# Can someone check my method please

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.

lightgrav
Homework Helper
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 .

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

lightgrav
Homework Helper
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.

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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lightgrav
Homework Helper
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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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

lightgrav
Homework Helper
okay, (sorry about the HTML code) .
What do you think Y "is" ?

ideasrule
Homework Helper
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.

I took y is youngs modulus. yes ive seen that equation, finding a approximaion for dl im dnt have a clue how to do.

lightgrav
Homework Helper
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?

lightgrav
Homework Helper
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?

yep, piR^2 = 0.196^2

lightgrav
Homework Helper
careful, those are mm^2 ... not m^2 .
now plug in F, Y and A to get dL/L .

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

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?

lightgrav
Homework Helper
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?

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.

lightgrav
Homework Helper
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.

sorry i cant vision it, slightly confused on what your trying to describe

lightgrav
Homework Helper
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.

<----- 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.