Three identical cars A,B and C are moving at the same speed on three bridges. The car A goes on a plane bridge, B on a bridge convex upward and C goes on a bridge concave upward. Let `F_A,F_B and F_C` be the normal forces exerted by the cars on the bridges when they are at the middle of bridges

Options

A. `F_A` is maximum of the three forces
B. `F_B` is maximum of the three forces.
C. `F_C` is maximum of the three forces.
D. `F_A=F_B=F_C`

Answer

To solve the problem, we need to analyze the forces acting on each car as it travels over the different types of bridges. The three cars A, B, and C are on a plane bridge, a convex bridge, and a concave bridge, respectively. We will denote the mass of each car as \( m \) and the gravitational force as \( F_g = mg \), where \( g \) is the acceleration due to gravity.

### Step-by-Step Solution:

1. **Car A on a Plane Bridge:**
- For car A, which is on a plane bridge, the only forces acting on it are the gravitational force downward and the normal force \( F_A \) exerted by the bridge upward.
- Since the car is not accelerating vertically, these forces must balance each other:
\[ F_A = mg \]

2. **Car B on a Convex Bridge:**
- For car B, which is on a convex bridge, the normal force \( F_B \) acts upward, and the gravitational force \( mg \) acts downward.
- Additionally, since the car is moving in a circular path, it experiences a centripetal acceleration. The net force towards the center of the circular path is provided by the difference between the normal force and the gravitational force:
\[ F_B - mg = \frac{mv^2}{R} \]
- Rearranging this gives:
\[ F_B = mg + \frac{mv^2}{R} \]

3. **Car C on a Concave Bridge:**
- For car C, which is on a concave bridge, the normal force \( F_C \) also acts upward, and the gravitational force \( mg \) acts downward.
- In this case, the normal force must provide the centripetal force required for circular motion, in addition to balancing the weight of the car:
\[ F_C - mg = \frac{mv^2}{R} \]
- Rearranging this gives:
\[ F_C = mg + \frac{mv^2}{R} \]

4. **Comparing the Normal Forces:**
- From the equations derived:
- \( F_A = mg \)
- \( F_B = mg - \frac{mv^2}{R} \)
- \( F_C = mg + \frac{mv^2}{R} \)
- We can now compare the normal forces:
- \( F_A < F_B < F_C \)
- Thus, the order of the normal forces is:
\[ F_C > F_B > F_A \]

### Conclusion:
The normal forces exerted by the cars on the bridges are ranked as follows:
\[ F_C > F_B > F_A \]