please pay special attention to rounding errors. 

A dam has the shape of an isosceles trapezoid with a height of 40 m. The width at the top is 80 m and the width at the bottom is 50 m. The surface of the water is 6 m below the top of the dam. Find the force on the dam due to hydrostatic pressure.

To find the force exerted on the dam by the hydrostatic pressure, we need to follow a step-by-step approach.

### Step 1: Formula for Hydrostatic Force

The force due to hydrostatic pressure is given by the equation:

\[
F = \int_0^h \rho g w(y) (h - y) \, dy
\]

Where:
- \( F \) is the total force.
- \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)).
- \( g \) is the acceleration due to gravity (\( 9.81 \, \text{m/s}^2 \)).
- \( w(y) \) is the width of the dam at depth \( y \).
- \( h \) is the depth of the water (34 m, since the water surface is 6 m below the top of the dam).
- \( (h - y) \) is the water depth at a point \( y \).

### Step 2: Width as a Function of Depth \( w(y) \)

We need to express \( w(y) \), the width of the dam at a depth \( y \), as a function of \( y \). The dam's top width is 80 m and the bottom width is 50 m, forming an isosceles trapezoid. The height of the dam is 40 m, so the change in width from bottom to top is:

\[
\Delta w = 80 - 50 = 30 \, \text{m}
\]

The width increases linearly, so at any depth \( y \), the width \( w(y) \) is given by:

\[
w(y) = 50 + \frac{30}{40} y = 50 + 0.75y
\]

### Step 3: Setting Up the Integral

The depth of the water is 34 m, so we integrate from \( y = 0 \) to \( y = 34 \). Plugging everything into the force equation:

\[
F = \rho g \int_0^{34} (50 + 0.75y)(34 - y) \, dy
\]

### Step 4: Solve the Integral

First, expand the integrand:

\[
(50 + 0.75y)(34 - y) = 50(34 - y) + 0.75y(34 - y) 
\]
\[
= 1700 - 50y + 25.5y - 0.75y^2
\]
\[
= 1700 - 24.5y - 0.75y^2
\]

Now, integrate term by term:

\[
F = \rho g \left[ \int_0^{34} 1700 \, dy - \int_0^{34} 24.5y \, dy - \int_0^{34} 0.75y^2 \, dy \right]
\]

### Step 5: Perform the Integrations

1. \(\int_0^{34} 1700 \, dy = 1700 \cdot 34 = 57800\)
2. \(\int_0^{34} 24.5y \, dy = 24.5 \cdot \frac{34^2}{2} = 24.5 \cdot 578 = 14161\)
3. \(\int_0^{34} 0.75y^2 \, dy = 0.75 \cdot \frac{34^3}{3} = 0.75 \cdot 13194.67 = 9896\)

Now, substitute these values back into the equation:

\[
F = \rho g (57800 - 14161 - 9896)
\]
\[
F = \rho g \cdot 33743
\]

### Step 6: Calculate the Force

Substitute \( \rho = 1000 \, \text{kg/m}^3 \) and \( g = 9.81 \, \text{m/s}^2 \):

\[
F = 1000 \cdot 9.81 \cdot 33743 = 330975183 \, \text{N}
\]

So, the hydrostatic force on the dam is approximately **330.98 MN** (meganeutons).

---

Would you like further details or have any questions about the steps?

### Related Questions:
1. How does the shape of the dam affect the hydrostatic force?
2. What if the water level was at the top of the dam—how would the force change?
3. How would the force change if the dam were wider at the bottom than at the top?
4. What would happen if the dam was taller, but the water level remained the same?
5. Can we apply this method to dams with curved shapes?

### Tip: 
Always check that your integral's limits reflect the actual physical scenario, especially when dealing with variable water depths.

A dam has the shape of an isosceles trapezoid with a height of 40 m. The width at the top is 80 m and the width at the bottom is 50 m. The surface of the water is 6 m below the top of the dam. Find the force on the dam due to hydrostatic pressure.

Learn how to calculate the hydrostatic force on a trapezoidal dam with a height of 40 m, top width of 80 m, and bottom width of 50 m. This problem involves the use of calculus, hydrostatic pressure, and integration to derive the force exerted on the dam. Suitable for advanced high school or college students.

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