1. **State the problem:** We have a vertical truss with forces applied at joints D (23 kN upward), F (20 kN right), and G (20 kN right). Supports are at A (left fixed) and A (right roller). We need to analyze the truss using the method of sections to find the sum of vertical forces ($\sum F_y$), horizontal forces ($\sum F_x$), and moments about point A ($\sum M_A$). 2. **Identify forces and dimensions:** - Vertical force at D: 23 kN upward - Horizontal forces at F and G: 20 kN each to the right - Distances: height between F and F joints = 4 m, width between A and G = 3 m, height between joints (assumed 4 m between levels), width between A and G = 3 m 3. **Sum of vertical forces ($\sum F_y$):** Considering upward positive, $$\sum F_y = R_{Ay} - 23 = 0 \implies R_{Ay} = 23 \text{ kN}$$ where $R_{Ay}$ is the vertical reaction at support A (left). 4. **Sum of horizontal forces ($\sum F_x$):** Considering right positive, $$\sum F_x = R_{Ax} - 20 - 20 = 0 \implies R_{Ax} = 40 \text{ kN}$$ where $R_{Ax}$ is the horizontal reaction at support A (left). 5. **Sum of moments about A ($\sum M_A$):** Taking moments counterclockwise positive, - Moment due to 23 kN at D (vertical force) at 4 m height: $23 \times 3 = 69$ kN·m (assuming horizontal distance 3 m) - Moment due to 20 kN at F (horizontal force) at 4 m height: $20 \times 4 = 80$ kN·m (vertical distance) - Moment due to 20 kN at G (horizontal force) at 0 m height: $20 \times 0 = 0$ Sum of moments: $$\sum M_A = 0 = -23 \times 3 + 20 \times 4 + 20 \times 0 + M_{R}$$ $$M_{R} = 23 \times 3 - 20 \times 4 = 69 - 80 = -11 \text{ kN·m}$$ Since the roller support at A (right) cannot resist moment, the moment is balanced by the fixed support at A (left). **Final answers:** - $\sum F_y = 0$ with $R_{Ay} = 23$ kN upward - $\sum F_x = 0$ with $R_{Ax} = 40$ kN right - $\sum M_A = 0$ with moment reaction $M_A = -11$ kN·m (counterclockwise) These reactions can be used to analyze internal forces in the truss members using the method of sections.