# A specific case of an exception to Archimedes' Principle?

• I
• Novus
In summary: If the object moves upwards, the pressure of the water on the bottom of the object will decrease since it is being forced up.In summary, the object can move up or down, depending on the pressure of the water below it.
Novus
TL;DR Summary
Recently I've become interested in the bottom and side case exceptions to Archimedes' Principle for which a limited number of interesting scientific papers can be found on-line.
First post on this forum, hopefully in the correct category and conforming to forum rules.

Will the object in below scenario move upwards, downwards or will it remain stationary?

An object is submersed in a container 'A' which contains a fluid 'B' (e.g. water)
The trapezium shaped object consists of two seperate parts which enclose each other in such a way that no water can seep in and, as a a result, can expand in a horizontal plane.
The submersed object consists of solid area's 'C' with the same density as water and a volume of air 'D' which can expand or compress in a horizontal plane. The volume of air 'D' is connected to the air outside of the container via air tubes.
It is given that, without going in further detail that "The sides of the object are 'embedded' water tight on a 'sliding system' inside the container wall" The object can move up or down whereby, as a result, the volume of air will either increase or decrease.
Each square in the picture is 1 cm2. The object has a width of 1 cm. The width of the container is > 1 cm.
The top of the object is at a dept of 4 cm and the bottom at a dept of 6 cm. The area of the top of the object is 10 cm2 and the area of the bottom of the object is 8 cm2.

The hydrostatic pressure on the bottom of the object will, as a result, exceed the hydrostatic pressure at the top of the object?

#### Attachments

• Buoyancy scenario_question.JPG
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Welcome to PF.

-1- It looks like the object is wedged into the container, with no freedom to move downward. How could it move downward?

-2- What makes you think that this would be an "exception to Archimedes' Principle"? What unusual feature are you relying on to cause that?

berkeman said:
Welcome to PF.

-1- It looks like the object is wedged into the container, with no freedom to move downward. How could it move downward?

-2- What makes you think that this would be an "exception to Archimedes' Principle"? What unusual feature are you relying on to cause that?
Hi Berkeman, thanks for taking an interest in this topic.

1. As I tried to explain the object can 'freely' move up- and downwards since:
- "The trapezium shaped object consists of two seperate parts which enclose each other in such a way that no water can seep in and, as a a result, can expand in a horizontal plane."
- "The object can move up or down whereby, as a result, the volume of air will either increase or decrease."

2. I' m not sure if this would qualify as a true exception to AP, hence the questionmark in the title. However in the side and bottomcase exception hydrostatic pressure is missing on part of the surface of the object which seems to be the case here as well.

I suspect there may be some handwaving magic here:

"The sides of the object are 'embedded' water tight on a 'sliding system' inside the container wall" The object can move up or down whereby, as a result, the volume of air will either increase or decrease."

russ_watters
Novus said:
TL;DR Summary: Recently I've become interested in the bottom and side case exceptions to Archimedes' Principle for which a limited number of interesting scientific papers can be found on-line.

First post on this forum, hopefully in the correct category and conforming to forum rules.
Nearly: if you are going to mention that "a limited number of interesting scientific papers can be found on-line" then you should provide links.

Novus said:
Will the object in below scenario move upwards, downwards or will it remain stationary?
You don't need to calculate anything to determine this: what would happen to the water below the object if the object moved downwards? What would happen below the object if it moved upwards?

Novus said:
The hydrostatic pressure on the bottom of the object will, as a result, exceed the hydrostatic pressure at the top of the object?
Do you mean hydrostatic pressure or hydrostatic force? What do you think the difference in force would be?
What is the difference in weight supported above the bottom and above the top of the object?

berkeman said:
-1- It looks like the object is wedged into the container, with no freedom to move downward. How could it move downward?
The object can "telescope up" horizontally and, if it were not otherwise constrained, move up or down freely (ignoring the changing pressure of the air inside it).

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The diagram shows no path for water to move from above the object to below the object. It would have to lift the water above it and make a vacuum under it to move.

That doesn't seem like bouyancy. That seems like an ordinary situation of water pressure holding a thing down where the water wants to flow. A styrofoam plate in a sink over the drain is held by the water pressure. As is a balloon.

If there was a side channel for the water to move from above to below, then the object would move up.

russ_watters and jbriggs444
What if you remove the water and push the object either up or down? Would it move? Have you figured this out, before adding water? Won't the behavior depend on angle of the wedges and the friction coefficient ( I mean for the dry run).

votingmachine said:
The diagram shows no path for water to move from above the object to below the object. It would have to lift the water above it and make a vacuum under it to move.
This.

