Problem 40 - Thomas Calculus

Question

Semicircular plate A semicircular plate 2 ft in diameter sticks straight down into freshwater with the diameter along the surface. Find the force exerted by the water on one side of the plate.

Step-by-Step Solution

Answer

The force exerted by the water is 41.6 pounds.

Step 1: Understand the Problem

We have a semicircular plate with a diameter of 2 feet that is submerged in water with its diameter aligned at the water's surface. The task is to find the total force exerted by the water on one side of the plate.

Step 2: Calculate the Area of the Semicircular Plate

The diameter of the semicircular plate is 2 ft, so the radius is 1 ft. The area of a full circle is \(\pi r^2\), and since our plate is semicircular, its area is \(\frac{1}{2} \cdot \pi \cdot (1)^2 = \frac{\pi}{2}\) sq ft.

Step 3: Determine the Centroid Depth

For a semicircular plate with its base on the surface, the centroid is located \(\frac{4r}{3\pi}\) ft below the diameter. Substituting \(r = 1\) ft, the centroid depth is \(\frac{4 \cdot 1}{3\pi} = \frac{4}{3\pi}\) ft.

Step 4: Calculate the Force Exerted by Water

The formula for the hydrostatic force is \(F = \rho \cdot g \cdot A \cdot h\), where \(\rho\) is the density of water (62.4 lb/ft³ for freshwater), \(A\) is the area of the submerged surface, and \(h\) is the depth of the centroid. Plug in the values: \[ F = 62.4 \cdot \frac{\pi}{2} \cdot \frac{4}{3\pi} = 62.4 \cdot \frac{2}{3} = 41.6 \text{ lb} \]

Step 5: Conclusion

The total force exerted by the water on one side of the semicircular plate is 41.6 pounds.

Key Concepts

Centroid Depth

The concept of centroid depth is central to calculating hydrostatic force on submerged objects. The centroid is essentially the "balance point" or the geometric center of a shape. For a semicircular plate submerged with its diameter along the water's surface, the centroid is not at the center of the plate but slightly deeper. This is due to the unique shape of the semicircle.

The formula to find the centroid depth of a semicircular shape is given by \( \frac{4r}{3\pi} \). Here, \( r \) is the radius of the semicircle. For our specific example with a radius of 1 ft, the centroid depth works out to be \( \frac{4}{3\pi} \) ft.

Knowing the precise depth of the centroid is crucial because it influences the pressure exerted by the fluid, which increases with depth. Hence, calculating the centroid depth accurately ensures that the hydrostatic force calculated reflects the actual pressure conditions experienced by the plate under water.

Semicircular Plate

The semicircular plate in question has a diameter of 2 feet, making it a half-circle when viewed from the side. Such shapes are common in practical applications, such as gates in dams or structural elements in water-retaining structures.

Calculating the area of a semicircular plate is a straightforward process. Since the full circle area formula is \( \pi r^2 \), for a semicircle, you take half of that. With a radius of 1 ft in this problem, the area becomes \( \frac{\pi}{2} \) square feet.

• The area is significant as it helps determine the total force exerted by the water.
• The larger the area, the more force the water can exert.

Understanding the geometry and properties of the semicircular plate allows us to apply the principles of physics correctly to find forces acting on submerged surfaces.

Hydrostatic Force Formula

When dealing with problems of submerged objects, the hydrostatic force formula becomes an essential tool. This formula can be written as:
\[ F = \rho \cdot g \cdot A \cdot h \]
where:

• \( F \) is the hydrostatic force.
• \( \rho \) is the density of the liquid (for freshwater, this is 62.4 lb/ft³).
• \( g \) is the acceleration due to gravity, though often incorporated into \( \rho \).
• \( A \) is the area of the submerged surface (from the semicircular plate).
• \( h \) is the centroid depth.

Given our semicircular plate, the formula helps calculate the force that water exerts on the submerged part. Plug in the known values: \( 62.4 \cos \frac{\pi}{2} \cdot \frac{4}{3\pi} \), simplifying to 41.6 lbs.

This formula accounts for how pressure varies with depth and ensures the force computation is both accurate and reliable. It's vital because it can help in designing safe and efficient water-holding structures.