Among the safety features on elevator cages are spring-loaded brake pads which g

Question

Among the safety features on elevator cages are spring-loaded brake pads which grip the guide rail if the elevator cable should break. Suppose that an elevator cage of 1600 kg has two such pads, arranged to press against opposite sides of the guide rail, each with a force of 1.20 105 N. The friction coefficient for the brake pads sliding on the guide rail is 0.15. Assume that the elevator cage is falling freely with an initial speed of 10.8 m/s when the brake pads come into action.

(a) How long will the elevator cage take to stop?

(b) How far will it travel?

(c) How much energy is dissipated by friction?

Explanation / Answer

Part A

Friction force by two pads

f =2* u*R

u = 0.15

reaction force R = 1.20*10^5 N

f = 2*0.15*1.20*10^5 = 0.306*10^5 N = 30600N

deceleration

a = f/m = 30600/1600 = 19.125m/s^2

now speed of elevator

V = 10.8m/s

time taken to stop.

Vf = Vi-at

0 = 10.8-19.125*t

t = 0.564 seconds

Part B

distance travelled

by third equation of motion

Vf^2 = Vi^2-2*a*d

0 = 10.8^2 - 2*19.125*d

d = 3.05 m

Part C

energy dissipated by friction

E = f*d

E = 30600*3.05

E = 93312 J

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