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Can someone check my method please

  • #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.
 

Answers and Replies

  • #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 .
 
  • #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
 
  • #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.
 
  • #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)
 
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  • #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?
 
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  • #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
 
  • #8
lightgrav
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okay, (sorry about the HTML code) .
What do you think Y "is" ?
 
  • #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.
 
  • #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.
 
  • #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?
 
  • #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?
 
  • #13
yep, piR^2 = 0.196^2
 
  • #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 .
 
  • #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
 
  • #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?
 
  • #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?
 
  • #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.
 
  • #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.
 
  • #20
sorry i cant vision it, slightly confused on what your trying to describe
 
  • #21
lightgrav
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a BB is a round piece of metal ; soft lead BB's can be cut half-way thru by a knife,
and the wire jammed into the slot that the knife made. squeeze tight with pliers.
 
  • #22
<----- regular fisher :P i know what your on about with the BB:P however don't understand the set up of the needle and how measurements can be made with it.
 

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