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A block of mass m is arranged on the wedge as shown in figure . The wedge angle is theta. If the masses of pulley and thread are negligible and friction is absent , find the acceleration of the wedge . |
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Answer» Solution :It is obvious that when block m moves downward along the incline of wedge, the wedge moves to the RIGHT . As the length of thread is constant, the distance traversed by wedge along the incline is equal to distance traversed by wedge to the right . This implies that acceleration of wedge to the right is equal to downward acceleration of wedge . Let a be the acceleration of wedge to the right . Then the force acting on the block m are (i) weight MG acting vertically downward (II) normal reaction `R_(1)` (iii)tension T (up the incline ) (IV)Fictitious force ma to the left .  The free body diagram of mass m is shoen in figure . For motion of block .m. on inclined plane `mg sin theta +ma cos theta -T = ma `....(1) As mass .m. does not breaks off the inclined plane , therefore for forces on .m. normal to inclined plane `R_(1) +ma sin theta =mg cos theta `....(2) The forces acting on the wedge are (i) weight Mg downward (ii) normal reactionof ground on wedge =`R_(2)` (iii) normal reaction of block on wedge =`R_(1)` (iv) tension (T,T) in string . The free body diagram of wedge is shown in figure . For motion of wedge in horizontal direction ` R_(1) sin theta +T-T cos theta = mg`....(4) From (1), `T=mg sin theta + ma cos theta -ma`....(5) From (2), `R_(1) =mg cos theta - ma sin theta`....(6) SUBSTITUTING these values in (3) , we get `(mg cos theta - ma sin theta ) sin theta +(mg sin theta +ma cos theta - ma ) ( 1-cos theta )=Ma` or `{M+m sin ^(2) theta +m(1-cos theta)^(2)}a` `=mg cos theta sin theta +mg sin theta (1-cos theta )` `a= (mg sin theta )/(M+m sin^(2) theta + m(1-2 cos theta+cos^(2) theta))=(mg sin theta)/ (M+m(sin^(2) theta +1 -2 cos theta+cos^(2) theta))` `:. a=(mg sin theta)/(M+2m(1-cos theta))` 
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