# Buoyancy and External Forces

Richard Spiteri
I am still intrigued by the neutral buoyancy of a body and have come up with some other questions. As previously encouraged, I attempted to answer my own questions (this is not home work, it is just me trying to wake up my brain in my retirement).

Fig 1 has no question but builds on what @kuruman and others showed me; Fig 2 and Fig 3 build on my understanding with some questions that I tried to answer myself

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Delta2

## Answers and Replies

Richard Spiteri
In the above, I am trying to understand the buoyancy of a body inside a buoyant body. Is the lack of replies or suggestions an indication that this is more difficult than I thought?

Mentor
Can you please say in words what you are trying to determine in each of these cases. I, for one, am unable to make sense out of it.

hutchphd
Gold Member
I am trying to understand the buoyancy of a body inside a buoyant body.
Topologically speaking, the water in the box is not ' inside' the air tank. It is joined to the water outside so the pressures at any level in the box are the same as the pressures at that level in the body of water. Any discussion about the proposed model could just as well apply to a simple air tank floating with neutral buoyancy and examining some of the surrounding water.

Richard Spiteri
Can you please say in words what you are trying to determine in each of these cases. I, for one, am unable to make sense out of it.
Topologically speaking, the water in the box is not ' inside' the air tank. It is joined to the water outside so the pressures at any level in the box are the same as the pressures at that level in the body of water. Any discussion about the proposed model could just as well apply to a simple air tank floating with neutral buoyancy and examining some of the surrounding water.
@sophiecentaur, I assume you are talking about Figure 2 not Figure 1, right?
In my calculations, I interpret it also holds for Figure 3 where the water-filled bags have the same pressure as the outside water but please see my reply to @Chestermiller where I elaborate a bit more.

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Richard Spiteri
Can you please say in words what you are trying to determine in each of these cases. I, for one, am unable to make sense out of it.
@Chestermiller, of course. Sorry I was trying to be clear but succinct. Here is an explanation in words and some of my thoughts.

Figure 1:
I basically assert that the net force upwards (denoted by the red person pushing in the opposite direction) is due to the buoyancy upwards minus the weight downwards. I calculate this using one of two ways to arrive at the same result:
(i) by using the total volume displaced - including the volume of contained water - to calculate upthrust and then subtracting the total weight - also including the weight of the water or equivalently,
(ii) by using the volume excluding water volume and the weight excluding the weight of water to arrive at a net force upwards of 3LWρAirg - MAirTankg - mWaterTankg in both cases.

I then try to make the point that any internal forces (e.g., weightless green person inside lifting the water tank) has no effect on the net forces of the entire model as all internal forces cancel one another. I do not have a question here, I am simply setting up the problem ...but feel free to comment.

Figure 2:
In this scenario, I open the water tank to the outside water pressure using two thin hollow tubes of negligible volume and weight. I assume that this tank can move relative to the Air Tank with no frictional forces (sorry, that is not clear in my original post).

This creates a scenario where, in my opinion, the force required to lift the water tank (assuming the Air Tank is held stationary) is different from Fig 1 since the green person (regardless of whether he sits inside the air tank as in Fig 1 or else stands outside as depicted in Fig 2) has to lift only the weight of the tank material (but not the weight of the water it contains).

This question - in my opinion - also has a straightforward answer similar to the equations I show and similar to what @sophiecentaur confirmed. I use the analogue of pushing a straw upwards through water where a single water molecule in the middle of the straw remains immobile (ignoring laminar flow and friction) as the straw moves up.

Figure 3:
This figure takes it a step further and this is where it gets interesting for me. I am asking what happens if you isolate the outside water shown in Fig 2 with two flexible, water-filled bags as shown. My curiosity really lies in this scenario because I think the physics is identical to Fig 2 but I would like to hear from others.

For instance, when the red person pushes the contraption up towards the bottom red solid plate, how does a water molecule in the bottom water-filled bag flow if I am not lifting the water? Does a molecule of water at the centre of the tank "theoretically" stay in place with the tank moving relative to it? Or does the red person have to bear the weight of the entire model including the contained water as in Fig 1?

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Mentor
@Chestermiller, of course. Sorry I was trying to be clear but succinct. Here is an explanation in words and some of my thoughts.

