To determine the hydrostatic force on one end of the trough, we will follow these steps:

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Step 1: Understand the Geometry

The end of the trough is an equilateral triangle with a side length of $6 \, \text{m}$. The vertex is at the bottom.

• The height $h$ of the equilateral triangle can be calculated using the formula: $h = \frac{\sqrt{3}}{2} \cdot s$ where $s$ is the side length.

  Substituting $s = 6 \, \text{m}$: $h = \frac{\sqrt{3}}{2} \cdot 6 = 3\sqrt{3} \, \text{m} \approx 5.196 \, \text{m}.$

Step 2: Pressure as a Function of Depth

The hydrostatic pressure at a depth $y$ is given by: $P(y) = \rho g y$ where:

• $\rho = 815 \, \text{kg/m}^3$ (density of the liquid),
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity),
• $y$ is the depth from the surface.

Step 3: Area Element of the Triangle

At depth $y$, the width of the triangle $w(y)$ is proportional to $y$, since the triangle is equilateral. The proportionality can be determined from the geometry: $w(y) = \frac{s}{h} \cdot y = \frac{6}{3\sqrt{3}} \cdot y = \frac{2}{\sqrt{3}} \cdot y.$

The area of a thin horizontal strip at depth $y$ is: $\text{Area} = w(y) \, dy = \left(\frac{2}{\sqrt{3}} y\right) dy.$

Step 4: Hydrostatic Force on the Strip

The force on the strip is: $dF = P(y) \cdot \text{Area} = \rho g y \cdot \left(\frac{2}{\sqrt{3}} y \, dy\right).$

Simplify: $dF = \rho g \frac{2}{\sqrt{3}} y^2 \, dy.$

Step 5: Integrate Over the Depth

The total force is obtained by integrating from $y = 0$ (vertex of the triangle) to $y = h$ (top of the triangle): $F = \int_0^h \rho g \frac{2}{\sqrt{3}} y^2 \, dy.$

Substitute the known values ($\rho = 815 \, \text{kg/m}^3$, $g = 9.8 \, \text{m/s}^2$, $h = 3\sqrt{3}$): $F = \frac{2}{\sqrt{3}} \cdot 815 \cdot 9.8 \int_0^{3\sqrt{3}} y^2 \, dy.$

The integral of $y^2$ is: $\int y^2 \, dy = \frac{y^3}{3}.$

Evaluate: $\int_0^{3\sqrt{3}} y^2 \, dy = \frac{(3\sqrt{3})^3}{3} - \frac{0^3}{3} = \frac{27 \cdot 3\sqrt{3}}{3} = 27\sqrt{3}.$

Now substitute back: $F = \frac{2}{\sqrt{3}} \cdot 815 \cdot 9.8 \cdot 27\sqrt{3}.$

Simplify: $F = 2 \cdot 815 \cdot 9.8 \cdot 27 = 431874 \, \text{N}.$

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Final Answer:

The hydrostatic force on one end of the trough is: $\boxed{431,874 \, \text{N}}$

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Do you want further explanation or have questions?
Here are 5 related questions for deeper understanding:

1. How does the shape of the submerged surface affect hydrostatic force calculations?
2. What role does the density of the liquid play in determining hydrostatic pressure?
3. How would the force change if the liquid had a density of 1000 kg/m³ (water)?
4. What happens to the force if the trough is tilted, making the triangle no longer vertical?
5. How can the hydrostatic force be experimentally measured?

Tip: Always ensure to use consistent units when calculating forces involving pressure and density!