Calculate the magnitude and direction of the resultant of the system of forces shown below.

We'll use $\cos \theta$ for the x-component and $\sin \theta$ for the y-component, where $\theta$ is the angle measured counter-clockwise from the positive x-axis.

• Force 1 ($F_1 = 30$ N): $60^\circ$ from negative x-axis towards positive y-axis. $\theta_1 = 180^\circ - 60^\circ = 120^\circ$. $F_{1x} = 30 \cos(120^\circ) = 30(-0.5) = -15$ N $F_{1y} = 30 \sin(120^\circ) = 30(\frac{\sqrt{3}}{2}) = 15\sqrt{3}$ N

• Force 2 ($F_2 = 5$ N): Acting along negative y-axis. $\theta_2 = 270^\circ$. $F_{2x} = 5 \cos(270^\circ) = 0$ N $F_{2y} = 5 \sin(270^\circ) = -5$ N

• Force 3 ($F_3 = 10$ N): Acting along negative y-axis. $\theta_3 = 270^\circ$. $F_{3x} = 10 \cos(270^\circ) = 0$ N $F_{3y} = 10 \sin(270^\circ) = -10$ N

• Force 4 ($F_4 = 12$ N): Acting along positive x-axis. $\theta_4 = 0^\circ$. $F_{4x} = 12 \cos(0^\circ) = 12$ N $F_{4y} = 12 \sin(0^\circ) = 0$ N

• Force 5 ($F_5 = 20$ N): $45^\circ$ from positive x-axis towards positive y-axis. $\theta_5 = 45^\circ$. $F_{5x} = 20 \cos(45^\circ) = 20(\frac{\sqrt{2}}{2}) = 10\sqrt{2}$ N $F_{5y} = 20 \sin(45^\circ) = 20(\frac{\sqrt{2}}{2}) = 10\sqrt{2}$ N

• Force 6 ($F_6 = 10$ N): $60^\circ$ from positive x-axis towards negative y-axis. $\theta_6 = 360^\circ - 60^\circ = 300^\circ$. $F_{6x} = 10 \cos(300^\circ) = 10(0.5) = 5$ N $F_{6y} = 10 \sin(300^\circ) = 10(-\frac{\sqrt{3}}{2}) = -5\sqrt{3}$ N

• Force 7 ($F_7 = 40$ N): $30^\circ$ from negative x-axis towards positive y-axis. $\theta_7 = 180^\circ - 30^\circ = 150^\circ$. $F_{7x} = 40 \cos(150^\circ) = 40(-\frac{\sqrt{3}}{2}) = -20\sqrt{3}$ N $F_{7y} = 40 \sin(150^\circ) = 40(0.5) = 20$ N

• Force 8 ($F_8 = 50$ N): Acting along positive x-axis. $\theta_8 = 0^\circ$. $F_{8x} = 50 \cos(0^\circ) = 50$ N $F_{8y} = 50 \sin(0^\circ) = 0$ N

• Force 9 ($F_9 = 10$ N): $60^\circ$ from negative y-axis towards negative x-axis. $\theta_9 = 270^\circ - 60^\circ = 210^\circ$. $F_{9x} = 10 \cos(210^\circ) = 10(-\frac{\sqrt{3}}{2}) = -5\sqrt{3}$ N $F_{9y} = 10 \sin(210^\circ) = 10(-0.5) = -5$ N

$$R_x = -15 + 0 + 0 + 12 + 10\sqrt{2} + 5 - 20\sqrt{3} + 50 - 5\sqrt{3}$$ $$R_x = (-15 + 12 + 5 + 50) + 10\sqrt{2} + (-20\sqrt{3} - 5\sqrt{3})$$ $$R_x = 52 + 10\sqrt{2} - 25\sqrt{3}$$ $$R_x \approx 52 + 10(1.4142) - 25(1.7321)$$ $$R_x \approx 52 + 14.142 - 43.3025 \approx 22.8395 \text{ N}$$

$$R_y = 15\sqrt{3} - 5 - 10 + 0 + 10\sqrt{2} - 5\sqrt{3} + 20 + 0 - 5$$ $$R_y = (15\sqrt{3} - 5\sqrt{3}) + 10\sqrt{2} + (-5 - 10 + 20 - 5)$$ $$R_y = 10\sqrt{3} + 10\sqrt{2} + 0$$ $$R_y = 10\sqrt{3} + 10\sqrt{2}$$ $$R_y \approx 10(1.7321) + 10(1.4142)$$ $$R_y \approx 17.321 + 14.142 \approx 31.463 \text{ N}$$

$$R = \sqrt{R_x^2 + R_y^2}$$ $$R = \sqrt{(22.8395)^2 + (31.463)^2}$$ $$R = \sqrt{521.64 + 989.82}$$ $$R = \sqrt{1511.46}$$ $$R \approx 38.88 \text{ N}$$

$$\tan \alpha = \frac{R_y}{R_x}$$ $$\tan \alpha = \frac{31.463}{22.8395}$$ $$\tan \alpha \approx 1.3775$$ $$\alpha = \arctan(1.3775)$$ $$\alpha \approx 54.0^\circ$$

Since $R_x$ is positive and $R_y$ is positive, the resultant force is in the first quadrant, so the angle is measured counter-clockwise from the positive x-axis.

The magnitude of the resultant force is $\boxed{38.88 \text{ N}}$ and its direction is $\boxed{54.0^\circ}$ counter-clockwise from the positive x-axis.