Question

A block of mass $m$ is placed on a surface with a vertical cross-section represented by the equation $y=\frac{x^3}{6}$. If the coefficient of static friction is $0.5$, the maximum height above the ground at which the block can be placed without slipping is

• A. $\frac{1}{3}\text{m}$
• B. $\frac{1}{2}\text{m}$
• C. $\frac{1}{6}\text{m}$
• D. $\frac{2}{3}\text{m}$

Answer 1

The correct option is C $\frac{1}{6}\text{m}$

The equation of the surface is given by $y=\frac{x^3}{6}$.
Normal reaction on the block $\left(N\right)$ at a given point, will act $\perp$ to the tangent to the curve at that point.

Static friction $\left(f_s\right)$ will act in a tangential direction as shown in figure, to prevent the slipping of the block triggered by $mg\sin \theta$ component.


For maximum height of block on the surface without slipping, friction $f_s$ will act at its limiting value.
$f_s={\mu }_{s}N...\left(i\right)$

Applying equilibrium condition, along tangential and normal direction,
$f_s=mg\sin \theta ....\left(ii\right)$
$N=mg\cos \theta ...\left(iii\right)$

Hence, from the above equations,
${\mu }_{s}mg\cos \theta =mg\sin \theta$
$\Rightarrow \tan \theta ={\mu }_s...\left(iv\right)$

Slope of the surface represented by the equation $y=\frac{x^3}{6}$ is given by,
$\tan \theta =\frac{dy}{dx}$
Or, ${\mu }_s=\frac{1}{6}\times 3x^2$
$\Rightarrow \frac{0.5\times 6}{3}=x^2$
$\therefore x=\sqrt{1}=1\text{m}$

Now, the maximum height for the block on the surface will correspond to a value of $x=1\text{m}$
$y=\frac{x^3}{6}$
$\therefore y_{\text{max}}=\frac{(1{)}^3}{6}=\frac{1}{6}\text{m}$