Answer to Question #283064 in Physics for Hercules

Question

Two forces F1 and F2 act on a particle. F1 has magnitude 6N and in direction 45°, and F2 has magnitude 7N and in direction 154°. Find the magnitude and direction of their resultant

Answer 1

The resultant force

$${\bf F}={\bf F}_1+{\bf F}_2$$

In projections

$$F_x=F_{1x}+F_{2x}=F_1\cos \theta_1+F_2\cos\theta_2\\ F_y=F_{1y}+F_{2y}=F_1\sin \theta_1+F_2\sin\theta_2$$

$$F_x=6\cos 45^\circ+7\cos154^\circ=-2.0\\ F_y=6\sin 45^\circ+7\sin154^\circ=7.3$$

The magnitude of resultant force

$$F=\sqrt{F_x^2+F_y^2}=\sqrt{(-2.0)^2+7.3^2}=7.6$$

The direction of resultant force

$$\theta=\tan^{-1}\frac{F_y}{F_x}=\tan^{-1}\frac{7.3}{-2.0}=105^\circ$$