1. **Problem Statement:** Find the magnitude and position of the resultant force for the beam with loads: 10 kN/m uniform distributed load, point loads 40 kN and 60 kN downward, and reactions 30 kN and 20 kN upward, with distances 2, 1, 2, 3, 5 units along the beam. 2. **Formula and Rules:** - The resultant force of distributed loads is the total load (intensity \times length). - The position of the resultant of a distributed load is at its centroid (for uniform load, at the midpoint). - Sum all vertical forces to find the total resultant force. - Use moments about a reference point to find the position of the resultant force. 3. **Calculate resultant of distributed load:** $$F_{udl} = 10 \times (2+1+2) = 10 \times 5 = 50\ \text{kN}$$ Position of this load is at the midpoint of the 5 units segment: $$x_{udl} = 2 + \frac{1+2}{2} = 2 + 1.5 = 3.5\ \text{units}$$ 4. **Sum of point loads:** $$F_{point} = 40 + 60 = 100\ \text{kN}$$ Positions given as distances along the beam: 2, 1, 2, 3, 5 units. Assuming 40 kN at 2 units and 60 kN at 3 units (from left end). 5. **Sum of reactions:** $$F_{reaction} = 30 + 20 = 50\ \text{kN}$$ 6. **Total resultant force:** $$R = F_{udl} + F_{point} - F_{reaction} = 50 + 100 - 50 = 100\ \text{kN}$$ 7. **Calculate moment about left end (taking counterclockwise positive):** $$M = (50 \times 3.5) + (40 \times 2) + (60 \times 3) - (30 \times 0) - (20 \times 10)$$ $$M = 175 + 80 + 180 - 0 - 200 = 235\ \text{kN}\cdot\text{units}$$ 8. **Position of resultant force from left end:** $$x_R = \frac{M}{R} = \frac{235}{100} = 2.35\ \text{units}$$ **Final answer:** The magnitude of the resultant force is **100 kN** acting downward. The position of the resultant force is **2.35 units** from the left end of the beam.