# Hard physics question

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

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 ebaines Posts: 11,885, Reputation: 1274 Expert #2 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 Posts: 984, Reputation: 204 Senior Member #3 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 Posts: 145, Reputation: 3 Junior Member #4 Jun 16, 2011, 10:11 AM
Comment on jcaron2's post
Originally Posted by jcaron2
We can help, but have you worked out ANY of the answers yourself? Parts a and b, for example, are simple trigonometry.
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 Posts: 984, Reputation: 204 Senior Member #5 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 Posts: 145, Reputation: 3 Junior Member #6 Jun 16, 2011, 10:57 AM
Comment on jcaron2's post
Originally Posted by jcaron2
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?
my mistake 2.36 x 10^11 not -11

could you check if it is good
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #7 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 Posts: 984, Reputation: 204 Senior Member #8 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 Posts: 11,885, Reputation: 1274 Expert #9 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 Posts: 8,076, Reputation: 723 Uber Member #10 Jun 16, 2011, 11:49 AM

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

Yes, now it's good :)
 Roddilla Posts: 145, Reputation: 3 Junior Member #11 Jun 16, 2011, 11:52 AM
Comment on ebaines's post
Originally Posted by ebaines
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%.
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 Posts: 11,885, Reputation: 1274 Expert #12 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:

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

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 Posts: 984, Reputation: 204 Senior Member #13 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 Posts: 145, Reputation: 3 Junior Member #14 Jun 16, 2011, 10:10 PM
Comment on jcaron2's post
Originally Posted by jcaron2
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.
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 Posts: 145, Reputation: 3 Junior Member #15 Jun 16, 2011, 10:11 PM
Comment on jcaron2's post
Originally Posted by jcaron2
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.
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 Posts: 145, Reputation: 3 Junior Member #16 Jun 16, 2011, 10:24 PM
Comment on jcaron2's post
Originally Posted by jcaron2
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.
If you take 236N only then you must take one half of the wire only, no?
 Roddilla Posts: 145, Reputation: 3 Junior Member #17 Jun 17, 2011, 01:08 AM
Comment on ebaines's post
Originally Posted by ebaines
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:

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

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.
Why do you divide by two if the every tension is acting on a half of the string?
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #18 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... =/

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 Posts: 145, Reputation: 3 Junior Member #19 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 Posts: 145, Reputation: 3 Junior Member #20 Jun 17, 2011, 06:27 AM
Comment on Roddilla's post
Originally Posted by Roddilla
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
not 2 * 10^11 but 2.36 * 10^11

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