1. **State the problem:** We have a vertical truss with horizontal forces applied at points B (23 kN left), C (63 kN left), and D (63 kN right). The truss has vertical segments of 12 ft each. We need to find the sum of moments about point A ($\sum M_A$), the sum of vertical forces ($\sum F_y$), and the sum of horizontal forces ($\sum F_x$) using the method of sections. 2. **Identify forces and distances:** - Forces at B and C act left (negative x-direction): $F_B = -23$ kN, $F_C = -63$ kN. - Force at D acts right (positive x-direction): $F_D = +63$ kN. - Vertical segments are each 12 ft, so distances from A are: - B: 12 ft - C: 24 ft (since B to E is 12 ft and E to D is 12 ft, assuming C is at 24 ft height) - D: 36 ft 3. **Calculate $\sum M_A$ (sum of moments about A):** Moments caused by horizontal forces are force times vertical distance from A. $$\sum M_A = F_B \times 12 + F_C \times 24 + F_D \times 36$$ Substitute values: $$\sum M_A = (-23)(12) + (-63)(24) + (63)(36)$$ Calculate each term: $$-23 \times 12 = -276$$ $$-63 \times 24 = -1512$$ $$63 \times 36 = 2268$$ Sum: $$\sum M_A = -276 - 1512 + 2268 = 480 \text{ kN-ft}$$ 4. **Calculate $\sum F_y$ (sum of vertical forces):** No vertical forces are given, so: $$\sum F_y = 0$$ 5. **Calculate $\sum F_x$ (sum of horizontal forces):** Sum all horizontal forces: $$\sum F_x = F_B + F_C + F_D = -23 - 63 + 63 = -23 \text{ kN}$$ **Final answers:** $$\sum M_A = 480 \text{ kN-ft}$$ $$\sum F_y = 0$$ $$\sum F_x = -23 \text{ kN}$$