Problem 41 - Thomas Calculus

Question

New England Aquarium The viewing portion of the rectangular glass window in a typical fish tank at the New England Aquarium in Boston is 63 in. wide and roms from 0.5 in. below the water's surface to 33.5 in. below the surface. Find the fluid force against this portion of the window. The weight-density of seawater is 64 \(\mathrm{lb} / \mathrm{ft}^{3} .\) (In case you were wondering, the glass is 3\(/ 4\) in. thick and the tank walls extend 4 \(\mathrm{in.}\) above the water to keep the fish from jumping out.)

Step-by-Step Solution

Answer

The fluid force against the window is approximately 1309.5 pounds.

Step 1: Convert Measurements to Feet

To use the weight-density given in \(\text{lb/ft}^3\), first convert all the measurements from inches to feet. The width of the window is 63 inches, which converts to \( \frac{63}{12} = 5.25 \) feet. The vertical range of the window is from 0.5 inches below the surface to 33.5 inches below, which converts to \( \frac{0.5}{12} = 0.0417 \) feet and \( \frac{33.5}{12} = 2.7917 \) feet, respectively.

Step 2: Determine Depth of Water

For integration, define \( y \) as the depth below the surface of the water in feet, where \( y = 0.0417 \) feet is the top of the window and \( y = 2.7917 \) feet is the bottom. The depth of water above any point \( y \) is simply \( y \) because we're measuring from the surface downward.

Step 3: Set Up the Integral for Fluid Force

The fluid force on the window, \( F \), can be calculated using the integral \( F = \int_{a}^{b} \text{weight-density} \times \text{width} \times \text{depth} \times dy \). Substitute the given weight-density of seawater as \( 64 \text{ lb/ft}^3\), width as 5.25 feet, and depth as \( y \), with limits from 0.0417 ft to 2.7917 ft.

Step 4: Solve the Integral

The integral becomes \( F = \int_{0.0417}^{2.7917} 64 \times 5.25 \times y \times dy \). Compute this integral: \[ F = 64 \times 5.25 \times \left[ \frac{y^2}{2} \right]_{0.0417}^{2.7917} = 336 \times \left( \frac{2.7917^2}{2} - \frac{0.0417^2}{2} \right). \] Evaluate this expression to get the force.

Step 5: Compute Result

Calculate the result of the evaluated integral: \[ F = 336 \times \left( \frac{7.7909}{2} - \frac{0.0017}{2} \right) = 336 \times 3.8946 \approx 1309.5 \text{ lb}. \] Thus, the fluid force against the window is approximately 1309.5 pounds.

Key Concepts

Integral Calculus

Integral calculus is a powerful mathematical tool used to calculate areas, volumes, and other quantities that are described by continuously changing variables. In the context of fluid force, it's indispensable. Here, we have an aquarium's window subject to seawater pressure.

The fluid force on the window is calculated using an integral. This involves integrating the product of the weight-density of seawater, the width of the window, and the variable depth. Through this, we can sum up infinitesimal forces exerted over the window.

The integral formula used for calculating the total fluid force is: \( F = \int_{a}^{b} \text{weight-density} \times \text{width} \times \text{depth} \times dy \).

Each variable in this formula is essential to find the exact force against the aquarium window at various depths.

Weight-Density

Weight-density is a measure of how heavy a fluid is per unit volume, expressed in terms of pounds per cubic foot (lb/ft³) in this exercise. Knowing the weight-density of a fluid is crucial for calculating fluid force.

• For seawater, the weight-density is given as 64 lb/ft³, making it relatively dense compared to freshwater.
• This density can change with temperature and salt concentration, but 64 lb/ft³ is standard for calculations involving seawater.

This high density increases the exerted force on surfaces submerged in seawater, making correct calculations vital.

Seawater Pressure

Seawater pressure is the force exerted by the weight of seawater above a surface. It's directly linked to the depth and weight-density of the seawater. Unlike atmospheric pressure, it depends heavily on the volume of water directly above the area considered.

As depth increases, so does the pressure exerted. In our case, this means the lower part of the aquarium window experiences more pressure. This changing pressure across the depth requires an integral setup to accurately sum the forces against the window.

Understanding pressure in fluids is essential for designing aquariums and submerged structures, ensuring that they can withstand the forces exerted by fluid pressure.

Unit Conversion

In any physics-related calculation, unit conversion can play a pivotal role. It's crucial to maintain consistency in units throughout your math, especially when combining dimensions measured in different units.

• In this problem, we converted inches to feet to match the weight-density unit in lb/ft³. For example, the window width initially in inches was converted from 63 inches to 5.25 feet by dividing by 12 (since 12 inches = 1 foot).
• Likewise, the window elevation was also converted to feet before integration.

Without these conversions, the calculations would lead to incorrect results, making conversions a necessary part of the mathematical process.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. It increases proportionally with depth in the fluid. Hypothetically, if you were to dive deeper into the ocean, you'd feel more pressure, which is essentially what's being calculated against the aquarium window.

The formula for hydrostatic pressure is \( P = \text{density} \times \text{gravitational force} \times \text{depth} \).

In the aquarium exercise, hydrostatic pressure is integrated to compute the overall fluid force on the window. Understanding that pressure rises with depth illustrates why portions of submerged objects experience varying forces. This is essential for ensuring they are structurally sound and safe.