Figure 1:
I basically assert that the net force upwards (denoted by the red person pushing in the opposite direction) is due to the buoyancy upwards minus the weight downwards. I calculate this using one of two ways to arrive at the same result:
(i) by using the total volume displaced - including the volume of contained water - to calculate upthrust and then subtracting the total weight - also including the weight of the water or equivalently,
(ii) by using the volume excluding water volume and the weight excluding the weight of water to arrive at a net force upwards of 3LWρAirg - MAirTankg - mWaterTankg in both cases.
I get 3LWρWaterg - MAirTankg - mWaterTankg

Richard Spiteri and sophiecentaur
Mentor
For Fig.2, I get 4LWρWaterg - MAirTankg

Richard Spiteri and sophiecentaur
Gold Member
, I assume you are talking about Figure 2 not Figure 1, right?
Definitely 2 and, if the bags and rigid plates are massless, 3.
In the case of 1, if you don't change the depth then closing or opening the tap makes no difference.

You have to discuss the compressibility of the water (do you want to know the situation at great depth, for instance)? Are you assuming that tank is totally rigid? Will it be affected more or less than the water at any particular depth?

I find it hard to read the text in the copy you posted so I may have missed detailed specification of the situation but those details are crucial if you want to have an answer. Is there a link to the scenario at higher definition? My topological idea will apply to all three situations if the components are 'ideal'.

Gold Member
For Fig.2, I get 4LWρWaterg - MAirTankg
As I look at your two answers, I have to conclude that opening and closing the tap would affect the net weight of the tank. That upsets me a bit?

Mentor
As I look at your two answers, I have to conclude that opening and closing the tap would affect the net weight of the tank. That upsets me a bit?
I must be missing something. I don't see a tap anywhere.

Richard Spiteri
Richard Spiteri
I get 3LWρWaterg - MAirTankg - mWaterTankg
yes, of course, mine is a typo taken from Fig 1 second equation (it is the water that is displaced not the Air). My Fig 1 first equation is correct and they must match.

Richard Spiteri
Definitely 2 and, if the bags and rigid plates are massless, 3.
In the case of 1, if you don't change the depth then closing or opening the tap makes no difference.

You have to discuss the compressibility of the water (do you want to know the situation at great depth, for instance)? Are you assuming that tank is totally rigid? Will it be affected more or less than the water at any particular depth?

I find it hard to read the text in the copy you posted so I may have missed detailed specification of the situation but those details are crucial if you want to have an answer. Is there a link to the scenario at higher definition? My topological idea will apply to all three situations if the components are 'ideal'.
@sophiecentaur, I have re-created the diagrams at a larger scale and I will post them below one by one (separating the figures) as, you are right -- it is getting confusing (even for me) and I am not sure to which figure people are referring when they reply.

(there is no tap anywhere but any insight is appreciated if you find it necessary to include a tap)

Richard Spiteri

Figure 1:
I basically assert that the net force upwards (denoted by the red person pushing in the opposite direction) is due to the buoyancy upwards minus the weight downwards. I calculate this using one of two ways to arrive at the same result:
(i) by using the total volume displaced - including the volume of contained water - to calculate upthrust and then subtracting the total weight - also including the weight of the water or equivalently,
(ii) by using the volume excluding water volume and the weight excluding the weight of water to arrive at a net force upwards of 3LWρWaterg - MAirTankg - mWaterTankg in both cases.

I then try to make the point that any internal forces (e.g., weightless green person inside lifting the water tank) has no effect on the net forces of the entire model as all internal forces cancel one another. I do not have a question here, I am simply setting up the problem ...but feel free to comment.

Richard Spiteri

Figure 2:
In this scenario, I open the water tank to the outside water pressure using two thin hollow tubes of negligible volume and weight. I assume that this tank can move relative to the Air Tank with no frictional forces (sorry, that was not clear in my original post).

This creates a scenario where, in my opinion, the force required to lift the water tank (assuming the Air Tank is held stationary) is different from Fig 1 since the green person (regardless of whether he sits inside the air tank as in Fig 1 or else stands outside as depicted in Fig 2) has to lift only the weight of the tank material (but not the weight of the water it contains).

This question - in my opinion - also has a straightforward answer similar to the equations I show and similar to what @sophiecentaur confirmed. I use the analogue of pushing a straw upwards through water where a single water molecule in the middle of the straw remains immobile (ignoring laminar flow and friction) as the straw moves up.

Richard Spiteri

Figure 3:
This figure takes it a step further and this is where it gets interesting for me. I am asking what happens if you isolate the outside water shown in Fig 2 with two flexible, water-filled bags as shown. My curiosity really lies in this scenario because I think the physics is identical to Fig 2 but I would like to hear from others.

