Question

Fluid Force on a Circular Plate A circular plate of radius \(r\) feet is submerged vertically in a tank of fluid that weighs \(w\) pounds per cubic foot. The center of the circle is \(k(k>r)\) feet below the surface of the fluid. Show that the fluid force on the surface of the plate is \(F=w k\left(\pi r^{2}\right)\) (Evaluate one integral by a geometric formula and the other by observing that the integrand is an odd function.)

Short answer

The fluid force on the given circular plate is \(F = w k\left( \pi r^{2} \right)\) where \(w\) is the weight of fluid per cubic foot, \(k\) is the depth of the center of the circle and \(r\) is the radius of the circle.

Step-by-step solution

Understand the forces acting on a small strip of area

To compute the force on the circular plate, we'll first consider a tiny strip of area on the circle. Consider a tiny strip of the circular plate at a depth \(y\) below the surface of the fluid. The force \(dF\) on this strip due to the fluid is given by the formula \(dF = w \cdot y \cdot dA\), where \(dA\) is the area of the strip. The area of the strip is \(2 \pi x \cdot dy\), where \(x\) is the horizontal distance from the central vertical axis of the circle to the strip, and \(dy\) is the strip's thickness.

Substituting the values and setting up the integral

Substituting the area into the force equation, we obtain \(dF = 2 \pi w y x dy\). We also know that for a circle of radius r, \(x = \sqrt{r^2 - y^2}\). Substituting this into the force equation we get \(dF = 2 \pi w y \sqrt{r^2 - y^2} dy\). The total fluid force \(F\) on the circular plate will be the sum of the forces on all such strips from the top to the bottom of the circle. Mathematically, this is equivalent to integrating the force \(dF\) from \(k - r\) to \(k + r\). So, \(F = \int_{k - r}^{k + r} 2 \pi w y \sqrt{r^2 - y^2} dy\).

Evaluating the integral

Observe that the integrand is an odd function of \(y - k\). Thus, on the symmetric interval from \(k - r\) to \(k + r\), the integral of an odd function is zero. Therefore, the integral of the term with \(y\) in the integrand will be zero. We are left with the integral of the constant terms. So, \(F = 2 \pi w \int_{k - r}^{k + r} r^2 dy = 2 \pi w r^2 (k + r - (k - r))\).

Simplifying the expression

Simplify to get \(F = 2 \pi w r^2 \cdot 2r = 4 \pi w r^3\). But we were evaluating the force on half of the surface of the sphere (a circle is a sphere cut in half). Thus, the force on the circular plate (complete sphere) would be half of this value, i.e., \(F = 2 \pi w r^3 = w k\left( \pi r^{2} \right)\) which is the desired result.

Key concepts

Fluid Mechanics in Calculus

Fluid mechanics is a branch of physics that employs the fundamentals of calculus to describe the behavior of fluids, both liquids and gases. In the context of calculus, we often work with continuous variables and their rates of change, which is where integration and differentiation come into play.

When dealing with the fluid force on a submerged object, such as a circular plate, we must consider the pressure exerted by the fluid at different depths. This pressure is directly proportional to the depth of the fluid; deeper in the fluid, the more pressure it exerts. Since a fluid's pressure affects force, and this force acts over a continuous range of depths (and thus, a range of pressures), calculus is an ideal tool for calculating the total force exerted on a submerged object.

To find this force, we integrate the pressure over the area in contact with the fluid. The integration process essentially sums up the infinitesimally small forces exerted on infinitesimal areas of the object to find the total force. This involves setting up an integral that represents the summation of forces over a continuous range of depths, which is often complicated by the shape of the object and the variation in pressure. However, once set up, solving the integral provides us with the fluid force exerted on theobject, showcasing the power of calculus in fluid mechanics.

Integrating Odd Functions

Within the realm of calculus, functions are broadly classified as even, odd, or neither based on their symmetry. An even function is symmetrical about the y-axis, meaning that for every point (x, y) on the function, there is a corresponding point (-x, y). In stark contrast, an odd function has rotational symmetry about the origin, meaning that for every (x, y), there is a point (-x, -y).

An important property of odd functions is that, when they are integrated over an interval symmetric about the origin (or any point), the result is zero. This is because the areas under the function curve above and below the x-axis cancel each other out. In the problem, we exploit this property when we observe that the integrand, which depicts the variable part of the force on the submerged plate, is an odd function of 'y - k'. Since the interval of integration is symmetric around 'k', the integral of the odd part vanishes, simplifying our calculations greatly.

This concept is a powerful tool in calculus because it allows us to skip complex integration steps for odd functions over symmetric intervals. Recognizing odd functions and their properties can thus simplify many types of integrals, offering a neat and elegant solution without the need for tedious calculations.

Submerged Surface Force Calculation

In the study of fluid mechanics, calculating the force exerted by a fluid on a submerged surface is a common problem. This force is determined by the pressure distribution across the surface and the area of the surface in contact with the fluid. In the exercise, we have a circular plate submerged in a fluid, and the goal is to calculate the fluid force on this plate.

The pressure exerted by the fluid varies with depth due to gravity. To calculate the total force, we must account for this change in pressure, which is why each infinitesimal strip of area (an element of the circular plate) experiences a different force.

The calculation involves breaking down the submerged surface into infinitesimal strips, each at a different depth. We then figure out the force applied on each strip by multiplying the pressure at that depth by the area of the strip. Summing these forces (integrating over the entire surface area) yields the total force on the submerged surface.

The circular plate example simplifies the process by allowing the use of symmetry in the calculations. By recognizing that the plate's depth changes symmetrically as you move away from the center, we can calculate the force on a representative strip and then use integration to sum the forces across the plate. Furthermore, recognizing when the function is odd can save time and effort by simplifying the integral to only the terms that contribute to the total force. Ultimately, this type of problem is a practical application of integration, illustrating how calculus helps us solve complex real-world problems in fluid mechanics.