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Homework Help: Youngs Modulus of copper wire experiment

  1. Jan 18, 2012 #1
    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. jcsd
  3. Jan 18, 2012 #2
    You have an error in your strain computation. You seem to be dividing mm/m.
     
  4. Jan 18, 2012 #3
    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
     
  5. Jan 18, 2012 #4
    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
     
    Last edited: Jan 18, 2012
  6. Jan 18, 2012 #5
    im getting 68077.97019246913753978559011651

    couldnt be right
     
  7. Jan 18, 2012 #6
    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.
     
  8. Jan 18, 2012 #7
    so would be answer of 68077.97019246913753978559011651 be right ?
     
  9. Jan 18, 2012 #8
    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.
     
  10. Jan 18, 2012 #9
    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.
     
  11. Jan 19, 2012 #10
    can anybody else hint to me where im going wrong as i see Cu is suppost to be 117GPa
     
  12. Jan 19, 2012 #11
    If you plot your data, the curve looks ok. Are you certain you measured the diameter and wire length correctly?
     
  13. Jan 22, 2012 #12
    Hi there - My mistake was the diameter - it should have been calculated as 0.000573mm or 5.73x10-3

    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.
     
    Last edited: Jan 22, 2012
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