# Hard physics question

Asked Jun 16, 2011, 08:19 AM — 31 Answers
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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 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #21 Jun 17, 2011, 07:31 AM
No no no, I'm saying

I got 1.22338 10^11 Nm^-2

Hmm... I forgot one detail >.<

F = ke

If the length is doubled (the k constant doubles), the extension is doubled, and doubling the force too is wrong.

Hence, if you use the total length of the wire, you don't double the tension in the wire. I'm changing it now.

To be sure, just take the part of the wire that is under the tension which you got. You got the tension in half the wire, take the length and extension of that half wire.
 jcaron2 Posts: 983, Reputation: 1034 Senior Member #22 Jun 17, 2011, 07:40 AM
Quote:
 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
Let's talk through this one more time: Instead of a single wire spanning across from left to right, picture the weight suspended from two separate wires, which happen to be tied to the weight at the same point. The forces, tensions, etc. Are all exactly the same in this scenario as in the given one where the wire is continuous. Now, however, it's a little more intuitive to just consider one of the wires. Whatever tension, stress, elongation, and modulus we calculate for one wire will be exactly the same for the other wire, since everything is symmetrical in this problem.

Since the weight is held up by two wires, each one carries half the weight, or 87.5N. This 87.5N of upward force provided by the wire is accompanied by a 218.75N horizontal force pulling the weight toward the wall. This equates to a tension of ~235.6N in the wire. Given the cross-sectional area of the wire, this equates to a stress of around 9.424 GPa. Meanwhile, the strain of the wire is 0.077. Thus, the Young's modulus is around 122 GP.
 Roddilla Posts: 144, Reputation: 10 Junior Member #23 Jun 18, 2011, 12:14 AM
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Quote:
Originally Posted by jcaron2
Quote:
 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
Let's talk through this one more time: Instead of a single wire spanning across from left to right, picture the weight suspended from two separate wires, which happen to be tied to the weight at the same point. The forces, tensions, etc. Are all exactly the same in this scenario as in the given one where the wire is continuous. Now, however, it's a little more intuitive to just consider one of the wires. Whatever tension, stress, elongation, and modulus we calculate for one wire will be exactly the same for the other wire, since everything is symmetrical in this problem.

Since the weight is held up by two wires, each one carries half the weight, or 87.5N. This 87.5N of upward force provided by the wire is accompanied by a 218.75N horizontal force pulling the weight toward the wall. This equates to a tension of ~235.6N in the wire. Given the cross-sectional area of the wire, this equates to a stress of around 9.424 GPa. Meanwhile, the strain of the wire is 0.077. Thus, the Young's modulus is around 122 GP.
The one thing which i still cannot understand is why working with 1 piece gives me one answer but working with two pieces the answer is different

1 half is experiencing 1 force and the other half is experiencing another force with same value
 Roddilla Posts: 144, Reputation: 10 Junior Member #24 Jun 18, 2011, 12:19 AM
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Quote:
 Originally Posted by Unknown008 No no no, I'm saying I got 1.22338 10^11 Nm^-2 Hmm... I forgot one detail >.< F = ke If the length is doubled (the k constant doubles), the extension is doubled, and doubling the force too is wrong. Hence, if you use the total length of the wire, you don't double the tension in the wire. I'm changing it now. To be sure, just take the part of the wire that is under the tension which you got. You got the tension in half the wire, take the length and extension of that half wire.
But if one each half there is a force of 236N acting, on the whole wire the total force acting is 236 x 2
 Roddilla Posts: 144, Reputation: 10 Junior Member #25 Jun 18, 2011, 12:20 AM
THANK you all for your patience but I am not yet getting the point
 Roddilla Posts: 144, Reputation: 10 Junior Member #26 Jun 18, 2011, 12:26 AM
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Quote:
 Originally Posted by Roddilla THANK you all for your patience but I am not yet getting the point
The point that working with one half gives me one answer and working with whole wire gives me a different answer
 jcaron2 Posts: 983, Reputation: 1034 Senior Member #27 Jun 18, 2011, 06:39 AM
Don't worry; you're not the first person to be confused by the concept of tension. It's confusing because intuitively it always seems like it should be double. I'll give another example:

Suppose you have a wire connected to an immovable object. You pull on the end of the wire with 200N of force. The tension in the wire is then 200N, of course. If you were to focus on one specific point on the wire (let's pick the middle), however, you'd find that it was experiencing a force of 200N in one direction as a result of you pulling on it, and it was experiencing a force of 200N in the opposite direction as a result of the equal and opposite reaction from the wall (Newton's third law). So it seems like each point on the wire is getting pulled by 200N in both directions. But by definition, that's still 200N of tension.

So in your case, yes, it seems that the wire is getting pulled by 236N in both directions, but that, by definition, is 236N of tension.

For a more familiar analogy, let's say you weigh around 70kg. That means that your body presses down on the bottoms of your feet with a force of around 700N. Meanwhile, the ground is pushing up on your feet with an equal and opposite reaction of 700N. Does that mean the bottoms of your feet are experiencing 1400N of total compression? No. Pushing down on something with 700N of force results in 700N of compression. Tension is the same; pulling on something with 236N of force results in 236N of total tension.

I hope that makes it a little clearer.
 Roddilla Posts: 144, Reputation: 10 Junior Member #28 Jun 18, 2011, 07:13 AM
Comment on jcaron2's post
Quote:
 Originally Posted by jcaron2 Don't worry; you're not the first person to be confused by the concept of tension. It's confusing because intuitively it always seems like it should be double. I'll give another example: Suppose you have a wire connected to an immovable object. You pull on the end of the wire with 200N of force. The tension in the wire is then 200N, of course. If you were to focus on one specific point on the wire (let's pick the middle), however, you'd find that it was experiencing a force of 200N in one direction as a result of you pulling on it, and it was experiencing a force of 200N in the opposite direction as a result of the equal and opposite reaction from the wall (Newton's third law). So it seems like each point on the wire is getting pulled by 200N in both directions. But by definition, that's still 200N of tension. So in your case, yes, it seems that the wire is getting pulled by 236N in both directions, but that, by definition, is 236N of tension. For a more familiar analogy, let's say you weigh around 70kg. That means that your body presses down on the bottoms of your feet with a force of around 700N. Meanwhile, the ground is pushing up on your feet with an equal and opposite reaction of 700N. Does that mean the bottoms of your feet are experiencing 1400N of total compression? No. Pushing down on something with 700N of force results in 700N of compression. Tension is the same; pulling on something with 236N of force results in 236N of total tension. I hope that makes it a little clearer. :)
very helpful indeed - you have solved a problem which i had ages ago as well I think
So the fact that half of the wire is getting pulled by 200N of force and the other half getting pulled by 200N of force still means that the wire is getting pulled by 200N of force overall
 Roddilla Posts: 144, Reputation: 10 Junior Member #29 Jun 18, 2011, 07:15 AM
Jcaron - big thanks for your ingenious explanation
 jcaron2 Posts: 983, Reputation: 1034 Senior Member #30 Jun 18, 2011, 07:35 AM
I'm glad that helped. It's not a very intuitive concept!

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