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Multiple Movable Pulley

An elevator w...

Question

An elevator which can carry a maximum load of 1800kg (elevator+passengers)is moving up with a constant speed of 2m/s. The frictional force opposing the motion is 4000N. Determine the minimum power delivered by the motor to the elevator in watt and horsepower .

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Solution

Dear Student,
$1 . w e i g h t i n t h e e l e v a t o r, w = m g w = 1800 \times 9.8 = 17640 N t o t a l f o r c e a c t i n g d o w n w a r d = w e i g h t + f r i c t i o n a l f o r c e F = w + f F = 17640 + 4000 F = 21640 N p o w e r d e l i v e r e d b y m o t o r = f o r c e \times v e l o c i t y P = F \times v P = 21640 \times 2 P = 43280 W a t t s$

We know that 1 horse power = 746 Watt

so Power = 43280/746 = 58.01 Hp

Regards

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