factor here.

To counter this we could say the weight could be made heavier and heavier and heavier.  And that would work.  But that then runs into the second thing I thought about this.

pmm03.jpg.d91b7b83c631d0a3ed9502dd68d20e2b.jpg

When the weight falls down on the lower left, this means that this weight has to be lifted a longer distance to the top.  And it takes more energy to lift a weight a longer distance.

And when the weight falls down the upper right, this means that this weight will move down along this side a shorter distance and so it has less energy to lift the weights up on the other side.

[ When the weight falls down on the lower left side this will impact the system and turn it in the wrong direction while when the weight falls down on the upper right side this will impact the system and turn it in the right direction.  And taken together they end up having no net impact. ]

And so returning to the first thing of “pressure increases with depth”, if we make the weights heavier and heavier in order to make this system taller and taller, the heavier the weight the greater the impact these two distance changes (of having to be raised more on the one side and of only getting to move down less on the other side) would have on the system.

And so, sadly if we want this to succeed, another limiting factor.

pmm04.jpg.b533d4d762481595c20f7850cc4fc9e5.jpg

And so it would come down to whether the extra buoyant force turning this clockwise is greater or lesser or equal to the energy needed to raise the weights more than they are lowered.

My guess without having done any calculations is that if we assume no friction then there is probably perfect balance.  But just an uncalculated guess.

Take care and cheers. : ) !!!

 

[ Edit: My uneducated side is showing.  I wrote: "And so it would come down to whether the extra buoyant force turning this clockwise is greater or lesser or equal to the energy needed to raise the weights more than they are lowered."  And I hope you know what I meant even though this doesn't really make sense.  I'm putting up "force" against "energy".  I really should have have said something more like "is the clockwise kinetic energy from the extra buoyant force on the right greater or lesser or equal to the energy needed to raise the weights more than they lowered".  Ugh!  Apologies.  : ) !!! ]

Edited by zetetic56
Posted
25 minutes ago, zetetic56 said:

I think the pressure change stuff I’ve been asking about here is totally different from the proposed perpetual motion machine in the other thread on a mechanical level.  But I have been butting up against conservation of energy questions here and all pmms do the same and so I can see the similarities in that aspect.  I know most people hate perpetual motion machines, but I love analyzing them.  As you know I am self-educated in Physics and so take all that follows with a truck load of salt.  And if I’m repeating anything that was already said in the other thread, I apologize.  : )

---

pmm01.jpg.cea8466b16afb813d108497a249fef2b.jpg

Fascinating perpetual motion machine.  I haven’t seen this one before.

pmm02.jpg.532c37c23cfd3e2778a49a61e9a9b707.jpg

The first thing I noticed about this has to do with the basic idea of “pressure increases with depth”.

The containers either have a vacuum inside or a compressible/decompressible fluid inside.  And so after the collapsed containers have reached the bottom and are then on the other side where they expand, when the weight falls and they expand this either creates a larger vacuum or decompresses the fluid inside.  It takes energy to either one of these.  And this energy comes from the falling weight and the loss of gravitational potential energy.

The point is that the deeper this occurs in the column of fluid the greater the pressure the falling weight has to push back against to create a larger vacuum or more decompressed fluid.

And so, this is a limiting factor.  At some depth the pressure will be too great and the force of the weight will not be able to overcome the force from the pressure of the surrounding fluid.  And so, sadly if we want this pmm to succeed, this system cannot be made taller and taller and taller.  There is limiting factor here.

To counter this we could say the weight could be made heavier and heavier and heavier.  And that would work.  But that then runs into the second thing I thought about this.

pmm03.jpg.d91b7b83c631d0a3ed9502dd68d20e2b.jpg

When the weight falls down on the lower left, this means that this weight has to be lifted a longer distance to the top.  And it takes more energy to lift a weight a longer distance.

And when the weight falls down the upper right, this means that this weight will move down along this side a shorter distance and so it has less energy to lift the weights up on the other side.

[ When the weight falls down on the lower left side this will impact the system and turn it in the wrong direction while when the weight falls down on the upper right side this will impact the system and turn it in the right direction.  And taken together they end up having no net impact. ]

And so returning to the first thing of “pressure increases with depth”, if we make the weights heavier and heavier in order to make this system taller and taller, the heavier the weight the greater the impact these two distance changes (of having to be raised more on the one side and of only getting to move down less on the other side) would have on the system.

And so, sadly if we want this to succeed, another limiting factor.

pmm04.jpg.b533d4d762481595c20f7850cc4fc9e5.jpg

And so it would come down to whether the extra buoyant force turning this clockwise is greater or lesser or equal to the energy needed to raise the weights more than they are lowered.

My guess without having done any calculations is that if we assume no friction then there is probably perfect balance.  But just an uncalculated guess.

Take care and cheers. : ) !!!

