Min Steel Wire Diam for 390N: 0.296mm

[johntuan2009]

A steel wire of length 1.92 m with circular cross section must stretch no more than 0.210 cm when a tensile (stretching) force of 390 N is applied to each end of the wire.

What minimum diameter is required for the wire?

Express your answer in millimeters. Take Young's modulus for steel to be Y = 2.00×10^11 .

 

[PhanthomJay]

Hello, johntuan! Welcome to PF! But per forum rules, before you receive help, you must show an attempt at a solution . Then we can provide assistance to help you obtain the solution, by seeing were you may have gone wrong. You should be familiar with the stress vs. strain or load versus elongation relationship for wires with given parameters of length, load, area, and Young's modulus.

 

[johntuan2009]

E=Stress/Strain
E=(force/area)/(Change in length/original length)
E*(change in length/original length)= force/area
area=force/(E*(change in length/original length))
PiR^2=(390N)/((2.0*10^11)/(0.0021m/1.92m))
R^2=((390N)/((2.0*10^11)/(0.0021m/1.92m)))/Pi
R=sqrt{((390N)/((2.0*10^(11))/(0.0021m/1.92m)))/Pi
d=2R
then change into mm

 

[PhanthomJay]

Well, that is a lot better, and correct. Nice work!

 

[rwisz]

Based on the given information, the minimum diameter required for the steel wire can be calculated using the formula for Young's modulus:

E = (F * L) / (pi * r^2 * delta)

Where:
E = Young's modulus
F = applied force
L = length of wire
r = radius of wire
delta = maximum allowed stretch

Plugging in the values:
2.00×10^11 = (390 * 1.92) / (pi * r^2 * 0.210)

Solving for r, we get:
r = 0.296 mm

Therefore, the minimum diameter required for the wire is 0.296 mm. This diameter will ensure that the wire does not stretch more than 0.210 cm when a force of 390 N is applied to each end.