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If a car is moving on a banked frictionless road one of the component of normal reaction force acts as the centripetal force required for the turn. But normal reaction force is a reaction force. Technically seeing, the the component of normal reaction force should be the result of the centripetal force required for the turn. And turning needs some friction. So how can the road be frictionless? Please explain.

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Roads are banked so that we get a component of normal reaction to assist in the turn if friction is not sufficient. For example on very sharp curves on road, in absence of banking, the tires can skid and the results can be deadly. Now if friction is absent, the only force which can help us in turning is the normal reaction. But normal reaction is perpendicular to the road and we want a component of it in the radial direction. So we bank the roads to provide us with the necessary centripetal acceleration.

And normal reaction is a reaction by surface of the road to the weight of the car as well as the centripetal acceleration needed to keep the car in circular motion.

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  • $\begingroup$ But as you can see the only the Ncostheta component is sufficing the weight...what about the Nsintheta....what is causing it? $\endgroup$ – Feynman 16 Oct 23 '18 at 3:19
  • $\begingroup$ @Feynman16 The reactive force being inclined is greater in magnitude than the weight because only one of its components Ncostheta is balancing the weight. The other component Nsintheta provides the centripetal acceleration. $\endgroup$ – Mechanic7 Oct 23 '18 at 4:54
  • $\begingroup$ In the diagram, N is the normal force of the road on the car. Ncostheta and Nsintheta are the components of that force. (See how they are dashed, whereas the other two vectors are solid?) Ncostheta is balanced by the weight mg. The Nsintheta component is not balanced, so the net force equals that amount. This net force is a centripetal force, which causes the car to travel in a circle. $\endgroup$ – DrSheldon Oct 23 '18 at 6:25
  • $\begingroup$ But if the car was still won't it be sliding down...as mgcosthea component will cancel out with N and mgsintheta will act. What exactly I am trying to say is the normal reaction force comes in a pair....what is causing the Nsintheta component of normal reaction force when the car is turning....isn't it the centripetal force...normal reaction force can't just appears to be greater than mg until there is some other force causing it? $\endgroup$ – Feynman 16 Oct 23 '18 at 15:08
  • $\begingroup$ Normal reaction is inclined. The cos component has to balance it. So normal reaction has to necessarily be of greater magnitude as cos theta ranges from -1 to 1 so that the car has zero acceleration in the downward direction. But the sin component is unbalanced and causes centripetal acceleration which enables the body to turn. $\endgroup$ – Mechanic7 Oct 23 '18 at 15:26

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