# please explain the mechanism and direction of force of friction in circular roads where vehicles move without banking

Aman Bansal
592 Points
11 years ago

Dear Argha,

When a vehicle goes round a curved road, it requires some centripetal force. While rounding the curve, the wheels of the vehicle have a tendency to leave the curved path and regain the straight line path. Force of friction between wheels and the roads opposes this tendency of the wheels. This force of friction therefore, acts towards the centre of circular track and provides the necessary centripetal force.

Fig. (1) Vehicle moving on level road

In fig (1), it is shown that a vehicle of weight ‘mg’ (acts vertically downwards) is moving on a level curved road. R1 and R2 are the forces of normal reaction of the road on the wheels. These are vertically upward since road is leveled. Hence,

R1 + R2 = mg

Let F1 & F2 are forces of frictions between tyre and road directed towards centre of curved road.

∴ F1 = μ R1

And

F2 =  μ R2

where μ is coefficient of friction between tyres and road.

If ‘v’ is the velocity of the vehicle while rounding the curve, the centripetal force required is mv²/r. As this force is provided by the force of friction therefore

Hence the maximum velocity with which a vehicle can go round a level curve; without skidding is

### Banking of Roads

In the above discussion, we see that the maximum  permissible velocity with which a vehicle can go round a level curved road depends on μ, the coefficient of friction between tyres and road. The value of  μ decreases when road is wet or extra smooth or tyres of the vehicle are worn out. Thus force of friction is not a reliable source for providing the required centripetal force to the vehicle. Especially in hilly areas where the vehicle has to move constantly along the curved track, the maximum speed at which it can run will be very low. If any attempt is made to run it at a greater speed, the vehicle is likely to skid and go out of track. In order that the vehicle can go round the curved track at a reasonable speed without skidding, the sufficient centripetal force is managed for it by raising the outer edge of the track a little above the inner edge. It is called banking of the circular track or Banking of Roads.

Consider a vehicle of weight  ‘Mg’ moving round a curved path of radius ‘r’ with speed ‘V’ on a road banked through angle θ. If OA is banked road and OX is horizontal line, then ∠AOX = θ is called angle of banking of road. Refer Fig (2)

Fig.(2) Vehicle moving on Banked Road

#### Following forces are involved:

1. The weight ‘Mg’ acting vertically downwards
2. The reaction ‘R’ of the ground to the vehicle acting along normal to the banked road OA in upward direction
3. The vertical component R.Cos θ of R will balance the weight of the vehicle.
4. The horizontal component R.Sin θ of R will provide necessary centripetal force to the vehicle.

Thus,

R.cos θ  = Mg            ….(1)

And

On dividing equation (1) & equation (2),we get

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Thanks

Aman Bansal