Question

Find minimum value of ${\mu }_{static}$ for which the block remains at rest on the stationery wedge.

Answer 1

Normal force ($N$) is

$N=mg\cos \theta +12$$=mg\cos {37}^{\circ }+12=1\times 10\times \frac{4}{5}+12=8+12=20N$

frictional force,

$f={\mu }_{s}N$

$f=mg\sin {37}^{\circ }$

${\mu }_{s}N$$=mg\sin {37}^{\circ }$

${\mu }_s\times 20=1\times 10\times \frac{3}{5}$

$\Rightarrow {\mu }_s\times 20=6$

$\Rightarrow {\mu }_s=\frac{6}{20}$$=\frac{3}{10}=0.3$