Question

A block of mass $m=2\text{kg}$ is placed on an inclined surface, which is placed inside a truck as shown in the figure. The coefficient of static friction ($\mu$) between the block and the inclined plane is $0.25$. If the truck accelerates at the rate of $2{\text{m/s}}^2$, what should be the maximum angle of the inclined surface so that the block does not slide down the incline? Take $g=10{\text{m/s}}^2$ .

• A. ${\tan }^{-1}\left(0.047\right)$
  
• B. ${\tan }^{-1}\left(7.47\right)$
  
• C. ${\tan }^{-1}\left(0.07\right)$
  
• D. ${\tan }^{-1}\left(0.56\right)$
  

Answer 1

The correct option is A ${\tan }^{-1}\left(0.047\right)$



From the FBD of block, let the truck be accelerated in direction as shown.
$\Rightarrow$ To prevent sliding of block down the incline, it must be in equilibrium from frame of reference of truck.
Hence $F_{\text{pseudo}}=ma$ will act in opposite direction of $a$, static friction ($F_f$) will act up the incline at maximum value.
$\begin{array}{}{F}_f=\mu N......& \text{(1)}\end{array}$
From the FBD, applying equilibrium condition;
Perpendicular to inclined plane:

$\begin{array}{}N+ma\sin \theta =mg\cos \theta ........& \text{(2)}\end{array}$
$\begin{array}{}\therefore F_f=\mu (mg\cos \theta -ma\sin \theta )....& \text{(3)}\end{array}$
Along the inclined plane:
$\begin{array}{}ma\cos \theta +mg\sin \theta =\mu (mg\cos \theta -ma\sin \theta )....& \text{(4)}\end{array}$
$\Rightarrow m\cos \theta (a+g\tan \theta )=\mu m\cos \theta (g-a\tan \theta )$
$\Rightarrow \tan \theta (g+\mu a)=\mu g-a$
$\therefore \tan \theta =\frac{\mu g-a}{g+\mu a}$

Or, $\tan \theta =\frac{(0.25\times 10)-2}{10+(0.25\times 2)}$
$\therefore \theta ={\tan }^{-1}\left(0.047\right)$

Hence obtained value of $\theta$ corresponds to ${\theta }_{\text{max}}$, for which block will not slip.