1. **State the problem:** Calculate the liquid pressure on a rectangular observation window submerged in water. The window is 6 ft wide and 4 ft high, with its top edge 3 ft below the water surface. 2. **Formula and concepts:** The pressure at a depth $h$ in a fluid is given by $P = \rho g h$, where $\rho$ is the fluid density and $g$ is acceleration due to gravity. Here, pressure increases linearly with depth. The total force (liquid pressure) on the window is the integral of pressure over the window area. Since pressure varies with depth, we use the average pressure at the window's midpoint depth. 3. **Calculate depths:** - Top of window depth: $h_1 = 3$ ft - Bottom of window depth: $h_2 = 3 + 4 = 7$ ft - Midpoint depth: $h_m = \frac{h_1 + h_2}{2} = \frac{3 + 7}{2} = 5$ ft 4. **Calculate pressure at midpoint:** $$P_m = \rho \times h_m = 62.4 \times 5 = 312 \text{ lb/ft}^2$$ 5. **Calculate window area:** $$A = \text{width} \times \text{height} = 6 \times 4 = 24 \text{ ft}^2$$ 6. **Calculate total force (liquid pressure) on window:** $$F = P_m \times A = 312 \times 24 = 7488 \text{ lb}$$ 7. **Round to nearest pound:** $$\boxed{7488 \text{ lb}}$$ This is the total liquid pressure force on the window.