B What Factors Influence Neutral Buoyancy in Water?

[Chestermiller]

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.

 

[Chestermiller]

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.

 

[Chestermiller]

@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]

I don't think we are on the same wavelength.

We certainly weren't but that statement (thank you) gave me the required kick up the backside to thing about the situation properly.
Obviously (now!) the upwards force in the water tank is less than the downwards force because of the pressure difference. The water in the tank is only 'supported' by the air pressure difference which is tiny. etc. etc. . . . . the rest of what you say makes perfect sense and I have to conclude that a small enough tube will make the water tank appear to have the same weight as without the tube.

 

[Richard Spiteri]

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.

We all agree on Fig 1 as, to me, this is equivalent to the Green Man standing in a submarine lifting a jug of water. The total buoyancy is unaffected by the internal forces of lifting the jug of water and energy spent by the Green Man.

 

[Richard Spiteri]

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.

@Chestermiller , This is your equation as relates to my original Fig 2.

F = M_{water tank}g + L(A-a)ρg​

(Note, @sophiecentaur , that I changed W to A and w to a so there is no confusion with upper case and lower case fonts and to reflect 'areas' rather than '2D widths'.)

I "plotted" (more like sketched) the net force that the Green man exerts to show that it changes from lifting the total weight of the tank material plus water (water tank on the left) to the weight of the material with no water (straw on the right).

I now understand this problem and I agree (thank you @Chestermiller ).

Please see my next post (#37) as relates to my original Fig 3 with water bags isolating the hollow tubes.

 

[Richard Spiteri]

The figure above, is my original Fig 3 reposted for clarity

In Fig 3, I isolated the hollow tubes from the outside water using the water bags. My question was "what force is the Green man exerting in each of the cases below" as the width a of the hollow tube increases from 0 to A?

The figure above with the accompanying 'sketched' plot shows my thoughts on the net force F that the Green guy exerts versus the width a of the closed hollow tube. I believe this is correct (i.e., the same results as Fig 2). Is this right?

 

[Richard Spiteri]

Also, what is the upthrust U (that the Red man must resist) on the Air Tank in #37?

Personally, I think it is a constant equal to the volume of water that the Air Tank displaces

U = L_{Air Tank} . A_{Air Tank})ρ_waterg​

if you assume that the Air tank dimensions are L_{Air Tank} by A_{Air Tank}.

Or am I counting the volume contributions of air tank and water tank twice (ref @kuruman where we must either include the volume and weight of the water everywhere or we must exclude the volume and weight of the water everywhere)?

Also, since I assumed that the relation between the Air Tank and the Water Tank is frictionless and water tight, then the net upwards force on the Air tank and Water Tank together must be equal to the difference in the two forces on my plot (actually the sum but taking into account that they are in opposite directions).

 

[Chestermiller]

I get that the downward force that the red guy must exert is $$F=L(A_{air\ tank}-a)\rho_{water}g-M_{air\ tank}g$$Does this make sense?