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Find M (in kg) for which m2 will start sliding over m1. Table on which m1 and m2 areplaced is smooth, u is the friction coefficientbetween my and m2.(Given : m1 = 2 kg, m2 = 1 kg, u = 0.5, g = 10m/s2)m2m,M |
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Answer» Gravitational Force on m 2 ALONG the inclined PLANE in downward direction is,F G2 =m 2 gsin60 0 =1×G× 23 = 23 g Similarly, Gravitational Force on m 1 along the inclined plane in downward direction is,F G1 =m 1 gsin30 0 = 3 ×g× 21 = 23 g So net external force on the tow-block system is F G2 −F G1 =0So ACCELERATION of the blocks is also zero.So Tension in the string balances the gravitational force on blocks along the inclined plane. So tension, T=F G2 =F G1 = 23 g Explanation:hope it is HELPFUL for you |
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