Sum of torque

[EOW]

The goal is always the same: have a torque on the red object and no torque on the support. The support turns around C1. C1 is fixed to the ground. The red object turns around C2. C2 is fixed to the support. I attrack balls from C2, all balls are attracked from C2. I want to have F1=F2 and the sum of forces of attraction from part1 =  the sum of forces of attraction from part2. For that I use the left part with a higher radius than the right part (ratio 9/7). But I need to have F1=F2, so it's important to change the pressure in the same time. When the radius at left increases, the volume of balls increases too, so I can cancel the forces of attraction left/right. If I take the good value of the pressure I can cancel in the same time F1+F2 on C2. One end of each spring is attached on C2, the other end of the spring is attached on a ball. There is one spring for each ball.

Like that the red object has a torque on it. It's unstable like before, but I accelerate more and more the support in the same time. I win the energy of the rotation of the red object. The center of gravity of the red object is on C2.

[EOW]

I changed the radius of the left part for have a counterclockwise torque on the support.

With the radius at 5.5 at left and the radius 3.5 at right I have :

The vector at left :
x=-6.27
y=1.49

The vector at right:
x=5.21
y=-1.9

The sum is :
x=-1.06
y=-0.41

So the force give a counterclockwise torque on the support.

The force on C2 from F1 and F2 is 0 because I take the pressure at 3.5 for the right and 2.22 for the left.

[EOW]

If I attrack balls at outer circle I can have F1=F2 and a counterclockwise torque on the support, no ?

[EOW]

Here I have a counterclockwise torque on the support and each red object receives a counterclockwise torque too.

[EOW]

Black lines = to the center C1. Red line = to the center C2.
Attraction: height = pressure.

Torque1=2.66
Torque2=-4
Torque3=-1
Torque4=+1
Torque5=+0.166
Torque6=-0.833

Sum of torques=-2

Force3+Force4=4
Force5+Force6=2

Sum of F3+F4+F5+F6 (vectors)=sqrt(4²+2²)=4.47

equation of the line of the sum of forces F3+F4+F5+F6 on C2: y=-2x+1; Equation of perpendicular: y=0.5x; Intersection: x=0.4,y=0.2; Distance =sqrt(0.4²+0.2²)=0.447

Torque from forces on C2 from F3+F4+F5+F6 = 4.47 * 0.447 =2