1. **Problem Statement:** We have a tractor weighing 8400N acting vertically downward at point C, and an additional weight of 3600N acting vertically downward at point D. The tractor rests on two wheels at points A (front) and B (rear). We need to find the reaction forces $R_A$ and $R_B$ at the wheels in equilibrium. 2. **Given Data:** - Weight of tractor, $W_T = 8400N$ at point C - Additional weight, $W_D = 3600N$ at point D - Distance between wheels A and B, $AB = 2m$ - Distance from B to D, $BD = 1m$ - Distance from D to vertical line through C, $DC = \frac{1}{3}m$ 3. **Assumptions and Approach:** - The system is in static equilibrium. - Sum of vertical forces equals zero. - Sum of moments about any point equals zero. 4. **Step 1: Define reaction forces at wheels:** Let $R_A$ be the reaction force at wheel A and $R_B$ at wheel B. 5. **Step 2: Sum of vertical forces:** $$R_A + R_B = W_T + W_D = 8400 + 3600 = 12000N$$ 6. **Step 3: Calculate moments about point A to find $R_B$:** Taking moments about A (counterclockwise positive): - Moment due to $R_B$ at distance $AB = 2m$ is $R_B \times 2$ - Moment due to tractor weight $W_T$ at distance from A to C: Distance $AC = AB - BC$, but BC is not given directly. Since $D$ is 1m from B and $D$ is $\frac{1}{3}m$ from C, then $BC = BD + DC = 1 + \frac{1}{3} = \frac{4}{3}m$ So, $AC = 2 - \frac{4}{3} = \frac{2}{3}m$ - Moment due to $W_T$ is $8400 \times \frac{2}{3} = 5600 Nm$ (clockwise, so negative) - Moment due to $W_D$ at distance from A: Distance $AD = AB - BD = 2 - 1 = 1m$ Moment due to $W_D$ is $3600 \times 1 = 3600 Nm$ (clockwise, so negative) Set sum of moments about A to zero: $$R_B \times 2 - 5600 - 3600 = 0$$ $$2R_B = 9200$$ $$R_B = 4600N$$ 7. **Step 4: Calculate $R_A$ using vertical force equilibrium:** $$R_A + 4600 = 12000$$ $$R_A = 12000 - 4600 = 7400N$$ 8. **Step 5: Check options:** None of the options exactly match $R_A = 7400N$ and $R_B = 4600N$. Let's verify the moment arm calculations carefully. 9. **Re-examining distances:** - Given $BD = 1m$ - Given $DC = \frac{1}{3}m$ - So $BC = BD + DC = 1 + \frac{1}{3} = \frac{4}{3}m$ - Distance $AC = AB - BC = 2 - \frac{4}{3} = \frac{2}{3}m$ 10. **Moments about B:** Alternatively, take moments about B to find $R_A$: - Moment due to $R_A$ at distance $AB = 2m$ (counterclockwise positive) - Moment due to $W_T$ at distance $BC = \frac{4}{3}m$ (clockwise negative) - Moment due to $W_D$ at distance $BD = 1m$ (clockwise negative) Sum moments about B: $$R_A \times 2 - 8400 \times \frac{4}{3} - 3600 \times 1 = 0$$ $$2R_A = 11200 + 3600 = 14800$$ $$R_A = 7400N$$ Then, $$R_B = 12000 - 7400 = 4600N$$ 11. **Conclusion:** The calculated reactions are $R_A = 7400N$ and $R_B = 4600N$, which do not match any given options exactly. However, the closest option by magnitude is (a) $R_A = 9400N$, $R_B = 2600N$ or (b) $R_A = 2600N$, $R_B = 9400N$. Since the problem likely expects the sum of reactions to equal total weight, and the moments to balance, the correct answer based on calculations is: $$R_A = 7400N, \quad R_B = 4600N$$ **Final answer:** $$R_A = 7400N, \quad R_B = 4600N$$ This does not match any provided options exactly, so please verify the problem data or options.