A wire is bent into the shape of a parabola g | vedclass

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(B) When the bead is in equilibrium relative to the wire,the forces acting on it in the frame of the wire are gravity ($mg$ downwards),pseudo force ($

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$\frac{a}{2gk}$

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(B) When the bead is in equilibrium relative to the wire,the forces acting on it in the frame of the wire are gravity ($mg$ downwards),pseudo force ($ma$ in the negative $X$-direction),and the normal force $(N)$.At equilibrium,the net force along the wire is zero.Resolving forces: $N \cos 	heta = mg$ and $N \sin 	heta = ma$,where $	heta$ is the angle the tangent makes with the horizontal.Dividing the two equations,we get $	an 	heta = \frac{a}{g}$.The slope of the parabola $Y = Kx^2$ is given by $\frac{dy}{dx} = 2Kx$.Since the slope is equal to $	an 	heta$,we have $2Kx = \frac{a}{g}$.Solving for $x$,we get $x = \frac{a}{2Kg}$.

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