Incline Plane versus Banked Curve

[TechCS]

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 anyone explain?

 

[member 553083]

The normal force is the force that is perpendicular to the surface. In an inclined plane, the normal force is equal to the weight of the object multiplied by the cosine of the angle of incline. However, in a banked curve, the normal force is not only perpendicular to the surface but also acts as the centripetal force, meaning it has both a vertical component (equal to the weight of the object multiplied by the cosine of the angle of incline) and a horizontal component (equal to the weight of the object multiplied by the sine of the angle of incline). The combination of these two components provides the centripetal force needed to keep the object on a circular path.