What is the normal force on the driver?

[Shatzkinator]

Homework Statement

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??

Homework Equations

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

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

 

[LowlyPion]

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

 

[Shatzkinator]

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

Fg = 9.8 * 70 = 686

 

[LowlyPion]

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

 

[Shatzkinator]

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

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

 

[LowlyPion]

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

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

 

[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.

 

[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