A gearbox is being driven by a power source of 4ft/lb of torque.....?

Output is 40 rpm. Because torque is halved, speed must be doubled.

Basically, the power input less a little bit of fiction loss equals the power output.

So T1R1 = T2R2 4*20 = 2*R2 ; R2 = 80/2 = 40 rpm

Also the gear ration will be applied both to the torque and to the speed, but inversely. That gear ratio is 2/4 = 0.5 , so 20 /0.5 = 40 rpm

For your gearbox the torque is halved and the speed is doubled.

If the gearbox were 100% efficient there'd be no drag in the gear box to absorb the input power. then input power would equal output power.

as shaft power is some kind of unit of torque times some kind of rate of revolutions. you can define power to be torque in ft/lbs multiplied by rpm. That's not a standard unit of power, but it'll work perfectly well here as the units cancel out to leave the units of the measure you want.

Power_in*efficiency=Power_Out

Rpm_in *torque_in*efficiency= Rpm_out*torque_out

rearranging for torque out

Rpm_out=Rpm_in *torque_in*efficiency/torque_out

plugging the numbers

Rpm_out=20*4*1/2

=40

units are the same as the input speed ie ft/lbs.

It's not entirely spelt out that you want the output shaft speed under a load of 2ft/lb. That isn't important for a gearbox operating at a fixed ratio, but with a If the gear box is a self adjusting constantly variable gear box you'd need to know what the actual output torque is under load to find the speed for that load.

(a pretty simple "gearbox" that'll self adjust it's ratio depending on load...

http://www.rexresearch.com/constran/1constran.htm )

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opps forgot to say it's probably safe to assume efficiency is 100% for a first analysis. That's where the 1 came from in my calculations.

real efficiency depends on all kind of things. for a simple low ratio gear box using spur gears (like this one almost certainly would be) I'd expect it to be above 95% ie 0.95