Three identical cars A, B and C are moving at the same speed on three bridges. The car A goes on a place 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 car on the bridges when they are at the middle of bridges.
(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.

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Solution

(c) F_C is maximum of the three forces.

At the middle of bridge, normal force can be given as:
$N_{A} = m g$
$N_{B} = \frac{m v^{2}}{r} - m g N_{C} = \frac{m v^{2}}{r} + m g$

So, F_C is maximum.

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(c) FC is maximum of the three forces. At the middle of bridge, normal force can be given as: NA=mg NB=mv2r-mgNC=mv2r+mg So, FC is maximum.

(c) F_C is maximum of the three forces.

At the middle of bridge, normal force can be given as:
$N_{A} = m g$
$N_{B} = \frac{m v^{2}}{r} - m g N_{C} = \frac{m v^{2}}{r} + m g$

So, F_C is maximum.