Need help with gravity/buoyancy design

[Novus]

Container 'A' consists of tubes with different diameters (4,6 en 8 squares). Container 'B' is a tube with a diameter of 6 squares.
Container 'A' is equal in weight to container 'B'
Both containers are at the bottom connected via a small tube.
Both containers are at the top connected via strings on a pulley.

Mass 'M' is a solid cylindrical object which loosely fits in the smallest diameter of container 'A' (e.g. slightly smaller than '4')

Blue squares represent a fluid.
Green squares represent the part of mass 'M' which is not submerged in the fluid.
Orange squares represent the part of mass 'M' submerged in the fluid.

On the left side the blue squares for 'A' and 'B' are counted and added up on the top resulting in a total nbr. of squares of fluid of 372
For the weight calculation of 'A' and 'B' de portion of 'M' submerged (orange) is added, as per Archimedes, to the volume of fluid.

Counterweights on the pulley cancel the weight differences between 'A' and 'B' (60 and minus 60)

The 3 positions (1A/1B, 2A/2B en 3A/3B) are each at equilibrium, therefore a small additional force applied to the right side of  the pulley will transition 1A/1B clockwise via 2A/2B to position 3A/3B.

The end result is that mass 'M' is submerged in the fluid while maintaining the same level of fluid in containers 'A' and 'B'

[Novus]

Attached scenario2 is basically the same as scenario1 from the previous post.
The only difference is the addition of a solid cylindrical mass 'V' which is suspended above and which loosely fits in container 'B'

Containers 'A' and 'B' and the solid cylindrical masses 'M' and 'V' could be constructed from clear acrylic with a specific density of 1.19 in which case the fluid could be a salt water solution with the same density of 1.19   

[Novus]

Attached scenario3 starts with the ending position 3A/3B of the previous scenario2

Container 'A' and 'B' remain in the same position (pulley is kept in fixed position)
Mass 'V' with the same density as the fluid is slowly submerged.
The portion of 'V' which is not submerged (green) is kept in balance with a changing counterweight and is therefore weightless/exerts no force (how this can be achieved will be expained later)

Each position 3A/3B, 4A/4B, 5A/5B is at equilibrium therefore a small force at mass 'V' will complete the transition from 3A/3B to 5A/5B

As a result the volume of fluid in container 'A' will move upwards.
Mass 'M' in container 'A' which has the same density as the fluid will float upwards.

[Novus]

Attached scenario4 starts with the ending position 5A/5B from scenario3.

Mass 'V' remains in the same position. Container 'A' moves upwards and container 'B' moves downwards.

Each position is at equilibrium as a result of the counterweights on the pulley. As a result a small force on the left side of the pulley will transition 5A/5B anti clockwise to 6A/6B and 7A/7B

Mass 'M' is 'pulled up' against gravity untill no longer submerged (which is cancelling out the positive gravitational force from scenario2 where mass 'M' was submerged in container 'A' )

[Novus]

Attached scenario5 starts with the ending position 7A/7B from scenario4 as per the previous post.

The pulley is fixed and 7A/7B, 8A/8B and 9A/9B stay in the same position.

Mass 'V' is moving upwards whereby the portion which is not submerged (green) is kept in balance at each stage with a counterweight (how this can be achieved will be explained later) As a result the not submerged (green) part of mass 'V' should be considered as being weightless.

Position 7A/7B, 8A/8B and 9A/9B are each at equilibrium. A small negative force on 'V' will complete the transition.

Mass 'M' is fixed at the top of container 'A'