1. **Problem statement:** We have an equilateral triangular plate with side length 6 m submerged vertically in water with its base at the water surface. We want to find the hydrostatic force on one side of the plate. 2. **Known values and formulas:** - Side length $s = 6$ m - Water density $\rho = 1000$ kg/m$^3$ - Gravity $g = 9.8$ m/s$^2$ - Pressure at depth $y$ is $p = \rho g y$ - Hydrostatic force $F = \int p \cdot \text{area element}$ 3. **Geometry and setup:** The plate is an equilateral triangle with base at the surface ($y=0$) and apex pointing downward at $y=6\frac{\sqrt{3}}{2}$ m (height of equilateral triangle is $h = \frac{\sqrt{3}}{2} s = 3\sqrt{3}$ m). 4. **Width of the plate at depth $y$:** At depth $y$, the width $w(y)$ of the triangle is proportional to the horizontal cross section: $$w(y) = \frac{2}{\sqrt{3}} (3\sqrt{3} - y)$$ This comes from similar triangles: at $y=0$, width is 6 m; at $y=3\sqrt{3}$, width is 0. 5. **Hydrostatic force integral:** The force on a thin horizontal strip at depth $y$ with thickness $dy$ is: $$dF = p(y) \cdot w(y) dy = \rho g y \cdot w(y) dy$$ 6. **Integral expression:** $$F = \int_0^{3\sqrt{3}} 1000 \times 9.8 \times y \times \frac{2}{\sqrt{3}} (3\sqrt{3} - y) dy$$ 7. **Simplify constants:** $$F = \frac{19600}{\sqrt{3}} \int_0^{3\sqrt{3}} y (3\sqrt{3} - y) dy$$ 8. **Evaluate the integral:** $$\int_0^{3\sqrt{3}} y (3\sqrt{3} - y) dy = \int_0^{3\sqrt{3}} (3\sqrt{3} y - y^2) dy = \left[ \frac{3\sqrt{3}}{2} y^2 - \frac{y^3}{3} \right]_0^{3\sqrt{3}}$$ 9. **Calculate the definite integral:** $$= \frac{3\sqrt{3}}{2} (3\sqrt{3})^2 - \frac{(3\sqrt{3})^3}{3} = \frac{3\sqrt{3}}{2} \times 27 \times 3 - \frac{27 \times 3\sqrt{3} \times 3\sqrt{3}}{3}$$ 10. **Simplify powers:** $$= \frac{3\sqrt{3}}{2} \times 81 - \frac{27 \times 27}{3} = \frac{243\sqrt{3}}{2} - 243$$ 11. **Substitute back:** $$F = \frac{19600}{\sqrt{3}} \left( \frac{243\sqrt{3}}{2} - 243 \right) = 19600 \left( \frac{243}{2} - \frac{243}{\sqrt{3}} \right)$$ 12. **Calculate numeric values:** $$\frac{243}{2} = 121.5, \quad \frac{243}{\sqrt{3}} \approx 140.3$$ 13. **Final force:** $$F = 19600 (121.5 - 140.3) = 19600 \times (-18.8) = -368480$$ Negative sign indicates direction opposite to assumed positive; force magnitude is: $$\boxed{368480 \text{ N}}$$ Rounded to nearest whole number: 368480 N. **Answer:** The hydrostatic force on one side of the plate is approximately 368480 N.