Question Number 178877 by Tawa11 last updated on 22/Oct/22
Two forces F1 and F2 of magnitude 3 and 4 Newtons are inclined at angles 150° and 60° to the positive x-axis respectively . Find the resultant force?
Answered by ARUNG_Brandon_MBU last updated on 22/Oct/22
$$\Sigma{F}_{{x}} :\:{F}_{\mathrm{2}} \mathrm{cos60}°−{F}_{\mathrm{1}} \mathrm{cos30}°=\mathrm{2}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Sigma{F}_{\mathrm{y}} :\:{F}_{\mathrm{2}} \mathrm{sin60}°+{F}_{\mathrm{1}} \mathrm{sin30}°=\mathrm{2}\sqrt{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${F}=\sqrt{{F}_{{x}} ^{\mathrm{2}} +{F}_{\mathrm{y}} ^{\mathrm{2}} } \\ $$$$\:\:\:=\sqrt{\mathrm{4}−\mathrm{6}\sqrt{\mathrm{3}}+\frac{\mathrm{27}}{\mathrm{4}}+\mathrm{12}+\mathrm{6}\sqrt{\mathrm{3}}+\frac{\mathrm{9}}{\mathrm{4}}} \\ $$$$\:\:\:=\sqrt{\mathrm{16}+\mathrm{9}}=\mathrm{5}\:\mathrm{units} \\ $$$$\mathrm{tan}\theta=\frac{{F}_{\mathrm{y}} }{{F}_{{x}} }=\frac{\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{3}}{\mathrm{4}−\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\left(\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{3}\right)\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{16}−\mathrm{27}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{12}+\mathrm{25}\sqrt{\mathrm{3}}+\mathrm{108}}{−\mathrm{11}}=−\frac{\mathrm{120}+\mathrm{25}\sqrt{\mathrm{3}}}{\mathrm{11}} \\ $$$$\theta=\mathrm{180}°−\mathrm{arctan}\left(\frac{\mathrm{120}+\mathrm{25}\sqrt{\mathrm{3}}}{\mathrm{11}}\right) \\ $$$$\mathrm{Resultant}\:\mathrm{force}\:\mathrm{is}\:\mathrm{of}\:\mathrm{magnitude}\:\mathrm{5}\:\mathrm{units}\:\mathrm{inclined} \\ $$$$\mathrm{at}\:\mathrm{angle}\:\theta\:\mathrm{from}\:\mathrm{positive}\:{x}-\mathrm{axis}. \\ $$
Commented by Tawa11 last updated on 22/Oct/22
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$