# Normal force on a car rounding a banked curve

1. Jul 29, 2013

### junaid314159

This is my first thread on the forum. I'm really glad to have found this wonderful resource. I have a lot of intuition in mathematics and one of my dreams has been to have the same level of intuition in physics. With your help I hope I can achieve that.

I was reading an example in Physics (P&A) - Giancoli - 6th Edition and wanted to try to get a deeper understanding of something. Specifically, I am looking at pg. 114 - Example 5-7 if anyone has the textbook handy. I will explain the question here so that you can understand it on its own.

When a car is rounding a banked curve, the normal force on the car from the road is:
Fn = mg/cos θ, where θ is the banking angle of the road and the center of the car is going in a circle through a horizontal plane.

I wanted to get a deeper understanding as to why the normal force was not what it usually is on an inclined plane, namely Fn = mg cos θ.

I understand that the net force on a car resting on an inclined plane is downward and angled therefore the normal force has no component in the direction of the net force, while in the case of a car going around a banked curve, the net force is horizontal towards the center of the circle in which case the normal force does have a component in the direction of the net force. I'm just trying to better understand what the unique differences are between this situation and the inclined plane which lead the normal force being mg/cos θ.

When you look at the diagram on pg. 114, it is a single cross-section of an inclined plane so one would imagine that if the speed of the car was slow enough, near enough to zero, all you would have is a car on an inclined plane and then the normal force should be close to mg cos θ as it would be exactly that if the speed of the car were zero. Does the normal force then abruptly go from mg/cos θ to mg cos θ when the car stops?

Sincerely,
Junaid

2. Jul 29, 2013

### technician

There are essentially only 2 forces acting on the car
1) the weight mg acting vertically down
2) the normal reaction, Fn from the track
( this is assuming that friction can be ignored )
The normal reaction has 2 components....the vertical component FnCosθ must equal the weight mg
So FnCosθ = mg
The horizontal component is directed towards the centre of the circle and is equal to the centripetal force.

3. Aug 2, 2013

Thanks