1. **Problem Statement:** Given lengths $L_1=2$ in, $L_2=3$ in, $L_3=6$ in, and $L_4=4$ in, and a downward force $F$ applied at point C, find the reactions at supports A and D for the structural system described. 2. **Understanding the Structure:** - The top horizontal bar extends from A to C, with segments $L_1$ and $L_2$. - The bottom bar extends vertically under B to D, with segments $L_3$ and $L_4$. - Force $F$ acts downward at C. - Supports at A and D provide reaction forces. 3. **Assumptions and Formulas:** - Assume the structure is in static equilibrium. - Sum of vertical forces must be zero: $$\sum F_y = 0$$ - Sum of moments about any point must be zero: $$\sum M = 0$$ 4. **Define Reaction Forces:** - Let $R_A$ be the vertical reaction at support A. - Let $R_D$ be the vertical reaction at support D. 5. **Calculate Total Lengths:** - Total length of top bar $AC = L_1 + L_2 = 2 + 3 = 5$ in. - Total length of bottom bar $BD = L_3 + L_4 = 6 + 4 = 10$ in. 6. **Sum of Vertical Forces:** $$R_A + R_D - F = 0 \implies R_A + R_D = F$$ 7. **Sum of Moments about A:** Taking moments about point A to eliminate $R_A$: - Moment arm of $F$ is $L_1 + L_2 = 5$ in. - Moment arm of $R_D$ is total horizontal distance from A to D. Since D is at the bottom bar end, horizontal distance is $L_1 + L_2 = 5$ in (assuming vertical bar under B does not affect horizontal distance). Moment equilibrium: $$-F \times 5 + R_D \times 5 = 0$$ 8. **Solve for $R_D$:** $$R_D \times 5 = F \times 5 \implies R_D = F$$ 9. **Solve for $R_A$:** From step 6: $$R_A + R_D = F \implies R_A + F = F \implies R_A = 0$$ 10. **Interpretation:** - Reaction at A is zero. - Reaction at D equals the applied force $F$. **Final answer:** $$R_A = 0, \quad R_D = F$$