A block of mass `'m'` is placed on a rough surface with a vertical cross section of `y =(x^(3))/(6)`. If the coefficient of friction is `0.5`, the maximum height above the ground at which the block can be placed without slipping is .

Explanation

`"Tan" theta = (dy)/(dx) =(d)/(dx) ((x^(3))/(6)) rArr Tantheta = (x^(2))/(2)`
At limiting equilibrium, we get `mu =T antheta rArr 0.5 =(x^(2))/(2)`
`x^(2) =1 rArr x = pm 1`
Now putting the values of 'x' in `y = (x^(3))/(6)` we get
When `x =1 rArr y = (1)/(6) , x = -1 rArr y = - (1)/(6)`
So the maximum height above the ground at which the block can be placed without slipping is `y =(1)/(6)m` .