How much will the aluminum wire stretch under a 3.57 kg load?

[Schu]

How much will the aluminum stretch

a 3.57 kg mass is supported by an aluminum wire with a length of 2.43 m and a diameter of 2.01mm. How much will the wire stretch?
Young's Modolus for Al= 6.9 * 10^10

First get the cross sectional area of the Al

then plug it into the formula
Y = Fl/A(delta)l

I come out with like 2572 m which just doesn't seem right. Where am I going wrong? I have checked the math and averything is in m and kg.

HELP​

 

[wisky40]

I found a result of 3.8830...10^-4m~.4mm.
delta(l)=(F*l)/(A*Y) with a force of mg=3.57kg(9.8m/s^2)=34.986N*m--->
F*l=34.986*2.43=85.01598 and A*Y=pi*((2.01/2)*10^-3)^2)(m^2)*(6.9*10^10)*(N/m^2)=218943.0113.. N ---->
delta(l)=85.01598N*m/218943.0113...N=3.8830...*10^-4m~.4mm
--->delta(l)~.4mm. I think you made a mistake somewhere,check everything again.

 

[wisky40]

You see I also made a mistake in the units, when I wrote mg=34.986N*m, I think that we should be more careful.

wisky40

 

[Doc Al]

Y = Fl/A(delta)l

Nothing wrong with this formula, but you'll have to rearrange it to find \Delta L:
\Delta L = \frac {L F}{Y A}

I come out with like 2572 m which just doesn't seem right. Where am I going wrong? I have checked the math and averything is in m and kg.

I'm glad you realize that your answer doesn't make sense. The only thing to do is to show exactly what numbers you plugged in, so we can check for an error.

wisky40's answer of about 0.4 mm is correct.

 

[Schu]

I went through it again after rearranging the formula, I'm not sure why I got what I did before but I came out with about 3.88 mm

THanks for the help.