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Roddilla · Jun 16, 2011, 08:19 AM

Could you answer this question for me please? It is the hardest question one may get about young modulus, elasticity ecc.

ebaines · Jun 16, 2011, 08:38 AM

Roddilla - we are not going to take your exam for you! I suggest you work through each step of the problem, and if you get stuck along the way please show us what you tried and how you got stuck - then we can help point the way.

jcaron2 · Jun 16, 2011, 09:42 AM

We can help, but have you worked out ANY of the answers yourself? Parts a and b, for example, are simple trigonometry.

Roddilla · Jun 16, 2011, 10:11 AM

Yes I worked it all out and in fact got a value of 2.36 x 10^-11 for Young's Modulus but I don't know if I worked it out correctly

jcaron2 · Jun 16, 2011, 10:53 AM

Your value for Young's Modulus should not have a negative sign in the exponent. It should be on the order of the reciprocal of what you got. What did you calculate for the stress and elongation?

Roddilla · Jun 16, 2011, 10:57 AM

my mistake 2.36 x 10^11 not -11

could you check if it is good

Unknown008 · Jun 16, 2011, 11:03 AM

Hm... may I ask you to post the stress and extension you got?

I'm not getting what you got.

jcaron2 · Jun 16, 2011, 11:15 AM

That seems just about right. I get the same answer if I round the extension up to 0.08, but I would suggest you keep at least one more significant digit.

Jerry, how different was your answer? I think I'm right, but I must admit I'm not entirely sure if I'm off by a factor of two. You've probably done this sort of problem much more recently than I have! ;)

ebaines · Jun 16, 2011, 11:43 AM

JC: I Think you're correct. I'm getting E=2.45 x 10^11 Pa, or 245 GPa. By the way, Young's Modulus for steel is on the order of 200 GPa (depending on the particular alloy), so this seems reasonable.

I calculate a stress of 18.8 GPa and strain of 7.7%.

Unknown008 · Jun 16, 2011, 11:49 AM

Oh, you had more posts going in the meantime :p

Yes, the last post I saw was the negative power :o

Yes, now it's good :)

Roddilla · Jun 16, 2011, 11:52 AM

So for the force which is extending the wire you take the tension of each half of the wire
My teacher is saying that in order to calculate extension you have to use the tension of one half of the wire only which surely doesn't make sense

Could you plase post the working of the question so that I can compare it to mine

ebaines · Jun 16, 2011, 12:04 PM

Originally Posted by Roddilla

So for the force which is extending the wire you take the tension of each half of the wire
My teacher is saying that in order to calculate extension you have to use the tension of one half of the wire only which surely doesn't make sense

OK, now I'm embarrassed. I'm afraid I forgot to divide by 2 and I'm off by a factor of 2.

From the symmetry of the problem you can see that half of the 175N is carried by each half of the wire. So the tension in the wire is found from:

$<br /> \frac 1 2 W= T sin(\theta)\\<br /> \frac 1 2 175 N = T sin(21.8 deg)<br />$

So T = 235.6N, and Stress = 235N/(.025 mm^2) x 10^6 mm^2/m^2 = 9.42 x 10^9 Pa

Strain is 7.7%, so

E = Stress/Strain = 9.42 x 10^9 Pa/0.077 = 122 GPa.

Sorry for the previous error.

jcaron2 · Jun 16, 2011, 12:20 PM

EB, that was exactly the factor of two I was unsure about. In fact, that was my first way of calculating it, but then I looked up the Young's Modulus of steel (and saw Rodilla's answer) and decided it must be wrong.

I definitely agree with your answer now though. And the fact that it's low compared to the book value for E is not surprising since the wire is assumed to have gone beyond its elastic limit.

Roddilla · Jun 16, 2011, 10:10 PM

But if two tensions are acting from the centre, why don't you take the force acting on the wire as 235 x 2?

Roddilla · Jun 16, 2011, 10:11 PM

In my opinion it is like 1 half of the wore is receiving 236N of force while the other half is also receiving 236N of force so the total length of the wore is receiving 236*2

Roddilla · Jun 16, 2011, 10:24 PM

If you take 236N only then you must take one half of the wire only, no?

Roddilla · Jun 17, 2011, 01:08 AM

Why do you divide by two if the every tension is acting on a half of the string?

Unknown008 · Jun 17, 2011, 05:13 AM

Darn I must have been asleep while doing this :(

I got 1.22338... × 10^11 and saw the same thing on the screen... =/

Okay Rodilla, for your comments now (please, use the answer box below the page instead of using the comments options please.

In my opinion it is like 1 half of the wore is receiving 236N of force while the other half is also receiving 236N of force so the total length of the wore is receiving 236*2

No, if you use the whole wire and double the extension the tension that you use has to be the same.

If you take 236N only then you must take one half of the wire only, no?

As per above, you take 236 N when taking the whole wire too.

why do you divide by two if the every tension is acting on a half of the string?

The tension is divided by two because the portion of the wire that he is working with is half the total length of the wire.

Roddilla · Jun 17, 2011, 06:26 AM

SO as you are saying unknown008 the answer would still come to be 2 * 10^11 approximately since total extension of wire is 0.008, the original length is 0.10, the total force acting on the whole wire is 236 * 2 and the area is 2.5 * 10^-8
E = 1.89 * 10^10 (stress) * 0.10/0.008 = 2.36 * 10^11

Roddilla · Jun 17, 2011, 06:27 AM

not 2 * 10^11 but 2.36 * 10^11