1. **Problem Statement:** We have a beam AB with a total length of 3 meters, divided into two segments: 2 meters from A to the midpoint, and 1 meter from the midpoint to B. A clockwise moment of 3 KN·m is applied at the midpoint. Point A is a fixed support, and point B is a roller support. 2. **Objective:** Determine the reactions at supports A and B due to the applied moment. 3. **Key Concepts:** - A fixed support can resist vertical and horizontal forces and a moment. - A roller support can resist only vertical forces. - The beam is in static equilibrium, so the sum of forces and moments must be zero. 4. **Equations of Equilibrium:** - Sum of vertical forces: $$\sum F_y = 0$$ - Sum of horizontal forces: $$\sum F_x = 0$$ (no horizontal loads here) - Sum of moments about any point: $$\sum M = 0$$ 5. **Define Reactions:** - At A: vertical reaction $$R_A$$ and moment reaction $$M_A$$ - At B: vertical reaction $$R_B$$ 6. **Apply Equilibrium Equations:** - Vertical forces: $$R_A + R_B = 0$$ (no vertical loads other than reactions) - Moments about A (taking clockwise as positive): $$M_A - 3 + R_B \times 3 = 0$$ 7. **Solve for Reactions:** From vertical forces: $$R_A = -R_B$$ From moments about A: $$M_A - 3 + 3R_B = 0$$ Since A is fixed, it can resist moment, so rearrange: $$M_A = 3 - 3R_B$$ Substitute $$R_A = -R_B$$ into vertical forces: $$-R_B + R_B = 0$$ which is always true. 8. **Interpretation:** The system has infinite solutions for $$R_B$$ and $$M_A$$ satisfying $$M_A = 3 - 3R_B$$ and $$R_A = -R_B$$. 9. **Conclusion:** The fixed support at A provides a moment reaction and vertical force balancing the applied moment and vertical reactions at B. The roller at B provides vertical reaction balancing the system. **Final answer:** $$R_A = -R_B$$ $$M_A = 3 - 3R_B$$ where $$R_B$$ is the vertical reaction at B, which can be determined if additional loads are applied.