X - Solving for Load in Young's Modulus Problem

[greeto8]

Had a problem with this question in my (Wilson Buffa Lou) College Physics book.

The problem says: A copper wire has a length of 5.0m and a diameter of 3mm. Under what load will its length increase by 0.3mm?
The answer in the back says 47N.

This is my work through the problem and I wanted to see where I went wrong.

delta L=0.3 x 10^-3m A=(3x10^-3)^2=A=(9x10^-6m^2) Lo=5.0m
Copper Wire=Y=11x10^10

ok... So...Y=FLo/(delta)LA

F(5.0m)/(0.3x10^-3)(9x10^-6m^2)=(11x10^10)

F(5.0m)/(2.7x10^-9)=11x10^10

multiplying both sides by 2x10^-9

297N/m=F(5.0m) 297N/m/5.0m=59.4N My answer: 59.4N Any help greatly appreciated. Is my work getting close? Cant find enough examples for Young's Modulus, any direction in that matter is greatly appreciated too. Thank you for your time.
Ed

 

[method_man]

A=(3x10^-3)^2=A=(9x10^-6m^2)

pi?
A=r²*pi
I just figured it out. In your formula you have to divide 3x10^-3 by 2 because you' re dealing with diameter.
I tried to solve it, and I get around 47 N

 

[tomcue]

win

Hello Edwin,

It looks like you are on the right track with your calculations. However, there are a few mistakes that may have led to your incorrect answer.

Firstly, in the equation Y=FLo/(delta)LA, the "A" should not be squared. It should just be A=9x10^-6m^2. This is because the area of a circle is calculated by A=πr^2, where r is the radius of the circle (in this case, the diameter of the wire is given, so you just need to divide it by 2 to get the radius).

Secondly, when you multiplied both sides by 2x10^-9, you should have also multiplied F(5.0m) by 2x10^-9. This would give you F=59.4N, which is the correct answer.

So, in summary, your mistake was mainly in squaring the area and not multiplying F(5.0m) by 2x10^-9. Keep in mind that when solving for a variable, you must perform the same operation on both sides of the equation.

I hope this helps clarify your doubts. Keep practicing and you will get more familiar with Young's Modulus and other concepts in physics. Good luck!