A plumb bob is hanging from the ceiling of a | vedclass

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(A) When the car moves with an acceleration $a$ in the forward direction,a pseudo force $F_p = ma$ acts on the plumb bob in the backward direction in

Vedclass · Accepted Answer

$	an ^{-1}\left(\frac{a}{g}ight)$

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(A) When the car moves with an acceleration $a$ in the forward direction,a pseudo force $F_p = ma$ acts on the plumb bob in the backward direction in the non-inertial frame of the car.Let $m$ be the mass of the bob. The forces acting on the bob are:$1$. Weight $mg$ acting vertically downwards.$2$. Pseudo force $ma$ acting horizontally backwards.$3$. Tension $T$ in the string.In the equilibrium position relative to the car,the net force on the bob is zero. Resolving the forces,we have:$T \sin 	heta = ma$$T \cos 	heta = mg$Dividing the two equations,we get:$\frac{T \sin 	heta}{T \cos 	heta} = \frac{ma}{mg}$$	an 	heta = \frac{a}{g}$Therefore,the angle made by the string with the vertical is $	heta = 	an ^{-1}\left(\frac{a}{g}ight)$.

$A$ plumb bob is hanging from the ceiling of a car. If the car moves with an acceleration $a$,the angle made by the string with the vertical is

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