Question Number 74418 by peter frank last updated on 24/Nov/19
Answered by ajfour last updated on 24/Nov/19
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{\:^{\overset{{N}} {\nwarrow}} \underset{\underset{{mg}} {\downarrow}} {\mathbb{C}}\rightarrow_{{mv}^{\mathrm{2}} /{R}} } {\diagup} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\diagup \\ $$$$\:\:\:\underset{\left.\centerdot\right)−^{\theta} −} {\diagup} \\ $$$${mg}\mathrm{cos}\:\theta+\frac{{mv}^{\mathrm{2}} }{{R}}\mathrm{sin}\:\theta={N} \\ $$$$\mu{N}+{mg}\mathrm{sin}\:\theta=\frac{{mv}^{\mathrm{2}} }{{R}}\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\mu{g}\mathrm{cos}\:\theta+\frac{\mu{v}^{\mathrm{2}} }{{R}}\mathrm{sin}\:\theta+{g}\mathrm{sin}\:\theta=\frac{{v}^{\mathrm{2}} }{{R}}\mathrm{cos}\:\theta \\ $$$$\:\:\:{v}^{\mathrm{2}} =\frac{{gR}\left(\mu+\mathrm{tan}\:\theta\right)}{\left(\mathrm{1}−\mu\mathrm{tan}\:\theta\right)} \\ $$$$\Rightarrow\:\:\:{v}=\sqrt{\frac{{gR}\left(\mu+\mathrm{tan}\:\theta\right)}{\left(\mathrm{1}−\mu\mathrm{tan}\:\theta\right)}}\:. \\ $$
Commented by $@ty@m123 last updated on 24/Nov/19
$$\:^{\nwarrow} \underset{\downarrow} {\mathbb{C}}\rightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\diagup \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\diagup \\ $$$$\:\:\:\:\:\diagup \\ $$$$\measuredangle\_\theta\_\_\_ \\ $$$${Wow}… \\ $$$${Nice}\:{drawing}\:{using}\:{this}\:{app}. \\ $$$${Quite}\:{surprising}. \\ $$$${Where}\:{there}\:{is}\:{will}\:{there}\:{is}\:{a}\:{way}. \\ $$$$ \\ $$
Commented by peter frank last updated on 24/Nov/19
$${thank}\:{you}\:{sir}\:{ajfour} \\ $$
Commented by peter frank last updated on 07/Dec/19
$${please}\:{sir}\:{help}\:{Qn}\:\:\mathrm{74419} \\ $$