1. **Problem Statement:** Calculate the reactions at supports A and B for a simply supported beam with a uniformly distributed load $w=10$ kN/m over 2 m and a point load of 50 kN at support B. 2. **Given:** - Distributed load $w=10$ kN/m over length $L=2$ m - Point load $P=50$ kN at B 3. **Step 1: Calculate total distributed load** $$W = w \times L = 10 \times 2 = 20 \text{ kN}$$ 4. **Step 2: Locate the resultant of distributed load** The resultant acts at the midpoint of the distributed load, which is 1 m from A. 5. **Step 3: Apply equilibrium equations** Sum of vertical forces: $$R_A + R_B = W + P = 20 + 50 = 70$$ Sum of moments about A (taking counterclockwise as positive): $$\sum M_A = 0 = -W \times 1 - P \times 2 + R_B \times 2$$ $$0 = -20 \times 1 - 50 \times 2 + R_B \times 2$$ $$0 = -20 - 100 + 2R_B$$ 6. **Step 4: Solve for $R_B$** $$2R_B = 120$$ $$R_B = \frac{120}{2} = 60 \text{ kN}$$ 7. **Step 5: Solve for $R_A$** $$R_A = 70 - R_B = 70 - 60 = 10 \text{ kN}$$ **Final answer:** $$R_A = 10 \text{ kN}, \quad R_B = 60 \text{ kN}$$