1.

A prism of mass M is placed on a horizontal surface. A block of mass m slides on the prism, which in tum slides on thehorizontal surface. Assuming all surfaces to be frictionless, find the acceleration of block with respect to prism.

Answer» <html><body><p></p>Solution :Let `a_0` be the acceleration of prism in the backward direction. The <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> .m. is placed on non-inertial frame, <a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> it is acted on by fictitious force `ma_0` in the forward direction. The forces acting on massm are <br/>i) Weight mg acting downward,<br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_AI_PHY_V01_P1_C06_SLV_040_S01.png" width="80%"/> <br/> ii) Normal reaction R, <br/>iii) Fictitious force `ma_0`<br/> According free body diagram of .m. is shown in <a href="https://interviewquestions.tuteehub.com/tag/figure-987693" style="font-weight:bold;" target="_blank" title="Click to know more about FIGURE">FIGURE</a>. Suppose block m slides down the prism with acceleration a. The equation of motion of .m. <a href="https://interviewquestions.tuteehub.com/tag/parallel-1146369" style="font-weight:bold;" target="_blank" title="Click to know more about PARALLEL">PARALLEL</a> to incline is<br/> `ma_0 cos theta + mg sin theta = ma ` <br/> `implies a = a_0 cos theta + g sin theta ` ........... (1)<br/> The block is in equilibrium perpendicular to incline, so resolving force perpendicular to incline <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_AI_PHY_V01_P1_C06_SLV_040_S02.png" width="80%"/><br/> `R_1 + ma_0 sin theta = mg cos theta ` ........ (2) <br/> Now we consider the forces acting on the prism. As prism is on <a href="https://interviewquestions.tuteehub.com/tag/ground-1013213" style="font-weight:bold;" target="_blank" title="Click to know more about GROUND">GROUND</a> (inertial frame) no fictitious force acts on it. The forces on prism are<br/> i) Weight (Mg) downward.<br/> ii) Normal force `R_1` exerted by block <br/>iii) Normal reaction `R_1` exerted by ground <br/> For horizontal motion of prism <br/> `R _1sin theta = Ma_0 implies R_1 = (Ma_0)/(sin theta ) `......... (3) <br/> substituting this value in (2) , <br/> `(Ma_0)/(sin theta) + ma_0 sin theta = mg cos theta ` <br/> This gives ` a_0 = ( mg sin theta cos theta)/( M + m sin^(2) theta ) `............ (4) <br/> `:. ` From (1) , ` a=(mg sin theta cos^(2) theta )/( M + m sin^(2) theta ) + g sin theta ` <br/> `= ((M+m) g sin theta )/( M + m sin^2 theta )`</body></html>


Discussion

No Comment Found

Related InterviewSolutions