Question Number 110860 by I want to learn more last updated on 31/Aug/20
Answered by mr W last updated on 31/Aug/20
Commented by mr W last updated on 31/Aug/20
$${F}_{{B}} \mathrm{cos}\:\mathrm{60}°={Mg} \\ $$$$\Rightarrow{F}_{{B}} =\frac{{Mg}}{\mathrm{cos}\:\mathrm{60}°} \\ $$$${F}_{{A}} ={F}_{{B}} \mathrm{sin}\:\mathrm{60}°={Mg}\:\mathrm{tan}\:\mathrm{60}° \\ $$
Commented by I want to learn more last updated on 31/Aug/20
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}. \\ $$$$\mathrm{Should}\:\mathrm{i}\:\mathrm{put}\:\:\:\mathrm{Mg}\:\:=\:\:\mathrm{500N}\:\:??? \\ $$$$\mathrm{And}\:\mathrm{are}\:\mathrm{we}\:\mathrm{not}\:\mathrm{using}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{sir}? \\ $$
Commented by mr W last updated on 31/Aug/20
$${yes} \\ $$
Commented by I want to learn more last updated on 31/Aug/20
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$