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In the above problem, the maximum length of the hanging part to prevent the from sliding down the incline is |
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Answer» Solution :`m_(1)G SIN 30=m_(2)g+mu m_(1)g COS 30` `(L-x)g.(1)/(2)=x g mu(L-x)g sqrt(3)//2` `(L-x)(1)/(2)=x+mu(L-x)(sqrt(3))/(2)` `therefore x=((1-sqrt(3)mu)L)/((3-sqrt(3)mu))` |
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