Let's solve the problem step by step: --- ## **Step 1: Understanding the Problem** We have a bin filled with carbon tetrachloride (CT), with density \(\rho_{CT} = 3.09 \, \text{slug/ft}^3\). We are to find the resultant hydrostatic force and center of pressure **on each side plate** (AFEB and BEDC) measured from BE. --- ## **Step 2: Plate Dimensions** Both side plates (AFEB and BEDC) are identical rectangles. - **Base width (AF or BE):** \(6\) ft - **Height (AB or ED):** \(3 + 2.2 = 5.2\) ft --- ## **Step 3: Hydrostatic Pressure at Depth** The pressure at depth \(h\) is: \[ p = \gamma h \] where - \(\gamma = \rho g\) = specific weight - \(\rho = 3.09\, \text{slug/ft}^3\) - \(g = 32.2\, \text{ft/s}^2\) So, \[ \gamma = 3.09 \times 32.2 = 99.\, \text{lb/ft}^3 \] --- ## **Step 4: Resultant Force on a Vertical Plate** For a vertical rectangular plate submerged in a fluid: - **Area \(A\):** \[ A = \text{width} \times \text{height} = 6 \times 5.2 = 31.2\, \text{ft}^2 \] - **Depth to centroid (\(h_c\)):** - The centroid is at half the height: \[ h_c = \frac{5.2}{2} = 2.6\, \text{ft} \] - **Resultant force (\(F_R\)):** \[ F_R = \gamma h_c A = 99. \times 2.6 \times 31.2 = 8,037.1\, \text{lb} \] --- ## **Step 5: Location of Center of Pressure** For a vertical plate: \[ h_{cp} = \frac{I_G + A h_c^2}{A h_c} \] where - \(I_G\) = moment of inertia about the horizontal axis through centroid: \[ I_G = \frac{b h^3}{12} = \frac{6 \times (5.2)^3}{12} = \frac{6 \times 140.608}{12} = 70.304\, \text{ft}^4 \] - \(A = 31.2\, \text{ft}^2\) - \(h_c = 2.6\, \text{ft}\) Plug in values: \[ h_{cp} = \frac{70.304 + 31.2 \times (2.6)^2}{31.2 \times 2.6} \] \[ (2.6)^2 = 6.76 \] \[ 31.2 \times 6.76 = 210.912 \] \[ \text{Numerator: } 70.304 + 210.912 = 281.216 \] \[ \text{Denominator: } 31.2 \times 2.6 = 81.12 \] \[ h_{cp} = \frac{281.216}{81.12} = 3.47\, \text{ft} \] This is **measured from BE** (bottom edge). --- ## **Step 6: Final Answers** ### **Resultant Force** \[ \boxed{F_R = 8,040\, \text{lb} \quad \text{(rounded to 3 significant figures)}} \] ### **Center of Pressure (from BE)** \[ \boxed{h_{cp} = 3.47\, \text{ft} \quad \text{(from BE)}} \] --- ## **Summary Table** | Plate | Force \(F_R\) (lb) | Center of Pressure (ft from BE) | |----------|:------------------:|:-------------------------------:| | AFEB | 8,040 | 3.47 | | BEDC | 8,040 | 3.47 | --- ### **Explanation** - The force is calculated using hydrostatic pressure at the centroid. - The center of pressure is slightly below the centroid, as expected for hydrostatic forces. - Both plates are identical, so the results are the same for each. --- **If you need the force broken down for each surface (e.g., triangle and rectangle separately), let me know!**