For instance, assume the red man holds the Air Tank in place as shown. When the green person pushes the contraption up against the bottom red solid plate, how does a water molecule in the bottom water-filled bag flow if I am not lifting the water (as in Fig 2)? Does another molecule of water at the centre of the tank "theoretically" stay in place as the tank moves relative to it?

Or is it that the green person has to bear the weight of the Water Tank including the trapped water (as in Fig 1)? @sophiecentaur this upsets me too as the physics looks identical to Figure 2 but the mathematics looks like it could be Figure 1.

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Mentor
View attachment 285456
Figure 2:
In this scenario, I open the water tank to the outside water pressure using two thin hollow tubes of negligible volume and weight. I assume that this tank can move relative to the Air Tank with no frictional forces (sorry, that was not clear in my original post).

This creates a scenario where, in my opinion, the force required to lift the water tank (assuming the Air Tank is held stationary) is different from Fig 1 since the green person (regardless of whether he sits inside the air tank as in Fig 1 or else stands outside as depicted in Fig 2) has to lift only the weight of the tank material (but not the weight of the water it contains).
Of course, the green guy needs to exert an upward force that includes the weight of water contained in the water tank. The water is exerting a downward force on the bottom of the water tank equal to the weight of the water in the tank (and this is transmitted to the lower straw).

Richard Spiteri
Richard Spiteri
@Chestermiller, I am a bit sceptical about that. The pressure field is continuous in Fig 2 and, therefore, I think the Green man only lifts the weight of the Water Tank material.

The Water Tank displaces a 'volume' (in my 2.5 dimensional model) of LW of air while the Air Tank, in turn, displaces a similar volume of water (plus another 3LW). The downward force that you mention is distributed as Pressure.Area everywhere.

I am curious what others think about this because Fig 1 was the obvious one; and Fig 2 was 'the easy one' in my set of three scenarios; and Fig 3 is where it gets confusing.

Mentor
In my judgment (and considerable fluid mechanics experience), your fundamental understanding of the basic concepts of hydrostatics needed to solve this problem are seriously flawed.
@Chestermiller, I am a bit sceptical about that. The pressure field is continuous in Fig 2 and, therefore, I think the Green man only lifts the weight of the Water Tank material.
The pressure field is not continuous in Fig. 2. The pressure changes discontinuously across the wall of each tank, and the vertical pressure gradient also changes discontinuously across the wall of each tank.
The Water Tank displaces a 'volume' (in my 2.5 dimensional model) of LW of air while the Air Tank, in turn, displaces a similar volume of water (plus another 3LW).

Actually, the air tank displaces an external volume of 4LW. It is only the external volume displaced that contributes to the buoyant force on the air tank. The buoyant force is determined by the external pressure distribution on a solid body. The external pressure acts normal to the surface of the body, and the buoyant force is equal to the integral of the vertical component of this pressure force, integrated over the external surface of the body. This is just equal to the weight of the displaced external volume of fluid.
By the same token, the buoyant force exerted by the air on the water tank is essentially zero, because the weight of the displaced air is essentially zero.

Richard Spiteri
Richard Spiteri
Thank you, @Chestermiller, this is really interesting for me and, you are right, I joined PF to learn as I pose these kinds of hypothetical questions (I learn best this way)!

So, to be clear, by stopping the Air Tank (red man) from rising any further as shown in Fig 2, you are saying that he has the effect of opposing the full buoyancy caused by displacing the 4LW (external) volume, such that the Water Tank is not neutrally buoyant but a 'dead weight full of water'?

So, when you say "The external pressure acts normal to the surface of the body, and the buoyant force is equal to the integral of the vertical component of this pressure force, integrated over the external surface of the body", does this integral of the vertical component cancel everywhere by symmetry except at the top and bottom openings of the hollow tubes (i.e., at different depths) and at the [negligible] top and bottom of the Water Tank sitting in air?

Mentor
Thank you, @Chestermiller, this is really interesting for me and, you are right, I joined PF to learn as I pose these kinds of hypothetical questions (I learn best this way)!

So, to be clear, by stopping the Air Tank (red man) from rising any further as shown in Fig 2, you are saying that he has the effect of opposing the full buoyancy caused by displacing the 4LW (external) volume, such that the Water Tank is not neutrally buoyant but a 'dead weight full of water'?
In essence, yes.
So, when you say "The external pressure acts normal to the surface of the body, and the buoyant force is equal to the integral of the vertical component of this pressure force, integrated over the external surface of the body", does this integral of the vertical component cancel everywhere by symmetry except at the top and bottom openings of the hollow tubes (i.e., at different depths) and at the [negligible] top and bottom of the Water Tank sitting in air?
Yes.

