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Q.
A block pressed against the vertical wall is in equilibrium. The minimum coefficient of friction is:
Laws of Motion
Solution:
Here net driving force
$=mg-\frac{mg}{2}=\frac{mg}{2}$ down ward
Hence friction will act upward and its magnitude should be $f=\frac{mg}{2}$
If the block ‘m’ is stationary the friction between $m$ and the wall should be static
$f \,\le\,f_{\text{lim}} $
$\frac{mg}{2}\le\,\mu\cdot N$
$\Rightarrow \frac{mg}{2} \le\,\mu \left(mg\right)$
$\Rightarrow \mu=\frac{1}{2}$
