1. **Problem Statement:** Calculate the hydrostatic force on a trapezoidal dam with water level 6 m from the top. 2. **Given Data:** - Height of dam: 30 m - Width at top: 70 m - Width at bottom: 40 m - Water depth: 24 m (since water level is 6 m from top) 3. **Formula for hydrostatic force:** $$F = \rho g \int_0^{24} A(x) \, dx$$ where $A(x)$ is the width of the dam at depth $x$ times the strip height $dx$. 4. **Width function $x_i$ from similar triangles:** Given $$-\frac{5}{25} = \frac{x_i - 40}{20}$$ Solving for $x_i$: $$-\frac{5}{25} = \frac{x_i - 40}{20} \Rightarrow -\frac{1}{5} = \frac{x_i - 40}{20}$$ Multiply both sides by 20: $$20 \times -\frac{1}{5} = x_i - 40 \Rightarrow -4 = x_i - 40$$ Add 40 to both sides: $$x_i = 36$$ (Note: The problem states $x_i = 2x_i + 40$ which seems inconsistent; we use the correct derived value.) 5. **Area of strip $A_i$:** $$A_i = 40x_i + \frac{x_i^2}{2}$$ 6. **Hydrostatic force integral:** $$F = 1000g \int_0^{24} \left(40x + \frac{x^2}{2}\right) dx$$ Using $g = 9.8$ m/s²: $$F = 1000 \times 9.8 \int_0^{24} \left(40x + \frac{x^2}{2}\right) dx$$ 7. **Evaluate the integral:** $$\int_0^{24} 40x \, dx = 40 \times \frac{24^2}{2} = 40 \times 288 = 11520$$ $$\int_0^{24} \frac{x^2}{2} \, dx = \frac{1}{2} \times \frac{24^3}{3} = \frac{1}{2} \times \frac{13824}{3} = \frac{1}{2} \times 4608 = 2304$$ 8. **Sum of integrals:** $$11520 + 2304 = 13824$$ 9. **Calculate total force:** $$F = 1000 \times 9.8 \times 13824 = 135475200 \text{ N}$$ 10. **Final answer rounded:** $$F \approx 135475200 \text{ N}$$