Question Number 209597 by peter frank last updated on 16/Jul/24
Commented by peter frank last updated on 18/Jul/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\ $$
Answered by mr W last updated on 16/Jul/24
$${ma}={mg}\:\mathrm{sin}\:\theta−\mu{mg}\:\mathrm{cos}\:\theta \\ $$$$\frac{{a}}{{g}}=\mathrm{sin}\:\theta−\mu\:\mathrm{cos}\:\theta \\ $$$$\mu=\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\left(\mathrm{sin}\:\theta−\frac{{a}}{{g}}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{60}°}\left(\mathrm{sin}\:\mathrm{60}°−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:=\sqrt{\mathrm{3}}−\mathrm{1}\:\checkmark \\ $$
Answered by Spillover last updated on 16/Jul/24
