This is a college question. I am studying for the Certificate in Electrical Services Eng in Ireland I think you call it the HNC. If this was put into practice then there would probably be gearing and inverter used. But to size the motor, you need to know the mechanical power necessary to drive the hoist. I calculated this using the method at the start of the thread. So you will need a motor with greater power than the mechanical power. This is to allow for losses in the motor, inefficiencies, safety factor, etc. Regardless of gearing, pulleys, etc, you size the motor to suit the actual requirements of the hoist. I think it's something to do with energy is neither created or lost, it is put from one form to another. So, if you need 56kW to drive hoist, then you will need 56kW + losses + inefficiencies to calculate electrical input of motor. That's my understanding of it and i'm open to correction cause i 'aint no Einstein! Hope this helps?