Suppose there is a banked road on which a body is placed as shown in the figure.

enter image description here

Now to derive the relation between the velocity and the angle of inclination of the slope we do the following:-

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

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

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

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

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

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

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

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

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

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 have found many other similar questions in this site but none of the answers were quite satisfying.

Please don't flag the question as duplicate because I have been struggling a long to find the answer.

Any help is highly appreciated.


2 Answers 2


Your equation (2) is wrong because you use Newton's 1st law in the tilted direction.

In that direction there is a component of acceleration, so Newton's 2nd law should have been used, not the 1st.

The point is that the horizontal acceleration can be split into

  • a component along with (parallel to) the slope and
  • a component perpendicular to the slope.

Choosing a coordinate system in the tilted direction perpendicular to the slope is thus a direction in which there is an acceleration component (the component perpendicular to the slope). This acceleration component must be included in any Newton's laws you set up - meaning, you now must use Newton's 2nd law $\sum F=ma_\perp$ (with $a_\perp$ being the component of $a$ in that direction) rather than Newton's 1st law $\sum F=0$ since there is an acceleration component present.

Since we are more used to work with force components than with acceleration components, you will often see teachers and answer sheets pick a coordinate system that fits the acceleration - and thus not necessarily one which is tilted along with the slope. In that way they avoid acceleration components and only have the full acceleration along one axis and none along the other.

You don't have to choose such a fitting coordinate system, of course. You can choose the tilted one, if you like - but then you just must include this acceleration component along with both tilted axis directions. That means Newton's 2nd law in both directions.

  • $\begingroup$ Can you please elaborate what you are trying to say $\endgroup$ Jun 6, 2018 at 7:20
  • $\begingroup$ So should it be $N=mgcos\theta + ma$ where a is the component of acceleration in that direction. $\endgroup$ Jun 6, 2018 at 7:44
  • 1
    $\begingroup$ @TheMathemagician The acceleration is horizontally sideways, right? Therefore, a horizontal direction is parallel to and must include the acceleration, while a vertical direction is perpendicular to and does not include any acceleration. If this is not clear, let me know. Now, any "tilted" non-vertical direction contains some acceleration component. Not the full acceleration as if parallel but also not zero as if perpendicular. But a "bit" of it, so to say. A component of it. The direction perpendicular to the slope is exactly such a tilted non-vertical direction. $\endgroup$
    – Steeven
    Jun 6, 2018 at 8:39
  • 1
    $\begingroup$ @TheMathemagician Yes. Resting means no acceleration, I assume, so Newton's 1st law would be valid. (That second equation is derived via Newton's 1st law; that's the whole point.) $\endgroup$
    – Steeven
    Jun 6, 2018 at 10:23
  • 1
    $\begingroup$ @TheMathemagician I am definitely not irritated 🙂 Bring on the questions 💪 Ah, I might have been unclear. The equation itself will naturally not be the exact same on - because that force, which holds the object stationary must be included as well. What I simply meant was that this method of reaching the equation, namely via Newton's 1st law, is correct when at rest. $\endgroup$
    – Steeven
    Jun 6, 2018 at 15:14

Your eq.1 should read

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

  • $\begingroup$ Thats what is written in the question. $\endgroup$ Jun 6, 2018 at 7:03
  • $\begingroup$ Note the units of $ {mv\over r^2}$ is mass/(lemgth-time) which are not those of force. So its wrong. $\endgroup$
    – SAKhan
    Jun 6, 2018 at 7:25
  • $\begingroup$ @TheMathemagician It seems that SAKhan was correct, the question has been changed to mach SAKhan 's assertion. I don't understand the logic for the negative vote by whoever did it. $\endgroup$
    – Dlamini
    Jun 7, 2018 at 0:06
  • $\begingroup$ I too dont understand the negative vote Dlamini. $\endgroup$
    – SAKhan
    Jun 7, 2018 at 2:55

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