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I have a doubt which i will try to explain by quoting a problem.

Sorry for not being able to pinpoint the exact concept, though i tried my best to do so.

Question

A car with mass $m$ is moving at velocity $v$ took a turn of radius $r$ on a banked path. Find the inclination $(\theta)$ if the path is smooth and the horizontal force $(f_c)$ is provided only by normal reaction $N$.

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Taking horizontal component of normal reaction and equating it to centripetal force.

$$N \sin(\theta) = f_c = {mv\over r^2}\qquad (1)$$

equating normal reaction to component of weight, co-linear to normal reaction.

$$N = mg \cos(\theta)\qquad (2)$$

Substituting $(2)$ in $(1)$

$$mg\cos(\theta)\sin(\theta) = {mv\over r^2}$$ $$\sin(2\theta) = {2v\over gr^2}$$ $$\theta = \large{\arcsin\left({2v\over gr^2}\right)\over 2}$$


But in solution set, they took $N \cos(\theta) = mg \qquad (3)$

Dividing $(1)$ by $(3)$

$$\tan (\theta) = {v\over gr^2}$$ $$\theta = \arctan\left({v\over gr^2}\right)$$


From $(3)$, $N =\large{mg \over \cos(\theta)}$, whereas from $(2)$ , $mg\cos(\theta) = N$. Now both of these can't be true. So why is $(2)$ false and $(3)$ true ?

I guess it is something to do with the fact that the objecting is moving in circular banked path, and not in sliding in still slope.

Why, for a object moving in a circular banked path normal reaction is not equal to the component of weight co-linear to normal reaction ?

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    $\begingroup$ A hint: your vehicle is moving in a horizontal circle. What is its acceleration? $\endgroup$ – rob Oct 19 '16 at 20:04
  • $\begingroup$ @rob centripetal and radial. $$\large{a = a_c \mathbf{e_r} + {dv\over dt}\mathbf{e_t}}$$ right ? $\endgroup$ – A---B Oct 19 '16 at 20:11
  • $\begingroup$ Eqn 2 is wrong. $N\cos\theta=mg$ because the vehicle is not accelerating vertically, but there is a net horizontal force causing centripetal acceleration. $\endgroup$ – sammy gerbil Oct 19 '16 at 20:12
  • $\begingroup$ @sammygerbil That's an answer, rather than a comment. $\endgroup$ – rob Oct 19 '16 at 20:14
  • $\begingroup$ @sammygerbil Thanks. I feel dumb now, i should have thought about it. :-). $\endgroup$ – A---B Oct 19 '16 at 20:14

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