Archimedes principle can be derived from Pascal's principle. Pascal's principle relies on the fluid being connected. That is, there needs to be a fluid-filled path available from every point in the container to every other point.

Once you seal off the top volume of water from the bottom volume, the pressure within the bottom volume is no longer necessarily equal to ##\rho g h##. This means that the net force from fluid pressure on the two faces of a seal with volume ##V## need not be given by ##\rho g V##.

pbuk said:
Nearly: if you are going to mention that "a limited number of interesting scientific papers can be found on-line" then you should provide links.
https://arxiv.org/pdf/1110.5264.pdf

pbuk said:
You don't need to calculate anything to determine this: what would happen to the water below the object if the object moved downwards? What would happen below the object if it moved upwards?
If the object moves downward the volume of water below the object would decrease. If it moves upwards the volume of water would increase. Not sure how this relates to the initial question if the object will move up, down or remain stationary?
pbuk said:
Do you mean hydrostatic pressure or hydrostatic force? What do you think the difference in force would be?
You are correct. Hydrostatic force rather than hydrostatic pressure.
Hydrostatic forces are the resultant force caused by the pressure loading of a liquid acting on submerged surfaces. Calculation of the hydrostatic force and the location of the center of pressure are fundamental subjects in fluid mechanics. The center of pressure is a point on the immersed surface at which the resultant hydrostatic pressure force acts.

Logically I think the force on the bottom would be higher than on the top based on multiplying the area's and dept.

pbuk said:
What is the difference in mass supported above the bottom and above the top of the object?
pbuk said:
The object can "telescope up" horizontally and, if it were not otherwise constrained, move up or down freely (ignoring the changing pressure of the air inside it).

votingmachine said:
The diagram shows no path for water to move from above the object to below the object. It would have to lift the water above it and make a vacuum under it to move.

That doesn't seem like bouyancy. That seems like an ordinary situation of water pressure holding a thing down where the water wants to flow. A styrofoam plate in a sink over the drain is held by the water pressure. As is a balloon.

If there was a side channel for the water to move from above to below, then the object would move up.
In the case description it was mention that "The object has a width of 1 cm. The width of the container is > 1 cm." Water can therefore move from below the object to below the object and vice versa.

nasu said:
What if you remove the water and push the object either up or down? Would it move? Have you figured this out, before adding water? Won't the behavior depend on angle of the wedges and the friction coefficient ( I mean for the dry run).
Since the density of the object > density of air it would move down if there is no water in the container.

Novus said:
If the object moves downward the volume of water below the object would decrease.
Where is it going to go?

Novus said:
If it moves upwards the volume of water would increase.
Where is it going to come from?

votingmachine said:
The diagram shows no path for water to move from above the object to below the object. It would have to lift the water above it and make a vacuum under it to move.

That doesn't seem like bouyancy. That seems like an ordinary situation of water pressure holding a thing down where the water wants to flow. A styrofoam plate in a sink over the drain is held by the water pressure. As is a balloon.

If there was a side channel for the water to move from above to below, then the object would move up.
"If there was a side channel for the water to move from above to below, then the object would move up."
Anyone else agrees based on the understanding that in this case the volume of air contained as part of the object would increase?

Novus said:
In the case description it was mention that "The object has a width of 1 cm. The width of the container is > 1 cm." Water can therefore move from above the object to below the object and vice versa.

pbuk said:
Where is it going to go?Where is it going to come from?
I understand that the objective of the forum is to educate the user by asking questions. Based on the scenario the object would move upwards since (i) the forces on the top and bottom of the object are directly related to surface area * dept and (ii) the density of the object < density of water? However the consequency of the object moving upwards is that the volume of air increases and, as a result the water level on top of the container moves upwards which seems to be counterintuitive?

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Novus said:
In the case description it was mention that "The object has a width of 1 cm. The width of the container is > 1 cm." Water can therefore move from below the object to below the object and vice versa.
Oh I now understand what you mean: you are measuring "width" perpendicular to the plane of the diagram.

pbuk said:
Oh I now understand what you mean: you are measuring "width" perpendicular to the plane of the diagram.
Correct, apologies if this was not clear - probably as a result of the fact that english is not my native language.

