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A 4 ky block A is placed on the top of 8 kg block B which rests on a smooth table. A just slipson B when a force of 12 N is applied on A. Then the maximum horizontal force F applied on Bto make both A and B move together, is x N. Find value of x159 |
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Answer» Since limiting friction is 12 N, Maxm. Static friction will be 12 N. When force slightly less than 12 N is applied on block A ,block B will move together with block A . The accln.of the block together will be 12/12 = 1 m/s^2 when the force is applied on A. If the force is applied on B,Accln. of B will be F/12(if F is the force.) Now let us consider blockA Force on the blockA will be( F*4/12 -12)N If both the blocks are to move together this must will be equal to zero F*4/12–12must be equal toor less than zero F*4/12–12= 0 4F=144 F= 144/4= 36N( F must be less than 36N. the answer given at back is 24N |
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