I get the answer, I just don't really understand why

[A(s)]

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?

Answer:

Spoiler

ok, well the answer is 2x gravity.

and you can reason it out, saying, "well if the force from the first circle (F1) is enough to cancel gravity, then the normal force from the second circle must be equal to gravity (Fn = Fg), and the Fn +Fg = Fg + Fg = Fnet = 2Fg."

What I don't understand though, is what do you call the force when the car is traveling on the outside of the first circle that counteracts gravity?(F1)


Thanks in Advance,

Justin

 

[Andrew Mason]

It is the centripetal force. - the force needed to provide the change in velocity at the top of the curve.

Draw a free-body diagram of the forces on the car at both points. The sum of all forces has to equal the mass x acceleration. The only acceleration is the centripetal acceleration.

In the first case, the only force is the force of gravity down. There is no normal force. So F1 = mg

In the second case, the forces are gravity and the normal force. The Normal force acts in the opposite direction as mg. The centripetal force/acceleration is in the opposite direction. So F2 = -F1 = mg-N.

Since F1 = N - mg AND F1 = mg ...

AM

 

[A(s)]

The only thing that bothers me is that in the first hill, Fn = Mg + X
there must be some x that counters gravity to make the Fn = 0
What is x?

Fn = Mg would be a car traveling on a flat hill.

 

[A(s)]

lol flat hill... i meant flat ground

 

[cepheid]

The only thing that bothers me is that in the first hill, Fn = Mg + X
there must be some x that counters gravity to make the Fn = 0
What is x?

Fn = Mg would be a car traveling on a flat hill.

Fn = 0 because at the crest of the hill, both the driver and car are accelerating downwards at g. It's like being in a free fall situation -- like in an elevator if the cable suddenly snapped. I think you can see that if both the driver and the car are being pulled down by gravity, then obviously the seat is not going to be pressing up against the driver, counteracting his weight. That's why there is zero normal force on him, and he feels weightless.

 

[A(s)]

true, that makes sense. Thank you

 

[Andrew Mason]

The only thing that bothers me is that in the first hill, Fn = Mg + X
there must be some x that counters gravity to make the Fn = 0
What is x?

Fn = Mg would be a car traveling on a flat hill.

You avoid this conceptual problem if you draw a free body diagram. The sum of the forces acting on the car has to equal the mass x acceleration. The reason the normal force is 0 because gravity is exactly equal to the mass x acceleration. There can be no other force acting.

AM