A popular physics page explaining how a car can make a banked turn even with no friction :

car on a banked turn
(source: k12.in.us)

gives the following free body diagram

(source: k12.in.us)

and then says

In the vertical direction there is no acceleration, and $N\cos(\theta) = mg$

What I'm unclear about is how do we (apriori) know this fact? Maybe the car will slide down the plane, and there will be a vertical acceleration component!

In fact, isn't the free body diagram for this problem identical to the one for a regular inclined plane? In both cases there are only two force vectors - $mg$ and $N$, both pointing in the same directions (vertically down and normal to the incline, respectively). The only difference between the two problems is that in this one the car has a velocity along the track (i.e. into the page), whereas in the inclined plane problem that velocity is 0. But otherwise, since all the forces are the same between the two problems, shouldn't what actually happens (i.e. the acceleration of the car) also be identical? Yet when we solve the "car on the inclined plane" problem, the car accelerates down the slope, but in this case it does not!

In the inclined plane case we resolve $mg$ into components $(mg)_\parallel$ parallel and $(mg)_\bot$ perpendicular to the incline, and we make the assumption that the N force is exactly equal $(mg)_\bot$. Then $(mg)_\bot$ is exactly canceled by the N force, leaving only $(mg)_\parallel$ to provide acceleration down the incline. So why don't we make the same assumption in this (banked turn) problem? Why do we instead make the $Ncos(\theta) = mg$ assumption?

A note on the site actually addresses this question somewhat and says :

Your initial thought might have been to resolve the weight vector parallel and perpendicular to the road - after all, that is what we did for all of those lovely inclined plane problems, remember? The difference is that we expected the object to accelerate parallel to the incline, so it made sense to have the vectors pointing parallel and perpendicular to the incline. Here, though, the acceleration is horizontal - toward the center of the car's circular path - so it makes sense to resolve the vectors horizontally and vertically.

But how do we know, before solving the problem, that the acceleration will be horizontal? Since the motion of the car is fully determined by the forces on it, it seems like that information should come out of solving the equations of motion based on those forces, and not be assumed beforehand...


4 Answers 4


It's an assumption.

If you know the velocity, but don't know if the car slides or not, then you would go about solving this problem differently.

In this case, you're solving for the velocity at which the car will not slide. Thus you can start with the assumption that the car will not accelerate vertically.

The problem is nearly as simple if you sum forces tangent to the road surface:


Simplifies to:


Then if you assume the acceleration is zero, you can solve for the velocity, or if you have the velocity you can solve for the acceleration. If you have both the velocity and you want to find the theta that give you zero acceleration you can do that as well.


What I'm unclear about is how do we (apriori) know this fact?

In the general case it might not be true. The forces will depend on the speed of the car, the incline of the road, and the curvature of the bend. However the problem is saying (or asking) whether the car can make this turn. We are defining the car to have no unwanted accelerations, and we are checking whether it is possible to arrange the other forces to make it work. We know (assume) the car isn't accelerating up into the air or down into the ground, so we know (assume) the gravitational force down and the ground force up are equal, then we compute what other forces are needed to make it work. (If we can compute those forces, then it's possible. If we can't then it's impossible.)

However I think you might be overlooking the curvature of the road and the force associated with that. Instead try imagining a car hanging on the end of a rope, and the rope is fixed to the ceiling. We could push that car so that it swings in a circle. The rope is pulling up on the car, and it is also pulling inwards on the car making it go in a circle. We can replace the rope with the inclined circular road (if it has the right slope). The car is following the identical path, so the forces on the car are the same as before. The road is pushing up with the same force that the rope was pulling up, and the curvature of the road is pushing inwards with the same force that the rope pulled inwards.

The point to the rope example is to help understand the forces involved. The values will be exactly what I claim they are because we assume those things when we set up the problem - then we do the math to check whether it is valid. Whether it is a car on a banked road or a car hanging from a rope, there are three variables. (1) The speed of the car, (2) the radius of the circle, and (3) the slope of the rope-or-road. If we know two of those values, we can compute a valid solution for the third value.

Note that the slope of the road and the slope of the rope would be exactly perpendicular, to get a car to go in the identical circle.


This is a classic curved bank problem. The assumption that the car doesn't slide is made to simplify analysis, to illustrate how the problem work. There is no a priori reasoining. For more complex problems the assumption will have to be abandoned.

  • $\begingroup$ But it should be noted that the bank can be made with a variable pitch, steeper at the top, shallower at the bottom, so that it's essentially "self-adjusting" -- the car will slide up/down the bank until it "finds" a spot where the forces balance out. $\endgroup$
    – Hot Licks
    Commented Mar 31, 2015 at 4:01
  • $\begingroup$ But suppose that all you are given is just the following : "There is a car on a curved bank which is θ degrees with respect to the ground". Nothing about the car's velocity. Can you then deduce that the car doesn't slide? Of course not, because it will slide - this is then just an instance of an object-on-an-incline problem! So, in fact, the assumption that the car doesn't slide is indeed an a priori assumption - and it seems that the only physical variable that differentiates this problem from just a (stationary) car on the inclined plane is the existing (non zero) velocity of the car. $\endgroup$
    – User9808
    Commented Mar 31, 2015 at 4:44
  • $\begingroup$ This suggests that it is that existing velocity that somehow is responsible for the force which keeps the car from sliding down. Actually, this raises a bigger question - since the gravity force (mg) is completely cancelled by the vertical component of N = N cos(theta), and yet there is still a non zero horizontal component of N = N sin (theta), this means that |N| > |mg|. But how could that be if N is supposed to be just a reaction force from the banked incline - shouldn't it be at most as large as |mg| (i.e. equal and opposite to the perpendicular component of mg)? How can it be greater? $\endgroup$
    – User9808
    Commented Mar 31, 2015 at 4:55
  • $\begingroup$ @User9808 The normal fore is a reaction force to both gravity and the sideways force that the car exerts on the road in order to accelerate to maintain it's path on the curve. Thus the sum of those two vectors gives it a magnitude greater than the weight of the car. It's just like in take and field, if you spin a hammer around, the pull of the hammer is much stronger than its weight. $\endgroup$
    – Rick
    Commented Apr 3, 2015 at 15:09

If it helps, you're mixing up cause and effect, and missing one part of the force diagram.

Your diagram applies to any vehicle on a banked turn. However, it is missing a force applied horizontally to the left, which opposes $N sin{\theta}$. This is, of course, the centrifugal force produced by the motion of the vehicle, and has the value $\frac{mv^2}{R}$. And you are correct in thinking that an unopposed horizontal component of N will cause the vehicle to slide down the track. The point of the discussion is to set the conditions for no sliding - when the centrifugal force equals the horizontal component of the normal. If the two are not perfectly equal, you get horizontal motion, assuming the track is frictionless.

So while it's true that the horizontal component is what's important, to say that "But how do we know, before solving the problem, that the acceleration will be horizontal?" misses the point. By separating the normal force into vertical and horizontal components, and then applying a counterbalancing horizontal centrifugal force, it is possible to guarantee that the vehicle does not slip down the track.

  • $\begingroup$ I am strongly opposed to introducing inertial pseudo forces as if they were real and without a discussion of inertial frames or whether or not the car is actually in equilibrium or not. Certainly this particular problem Is more tractable in the rotating frame but students must learn to distinguish if the car is in inertial motion or not if they are to succeed in mechanics. $\endgroup$ Commented Sep 11, 2019 at 23:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.