
#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. 


Register to reply 
Related Discussions  
Youngs Modulus of Copper  Materials & Chemical Engineering  2  
Youngs Modulus. Copper wire experiment  Introductory Physics Homework  7  
Youngs modulus for a wire  Introductory Physics Homework  3  
Youngs modulus of copper  Advanced Physics Homework  4  
Youngs Modulus and Elastic Modulus  Introductory Physics Homework  6 