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Question
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 FA, FB and FC be the normal forces exerted by the car on the bridges when they are at the middle of bridges.
Options
FA is maximum of the three forces.
FB is maximum of the three forces.
FC is maximum of the three forces.
FA = FB = FC.
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Solution
FC is maximum of the three forces.
At the middle of bridge, normal force can be given as : \[\text{N}_\text{A} = \text{mg}\]
\[\text{N}_\text{B} = \frac{\text{mv}^2}{\text{r}} - \text{mg}\]
\[ \text{N}_\text{C} = \frac{\text{mv}^2}{\text{r}} + \text{mg}\]
So, FC is maximum.
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