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

$\tan \theta =\frac{dy}{dx}\frac{d\left(\frac{x^3}{6}\right)}{dx}=\frac{x^2}{2}$
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 $1/6m$.