Question

You are investigating an elevator accident which happened in a tall building. An elevator in this building is attached to a strong cable which runs over a pulley attached to a steel support in the roof. The other end of the cable is attached to a block of metal called a counterweight which hangs freely. An electric motor on the side of the elevator drives the elevator up or down by exerting a force on the side of the elevator shaft. You suspect that when the elevator was fully loaded, there was too large a force on the motor . A fully loaded elevator at maximum capacity weighs 2400 lbs. The counterweight weighs 1000 lbs. The elevator always starts from rest at its maximum acceleration of g/4 whether it is going up or down.A. What force does the wall of the elevator shaft exert on the motor if the elevator starts from rest and goes up?B. What force does the wall of the elevator shaft exert on the motor if the elevator starts from rest and goes down?C. What should be the design limit for the replacement motor? Assume a 20% margin above the maximum load expected.

Answer 1

Answer:

a) F = 2250 Ib

b) F = 550 Ib

c) new max force ( F newmax ) = 2850 Ib

Explanation:

A) The force the wall of the elevator shaft exert on the motor if the elevator starts from rest and goes up

max capacity  of elevator = 24000 Ibs

counterweight = 1000 Ibs

To calculate the force (F) :

we first calculate the Tension using this relationship

Counterweight (1000) - T =  ( 1000 / g ) ( g/4 )

Hence T = 750 Ib

next determine F

750 + F - 2400 = 2400 / 4

hence F = 2250 Ib

B ) calculate Tension first

T - 1000 = ( 1000/g ) ( g/4)

T = 1250 Ib

F = 2400 -1250 - 2400/ 4

F = 550 Ib

C ) determine design limit

Max = 2400 * 1.2 = 2880 Ib

750 + new force - 2880 = 2880 / 4

new max force ( F newmax ) = 2850 Ib