Question Number 227085 by Rojarani last updated on 29/Dec/25
Commented by mr W last updated on 30/Dec/25
$${to}\:{wabi}\:{sir}: \\ $$$${please}\:{don}'{t}\:{put}\:{your}\:{new}\:{question} \\ $$$${as}\:{comment}\:{to}\:{an}\:{other}\:{question}! \\ $$$${please}\:{post}\:{it}\:{as}\:{new}\:{question}! \\ $$$${btw},\:{all}\:{options}\:{are}\:{wrong}.\:{the} \\ $$$${correct}\:{answer}\:{should}\:{be} \\ $$$$\theta\leqslant\mathrm{tan}^{−\mathrm{1}} \left(\mu\right) \\ $$
Commented by wabi last updated on 30/Dec/25
Commented by fantastic2 last updated on 30/Dec/25
$${downward}\:{force}={mg}\mathrm{sin}\:\theta \\ $$$${upward}\:{force}=\mu{mg}\mathrm{cos}\:\theta \\ $$$${for}\:{eqilibrium} \\ $$$$\mu{mg}\mathrm{cos}\:\theta\geqslant{mg}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\mathrm{tan}\:\theta\leqslant\mu \\ $$$${for}\:{max}\:{inclination} \\ $$$$\theta=\mathrm{tan}^{−\mathrm{1}} \mu \\ $$