To calculate the hydrostatic force on the dam, we use the following principles:

• Hydrostatic pressure at depth $h$ is given by: $P(h) = \rho g h$ where $\rho$ is the density of water ($1000 \, \text{kg/m}^3$), $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$), and $h$ is the depth below the surface of the water.

• Hydrostatic force is the integral of the pressure over the surface area of the dam submerged in water.

Step 1: Geometric Analysis

The dam has the shape of an isosceles trapezoid. The water level is $6 \, \text{m}$ below the top of the dam, meaning the height of the submerged portion of the dam is: $h_{\text{submerged}} = 40 \, \text{m} - 6 \, \text{m} = 34 \, \text{m}.$

The dam’s top width is $70 \, \text{m}$ and bottom width is $60 \, \text{m}$, so the width of the dam as a function of height $h$ below the top can be described by a linear function. The change in width per meter of height is: $\text{Slope} = \frac{70 - 60}{40} = \frac{10}{40} = 0.25.$ Thus, the width $w(h)$ at depth $h$ meters below the top is: $w(h) = 60 + 0.25 h.$

Step 2: Hydrostatic Force Calculation

The differential force on a small horizontal strip of the dam at depth $h$ with width $w(h)$ and thickness $dh$ is: $dF = P(h) \cdot w(h) \cdot dh = \rho g h \cdot (60 + 0.25 h) \cdot dh.$

The total force is the integral of $dF$ from $h = 6 \, \text{m}$ to $h = 40 \, \text{m}$: $F = \int_{6}^{40} \rho g h (60 + 0.25 h) \, dh.$

Expanding the integrand: $F = \rho g \int_{6}^{40} \left( 60h + 0.25 h^2 \right) \, dh.$

Step 3: Solving the Integral

Now, we compute the integral: $\int 60h \, dh = 30h^2,$ $\int 0.25 h^2 \, dh = \frac{0.25}{3} h^3 = \frac{h^3}{12}.$ Thus, the total force is: $F = \rho g \left[ 30h^2 + \frac{h^3}{12} \right]_{6}^{40}.$

First, evaluate at $h = 40$: $30(40)^2 + \frac{(40)^3}{12} = 30 \times 1600 + \frac{64000}{12} = 48000 + 5333.33 = 53333.33.$

Next, evaluate at $h = 6$: $30(6)^2 + \frac{(6)^3}{12} = 30 \times 36 + \frac{216}{12} = 1080 + 18 = 1098.$

The difference is: $53333.33 - 1098 = 52235.33.$

Step 4: Multiply by Constants

Now, multiply by $\rho g$: $F = 1000 \times 9.8 \times 52235.33 = 511904234 \, \text{N}.$

Step 5: Scientific Notation

Convert the result into scientific notation: $F \approx 5.1 \times 10^8 \, \text{N}.$

Thus, the force on the dam due to hydrostatic pressure is approximately: $\boxed{5.1 \times 10^8 \, \text{N}}.$

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Would you like to see more details or explanations on any part of the process? Here are five related questions for further exploration:

1. How does the hydrostatic pressure change with depth?
2. Why is the trapezoidal shape of the dam important in this calculation?
3. How would the result differ if the dam had a constant width?
4. What happens to the force if the water level rises to the top of the dam?
5. How would this calculation change for seawater, which has a different density?

Tip: Always express results in scientific notation when dealing with very large or small numbers for clarity and precision!