How is the normal force on a banked track larger than gravity?

Question

The formula of the vertical force of an object that is not accelerating up or down on a banked curve is: Ncos(θ) = mg

There is also a centripetal force component which shows that the normal force is larger than the weight of an object. How is this possible. Aren't forces only equal and opposite? Where is the source of the rest of the force?

Answer 1

We take the case where a vehicle is going around the banked track (bank angle = theta) at just that speed at which it tends neither to fly off the track outwards nor fall inwards.

Since it is turning in a circle, its velocity vector is changing direction, which requires an acceleration towards the center point of that circle. The acceleration is furnished by the normal force developed at the tire contact points.

Since the track is angled, so is the normal force, which points upwards (due to the reaction force furnished by gravity) and inwards (due to the radial component of acceleration, which is required to turn the vehicle in a circular path).

The resultant acceleration vector is greater than that furnished by gravity alone because it is the vector sum of gravity and the radially-inward directed acceleration which is turning the vehicle's velocity vector.

Answer 2

The special thing here is that road is banked.Imagine a vehicle moving on a banked road. Though we actually calculate the velocity perpendicular to the centripetal force the direction of velocity acting isn't perpendicular to the centripetal force ,only the direction of motion is perpendicular to the centripetal force. The reason for this is that the banking of the road makes the vehicle angled and therefore the velocity direction directed by the vehicle acts perpendicular to the normal reaction force (not perpendicular to the centripetal force) exerted on it.This makes an additional reaction force on road.

To understand this some imagination as well as geometrical skills will be needed