Using Normal Force on inclined plane vs banked curve

[Rick4]

On an inclined plane, "$$\Sigma$$F_y = N - mg cos ø = ma = 0".
And "$$\Sigma$$F_x = mg sin ø = ma" (frictionless).
So, in the y direction, "N = mg cos ø".


But on a banked curve, "$$\Sigma$$F_y = N cos ø - mg = ma= 0".
And "$$\Sigma$$F_x = N sin ø = ma = mv²/r" (frictionless).
So, in the y direction, "N cos ø = mg".
I know how to get "tan ø = v²/rg".
The two free body diagrams look identical to me so why doesn't "N = mg cos ø" (from inclined plane) work to get "tan ø = v²/rg"?

 

[Kurdt]

On a banked curve, circular motion is normally involved and so there is an extra centrifugal force. I don't see the problem except you've neglected to take that into account on the free body diagram.