It requires concept from Newton's laws of motion and circular motion. While solving for net force on a body on an inclined plane , we always resolve mg (weight) parallel and perpendicular to the plane but in the banking of road concept, we resolve normal reaction. Why don't we simply resolve weight and then equate it with normal reaction?

  • $\begingroup$ Weight and the normal reaction are not (anti)parallel to each other on a banked curve, so one or the other has to be broken down into components. $\endgroup$ – probably_someone May 25 '18 at 17:33

As a general rule, resolve force vectors into components that are parallel to and perpendicular to the acceleration of the object. In the case of a vehicle driving around a banked curve the acceleration is horizontal, towards the center so it's the normal force (and friction if any) that you would resolve into components.


What you solve for depends on the coordinate system you choose.

  • Choose a tilted coordinate system along the incline, and you must resolve the weight into components (because it is not parallel to any axis) but not the normal force (which is parallel to the y-axis).
  • Choose a horizontal/vertical coordinate system, and you don't resolve the weight (it is already parallel to the y-axis) but you do resolve the normal force (since it is not parallel to any axis).

So, it is all about the choice of coordinate system.

People often choose it to fit along the acceleration. Not because it has to, but because it can be easier to handle.

  • For an object sliding down an incline, the acceleration is parallel to the incline. Thus the choice of tilted coordinate system that follows the incline.

  • For an object in horizontal circular motion, the acceleration is horizontal towards the centre of the circle (not along the incline). We can then pick the horizontal/vertical coordinate system here.

  • $\begingroup$ Do we resolve forces here knowing the object is going In a circular motion and should have a centripetal force......also where is the tangential acceleration $\endgroup$ – Abhinav Jul 20 '18 at 10:02
  • $\begingroup$ @Abhinav Yes, we know that there is horizontal circular motion, meaning horizontal centripetal acceleration and force. So we can resolve components according to this. $\endgroup$ – Steeven Jul 20 '18 at 11:55
  • $\begingroup$ The tangential acceleration will be into the paper / perpendicular to the coordinate system in this example. $\endgroup$ – Steeven Jul 20 '18 at 11:56
  • $\begingroup$ So any chance any force will be providibg this force....I don't see any force in this direction and how did you determine the direction of tangential acceleration....can you pls tell me these $\endgroup$ – Abhinav Jul 21 '18 at 11:11
  • $\begingroup$ @Abhinav For a car driving on a banked road, the centripetal force comes from static friction (which prevents the car from sliding sideways up the incline) and the normal force. Both of these have a horizontal component, and the sum of these is the centripetal force. $\endgroup$ – Steeven Jul 21 '18 at 12:29

While solving for net force on a body on an inclined plane, we always resolve mg (weight) parallel and perpendicular to the plane but in the banking of road concept, we resolve normal reaction. Why don't we simply resolve weight and then equate it with normal reaction?

Because the acceleration is horizontal - toward the center of the car's circular path.

A car's tires require friction to turn so you want to allow for various weights of vehicles and varying road conditions.

The angle of the banking can not be changed once it is set (the road construction is completed) thus it is calculated so mass cancels (various vehicle weights are permitted) because friction is assumed proportional to the normal force, which in turn is proportional to mass.

The road can be dry or icy. It is hoped that people will navigate the road at appropriate speeds for the conditions. All those variables are set to allow for the greatest range of reasonable conditions, with vehicle weight and friction of the road being the most variable.

See: "Uniform Circular Motion and Gravitation - Centripetal Force" or "A Banked Turn - No Friction".

Car on a banked turn

Calculator at hyperphysics.phy-astr.gsu.edu - Maximum speed on banked roadway.

Wikipedia - Superelevation (Bank angle vs. curve radius).

Engineering.stackexchange.com - How are maximum speed limits determined for banked curves?

South Dakota DOT - Road Design Manual - Chapter 5 - Horizontal Alignment (.PDF)

Wyoming DOT - Road Design Manual - Chapter 3 - Elements of Design - 3-02 Horizontal Alignment and Superelevation (878.4 KB) - (.PDF)

  • $\begingroup$ So a driver rounding a banked curve who sees a patch of ice and slows down is setting himself up for sliding into the ditch on the inside of the curve... $\endgroup$ – DJohnM May 25 '18 at 20:47

If you think it would make the problem easier to solve then you can do it any way you want.

However, one reason we might resolve the normal force rather than the weight is because we are usually interested in the resulting centripetal force, which is horizontal. Since the weight is already perpendicular to the centripetal direction resolving the weight would add complication.


protected by Qmechanic May 25 '18 at 18:48

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