Question

A 2000 meter long steel wire is suspended vertically in the lab. When a 2.000 kg hooked weight is added to the tower end of the wire it stretches by 1.000 mm. Find the diameter of the wire in mm.

Answer 1

given
length of steel wire, l = 2000 m
suspended vertically
added weight,m = 2 kg
dx = 1 mm
diamter of wire = d

now, youngs modulus = E = 200 GPa
also, density of steel = rho = 8050 kg/m^3

now, consider a point at a distacne y from the bottom of the wire
at this point
stress = mg/A + rho*yg
A = pi*d^2/4
ds = 4mg/pi*d^2 + rho*y*g
let the strain be dx
ds/dx = E
(4mg/pi*d^2 + rho*y*g)/dx = E
dx = (4mg/pi*d^2 + rho*y*g)/E
now
dx = d(dy)/dy
then
d(dy) = (4mg/pi*d^2 + rho*y*g)dy/E
integrating form y = 0 to y = L we get
d(dy) integral will be total change in length = 1 mm
hence
200*10^9/1000 = 4*2*9.81*2000/pi*d^2 + 8050*2000^2*9.81/2
hence
d is imaginary, this measn the data did not consider weight of the steel wire
hence

4mg*2000 m/pi*d^2*1 mm = 200*10^9
hence
d = 0.01580536613 m = 15.805366135494 mm