Richard Spiteri
Gold Member
(there is no tap anywhere but any insight is appreciated if you find it necessary to include a tap)
Thanks for the improved graphics. They helped me to make sense of the first diagram which looked like a tube to the inside with a tap / plug, connecting the inside water to the outside.
I can now give the problem some proper thought.
The man in 1 is just lifting the weight of the water tank. Nothing more to be said about that as long as the air tank is not accelerating up or down.
In 2, the water in the tank is not affected by the air tank at all (that is just floating but will only be neutral if the volume of air is appropriate - i.e. less than the volume in 1 because it doesn't support the water tank). I can see no reason to think that the water tank is not just floating (neutral) in the surrounding water (double negative I'm afraid) so, as you say, the man (presumably standing on the bottom of the ocean) is just supporting the metal of the water tank.
The coupled water bags in 3 have no effect at all afaics - it's just a volume of water being supported by the surrounding water. The green man will only feel the weight of the flat plates and the water tank material; I can't see that the presence of the plates needs to be relevant to the situation. The upthrust on the air tank is just due to the volume of air.
physics looks identical to Figure 2 but the mathematics looks like it could be Figure 1.
I think the source of the apparent paradox is that there are two different amounts of water displaced. In 1, the water tank volume needs to add to the air volume for neutral buoyancy but in the other two cases the air in the tank needs only to displace enough to support the air tank.

Richard Spiteri
Richard Spiteri
In essence, yes.

Yes.
Sorry, @Chestermiller, but I am still not convinced.

Take the two limits of the Water Tank in Fig 2 where (i) it is infinitesimally smaller than the Air Tank (i.e., 4LW), and (ii) where L remains the same but W shrinks to the width of the Hollow Tube.

We know that in (i) the whole system is neutrally buoyant with an upthrust of 4LW.ρWaterg and an opposing weight of (4LW.ρWater + MAirTank + mWaterTank).g consistent with my original equation in my thread, and

The "Water Tank" in (ii) reduces to a "Straw" which we agreed is also neutrally buoyant as it is free to slide up and down (assuming as before that there is no friction between the Straw and the Air Tank). The Green man experiences a net downwards weight of mAirTankg.

In both cases, the Green Man only needs to lift the weight of the material of the Water Tank and Air Tank for (i) and the Straw for (ii) as the water is neutrally buoyant.

It stands to reason, therefore, that the Water Tank of dimensions L x W must follow the same physics (and hence the same equations) as I have shown. In my argument, a single equation satisfies Fig 2, Fig 2(i), and Fig 2(ii)

Can you show me how you arrive at your equations based on your argument or tell me where I am wrong?

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Richard Spiteri
Thanks for the improved graphics. They helped me to make sense of the first diagram which looked like a tube to the inside with a tap / plug, connecting the inside water to the outside.
I can now give the problem some proper thought.
The man in 1 is just lifting the weight of the water tank. Nothing more to be said about that as long as the air tank is not accelerating up or down.
In 2, the water in the tank is not affected by the air tank at all (that is just floating but will only be neutral if the volume of air is appropriate - i.e. less than the volume in 1 because it doesn't support the water tank). I can see no reason to think that the water tank is not just floating (neutral) in the surrounding water (double negative I'm afraid) so, as you say, the man (presumably standing on the bottom of the ocean) is just supporting the metal of the water tank.
The coupled water bags in 3 have no effect at all afaics - it's just a volume of water being supported by the surrounding water. The green man will only feel the weight of the flat plates and the water tank material; I can't see that the presence of the plates needs to be relevant to the situation. The upthrust on the air tank is just due to the volume of air.

I think the source of the apparent paradox is that there are two different amounts of water displaced. In 1, the water tank volume needs to add to the air volume for neutral buoyancy but in the other two cases the air in the tank needs only to displace enough to support the air tank.
@sophiecentaur, thanks for your explanation. I should clarify (my omission!) that the solid red plates in Fig 3 are fixed (they do not move). I added the red plates to give the Water Bags some structure and to limit the motion of the rising Air Tank (provided the Red man doesn't hit his head first!).

Can anyone comment on my Fig 3 molecule question?

If the Green Man lifts the lower Water Bag upwards to the limit of the bottom solid plate, the molecules in the lower Water Bag must exit, travel up into the Water Tank and into the top Water Bag.

This concerns me. Is the outside water (at equal pressure for every depth) providing the work necessary for the molecules to travel upwards?

