# Brilliant oriented solution manual

So these days I have been working through the problems contained in the book Introduction to classical mechanics by D. Morins. I will say I really liked the problems. Apart from some problems with solutions, the book contains over 350 exercise problems with no solutions. So I was thinking we could make a solution manual for the book.

The problems in the book, have different stars labeled on them;

• 1-star(plenty) - requires good concept of the subject to solve.
• 2-star(plenty) - excellent grasp of the topic
• 3-star(scarce) - profound mastery of the subject
• 4-star(very rare) - the Harvard professors who created the book, won't give them to their students.

The problems are excellent, and have the Brilliant.org feel to them. I was hoping that every week, I will post a couple of the problems, and after gathering ideas and solutions, One solution will be selected (maybe more if they have different approaches), for that particular problem. In a sense it would be a Brilliant.org solution manual.

Please comment below, if you agree with the idea.

Mardokay has shared a link to the book.

And I have worked my way through some of the book, and if you have ideas not clear I would be happy to help.

Note by Beakal Tiliksew
4 years, 3 months ago

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2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

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I think it is a good idea, by the way it is a good book thanks for sharing. I found an online pdf format of this book yay we don't have to buy it Introduction to classical mechanics.I will re-share this.

- 4 years, 3 months ago

Glad you think so, Thanks for sending the link to the book, now everybody can join.

- 4 years, 3 months ago

But the pdf is with solutions...

- 4 years, 3 months ago

The exercise problems have no answer.

- 4 years, 3 months ago

Ok...I have submitted the solution to your third problem...Please check and see if the answer is correct..If not, I will make the necessary changes..

- 4 years, 3 months ago

Yeah, it looks correct. That is the answer I got. The thing is no one knows the answer to this problems,

- 4 years, 3 months ago

Could you check the second solution?

- 4 years, 3 months ago

Don't you think there should be tension force in the rope? Because friction cannot act on the hanging part as it can act only on surfaces in contact and only tension can pull it and support its weight.

- 4 years, 3 months ago

I think that that tension is developed due to the friction..If there were no friction, then there would be no tension in the string...Also, since there is equilibrium, we can say that the friction force (total) is equal to the tension in the string (look at the last piece of the rope, which is just about to leave contact from the incline)

- 4 years, 3 months ago

No, I did not mean to say in that manner. It is true that that tension is developed only due to friction. However, your solution considers that the entire friction is acting on the point which is just about leave contact from the incline. If you consider the FBD of the part lying on the incline then there would be three forces along the incline acting on the part which is not hanging namely, friction, tension and the component of its weight. As their resultant is zero, it is evident that tension will differ in magnitude from friction. This is what I feel.

- 4 years, 3 months ago

If the tension in the string is $$\displaystyle T$$, then it is evident that :

$2T\sin\theta = \lambda (fl)g$

As you mentioned, the tension will be different from the friction...That is true..Considering the forces for the part of the rope which is in contact with the incline,

$F = \frac{1-f}{2}mg\sin\theta + T$

And we know that, the friction force $$\displaystyle F$$ is simply $$\displaystyle \mu N = 1\cdot \frac{1-f}{2}mg\cos\theta$$.

From these equations, we get,

$f = \frac{\cos\theta\sin\theta - \sin^2\theta}{\cos\theta\sin\theta + \cos^2\theta}$

And, we can find the maximum value for this function easily..Is this correct?

- 4 years, 3 months ago

Yeah, I think this is correct. The maximum value is attained at $$\frac{\pi }{8}$$ as @Mardokay Mosazghi mentioned in the note with the question, right?

- 4 years, 3 months ago

Wait that it right, i was convinced that it was$$pi/4$$ as @Anish Puthuraya said @Sudeep Salgia

- 4 years, 3 months ago

Do you want to say that the original solution which Anish had posted initially was correct and not this one?

- 4 years, 3 months ago

I think he meant to say that he was convinced that my original solution was correct, and that the answer was $$\frac{\pi}{4}$$

- 4 years, 3 months ago

yes that is what i meant

- 4 years, 3 months ago

Oh... Okay.

- 4 years, 3 months ago

Yes..

- 4 years, 3 months ago

Yeah, that seems correct...you have any suggestions as to how to approach it?

- 4 years, 3 months ago

Sorry, that I have become unresponsive, I am flying out in an hour to the US, I am going to be out for a week.

- 4 years, 3 months ago

Probably @Anish Puthuraya will keep a lot solutions ready by then, making your work easier. :D

- 4 years, 3 months ago

I agree.

- 4 years, 3 months ago

Attending the event, are you?

- 4 years, 3 months ago

I seem to have solved it...Ill reply to you in a minute

- 4 years, 3 months ago

Nice, thats what I got before posting them, I agree, lets wait for an objective , then take it as a solution

- 4 years, 3 months ago

Well, if others can confirm it, then there is no doubt that the solution is correct.

- 4 years, 3 months ago

Hey thanks for the link. BTW, can you give the E&M by david morin too?

- 3 years, 10 months ago

Sorry @Kartik Sharma there is no pdf for that one i searched and searched but couldnot find it,try it and tell me

- 3 years, 10 months ago

Let's do it

Staff - 4 years, 3 months ago

Great lets just wait a couple of days, for people to see this, and start.

- 4 years, 3 months ago

It really sounds interesting. I would love to join in.

- 4 years, 3 months ago

Same here : )

- 4 years, 3 months ago

Me too

- 4 years, 3 months ago

Glad you think so, i have included some of the statics problem, as a sample, check it out here

This is not going to be the real thing.

- 4 years, 3 months ago

Good idea.........

- 3 years, 8 months ago

Good idea. I will try to help.

- 3 years, 9 months ago

hey

- 3 years, 9 months ago

How many per week, would you guys say is optimal?

- 4 years, 3 months ago

Personaly I am here everyday so the planning goes for others

- 4 years, 3 months ago