 

[ Edit: My uneducated side is showing.  I wrote: "And so it would come down to whether the extra buoyant force turning this clockwise is greater or lesser or equal to the energy needed to raise the weights more than they are lowered."  And I hope you know what I meant even though this doesn't really make sense.  I'm putting up "force" against "energy".  I really should have have said something more like "is the clockwise kinetic energy from the extra buoyant force on the right greater or lesser or equal to the energy needed to raise the weights more than they lowered".  Ugh!  Apologies.  : ) !!! ]

You are right. Very good analysis. +1

Posted
21 hours ago, Genady said:

... its weight is not balanced by the decreased buoyancy ...

I understand your reasoning that if we think in terms of weight and buoyance that the boat can’t rise.

What do you think about the question stated this way:

apaddlewheelinaboat01.jpg.51b0ffcf940a66cd5d8872b5b8f039df.jpg

There is a U shaped tube filled with a fluid.

And there are two fluid-filled open-topped flat-bottom boats with paddle wheels inside them at the top of each side of the tube.

[ The fluid inside and outside the boats is the same fluid and the densities of the paddle wheels and the boats are the same as the density of the fluid. ]

apaddlewheelinaboat02.jpg.5c19bb4587eb2d604fc0211fd02f34d8.jpg

The flat bottoms of the boats have pressure pushing down on them from the fluid above and pressure pushing up on them from the fluid below.

apaddlewheelinaboat05.jpg.ee30034faae0cd41f6176ce7f6e3884c.jpg

The paddle wheel on the right is set into motion.  And so the fluid moves around inside the boat.  And so the pressure pushing down on the flat-bottom of the right boat decreases.

The rest of the fluid remains static.  And so the pressure pushing up on the flat-bottom of the boat on the right remains the same.  And the pressures pushing down and up on the flat-bottom of the boat on the left remain the same.

apaddlewheelinaboat03.jpg.4623829df7e124a5803cd060d692ef65.jpg

Both the upward and downward pressures from the fluid on the right decrease.  But the tops of the boats are open and so there is not a decrease in upward pressure on the boat on the right.  (There is no atmosphere above the two columns of fluid.)

apaddlewheelinaboat06.jpg.96c10df152f1c2761902618afdcfa09f.jpg

If there is pressure on two sides of a moveable wall, and if there is balance, and then if the pressure on one side of that moveable wall decreases, it seems to me that that wall must move.  It cannot not move.  (Or, so it seems to me.)

apaddlewheelinaboat07.jpg.cd3267f0a44943e24dd23326bac95e56.jpg

If that is right (if my reasoning is right) then the fluid in the U shaped tube will shift from left to right and the boat on the left will fall while the boat on the right will rise.

What do ya’ think about my question stated this way?

Thank you and cheers. : ) !!!

---

apaddlewheelinaboat08.jpg.321304b6a5f755997e977e13db1c4f0e.jpg

And I suppose there is really no need for the second boat.  This would be the same question with just one boat on one side.  Since the paddle wheel and the boat are the same density as the fluid and since the fluid in the boat is the same as the fluid outside the boat, then when the paddle wheel is at rest the height of the fluid in the left tube would be the same as the height of the fluid at the top of the boat in the right tube.

Thank you for continuing to consider my question (if I haven’t totally bored you already with this by now : ) ).

Posted
2 hours ago, zetetic56 said:

The paddle wheel on the right is set into motion.  And so the fluid moves around inside the boat.  And so the pressure pushing down on the flat-bottom of the right boat decreases.

I don't think so. If the fluid motion is caused by the paddle wheel, i.e., by a shaft work, it will not cause a decrease in pressure.

Posted

My take on the original question is - there won't be a pressure ring (?) around the most constricted zone that follows the sinking ball; the zone that experiences the greatest increase in pressure is below the bottom of the ball and it will experience it equally.

ChangingPressure.png.580f793472c92a347e958b4010b606aa.png

 

 

Posted
54 minutes ago, Ken Fabian said:

My take on the original question is - there won't be a pressure ring (?) around the most constricted zone that follows the sinking ball; the zone that experiences the greatest increase in pressure is below the bottom of the ball and it will experience it equally.

ChangingPressure.png.580f793472c92a347e958b4010b606aa.png

 

 

I think I can consider, for simplicity, instead of a sinking ball, a piston with a narrow gap being pushed down. This makes it obvious (to me) that there will be an increased pressure under the piston. Why the pressure above the piston will be lowered?

Posted (edited)

@Genady Firstly I expect there is a kind of conservation of pressure applying - that the total pressure within the container whilst the ball (or piston) drops is equal to the total pressure at rest.

