1. Problem statement: A 5 m uniform wooden beam of weight 300 N is supported at both ends. A toolbox of 200 N is placed at 1.2 m from the left end and another tool of 150 N at 3.5 m from the left end. Determine the reaction forces at the left support A and the right support B using equilibrium. 2. Free-body diagram and notation: Let the left support reaction be $R_A$ and the right support reaction be $R_B$. The beam length is $5\text{ m}$ and the beam weight acts at its center at $2.5\text{ m}$ from the left end. 3. Equations of equilibrium and formula used: Use vertical force balance and moment balance about a convenient point. Vertical equilibrium: $$R_A + R_B = 300 + 200 + 150$$ Moment equilibrium about left support A: $$\sum M_A = 0:\quad R_B\times5 - 300\times2.5 - 200\times1.2 - 150\times3.5 = 0$$ 4. Compute moments to find $R_B$ (show intermediate arithmetic): Compute the moments of the loads about A. Beam weight moment: $$300\times2.5 = 750$$ Toolbox moment: $$200\times1.2 = 240$$ Second tool moment: $$150\times3.5 = 525$$ Sum of moments: $$750 + 240 + 525 = 1515$$ Solve for $R_B$ from the moment equation. $$R_B = \frac{1515}{5} = 303\ \text{N}$$ 5. Find $R_A$ from vertical equilibrium: Total load: $$300 + 200 + 150 = 650$$ From $$R_A + R_B = 650$$ and $R_B = 303$ we get. $$R_A = 650 - 303 = 347\ \text{N}$$ 6. Check and interpretation: Check sum: $$347 + 303 = 650$$ which equals the total applied load so equilibrium is satisfied. Interpretation: The left reaction is larger because the combined loads produce a larger moment about the right support, so more load is taken by the left support. 7. Final answer: Left support reaction: $$R_A = 347\ \text{N}$$ Right support reaction: $$R_B = 303\ \text{N}$$