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

Answer: 125

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