Novus said:
However the consequency of the object moving upwards is that the volume of air increases and, as a result the water level on top of the container moves upwards which seems to be counterintuitive?
It's more than counterintuitive, in order for the level to have been raised (and therefore potential energy increased) something must be doing work on the water: what could that be?

Consider instead the air "inside" the object: what is going to happen to that?

Novus said:
https://arxiv.org/pdf/1110.5264.pdf

If the object moves downward the volume of water below the object would decrease. If it moves upwards the volume of water would increase. Not sure how this relates to the initial question if the object will move up, down or remain stationary?
You are correct. Hydrostatic force rather than hydrostatic pressure.
Hydrostatic forces are the resultant force caused by the pressure loading of a liquid acting on submerged surfaces. Calculation of the hydrostatic force and the location of the center of pressure are fundamental subjects in fluid mechanics. The center of pressure is a point on the immersed surface at which the resultant hydrostatic pressure force acts.

Logically I think the force on the bottom would be higher than on the top based on multiplying the area's and dept.

pbuk said:
It's more than counterintuitive, in order for the level to have been raised (and therefore potential energy increased) something must be doing work on the water: what could that be?
Wouldn't the P.E. of the object, when moving upwards decrease in comparison to the initial lower position?
pbuk said:
Consider instead the air "inside" the object: what is going to happen to that?
The air is connected to the outside air, and since there are no lateral forces, nothing will happen to the air? In other words there are no forces, except for the atmospheric pressure of the ambient air which will remain the same if the object moves up, down or remain stationary?

Novus said:
"If there was a side channel for the water to move from above to below, then the object would move up."
Anyone else agrees based on the understanding that in this case the volume of air contained as part of the object would increase?
I think I was wrong. The water above is above the level it would be at if the air in the middle could be displaced. I think the air would be displaced by the accordion collapse of the middle chamber (if there was a connection tube top-to-bottom).

It seems to depend on the connection to the sides described as a water tight seal. Say the sides are pulled out by magnets ... then it would definitely slide down. Say the sides are just there due to a temporary bit of hydraulic pressure ... then it would let go of the sides and water flow around it.

It seems like the water-tight connection to the sides is a matter of interpretation. If there is some magnetic pull then the seal would never break and it would stay motionless, even though it is holding a bubble of air (but so does a submarine). If the two pieces can slide together, then they would. But the can only slide together if the hydraulic pressure of the lower chamber is released via a tube.

The telescoping thing needs better description of the water tight side system.

As I see the description, the thing is just locked in place by the downward force of the water above it.

votingmachine said:
I think I was wrong. The water above is above the level it would be at if the air in the middle could be displaced. I think the air would be displaced by the accordion collapse of the middle chamber (if there was a connection tube top-to-bottom).

It seems to depend on the connection to the sides described as a water tight seal. Say the sides are pulled out by magnets ... then it would definitely slide down. Say the sides are just there due to a temporary bit of hydraulic pressure ... then it would let go of the sides and water flow around it.

It seems like the water-tight connection to the sides is a matter of interpretation. If there is some magnetic pull then the seal would never break and it would stay motionless, even though it is holding a bubble of air (but so does a submarine). If the two pieces can slide together, then they would. But the can only slide together if the hydraulic pressure of the lower chamber is released via a tube.

The telescoping thing needs better description of the water tight side system.
I'll try to come up with a better description (which probably needs a picture) on how the sides of the trapezium shaped object are water tight connected to the side walls of the container and yet can freely move up and down.
votingmachine said:
As I see the description, the thing is just locked in place by the downward force of the water above it.
Why? The area * dept at the bottom > the area * dept at the top and the density of the object < density of water?

Novus said:
I'll try to come up with a better description (which probably needs a picture) on how the sides of the trapezium shaped object are water tight connected to the side walls of the container and yet can freely move up and down.

Why? The area * dept at the bottom > the area * dept at the top and the density of the object < density of water?
If it moves down, it is opposed by the resistance to compression of the bottom water. If it moves up it is opposed by the vacuum it creates underneath it and the force of lifting the water above it.

There are many ways to trap air underwater. You have come up with one.

Consider an upside down bucket of air under water. It wants to rise as the water wants to fall around it. Now put that bucket in a cylinder with the seal water tight. The bucket can't rise because the water cannot fall.

I'm woefully out of tune with Archimedes, but buoyancy depends on the water displacing the object until the weight of the volume displaced equals the weight of the object. So a boat rides above the water as it weighs less than the total displacement of the boat. But it depends on the water being able to fall downwards, around the object. If you step off a boat onto the dock the boat rises as the water falls.

