# How to predict direction of force of friction on car in banked roads?

Consider a car moving on a banked road which is at an angle say x with the horizontal. Now if we consider friction t9 be zero then the ideal velocity with which car should move is $√(rgtanx)$ .

But what about if it moves with velocity greater or less than this velocity ? How do we predict the frictional force direction then ?

Some say that according to normal reaction being more or less than required as centripetal car has different tendancies to move but I dont understand why also if we consider friction of the road then how can we find minimum velocity with which car should move in a circle of radius R ?

• – Farcher Apr 13 '17 at 19:37
• @Farcher Can you guve an intuitive explanation for the reason why the impending motion vhamges when velocity of object is more or less than ideal one and hence changes direction of friction ? – Matt Apr 13 '17 at 19:51

You are setting a constraint on the car which is that it has to go around a banked corner at a certain speed $v$ and given radius $r$ without a frictional force being present.

Applying that constraint you get the condition $v = \sqrt {rg\tan x}$ where $x$ the angle of banking.

The force to provide the necessary centripetal acceleration $\frac{v^2}{r}$ was supplied by the horizontal component of the normal reaction of the slope on the car.

If now the speed of the car is increased then that horizontal component of the normal reaction is not sufficient to provide the necessary centripetal acceleration to keep the radius of the car's path the same.

With no frictional force the only thing the car can do is start moving along a path with a larger radius.
Put another way the force provided by the horizontal component of the normal reaction cannot change the direction of motion of the car sufficiently so the car starts moving up the slope.
To prevent this happen and keep the car moving at a faster speed along a path with the same radius as before there must be an extra inward force to increase the centripetal acceleration.
That inward force can be provided by the horizontal component of a frictional force acting down the slope.

With the car slowing down the horizontal component of the normal reaction changes the direction of motion of the car too much and so the car would start moving down the slope, so there must be a frictional force up the slope.

The easy way to remember which direction the frictional force should action is to think of the car not moving on a (frictionless) slope and the direction of a frictional force which will stop it sliding down the slope.

• But Ncos(theta)=mg so if we introduce friction then Normal is gonna change right......wouldn't that provide too much acceleration like when it was doing when velocity is less than ideal banked case – Abhinav Jul 21 '18 at 11:51