Above the ball appears to me to be a case of the falling ball sucks - or perhaps better described as a case of reduced gravitational pull on that part of the water column whilst the obstruction below it is falling, a bit like we experience less weight in a lift as it falls. I'm not getting this from an external reference (or I would link) - it just seems (given the very basic things I have learned about gravity and fluids and pressure) logical to me - if wrong I am willing to be corrected.

Edited by Ken Fabian
Posted
7 minutes ago, Ken Fabian said:

@Genady Firstly I expect there is a kind of conservation of pressure applying - that the total pressure within the container whilst the ball (or piston) drops is equal to the total pressure at rest. Above the ball appears to me to be a case of the falling ball sucks - or perhaps better described as a case of reduced gravitational pull on that part of the water column whilst the obstruction below it is falling, a bit like we experience less weight in a lift as it falls. I'm not getting this from an external reference (or I would link) - it just seems (given the very basic things I have learned about fluids and pressure) logical to me - if wrong I am willing to be corrected.

I don't think it will happen. Certainly not the reduced gravitational pull - the latter requires an accelerated fall, but the ball/piston rather moves in the fluid with a constant speed except momentarily at the very beginning.

I think, there is a higher pressure under the piston, a normal pressure above it, and a pressure gradient in the gap where the fluid moves up.

Posted
5 minutes ago, Genady said:

I don't think it will happen. Certainly not the reduced gravitational pull - the latter requires an accelerated fall, but the ball/piston rather moves in the fluid with a constant speed except momentarily at the very beginning.

I think, there is a higher pressure under the piston, a normal pressure above it, and a pressure gradient in the gap where the fluid moves up.

I don't see how the overall pressure can be higher whilst the ball falls than when at rest. But I stand corrected; it is not like less weight in a falling lift - as you say that depends on accelerations ie occupants of a lift falling at a steady rate will experience normal gravity - but I do think it will create suction above as it falls.

Posted

I'll think about the suction, but regarding this:

10 minutes ago, Ken Fabian said:

how the overall pressure can be higher whilst the ball falls than when at rest

I suggest a mental experiment. What happens if there is vacuum above the piston? When it moves down, the pressure under it gets higher, but the pressure above remains zero. Thus, the overall pressure gets higher. 

Posted
43 minutes ago, Genady said:

I'll think about the suction, but regarding this:

I suggest a mental experiment. What happens if there is vacuum above the piston? When it moves down, the pressure under it gets higher, but the pressure above remains zero. Thus, the overall pressure gets higher. 

 

Interesting. Initially I thought that scenario was too different to be instructive. After thinking about it I decided it is too similar to be instructive. Posit a liquid filled open vessel in air or posit it in vacuum and they are much the same. A perfect piston would not sink but the point is it isn't perfect and it does sink; the liquid flows around it as it falls and there will be liquid above it, with pressure; it displaces vacuum (or air) above and it starts being liquid above as per the initial scenario - with the pressure within the water above the question. It may simplify things to have a flat top/bottom piston with vertical sides - there will be no pressure gradient zone; I think the pressure around it as it sinks will be "normal", ie what it would be in it's initial or steady state.

I still think what happens with an immersed piston in motion making positive pressure below is that the inverse happens above, like a piston working in reverse, lowering the pressure. Whilst the top of the vessel looks to be open the liquid is contained, by gravity and as long as the piston is moving/sinking the pressure above will be lowered and below will be raised.

Posted
7 minutes ago, Ken Fabian said:

 

Interesting. Initially I thought that scenario was too different to be instructive. After thinking about it I decided it is too similar to be instructive. Posit a liquid filled open vessel in air or posit it in vacuum and they are much the same. A perfect piston would not sink but the point is it isn't perfect and it does sink; the liquid flows around it as it falls and there will be liquid above it, with pressure; it displaces vacuum (or air) above and it starts being liquid above as per the initial scenario - with the pressure within the water above the question. It may simplify things to have a flat top/bottom piston with vertical sides - there will be no pressure gradient zone; I think the pressure around it as it sinks will be "normal", ie what it would be in it's initial or steady state.

I still think what happens with an immersed piston in motion making positive pressure below is that the inverse happens above, like a piston working in reverse, lowering the pressure. Whilst the top of the vessel looks to be open the liquid is contained, by gravity and as long as the piston is moving/sinking the pressure above will be lowered and below will be raised.

Let's look at it this way. The piston moves down with a constant velocity. Thus, any volume of fluid above it moves down with the same constant velocity. Thus, total force on that unit =0. The force down is its weight. The force up is the pressure difference. Thus, this pressure difference is equal its weight. This is exactly what the "normal" pressure is. Thus, the pressure above the piston is the "normal" pressure.

Posted

@Genady I need to think about it. I note there is pressure reduction behind an object moving through water, at the extreme you get cavitation, but I don't know, your reasoning seems sound too.

