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.