What we know:

P=V times I

P=12 V times 20A=240 W

P=14 V times 20A=280 W

80% efficient, 0.80 times 240 W=192 W

80% efficient, 0.80 times 280 W=224 W

½ HP is 372.8 W or 373 W

1 HP = 745.69987 W

http://www.magtrol.com/support/power_calc.htmlhttp://en.wikipedia.org/wiki/HorsepowerThe 400 W load is greater than the dynamo output between 192 W to 224 W. What is the torque? To figure out torque we need, the RPM of the dynamo and the torque in Newton meters to get power in Watts. This is called motor power calculations.

http://www.magtrol.com/support/motorpower_calc.htmlhttp://www.elec-toolbox.com/Formulas/Motor/mtrform.htmTorque=(HP times 5252)/RPM

Torque is pounds foot)

HP=(V times I times Eff)/746

Eff is efficiency from a percentage divided by 100% to get a decimal value for this equation.

I need to know the RPM of the dynamo, that is measured by a tachometer mounted on the drive shaft, careful with your eyes and fingers taking this measurement.

Let us assume that it is spinning at 500 RPM. 224 W to Horse Power is 224 W times (1 HP/ 745.69987 W) =0.300 HP

Torque=( 0.300 HP times 5252)/500 RPM=3.16 pounds – foot

I got 500 RPM from my imagination, not measured. You do not have enough information to solve this problem.

What about 400 W at 500 RPM motor?

400 W to HP; 400 W times (1HP/745.69987 W) =0.5364 HP

Torque=(0.5364 HP times 5252)/500 RPM=5.63 pounds - foot

The answer is no, because there will not be enough torque in the motor with an output at 3.16 pounds foot to spin a load torque of 5.63 pounds foot. We will need a stronger motor to loosen this 5.63 pounds foot fastener, screw thread, or hardware.