1. **Problem Statement:** Given a force vector $F = 20$ kN with components $F_x$ and $F_y$ such that $\frac{F_x}{4} = \frac{F_y}{3} = \frac{F}{5}$, find the components $F_x$, $F_y$, the resultant force magnitude $F_R$, and the angle $\theta$ it makes with the x-axis. 2. **Formula and Rules:** The force components relate to the total force by the ratios given, which come from the 3-4-5 right triangle relationship. The magnitude of the resultant force is given by the Pythagorean theorem: $$F_R = \sqrt{F_x^2 + F_y^2}$$ The angle $\theta$ is found using the inverse tangent function: $$\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)$$ 3. **Calculate Components:** Given: $$\frac{F_x}{4} = \frac{F_y}{3} = \frac{F}{5}$$ Calculate $F_x$: $$F_x = \frac{4}{5} F = \frac{4}{5} \times 20 = 16$$ Calculate $F_y$: $$F_y = \frac{3}{5} F = \frac{3}{5} \times 20 = 12$$ 4. **Calculate Resultant Force Magnitude:** $$F_R = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20$$ 5. **Calculate Angle $\theta$:** $$\theta = \tan^{-1}\left(\frac{12}{16}\right) = \tan^{-1}\left(0.75\right) \approx 36.87^\circ$$ **Final answers:** - $F_x = 16$ kN - $F_y = 12$ kN - $F_R = 20$ kN - $\theta \approx 36.87^\circ$