Asked by Sarah
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.
Answers
Answered by
drwls
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.
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.
Answered by
Sarah
Thanks so much!
Answered by
Washington
Can you explain a little bit further
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