Normal force in case of inclined plane and banked turn

[DrBanana]

This isn't really a question per se, but it recently just 'clicked' for me and I would like to share what made that do so.

There are dozens of threads answering this question, but in my opinion most (but not all; there are some good threads on this forum, I just want to show how I came to understand it) of them don't really resolve a significant misconception (or rather, I just didn't understand them fully).

First, the context: https://www.physicsforums.com/threads/normal-force-inclined-plane-vs-banked-turn.944330/

I think the usual explanation goes, in the inclined plane, we had one coordinate system, and in the banked turn, we had another. I don't think this explains the full picture, because the implicit question is, why there should be a discrepancy in the value of the normal force whether the body is in one type of motion as opposed to another. This 'change of coordinate system' explanation doesn't really address that.

I used to think the motion of a body is influenced by the forces acting on it, but the forces acting on it, act the same whether the body is in motion or not (in hindsight, I don't know why I thought this). But forces on a body do depend on its motion, at least in an implicit way. The 'aha!' moment was reading that paragraph in post #7 in the above linked thread, which I'll quote here:

"The normal force is the normal component of the force exerted by the surface. It's caused by the deformation of the surface which in turn is caused by both gravity and the motion. You should try working the problem where the inclined plane is formed by a wedge that's free to move in a horizontal direction."

So the motion of the body depends on the forces acting on it, the forces acted on other objects by the body depend on the body's motion, and because of Newton's third law, we can say finally that the forces acted on a body can also depend on that body's motion.

The crux of it is this: in the banked turn case we do the mathematics so that there is no acceleration component in the vertical direction, and this just happens to be $Rcos\theta=mg$. There is no acceleration in the vertical direction because the body is moving in such a way that there is no acceleration in the vertical direction. In other words, it was the constraints of the problem that allowed us to write $Rcos\theta=mg$. It could very well be that the car was going too fast or too slow, and then the mathematics would be different. In the inclined plane case we do the mathematics so that there is an acceleration component in the vertical direction. This is because the motion of the body is such that there is an acceleration in the vertical direction. Sorry if the last lines got tautological. Another way of summarizing is that, normal force is not inherent, and it can depend on a lot of things (again I know this is supposed to be common knowledge, I don't know why I just didn't realise this before).

Please tell me if I missed anything.

 

[TSny]

Your comments look good to me. Normal forces and tension forces are associated with deformations (strains) of materials. They “adjust” to the situation.

 

[kuruman]

Another way of summarizing is that, normal force is not inherent, and it can depend on a lot of things (again I know this is supposed to be common knowledge, I don't know why I just didn't realise this before).

You can extend this to other contact forces beyond the normal force such as tension and static (not kinetic) friction. In the sense used by @TSny in post #2, "contact forces adjust themselves to produce the observed acceleration." In other words, observe the acceleration and from it deduce the contact force using Newton's second law.

This explains your observation that the normal force is different when the mass is moving in a horizontal circle around a banked curve from when it just sliding down an incline. Different accelerations, different contact forces. Same idea when a mass, hanging from a string, is accelerated vertically. If the acceleration is "up" the tension is greater than the weight; if the acceleration is "down" the tension is less than the weight.

It could very well be that the car was going too fast or too slow, and then the mathematics would be different.

The mathematics would not be different. The components of the acceleration would be different.

 

[Lnewqban]

For same car and speed, the normal force when taking a banked curve (conical trajectory) will always have a greater value than when moving along an inclined plane on a horizontal trajectory.
Gravity and centripetal accelerations are vectorially added for the former case.

Sometimes the dynamic forces get balanced, but not always.
When not, a movement with vertical component is observed (car simultaneously slides up or down the banked curved).

Please, see how this work for flying:
https://en.m.wikipedia.org/wiki/Coordinated_flight