1. **Problem statement:** We have four parallel loads on a beam: three known loads and one missing load. The resultant of all four loads is 200 kN downward, located 6 m to the left of point B. We need to find the magnitude and position of the missing load. 2. **Given data:** - Load 1: 30 kN upward at 3 m from A - Load 2: 45 kN upward at 7 m from A (3 m + 4 m) - Load 3: 15 kN downward at B (10 m from A) - Load 4: Missing load $R$ downward at position $x$ m from A - Resultant load: 200 kN downward at position $P = 4$ m from A (since 6 m to the left of B, and B is at 10 m, so $10 - 6 = 4$ m) 3. **Sign convention:** Upward loads are positive, downward loads are negative. 4. **Step 1: Calculate the resultant force equation:** $$ R_{total} = 30 + 45 - 15 - R = 200 $$ Simplify: $$ 75 - 15 - R = 200 \\ 60 - R = 200 \\ -R = 140 \\ R = -140 $$ Since $R$ is downward, magnitude is positive 140 kN downward. 5. **Step 2: Calculate the moment about point A for all loads and set equal to moment of resultant:** $$ \text{Sum of moments} = 30 \times 3 + 45 \times 7 - 15 \times 10 - R \times x = 200 \times 4 $$ Substitute $R = 140$: $$ 90 + 315 - 150 - 140x = 800 $$ Simplify: $$ 255 - 140x = 800 \\ -140x = 545 \\ x = \frac{\cancel{-}545}{\cancel{-}140} = 3.893 $$ 6. **Final answers:** - Magnitude of missing load $R$ is 140 kN downward. - Position $x$ of missing load is 3.893 m from point A. These answers are rounded to the thousandth place as requested.