Sorry guys, but most of the answers in this thread are wrong/incomplete. It's not so much weight, but more where you put it. A 6 lb stand could be much more stable than a 30 lb stand if you put the weight in the right spots.
For example, imagine 2 situations:
1. Have a 6 lb stand where you have 3 legs of negligible weight and a top of negligible weight, where the legs have a base of length l. Now you put 6 lbs of sand into the legs at the very bottom of each leg, causing all of the mass to be at the bottom of the stand.
2. Have a 30 lb stand where you have a single column of negligible weight and base width l with a top weighing 30 lbs.
Obviously the first configuration is much more stable, despite having less mass. The idea behind stability is to have the stand be able to return to it's proper position after being deflected through a large enough angle. If you put the mass in the right spots, a stable stand will return to it's upright position after being knocked over, where as an unstable one will continue falling.
It's called stable static equilibrium
eg: imagine a small steel ball on a stick. You can balance the stick up on your finger with the ball directly above. It will be in equilibrium, but a small deflection will cause the system to continue away from it's original state. This is unstable static equilibrium. Now imagine the same setup, but you hold it upside down, like a pendulum. Now if you move the ball it will return to it's original position. This is stable static equilibrium.
The idea is to build a stand with the lowest centre of mass possible. If your stand has minimum base radius r (from centre of object to closest edge... sort of), and the centre of mass is situated at a height h above the ground directly above the centroid of your base, then the maximum tipping angle a is equal to the arctan of (r/h).
(I really hope I did all that geometry right in my head. If not, someone fix it please

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