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By what acceleration the boy must go up so that `100kg` block remains stationary on the wedge. The wedge is fixed and is smooth `(g =10m//s^(2))` . |
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Answer» For the block to remain stationary `T =Mg sin theta = 100 xx 10 xx sin53` ` =100 xx 10 xx (4)/(5) = 800N` For man `, T -mg =ma` `T =ma (g +a) rArr 800 = 50 (10 +a) a = 6m//s^(2)` . |
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