Question Number 209597 by peter frank last updated on 16/Jul/24
![](https://www.tinkutara.com/question/34439.png)
Commented by peter frank last updated on 18/Jul/24
![thank you both](https://www.tinkutara.com/question/Q209667.png)
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\ $$
Answered by mr W last updated on 16/Jul/24
![ma=mg sin θ−μmg cos θ (a/g)=sin θ−μ cos θ μ=(1/(cos θ))(sin θ−(a/g)) =(1/(cos 60°))(sin 60°−(1/2)) =(√3)−1 ✓](https://www.tinkutara.com/question/Q209605.png)
$${ma}={mg}\:\mathrm{sin}\:\theta−\mu{mg}\:\mathrm{cos}\:\theta \\ $$$$\frac{{a}}{{g}}=\mathrm{sin}\:\theta−\mu\:\mathrm{cos}\:\theta \\ $$$$\mu=\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\left(\mathrm{sin}\:\theta−\frac{{a}}{{g}}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{60}°}\left(\mathrm{sin}\:\mathrm{60}°−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:=\sqrt{\mathrm{3}}−\mathrm{1}\:\checkmark \\ $$
Answered by Spillover last updated on 16/Jul/24
![](https://www.tinkutara.com/question/34444.png)