# Conceptual Background for Banked Turn Problems?

I've read through the past posts on this topic and still am not finding an answer to this specific aspect. I get that in the "no friction" case, gravity pulls the car down on the track and the resulting Normal force provides both the centripetal force and the "y direction" force that counteracts gravity (which keeps car from sliding down track). What I don't get is what happens as car speeds up. Does the y-component of the normal force remain "mg" and just the x-component of the normal force increases as a reaction force to the car speeding up (and so, colloquially, wanting more and more to go in a straight line and so banked track "hits" it harder as it turns it). Or does the normal force in the direction perpendicular to the track increase as a reaction force to increased speed (and so the normal force in both the x and the y-direction bothincrease?)

• Once the car is going too fast for the ideal banking case it will try and climb up the track ie move a less curved trajectory and to prevent it doing so there must be a frictional force pointing down the banked track. A component of that frictional force will then provide some of the force needed to accelerate the car around the corner. – Farcher Apr 19 '18 at 22:05

Since the direction of the force is constant, that means you cannot change the $x$ component of the force and leave the $y$ component unchanged when both are non-zero. Increasing one requires increasing the other.
• Sure. Just imagine the limiting case where $v=0$. – BowlOfRed Apr 20 '18 at 18:52