1. **Problem Statement:** A 5 m uniform wooden beam weighing 300 N is supported at both ends. A toolbox of 200 N is placed 1.2 m from the left end, and another tool of 150 N is placed 3.5 m from the left end. Find the reaction forces at both supports using equilibrium conditions. 2. **Formula and Rules:** For a beam in equilibrium, the sum of vertical forces and the sum of moments about any point must be zero. - Sum of vertical forces: $$\sum F_y = 0$$ - Sum of moments about a point (e.g., left support): $$\sum M = 0$$ 3. **Define Variables:** - Let reaction at left support be $R_A$ and at right support be $R_B$. - Beam length $L = 5$ m. - Weight of beam $W_b = 300$ N acting at midpoint (2.5 m from left). - Toolbox weight $W_1 = 200$ N at 1.2 m. - Tool weight $W_2 = 150$ N at 3.5 m. 4. **Sum of vertical forces:** $$R_A + R_B - W_b - W_1 - W_2 = 0$$ $$R_A + R_B = 300 + 200 + 150 = 650 \text{ N}$$ 5. **Sum of moments about left support (taking counterclockwise as positive):** $$R_B \times 5 - 300 \times 2.5 - 200 \times 1.2 - 150 \times 3.5 = 0$$ Calculate moments: $$R_B \times 5 - 750 - 240 - 525 = 0$$ $$5R_B = 750 + 240 + 525 = 1515$$ $$R_B = \frac{1515}{5} = 303 \text{ N}$$ 6. **Calculate $R_A$:** $$R_A = 650 - R_B = 650 - 303 = 347 \text{ N}$$ 7. **Interpretation:** - The left support reaction force is 347 N. - The right support reaction force is 303 N. These forces balance the weights and keep the beam in equilibrium.