Question

A block rests on a fixed wedge inclined at an angle $\theta$. The coefficient of friction between the block and plane is $\mu$. The maximum value of $\theta$ for the block to remain motionless on the wedge is

• A. $\mu=\tan \theta$
• B. $\mu=\sin \theta$
• C. $\mu=\cos \theta$
• D. $\mu=\cot \theta$

Answer 1

Answer: (A)

The given condition can be shown with a proper $FBD$ of the block, as below

Let us suppose the mass of the block is $m$ and acceleration due to gravity is $g$.
Normal reaction, $N=m g \cos \theta$
Limiting friction, $(f r)_{\lim }=\mu_{s} N=\mu m g \cos \theta$
Net driving force, $(F)_{\text {net driving }}=m g \sin \theta$
If the block remain motionless on the wedge, then
$(F)_{\text {net driving }} \leq(f r)_{\text {lim }}$
$ \Rightarrow m g \sin \theta \leq \mu m g \cos \theta$
$\sin \theta \leq \mu \cos \theta $
$\Rightarrow \tan \theta \leq \mu $
$\Rightarrow \tan \theta_{\max }=\mu$