A nine-story building has an elevator whose cabin with its load weighs 2500N and

Question

A nine-story building has an elevator whose cabin with its load weighs 2500N and rated speed is 6 m / s. T which will force the suspension cable cabin ascensir exercises while climbing the first to the ninth floor of the subs and how?

a) Between the first and third floor the elevator boot undergoes a period
b) Between the third and sixth floors the elevator travels at the speed of regime
c) Between the sixth and ninth floors the elevator undergoes braking period. Suppose that in each of the periods starting and stopping the tour is 9 meters and the caseleracion is constant.
d) Calculate also the electric motor power drive, assuming that the total yield is 70%.

For simplicity consider the value of the acceleration due to gravity is g = 10 m / s

Explanation / Answer

during climbing from 1st floor to 3rd floor:- let time taken be t. it starts at speed 0 and reaches 6m/s in distance of 9 m so acceleration=(final speed^2-initial speed^2)/(2*distance)=2 m/s^2 so if tension is T, then T-m*g=m*a so T=m*(g+a)=3000 N work done=3000*9=27000 J b. during travel from 3rd floor and 6th floor,speed remains constant. so T=m*g=2500 N work done=2500*9=22500 J c. during travel from 6th floor to 9th floor, initial speed=6 m/s final speed=0 distance=9 m so deceleration=(final speed^2-initial speed^2)/(2*distance)=-2 m/s^2 so if tension is T, then T-m*g=m*a so T=m*(g+a)=2000 N work done=2000*9=18000 J so total work done=67500 J so power of electric motor drive=67500/0.7=96428.57 J

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