1. **State the problem:** We need to find the reaction forces at supports D (pinned) and B (roller) for the given truss with external loads: 800 lb downward at A and 200 lb rightward at C. 2. **Identify supports and reactions:** - Roller at B: vertical reaction $B_y$ only (roller allows horizontal movement). - Pinned at D: vertical and horizontal reactions $D_y$ and $D_x$. 3. **Apply equilibrium equations for the entire truss:** Sum of forces in horizontal direction ($\Sigma F_x = 0$): $$D_x + 0 + 200 = 0 \implies D_x = -200 \text{ lb (to the left)}$$ Sum of forces in vertical direction ($\Sigma F_y = 0$): $$B_y + D_y - 800 = 0 \implies B_y + D_y = 800$$ Sum of moments about point D ($\Sigma M_D = 0$): Taking counterclockwise as positive. - The 800 lb force at A is 4 ft horizontally from D (since A to B is 4 ft, and B to D is horizontal 4 ft, total horizontal distance from D to A is 4 ft). - The 200 lb force at C is 3 ft vertically from D (since B to C is 3 ft vertical, and D to B is horizontal 4 ft, vertical distance from D to C is 3 ft). - The vertical reaction at B ($B_y$) is 4 ft from D horizontally. Moment equation: $$-800 \times 4 + 200 \times 3 + B_y \times 4 = 0$$ $$-3200 + 600 + 4 B_y = 0$$ $$4 B_y = 2600$$ $$B_y = 650 \text{ lb (upward)}$$ 4. **Find $D_y$ using vertical force equilibrium:** $$B_y + D_y = 800 \implies 650 + D_y = 800 \implies D_y = 150 \text{ lb (upward)}$$ 5. **Summary of reactions:** - $D_x = -200$ lb (to the left) - $D_y = 150$ lb (upward) - $B_y = 650$ lb (upward) These are the reaction forces at supports D and B.