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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.7.65a
Chapter 6, Problem 6.7.65a

Critical depth 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 withstand the resulting force?

Verified step by step guidance
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Identify the physical principle involved: The force on the window is due to the hydrostatic pressure exerted by the water. Hydrostatic pressure increases linearly with depth according to the formula \(P = \rho g h\), where \(\rho\) is the density of water, \(g\) is the acceleration due to gravity, and \(h\) is the depth below the water surface.
Determine the depth range over which the window extends. The window is 2 m tall, with its lower edge 1 m from the bottom of the tank. Since the tank is filled to 4 m, the bottom of the tank is at 0 m depth, so the window extends from 1 m to 3 m below the water surface.
Calculate the pressure at the top and bottom edges of the window using \(P = \rho g h\). Let \(P_{top} = \rho g \times 1\) m and \(P_{bottom} = \rho g \times 3\) m. This gives the pressure distribution over the window height.
Find the average pressure on the window by taking the average of the pressures at the top and bottom edges: \(P_{avg} = \frac{P_{top} + P_{bottom}}{2}\).
Calculate the total force on the window by multiplying the average pressure by the area of the window: \(F = P_{avg} \times A\), where \(A = 2 \text{ m} \times 2 \text{ m} = 4 \text{ m}^2\). Compare this force to the maximum force the window can withstand (90,000 N) to determine if it will hold.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It increases linearly with depth and is calculated as P = ρgh, where ρ is fluid density, g is gravitational acceleration, and h is the depth below the surface.

Force on a Submerged Surface

The force exerted by a fluid on a submerged surface depends on the pressure distribution over the area. For a vertical surface, the total force is the pressure at the centroid multiplied by the area, considering that pressure varies with depth.
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Center of Pressure

The center of pressure is the point where the resultant hydrostatic force acts on a submerged surface. It lies below the centroid because pressure increases with depth, affecting the moment calculations for stability and force analysis.
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