Question

A block of mass $m$ is placed on a surface having coefficient of friction $0.25,$ with a vertical cross-section given by $y=\frac{x^{4}}{16}.$ The maximum height above the ground (in $cm$ ) at which block can be placed without slipping is $\alpha $ . Write value of $2\alpha $ .

Answer 1

At limiting equilibrium, $mgsin \theta=\mu mg \cos \theta$
$\Rightarrow \mu=\tan \theta$
But $\tan \theta=\frac{ d y}{ dx }=\frac{ d }{ dx } \frac{ x ^4}{16}=\frac{ x ^3}{4}$
$\Rightarrow 0.25=\frac{x^3}{4} $
$\therefore x^3=1 $
$\therefore x=1 $
$\therefore y=\frac{1^4}{16}=0.625\, m =62.5\, cm =\alpha$
$2 \alpha=125$