- #1
Rick4
- 2
- 0
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"?
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"?