Using Normal Force on inclined plane vs banked curve

  • Thread starter Rick4
  • Start date
  • #1
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"?
 

Answers and Replies

  • #2
Kurdt
Staff Emeritus
Science Advisor
Gold Member
4,812
6
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.
 

Related Threads on Using Normal Force on inclined plane vs banked curve

Replies
2
Views
1K
  • Last Post
Replies
9
Views
20K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
12
Views
9K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
2
Views
25K
Replies
12
Views
6K
Replies
3
Views
3K
Replies
3
Views
15K
  • Last Post
Replies
6
Views
4K
Top