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)

A block of mass $m$ is placed on a surface with a vertical cross-section, then

$\tan \theta=\frac{d y}{d x} \frac{d\left(\frac{x^{3}}{6}\right)}{d x}=\frac{x^{2}}{2}$
At limiting equilibrium, we get
$\mu =\tan \theta$
$\Rightarrow 0.5 =\frac{x^{2}}{2}$
$\Rightarrow x^{2} =1 $
$ x =\pm 1$
Now, putting the value of $x$ in $y=\frac{x^{3}}{6}$, we get
When $ x=1 $
$y=\frac{(1)^{3}}{6}=\frac{1}{6} $
When $x = -1$
$ y=\frac{(-1)^{3}}{6}=\frac{-1}{6}$
So, the maximum height above the ground at which the block can be placed without slipping is $1 / 6 m$.