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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
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Verified Answer
The correct answer is:
$\mu=\tan \theta$
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
$\begin{gathered}
(F)_{\text {net driving }} \leq(f r)_{\lim } \Rightarrow m g \sin \theta \leq \mu m g \cos \theta \\
\sin \theta \leq \mu \cos \theta \Rightarrow \tan \theta \leq \mu \Rightarrow \tan \theta_{\max }=\mu
\end{gathered}$
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
$\begin{gathered}
(F)_{\text {net driving }} \leq(f r)_{\lim } \Rightarrow m g \sin \theta \leq \mu m g \cos \theta \\
\sin \theta \leq \mu \cos \theta \Rightarrow \tan \theta \leq \mu \Rightarrow \tan \theta_{\max }=\mu
\end{gathered}$
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