• Support PF! Buy your school textbooks, materials and every day products Here!

Incline Plane versus Banked Curve

  • Thread starter TechCS
  • Start date
1
0
Okay, so I've been working on incline plane and then banked curve problems. From the incline plane problems and the definitions in my book I believed that normal force was just the y component of the weight, ie the weight times the cos of the angle of incline and no x component, ie parallel to the surface. This assumptions worked for the inclined plane problems, but when I got to the banked curve problems, a car going in a circular banked path with now friction, the centripetal force was now the normal force times sin of the angle of incline, the "x-component" of the normal force. Can any one explain?
 

Answers and Replies

Related Threads for: Incline Plane versus Banked Curve

Replies
1
Views
3K
Replies
2
Views
1K
  • Last Post
Replies
3
Views
621
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
1
Views
590
  • Last Post
Replies
5
Views
7K
Top