Novus said:
Wouldn't the P.E. of the object, when moving upwards decrease in comparison to the initial lower position?
No, if it moves upwards the PE will increase, as well as the PE of the water. This can't happen.

Novus said:
The air is connected to the outside air, and since there are no lateral forces,
Why do you think there are no lateral forces? Is there not a horizontal component of the normal force between the sloping ends of the object and the container?

Novus said:
nothing will happen to the air?
I think the air will be forced out and the object will sink until either it reaches the bottom or it reaches its minimum length, thus minimizing potential energy.

pbuk said:
No, if it moves upwards the PE will increase, as well as the PE of the water. This can't happen.
I agree that, obviously, P.E. can not increase without an external input of energy, however I'm not sure if this is neccesarily the case when the object would move upwards - see attached "P.E. scenario's picture"
pbuk said:
Why do you think there are no lateral forces? Is there not a horizontal component of the normal force between the sloping ends of the object and the container?
Please find attached picture ("example object movement") with one possibly way on how the object could move up or down in relation to the the walls of the container. I fail to see/understand why there would be a horizontal/lateral force?
pbuk said:
I think the air will be forced out and the object will sink until either it reaches the bottom or it reaches its minimum length, thus minimizing potential energy.
I agree that this is the most likely outcome from an P.E. perspective. However (i) the density of the object < the density of water (ii) the bottom area * depth > the top area * depth (iii) there is no difference between the pressure of the air pocket in the object compared to the ambient air, which leaves (iv) there are no lateral forces acting upon the object?

#### Attachments

• Example_object movement.JPG
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• P.E. scenario's.JPG
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Novus said:
I agree that, obviously, P.E. can not increase without an external input of energy, however I'm not sure if this is neccesarily the case when the object would move upwards - see attached "P.E. scenario's picture"
Yes, increasing the volume of a sphere that has fluid completely surrounding it increases buoyancy.

No, increasing the volume of a trapezium that has fluid on only two sides does not increase the pressure differential between those two sides.

The PE of the system containing the trapezium will include the PE of the water that is displaced. As the trapezium moves down and shrinks, this reduces the PE of the water. As the trapezium moves up and expands, this increases the PE of the water.

Novus said:
I agree that this is the most likely outcome from an P.E. perspective. However (i) the density of the object < the density of water (ii) the bottom area * depth > the top area * depth (iii) there is no difference between the pressure of the air pocket in the object compared to the ambient air, which leaves (iv) there are no lateral forces acting upon the object?
Because the object is not surrounded by fluid on all sides, Archimedes law does not apply. The net force from fluid pressure must be calculated from first principles rather than from the weight of the displaced fluid.

I agree that there are no lateral forces. And no vertical forces from the sealed and slanting sides.

jbriggs444 said:
Yes, increasing the volume of a sphere that has fluid completely surrounding it increases buoyancy.
Using the method in attached file ('buoyancy calculations') the resulting Fb for the scenario would be positive. F1 = h1pgA = 4pg10 = 40pg F2=h2pgA = 6pg8 = 48pg Fb = F2 - F1 = 8pg
jbriggs444 said:
No, increasing the volume of a trapezium that has fluid on only two sides does not increase the pressure differential between those two sides.
When the trapezium moves upwards 2 cm the resulting Fb would be F1 = h1pgA = 2pg12 = 24pg F2=h2pgA = 4pg10 = 40pg Fb = F2 - F1 = 16pg
jbriggs444 said:
The PE of the system containing the trapezium will include the PE of the water that is displaced. As the trapezium moves down and shrinks, this reduces the PE of the water. As the trapezium moves up and expands, this increases the PE of the water.
I agree that the fact that P.E. increases in case the object would move upwards answers the initial question; e.g. the object will move downwards.
jbriggs444 said:
Because the object is not surrounded by fluid on all sides, Archimedes law does not apply. The net force from fluid pressure must be calculated from first principles rather than from the weight of the displaced fluid.
Thanks for confirming that in this case AP does not apply. I'm still unclear on how the net force from fluid pressure can be calculated from first principles rather than from the weight of the displaced fluid?
I agree that there are no lateral forces. And no vertical forces from the sealed and slanting sides.
Thanks for your confirmation that there are no lateral forces or vertical forces from the sealed and slanting sides.

