Applying centrifugal forces 
Three identical cars A, B, and C are moving at the same speed on three bridges. Car A goes on a plane bridge, car B on a bridge concave downward, and car C goes on a bridge concave upward. Let $N_A$, $N_B$, and $N_C$ be the normal forces exerted by the cars on the bridges when they are in the middle of bridges. Choose the correct option.


Please refer to the FBD in the attached image. In the second free-body diagram of concave upwards, the centrifugal force is shown on the body in the upwards direction. If we are solving it in the ground frame (which is an inertial frame), how can we apply a centrifugal force, which is a pseudo force? Furthermore, why are we even applying centrifugal force? Why can't we just apply centripetal force in the opposite direction?
 A: Centrifugal and centripetal forces are a frame dependant forces. Centripetal force is considered when examining a system from outside system's frame. Centrifugal force is considered when viewing system from system's own frame.
The solution given in question according to me is woefully incomplete, as it doesn't care about frame from which we are observing, which creates confusion. Let's examine second case from both frames.
Ground/Outside Frame
The centripetal force must act inwards(down), in order for car to remain sticked to road, or else car will become airborne. Our forces acting are: $N_s$ upwards, $mg$ downwards. Their net (downwards) is our magnitude of centripetal force:
$$mg-N_s=\frac{mv}{r^2}$$
$$N_s=mg-\frac{mv}{r^2}$$
Car/Inside Frame
The observer will 'feel' pseudoforce acting upwards, as car goes around the curvature. Actually 'feel' is due to inertia of direction of motion, so don't confuse feeling to be caused by real force.
Pseudoforce is applied here because we now are in the car frame which is accelerating frame. Newton's second Law is applicable here only if we use Pseudoforce .
Now this upwards pseduoforce (net) is centrifugal force:
$$N_s-mg=-\frac{mv}{r^2}\,\,\, (\text{minus sign signifies opposite direction})$$
$$N_s=mg-\frac{mv}{r^2}$$
So from both frames, our final answer remains same. Its just way of writing the same equation in two ways if you can see it.
Hope it helps!
A: In the inertial frame the car is accelerating towards the middle of the circle, so we would do $$F=ma$$$$mg-N_B = \frac{mv^2}{R}$$ the last term is a centripetal force as you suggest and it gives the same result as the diagram.
