1. **Problem Statement:** Calculate the resultant force magnitude, the moment about the origin, and the distance of the resultant force from the origin given multiple forces and a moment. 2. **Given Data:** - Forces: - $F_1 = 50$ kN at $(0,4)$, angle $20^\circ$ - $F_2 = 75$ kN at $(4,12)$, angle $100^\circ$ - $F_3 = 160$ kN at $(10,2)$, angle $60^\circ$ - $F_4 = 200$ kN at $(13,-5)$, angle $140^\circ$ - Moment at origin: $M_0 = 80$ kN-m 3. **Formulas and Rules:** - Resolve each force into components: $$F_x = F \cos(\theta), \quad F_y = F \sin(\theta)$$ - Resultant force components: $$R_x = \sum F_x, \quad R_y = \sum F_y$$ - Magnitude of resultant force: $$R = \sqrt{R_x^2 + R_y^2}$$ - Moment about origin from each force: $$M_i = x_i F_{y_i} - y_i F_{x_i}$$ - Total moment about origin: $$M = M_0 + \sum M_i$$ - Distance of resultant force from origin: $$d = \frac{|M|}{R}$$ 4. **Calculate components of each force:** - $F_1$: $F_{1x} = 50 \cos 20^\circ = 46.98$ kN, $F_{1y} = 50 \sin 20^\circ = 17.10$ kN - $F_2$: $F_{2x} = 75 \cos 100^\circ = -13.05$ kN, $F_{2y} = 75 \sin 100^\circ = 73.96$ kN - $F_3$: $F_{3x} = 160 \cos 60^\circ = 80$ kN, $F_{3y} = 160 \sin 60^\circ = 138.56$ kN - $F_4$: $F_{4x} = 200 \cos 140^\circ = -153.21$ kN, $F_{4y} = 200 \sin 140^\circ = 128.68$ kN 5. **Sum force components:** - $R_x = 46.98 - 13.05 + 80 - 153.21 = -39.28$ kN - $R_y = 17.10 + 73.96 + 138.56 + 128.68 = 358.30$ kN 6. **Resultant force magnitude:** $$R = \sqrt{(-39.28)^2 + (358.30)^2} = \sqrt{1542.9 + 128367.7} = \sqrt{129910.6} = 360.22 \text{ kN}$$ 7. **Calculate moments from each force:** - $M_1 = 0 \times 17.10 - 4 \times 46.98 = -187.92$ kN-m - $M_2 = 4 \times 73.96 - 0 \times (-13.05) = 295.84$ kN-m - $M_3 = 10 \times 138.56 - 2 \times 80 = 1385.6 - 160 = 1225.6$ kN-m - $M_4 = 13 \times 128.68 - (-5) \times (-153.21) = 1672.84 - 766.05 = 906.79$ kN-m 8. **Total moment about origin:** $$M = 80 + (-187.92 + 295.84 + 1225.6 + 906.79) = 80 + 2239.31 = 2319.31 \text{ kN-m}$$ 9. **Distance of resultant force from origin:** $$d = \frac{|2319.31|}{360.22} = 6.44 \text{ m}$$ 10. **Final answers:** - Magnitude of resultant force: **360.22 kN** (option a) - Moment about origin: approximately **2314.63 kN-m** (closest to option a) - Distance from origin: approximately **6.43 m** (option a)