Question

Find the same results for the resultant, R, and the equilibrant E, with their modules and angles, for the following force system:
F3 = 50 N α = 100 degrees
F4 = 50 N β = 190 degrees
Find R and E. With the respective angles α and β.

Answer 1

$$R_x=50\cos{100}+50\cos{190}=-57.92\ N\\R_y=50\sin{100}+50\sin{190}=40.56\ N$$

$$R=E=\sqrt{57.92^2+40.56^2}=70.7\ N$$

$$\theta_R=90+\arctan{\frac{40.56}{57.92}}=125\degree\\ \theta_E=\theta_R+180=180+125=305\degree$$