Question # ( mathrm{A} ) bead of mass ( mathrm{m} ) is located on a parabolic wire with its axis vertical and vertex directed towards downward as in figure and whose equation is ( x^{2}= ) ay. If the coefficient of friction is ( mu, ) the highest distance above the ( x ) -axis at which the particle will be in equilibrium is

# ( mathrm{A} ) bead of mass ( mathrm{m} ) is located on a parabolic wire with its axis vertical and vertex directed towards downward as in figure and whose equation is ( x^{2}= ) ay. If the coefficient of friction is ( mu, ) the highest distance above the ( x ) -axis at which the particle will be in equilibrium is

(A) ( mu a )

(B) ( mu^{2} a )

(C) ( frac{1}{4} mu^{2} a )

(D) ( frac{1}{2} mu a )

Solution

tangent at any x distance would be ( => ) tanthetha ( =2 times / a )

Now

friction = umgcosthetha Balancing friction with ( mathrm{mg}^{*} ) sin(thetha) we get, ucosthetha = sinthetha

or

tan(thetha) ( =u ) ( 2 x / a=u )

( x=a u / 2 )

The highest point would be, ( y=(a u / 2)^{wedge} 2 / a=>a^{wedge} 2 u^{wedge} 2 / 4 a=>a u^{wedge} 2 / 4 )