Question and Answers Forum

All Question   Topic List

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?\\n

Answered by ARUNG_Brandon_MBU last updated on 22/Oct/22

ΣF_x : F_2 cos60°−F_1 cos30°=2−((3(√3))/2) ΣF_y : F_2 sin60°+F_1 sin30°=2(√3)+(3/2) F=(√(F_x ^2 +F_y ^2 )) =(√(4−6(√3)+((27)/4)+12+6(√3)+(9/4))) =(√(16+9))=5 units tanθ=(F_y /F_x )=((4(√3)+3)/(4−3(√3)))=(((4(√3)+3)(4+3(√3)))/(16−27)) =((12+25(√3)+108)/(−11))=−((120+25(√3))/(11)) θ=180°−arctan(((120+25(√3))/(11))) Resultant force is of magnitude 5 units inclined at angle θ from positive x-axis.

$$\Sigma{F}_{{x}} :\:{F}_{\mathrm{2}} \mathrm{cos60}°−{F}_{\mathrm{1}} \mathrm{cos30}°=\mathrm{2}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$ $$\Sigma{F}_{\mathrm{y}} :\:{F}_{\mathrm{2}} \mathrm{sin60}°+{F}_{\mathrm{1}} \mathrm{sin30}°=\mathrm{2}\sqrt{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $${F}=\sqrt{{F}_{{x}} ^{\mathrm{2}} +{F}_{\mathrm{y}} ^{\mathrm{2}} } \\ $$ $$\:\:\:=\sqrt{\mathrm{4}−\mathrm{6}\sqrt{\mathrm{3}}+\frac{\mathrm{27}}{\mathrm{4}}+\mathrm{12}+\mathrm{6}\sqrt{\mathrm{3}}+\frac{\mathrm{9}}{\mathrm{4}}} \\ $$ $$\:\:\:=\sqrt{\mathrm{16}+\mathrm{9}}=\mathrm{5}\:\mathrm{units} \\ $$ $$\mathrm{tan}\theta=\frac{{F}_{\mathrm{y}} }{{F}_{{x}} }=\frac{\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{3}}{\mathrm{4}−\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\left(\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{3}\right)\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{16}−\mathrm{27}} \\ $$ $$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{12}+\mathrm{25}\sqrt{\mathrm{3}}+\mathrm{108}}{−\mathrm{11}}=−\frac{\mathrm{120}+\mathrm{25}\sqrt{\mathrm{3}}}{\mathrm{11}} \\ $$ $$\theta=\mathrm{180}°−\mathrm{arctan}\left(\frac{\mathrm{120}+\mathrm{25}\sqrt{\mathrm{3}}}{\mathrm{11}}\right) \\ $$ $$\mathrm{Resultant}\:\mathrm{force}\:\mathrm{is}\:\mathrm{of}\:\mathrm{magnitude}\:\mathrm{5}\:\mathrm{units}\:\mathrm{inclined} \\ $$ $$\mathrm{at}\:\mathrm{angle}\:\theta\:\mathrm{from}\:\mathrm{positive}\:{x}-\mathrm{axis}. \\ $$

Commented byTawa11 last updated on 22/Oct/22

God bless you sir. I appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com