1. **State the problem:** Calculate the hydrostatic force $N$ on a vertical equilateral triangular plate submerged in water, where the force is given by the integral $$N = pg \int_0^{3\sqrt{3}} \frac{6 - \frac{x}{\sqrt{3}}}{2} \, dx.$$ 2. **Understand the integral:** The integrand represents half the horizontal length of the plate at depth $x$, multiplied by the weight density $pg$. The limits $0$ to $3\sqrt{3}$ correspond to the height of the triangle. 3. **Rewrite the integral:** $$N = pg \int_0^{3\sqrt{3}} \frac{6 - \frac{x}{\sqrt{3}}}{2} \, dx = pg \int_0^{3\sqrt{3}} \left(3 - \frac{x}{2\sqrt{3}}\right) dx.$$ 4. **Integrate term-by-term:** $$\int_0^{3\sqrt{3}} 3 \, dx = 3x \Big|_0^{3\sqrt{3}} = 3 \times 3\sqrt{3} = 9\sqrt{3}.$$ $$\int_0^{3\sqrt{3}} \frac{x}{2\sqrt{3}} \, dx = \frac{1}{2\sqrt{3}} \times \frac{x^2}{2} \Big|_0^{3\sqrt{3}} = \frac{1}{4\sqrt{3}} (3\sqrt{3})^2 = \frac{1}{4\sqrt{3}} \times 27 = \frac{27}{4\sqrt{3}}.$$ 5. **Simplify the second integral:** $$\frac{27}{4\sqrt{3}} = \frac{27}{4} \times \frac{\sqrt{3}}{3} = \frac{27\sqrt{3}}{12} = \frac{9\sqrt{3}}{4}.$$ 6. **Combine results:** $$\int_0^{3\sqrt{3}} \left(3 - \frac{x}{2\sqrt{3}}\right) dx = 9\sqrt{3} - \frac{9\sqrt{3}}{4} = \frac{36\sqrt{3}}{4} - \frac{9\sqrt{3}}{4} = \frac{27\sqrt{3}}{4}.$$ 7. **Final answer:** $$N = pg \times \frac{27\sqrt{3}}{4}.$$ This is the hydrostatic force on the plate.