1. **Stating the problem:** We have five forces acting at different angles from a single point with magnitudes and directions: - $4$ N at $120^\circ$ - $2\sqrt{5}$ N at $90^\circ$ - $4\sqrt{3}$ N at $60^\circ$ - $2$ N at $30^\circ$ - $2$ N at $0^\circ$ We need to find the resultant force vector (magnitude and direction). 2. **Formula and approach:** To find the resultant force, we resolve each force into its horizontal ($x$) and vertical ($y$) components using: $$F_x = F \cos \theta$$ $$F_y = F \sin \theta$$ where $F$ is the magnitude and $\theta$ is the angle. 3. **Calculate components for each force:** - For $4$ N at $120^\circ$: $$F_x = 4 \cos 120^\circ = 4 \times (-0.5) = -2$$ $$F_y = 4 \sin 120^\circ = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}$$ - For $2\sqrt{5}$ N at $90^\circ$: $$F_x = 2\sqrt{5} \cos 90^\circ = 0$$ $$F_y = 2\sqrt{5} \sin 90^\circ = 2\sqrt{5}$$ - For $4\sqrt{3}$ N at $60^\circ$: $$F_x = 4\sqrt{3} \cos 60^\circ = 4\sqrt{3} \times 0.5 = 2\sqrt{3}$$ $$F_y = 4\sqrt{3} \sin 60^\circ = 4\sqrt{3} \times \frac{\sqrt{3}}{2} = 6$$ - For $2$ N at $30^\circ$: $$F_x = 2 \cos 30^\circ = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$$ $$F_y = 2 \sin 30^\circ = 2 \times 0.5 = 1$$ - For $2$ N at $0^\circ$: $$F_x = 2 \cos 0^\circ = 2$$ $$F_y = 2 \sin 0^\circ = 0$$ 4. **Sum all components:** $$R_x = (-2) + 0 + 2\sqrt{3} + \sqrt{3} + 2 = (-2 + 2) + (2\sqrt{3} + \sqrt{3}) = 0 + 3\sqrt{3} = 3\sqrt{3}$$ $$R_y = 2\sqrt{3} + 2\sqrt{5} + 6 + 1 + 0 = 2\sqrt{3} + 2\sqrt{5} + 7$$ 5. **Calculate magnitude of resultant force:** $$R = \sqrt{R_x^2 + R_y^2} = \sqrt{(3\sqrt{3})^2 + (2\sqrt{3} + 2\sqrt{5} + 7)^2}$$ Calculate each term: $$ (3\sqrt{3})^2 = 9 \times 3 = 27 $$ Let $A = 2\sqrt{3} + 2\sqrt{5} + 7$. Calculate $A^2$: $$A^2 = (2\sqrt{3})^2 + (2\sqrt{5})^2 + 7^2 + 2 \times 2\sqrt{3} \times 2\sqrt{5} + 2 \times 2\sqrt{3} \times 7 + 2 \times 2\sqrt{5} \times 7$$ $$= 4 \times 3 + 4 \times 5 + 49 + 8 \sqrt{15} + 28 \sqrt{3} + 28 \sqrt{5}$$ $$= 12 + 20 + 49 + 8 \sqrt{15} + 28 \sqrt{3} + 28 \sqrt{5} = 81 + 8 \sqrt{15} + 28 \sqrt{3} + 28 \sqrt{5}$$ So, $$R = \sqrt{27 + 81 + 8 \sqrt{15} + 28 \sqrt{3} + 28 \sqrt{5}} = \sqrt{108 + 8 \sqrt{15} + 28 \sqrt{3} + 28 \sqrt{5}}$$ 6. **Calculate direction (angle) $\phi$ of resultant force:** $$\phi = \tan^{-1} \left( \frac{R_y}{R_x} \right) = \tan^{-1} \left( \frac{2\sqrt{3} + 2\sqrt{5} + 7}{3\sqrt{3}} \right)$$ This angle is measured from the positive x-axis. **Final answer:** The resultant force has magnitude: $$R = \sqrt{108 + 8 \sqrt{15} + 28 \sqrt{3} + 28 \sqrt{5}}$$ and direction: $$\phi = \tan^{-1} \left( \frac{2\sqrt{3} + 2\sqrt{5} + 7}{3\sqrt{3}} \right)$$ This completes the solution.