Problem 39 - 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 runs 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 in. above the water to keep the fish from jumping out.)

Step-by-Step Solution

Answer

The fluid force against the window is approximately 1309.375 pounds.

Step 1: Convert Units

First, notice that the dimensions given (0.5 in. and 33.5 in.) are in inches. To simplify calculations with fluid density, we need to convert these into feet. Thus, \(0.5 \text{ in} = \frac{0.5}{12} \text{ ft}\) and \(33.5 \text{ in} = \frac{33.5}{12} \text{ ft}\). This gives us \( h_1 = \frac{1}{24} \text{ ft} \) and \( h_2 = \frac{67}{24} \text{ ft} \).

Step 2: Determine Height Difference

Calculate the height of the window submerged under the water. This is the difference between the lower and upper bounds of the window: \( h = h_2 - h_1 = \frac{67}{24} - \frac{1}{24} = \frac{66}{24} = \frac{11}{4} \text{ ft}.\)

Step 3: Convert Width to Feet

The width of the window is given as 63 inches. Convert this to feet: \( 63 \text{ in} = \frac{63}{12} \text{ ft} = 5.25 \text{ ft}.\)

Step 4: Set Up the Fluid Force Formula

The fluid force \( F \) on a submerged surface is given by the integral \( F = \int_{a}^{b} \rho g(y) \text{ d}y \), where \( \rho \) is the fluid density, \( g(y) \) is the gravitational force at depth \( y \), and \( y \) ranges from \( h_1 \) to \( h_2 \). Here, \( \rho = 64 \text{ lb/ft}^3 \) and \( g(y) = y \), the depth. The width is constant, so it factors out.

Step 5: Integrate to Calculate Force

Evaluate the integral: \[ F = \int_{1/24}^{67/24} (64 \times 5.25) y \text{ d}y = 336 \int_{1/24}^{67/24} y \text{ d}y. \]This evaluates to:\[ F = 336 \left[ \frac{y^2}{2} \right]_{1/24}^{67/24} = 336 \cdot \left( \frac{(67/24)^2}{2} - \frac{(1/24)^2}{2} \right). \]

Step 6: Simplify the Integral Result

Calculate the definite integral:\[F = 336 \left( \frac{4489}{1152} - \frac{1}{1152} \right) = 336 \cdot \frac{4488}{1152} = 336 \times \frac{1122}{288} = 336 \times 3.895833. \] Finally, multiply these values to find \( F = 1309.375 \).

Step 7: Conclusion

After performing the calculations, we obtain that the force exerted by the water against the window is approximately 1309.375 pounds.

Key Concepts

Seawater Weight-Density

In fluid dynamics and in many problems related to fluid force, it is crucial to understand the concept of weight-density, especially of seawater. Seawater weight-density is essentially the weight per unit volume of seawater and is typically measured in pounds per cubic foot (lb/ft³). For most calculations regarding aquatic environments, the approximation used is 64 lb/ft³. This constant is used to calculate the force exerted by seawater on submerged objects, like the aquarium window.
Understanding weight-density helps in determining how heavy the fluid is over a certain area. In practical terms, it means if you have a cubic foot of seawater, it weighs about 64 pounds. This is important because the pressure exerted by a fluid is directly related to its weight-density multiplied by the depth of fluid above the point in question. Therefore, when you calculate fluid force, knowing the weight-density of the fluid—like seawater—is essential.

Integral Calculus

Integral calculus plays a vital role in calculating fluid force against surfaces submerged in a fluid. The fluid force can be computed using the integral of pressure over the area of interest. For our aquarium window example, we use the integration formula for fluid force: \[ F = \int_{a}^{b} \rho g(y) \underline{\phantom{xxx}} dy \] Here, \( F \) represents the total fluid force, \( \rho \) symbolizes the fluid's weight-density (such as 64 lb/ft³ for seawater), and \( g(y) \) represents the gravitational force at a given depth \( y \). The limits \( a \) and \( b \) correspond to the depths just below the water surface and even further down to the bottom of the window. The parameters \( y \) and \( dy \) represent integration with respect to the depth of the window submerged in water. By solving this integral, one can determine the total force exerted by the fluid across the submerged surface, giving insight into how the depth and size of the surface affect the force experienced by the glass window.

Unit Conversion

Converting measurements to consistent units is an essential skill, particularly in fluid dynamics problems. This skill ensures that measurements align correctly when substitutions are made into equations, avoiding errors in calculation. When solving the aquarium window problem, we convert inches to feet to match the unit of the seawater weight-density (lb/ft³) as much as are consistent in calculations.
This conversion involves recognizing that 12 inches make a foot, so to convert inches to feet, you simply divide by 12. For example, the upper boundary of the window at 0.5 inches below the surface converts to \( \frac{0.5}{12} \approx 0.0417 \) feet, and similarly, the lower boundary at 33.5 inches converts to \( \frac{33.5}{12} \approx 2.792 \) feet. By ensuring all dimensions—both vertical depths and horizontal spans—are in feet, the equations can be solved properly, leading to accurate fluid force calculations.

Fluid Dynamics

Fluid dynamics is the study of fluid flow, including the forces and effects of fluids in motion or at rest. In this context, understanding how fluids exert forces helps in correctly determining pressures and stresses on submerged surfaces. In the aquarium example, knowledge of fluid dynamics is used to compute the forces acting normally to the submerged window.
Fluid dynamics principles explain why fluid pressure increases with depth. The deeper you go, the higher the force due to the increased weight of the water above. In the equation for force \( F = \int_{a}^{b} \rho g(y) \underline{\phantom{xxx}} dy \), the function \( g(y) = y \) portrays how gravitational force increases linearly based on depth, underpinning this principle. Such insights from fluid dynamics guide the design and evaluation of systems interacting with fluids, ensuring structural integrity, safety, and efficiency.