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?)

  • $\begingroup$ 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. $\endgroup$
    – Farcher
    Apr 19 '18 at 22:05

A "normal" force is normal or orthogonal to the surface creating the force. In other words, the ground or track can only push straight out from the surface. (Forces along the surface would be due to friction, but you are not considering friction here).

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.

If increased speed increases the centripetal force, then it increases the upward force as well.

  • $\begingroup$ Thanks. So to be super-clear, the comment (I saw elsewhere) that the vertical component of the Normal Force is always equal to mg is wrong once the velocity of the car goes above the "ideal" tuned velocity? $\endgroup$ Apr 20 '18 at 1:24
  • $\begingroup$ That depends on the question. Most are set up that the car is in equilibrium (and ask you to find the speed where that holds). $\endgroup$
    – BowlOfRed
    Apr 20 '18 at 2:08
  • $\begingroup$ To be double super-clear-squared, my understanding is that in cases with friction, if speed is below "ideal" speed, then upward component of Normal force is greater than mg (and equilibrium is achieved through additional upward force of static friction, and vice versa in cases where speed is above "ideal" (?) $\endgroup$ Apr 20 '18 at 13:04
  • $\begingroup$ [Sorry, in the first case, I meant upward component of Normal force would be less than mg (when speed is lower than ideal] $\endgroup$ Apr 20 '18 at 13:36
  • $\begingroup$ Sure. Just imagine the limiting case where $v=0$. $\endgroup$
    – BowlOfRed
    Apr 20 '18 at 18:52

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