A steel wire 205 m long with circular cross section must str

A steel wire 2.05 m long with circular cross section must stretch no more than 0.250 cm when a 400.3 N weight is hung from one of its ends. Part A: What minimum diameter must this wire have?

Solution

stress = F/A =

Young\'s modulus is Y = 2*10^11 N/m^2

elongation = 0.25 *10^-2 m

length l = 2.05 m

strain = e/l = 0.0025/2.05 = 0.00125

stress = Y*strain = 2*10^11*0.00125 = 0.0025*10^11
F/A = 0.0025*10^11

A = 400.3/(0.0025*10^11) = 1.602*10-6

pi*(d/2)^2 = 1.602*10^-6

d/2 = Sqrt[(1.602*10^-6)/3.142] = 0.714 mm

d = 1.428 mm