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A rough inclined plane is inclined at 30^(@) to the horizontal as shown in the figure. A uniform chain of length L is partly on the inclined plane and partly hanging from the top of the incline. If the coefficient of friction between chain and inclined plane is mu, the maximum length of the hanging part to prevent the chain from falling vertically is |
Answer» Solution : `sigma xg = sigma (L-x)g SIN 30+mu sigma(L-x)g cos 30` `x=(L-x)(1)/(2)+mu(L-x)(sqrt(3))/(2)` `therefore x=((1+sqrt(3)mu)L)/(sqrt(3)(mu+sqrt(3)))` |
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