Question

Four forces F1,F2,F3,and F4 acts outward from the same point F2,F3,F4 makes angles of 60,160 and 240 degree with F1 if F1 =5.0N, F2=7.0N, F3=3.0N and F4=10.0N. determine the resultant forces

Answer 1

The resultant force

$$\bf R=F_1+F_2+F_3+F_4$$

Let $\bf F_1$ is directed toward positive x-axis. Then

$$R_x=F_1+F_2\cos 60^{\circ}+F_3\cos 160^{\circ}+F_4\cos 240^{\circ}\\ R_y=F_2\sin 60^{\circ}+F_3\sin 160^{\circ}+F_4\sin 240^{\circ}$$

$$R_x=5.0+7.0\cos 60^{\circ}+3.0\cos 160^{\circ}+10\cos 240^{\circ}\\ R_y=7.0\sin 60^{\circ}+3.0\sin 160^{\circ}+10\sin 240^{\circ}$$

$$R_x=0.68{\:\rm N}\\ R_y=-1.6\:\rm N$$

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

$$\theta=\tan^{-1}\frac{R_y}{R_x}=\tan^{-1}\frac{-1.6}{0.68}=-67^{\circ}$$