A dam has the shape of an isosceles trapezoid with a height of 40 m. The width at the top is 70 m and the width at the bottom is 50 m. The surface of the water is 8 m below the top of the dam. Find the force on the dam due to hydrostatic pressure.

Give your answer in scientific notation. If necessary, round the coefficient to two significant digits. Use multiplication in the notation. For example,
2.83*10^7 is the correct representation of 28,300,000.


To find the force on the dam due to hydrostatic pressure, we will follow these steps:

### Step 1: General Formula for Hydrostatic Force
The hydrostatic force on a submerged surface is given by the formula:

\[
F = \rho g \int_A y \, dA
\]

Where:
- \( \rho \) is the density of water, \( \rho = 1000 \, \text{kg/m}^3 \),
- \( g \) is the acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \),
- \( y \) is the depth of the surface at a point from the surface of the water,
- \( A \) is the area of the submerged surface.

The pressure at depth \( y \) is \( P = \rho g y \), and the force is the integral of this pressure over the submerged area.

### Step 2: Geometry of the Dam and Water Depth
- The height of the dam is 40 m.
- The top width is 70 m, and the bottom width is 50 m.
- The water surface is 8 m below the top, so the height of the submerged portion is \( 40 - 8 = 32 \, \text{m} \).

The dam's shape is an isosceles trapezoid, and we can express the width of the dam at any depth \( y \) as a linear function because the sides of the trapezoid are straight.

The width of the dam \( w(y) \) at a depth \( y \) from the top can be found using similar triangles:

\[
w(y) = 70 - \left( \frac{70 - 50}{40} \right) y = 70 - 0.5y
\]

### Step 3: Area Element \( dA \) and Hydrostatic Pressure Formula
At depth \( y \), the width is \( w(y) = 70 - 0.5y \), and the height element \( dy \) is infinitesimally small, so the area element \( dA \) is:

\[
dA = w(y) \, dy = (70 - 0.5y) \, dy
\]

The hydrostatic force on an infinitesimal strip of area at depth \( y \) is:

\[
dF = \rho g y \, dA = \rho g y (70 - 0.5y) \, dy
\]

### Step 4: Integrate to Find Total Force
We need to integrate from the water surface (depth 0) to the bottom of the submerged portion (depth 32 m):

\[
F = \int_0^{32} 1000 \cdot 9.8 \cdot y \cdot (70 - 0.5y) \, dy
\]

Let's compute this integral.The total hydrostatic force on the dam due to the water is approximately:

\[
F = 2.98 \times 10^8 \, \text{N}
\]

Would you like further details or have any questions?

Here are five related questions to expand on this topic:
1. How does the force on the dam change if the water level rises by 5 meters?
2. How would the force change if the shape of the dam was a rectangle instead of an isosceles trapezoid?
3. What role does the water density play in determining the hydrostatic force?
4. How would the force calculation differ if the dam were holding seawater, which is denser than freshwater?
5. How do engineers ensure that dams can withstand such large hydrostatic forces?

**Tip:** Hydrostatic pressure increases linearly with depth, which is why the pressure is greatest at the bottom of the dam.

A dam has the shape of an isosceles trapezoid with a height of 40 m. The width at the top is 70 m and the width at the bottom is 50 m. The surface of the water is 8 m below the top of the dam. Find the force on the dam due to hydrostatic pressure. Give your answer in scientific notation. If necessary, round the coefficient to two significant digits.

This problem demonstrates how to calculate the hydrostatic force on a dam shaped as an isosceles trapezoid with a height of 40 m, top width of 70 m, and bottom width of 50 m. The water level is 8 m below the top. Using hydrostatic pressure and integration, this problem provides a detailed solution, suitable for Grades 11-12 and college students.

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