Hmm, to be fair, I've done a little more looking around and found this in my book:

"Curved portions of highways are always banked (tilited) to prevent cars from sliding off the highway. When a highway is dry, the frictional force between the tires and the road surface may be enough to prevent sliding. When the highway is wet, however, the frictional force may be negligible, and banking is then essential. Figure 6-13a represents a car of mass m as it moves at a constant radius R = 190 m. (It is a normal car, rather than a race car, which means any vertical force from the passing air is negligible.) If the frictional force from the track is negligible, what bank angle

prevents sliding?

Key Idea: Unlike Sample Problem 6-9, the track is banked so as to tilt the normal force

on the car toward the center of the circle (Figure 6-13b). Thus

now has a centripetal component of magnitude

, directed inward along a radial axis r. We want to find the value of the bank angle

such that this centripetal component keeps the car on the circular track without need of friction.

Radial calculation: As Fig. 6-13b shows (and as you should verify), the angle that force

makes with the vertical is equal to the bank angle

of the track. Thus, the radial component

, is equal to

. We can now write Newton's second law for components along the r axis (

) as

.

This will be equation (6-25).

We cannot solve this equation for the value of

because it also contains the unknowns

and m.

Verticle calculations: We next consider the forces and acceleration along the y axis in Fig. 6-13b. The verticle component of the normal force is

, the gravitational force

on the car has the magnitude mg, and the acceleration of the car along the y axis is zero. Thus we can write Newton's second law for components along the ya axis (

) as

from which

This will be Equation (6-26).

Combining results: Equation 6-26 also contains the unknowns

and m, but note that dividing Eq. 6-25 by Eq. 6-26 neatly eliminates both those unknowns. Doing so, replacing

with

, and solving for

then yield

12 degrees."

I apologize for the lack of visuals which this text refers to.