Problem 61 - Calculus Early Transcendentals

Question

A large tank has a plastic window on one wall that is designed to withstand a force of 90,000 N. The square window is \(2 \mathrm{m}\) on a side, and its lower edge is \(1 \mathrm{m}\) from the bottom of the tank. a. If the tank is filled to a depth of \(4 \mathrm{m}\), will the window withstand the resulting force? b. What is the maximum depth to which the tank can be filled without the window failing?

Step-by-Step Solution

Answer

Answer: Yes, the window can withstand the force when the tank is filled to a depth of 4 meters. The maximum depth to which the tank can be filled without the window failing is approximately 4.59 meters.

Step 1: Calculate the hydrostatic pressure at the center of the window

In order to determine the force acting on the window, we need to find the hydrostatic pressure at the center of the window when it is submerged. The hydrostatic pressure (\(P_h\)) at a certain depth can be calculated using the equation: \(P_h=\rho g h\) where \(\rho\) is the density of the fluid, \(g=9.8\mathrm{m/s^2}\) is the acceleration due to gravity, and \(h\) is the depth of the center of the window. First, let's find the depth of the center of the window. The window is \(2 \mathrm{m}\) wide and its lower edge is \(1 \mathrm{m}\) above the bottom of the tank. Therefore, the center of the window is at a height of \(1 + (2 / 2) = 2 \mathrm{m}\) from the bottom. Given the tank is filled to a depth of \(4 \mathrm{m}\), the depth of the center of the window below the surface is \(4 - 2 = 2 \mathrm{m}\). Assuming the tank is filled with water, the density of water is \(1000 \mathrm{kg/m^3}\). Now, we can calculate the hydrostatic pressure: \(P_h = 1000\mathrm{kg/m^3} * 9.8\mathrm{m/s^2} * 2\mathrm{m} = 19600\mathrm{Pa}\) (Pascals)

Step 2: Calculate the force acting on the window

Now that we have the hydrostatic pressure at the center of the window, we can find the force acting on the area of the window. The equation to calculate force due to pressure is: \(F = P A\) where \(P\) is the pressure and \(A\) is the area. Given that the window is a square of side \(2 \mathrm{m}\), its area is: \(A = 2\mathrm{m} * 2\mathrm{m} = 4\mathrm{m^2}\) Now, we can calculate the force acting on the window: \(F = 19600\mathrm{Pa} * 4\mathrm{m^2} = 78400\mathrm{N}\)

Step 3: Determine if the window can withstand the force when the tank is filled to a depth of 4 meters

The window is designed to withstand a force of \(90,000\mathrm{N}\). We have calculated the force acting on the window to be \(78,400\mathrm{N}\) when the tank is filled to a depth of \(4\mathrm{m}\). Since the calculated force (\(78400\mathrm{N}\)) is less than the maximum force the window can withstand (\(90,000\mathrm{N}\)), the window will not fail when the tank is filled to a depth of \(4\mathrm{m}\).

Step 4: Find the maximum depth to which the tank can be filled without the window failing

For finding the maximum depth that the tank can be filled, we can use the following equation derived from the earlier formulas: \(F = \rho g (h_d + h_w / 2) A\) where \(F\) is the force the window can withstand, \(h_d\) is the maximum filling depth of the tank, \(h_w\) is the distance from the bottom to the center of the window, and \(A\) is the area of the window. Now we can rearrange the equation for \(h_d\): \(h_d = \frac{F}{\rho g A} - \frac{h_w}{2}\) The values are: \(F = 90,000\mathrm{N}\), \(\rho = 1000\mathrm{kg/m^3}\), \(g = 9.8\mathrm{m/s^2}\), \(A = 4\mathrm{m^2}\), \(h_w = 2\mathrm{m}\) Substitute these into the equation to find the maximum filling depth: \(h_d = \frac{90,000\mathrm{N}}{1000\mathrm{kg/m^3}*9.8\mathrm{m/s^2}*4\mathrm{m^2}} - \frac{2\mathrm{m}}{2} = 4.5918\mathrm{m}\) The maximum depth to which the tank can be filled without the window failing is approximately \(4.59\mathrm{m}\).

Key Concepts

Force Calculation

Force calculation in physics is crucial for understanding how a force acts on an object due to external pressure. In the context of hydrostatics, this involves finding the force exerted by a fluid on a submerged surface like the tank’s window. Pressure, denoted as \( P \), and area, \( A \), are components in the formula:

• \( F = P \times A \)
• \( F \) is the force acting on the surface.

To illustrate, consider a square window with sides measuring \(2 \mathrm{m}\). Its area is calculated as \( A = 2 \mathrm{m} \times 2 \mathrm{m} = 4 \mathrm{m^2} \). Given a calculated pressure of \( 19600 \mathrm{Pa} \) (Pascal), the force \( F \) exerted by the fluid is \( 19600 \mathrm{Pa} \times 4 \mathrm{m^2} = 78400 \mathrm{N} \). This demonstrates how pressure over a surface area translates into a force. Breaking calculations into parts and applying these formulas ensures accuracy in determining whether structures can withstand forces.

Pressure Analysis

Pressure analysis is pivotal in fluid mechanics, particularly when assessing the effect of fluid pressure at various depths. Hydrostatic pressure is the pressure exerted by a fluid due to its weight and is calculated with:

• \( P_h = \rho \times g \times h \)

Where:

• \( \rho \) represents the fluid density, typically \(1000 \mathrm{kg/m^3}\) for water.
• \( g \) is the gravitational acceleration, \(9.8 \mathrm{m/s^2}\).
• \( h \) is the depth below the fluid surface.

Analyzing the window situation, assume the center is \(2 \mathrm{m}\) beneath the surface, hence \( P_h = 1000 \mathrm{kg/m^3} \times 9.8 \mathrm{m/s^2} \times 2 \mathrm{m} = 19600 \mathrm{Pa} \). By examining these components, you gain insights into how depth impacts pressure, a critical factor in ensuring structural integrity under fluid forces.

Fluid Mechanics

Fluid mechanics explores the behavior of fluids, essential in comprehending how fluids exert pressures on submerged surfaces. This discipline covers the principles underlying fluid pressure, density, and forces, as observable in the provided exercise. Hydrostatic pressure, linked to a fluid's static condition, is a significant aspect, with calculations predicated on variables like density and depth.
Key principles include:

• The pressure at any given depth is uniform in a fluid at rest.
• Fluids exert equal pressure in all directions.

Considering our example, fluid mechanics principles elucidate why pressure grows with depth. Understanding these fundamentals, such as how hydrostatic pressure translates into a directed force onto surfaces like the tank window, is fundamental for ensuring systems are designed to withstand specific fluid-induced loads, adding both practical and theoretical value to fields like engineering and physics.