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Mentor
Sorry, @Chestermiller, but I am still not convinced.

Take the two limits of the Water Tank in Fig 2 where (i) it is infinitesimally smaller than the Air Tank (i.e., 4LW), and (ii) where L remains the same but W shrinks to the width of the Hollow Tube.

View attachment 285518
Sorry Richard, but I really don't understand what you are getting at.
We know that in (i) the whole system is neutrally buoyant with an upthrust of 4LW.ρWaterg and an opposing weight of (4LW.ρWater + MAirTank + mWaterTank).g consistent with my original equation in my thread, and
It is not clear in (i) if the two tanks are making contact with one another, or not. The answer depends on this.

The "Water Tank" in (ii) reduces to a "Straw" which we agreed is also neutrally buoyant as it is free to slide up and down (assuming as before that there is no friction between the Straw and the Air Tank). The Green man experiences a net downwards weight of mAirTankg.

It is my understanding that the green guy in this case supports the weight of the straw which is the same as the weight of the "water tank." The red guy in this case is pushing down with a force of ##4LW \rho_{water}g-M_{Air Tank}g##
In both cases, the Green Man only needs to lift the weight of the material of the Water Tank and Air Tank for (i) and the Straw for (ii) as the water is neutrally buoyant.
In case (i), it depends on whether the two tanks are making contact. In the actual Fig.2, they are not. The green guy supports the weight of the water in the water tank (minus a small correction proportional to the area of the straw divided by the area of the tank) plus the weight of the water tank. The red guy presses down with ##4LW\rho_{water}g-M_{air tank}g##
It stands to reason, therefore, that the Water Tank of dimensions L x W must follow the same physics (and hence the same equations) as I have shown. In my argument, a single equation satisfies Fig 2, Fig 2(i), and Fig 2(ii)

Can you show me how you arrive at your equations based on your argument or tell me where I am wrong?
Like I said, in (i), it depends on whether the two tanks are making contact with one another.

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Mentor
Thanks for the improved graphics. They helped me to make sense of the first diagram which looked like a tube to the inside with a tap / plug, connecting the inside water to the outside.
I can now give the problem some proper thought.
The man in 1 is just lifting the weight of the water tank. Nothing more to be said about that as long as the air tank is not accelerating up or down.
In 2, the water in the tank is not affected by the air tank at all (that is just floating but will only be neutral if the volume of air is appropriate - i.e. less than the volume in 1 because it doesn't support the water tank). I can see no reason to think that the water tank is not just floating (neutral) in the surrounding water (double negative I'm afraid) so, as you say, the man (presumably standing on the bottom of the ocean) is just supporting the metal of the water tank.
I disagree with this. The water in the tank is exerting a greater force on the bottom of the tank than on the top of the tank by an amount equal to the weight of the water in the tank ##\times [1-(area\ of\ straw)/(area\ of\ tank\ base]##. So the grey guy has to support the weight of water in the tank plus the weight of the tank. This follow from a force balance on the tank with straw extensions.

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Gold Member
I disagree with this.
Afaics, trying to approach this problem in terms of pressure just makes life harder. We can believe Archimedes' Principle and (validly) save a lot of time.

That's my fault for sloppy writing. In 1 I should have talked in terms of the "water tank plus the water". That could be replaced with anything of the same total weight and the air tank would still be displacing the same amount of water.

In 2 and 3 (certainly 2) the 'internal water' is not part of the air tank system as it could just as well be outside the tank but the steel tank and tube are connected to the air tank so add to its weight. The volume of contained air is still the same and there is less total effective weight (the mean density is lower) so my point was that there would now be more effective buoyancy unless the air tank volume were decreased (by the same volume as the water tank volume). Hence the apparent problem with different forces needed in 1 and 2.

I still can't be sure of what happens in 3 but the plates and water bags seem to me equivalent to surrounding water without them. The details of what the green guy is actually doing may make a difference but, as he doesn't seem to be touching the air tank, how can he affect it (sliding tube)?

Mentor
Afaics, trying to approach this problem in terms of pressure just makes life harder. We can believe Archimedes' Principle and (validly) save a lot of time.

That's my fault for sloppy writing. In 1 I should have talked in terms of the "water tank plus the water". That could be replaced with anything of the same total weight and the air tank would still be displacing the same amount of water.

