1. **Problem Statement:** We have three forces acting on a structure: $F_1 = 4$ kN directed along the positive x-axis, $F_3 = 5$ kN directed downward along the negative y-axis, and $F_2$ acting diagonally downward to the left at an angle $\theta$ from the horizontal axis. 2. **Goal:** To analyze the forces, we often want to find the components of $F_2$ along the x and y axes and possibly find the resultant force or equilibrium conditions. 3. **Formulas and Rules:** - The components of a force $F$ at angle $\theta$ from the horizontal are: $$F_x = F \cos\theta$$ $$F_y = F \sin\theta$$ - Since $F_2$ acts downward to the left, its x-component is negative and y-component is negative: $$F_{2x} = -F_2 \cos\theta$$ $$F_{2y} = -F_2 \sin\theta$$ 4. **Force Components:** - $F_1$ components: $F_{1x} = 4$ kN, $F_{1y} = 0$ - $F_3$ components: $F_{3x} = 0$, $F_{3y} = -5$ kN - $F_2$ components as above. 5. **Equilibrium Conditions:** For the structure to be in equilibrium, $$\sum F_x = F_{1x} + F_{2x} + F_{3x} = 0$$ $$\sum F_y = F_{1y} + F_{2y} + F_{3y} = 0$$ 6. **Setting up equations:** $$4 - F_2 \cos\theta + 0 = 0 \implies F_2 \cos\theta = 4$$ $$0 - F_2 \sin\theta - 5 = 0 \implies F_2 \sin\theta = -5$$ 7. **Finding $F_2$ and $\theta$:** - Square and add both equations: $$ (F_2 \cos\theta)^2 + (F_2 \sin\theta)^2 = 4^2 + (-5)^2 = 16 + 25 = 41$$ - Since $\cos^2\theta + \sin^2\theta = 1$, we get: $$F_2^2 = 41 \implies F_2 = \sqrt{41} \approx 6.4 \text{ kN}$$ 8. **Calculate $\theta$:** $$\tan\theta = \frac{F_2 \sin\theta}{F_2 \cos\theta} = \frac{-5}{4} = -1.25$$ $$\theta = \arctan(-1.25) \approx -51.34^\circ$$ Since $F_2$ acts downward to the left, the angle $\theta$ is about $51.34^\circ$ below the negative x-axis. **Final answers:** - $F_2 \approx 6.4$ kN - $\theta \approx 51.34^\circ$ downward from the horizontal axis to the left