# Homework Help: Using Normal Force on inclined plane vs banked curve

1. Oct 6, 2008

### Rick4

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

But on a banked curve, "$$\Sigma$$Fy = N cos ø - mg = ma= 0".
And "$$\Sigma$$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"?

2. Oct 7, 2008

### Kurdt

Staff Emeritus
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.