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 FA, FB and FC be the normal forces exerted by the cars on the bridges when they are at the middle of the bridges.
(a) FA is maximum of the three forces.
(b) FB is maximum of the three forces.
(c) FC is maximum of the three forces.
(d) FA = FB = FC
Answers
Follow the diagram attached below. After that Read the solution given below.
Answer ⇒ We know that Centripetal is not the new Force. It is Just the Net Force towards the center during the circular motion. Now, As you can see in the Diagram attached below, there will be no centripetal force since the surface of the Bridge is plane.
∴ Fa = mg.
But in second case, there will be centripetal force which will be acting twords the centre. By the definition of the Centripetal Force as I have written above,
mv²/r = mg - Fb.
⇒ Fb = mg - mv²/r
Similarly, For the third Car, ''C'', the Center is in opposite direction since it shape in concave upwards, therefore, Centripetal Force will be,
mv²/r = Fc - mg
⇒ Fc = mv²/r + mg.
From these Equations we can see that Fc, is highest and then Fa and then Fb.
This means Fc is maximum among three forces. Therefore, Option (c). is correct.
Hope it helps.
