InterviewSolution
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InterviewSolution
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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. |
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Answer» Solution :Let `a_0` be the acceleration of PRISM in the backward direction. The mass .m. is placed on non-inertial FRAME, therefore it is acted on by fictitious FORCE `ma_0` in the forward direction. The forces acting on massm are i) Weight mg acting downward,  ii) Normal REACTION R, iii) Fictitious force `ma_0` According free body diagram of .m. is shown in figure. Suppose block m slides down the prism with acceleration a. The equation of motion of .m. parallel to incline is `ma_0 cos theta + mg sin theta = ma ` `implies a = a_0 cos theta + g sin theta ` ........... (1) The block is in equilibrium perpendicular to incline, so resolving force perpendicular to incline  `R_1 + ma_0 sin theta = mg cos theta ` ........ (2) Now we consider the forces acting on the prism. As prism is on ground (inertial frame) no fictitious force acts on it. The forces on prism are i) Weight (Mg) downward. ii) Normal force `R_1` exerted by block iii) Normal reaction `R_1` exerted by ground For horizontal motion of prism `R _1sin theta = Ma_0 implies R_1 = (Ma_0)/(sin theta ) `......... (3) substituting this value in (2) , `(Ma_0)/(sin theta) + ma_0 sin theta = mg cos theta ` This gives ` a_0 = ( mg sin theta cos theta)/( M + m sin^(2) theta ) `............ (4) `:. ` From (1) , ` a=(mg sin theta cos^(2) theta )/( M + m sin^(2) theta ) + g sin theta ` `= ((M+m) g sin theta )/( M + m sin^2 theta )` |
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