Suppose a motor stalls at 20kg.cm and its shaft is .25" in diameter. What is the maximum load that can be exerted on a cord wound directly on the shaft? (Be nice; remember I am a marketing guy.)
Well I assume that the cord is wound round securely at one and and the other hangs down, supporting a pan. Then weights are added until it stalls and the question is what weight stalls it? So the shaft has diameter 0.25 inches = 6.35mm = .00635m So the radius is half this or .003175m. The cord hangs down with tension T equal to the weight in the pan. This develops a torque of T x radius which we are told stalls at 20 kgcm = 0.2 kgm So max T x R = 0.2 T = 0.2/0.003175 = 63 kg force, or 63 x g Newtons force = 618N. So your motor shaft will wind up a load of just under 63kg mass.
The only real thing is that kilograms force are a non standard unit. In the ISO metric system the kilogram is a unit of mass and the unit of force that corresponds is the newton. So working is kgcm is definitely likely to lead to trouble. since you had a 1/4 inch shaft and are in America, why not keep it imperial and use poundsinches poundsfeet?
I converted 63 kg to 139 pounds, which tells me what I need to know, which is that a 100 pound test cord would be suitable.
Glad it is sorted for you. I assume you know that if you wrapped the cord several times around the shaft the anchoring force needed at the fixed end would be a lot less than the lifting force available at the free end?