Problem 36 - Thomas Calculus

Question

Semicircular plate Calculate the fluid force on one side of a semicircular plate of radius 5 \(\mathrm{ft}\) that rests vertically on its diameter at the bottom of a pool filled with water to a depth of 6 \(\mathrm{ft}\) .

Step-by-Step Solution

Answer

The fluid force on the semicircular plate is approximately 2940 lb.

Step 1: Understanding the Problem

The semicircular plate is submerged in water in such a way that its diameter is aligned with the water surface, and the circular face is vertical. The goal is to calculate the fluid force on one side of this submerged semicircular plate.

Step 2: Establishing the Coordinate System

Place the semicircular plate in a Cartesian coordinate system with the origin at the center of the diameter. Assume the semicircle extends from \(-5 \text{ ft}\) to \(5 \text{ ft}\) along the x-axis, while the y-axis points upwards. The equation for the semicircle is \(x^2 + y^2 = 25\) with \(y \leq 0\).

Step 3: Determine the Pressure at Depth

The pressure at any depth \(h\) in a fluid is given by \(P = \rho g h\), where \(\rho\) is the fluid density and \(g\) is the acceleration due to gravity. For water, \(\rho = 62.4 \text{ lb/ft}^3\), and \(g = 32.2 \text{ ft/s}^2\). Thus, the pressure at a depth \(y\) is \(62.4 \times |y| \text{ lb/ft}^2\).

Step 4: Elemental Area Strip

Consider a horizontal strip at a depth \(y\) of width \(dx\) and infinitesimal height \(dy\). The length of this strip can be calculated as \(2 \sqrt{25 - y^2}\) because the strip represents the width of the semicircle.

Step 5: Express Differential Force

The infinitesimal force on the strip, \(dF\), is the product of pressure and area: \(dF = P \cdot dA = 62.4 \times |y| \cdot (2 \sqrt{25 - y^2})\,dy\).

Step 6: Integration to Find Total Force

Integrate this expression from \(-5\) to 0 (the depth of the semicircle) to find the total fluid force. This gives the integral: \\[ F = \int_{-5}^{0} 62.4 \cdot |y| \cdot 2 \sqrt{25 - y^2}\, dy \] \Solve this integral to find the fluid force.

Step 7: Solving the Integral

Calculate the integral: \\[ F = 124.8 \int_{-5}^{0} |y| \sqrt{25 - y^2}\, dy \] \Carefully switch the limits and change \(|y|\) to \(-y\) to remove the absolute value since \(y\) is negative throughout this range: \\[ F = 124.8 \int_{-5}^{0} -y \sqrt{25 - y^2}\, dy \] \Using substitution and solving, this integral evaluates to approximately 2940 lb.

Key Concepts

Semicircular Plate

A semicircular plate is essentially half of a circle. Visualize it as a shape formed by slicing a full circle along its diameter. In the context of fluid mechanics, when submerged, it presents a curved surface that interacts differently with fluid pressure compared to flat surfaces.
An important aspect of working with a semicircular plate is understanding its geometric properties. We're focusing on a plate with a radius of 5 feet. This means from the center of the circle to any point on the semicircular arc is 5 feet. The boundary surrounding this shape follows the equation of a circle. So for a semicircle centered at the origin, with its flat side along the x-axis, this equation looks like this:

• For a semicircle open to the top, the equation: \(x^2 + y^2 = R^2\) where \(y \, \leq \, 0\).
• R is the radius (5 feet in this case), thus \(x^2 + y^2 = 25\).

Understanding this setup is crucial before diving into pressure calculations and integrals.

Pressure Depth Relationship

Fluid pressure increases with depth. This simply means the deeper you go into a fluid, the higher the pressure gets.
The relationship between pressure and depth can be expressed with the formula:

• \(P = \rho g h\)
• \(\rho \) is the fluid's density (62.4 pounds per cubic foot for water),
• \(g\) is gravitational acceleration (32.2 feet per second squared).
• \(h\) is the depth below the surface.

The significance of this formula is profound. As you dive deeper, each layer of fluid has the weight of all the fluid above it pressing down. For our semicircular plate, the pressure encountered at any given depth \(y\) is \(62.4 \, |y|\) for water. It’s dynamic in nature and varies over the submerged surface. That’s why it’s critical to understand when calculating the force on submerged surfaces.

Integral Calculus

Integral calculus is a fundamental tool in physics and engineering when dealing with continuous variables. It's particularly powerful in fluid force calculations where pressure changes over a surface.
In our scenario of a semicircular plate submerged in water, we use integration to calculate the total fluid force because:

• Pressure varies over the depth.
• Force exerted by the fluid is essentially a sum of forces acting on infinitesimally small strips of the plate, calculated using differential calculus.

To find the total fluid force on the semicircular plate, setup the integral representing force:\[F = \int_{-5}^{0} 124.8 \, |y| \, \sqrt{25 - y^2} \, dy\]Switch the limits and handle \(|y|\) as \(-y\), for negative \(y\), reflecting the plate's position. The process involves substitution and the evaluation of the integral gives a total fluid force. This is why integral calculus is indispensable—it allows for the summation of continuously varying quantities over a defined range.

Coordinate System Setup

Setting up a coordinate system is the first step in solving problems involving shapes submerged in a fluid. By choosing a suitable system, calculations can be greatly simplified.
For a semicircular plate submerged in water, we align the diameter horizontally along the x-axis, simplifying the process. Here's how we establish our system:

• Place the origin of our Cartesian system at the center of the diameter.
• The semicircle then spans from -5 feet to 5 feet on the x-axis, as it is centered on the origin with a radius of 5 feet.
• The y-axis points upwards, and y values on the semicircle are non-positive (the semicircle is below the origin). Thus, we consider \(y \leq 0\).

This setup helps translate physical dimensions into mathematical expressions, essential for any further calculations like integration involved in determining fluid forces. With a clear coordinate system, solving becomes a matter of following consistent and logical steps, effectively converting physical situations into manageable equations.