Using Normal Force on inclined plane vs banked curve

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On an inclined plane, the normal force is calculated using the equation N = mg cos ø, with forces balanced in both the x and y directions. In contrast, a banked curve involves circular motion, leading to different force dynamics, where the equations include N cos ø = mg and N sin ø = mv²/r. The key distinction lies in the presence of centripetal force on the banked curve, which is not accounted for in the inclined plane scenario. This results in the relationship tan ø = v²/rg, highlighting the need to consider additional forces in curved motion. Understanding these differences is crucial for accurately applying the concepts of normal force in both scenarios.
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On an inclined plane, "\SigmaFy = N - mg cos ø = ma = 0".
And "\SigmaFx = mg sin ø = ma" (frictionless).
So, in the y direction, "N = mg cos ø".


But on a banked curve, "\SigmaFy = N cos ø - mg = ma= 0".
And "\SigmaFx = 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"?
 
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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.
 
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