In the given arrangement of a doubly inclined plane two blocks of masses M and m are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25. The value of m, for which M=10 kg will…

Question

In the given arrangement of a doubly inclined plane two blocks of masses M and m are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25. The value of m, for which M=10 kg will move down with an acceleration of 2 m/s² is (Take g=10 m/s² and tan 37°=3/4)

Accepted Answer

(3) 4.5 kg. Idea: two blocks, two equations. M slides down the steep 53° face and drags m up the 37° face, so friction opposes each of them in turn. From tan 37°=3/4 we have sin 37°=0.6, cos 37°=0.8, and the other face at 53° has sin 53°=0.8, cos 53°=0.6. Block M, moving down its face, with friction acting up it: Mg sin 53°-T-μ Mg cos 53°=Ma 10(10)(0.8)-T-0.25(10)(10)(0.6)=10(2) 80-T-15=20, so T=45 N. Block m, dragged up its face, with friction acting down it: T-mg sin 37°-μ mg cos 37°=ma 45-6m-0.25(8m)=2m 45-6m-2m=2m 45=10m, so m=4.5 kg. The two faces are complementary (37°+53°=90°), which is why the same pair of numbers 0.6 and 0.8 does all the work.

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