Question

An inclined plane is bent in such a way that the vertical cross-section is given by $y=\raisebox{1ex}{$x^2$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ where $y$ is in vertical and $x$ in horizontal direction. If the upper surface of this curved plane is rough with a coefficient of friction $\mu =0.5$, the maximum height in cm at which a stationary block will not slip downward is _____ $cm$.

Answer 1

Step1: Given data:

Vertical cross-section, $y=\raisebox{1ex}{$x^2$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

Coefficient of friction $\mu =0.5$

Assuming $x$ and $y$ in the equation is in meter.

Let the maximum height at which a stationary block will not slip downward $=hcm$

Step2: Finding the maximum height in cm at which a stationary block will not slip downward.

Since, at maximum height block will experience a maximum friction force.

Therefore,

At this height slope of the tangent as shown in figure $\left(a\right)$.

$\tan \left(\theta \right)=\frac{dy}{dx}$

Where $\theta$ is angle repose.

$\Rightarrow$$\tan \left(\theta \right)=\frac{d{\displaystyle \raisebox{1ex}{$x^2$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{dx}$

$\Rightarrow$$\tan \left(\theta \right)=\frac{2x}{4}=\frac{x}{2}$ ….$\left(i\right)$

Now,

From figure $\left(b\right)$

$mg\sin \left(\theta \right)=\mu mg\cos \left(\theta \right)$

$\Rightarrow$$\tan \left(\theta \right)=\mu$

$\Rightarrow$$\frac{x}{2}=\mu$ $\left[Fromequation\left(i\right)\right]$

$\Rightarrow$$x=2\times 0.5$

$\Rightarrow$$x=1m$ ….$\left(ii\right)$

And,

$\Rightarrow$ $y=\raisebox{1ex}{$x^2$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

$\Rightarrow$$y=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

$\Rightarrow$$y=0.25m=25cm$

Hence, the maximum height in cm at which a stationary block will not slip downward is $\mathbf{25}$ cm.