To reduce the power requirement of elevator motors, elevators are counterbalance
ID: 1272877 • Letter: T
Question
To reduce the power requirement of elevator motors, elevators are counterbalanced with weights connected to the elevator by a cable that runs over a pulley at the top of the elevator shaft. Neglect any effects of friction in the pulley. Assume a 1500 kg elevator that carries a maximum load of 800 kg is counterbalanced with a mass of 1900 kg.
What is the power provided by the motor when the elevator ascends fully loaded at a speed of 1.6 m/s. IN kW?
How much power is provided by the motor when the elevator ascends at 1.6 m/s without a load? In kW?
Explanation / Answer
Power is Force X Velocity so in this case
a) Force = (1500 + 800 - 1900) X 9.8 = 3920 Nt so the Power will be 3920 X 1.6 = 6272 Watts = 6.272 kW
b) Force = ( 1500 - 1900) X 9.8 = -3920 Nt so the Power will be -3920 X = -6272 Watts = - 6.272 kW , note in part b that power is actually being generated.
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