# Resolving forces on a banked track

This is basically the same question as Why resolve some forces into components and not others?

However, I don't understand how the preferance of coordinate systems affect the resolution of forces. In the diagram above, if we resolve the forces with respect to the horizontal axes, that is the one of the left, we get $$\begin{equation} F_N=\dfrac{mg}{\cos \theta} \end{equation}$$ whereas if we resolve the forces diagonally with respect to the incline, such as on the right, we get $$\begin{equation} F_N=mg\cos\theta \end{equation}$$ These two quantities are obviously different, and I fail to see how a choosing one coordinate system over another changes the value of $$F_N$$.

$$\underline{\textbf{My attempt at understanding}}$$

Apart from the axes argument, there is also this post Looking for an intuitive understanding of normal force that explains it (or so as I interpreted it) as a result of the direction of motion, because when an object slides down an incline, the motion of along the slope, thus $$F_N$$ cancels off forces in any other direction, whereas on a banked road the direction of motion is horizontal, namely towards the centre, so $$F_N$$ is such that the net force is in that direction.

This seems to make a bit more sense to me. But then again the choice of the direction of motion seems to be rather arbitrary, what if the direction is $$10^\circ$$ north of east, does $$F_N$$ change according to that?