In the above problem, the maximum length of the hanging part to prevent the chain from sliding down the incline is

In the above problem, the maximum length of the hanging part to preven

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