Question

One kind of elevator has the elevator car hanging from a cable, which then passes over a pulley and connects to a counterweight, as shown in (Figure 1). The mass of a fully-loaded elevator (that is, the elevator and all its passengers) is 1850 kg , the mass of the counterweight is 1500 kg , and the pulley has a radius of 0.4 m and a mass of 400 kg . A motor drives the motion of the elevator by exerting a torque on the pulley. Assume that friction is negligibe.

counter weight elevator

Part A.) What is the magnitude of the torque the motor must exert in order to lift a fully-loaded elevator with a constant acceleration of magnitude 1.0m/s2?

Part B.)What is the magnitude of the torque the motor must exert in order to lower a fully-loaded elevator with a constant acceleration of magnitude 1.0m/s2?

Part C.)A group of people get on a fully-loaded elevator and ride it up three floors (a distance of about 9 meters), where they exit the elevator. If it took 5 seconds for the elevator to travel up those three floors, how much power did the motor need to provide?

Part D.)To see the value of the counterweight, repeat the calculation in Part C if the elevator had no counterweight (or alternatively, if the mass of the counterweight were 0kg).

Answer 1

Mass of fully loaded elevatos, М - 19 Sova Mass of counter weight, m2 = 1500kg. Mass of pulley , M = tookg, Radius of pulley, R = 0.4n. Parta If a be the acceleration of upwards then , a=1.om/st. Mag. migo elevator

. T = Mig + M, a , tension on left string. & Too Mag - Maa , tension on right storing If I be the required torque needed for this acceleration then - Тро - т, е , те ei tea + 2 = - Id. where, ra torque due to t = TR. Te = torque due to tq = -T2 R. Cas in clockwise) I = I MR² = moment of inestia of pulley. a- a = angular acceleration of pulley. so. on, TR - TzR tr=-1 MR²x (migtm, a) R -( Mag - Mza) R = { mra. +r. " Z=- I MRa - m, (gta) R + m2 (g-a) putting all values, x = 2632 Moos - 2792 Nam, so needed torque is in clockwise having value reo & 26525N direction 2792 Min.

How to derator lower with the a = 1.00/st. T = Mig-m, a. T = m, (g-a). Ta = Mag + Mza. on, Tz = m2 (gra) het, e be the needed torque - et Ta + 2 = I2. os, TIR - Te R te= I MR² x a. = MPa or micg-a) R - m₂ (gta)r +2 = // mra. on, r = tmra - m. (g-a) R + m2 (gta) R. disection anticlockwise Nim. 1. Q = 48 Nm. So, Needed tosque is in having value = 48 sir Distance coresed, Time taken, If a be the needed d=an. t= 5 see. acceleration then - Cas uro = initial relocity) oh, as 0.72 in 152,

Now dorque required - - - Imra - m. (gra) + m2 (g-a)R. a = -2394-4 Nami For this doom lineas displacement, the angulas displacement is , 0 - ļoti į cen xe os, os 4x ( 075 J x 52. on o = 22.5 sad , so work done Wartol = 2394-4 x 22.5 J.. on, Wa 5387 4 J. powes 1 required, 53874 p = watt. = on, p= 1077 448 ratt. DS Fors, M₂=0, r'- - t mra -mi (gra) R. cia - 784 2-4 Nom. power needed, pla w p's 7842-4x 22-5. Iro! on, p= 35290.8 watt.