The maximum upward acceleration of a lift of mass 2500kg is 0.5m/s^2.The lift is supported by a steel cable *?

Question

* which has a maximum safe working stress of 1.0*10^8Pa.What is the minimum cross sectional area of the cable?(Assume g=10m/s^2)

Answer 1

F = ma = 2500*10.5

S = F/A

A = F/S = 2500*10.5/1.0*10^8 = 2.625*10^(-4) m^2

That's for an empty lift.

If the lift can carry a maximum load of five 100 kg passengers add 500 kg to the 2500 kg.

A = F/S = 3000*10.5/1.0*10^8 = 3.15*10^(-4) m^2