posted by sweety on .
A fully loaded, slow-moving freight elevator has a cab with a total mass of 1800 kg, which is required to travel upward 50 m in 2.6 min, starting and ending at rest. The elevator's counterweight has a mass of only 780 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable?
average velocity = 50m/156 s
= 0.3205 m/s
average force required = (1800 - 780)kg*9.81 m/s^2 = 10,006 N
Power (average) = force x velocity = 3207 W = 4.3 horsepower
The power will need to be higher during acceleration, and lower during deceleration, but 4.3 Hp is the average needed.