1. **State the problem:** We need to find the hydrostatic force on one end of a trough filled with water. The end is an inverted isosceles triangle with base 5 ft and height 6 ft. 2. **Relevant formula:** The hydrostatic force on a vertical surface submerged in a fluid is given by $$F = \rho g \bar{h} A$$ where $\rho g$ is the weight-density of the fluid, $\bar{h}$ is the depth of the centroid of the submerged surface from the free surface, and $A$ is the area of the surface. 3. **Calculate the area $A$ of the triangular end:** $$A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 6 = 15 \text{ ft}^2$$ 4. **Find the depth of the centroid $\bar{h}$:** For a triangle with height $h$, the centroid is located at $\frac{h}{3}$ from the base (the wider side). Since the triangle is inverted, the base is at the top (water surface), so the centroid is $\frac{2}{3}h$ from the water surface. Thus, $$\bar{h} = \frac{2}{3} \times 6 = 4 \text{ ft}$$ 5. **Weight-density of water:** Given as 62.4 lb/ft$^3$. 6. **Calculate the force:** $$F = 62.4 \times 15 \times 4 = 3744 \text{ lb}$$ 7. **Round to nearest integer:** $$F \approx 3744 \text{ lb}$$ **Final answer:** The force on each end of the trough when full of water is approximately 3744 lb.