Posted (edited)
On 6/16/2023 at 7:10 PM, Ken Fabian said:

I still think what happens with an immersed piston in motion making positive pressure below is that the inverse happens above, like a piston working in reverse, lowering the pressure. Whilst the top of the vessel looks to be open the liquid is contained, by gravity and as long as the piston is moving/sinking the pressure above will be lowered and below will be raised.

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If gas is in a chamber and a wall is pulled out then there will be less pressure.

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If there is a little hole in the moveable wall and if there is fluid on the other side of the wall, then since the fluid will not flow through the little hole and into the expanding chamber instantaneously, it will take a moment, it seems to me like for a moment (a very brief moment) there will be less pressure in the expanding chamber.

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If I understand your point correctly you were thinking about something like this but falling due to gravity.  (?)

If so, I think the same logic would hold, and there would be a brief moment of less pressure above the falling piston.

Well, that's my guess.

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On 6/14/2023 at 11:50 AM, Genady said:

BTW ...

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This is my favorite perpetual motion machine.  When I first saw it I thought it worked.  But then I learned some basic physics and realized why it doesn't work.

There is a wheel resting on a central axis that it can rotate around.  The wheel is half submerged in a fluid and is half outside the fluid.

halfsubmergedwheel02.jpg.9e8b170f6239d722e6b74ec39c58f604.jpg

And so there is an upward buoyant force on one side of the wheel and not on the other.

halfsubmergedwheel03.jpg.77b338506dee0c024c9e0b739d01cc25.jpg

And so it seems like it will rotate counter clockwise and it will keep rotating forever.

 

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But then I learned some basic Physics and found out that the pressure from a fluid on a submerged body has direction and its direction is perpendicular to the surface of the body.

And so, while there is more pressure on the right half of the wheel at its bottom and less pressure on its top, all of this pressure, because this is a circle, is directed at the central axis.  There is an overall upward buoyant force on the right side of this wheel, but there is no torque, there is no twisting force around the central axis.

And so, my favorite pmm does not work.

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Question:

sfq04.jpg.4ee2bfd42fba33bdb0b4d38f23f04482.jpg

On the left, there is a open topped container filled with a fluid and it has a moveable wall on its side.  On the other side of the moveable wall is a compressible spring.

I think in the left case we would say where are measuring the "pressure" of the fluid against the wall.

And then on the right there is another open topped container fill with a fluid and a moveable wall.  And this moveable wall is the floor.  And underneath the moveable floor is a compressible spring.

And I would think we would say in the second case we are measuring the "weight" of the fluid.  Right?

Question: Can we measure the "downward pressure" from a fluid on the bottom of an open topped container or would we always end up measuring "weight"?

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Cheers. : ) !

 

[ Edit:  Above I wrote "If so, I think the same logic would hold, and there would be a brief moment of less pressure above the falling piston."  And didn't express myself right.  As the tube falls and the chamber above expands, there will be brief moment followed by brief moment following by brief moment.  And so it would be this slight less pressure above the piston as the piston falls.  The "brief moment" of the fluid moving through the hole and into the expanding upper chamber will happen again and again and again.  Sorry I didn't really say it right the fist time.  I think this is better.  : ) ]

Edited by zetetic56
Posted
1 hour ago, zetetic56 said:

Can we measure the "downward pressure" from a fluid on the bottom of an open topped container or would we always end up measuring "weight"?

We measure the downward pressure times the area and this is equal the weight, if the wall moves with a constant velocity.

 

1 hour ago, zetetic56 said:

And so it would be this slight less pressure above the piston as the piston falls.  The "brief moment" of the fluid moving through the hole and into the expanding upper chamber will happen again and again and again.

The brief moment of the reduced pressure will last while the wall accelerates from the resting state to a constant velocity. When the velocity is constant, the pressure will return to be equal the weight of the fluid divided by the area of the wall, i.e., not reduced.

Posted

 

12 minutes ago, Genady said:

The brief moment of the reduced pressure will last while the wall accelerates from the resting state to a constant velocity. When the velocity is constant, the pressure will return to be equal the weight of the fluid divided by the area of the wall, i.e., not reduced.

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I think I understand what you're saying.

What I was thinking about was a case where the fluid's ability to flow upwards is in some way restricted.  But maybe I just misunderstood the conversion you two were having.  Or maybe such a restriction is irrelevant and I need some more Physics studying.  : )  Apologies for weighing in and not adding to the conversation.

 

7 minutes ago, Genady said:

We measure the downward pressure times the area and this is equal the weight, if the wall moves with a constant velocity.

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Thank you.  Downward pressure (amount x area) is equal to the weight, cool, that makes sense.  Weight and downward pressure are equal amounts.

What I'm wondering is this: There is a force pushing down on the spring.  Would we say "this force is from the downward pressure from the fluid" or would we say "this force is from the gravitational attraction between the fluid and the Earth below"?

Thank you. : )

 

 

 

 

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