Explaining the Mechanics of Banking a Curve on an Inclined Path

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When banking a curve, the normal force can provide the necessary centripetal force for turning, eliminating the need for friction at certain speeds. The weight of the car acts vertically and does not have an x component, which is why it is not projected into x and y components like in inclined plane problems. In an unbanked turn, friction is the primary source of centripetal force since the normal force does not contribute horizontally. Conversely, a banked turn allows the normal force to contribute to the horizontal force required for the turn. Understanding these dynamics is crucial for analyzing vehicle motion on different road types.
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can someone please explain why when banking a curve at an angle so that no friction at all is needed to maintain the car's turning radius , we project the normal into an x and y component and we don't project the weight into an x and y component as we do for problems involving motion over an inclined path ? thanks for any help it will be really appreciated .
 
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Weight always acts vertically, so it has no x component.

To make the turn, you need some centripetal force (in the x direction). That could be supplied by the normal force or friction or both. If the road is unbanked, the normal force has no horizontal component, so friction is all you have. But if the road is banked, the normal force can supply the horizontal force; for some speeds, you won't need any friction at all to make the turn.
 
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