Answer in Physics for Hennest #226819
January 24th, 2023
The resultant force
“bf R=F_1+F_2+F_3+F_4”
Let “bf F_1” is directed toward positive x-axis. Then
“R_x=F_1+F_2cos 60^{circ}+F_3cos 160^{circ}+F_4cos 240^{circ}\nR_y=F_2sin 60^{circ}+F_3sin 160^{circ}+F_4sin 240^{circ}”
“R_x=5.0+7.0cos 60^{circ}+3.0cos 160^{circ}+10cos 240^{circ}\nR_y=7.0sin 60^{circ}+3.0sin 160^{circ}+10sin 240^{circ}”
“R_x=0.68{:rm N}\nR_y=-1.6:rm N”
The magnitude of resultant force
“R=sqrt{R_x^2+R_y^2}=sqrt{0.68^2+(-1.6)^2}=1.7:rm N”
The direction of resultant force
“theta=tan^{-1}frac{R_y}{R_x}=tan^{-1}frac{-1.6}{0.68}=-67^{circ}”
