Thanks for the link.

I enlisted on a science forum and posted below question;

Will the object in below scenario (see picture 1) 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 separate parts which enclose each other in such a way that no water can seep in and, as 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 depth of 4 cm and the bottom at a depth 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 answers as a result of the ensuing discussion can be summarized as follows;
* ‘I agree that there are no lateral forces. And no vertical forces from the sealed and slanting sides.’
* ‘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.’
* ‘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…’ ‘When air bubbles rise they also expand and so does the water level, but the total PE of the water decreases’. (I’ve calculated the water’s center of mass for the scenario and for the object having moved upwards 2 cm – including a subsequent rise in the water level - and found that the center of mass for the water (and thus the P.E.) decreased)

From scientific articles on the bottom-case scenario we find that;
...However, this law, also known as Archimedes’ principle (AP), does not yield the force observed when the body is in contact to the container walls, as is more evident in the case of a block immersed in a liquid and in contact to the bottom, in which a downward force that increases with depth is observed.
https://www.researchgate.net/publication/51958652_Using_surface_integrals_for_checking_Archimedes'_law_of_buoyancy

...A question that normally comes up during discussions of Archimedes’ principle is that when an object in the form of a rectangular block rests on the bottom of a container with no fluid under it, where does the upward buoyant force come from? In fact, in this case because of the fluid pressure on top of the block, the net hydrostatic force on it would be downward, resulting in the apparent weight of the block to be greater than its true weight. But this conclusion is in complete contradiction with all observations since even in this case the apparent weight of the block is less than its true weight by the weight of the fluid displaced.
https://file.scirp.org/Html/1-1720827_75679.htm

The question as stated in the forum can now be answered as follows (see also picture 2);
Solving Fb for the total volume of the object (note that in this case calculating Fb as the difference between the forces on the top and bottom area’s does not work); Fb = V * p * g = 18 cm3 * 1 kg/m3 * 9.81 m/s2 = 0.1766 N. For the downwards hydrostatic force on top of the slanted area's; Hf = p* g * h * A = 1 kg/m3 * 9.81 m/s2 * 4 cm * 2 cm2 = 0.0785 N. The resulting Fb(res) = Fb - Hf = 0.0981 N.  W = m * g = 0.014 kg* 9.81 m/s2 = 0.1373 N. The object will therefore move in a downwards direction.

Note that this is my calculation and conclusion as posted on the forum which was neither confirmed nor rejected possibly because at the same time questions were being raised if this was not connected to an attempt ‘to build a PMM which is against the forum’s rules’.

If above statements and calculations are correct…

It is probably easier to understand that the scenario is in equilibrium by increasing the mass of the solid area's of the object (currently 1000 kg/m3) with a factor of 1.6666 after which we get;
1. F = -/- 0.07848N
2. F = +   0.07848N
3. F = 0N (neutrally buoyant)
4. F = 0N (neutrally buoyant)

As for the change in force from 4 (F = -/- 0.07848N) to 1 (F = 0N) by closing the seal - see below simplified bottom-case example.

The compression of the object between 2 and 3 generates 'free energy' (depending on the question if the forces are calculated correctly)
 

« Tarsier  Reply #15 on: May 04, 2023, 03:26:26 AM »

Novus. A big issue is the displacement of water. If your "exception" mechanism sits at the bottom position sealed at both sides, lets pretend that it is positively buoyant looking at your fig1 and fig2. As it rises, we have about 14 units of water that drop 2 spaces and 4 units that are displaced all the way to the top of the water surface, about 5 or 6 spaces. So around 22 units of Potential energy working against you vs 14 units moving down 2 spaces...28 units of PE. That is if the buoyant container volume is equal to air. If it weighs less than 6 units of water, it will float, any more (ie if it is neutrally buoyant), it will not rise. I don't think this "exception" will overcome this, but you would need to test.

In the revised scenario (post 34)  8 cm3 of water moves to the top of the container as a result of the expanding/upwards movement of the object between 1 and 2, however the water's center of mass (and thus PE) drops by 2.29 cm.