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

Answer: (A)

$\frac{1}{6}m$

Answer 2

$\tan \theta=\frac{dy}{dx}=\frac{x^{2}}{2}$
At limiting equilibrium,
$\mu=\tan \theta$
$0.5=\frac{x^{2}}{2}$
$\Rightarrow x=\pm 1$
Now, $y=\frac{1}{6}$