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Physics Question about an elevator?

An elevator company is in the process of designing an emergency
brake for their new 1000 kg elevator which will act automatically in
the event of the cable breaking. The brake consists of hydraulic rams
on either side of the lift which press brake pads onto guide rails on the
lift shaft wall. (Figure QD1). Once a cable break is sensed it takes 0.5
seconds before the brake is deployed. Following this the system is to
be designed to decelerate the lift at a constant maximum rate of ‘2g’
(19.62 m/s2) – a rate designed to limit passenger injury.
a) The elevator is initially at rest when the cable breaks. Ignoring drag
and other resistance, determine the distance it falls and the speed it
attains just before the brake is applied.
(5)
b) With the aid of Free Body Diagrams where necessary, determine the
friction force required from the brake system to decelerate the falling
elevator at the maximum rate allowed. How long does it take for the
lift to stop? Assuming the coefficient of kinetic friction between the
brake pads and the guide rail is 0.8, how much force must the rams
together provide to create this level of friction.
(9)
c) What is the minimum height above the ground required for the
emergency brake to work without the elevator hitting the bottom of the
lift shaft? What simple safety feature would you recommend for
heights less than this?

Public Comments

1. Do your own homework.

2. i have no time to do your homework for you but i'll tell you i recommend ceramic breaks as a safety feature because ceramic works better when heated and it's used around the world it's the best....the math do it yourself, it's your homework not mine.

3. do your own homework

4. a) 0.5 seconds
s = ut + 0.5gt^2 = 0.5 x 9.81 x 0.5^2 = 1.226m
v = u +gt = 9.81 x1.226 = 12.03 m/s