Question

In the figure shown, find the angle $\theta$ for which the pulling force is $\mu mg$ where $\mu$ is the coefficient of friction between the two surfaces

• A. ${\tan }^{-1}\left(\mu \right)$
• B. $2{\tan }^{-1}\left(\mu \right)$
• C. ${\cot }^{-1}\left(\mu \right)$
• D. $2{\cot }^{-1}\left(\mu \right)$

Answer 1

The correct option is B $2{\tan }^{-1}\left(\mu \right)$

FBD of the block can be drawn as

From the equation of force we get
$N=mg-F\sin \theta$
$F\cos \theta =f=\mu N$
solving above two equations we get,
Pulling force $\left(F\right)=\frac{\mu mg}{\cos \theta +\mu \sin \theta }$
For it to be equal to $\mu mg$
$\Rightarrow \cos \theta +\mu \sin \theta =1$
$\Rightarrow \mu \sin \theta =1-\cos \theta$
from trigonometry, we know that
$[\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2};1-\cos \theta =2{\sin }^2\frac{\theta }{2}]$
$\therefore$ we have,
$2\mu \sin \frac{\theta }{2}\cos \frac{\theta }{2}=2{\sin }^2\frac{\theta }{2}$
$\Rightarrow \mu \cos \frac{\theta }{2}=\sin \frac{\theta }{2}$
$\Rightarrow \tan \frac{\theta }{2}=\mu$
$\Rightarrow \theta =2{\tan }^{-1}\left(\mu \right)$