Frictionless Bank: Normal Force vs. Weight

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In a frictionless banked curve, the horizontal component of the normal force is responsible for holding the car in place, not the weight. The weight acts vertically and does not contribute a horizontal component. The normal force, which is perpendicular to the banked surface, has both vertical and horizontal components due to the angle of the bank. This scenario involves circular motion and centripetal acceleration, highlighting the importance of the normal force in maintaining the car's trajectory. Understanding these forces is crucial for analyzing motion on banked curves.
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Which force is responsible for holding a car in a frictionless banked curve?

The answer is the horizontal component of the normal force. But could anyone explain why that is?

I thought it would be the horizontal component of the weight since normal force is vertical?
 
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Rawr said:
I thought it would be the horizontal component of the weight since normal force is vertical?
The weight acts vertically--it has no horizontal component. The normal force is perpendicular to the road surface; since the road is banked (at an angle) that normal force is not vertical.
 
This is a rotational velocity problem. There are two kinda 'normal' forces here. The real vertical normal force to resist gravity, and the horizontal force in the 'normal' directionto accelerate the object along a curve.
 
Last edited:
This is a problem involving circular motion and thus centripetal acceleration. There is only one normal force, which is perpendicular to the road surface. That normal force has vertical and horizontal components.
 

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