t_cam = 6.92 · 0.155 » 1.07 Nm

Next, the plates. They are located 100 + 5 = 105 mm = 0.105 meter from the center of gravity:

t_plates = 1.70 · 0.105 » 0.18 Nm

And last, the post. Now, although only a part of the post is actually affecting the counter-clockwise torque, the post's gravity force is the same along the entire post (since it's symmetric). It's not actually in balance by itself, but the force is constant, so we can still calculate the torque at 100 mm = 0.1 meter:

t_{post_left} = 0.78 · 0.1 » 0.08 Nm

Now, we sum the torque and get the total, counter-clockwise (left) torque:

t_left = t_cam + t_plates + t_{post_left}
t_left = 1.07 + 0.18 + 0.08
t_left = 1.33 Nm

We continue with the clockwise torque, (right half). Those forces consist of a part of the post and the counterweight. The post sticks out 300 mm = (0.3 meter) on the right side, so its torque is:

t_{post_right} = 0.78 · 0.3 » 0.23 Nm

Now, we don't know how many washers we'll use, so I'm gonna guess how far out the center of gravity is. Each washer is 3 mm thick, and if we use 3 x 2 washers the center of gravity would be at about 300 + 5 = 305 mm = 0.305 meter:

As first described, the left side torque and right side torque must be equal to keep the entire rig balanced, so here's how we solve for F₄:

t_left = t_{post_right} + (F₄ · 0.305)
1.33 = 0.23 + (F₄ · 0.305)
F₄ · 0.305 = 1.33 - 0.23
F₄ = (1.33 - 0.23) / 0.305
F₄ = 3.60 N

The needed force to keep the rig balanced is 3.60 N. Converted to kilo that is: 3.60 / 9.81 = 0.37 kg = 370 g. Each washer was estimated to weight 55 g so we'll need: 370 / 55 = 6.7, i.e. 7 washers!

We can now also check how much the entire rig would weight:

Weight = (F₁ + F₂ + F₃ + F₄) / g
Weight = (6.92 + 1.70 + 0.78 + 3.60) / 9.81
Weight = 1.32 kg

So, the shaky cam by itself weights about 0.7 kg. With the stabilizing system it would weight about 1.3 kg. Almost twice as much, but still manageable and hopefully much more stable. I made a drawing, using QCad.

Here's some pictures from it »