Q: A block of mass m is placed on a surface with a vertical cross-section given by y = x3/6. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is
(a) 1/6 m
(b) 2/3 m
(c) 1/3 m
(d) 1/2 m
Ans: (a)
Solution: $ \displaystyle y = \frac{x^3}{6} $
$\displaystyle \frac{dy}{dx} = \frac{3x^2}{6} = \frac{x^2}{2} $
$\displaystyle tan\theta = \frac{x^2}{2} $
$ \displaystyle \mu = \frac{x^2}{2} $
$ \displaystyle 0.5 = \frac{x^2}{2} $
x = 1
Hence y = 1/6 m