Using Normal Force on inclined plane vs banked curve

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SUMMARY

The discussion clarifies the differences between the normal force on an inclined plane and a banked curve in physics. For an inclined plane, the equations governing the forces are "\SigmaFy = N - mg cos ø = 0" and "\SigmaFx = mg sin ø = ma", leading to "N = mg cos ø". In contrast, for a banked curve, the equations are "\SigmaFy = N cos ø - mg = 0" and "\SigmaFx = N sin ø = mv²/r", resulting in "N cos ø = mg". The key distinction lies in the presence of circular motion on a banked curve, which introduces additional forces not accounted for in the inclined plane scenario.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with free body diagrams
  • Knowledge of circular motion dynamics
  • Basic trigonometry, particularly sine and cosine functions
NEXT STEPS
  • Study the derivation of centripetal force in circular motion
  • Learn about the effects of friction on banked curves
  • Explore the concept of centrifugal force in non-inertial frames
  • Investigate real-world applications of banked curves in road design
USEFUL FOR

Physics students, educators, and anyone interested in understanding the mechanics of inclined planes and banked curves in dynamics.

Rick4
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On an inclined plane, "[tex]\Sigma[/tex]Fy = N - mg cos ø = ma = 0".
And "[tex]\Sigma[/tex]Fx = mg sin ø = ma" (frictionless).
So, in the y direction, "N = mg cos ø".


But on a banked curve, "[tex]\Sigma[/tex]Fy = N cos ø - mg = ma= 0".
And "[tex]\Sigma[/tex]Fx = 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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