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Homework Help: Can someone check my method please

  1. Mar 7, 2010 #1
    This is my method for finding youngs modulus of copper wire. I have tryed to make it as accurate and fair as possible. This is just the test, not the analysis of the data. Please be critical.

    -Measure Out 1m of wire
    -close a vernier scale and measure when completly shut to calculate the systematic error.
    - Calculate the area by taking measurements of the wire with a venier scale every 10cm (calculate the average)
    -Set up apparatus
    -weigh weights to make sure they are what they say they are. E.G a 200g weight really weighs 200g.
    -Apply a 300g test weight to straighten out the wire, leave for 2 minutes for the weights to have effect
    - Mark out a section of wire
    -measure the distance between the fixed end of the wire and the marker
    -Apply weights in 200g intervals till reaching 1000g above test weight. Leave for 2 minutes for weight to have effect. Then see how much the marker has moved using a ruler.
    The extention is the difference between that reading and the one between the fixed end of the wire and the marker we did earlier on.

    [ P.S can anyone tell me how to calulate uncertainys, I see in testresualts of a percentage of uncertianty]

    P.S.S sorry about spelling im dyslexic.
     
  2. jcsd
  3. Mar 7, 2010 #2

    lightgrav

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    if you're actually going to try this in a lab, you will find that the wire with 10 Newtons hanging from it is not much longer than the original 1 meter (even with small Area) , maybe only an extra 3/4 millimeter. Using a "ruler" (meterstick?) , you might be able to estimate this length to the nearest ¼ mm , which would mean distinguishing it from 1m+½mm , and from 1m+1mm . Even though the Length is measured to 3½ digits precision, your "stretch" of the wire would only be measured as 3 of your "¼mm" marks ... maybe only 2 of those marks, maybe 4 of them. If your uncertainty is plus or minus one mark, and you only measure 3 marks, your percent uncertainty is plus or minus 1 out of 3 ... that is 1/3 , or 30% uncertainty .
     
  4. Mar 7, 2010 #3
    Sorry? what?!

    Im sorry i don't understaand what your trying to say. Yes i will be doing it in a lab, and i understood the bit when you said the difference in 1m +10N would be small. (should i use a longer piece?)

    But i can't understand anything past that. Maybe a couple of typos?

    hope you could re explain for me.

    Thankyou
     
  5. Mar 7, 2010 #4

    lightgrav

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    do you know what your wire Area is going to be? and the "accepted" value of Y?
    (if you don't, thin copper wire is about ½ millimeter diameter)
    use these to estimate "about" how much it is going to stretch.
    envision yourself using a ruler to measure that tiny "extra" length.
     
  6. Mar 7, 2010 #5
    No i don't know the area of the wire. nor how to estimat how much it will stretch :(

    As for the measuring with a small tiny length. I see your point. Would having 2 markers maybe 1 or 2 cm apart then measure the difference with a vernier scale work? possibly with an electic one with i think can measure to 2 dp of a mm (the ones we sue anyway)
     
    Last edited: Mar 7, 2010
  7. Mar 7, 2010 #6

    lightgrav

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    the vernier can probably only measure 150mm , so .02mm/150mm = 133 parts per million

    Y for a "soft" metal is 70 x 10^9 N/m^2 ... what's the formula for the stretch, in parts per million?
     
    Last edited: Mar 7, 2010
  8. Mar 7, 2010 #7
    This is miles over my level im doing AS physics OCR B. Iv havent yet come across anything like that. I think thats where im finding differculty understandinh :( or maybe its me having a brain fart but i don't understnad
     
  9. Mar 7, 2010 #8

    lightgrav

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    okay, (sorry about the HTML code) .
    What do you think Y "is" ?
     
  10. Mar 7, 2010 #9

    ideasrule

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    Have you ever seen the formula Y = (F / A) / (dL / L) or anything like it? If so, calculate an approximate value for dL using the accepted value for Y and the F, A, and L you plan to be using.
     
  11. Mar 7, 2010 #10
    I took y is youngs modulus. yes ive seen that equation, finding a approximaion for dl im dnt have a clue how to do.
     
  12. Mar 7, 2010 #11

    lightgrav

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    you get to choose L , maybe 1m , maybe 150 mm .
    You planned to use F = 10N
    Y is probably going to be about 70 x 10^9 N/m^2 (or google for it)
    your wire diameter is likely going to be 0.5 mm ... do you know what Area that would have?
     
  13. Mar 7, 2010 #12

    lightgrav

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    you get to choose L , maybe 1m , maybe 150 mm .
    You planned to use F = 10N
    Y is probably going to be about 70 x 10^9 N/m^2 (or google for it)
    your wire diameter is likely going to be 0.5 mm ... do you know what Area that would have?
     
  14. Mar 7, 2010 #13
    yep, piR^2 = 0.196^2
     
  15. Mar 7, 2010 #14

    lightgrav

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    careful, those are mm^2 ... not m^2 .
    now plug in F, Y and A to get dL/L .
     
  16. Mar 7, 2010 #15
    10N * 0.000000196 / 70 x 10^9 N/m^2 = 2.8 x 10^-17 ???

    Knowing length of 1m. strain = dl/l
    =dl=strain x l = 2.8x 10^-17
     
  17. Mar 7, 2010 #16
    Ah i think im waking up now. Just thought about uncertaintys. So if you could measure to 1/4 mm by eye with a ruler. the certainty would be +- 1/4 mm. so in 1mm it would be a 25% uncertainty. And ofc if a electronic venier scale could measure to 2 dp mm or +-0.01????

    correct?
     
  18. Mar 7, 2010 #17

    lightgrav

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    F/A , not F*A ... but still less than 1 part per thousand
    ... less than 1 mm per meter , about 0.1mm in 150mm (for the vernier)
    how about using a long needle-pointer as a lever?
     
  19. Mar 7, 2010 #18
    duh ofc its divided my mistake. I think that doing this by eye using such small numbers is almost impossible. Im not convinced a needle would work. To make it accurate i think i will need something with smaller resoulution than the human eye. I still think an electronic vernier scale needs to be placed in the experiment somewhere.
     
  20. Mar 7, 2010 #19

    lightgrav

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    vernier only gets 0.02 mm precision if it has hard surfaces to measure from;
    you could squeeze 2 "split BB's" (for fishing tackle) onto the wire.
    If the back of the needle is held in a pivot, 10mm behind the wire,
    the needle could rest on one of these BB's ... and stick out 100 mm.
     
  21. Mar 7, 2010 #20
    sorry i cant vision it, slightly confused on what your trying to describe
     
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