At the top of the bridge, the car is undergoing uniform circular motion. The forces acting on the car are the gravitational force $F_g = mg$ (downwards) and the normal force $N$ from the bridge (upwards). The net force provides the required centripetal force, $F_c = mv^2/R$, which is directed downwards towards the center of the circular arc.

Applying Newton's second law in the vertical direction (taking downwards as positive):

$$ \sum F = ma_c $$ $$ mg - N = \frac{mv^2}{R} $$

Solving for the normal force $N$ exerted by the bridge on the car:

$$ N = mg - \frac{mv^2}{R} $$

By Newton's third law, the force exerted by the car on the bridge, $F_{pressure}$, is equal in magnitude and opposite in direction to the normal force $N$.

$$ F_{pressure} = N = mg - \frac{mv^2}{R} = m \left( g - \frac{v^2}{R} \right) $$