In this problem:

A car of mass $430\ \mathrm{kg}$ travels around a flat, circular race track of radius $178\ \mathrm{m}$. The coefficient of static friction between the wheels and the track is $0.266$.

The same car now travels on a straight track and goes over a hill with radius $178\ \mathrm{m}$ at the top.

What is the maximum speed that the car can go over the hill without leaving the road?

Correct answer: $41.766\ \mathrm{m/s}$.


$$\frac{mv^2}{r} = mg - N$$ where $N$ is the normal force acting on the car from the ground. The car will fly off the ground just when $N = 0$ so the maximum speed allowed will be $$\begin{align}v_{\text{max}} &= \sqrt{gr} \\ &= \sqrt{(9.8\ \mathrm{m/s^2})(178\ \mathrm{m})} \\ &= 41.766\ \mathrm{m/s}.\end{align}$$

The car is not driving upside down! So why is the force of gravity positive and the normal force considered negative in this problem?

  • $\begingroup$ Gravity pulls the car down. The normal force from the road pushes the car up. Whatever coordinate system you choose, these forces have opposite sign. Whoever wrote the question just chose a coordinate system where down is positive and up is negative. $\endgroup$
    – The Photon
    Commented Jul 18, 2016 at 19:22
  • $\begingroup$ @ThePhoton that should have been an answer $\endgroup$
    – David Z
    Commented Jul 18, 2016 at 19:25
  • $\begingroup$ @DavidZ I assumed it would get closed as a duplicate. $\endgroup$
    – The Photon
    Commented Jul 18, 2016 at 19:26
  • 1
    $\begingroup$ @ThePhoton Well... okay, what I really mean is that it shouldn't have been posted as a comment. It's an answer, so either post it as an answer, or not at all. (You may be right, this could easily be a duplicate, but I didn't find one from a cursory search.) $\endgroup$
    – David Z
    Commented Jul 18, 2016 at 19:28
  • $\begingroup$ In addition to other comments on this subject, you are free to define the positive direction. Up can be positive OR negative, depending on how you want to set up the problem. You will get the correct answer as long as your sign convention is totally consistent. Also note that when you take the square root to arrive at velocity, both the negative and positive roots may be valid answers. $\endgroup$ Commented Jul 18, 2016 at 20:14

2 Answers 2


The centripetal acceleration always points toward the center of the circle. In this case, the center of the circle is below the car, so the centripetal acceleration points downward.

Now, you'll notice that, in the given solution, the centripetal acceleration term is positive. That means the writer has chosen a coordinate system where positive is downward, and negative is upward. Any force that acts downward (like gravity) will be represented by a positive term ($+mg$), and any force that acts upward (like the normal force) will be represented by a negative term ($-N$).

Alternatively, you could make the opposite choice: pick positive to be up, and negative to be down. In that case, the centripetal acceleration, since it points downward, would be represented by a negative term, $-\frac{mv^2}{r}$. Same for gravity ($-mg$). And the normal force, pointing upward, would be represented by a positive term, $+N$.


Which way does the y axis point?

  • If the y axis is chosen to point up, (having the positive direction upwards) then you are right, normal force should be positive (it points upwards as well) and weight negative.
  • Is it chosen to point down, then normal force is negative (points in the negative direction along the axis) and weight positive.

Remember that the acceleration will also have a sign according to this.

The point is that coordinate systems can be placed as you want - if it doesn't confuse, it is smart to place it in a way that gives the fewest symbols and parameters; fx downwards so not too many properties have a negative sign.


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