# Centripetal force due to banked curves

1. Nov 9, 2015

### Mr Davis 97

I don't really understand how the centripetal force arises from driving on a banked curve. There are three forces acting on a car in circular motion on a banked curve: normal force, gravity, and applied force. Let's assume that there is no friction, and that somehow the car is already at a constant speed. In this case, there are only two forces; gravity and the normal force. Essentially, what allows the normal force to keep the car moving on the ramp? Assuming no friction, gravity will always pull the car down the ramp if its velocity is zero. However, when it's greater than some threshold, the car maintains its motion on the ramp. How does this work? Why does changing the velocity change your ability to stay on the ramp in constant motion? Why doesn't gravity just pull down in every case?

2. Nov 9, 2015

### mathman

Car's inertia - tendency to go straight. Normal force keeps it on curve.

3. Nov 10, 2015

### Mr Davis 97

But whether the car is moving 0 m/s or 100 m/s does not change the free body diagram of the car. The normal force provides a centripetal force when the car goes a certain speed, but does not when it slows down; yet, the FBD is the same. This is what confuses me.

EDIT: Oh wait, I think I see... Is the situation analogous to an orbiting satellite? You must have a velocity (inertia) in order to not "fall" in ?

4. Nov 10, 2015

### A.T.

The motion isn't constant. The direction changes which means acceleration and requires a centripetal force.

Going down also reduces the path radius, which requires more horizontal centripetal force, which gravity cannot provide.

5. Nov 10, 2015

### Mr Davis 97

I need further elaboration on this. In the case where a car is just moving straight at a constant velocity with no friction, there are two forces which cancel out: gravity and the normal force. When the car reaches a banked curve and starts moving in uniform circular motion, the normal force cancels out gravity and provides the centripetal force. How is this possible. Why does the magnitude of the normal vector increase? What causes it to increase? Additionally, why does it cancel out gravity while proving a centripetal force? Why doesn't gravity just pull it down the curve?

6. Nov 10, 2015

### jbriggs444

Yes, normal force increases. That is because it is providing both the centripetal force and the supporting force that balances against gravity. Assuming that the curve is banked exactly right, the normal force is the vector sum of the two.

If the curve is not banked exactly right, friction makes up the difference.

Why does the magnitude of the normal vector increase? For the same reason that a table provides exactly the right support force to hold your bag of groceries in position. If it provided any less force, the groceries would move downward into the table. This increases the normal force. If it provided any more force, the groceries would rise off of the table, thus decreasing the normal force. A stable equilibrium is quickly (almost immediately) attained.

7. Nov 10, 2015

### Staff: Mentor

And for the last question: whether the car slides down or up in the curve depends on whether the forces actually balance: if the curve is banked to an angle that provides such a balance.

8. Nov 10, 2015

### Mr Davis 97

What is the physical cause for the increase in the normal force if a car is in uniform circular motion on a banked ramp as compared to a straight road? Why does this increase not occur when the car is on the banked ramp but with velocity of zero?

9. Nov 10, 2015

### Staff: Mentor

You seem to be asking what the cause is of the centripetal force: it is centripetal acceleration. F=ma.