I finally figured it out!!! The equation to find the torque needed to begin dumping a bed is as follows...
(1/2 the length of the bed) X (weight of the load) = (initial torque needed)
I know this is the same thing that DumbDozer said, but I proved it to myself with multiple calculations, thought experiments, etc.
This equation assumes three things... The truck is level, the hinge is at the very back of the bed, and the load is evenly distributed. Any weight overhanging the back of the bed can be subtracted according to the same equation.
So, In PowerStroke's example with a 6500 pound load and a ten foot bed (ignoring the 6" overhang for now) the torque needed is 1/2 of 10' X 6500 lbs. = 32,500 ft.lbs. of torque. Where the ram is located 9' in front of the hinge, the vertical force needed at that point is 32,500/9 or about 3,611 pounds of vertical force.
To LD1's point, this is only the force in the direction of gravity. A few trigonometry calculations will get you the portion of that force that is tangent to the effective angle created by the hinge being lower than the top of the ram. I think I'm right. Correct me if I'm wrong.