If the elastic limit of steel is 5.0 multiplied by 108 Pa, determine the minimum diameter a steel wire can have if it is to support a 79 kg circus performer without its elastic limit being exceeded.

Answer 1

Maximum stress = (Force/Area)
5*10^8 N/m^2 = (79*9.8 N)/[(pi/4)D^2]

The mimimum diameter, D, will be in meters

pi/4 * D^2 = 1.55*10^-6 m^2

D^2 = 1.97*10^-6 m^2
D = 1.4*10^-3 m = 1.4 mm.

Answer 2

Thanks so much!

Answer 3

Can you explain a little bit further