Circular Banked Track Friction 
I've tried answering this question by resolving forces, then finding an expression for friction and inserting the given data so I can prove $F = 0$. However, I never get an answer of $0$. 
How do I do it?

Notice, mass of the car $m=1\ t=1000\ kg$,   
speed of the car $v=72\ km/hr=20\ m/sec$,  
radius of the circular path $r=160\ m$ 
$$\tan\alpha=\frac{1}{4}\iff \sin\alpha=\frac{1}{\sqrt {17}}, \cos \alpha=\frac{4}{\sqrt{17}}$$
$\color{red}{\text{Method 1}}$:Resolving components
Component of centrifugal force acting on the center of gravity of car parallel to & up the plane $$F_1=\frac{mv^2}{r}\cos \alpha$$
Component of gravitational force ($mg$) acting on the center of gravity of car parallel to & down the plane $$F_2=mg\sin \alpha$$
The lateral frictional force, $\color{red}{F_f}$ (parallel to & down the plane) acting between the tyres & the track $$F_f=F_1-F_2$$  $$=\frac{mv^2}{r}\cos \alpha-mg\sin \alpha=m\left(\frac{v^2}{r}\cos \alpha-g\sin \alpha\right)$$
$$=1000\left(\frac{20^2}{160}\frac{4}{\sqrt{17}}-10\frac{1}{\sqrt{17}}\right)$$
$$=1000\left(\frac{10}{\sqrt{17}}-\frac{10}{\sqrt{17}}\right)=\color{red}{0}$$
Hence, there is no lateral frictional force between tyres & the track. 
$\color{red}{\text{Alternative method}}$: There is direct condition for zero frictional force $$\tan \alpha=\frac{v^2}{rg}$$  Setting the corresponding values values, we get 
$$\frac{1}{4}=\frac{20^2}{160\times 10}$$ $$\frac{1}{4}=\frac{1}{4}$$ The condition is satisfied hence, there is no lateral frictional force between tyres & the track. 
