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Re: [Fhsst-physics] False Information in an example?

From: Lucas Vogelsang
Subject: Re: [Fhsst-physics] False Information in an example?
Date: Sat, 21 Oct 2006 12:01:12 +0200
User-agent: Thunderbird (X11/20060918)


Ok, thanks!

I am actually learning for an ordinary exam. About electronics and magnetism. I was just a little too lazy during the lessons to write down what the teacher said. I am going to the Gymnasium Hohe Promenade in Zurich.

I am half american so reading & writing in english isn't a big problem.

Why I wrote "even though I am from Switzerland". At our school we have to pay for the books. I bought a physics book, but I don't really find it helpfull. The book you wrote, teachs physics exactly in the way my teacher did and I find it quite helpful.

I think one should publish this book not only in South Africa. We have to pay about 400 CHF(roughly 300$) each year for books. For most of the pupils, this isn't a problem, but I still think, that it is a waste and could be cheaper if we were using open contents.

Je préfère écrire en anglais, c'est plus facile. :)

Hope you have fun at the CERN. You're have already several more years of education on your back. :) I hope I'll be able to work at such an institute as well.

Greetings from Zurich

Markus Oldenburg wrote:
Hello Lucas,

thanks for coming back to us with your findings. It is great to hear that you can make good use of the book 'even in Switzerland'. ;-)

I checked the worked example 75 and you are right. There is an error with the sin and the cos. Unfortunately there is another subsequent error, which renders the whole solution wrong (numerically). We will change the book accordingly.

For now here is how it is supposed to be:
[Remark: All the text stays the same (except for the last sentence), just the equations change.]

First change: Step 3
F_E = T * cos (60 degrees)      F_g = T * sin (60 degrees)
This is the error you spotted.
Explanation: T is the 60 degree projection on F_E, therefore it has to be the cos. On F_g instead T is the 30 degree (90 degrees - 60 degrees) projection, which leaves us with cos (30 degrees) = sin (60 degrees).

Second problem: Step 3, second equation:
This equation is mathematically correct, but the result is wrong. It should be T=115.5 N (not 1155 N).

Subsequently, step 3, third equation (the cos (60 degrees) was and is correct here, the typo happened earlier as mentioned above): Putting the correct result for T from the last equation here gives us: F_E = 115.5 N * cos (60 degrees) = 57.75 N.

Finally, step 4:
|Q_x| is numerically wrong, due to the error introduced earlier. The correct value is: |Q_x| = sqrt( F_E * r^2 / k) = sqrt ( (57.75 N) * (0.5 m)^2) / ( 8.99 * 10^9 N * m^2 / C^2) )
         = 5.67 * 10^(-5) C
"This the charge on X is -5.67 * 10^(-5) C."

Thanks again for finding this inconsistency. Please let me know if everything is clear to you now. What physics test are you learning for? Where in Switzerland are you? You can also contact me in German (not in French, yet ;-) ) if this is easier for you.

Greetings from Geneva,


Lucas Vogelsang wrote:

I have found your book via Wikibooks. I am currently learning for a physicstest and think that this book is quite helpfull for me, even though I am from Switzerland.

However one thing confused me:

The worked example 75 seems to be wrong.
It first says:

F_{E} = T*sin(60) and F_{g} = T*sin(60)

next time they refer to F_{E} as F_{E} = T* cos(60) = ...

So what is correct?

I have found this in a pdf. It is on the pdf page 230 or the page 222 according to the documents page numbers.


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