# Car Driving over a hill

1. Feb 22, 2009

### Shatzkinator

1. The problem statement, all variables and given/known data
A car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill the normal force on the driver from the car seat is 0. The driver's mass is 70.0 kg. What is the magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley??

2. Relevant equations
Fg - Fn = 0 ?
F = mv^2/r

3. The attempt at a solution
Stumped. I looked at this question and was like... wtf?

Only thing I can think of is Fg - Fn cannot equal zero because there is always Fg and if Fn is 0, then that equation doesn't make sense. But I don't know any better than this equation. I'm so confused!! =S

2. Feb 22, 2009

### LowlyPion

At the top of the hill if his normal force is 0, then what must his centripetal acceleration be?

3. Feb 22, 2009

### Shatzkinator

Fg = 9.8 * 70 = 686

4. Feb 22, 2009

### LowlyPion

How is the direction of the centripetal acceleration different at the top as opposed to the valley?

5. Feb 22, 2009

### Shatzkinator

At the top, it is directed downwards toward the centre, at the bottom is directed up toward the centre...

6. Feb 22, 2009

### LowlyPion

... and you still can't answer the original question?

7. Feb 22, 2009

### Shatzkinator

Fn = Fg .. which would be al;so 686 at the bottom of the hill.

The answer should be 1.37 x 10^3 N

=P

No I still can't answer the original question.

Edit: and yes I see the answer is 2* 686... I have no idea why.

8. Feb 22, 2009

### LowlyPion

In the first case Fn = 0 = mg - mv2/r

That means that mv2/r = mg.

In the valley

Fn = mg + mv2/r

But you just figured out that mg = mv2/r so

Fn = 2*mg