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

[Novus]

Wikipedia: There is an exception to Archimedes' principle known as the bottom (or side) case. This occurs when a side of the object is touching the bottom (or side) of the vessel it is submerged in, and no liquid seeps in along that side.

Picture 1 
A ('brown') container contains a fluid.
Submerged in de container is a trapezium shaped mass 'A'
The trapezium consists of two separate parts which fit air tight and can slide laterally. As a result mass 'A' can change in volume.
The sides of 'A' are kept airtight to the wall of the 'brown' container by some kind of guiding rail system without liquid seeping in between (= exception to Archimedes' principle)
Tube 'B' connects 'A' with the air outside of the container.
In picture 1 the combined density of the walls of mass 'A' and the volume of air trapped inside is buoyancy neutral.
The width of 'A' is 1. The width of the 'brown' container is more than 1.
The downward buoyance force on top of 'A' =  Fb1 = h(height)*p(density)*g(gravitation constant)*a(area) = 6 * 1 * 9.8 * 8 = 470.4
The upward buoyance force on the bottom of 'A' =  Fb2 = h(height)*p(density)*g(gravitation constant)*a(area) = 8 * 1 * 9.8 * 6 = 470.4
Fb1 = Fb2

Picture 2 
'A' has increased in volume. The combined density of the walls of mass 'A' and the volume of the air trapped inside is now positive buoyant.
The downward buoyance force on top of 'A' =  Fb1 = h(height)*p(density)*g(gravitation constant)*a(area) = 4 * 1 * 9.8 * 10 = 392
The upward buoyance force on the bottom of 'A' =  Fb2 = h(height)*p(density)*g(gravitation constant)*a(area) = 6 * 1 * 9.8 * 8 = 470.4
Fb2 > Fb1 + positive buoyancy results in 'A' moving upwards.
Each interval between picture 1 en 2 will results in an increase in positive upwards buoyancy force.

Picture 3 
The 'guiding rail system' is disconnected and the sides of 'A' are no longer kept airtight to the wall of the brown container.
The two separate parts of trapezium 'A' are locked.
Fb2 > Fb1 + positive buoyancy results in 'A' moving upwards.

Picture 4 
The two separate parts of trapezium 'A' are unlocked.
The force on the sides of 'A' reduces the volume of the trapezium to the same size as in picture 1.
The combined density of the walls of mass 'A' and the volume of air trapped inside is buoyancy neutral.
'A' will 'float' down to the same position as in picture 1.

[sm0ky2]

How are you changing the volume of 'A'?


Without some mechanism (and energy input) the transition between steps 2 & 3 will not occur in the manner as stated.


"unlocking" the width of A, will simply compress 'A' to its' minimum width at any point in the diagram. Water pressure at every depth = > 1 ATM
This forces air out through the tube and 'A' will remain at its minimum volume.

[Novus]

The key point, as per the exception to the Archimes principle, is that there are no lateral forces in step 1 and 2 and therefore object A will not compress to its minimum volume even when the 2 parts of the trapezium A are not locked.

[Novus]

In your opinion will object A move up, down or remain stationary based on calculating Fb1, Fb2 and any other factors which you believe are relevant?

[panyuming]

Thanks to Novus for reminding Archimedes buoyancy calculation.
Imagine that this graph can rotate itself?
Thank you!