Question

An inclined plane is bent in such a way that the vertical cross-section is given by $y=\frac{x^2}{4}$ where $y$ is in vertical and $x$ is in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction $\mu =0.5$, the maximum height in $\text{cm}$ at which a stationary block will not slip downward is _____ $\text{cm}$.

Answer 1

Given:
$y=\frac{x^2}{4}$
$\mu =0.5$

The condition for which the block will not slip downward is -
$mg\sin \theta =\mu mg\cos \theta$

$\Rightarrow \tan \theta =\mu$

$\Rightarrow \frac{dy}{dx}=\mu$

$\Rightarrow \frac{x}{2}=0.5$

$\Rightarrow x=1\text{m}$

Now, $y=\frac{x^2}{4}=\frac{1^2}{4}=0.25\text{m}=25\text{cm}$