[Physics] Why doesn’t centrifugal force appear on an FBD for a car on a banked track

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Suppose you wish to find the max speed at which a car can travel on a banked curve of radius R and angle theta to the horizontal and coeff. of static friction mu.

I don't understand the relevant free body diagram.

We should have $F_g$ downwards, $F_f$ pointing towards the center of the track parallel to the track, and $F_n$ perpindicular to the track. The net force in the direction parallel to the track points towards the center of the track, so the car should accelerate down the track right (towards the center), right? What is keeping the track up – its the centrifugal force, right? But why don't we draw this on the FBD; what is providing the centrifugal force?

Best Answer

As you seem (correctly) to understand, your free body diagram for the car should be as in the left of the drawing below. The force summation does not close to a polygon.

Banked Curve FBD

You know neither $F_N$ nor $F_F$, the components of the force on the car from the road. But you know their directions ($\theta$ is the banking angle) and you know that all the forces must sum, as in the right of my drawing, to a net force (shown in red) pointing towards the centre of curvature, and whose magnitude is $m\,v^2/R$, where $v$ is the car's speed and $R$ the radius of curvature. So there is a net unbalanced force on the car. This is the centripetal force that accelerates the car to keep it moving along the curved track. You can work out $F_N$ and $F_F$ by trigonometry: from my force diagram it is readily seen that:

$$F_N\,\cos\theta - F_F\sin\theta = \frac{m\,v^2}{R}$$ $$F_N\,\sin\theta + F_F\,\cos\theta = m\,g$$

which is an easy set of equations to invert, given that the inverse of the rotation matrix is easily found

$$\left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)^{-1}=\left(\begin{array}{cc}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{array}\right)$$

Note that $F_F$ can be positive or negative: at a critical speed given by $v^2\,\tan\theta/R = g$ it will be zero. Your last step is to check that $|F_F|\leq\mu\,F_N$; if this condition doesn't hold, then your car will slip (either upwards if its going too fast, or downwards if it's going too slow) and uniform circular motion around the track at the speed in question is thus not possible.