1. **Problem Statement:** A 12 m long beam with negligible weight has a 13 kN crate at point C, which is 2 m from the right end (point B). We need to find the reaction forces at point A. 2. **Setup and Known Values:** - Length of beam, $L = 12$ m - Load at point C, $F = 13$ kN - Distance from point C to point B, $d_C = 2$ m - Distance from point C to point A, $L - d_C = 12 - 2 = 10$ m 3. **Assumptions:** - Beam is simply supported at points A and B. - Weight of beam is negligible. - Reaction forces at A and B are vertical and denoted as $R_A$ and $R_B$ respectively. 4. **Equilibrium Equations:** - Sum of vertical forces: $$R_A + R_B = 13$$ - Taking moments about point A (counterclockwise positive): $$R_B \times 12 = 13 \times 10$$ 5. **Calculate $R_B$:** $$R_B = \frac{13 \times 10}{12} = \frac{130}{12} = 10.8333$$ kN 6. **Calculate $R_A$ using vertical forces sum:** $$R_A = 13 - R_B = 13 - 10.8333 = 2.1667$$ kN 7. **Round to tenths:** $$R_A = 2.2 \text{ kN}$$ **Final answer:** The reaction force at point A is $\boxed{2.2}$ kN.