Question

A block of mass m is placed on a surface with a vertical cross-section given by $y=\frac{x^3}{6}$. 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. $\frac{1}{6}m$
• B. $\frac{2}{3}m$
• C. $\frac{1}{3}m$
• D. $\frac{1}{2}m$

Answer 1

The correct option is A

$\frac{1}{6}m$

For equilibrium, $mgsin\theta =\mu mgcos\theta \Rightarrow tan\theta =\mu$
Now, slope $=tan\theta =\frac{dy}{dx}=\frac{d}{dx}\frac{x^3}{6}=\frac{x^2}{2}\Rightarrow \mu =0.5=\frac{x^2}{2}\Rightarrow x=1\therefore y=\frac{1}{6}m$