A lift is tied with a thick iton wires and its mass is 800 kg. if the maximum ac celeration of lift is `2.2 ms^(-2)` and the maximum safe stress is `1.4 xx 10^8 Nm^(-2)` find the minimum diameter of the wire take `g =9.8 ms^(-2)`

Answer

9.34 mm

Explanation

Tension in the wire, F = m(g+a) = 800 (9.8 + 2.2) = 9600 N Stress, `S = (F)/(pi(D//2))^2 = (4F)/(pi D^2)`
or `D = sqrt((4F)/(pi S)) = sqrt((4xx9600xx7)/(22xx1.4 xx 10^8)`
` =9.34 xx 10^(-3) m = 9.34 mm`