Problem 39 - Thomas Calculus in SI Units

Question

The viewing portion of the rectangular glass window in a typical fish tank at the New England Aquarium in Boston is \(1.6 \mathrm{m}\) wide and runs from \(0.01 \mathrm{m}\) below the water's surface to \(0.85 \mathrm{m}\) below the surface. Find the fluid force against this portion of the window. The weight-density of seawater is \(10,050 \mathrm{N} / \mathrm{m}^{3} .\) (In case you were wondering, the glass is \(2 \mathrm{cm}\) thick and the tank walls extend \(10 \mathrm{cm}\) above the water to keep the fish from jumping out.)

Step-by-Step Solution

Answer

The fluid force against the window is approximately 5692.37 N.

Step 1: Understand the Problem Context

We need to find the fluid force against a portion of a rectangular glass window submerged in seawater. The width of the window is given as 1.6 meters, and it extends vertically from 0.01 meters to 0.85 meters below the water surface. The weight-density of the seawater is 10,050 N/m³.

Step 2: Determine the Depth Range of Pressure Application

The pressure on the window increases as the depth increases. The depth of the window ranges from 0.01 meters to 0.85 meters below the water's surface. Therefore, we need to consider how pressure varies over this depth of 0.84 meters.

Step 3: Calculate the Hydrostatic Pressure Formula

Hydrostatic pressure at a particular depth is given by \( P = \rho gh \), where \( \rho \) is the density of seawater (10,050 N/m³), \( g \) is the acceleration due to gravity (approximately 9.81 m/s²), and \( h \) is the depth.

Step 4: Set up the Integral for Fluid Force

The force on a horizontal strip at depth \( h \) is given by \( dF = P \times A = \rho gh \times (width \times dh) \). Since the width is constant at 1.6 meters, we set up the integral to find the total force as \[ F = \int_{0.01}^{0.85} 10050 \times 9.81 \times h \times 1.6 \, dh \].

Step 5: Evaluate the Integral

Evaluate the integral \[ F = 10050 \times 9.81 \times 1.6 \int_{0.01}^{0.85} h \, dh \] which simplifies to \[ F = 10050 \times 9.81 \times 1.6 \left[ \frac{1}{2}h^2 \right]_{0.01}^{0.85} \].

Step 6: Calculate and Solve

Calculate \( \frac{1}{2}h^2 \) from 0.01 to 0.85: \[ \left(\frac{1}{2} \times 0.85^2\right) - \left(\frac{1}{2} \times 0.01^2\right) = \frac{1}{2} \times (0.7225 - 0.0001) = 0.3612 \]. Now, substitute to get \[ F = 10050 \times 9.81 \times 1.6 \times 0.3612 \] to find the total force.

Step 7: Final Calculation

Perform the final multiplication: \[ F = 10050 \times 9.81 \times 1.6 \times 0.3612 = 5692.37 \] N. Thus, the fluid force against the window is approximately 5692.37 N.

Key Concepts

Understanding Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. Here, we look at water pressure exerted on part of a fish tank window submerged in seawater.
A key factor in hydrostatic pressure is depth. The deeper you go, the greater the pressure due to the increasing weight of the water above. This pressure difference is what makes ears pop when diving underwater.
Mathematically, we calculate hydrostatic pressure using the formula:\[ P = \rho gh \]Where:

• \( P \) is the hydrostatic pressure.
• \( \rho \) is the fluid's density.
• \( g \) is the acceleration due to gravity (approximated as 9.81 m/s² on Earth).
• \( h \) is the depth below the surface.

For a fish tank window, this pressure varies from the top to the bottom of the window because the water depth is different across the window's height.

Calculating Fluid Force

When we talk about fluid force on a surface submerged in a fluid, we are referring to the total force exerted by the fluid on that surface.
This force is a result of the hydrostatic pressure distributed over the area of the surface. Fluid force \( (F) \) is calculated by multiplying the pressure by the area, expressed as:\[ F = P \times A \]
This concept is particularly relevant when considering objects like the glass window of a fish tank. In this scenario, the area of interest is rectangular and extends vertically.
The pressure, as established, varies with depth, so we integrate the varying pressure over the window's area. To find the total force, use the formula:\[ F = \int_{h_1}^{h_2} \rho gh \times \text{width} \times dh \]Where:

• \(h_1\) and \(h_2\) are the depths at the top and bottom of the window, respectively.

Thus, for our aquarium, this yields a comprehensive measure of force impacting the window.

Integration of Pressure over Depth

Integration comes into play when we have to account for varying conditions, such as depth-dependent pressure in our example.
To determine the total fluid force on a submerged surface, you take the integral of pressure over the area dependent on depth. The math behind this requires breaking down the force into tiny strips covering the small variable distances from the top to bottom of the window.
Each strip at depth \(h\) experiences a force \(dF \), calculated as:\[ dF = \rho gh \times \text{(strip width)} \times dh \]
Then, sum up these tiny forces over the depth range by integrating from \(h_1\) = 0.01 m to \(h_2\) = 0.85 m. Solve:\[ F = 10050 \times 9.81 \times 1.6 \int_{0.01}^{0.85} h \, dh \]The result of the integral gives the total fluid force exerted on the fish tank window.
The integration simplifies calculations by turning what is conceptually a summation into a single mathematical expression that accounts for changes in depth across the window.