In 2 and 3 (certainly 2) the 'internal water' is not part of the air tank system as it could just as well be outside the tank but the steel tank and tube are connected to the air tank so add to its weight. The volume of contained air is still the same and there is less total effective weight (the mean density is lower) so my point was that there would now be more effective buoyancy unless the air tank volume were decreased (by the same volume as the water tank volume). Hence the apparent problem with different forces needed in 1 and 2.
I have no idea what you are saying here with respect to the water tank and tube. As far as the OP problem statement goes, the water tank and tube have no mechanical connection to the air tank because of negligible friction.

Please see my next post for questions that I have posed for Richard and yourself.

Mentor
Here are some question regarding the arrangement in Fig. 2 for Richard and @sophiecentaur to consider:

1. Is the water pressure at the floor of the water tank higher than the water pressure at the ceiling of the water tank?

2. For configurations in which the water tank is wider than the tubes leading into and out of the water tank, does the water in the tank exert a larger force on the base of the water tank than on the ceiling of the water tank?

3. For configurations in which the water tank is wider than the tubes leading into and out of the water tank, des the water exert a net downward force on the water tank?

4. Does the net downward force of the water on the water tank have to be supported by the lower tube leading into the tank, and, from that, ultimately, does the grey guy have to support this force.

Gold Member
Here are some questions
1 - do you mean the hydrostatic pressure on the outside ? Yes the pressure underneath is greater - producing upthrust. Upthrust is due to displaced volume and, for anything other than a basic cylinder / prism shape, the net upwards force is a pain to work out. What happens inside must be independent of shape (?? why not ??).
Did you ask this first question with something else in mind?

2 - The pressures in an out or at the top and the pressures in and out at the bottom must be equal or water would be flowing so the weight of the water in the tank / tubehttps://www.physicsforums.com/help/bb-codes/s displaces its own weight of water (Archimedes). The weight of the metalwork will be the only thing that will contribute to what the green man feels. The question is whether the air pressure affects the net weight of the water tank. It the air is replaced by water then the water displaced by the tank metal would reduce the force as for a lump of the same metal under water. Perhaps it is in fact necessary to consider the pressures here and that would suggest that the apparent weight felt by the green man would be greater.
But I have a difficulty here because the actual shape of the tank would appear to be relevant and can that be?. A thick walled tank of less dense material would be affected differently from a steel tank. Are we looking at a 'mean density' of the water and air - say half the density of water??
3 & 4 - The water has neutral buoyancy and I can't see the presence of the bags can have any effect; water will always flow to equalise pressures on either sides of the bag fabric.

Mentor
1 - do you mean the hydrostatic pressure on the outside ? Yes the pressure underneath is greater - producing upthrust. Upthrust is due to displaced volume and, for anything other than a basic cylinder / prism shape, the net upwards force is a pain to work out. What happens inside must be independent of shape (?? why not ??).
Did you ask this first question with something else in mind?
Yes. I was referring the the pressures inside the water tank.
2 - The pressures in an out or at the top and the pressures in and out at the bottom must be equal or water would be flowing so the weight of the water in the tank / tubehttps://www.physicsforums.com/help/bb-codes/s displaces its own weight of water (Archimedes). The weight of the metalwork will be the only thing that will contribute to what the green man feels. The question is whether the air pressure affects the net weight of the water tank. It the air is replaced by water then the water displaced by the tank metal would reduce the force as for a lump of the same metal under water. Perhaps it is in fact necessary to consider the pressures here and that would suggest that the apparent weight felt by the green man would be greater.
But I have a difficulty here because the actual shape of the tank would appear to be relevant and can that be?. A thick walled tank of less dense material would be affected differently from a steel tank. Are we looking at a 'mean density' of the water and air - say half the density of water??
3 & 4 - The water has neutral buoyancy and I can't see the presence of the bags can have any effect; water will always flow to equalise pressures on either sides of the bag fabric.
I don't think we are on the same wavelength.

sophiecentaur
Mentor
What I'm saying, to be more precise, is that the green guy exerts an upward force given by:
$$F=M_{water\ tank}g+L(W-w)\rho_{water}g$$where w is the width of the tube opening (where the water is present in each tube). In this way, only when W = w (the tank width is equal to the tube width), ##F=M_{water\ tank}g##, but otherwise, no.

Richard Spiteri and sophiecentaur
Mentor
@sophiecentaur Suppose that you determined the upward force that the green guy exerts by applying a simple static force balance to the solid assembly consisting of the water tank metal shell and connected inlet and outlet tubes, and you got a different result than you obtained from what you had believed to be a proper application of Archimedes principle. What would conclude? What would be your next step?

sophiecentaur