# Homework Help: Normal force at the top of a hill

1. Sep 20, 2009

### Tin Man

I'm bypassing the normal questions, only because I have already solved the question, but want help figuring out why the answer is what it is (without incorporating numbers into it at all).

The idea is that a car is at the top of a hill going a constant speed. We are to treat the hill as having a radius, allowing us to use the centripetal force to find out the normal force.

My diagram I drew out shows a normal force acting upward on the car, and both the gravity of the car and the centripetal force of the car acting downward toward the radius of the hill (since the car is at the very top).

I looked around and was able to figure out the following equation based off a similar question asked on Yahoo...

Fn: Normal force, Fg: Gravity, Fr: centripetal force

Fn = Fg - Fr

What I can't figure out is WHY this is the case. I know it's the correct equation because it led me to the correct answer, but I was under the impression that if the normal force was a reaction of the road to the force of the car, then the normal force would be a sum of the gravity and the centripetal force, not a difference. Why are we not adding the two downward forces, and saying that the normal force is the negative of the two downward forces?

2. Sep 21, 2009

### carbon55

That is a great question. The answer involves centrifugal force. Yes, it's not a true force, but you could use it to understand this problem. The inertia of the car is the centrifugal force. When it travels in a circle, such as implemented at the top of the hill, the car resists the acceleration, as all objects with mass do, caused by changing direction. Think about swinging a weight at the end of a string in circles. If the speed is sufficient, the string will be drawn taught. This outward pull you would feel comes from the weight's resistance to centripetal acceleration. Objects with mass don't want to travel in circles, but straight lines. This is the same concept with the car. It wants to fly off tangent to the hill's circle, but gravity and friction are keeping it on the ground. I hope this explanation helps.

3. Sep 21, 2009

### Tin Man

What's confusing me is that both forces are supposed to be pointed downward toward the middle of the hill. So we're treating centripetal force as going back out to the car? I know that intuitively the car will want to (if it's going fast enough) lift off the hill, therefore reducing the weight. I'm just having a hard time justifying that in my head with the calculations, since I know that the centripetal force is ALWAYS going toward the middle of the circle, that gravity is ALWAYS pointing to the ground, and that the normal force is ALWAYS perpendicular to the ground.

4. Sep 21, 2009

### Staff: Mentor

Only two forces act on the car: Gravity and the normal force. "Centripetal force" is just the label we give to the net force in the radial direction; it should not appear on a free body diagram.

So ΣF = Fc = Fg - Fn

The gravitational force is fixed, of course, but the normal force is not. The normal force will be whatever it needs to be to prevent the car from sinking into the ground. You can think of gravity as providing the centripetal force and more--that extra force needs to be balanced by the normal reaction force of the ground. The "extra force" equals Fg - Fc. The faster the car is going, the less "extra force" there is.