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 and C goes on a concave. Let FA, FB and FC be the normal forces exerted by the cars on the bridges when they are at the middle of bridges.
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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
answer is C.
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Detailed Solution
(i) For Plane BridgeFA = N = Mg(ii) For Convex Upward BridgeFB = N = Mg-Mv2r(iii) For Concave Downward BridgeN-Mg = Mv2rFC = Mv2r+ MgFC > FA > FB