A rectangular dam is 40 ft high and 60 ft wide. Compute the total force on the dam for each of the following scenarios.First,  The surface of the water is at the top of the dam. Then,  The surface of the water is halfway down the dam. 

To compute the total force exerted by the water on the dam, we will use the hydrostatic force formula, which accounts for water pressure increasing with depth.

The formula for the hydrostatic force \( F \) on a vertical surface is:

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

Where:
- \( \rho \) is the density of water (approximately \( 62.4 \, \text{lb/ft}^3 \)),
- \( g \) is the acceleration due to gravity (but since we're using pounds and feet, \( g \) is already accounted for in \( \rho \)),
- \( w \) is the width of the dam (in feet),
- \( h \) is the height of the dam (in feet),
- \( y \) is the depth variable from the top of the water surface to the base.

Let's compute the total force for the two scenarios:

### 1. **Water at the top of the dam**

The water reaches the full height of the dam, so \( h = 40 \, \text{ft} \), and the width of the dam \( w = 60 \, \text{ft} \).

The force on the dam is given by:

\[
F_1 = \rho \cdot w \int_0^{40} (40 - y) \, dy
\]

Solving the integral:

\[
\int_0^{40} (40 - y) \, dy = \left[ 40y - \frac{y^2}{2} \right]_0^{40} = 40(40) - \frac{40^2}{2} = 1600 - 800 = 800
\]

Thus, the force is:

\[
F_1 = 62.4 \cdot 60 \cdot 800 = 2,995,200 \, \text{lb}
\]

### 2. **Water halfway down the dam**

In this case, the water reaches halfway down, so \( h = 20 \, \text{ft} \), and the width of the dam is still \( w = 60 \, \text{ft} \).

The force on the dam is now:

\[
F_2 = \rho \cdot w \int_0^{20} (20 - y) \, dy
\]

Solving the integral:

\[
\int_0^{20} (20 - y) \, dy = \left[ 20y - \frac{y^2}{2} \right]_0^{20} = 20(20) - \frac{20^2}{2} = 400 - 200 = 200
\]

Thus, the force is:

\[
F_2 = 62.4 \cdot 60 \cdot 200 = 748,800 \, \text{lb}
\]

### Summary of Forces:
- When water is at the top of the dam: \( 2,995,200 \, \text{lb} \),
- When water is halfway down the dam: \( 748,800 \, \text{lb} \).

Would you like any further details or clarifications?

Here are five related questions for deeper exploration:
1. How does the water pressure vary with depth along the dam?
2. What is the force on the dam if the water level drops by one-quarter?
3. How does the width of the dam affect the total hydrostatic force?
4. What happens to the force if the dam height is doubled, but the water level remains at the original height?
5. How would the force change if the water density increases (e.g., saltwater)?

**Tip:** Hydrostatic force always increases with depth because pressure is proportional to the depth below the surface.

A rectangular dam is 40 ft high and 60 ft wide. Compute the total force on the dam for each of the following scenarios. First, the surface of the water is at the top of the dam. Then, the surface of the water is halfway down the dam.

Calculate the total hydrostatic force on a dam that is 40 ft high and 60 ft wide when the water is at full height and halfway down. Learn about hydrostatics, fluid pressure, and integral calculus with detailed solutions for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation