Question #22263

A large elevator uses a diesel engine with lift power P of 500kw to pull a cable lift which has a 1000 newton pull Fdrag when empty. The total vertical lift distance (d) is 400 meters. How many 60kg boxes can the elevator accommodate at one time if the vertical speed (v) of the elevator is 2m/s®?

Expert's answer

The power of engine is equal to P = F * d / t,t = d / V, where d = 400 m and V = 2 m/s.

t = 400 / 2 = 200 s

The maximum force that could be applied to a cable is:

Fmax = P * t / d = 500000 * 200 / 400 = 250000 N

The number of boxes can the elevator accommodate atone time is n = (Fmax - Fempty) / (m0 * g), where m0 is the mass of single box.

n = (250000 - 1000) / (60 * 10) = 249000 / 600 = 2490 / 6 = 415 boxes.

t = 400 / 2 = 200 s

The maximum force that could be applied to a cable is:

Fmax = P * t / d = 500000 * 200 / 400 = 250000 N

The number of boxes can the elevator accommodate atone time is n = (Fmax - Fempty) / (m0 * g), where m0 is the mass of single box.

n = (250000 - 1000) / (60 * 10) = 249000 / 600 = 2490 / 6 = 415 boxes.

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