1. **Problem Statement:** A tractor weighing 8400 N acts vertically downward at point C. It raises a weight of 3600 N acting vertically upward at point D. Points A and B are the wheels on the horizontal line. Distances are: AB = 2 m, BC = 1 m, CD = 1/3 m. We need to find the reaction forces $R_A$ and $R_B$ at wheels A and B in equilibrium. 2. **Formulas and Rules:** - The system is in equilibrium, so the sum of vertical forces is zero: $$\sum F_y = 0$$ - The sum of moments about any point is zero: $$\sum M = 0$$ - Taking moments about point A to eliminate $R_A$ and solve for $R_B$. 3. **Set up equations:** - Sum of vertical forces: $$R_A + R_B = 8400 + 3600 = 12000$$ - Taking moments about A (counterclockwise positive): $$-8400 \times 3 + 3600 \times \left(3 + \frac{1}{3}\right) + R_B \times 2 = 0$$ Here, distances from A: - Point C is at $3$ m (AB + BC = 2 + 1) - Point D is at $3 + \frac{1}{3} = \frac{10}{3}$ m 4. **Calculate moments:** $$-8400 \times 3 + 3600 \times \frac{10}{3} + 2 R_B = 0$$ $$-25200 + 12000 + 2 R_B = 0$$ $$2 R_B = 25200 - 12000 = 13200$$ $$R_B = \frac{13200}{2} = 6600$$ 5. **Find $R_A$:** $$R_A = 12000 - R_B = 12000 - 6600 = 5400$$ 6. **Interpretation:** The calculated reactions are $R_A = 5400$ N and $R_B = 6600$ N, which do not match any of the given options exactly. However, the closest option considering the problem setup is not listed, so the problem might have an error or require rechecking distances. **Final answer:** $$R_A = 5400\text{ N}, \quad R_B = 6600\text{ N}$$