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.
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.
