Question

Fluid Force Find the fluid force on a circular observation window of radius 1 foot in a vertical wall of a large water-filled tank at a fish hatchery when the center of the window is (a) 3 feet and (b) \(d\) feet \((d>1)\) below the water's surface. Use trigonometric substitution to evaluate the one integral.

Short answer

The fluid force on the circular window when the center is 3 feet below the surface is approximately 797.4 lbs and when it's \( d \) feet below is \( 124.7d \) lbs.

Step-by-step solution

Determine the fluid force when center of the window is 3 feet below the water's surface

We know that force on a differential element \( dA \) at a depth of \( h \) ft is \( dF= \rho g h dA \), where \( \rho \) is the water density in lbs/ft^3, \( g \) is the acceleration due to gravity in ft/s^2, and \( h \) is the depth in feet. For water, \( \rho \)= 62.4 lbs/ft^3 and \( g \)= 32.2 ft/s^2. The force on the differential rectangular section is given by \( dF= \rho g (3-y) dx dy \), where \( y \) varies from -1 to 1. This force needs to be integrated over the area of the circle (radius 1 ft), and we use trigonometric substitution \( y= \sin \theta \). After solving the integral, we get the force \( F = 797.4 lbs \).

Determine the fluid force when center of the window is d feet below the water's surface

By following the same steps as before, the only difference this time is in the bounds of \( y \) in our force \( dF= \rho g (d-y) dx dy \), where \( y \) varies from \( d-1 \) to \( d+1 \). We use again the trigonometric substitution \( y= d + \sin \theta \). After solving the integral, we obtain the force \( F = 124.7d lbs \).

Key concepts

Circular Observation Window

Understanding the fluid force acting on structures such as circular observation windows is a practical application of calculus and physics. Consider a circular window submerged in a liquid, like those often found in aquariums or fish hatcheries. The force exerted by the fluid on the window depends on factors such as the depth of the window below the surface, the density of the fluid, and gravitational acceleration.

When calculating the force on such a window, we divide the window into infinitesimally small elements for which the force can be more easily calculated. Then, we sum up the individual forces over the entire area of the window to get the total force. This is done using the concept of integrals, which allows us to account for the variation of pressure with depth. The deeper parts of the window endure greater forces due to the increased water pressure at those depths. By integrating these differential forces across the entire surface area of the window, we obtain the total fluid force the window must withstand.

Trigonometric Substitution

Trigonometric substitution is a technique used in calculus to simplify the integration process, especially when dealing with integrals involving square roots and other expressions that can complicate direct integration.

In our example dealing with the circular window, we define a differential area element whose position is described by the variable 'y'. However, since the window is circular, using 'y' directly in our calculations would necessitate integrating over a square root, reflecting the circular shape of the window. To avoid this, we use a trigonometric substitution such as \( y = \sin(\theta) \) or \( y = d + \sin(\theta) \), transforming the integrand into a trigonometric function that is easier to integrate.

This substitution utilizes the circular nature of trigonometric functions to mirror the circular nature of the window, resulting in a more approachable integral. The boundaries of integration also change accordingly, typically running from \(-\pi/2\) to \(\pi/2\), corresponding to the full range of the sine function from its minimum to maximum values.

Integral Evaluation

Integral evaluation is the final step in calculating the fluid force on the observation window. After setting up the integral with our differential element and applying trigonometric substitution, we must evaluate this integral to obtain a numeric value for force.

Evaluating an integral involves finding the antiderivative of the integrand and then applying the Fundamental Theorem of Calculus, which relates antiderivatives to definite integrals. For the fluid force problem, once the trigonometric substitution is applied, we find the antiderivative of the resulting trigonometric function, and then calculate the difference between its values at the upper and lower limits of integration.

The result gives the total force acting on the circular window at a specific depth. By thoroughly understanding each step of this process, students can tackle a variety of problems involving fluid forces on different shapes and at various depths, an invaluable skill in fields such as engineering and physics.