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Youngs Modulus of copper wire experiment 
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#1
Jan1812, 12:32 PM

P: 18

1. The problem statement, all variables and given/known data
I’m doing some basic young’s modulus homework and I think I’ve solved it but I want other people to give their answers to see if the match my own. Its young are of copper wire. 2. Relevant equations E = stress over strain Stress = applied force (F) over Cross sectional area (A) Strain = Extension (e) over Original length (Lo) A = 0.27mm F = mxg Lo = 1.788m e = variable My table of results are: M / kg e / mm 0.100 0.45 0.200 0.94 0.300 1.40 0.400 1.97 0.500 2.46 0.600 3.29 0.700 4.12 0.800 5.00 0.900 6.36 1.000 8.22 3. The attempt at a solution I’ve worked it out with 0.100kg M (0.100) x g (9.81) = 0.981 so F = 0.981 Pi x 0.27 squared over 4 = 0.05725566 so A = 0.05725566 So stress = 17.13371733038794867967 e = 0.45 Lo = 1.788 So strain = 0.25167785234899 Young’s Modulus = 68.0779701927423389726993185 I know I’m going wrong somewhere so any help would be appreciated 


#2
Jan1812, 02:13 PM

P: 1,195

You have an error in your strain computation. You seem to be dividing mm/m.



#3
Jan1812, 02:21 PM

P: 18

Ohh yes i see that  i know im going wrong somehwere and i cant get my head fully around that.
Why would the strain be such a low figure but the stress be so high Im sure 00025167785235 is to low for strain 


#4
Jan1812, 02:22 PM

P: 18

Youngs Modulus of copper wire experiment
Ohh yes i see that  i know im going wrong somewhere and i cant get my head fully around that.
Why would the strain be such a low figure but the stress be so high Im sure 00025167785235 is to low for strain 


#5
Jan1812, 02:34 PM

P: 18

im getting 68077.97019246913753978559011651
couldnt be right 


#6
Jan1812, 02:38 PM

P: 1,195

Because
sigma = e * E where E is Young's Modulus, a large number. Strain, e, is small. For steel, the modulus is 30X10^6 psi. sigma is stress. So as an example, for a strain of 0.00025, the stress is 7500 psi. Reasonable stress example. 


#7
Jan1812, 02:42 PM

P: 18

so would be answer of 68077.97019246913753978559011651 be right ?



#8
Jan1812, 02:51 PM

P: 1,195

Units would be N/mm^2. Watch your significant figures. Based on your first set of data, you did it correctly but consider the comment below.
You say A is area and its value is 0.27 mm. Is that a typo? Should 0.27 be the diameter? That is how you used it. I looked up the modulus for Cu. It's somewhat higher than what you calculate. 


#9
Jan1812, 03:02 PM

P: 18

Yes the 0.27 is the diameter which is the cross sectional area (A)
I calculated that as Pi x 0.27squared / 4 Thanks very much for your help so far. 


#10
Jan1912, 04:05 AM

P: 18

can anybody else hint to me where im going wrong as i see Cu is suppost to be 117GPa



#11
Jan1912, 08:05 AM

P: 1,195

If you plot your data, the curve looks ok. Are you certain you measured the diameter and wire length correctly?



#12
Jan2212, 07:40 AM

P: 18

Hi there  My mistake was the diameter  it should have been calculated as 0.000573mm or 5.73x103
so the result in getting now is 6.03 x 10(10) Pa as the youngs modulus. Im also working out the percentage error; ive worked out the length as 1.788m +/ 0.01m so 0.01/1.788 x 100 = +/ 0.56% does this look correct? My next task is to work out the percentage uncertainty of the cross section area and the gradient of the graph. 


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