1. **State the problem:** We have a crane base with two supports at points A and B. The reaction force at A, $R_A$, is given as 0. There are two downward forces: $W_c = 90.0$ kN at point C (above A) and $W_D = 30.0$ kN at point D (to the right). Distances are $L_A = 1.0$ m, $L_B = 1.5$ m, and $L_D$ is unknown. We want to find the reaction force at B, $R_B$, and possibly the distance $L_D$ if needed. 2. **Write the equilibrium equations:** For vertical forces, the sum must be zero: $$R_A + R_B - W_c - W_D = 0$$ Since $R_A = 0$, this simplifies to: $$R_B = W_c + W_D$$ 3. **Calculate $R_B$:** $$R_B = 90.0 + 30.0 = 120.0$$ kN 4. **Write the moment equilibrium about point A:** Taking moments about A (counterclockwise positive): $$\sum M_A = 0 = R_B \times (L_A + L_B) - W_c \times 0 - W_D \times (L_A + L_B + L_D)$$ Note: $W_c$ acts at A, so moment arm is 0. 5. **Solve for $L_D$:** Rearranged: $$R_B \times (L_A + L_B) = W_D \times (L_A + L_B + L_D)$$ Substitute known values: $$120.0 \times (1.0 + 1.5) = 30.0 \times (1.0 + 1.5 + L_D)$$ $$120.0 \times 2.5 = 30.0 \times (2.5 + L_D)$$ $$300 = 30.0 \times (2.5 + L_D)$$ Divide both sides by 30.0: $$\cancel{30.0} \times 10 = \cancel{30.0} \times (2.5 + L_D)$$ $$10 = 2.5 + L_D$$ 6. **Calculate $L_D$:** $$L_D = 10 - 2.5 = 7.5$$ m **Final answers:** - Reaction force at B: $R_B = 120.0$ kN - Distance $L_D = 7.5$ m