#### Attachments

• Buoyancy_calculation.jpg
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jbriggs444 said:
As the trapezium moves up and expands, this increases the PE of the water.
As the trapezium moves up and expands, water is also flowing down, from above to under the trapezium. Just because the water level rises, doesn't necessarily mean that the water's center of mass (and thus PE) also rises.

jbriggs444
Novus said:
Thanks for confirming that in this case AP does not apply. I'm still unclear on how the net force from fluid pressure can be calculated from first principles rather than from the weight of the displaced fluid?
That is straightforward as long as one makes the assumption that the trapezium moves slowly enough that the fluid remains at approximate equilibrium at all times.

The fluid is free to flow up and down past the trapezium. So Pascal's principle applies. Pressure is given by ##P = \rho g h##. The total force on the top and bottom surface is easily calculated: ##F=PA##. The difference between those forces is the net force on the trapezium. We can ignore the forces on the slanted faces on the right and left because those are both zero. We can ignore the forces on the flat faces on the front and back because those cancel with each other.

Novus said:
I agree that the fact that P.E. increases in case the object would move upwards answers the initial question; e.g. the object will move downwards.
I would not be so sure about the change in PE. See my post #27

Novus said:
I'm still unclear on how the net force from fluid pressure can be calculated from first principles rather than from the weight of the displaced fluid?
By adding up the pressure forces and weight to get the net force. This result should be consistent with the result from PE change in terms of movement direction.

jbriggs444
A.T. said:
As the trapezium moves up and expands, water is also flowing down, from above to under the trapezium. Just because the water level rises, doesn't necessarily mean that the water's center of mass (and thus PE) also rises.
Good point. See attached calculations for the water's center of mass for the original scenario (3.324 cm from the top) and for a second scenario where the object has moved upwards by 2cm resulting in a shift of the center of mass to 3.831 cm from te top down. If these calculations are correct P.E. will decrease with an upwards motion of the trapezium shaped object.

#### Attachments

• Center of mass2.JPG
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• Center of mass1.JPG
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A.T.
jbriggs444 said:
That is straightforward as long as one makes the assumption that the trapezium moves slowly enough that the fluid remains at approximate equilibrium at all times.

The fluid is free to flow up and down past the trapezium. So Pascal's principle applies. Pressure is given by ##P = \rho g h##. The total force on the top and bottom surface is easily calculated: ##F=PA##. The difference between those forces is the net force on the trapezium. We can ignore the forces on the slanted faces on the right and left because those are both zero. We can ignore the forces on the flat faces on the front and back because those cancel with each other.
This would than be the same calculation as per my suggestion in post 26 I believe.

Novus said:
This would than be the same calculation as per my suggestion in post 26 I believe.
Yes, but note that's it's not enough to show positive buoyancy. In order for the object to rise, the buoyancy also has to be greater than the weight of the object.

pbuk and jbriggs444
Novus said:
If these calculations are correct P.E. will decrease with an upwards motion of the trapezium shaped object.
Makes sense to me. When air bubbles rise they also expand and so does the water level, but the total PE of the water decreases.

A.T. said:
Yes, but note that's it's not enough to show positive buoyancy. In order for the object to rise, the buoyancy also has to be greater than the weight of the object.
In post 26 I calculated positive Fb for both scenarios. In addition, as per the original description, the object consists of solid area's with the same density as water and a 'pocket' of air, therefore the overall density of the object will be less than the density of the surrounding water. I'm still uncertain if the object will move up or down, you're example of air bubbles rising and expanding doesn't apply here I believe, since the bubbles expand as a result of a higher initial internal pressure. In the scenario being discussed the air volume is directly connected to the outside ambient air and therefore will maintain the same pressure when moving up or down.

A.T. said:
Yes, but note that's it's not enough to show positive buoyancy. In order for the object to rise, the buoyancy also has to be greater than the weight of the object.
Novus said:
In post 26 I calculated positive Fb for both scenarios. In addition, as per the original description, the object consists of solid area's with the same density as water and a 'pocket' of air, therefore the overall density of the object will be less than the density of the surrounding water.
Lower density doesn't necessarily mean it will rise, because not all sides are subject to hydrostatic pressure. If you do the force analysis, you add up all forces, including weight, and look at the direction of the net force. Comparing densities is not part of that analysis.

Novus said:
you're example of air bubbles rising and expanding doesn't apply here I believe,
I didn't mean to imply it's exactly the same. It's just a more natural example which shows that raising the water level doesn't imply increasing the water's PE.

Note the in the energy analysis you also have to take the change of the object's PE into account as well. The total PE energy of water and object must decrease.

pbuk

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