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I'm looking at a basic free body diagram of a car travelling on a banked (inclined) road with inclination theta. Its mass is $m$. Here the diagram shows weight ($mg$), normal reaction force ($N$), and centripetal force ($mv^2/r$). They have resolved $N$ as $N\cosθ$ upwards and $N\sinθ$ horizontally: $$\begin{align} N\cosθ &= mg \\ N\sinθ &= \frac{mv^2}{r} \end{align}$$ My question is, why can't we resolve $mg$ as $mg\cosθ$, and equate it to $N$, normal to the plane? Like $mg\cosθ=N$. Why is this logic wrong?

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You can, but the acceleration $v^2/R$ is also diagonal in that coordinate system, so it has a component in the "normal" direction.

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  • $\begingroup$ I guess that's it. So is it an unspoken rule to always resolve the forces either along the created circle or normal to it? So as to avoid other forces? $\endgroup$ – Aman Thakkar Dec 23 '15 at 14:01
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    $\begingroup$ Pick a coordinate system that makes the analysis easier. $\endgroup$ – Spirko Dec 23 '15 at 14:06
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You can resolve forces in any direction you wish and then apply Newton's laws to solve the problem, but we resolve the forces in horizontal direction in the case of banking as centripetal Force is horizontal, which makes calculations easier. It is easy to resolve forces in the direction of acceleration, in this case it is $\frac{mv^2}{r}$ in horizontal direction, which would also have to resolved in the coordinate parallel to inclined plane, thus increasing complexity.

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To make the car go around the curve, you need centripetal force, which acts toward the center of the circular path that the car is on. Due to this, the centripetal force is horizontal, rather than parallel to the banked curve. This implies that you don't want to rotate horizontal and vertical axes for this problem. And, as you have stated, the car's weight must be supported by mgcos(theta).

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The physical problem you are solving with these kind of questions is F=ma, where F is the sum of forces in a certain direction, and a is the acceleration in that direction. In many cases the acceleration in a direction is zero, and it reduces to F=0 (as in your first equation). Unfortunately many people don't write down the acceleration, and explain why it has to be zero. This leads to many mistakes. If you add the inertia term (ma) to your equation is is correct.

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I think most of the answrs here are very well written.But considering you a begginer i would just like to say that take this as a thumb rule that u need to resolve the forces along and perpendicular to where the body accelerates.T his will make the analysis easier. P.S: You first need to make an intelligent guess about the direction of acceleration which in most cases is not a problem.

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