Understanding the Horizontal Component of Normal Force in Banked Curves

marlasca23
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Why during a banked curve is the horizontal component of the normal force considered as the net force and the centripetal force? The horizontal component of Normal force, N·sinθ, is not even ponting towards the centre of the curvature.
 
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marlasca23 said:
The horizontal component of Normal force, N·sinθ, is not even ponting towards the centre of the curvature.
It is pointing toward the center of curvature of the trajectory of the vehicle.

Edit: A geodesic on the surface would curve otherwise, but gravity affects the vehicle's path.
 
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marlasca23 said:
Why during a banked curve is the horizontal component of the normal force considered as the net force and the centripetal force?
Because the vertical component is canceled by gravity.
 
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jbriggs444 said:
It is pointing toward the center of curvature of the trajectory of the vehicle.

Sorry, I don't understand that. Could you explain it more in depth, please?
 
The vehicle travels in a circular path around the banked curve. The circle is in a horizontal plane. The center of the circle is in that plane. The net force on the vehicle is toward that center.

The "center of curvature" of a circular arc is the center of the circle containing that arc.
 
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