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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 .

Answer» <html><body><p></p>Solution :It is obvious that when block m moves downward along the incline of wedge, the wedge moves to the <a href="https://interviewquestions.tuteehub.com/tag/right-1188951" style="font-weight:bold;" target="_blank" title="Click to know more about RIGHT">RIGHT</a> . 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 . <br/> Let a be the acceleration of wedge to the right . <br/> Then the force acting on the block m are <br/>(i) weight <a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a> acting vertically downward <br/>(<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) normal reaction `R_(1)`<br/> (iii)tension T (up the incline )<br/>(<a href="https://interviewquestions.tuteehub.com/tag/iv-501699" style="font-weight:bold;" target="_blank" title="Click to know more about IV">IV</a>)Fictitious force ma to the left .<br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AI_PHY_XI_V01_A_C06_SLV_037_S01.png" width="80%"/><br/> The free body diagram of mass m is shoen in figure .<br/> For motion of block .m. on inclined plane <br/> `mg sin theta +ma cos theta -T = ma `....(1) <br/> As mass .m. does not breaks off the inclined plane , therefore for forces on .m. normal to inclined plane <br/>`R_(1) +ma sin theta =mg cos theta `....(2) <br/> The forces acting on the wedge are <br/> (i) weight Mg downward <br/>(ii) normal reactionof ground on wedge =`R_(2)`<br/>(iii) normal reaction of block on wedge =`R_(1)` <br/> (iv) tension (T,T) in string . <br/> The free body diagram of wedge is shown in figure . <br/> For motion of wedge in horizontal direction <br/>` R_(1) sin theta +T-T cos theta = mg`....(4)<br/> From (1), <br/> `T=mg sin theta + ma cos theta -ma`....(5) <br/> From (2), `R_(1) =mg cos theta - ma sin theta`....(6) <br/> <a href="https://interviewquestions.tuteehub.com/tag/substituting-1231652" style="font-weight:bold;" target="_blank" title="Click to know more about SUBSTITUTING">SUBSTITUTING</a> these values in (3) , we get <br/> `(mg cos theta - ma sin theta ) sin theta +(mg sin theta +ma cos theta - ma ) ( 1-cos theta )=Ma`<br/> or `{M+m sin ^(2) theta +m(1-cos theta)^(2)}a`<br/> `=mg cos theta sin theta +mg sin theta (1-cos theta )`<br/> `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))`<br/>`:. a=(mg sin theta)/(M+2m(1-cos theta))`<br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AI_PHY_XI_V01_A_C06_SLV_037_S02.png" width="80%"/></body></html>


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