Question

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

Answer 1

At limiting equilibrium,
$m g \sin \theta=\mu m g \cos \theta$
$\Rightarrow \mu=\tan \theta$
But $\tan \theta=\frac{ dy }{ dx }=\frac{ d }{ dx }\left(\frac{ x ^{4}}{16}\right)=\frac{ x ^{3}}{4}$
$\Rightarrow 0.25=\frac{x^{3}}{4}$
$\therefore x ^{3}=1$
$\therefore x=1$
$\therefore y=\frac{1^{4}}{16}=0.625\, m =62.5\, cm$