Question

A block of mass m is placed on a suface with a vertical cross section given by $y=\frac{x^3}{6}.$ If the coefficient of 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 $\frac{1}{6}\text{m}$


$y=\frac{x^3}{6}$
Slope $\frac{dy}{dx}=\frac{3x^2}{6}=\frac{x^2}{2}=tan\theta$
At equilibrium the frictional force will be balanced by the component of gravity acting down the incline and tangent to the point of the curve.
$f=\mu mgcos\theta =mgsin\theta$
$\mu =tan\theta$
$0.5=\frac{x^2}{2}\Rightarrow x=1$
$y=\frac{x^3}{6}=\frac{1}{6}m$