Question involving Young's modulus (mass hanging from a wire)

AI Thread Summary
The discussion centers on a physics homework problem involving Young's modulus and the calculation of stress from a hanging mass. Participants express confusion over discrepancies in their calculations, particularly a factor of ten difference in stress results. It is noted that the provided modulus of copper in the question is incorrect, as it suggests an unrealistic elongation of the wire under a 2kg weight. The conversation highlights the importance of accurately presenting problems and calculations to facilitate better assistance. Overall, the thread emphasizes the need for clarity and correctness in physics problem statements.
chriscarson
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Homework Statement
finding the stress
Relevant Equations
force over area
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so what I did wrong?
 
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chriscarson said:
Homework Statement:: finding the stress
Relevant Equations:: force over area

View attachment 269268

View attachment 269269

so what I did wrong?
(Please get in the habit of typing the problem and your work into the forum window. It makes it so much easier to try to quote your work and help you. Also, see the LaTeX tutorial in the lower left of the Edit Window. Thank you.)

I'm not sure how they got their answer. Mine is similar to yours, but different by a factor of 10 for some reason.

$$F = 2kg * 9.8m/s^2 = 19.6N$$
$$A = \pi r^2 = \pi * (0.002m)^2$$
$$Stress = \frac{F}{A} = \frac{2kg * 9.8m/s^2}{\pi * (0.002m)^2} = \frac{19.6N}{12.56 * 10^{-6}m^2} = 1.56 * 10^6 N/m^2$$
 
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berkeman said:
(Please get in the habit of typing the problem and your work into the forum window. It makes it so much easier to try to quote your work and help you. Also, see the LaTeX tutorial in the lower left of the Edit Window. Thank you.)

I'm not sure how they got their answer. Mine is similar to yours, but different by a factor of 10 for some reason.

$$F = 2kg * 9.8m/s^2 = 19.6N$$
$$A = \pi * r^2 = \pi * (0.002m)^2$$
$$Stress = \frac{F}{A} = \frac{2kg * 9.8m/s^2}{\pi * (0.002m)^2} = \frac{19.6N}{12.56 * 10^{-6}m^2} = 1.56 * 10^6 N/m^2$$

(the problem is that I m not understanding how you ask a question here )

Probably because I typed 19.6133 instead of 19.6 like you I have the answer different from yours, so another mistake by the teacher, that s bad , ok thanks for now
 
I press

forums,

Introductory Physics Homework Help

1599940001150.png


and I fill up this , have to be different as where to write ?
 
chriscarson said:
I press

forums,

Introductory Physics Homework Help

View attachment 269270

and I fill up this , have to be different as where to write ?
Sorry, are you asking where the LaTeX Guide / tutorial is? It's in the lower left of the Edit Window, as shown in the screenshot below. It is also available by clicking INFO, Help at the top of the page. :smile:

1599940177498.png
 
Also, when you clicked "Reply" to my post above with the LaTeX math in it, you could see what I typed to generate that LaTeX math rendering. :smile:
 
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is this place ?
 
chriscarson said:
View attachment 269273

is this place ?
Interesting, I think I see your point. When starting a new thread, the LaTeX Guide link does not seem to show up at the bottom of the Edit Window. That looks like a bug to me, so I'll report it.

Here is a direct link to the LaTeX tutorial, which you can always reach by clicking on INFO, Help at the top of the page:

https://www.physicsforums.com/help/latexhelp/
 
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berkeman said:
Interesting, I think I see your point. When starting a new thread, the LaTeX Guide link does not seem to show up at the bottom of the Edit Window. That looks like a bug to me, so I'll report it.

Here is a direct link to the LaTeX tutorial, which you can always reach by clicking on INFO, Help at the top of the page:

https://www.physicsforums.com/help/latexhelp/

still I was not going to use it because I don t know the purpose but next time I will write in the last screen shot I sent you .hope I don t cause you trouble .
 
  • #10
No trouble. We just like to help folks learn how to use the PF tools to make better posts, which helps you to get better replies (and helps us to reply to you). Feel free to send me a PM (private message, just click on my Avatar and "Start a Conversation) if you have further questions about how best to use the PF. :smile:
 
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  • #11
The question is a mess.
If my eyes do not deceive me, it gives the answer as 2.9 x 10-4Nm-2, ten orders of magnitude out.
And it states the modulus of copper as 1.6x106Nm-2. Were that so, a 2kg weight should double the length of the wire, which defies my intuition. In fact, the modulus is around 1011 Nm-2.
Note the coincidence between the erroneous modulus value and the correct answer to the stress question.
 
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  • #12
berkeman said:
No trouble. We just like to help folks learn how to use the PF tools to make better posts, which helps you to get better replies (and helps us to reply to you). Feel free to send me a PM (private message, just click on my Avatar and "Start a Conversation) if you have further questions about how best to use the PF. :smile:

Thanks 🙂
 
  • #13
haruspex said:
The question is a mess.
If my eyes do not deceive me, it gives the answer as 2.9 x 10-4Nm-2, ten orders of magnitude out.
And it states the modulus of copper as 1.6x106Nm-2. Were that so, a 2kg weight should double the length of the wire, which defies my intuition. In fact, the modulus is around 1011 Nm-2.
Note the coincidence between the erroneous modulus value and the correct answer to the stress question.

So there is a double mistake in question and answer ?
 
  • #14
chriscarson said:
So there is a double mistake in question and answer ?
Yes. The value it provides for the modulus is actually the answer to the first part of the question.
I assume the next part of the question asks you to find the extension. I don't think this will be 2.9 x 10-4m, but within an order of magnitude of that. Depends what value they intended to provide for the modulus.
 
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  • #15
yes next question is find the elongation.(answer 3.61m )
 
  • #16
chriscarson said:
yes next question is find the elongation.(answer 3.61m )
Fascinating. Roughly doubling the length would be the expected consequence of using the stress as also the modulus.
 
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  • #17
haruspex said:
Fascinating. Roughly doubling the length would be the expected consequence of using the stress as also the modulus.

Hope it was one of a kind with this question and answer.
 

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