Question Number 99982 by Dwaipayan Shikari last updated on 24/Jun/20
Commented by mr W last updated on 24/Jun/20
$${a}=\frac{{g}}{\mu}={g}\:\mathrm{tan}\:\theta \\ $$$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mu}=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \mu \\ $$
Commented by Dwaipayan Shikari last updated on 24/Jun/20
$${f}=\mu{ma} \\ $$$${f}={mg} \\ $$$$\frac{{a}}{{g}}={tan}\theta\:\:,\:\:\:{a}=\frac{{g}}{\mu}\:\:,{tan}\theta=\frac{\mathrm{1}}{\mu}\:\:\:\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mu}\:\:\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$