Youngs Modulus. Copper wire experiment

1. Dec 9, 2008

Gregg

1. The problem statement, all variables and given/known data
Calculate Youngs Modulus for the copper wire

We have done the experiment today, here is the data:

(KG)/F(N)/x(M)

0.1/0.1g/0.0
0.2/0.2g/0.0
0.3/0.3g/0.001
0.4/0.4g/0.001
0.5/0.5g/0.002
0.6/0.6g/0.003
0.7/0.7g/0.004
0.8/0.8g/0.005
0.9/0.9g/0.006
1.0/1.0g/0.009
1.1/1.1g/0.029
1.2/1.2g/0.053
1.3/1.3g/0.089
1.4/1.4g/0.160

Diameter = 0.27mm = 2.7x10-4

Natural Length = 1m

2. Relevant equations

E = FL/AX

3. The attempt at a solution

A = (pi)(1.4x10-4)2

So i take the elastic region to be up to the 1kg load. The gradient of the line is (1g/0.009) = 1090.

The beginning length of the copper wire was 1m

E = 1090 x 1 / A

A = (pi)(1.4x10^-4)^2

E = 1090/(pi)(1.4x10^-4)^2

E = 1.77x10^10 Pa

= 17.7 GPa

According to the internet the young modulus is about 10 times larger than this. Have i gone wrong somewhere?

2. Dec 9, 2008

LowlyPion

Here is an online lecture that performs that very experiment.

3. Dec 9, 2008

Gregg

Ah, brilliant.

4. Dec 9, 2008

LowlyPion

I would have chosen another point further back because you might notice that the 1 kg point is already into the elastic region as the slope of the curve has changed.

5. Dec 9, 2008

Gregg

Yeah I have drawn a graph and can see this, also. Still strange to be out by a factor of 10, though.

6. Dec 9, 2008

LowlyPion

Using the .8/.005 point I get F/A as 1.371*108

divide by .005 and that yields 27.4 GPa

Copper looks like 110 to 130.

Do an error propagation analysis of the measurements. You're only a factor of 3 to 4 off. And a small measurement uncertainty in A or in ΔL can be pretty substantial.

7. Dec 2, 2009

queend4eva22

you need to convert your mass to (N) newtons

8. Dec 2, 2009

queend4eva22

oh n your lengths should be in meters (m) too stick with the metric measurements