High Banked Turn

While negotiating the banked turn, there are 3 forces acting on the car: gravity, centrifugal force, and the ground reaction forces the track surface exerts on the tires.

Gravity exerts a force of $m g$ and always acts vertically downward.
Centrifugal force is given by mass $\frac{m v^2} {\text{R}}$ and acts radially outward in the plane of motion, the horizontal plane in this problem.
Ground reaction forces (GRF) act in 3 dimensions at each tire contact with the track. We will only consider GRF normal to track surface in this problem.

The mass of the car is $\SI{1400}{ \kilo \gram}$ so if $g=10 \text{m/s}^2$ , then gravitational force is $\SI{14000}{\newton}$ .
If $v$ is $40 \text{m/ s}$ and $R$ is $100$ meters then centrifugal force is $\SI{22400}{\newton}$ .

See figures.

The next step is to resolve each of these vector quantities into components normal to and parallel to the track surface.

The gravitational force has a component normal to track surface of $14000 \times \cos 30^\circ = \SI{12124}{\newton}$ . The gravitational force has a component parallel to track surface of $14000 \times \sin 30^\circ = \SI{7000}{\newton}$ . Centrifugal force has a component normal to track surface of $22400 \times \sin 30^\circ = \SI{11200}{\newton}$ . Centrifugal force has a component parallel to track surface of $22400 \times \cos 30^\circ = \SI{19400}{\newton}$ .

Combining these components, we have a normal component of $\SI{23324}{\newton}$ acting into the track surface and a parallel component of $\SI{12400}{\newton}$ acting up the bank.

See figures.

Now to find the GRF acting on the right side tires we will balance the moments about the point of contact of the left side tires with the track surface. This is point $P$ in the drawing. Gravity and centrifugal force act through the center of mass.

The normal force is $170/2$ or $\SI{85}{\centi\meter}$ to the right of point $P$ . It generates a clockwise moment of $\SI{19825}{\newton \meter}$ about point $P$ . The parallel component acts through the center of mass which is $\SI{50}{\centi \meter}$ above point $P$ . It generates a clockwise moment of $\SI{6200}{\newton \meter}$ about point $P$ .

There is a total clockwise moment of $\SI{26025}{\newton\meter}$ about $P$ . The right side normal GRF x 1.70 must generate a counterclockwise moment of magnitude $\SI{26025}{\newton\meter}$ . So, $\dfrac{26025}{ 1.7} = \SI{15310}{\newton}$