What is the Net Force on a Banked Curve?

[Abu]

I understand why when deriving the formulas for cars on banked curves, the net force in the Y direction is zero. However, when I google how to derive them, people say that there is a net force greater than zero in the X direction. This is not what my professor says in his explanations however: he says that the net force in the X direction is also zero, and that this is because there is a centrifugal force equal to mv^2/r pointed to the left that cancels Fnsintheta..

I will attach the following diagram to better explain my question. Note the mv^2/r that is pointing to the left, where typically other diagrams do not include that.

All in all, why is the net force in the X direction for my professors explanation equal to zero, where other explanations say it is not?

 

[Orodruin]

This depends on which reference frame you consider. If you consider the ground inertial frame, there is a net force. If you consider the car's accelerated reference frame there is an additional inertial force, the centrifugal force.

 

[Abu]

This depends on which reference frame you consider. If you consider the ground inertial frame, there is a net force. If you consider the car's accelerated reference frame there is an additional inertial force, the centrifugal force.

Oh okay, so let's say there is a string tied to rock spinning in a vertical motion. If I am looking at the scenario as an outside observer, the forces include the tension that acts as the centripetal force and the force of gravity (depending on where the rock is in its rotating motion), there is no centrifugal force that needs to be included. But if I am looking it at as if I am the rock, then there is a centrifugal force that will be equal to the tension? Am I correct in this regard?

 

[Orodruin]

In essence, yes.