# Frictionless banked turn, not sliding down an incline?

I was thinking about the problem in terms of defining the axes as follows in this answer, where the author states that:

$$mg \sin\theta = (\text{Centripetal Force})\times \cos\theta$$

However, I don't see how the forces acting along the incline end up cancelling in this situation. Both $mg \sin\theta$ and $F_N(\sin\theta)(\cos\theta)$ are working in the same direction. They are both working to bring the car down the incline. So shouldn't the car thus be sliding down the incline because both a component of the force of gravity down the incline and a component of the centripetal force are bringing the object down the incline? I know that this is not the standard way to think about it as defining the axes as such nontraditionally. However, I want to see how it works even despite this.

This answer begins to address my misunderstanding, however it uses the idea of centrifugal force which I come to understand as fictitious. Thank you.

• Have you drawn a free body diagram showing the actual forces acting on the car? Oct 8 '16 at 11:44 The vehicle is moving in a horizontal circle with a constant speed. That means it is constantly accelerating towards the centre of this circle. (Acceleration does not have to be a change of speed; it can be a change of direction, or both.) The acceleration is $a=v^2/r$.

Newton's 2nd Law $F=ma$ applies here; $F$ is the net force on the vehicle. There are not 2 horizontal forces acting on the vehicle (centripetal force, component of normal reaction). There is only 1 force (component of normal reaction, which is $N\sin\theta$). This force is the centripetal force. $F=ma$ tells you how this force $N\sin\theta$ is related to the centripetal acceleration $a=v^2/r$.

In response to your query about taking components parallel and perpendicular to the incline, these are $W\sin\theta$ and $N-W\cos\theta$ (see diagram below). The resultant R must be horizontal, because the condition is that the car moves in a horizontal circle. So
$\frac{N-W\cos\theta}{W\sin\theta}=\tan\theta=\frac{\sin\theta}{\cos\theta}$
$N\cos\theta-W\cos^2\theta=W\sin^2\theta$
$N\cos\theta=W(\sin^2\theta+\cos^2\theta)=W$.

The magnitude of the resultant force R is such that
$R^2=(N-W\cos\theta)^2+(W\sin\theta)^2=N^2-2WN\cos\theta+W^2=N^2-W^2$
$=N^2-N^2\cos^2\theta=N^2\sin^2\theta$
$R=N\sin\theta$
as before.

This calculation was much more difficult than in the 1st diagram. Which shows how much simpler problems can be when you choose an appropriate co-ordinate system.

• Thanks for answering. However, I don't see how we can just neglect the force of gravity that would make it slide down the incline. Oct 9 '16 at 4:07
• We don't neglect gravity. The force of gravity is balanced by the vertical component of the normal reaction force : $mg=N\cos\theta$. See diagram. The net force is $N\sin\theta$. Combined with the inertia of the vehicle, it causes motion in a circle. Oct 9 '16 at 17:08
• Thank you! Except, if you break it up into components parallel to the road and perpendicular, will the vertical forces still cancel? Oct 13 '16 at 6:20
• Why not draw a FBD and see try it for yourself? Oct 13 '16 at 22:20
• done - and it worked out when it came to the vertical forces! However, I realized something - there is a gravitational component pointing horizontally too which is equal to $Nsin\theta$ so now I think the centripetal net force must be $2Nsin\theta$! If you draw the components for $mg$, there is a component of $mgsin\theta$ which is directed centripetally and equal to in magnitude and direction to $Nsin\theta$. How can this be? Oct 14 '16 at 2:14