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

Answer: 62.5

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