This question , which I've asked, I got two close votes on it already but I can not understand why. I think having a complete discussion of my question/ answer to it would benefit the community as a whole because if my question is solved then there is a large category of problems which could be solved using the techniques I've discussed in the question. Further, I've put in some effort of my own by analysing the problem to the best of my knowledge. So, I can not understand where exactly I'm going wrong/ making people think this is some homework question

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  • $\begingroup$ BTW I don't think your technique would be very useful for finding equilibrium of a particle moving along a plane curve. You may want to think back on some of the assumptions you made in those equations to see if they apply. $\endgroup$ – JMac Jul 27 at 19:47
  • $\begingroup$ Other than gravity being the most direct one and friction being uniform, I don't see anything else $\endgroup$ – Buraian Jul 27 at 20:04
  • $\begingroup$ You're using equations as if it's an inclined plane problem, but there's an extra factor you're missing. $\endgroup$ – JMac Jul 27 at 20:07
  • $\begingroup$ I can't see it... can you say specifically? $\endgroup$ – Buraian Jul 27 at 20:09
  • $\begingroup$ If the path is curved you need more forces to follow that curve. $\endgroup$ – JMac Jul 27 at 20:13
  • $\begingroup$ Basically the plane curve is like a race track of sorts, imagine that curve being like a structure which the particle could move through. Would I still need application of centripetal force? $\endgroup$ – Buraian Jul 27 at 20:13
  • $\begingroup$ en.wikipedia.org/wiki/Centripetal_force $\endgroup$ – JMac Jul 27 at 20:14
  • $\begingroup$ Ok, I think I'm starting to realize my mistake, How do I get equilibrium when I include centripetal force? Do I just equate component of normal as centripetal force? $\endgroup$ – Buraian Jul 27 at 20:17
  • $\begingroup$ I have an extra unknown because I do not know the radius of circle in the centripetal acceleration formula at some point on curve $\endgroup$ – Buraian Jul 27 at 20:18
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    $\begingroup$ Hey all, the comment section here isn't really the place to discuss how to solve the original question. Could you take it to Physics Chat? If you like, I can migrate the existing comments to a new chat room. $\endgroup$ – David Z Jul 27 at 20:34

Speaking for myself:

I do not understand what you mean by " find the points on the curve where the particle would become at equilibrium." Presumably that would be "reach equilibrium" and not "become at equilibrium".

Next, the notation is unclear. Your curve is parametrized by $(x,g(x))$ which is a bad choice because you also use $g$ which I presume is gravity. In your text you use $`\mu$' when I think you meant "$\mu$" in quotation mark. Or did you? Do you mean to use $\mu^\prime$ as your coefficient of friction? Why use `$\mu$' or quotation mark? Is this static or kinetic friction? (I presume static as I can't see how kinetic friction would enter in equilibrium.)

Next, you have an inequality $f_{\mathrm{max}}\ge mg\sin\theta$ but the inequality sign disappears afterwards. I suspect that $\mu=g'(x)$ should be something like $\mu\ge g'(x)$ but I'm not sure, all the more so as I suppose $g'(x)$ is the derivative of $g(x)$ or is it related to the gravitational constant $g$ (which doesn't make sense).

Next, you seem to treat the problem as in rotational motion, introducing $mv^2/r$. As you admit yourself, you cannot find a meaning for $r$ which I suppose would be the local radius of curvature at $x$, i.e. find the osculating circle at $(x,g(x))$ and use the radius for that.

Next, how to you define equilibrium when there's friction? Do you mean if you place a particle there it would remain there? Then surely there might be a range for if I imagine the slope of the curve is very small everywhere then a particle might be stuck anywhere along the curve and not overcome static friction. If you define equilibrium position as the position $x$ where the particle would come to rest, then it's intuitively clear that this must be the bottom of your curve (or at least a local minimum) since the particle would slowly dissipate energy until it reaches that point with no more energy, and would stop there.

I could go on but if you start with these it will immediately improve the question, although i must say it does look like a homework question.

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  • $\begingroup$ I tried to fix everything you noted to the best of my ability , if there is anything then pls inform $\endgroup$ – Buraian Jul 27 at 21:47

The current version (v11) has several problems. It is not focused, as evident by the plethora of questions in the post. And you're really not asking about any physics concepts per se, you're just asking for the results of physics calculations:

Basically I want to find a description of points of equilibrium for which the body was moving already (kinetic friction ).

And you're also somewhat expecting others to check your work:

Would there exist no points of equilibria by this derivation or would I have to account for something else there?

So in summary it looks like the question boils down to "Hey, I have this interesting problem, and I don't really know how to analyze it. What should I add to my analysis? ... What about this point? ... And what if this were the case? ... If this was the case then what would I do differently?" This, to me, is definitely not on topic for PSE.

To answer the title question of "How do I prevent general analysis questions from being close voted?" I will say that the answer has nothing to do with the fact that the question is about "general analysis" and more about just the quality of the post.

For v12 you are still just asking how to do a calculation/solve a problem. Therefore, I still think it shouldn't be reopened yet.

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