To solve this problem, we need to calculate the hydrostatic force $F$ acting on the dam due to water pressure. Here's a step-by-step solution:

1. Key Information:

• The dam is in the shape of an isoceles trapezoid.
• Height of the dam $H = 40 \, \text{m}$.
• Width at the top $W_{\text{top}} = 70 \, \text{m}$.
• Width at the bottom $W_{\text{bottom}} = 50 \, \text{m}$.
• Water surface is 6 m below the top of the dam, so the height of water $h_{\text{water}} = 40 - 6 = 34 \, \text{m}$.

The trapezoidal shape complicates things slightly, but we can proceed by integrating the pressure over the submerged area.

2. Hydrostatic Pressure:

The pressure at a depth $y$ below the surface of the water is given by:

$P(y) = \rho g y$

Where:

• $\rho = 1000 \, \text{kg/m}^3$ (density of water),
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity),
• $y$ is the depth below the water surface.

3. Force on the Dam:

The total hydrostatic force on the dam can be calculated using:

$F = \int_{0}^{h_{\text{water}}} P(y) \cdot w(y) \, dy$

Where:

• $w(y)$ is the width of the dam at depth $y$,
• $P(y) = \rho g y$.

4. Expression for Width $w(y)$:

The dam is an isosceles trapezoid, so the width $w(y)$ varies linearly with depth $y$. The width at depth $y$ is:

$w(y) = W_{\text{bottom}} + \left( \frac{W_{\text{top}} - W_{\text{bottom}}}{H} \right) \cdot y$

Substitute the given values: $w(y) = 50 + \left( \frac{70 - 50}{40} \right) y = 50 + 0.5y$

5. Setting Up the Integral:

The total force $F$ is:

$F = \int_{0}^{34} \rho g y \cdot w(y) \, dy$

Substitute $w(y) = 50 + 0.5y$ and constants $\rho = 1000$ and $g = 9.8$:

$F = \int_{0}^{34} 1000 \cdot 9.8 \cdot y \cdot (50 + 0.5y) \, dy$

Simplify constants:

$F = 9800 \int_{0}^{34} y (50 + 0.5y) \, dy$

Distribute $y$:

$F = 9800 \int_{0}^{34} (50y + 0.5y^2) \, dy$

6. Solving the Integral:

Evaluate the integral term by term:

$\int (50y + 0.5y^2) \, dy = \frac{50y^2}{2} + \frac{0.5y^3}{3}$

Simplify:

$\int (50y + 0.5y^2) \, dy = 25y^2 + \frac{y^3}{6}$

Now evaluate this from $0$ to $34$:

$\left[ 25y^2 + \frac{y^3}{6} \right]_{0}^{34}$

At $y = 34$:

$25(34)^2 + \frac{(34)^3}{6}$

Calculate step by step:

• $34^2 = 1156$,
• $25(34)^2 = 25 \cdot 1156 = 28,900$,
• $34^3 = 39304$,
• $\frac{34^3}{6} = \frac{39304}{6} \approx 6550.67$.

Thus:

$25(34)^2 + \frac{34^3}{6} = 28,900 + 6550.67 = 35,450.67$

7. Multiply by $9800$:

Now multiply the result by $9800$ to get the total force:

$F = 9800 \cdot 35,450.67 \approx 3.475 \times 10^8 \, \text{N}$

8. Final Answer:

The force on the dam due to hydrostatic pressure is approximately:

$F \approx 3.5 \times 10^8 \, \text{N}$

Do you want further clarification on any step? Let me know!

Related Questions:

1. How is hydrostatic pressure distributed across a submerged surface?
2. How do you derive the formula for force on a trapezoidal dam?
3. What happens to the hydrostatic force if the water depth increases?
4. How do you calculate the center of pressure on a dam?
5. How does the shape of the dam affect the hydrostatic force?

Tip: For problems involving variable width, express the width as a function of depth and integrate systematically!