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Chapter 6: Problem 65

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 m on a side, and its lower edge is 1 m from the bottom of the tank. a. If the tank is filled to a depth of 4 m, will the window with-stand the resulting force? b. What is the maximum depth to which the tank can be filled without the window failing?

Short Answer

Expert verified
A square plastic window in a large tank can withstand a force of 58,860 N when the tank is filled to a depth of 4 meters, which is less than its maximum force capacity of 90,000 N. The maximum depth the tank can be filled without the window failing is approximately 1.79 meters.

Step by step solution

01

Calculate the pressure exerted by the water at a depth of 4 meters

Use the fluid pressure equation with the given depth: \(P = \rho g h = (1000\,\text{kg/m}^3)(9.81\,\text{m/s}^2)(4\,\text{m}) = 39240\,\text{N/m}^2\)
02

Determine the mid-depth of the window

We will assume uniform pressure on the entire window since it is a relatively small area compared to the depth of the tank. Calculate the mid-depth of the window (1.5 m), where pressure would be representative of the overall pressure on the window: \(P_{mid} = \rho g h_{mid} = (1000\,\text{kg/m}^3)(9.81\,\text{m/s}^2)(1.5\,\text{m}) = 14715\,\text{N/m}^2\)
03

Calculate the force exerted on the window at 4-meter depth

Now, use the force exerted by a fluid formula, with the mid-depth pressure and the area of the window: \(F = P_{mid} \cdot A = (14715\,\text{N/m}^2)(2\,\text{m} \times 2\,\text{m}) = 58860\,\text{N}\)
04

Answer Part a

Compare the calculated force to the maximum force the window can withstand (90,000 N): Since \(58860\,\text{N} < 90000\,\text{N}\), the window will withstand the resulting force at a depth of 4 meters.
05

Find the pressure at which the window fails

We now need to find the pressure at which the window will fail. Use the force exerted by a fluid formula, solving for \(P_{fail}\): \(P_{fail} = \frac{F_{max}}{A} = \frac{90000\,\text{N}}{2\,\text{m} \times 2\,\text{m}} = 22500\,\text{N/m}^2\)
06

Determine the maximum depth without window failing

Now, use the fluid pressure equation, solving for the maximum depth (\({h}_{max}\)): \({h}_{max} = \frac{P_{fail}}{\rho g} = \frac{22500\,\text{N/m}^2}{(1000\,\text{kg/m}^3)(9.81\,\text{m/s}^2)} \approx 2.29\,\text{m}\) So the maximum depth from the surface to the mid-point of the window without failing is 2.29 m. Since the bottom edge is 1 m from the bottom, the maximum depth the tank can be filled without the window failing is approximately: \({h}_{max\_tank} = {h}_{max} - 0.5\,\text{m} = 2.29\,\text{m} - 0.5\,\text{m} \approx 1.79\,\text{m}\) So, the maximum depth the tank can be filled without the window failing is around 1.79 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Calculation
Pressure in a fluid is an important concept that helps determine how forces are distributed within it. When calculating pressure exerted by a fluid, we use the formula: \[ P = \rho g h \]where:
  • \( P \) is the pressure exerted by the fluid (in N/m虏 or Pascals)
  • \( \rho \) is the fluid density (for water, \( 1000 \, \text{kg/m}^3 \))
  • \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
  • \( h \) is the depth of the fluid
By inputting the known values of depth, density, and gravity for a tank filled to 4 meters, we calculate the pressure at any given depth. The depth directly impacts the pressure, as pressure increases linearly with depth. In practical terms, the deeper the water, the greater the pressure exerted at that point.
Force Exerted by a Fluid
Once we know the pressure exerted by a fluid, it is possible to calculate the force on a particular surface or area. The force is essentially the pressure acting over the area. This is given by the equation:\[ F = P \cdot A \]where:
  • \( F \) is the force exerted (in Newtons)
  • \( P \) is the pressure (calculated earlier, in N/m虏)
  • \( A \) is the area of the window (in square meters)
Calculating the force helps ensure that structures, like our tank window, can withstand the pressure imposed by the fluid above them. This aspect is critical for safety as inadequate force resistance can lead to structural failure, potentially causing harm.
Fluid Pressure Equation
The Fluid Pressure Equation is essential for understanding how to compute the pressure within a fluid at various depths. Applying these calculations ensures accurate assessments of the pressure at any point in the fluid. The standard equation utilizes the relationship between density, gravitational force, and height or depth as previously noted: \[ P = \rho g h \]It's important to understand that pressure is not constant throughout a fluid but varies with depth. Therefore, identifying the representative pressure at the mid-depth of a window, like in this exercise, gives a practical average for ensuring we account for all forces exerted by the fluid at that level.
Window Integrity Under Pressure
Ensuring that a window or any solid structure can withstand the pressure exerted by fluid is crucial for maintaining structural integrity. For the window in our tank, understanding the maximum force it can endure helps us set limits on operational depths. The force tolerance of the window at \( 90,000 \, \text{N} \) dictates how much pressure it can withstand before failing.From the pressure and force calculations, the maximum operational depth can be safely determined. The calculated maximum depth the window can handle without failing (approximately 1.79 meters in this example) provides a valuable benchmark for safety and operational guidelines.By keeping depths within this safe limit, designers and engineers ensure the durability and safety of the tank system.

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Most popular questions from this chapter

Surface-area-to-volume ratio (SAV) In the design of solid objects (both artificial and natural), the ratio of the surface area to the volume of the object is important. Animals typically generate heat at a rate proportional to their volume and lose heat at a rate proportional to their surface area. Therefore, animals with a low SAV ratio tend to retain heat, whereas animals with a high SAV ratio (such as children and hummingbirds) lose heat relatively quickly. a. What is the SAV ratio of a cube with side lengths \(R ?\) b. What is the SAV ratio of a ball with radius \(R ?\) c. Use the result of Exercise 38 to find the SAV ratio of an ellipsoid whose long axis has length \(2 R \sqrt[3]{4},\) for \(R \geq 1\) and whose other two axes have half the length of the long axis. (This scaling is used so that, for a given value of \(R,\) the volumes of the ellipsoid and the ball of radius \(R\) are equal.) The volume of a general ellipsoid is \(V=\frac{4 \pi}{3} A B C,\) where the axes have lengths \(2 A, 2 B,\) and \(2 C .\) d. Graph the SAV ratio of the ball of radius \(R \geq 1\) as a function of \(R\) (part (b)) and graph the SAV ratio of the ellipsoid described in part (c) on the same set of axes. Which object has the smaller SAV ratio? e. Among all ellipsoids of a fixed volume, which one would you choose for the shape of an animal if the goal were to minimize heat loss?

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