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.

Facts about 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.

## Comments

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TopNewestI 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. – Mardokay Mosazghi · 2 years, 9 months ago

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– Beakal Tiliksew · 2 years, 9 months ago

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

– Anish Puthuraya · 2 years, 9 months ago

But the pdf is with solutions...Log in to reply

– Beakal Tiliksew · 2 years, 9 months ago

The exercise problems have no answer.Log in to reply

– Anish Puthuraya · 2 years, 9 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..Log in to reply

– Beakal Tiliksew · 2 years, 9 months ago

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

– Anish Puthuraya · 2 years, 9 months ago

Could you check the second solution?Log in to reply

– Sudeep Salgia · 2 years, 9 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.Log in to reply

– Anish Puthuraya · 2 years, 9 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)Log in to reply

– Sudeep Salgia · 2 years, 9 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.Log in to reply

\[2T\sin\theta = \lambda (fl)g\]

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? – Anish Puthuraya · 2 years, 9 months ago

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@Mardokay Mosazghi mentioned in the note with the question, right? – Sudeep Salgia · 2 years, 9 months ago

Yeah, I think this is correct. The maximum value is attained at \(\frac{\pi }{8} \) asLog in to reply

@Anish Puthuraya said @Sudeep Salgia – Mardokay Mosazghi · 2 years, 9 months ago

Wait that it right, i was convinced that it was\(pi/4\) asLog in to reply

– Sudeep Salgia · 2 years, 9 months ago

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

– Anish Puthuraya · 2 years, 9 months ago

I think he meant to say that he was convinced that my original solution was correct, and that the answer was \(\frac{\pi}{4}\)Log in to reply

– Mardokay Mosazghi · 2 years, 9 months ago

yes that is what i meantLog in to reply

– Sudeep Salgia · 2 years, 9 months ago

Oh... Okay.Log in to reply

– Anish Puthuraya · 2 years, 9 months ago

Yes..Log in to reply

– Anish Puthuraya · 2 years, 9 months ago

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

– Beakal Tiliksew · 2 years, 9 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.Log in to reply

@Anish Puthuraya will keep a lot solutions ready by then, making your work easier. :D – Sudeep Salgia · 2 years, 9 months ago

ProbablyLog in to reply

– Mardokay Mosazghi · 2 years, 9 months ago

I agree.Log in to reply

– Anish Puthuraya · 2 years, 9 months ago

Attending the event, are you?Log in to reply

– Anish Puthuraya · 2 years, 9 months ago

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

– Beakal Tiliksew · 2 years, 9 months ago

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

– Anish Puthuraya · 2 years, 9 months ago

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

– Kartik Sharma · 2 years, 4 months ago

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

@Kartik Sharma there is no pdf for that one i searched and searched but couldnot find it,try it and tell me – Mardokay Mosazghi · 2 years, 4 months ago

SorryLog in to reply

Let's do it – Josh Silverman Staff · 2 years, 9 months ago

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– Beakal Tiliksew · 2 years, 9 months ago

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

It really sounds interesting. I would love to join in. – Sudeep Salgia · 2 years, 9 months ago

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– Karthik Kannan · 2 years, 9 months ago

Same here : )Log in to reply

– Anish Puthuraya · 2 years, 9 months ago

Me tooLog in to reply

here

Glad you think so, i have included some of the statics problem, as a sample, check it outThis is not going to be the real thing. – Beakal Tiliksew · 2 years, 9 months ago

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Good idea......... – Vishal Bambhaniya · 2 years, 2 months ago

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Good idea. I will try to help. – Chew-Seong Cheong · 2 years, 3 months ago

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hey – Jatasia Hudson · 2 years, 3 months ago

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How many per week, would you guys say is optimal? – Beakal Tiliksew · 2 years, 9 months ago

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– Mardokay Mosazghi · 2 years, 9 months ago

Personaly I am here everyday so the planning goes for othersLog in to reply