Banked Turns Part II: What Force Causes a Car to Creep up a Wall

[jon4444]

I wanted to dig deeper on previous question regarding Normal forces and banked turns vs. inclined planes. Intuitively, when a car going around a banked turn is going faster than the "tuned" speed (i.e., the speed at which no friction is required to keep it on the track), it will need friction to prevent it from moving up the wall. I get that this motion ultimately comes from the fact that the car wants to move in a straight line, but how do you conceptualize this motion in terms of forces? And is the Normal force the same at all speeds?

 

[Dale]

Take the frictionless case for simplicity. Without friction there are two forces, the normal force and the weight. The weight points down. The normal force points in and up. If the speed is just right then the normal force cancels gravity and accelerates you around the track perfectly. If the speed is too fast that means the normal force is too vertical so you accelerate up and do not turn enough, so you go up and off the outside. If the speed is too low then the normal force is too horizontal so you accelerate down and turn too much, so you go down and off the inside.

 

[sophiecentaur]

how do you conceptualize this motion in terms of forces?

The force is called a Centripetal force. Take the car and put it in space on the end of a rope. The tension in the rope needs to be the same value force that keeps the car going in a circle of same radius at the same speed on the track. Now, a separate calculation will tell you the angle of track that will produce equilibrium with that force, applied horizontally against the track. (If you allow some lateral friction, the equation becomes a bit more complicated.) I think putting it that way 'explains' the situation usefully but it doesn't give you the answer yet. Another step is needed to tell you the speed needed on a track with a given slope. The Circular Motion equation and the Slope equation (same force in each equation so you can equate them) can be combined and re-arranged to give you the necessary speed.

If the speed is just right then the normal force cancels gravity and accelerates you around the track perfectly.

That is perfectly true but I think you also need to point out that it is the horizontal component of this force that is the required centripetal force. That may not be obvious to someone who is not familiar with 'resolving forces'.

Our friendly Hyperphysics Website gives the explanation of the relationships between all the factors. Put in your own figures (including frictionless option) and use the ready made calculator.

 

[jon4444]

If the speed is too fast that means the normal force is too vertical so you accelerate up and do not turn enough, so you go up and off the outside. If the speed is too low then the normal force is too horizontal so you accelerate down and turn too much, so you go down and off the inside.

I think the conceptual aspect of the situation I want to be clear on is that the Normal force increases as the speed increases, correct? And explaining why this is, is what I would like to be clear on. What I might say is that the Normal force in the situation where the speed is "too fast" is made up of a reaction force to gravity pulling down and a separate reaction force to the car banging against the wall (since the car wants to go straight, but the wall wants to turn it.)
Can you clarify that explanation any or provide better language?
Thx.

 

[Dale]

Normal force in the situation where the speed is "too fast" is made up of a reaction force to gravity pulling down

The normal force is never a reaction force to gravity. Action-reaction pairs are always of the same type. So the reaction force to the car’s weight is the upwards gravitational pull of the car on the earth.

a separate reaction force to the car banging against the wall (since the car wants to go straight, but the wall wants to turn it.)

This is closer. The reaction is to the contact force of the car against the road. However, the whole action reaction framework here is not terribly helpful. Yes, if you know the force of the road on the tires then that is equal and opposite to the force of the tires on the road. But that doesn’t tell you much since the force of the road on the tires is no more nor less confusing than the force of the tires on the road. I would probably avoid the third law here.

The key is to recognize that the normal force is whatever value it needs to be to prevent the car from sinking into the road. That value depends on the speed. At high speed the acceleration required to not sink is higher so the force is correspondingly higher.

 

[A.T.]

Normal force increases as the speed increases, correct? And explaining why this is,

Because the acceleration required to avoid sinking into the road increases with speed.