Buoyancy calculations – making use of an exception to Archimedes' principle?

[Novus]

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)
 

[Novus]

As part of the discussion on the science forum I received multiple questions on how the object could be sliding up and down while expanding/decreasing in volume. Below a possible solution as posted on this forum.

The airtube should probably be in a vertical direction rather then horizontally through the container wall.

As for achieving a water-thight 'seal' this becomes less of a challenge when we realize that rather than trying to prevent any water seeping through the waterpressure needs to be removed.

The pulley/counterweight construction is probably not necessary by properly calibrating the combined density of the sliding panel and the object.

[Novus]

« 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.