Q: A block of mass *m* is placed on a surface with a vertical cross-section given by 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

(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