[![Car on a slope.][1]][1]

Place the car's centre of gravity at the origin of a Cartesian coordinate system $x,y$, as shown above.

Now determine the balance of forces along the $x$ axis. Assume the car will slide (up or down), as there's no friction. The equation of motion is:

$ma_x=mg\sin\alpha-F_c\cos\alpha$ (Eq.1).

$F_c=\frac{mv^2}{R}$ with $v$ the velocity of the car and $R$ the radius of the turn. Insert into Eq.1, drop $m$ and solve for $a_x$:

$a_x=g\sin\alpha-\frac{v^2}{R}\cos\alpha$.

If $g\sin\alpha>\frac{v^2}{R}\cos\alpha$, then $a_x>0$ and the car will sliding 'down' (acceleration in the $x$ direction).

If $g\sin\alpha<\frac{v^2}{R}\cos\alpha$, then $a_x<0$ and the car will sliding 'up' (acceleration in the opposite of the $x$ direction).

Note that this is a very unrealistic case: a road that offers no friction at all. [1]: https://i.sstatic.net/Gr0lm.png

Place the car's centre of gravity at the origin of a Cartesian coordinate system $x,y$, as shown above.

Now determine the balance of forces along the $x$ axis. Assume the car will slide (up or down), as there's no friction. The equation of motion is:

$ma_x=mg\sin\alpha-F_c\cos\alpha$ (Eq.1).

$F_c=\frac{mv^2}{R}$ with $v$ the velocity of the car and $R$ the radius of the turn. Insert into Eq.1, drop $m$ and solve for $a_x$:

$a_x=g\sin\alpha-\frac{v^2}{R}\cos\alpha$.

If $g\sin\alpha>\frac{v^2}{R}\cos\alpha$, then $a_x>0$ and the car will sliding 'down' (acceleration in the $x$ direction).

If $g\sin\alpha<\frac{v^2}{R}\cos\alpha$, then $a_x<0$ and the car will sliding 'up' (acceleration in the opposite of the $x$ direction).

Note that this is a very unrealistic case: a road that offers no friction at all.