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?

 

[Nate810]

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.

 

[m2003]

I can provide an explanation for the difference in normal force calculations between an incline plane and a banked curve. In an incline plane, the normal force is equal to the component of the weight that is perpendicular to the surface. This is because the incline plane is at an angle, so the weight is not directly acting downwards but is instead divided into two components: one perpendicular to the surface and one parallel to the surface. Therefore, the normal force is only equal to the perpendicular component of the weight.

On the other hand, in a banked curve, the normal force is the force that keeps the object moving in a circular path. This means that the normal force is not just acting perpendicular to the surface, but it also has a component that is pointing towards the center of the circle. This component is necessary to provide the centripetal force that keeps the object in its circular motion. This is why the normal force is now equal to the component of the weight that is parallel to the surface, which is the x-component in this case.

In summary, the difference in normal force calculations between an incline plane and a banked curve is due to the different forces at play. In an incline plane, the normal force is only needed to counteract the perpendicular component of the weight, while in a banked curve, the normal force is also responsible for providing the centripetal force required for circular motion. I hope this explanation helps clarify the difference between the two scenarios.