How much will it stretch

1. Aug 12, 2004

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

Last edited: Aug 12, 2004
2. Aug 13, 2004

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.

3. Aug 13, 2004

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

4. Aug 13, 2004

Staff: Mentor

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

5. Aug 13, 2004

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.

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