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

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?

• A vehicle on a banked track is usually accelerating. – Whit3rd Jan 15 '19 at 2:19

## 1 Answer

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.