Calculate the fluid force on one side of a semicircular plate of radius 8 ft that rests vertically on its diameter at the bottom of a pool filled with water to a depth of 12 ft. Assume the​ weight-density of water is 62.4 lb divided by ft cubed.

To calculate the fluid force on one side of a semicircular plate submerged vertically in water, we need to use the following formula:

\[
F = \int_a^b \rho \cdot w(y) \cdot y \, dy
\]

Where:
- \( \rho = 62.4 \, \text{lb/ft}^3 \) is the weight-density of water.
- \( w(y) \) is the width of the plate at depth \( y \).
- \( y \) is the depth below the surface of the water.
- The integration is performed over the depth of the plate from the bottom of the pool to the top of the plate.

### Step-by-Step Breakdown:

1. **Geometry of the Plate**:  
   The plate is a semicircle with a radius of 8 ft, resting on its diameter at the bottom of the pool. Therefore, its diameter is along the \( x \)-axis (horizontal axis), and the top of the plate is at 4 feet below the surface of the water, since the total depth of water is 12 feet.

   The equation of a semicircle of radius 8 ft is:
   \[
   x^2 + y^2 = 8^2
   \]
   This gives us the width \( w(y) \) at a depth \( y \).

2. **Limits of Integration**:  
   The water covers the semicircular plate from the top (4 feet below the surface) to the bottom of the pool (12 feet below the surface). So, the integral limits will range from \( 4 \) feet to \( 12 \) feet.

3. **Width of the Plate \( w(y) \)**:  
   At a depth \( y \), the width of the plate is twice the \( x \)-coordinate from the center of the semicircle. From the equation of the semicircle, we get:
   \[
   w(y) = 2\sqrt{8^2 - (y - 4)^2}
   \]
   because \( y - 4 \) is the distance from the center of the semicircle.

4. **Setting up the Integral**:  
   The fluid force is given by the integral:
   \[
   F = \int_4^{12} 62.4 \cdot 2\sqrt{64 - (y - 4)^2} \cdot y \, dy
   \]

Let's compute this integral to find the fluid force.The integral for the fluid force can be expressed as:

\[
F = 124.8 \cdot \int_4^{12} y \cdot \sqrt{12 - y} \cdot \sqrt{y + 4} \, dy
\]

This integral requires numerical methods to compute the exact value. Let's now evaluate it to find the fluid force.It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, I can guide you through the steps to calculate this fluid force manually, or assist with another question.

Would you like further details, or do you have any other questions?

Here are 5 related questions you may find interesting:
1. How does the depth of water affect the force on the plate?
2. What would change if the plate were fully submerged?
3. How does the shape of the plate (e.g., rectangular) affect the force calculation?
4. What if the fluid was not water, but another liquid with a different density?
5. How would the result change if the plate was tilted instead of vertical?

**Tip:** When integrating functions involving radicals, substituting variables to simplify the expression can make evaluation easier!

Calculate the fluid force on one side of a semicircular plate of radius 8 ft that rests vertically on its diameter at the bottom of a pool filled with water to a depth of 12 ft. Assume the weight-density of water is 62.4 lb divided by ft cubed.

Learn how to calculate the fluid force on a semicircular plate with a radius of 8 ft submerged in water to a depth of 12 ft. This problem involves fluid mechanics, integral calculus, and geometry, suitable for undergraduate engineering or physics students.

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