Ch 14: Fluids and Elasticity

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 14: Fluids and Elasticity

Problem 47b

Chapter 14, Problem 47b

An aquarium of length L, width (front to back) W, and depth D is filled to the top with liquid of density ρ. Find an expression for the force of the liquid on the front window of the aquarium.

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The force exerted by a liquid on a surface is due to the pressure of the liquid. The pressure at a depth \( h \) in a liquid is given by \( P = \rho g h \), where \( \rho \) is the density of the liquid, \( g \) is the acceleration due to gravity, and \( h \) is the depth below the surface.

The pressure varies with depth, so we need to calculate the total force by integrating the pressure over the area of the front window. The front window has an area \( A = W \cdot D \), where \( W \) is the width and \( D \) is the depth.

Consider a horizontal strip of the front window at a depth \( h \) with a small thickness \( dh \). The pressure at this depth is \( P = \rho g h \), and the force on this strip is \( dF = P \cdot dA \), where \( dA = W \cdot dh \).

Substitute \( P \) and \( dA \) into the expression for \( dF \): \( dF = (\rho g h) (W \cdot dh) = \rho g W h \cdot dh \). To find the total force, integrate \( dF \) over the depth of the window from \( h = 0 \) to \( h = D \).

The total force is \( F = \int_0^D \rho g W h \; dh \). Solve the integral: \( F = \rho g W \int_0^D h \; dh = \rho g W \left[ \frac{h^2}{2} \right]_0^D = \frac{1}{2} \rho g W D^2 \). This is the expression for the force of the liquid on the front window of the aquarium.

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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 with depth and is given by the formula P = pgh, where P is the pressure, p is the fluid density, g is the acceleration due to gravity, and h is the depth of the fluid. This concept is crucial for understanding how pressure varies in a liquid and how it affects forces on submerged surfaces.

Force on a Surface

The force exerted by a fluid on a surface is calculated by integrating the pressure over the area of the surface. For a vertical surface like the front window of the aquarium, the total force can be found using F = ∫ P dA, where dA is an infinitesimal area element. This concept is essential for determining how the pressure from the liquid translates into a net force acting on the aquarium's window.

Center of Pressure

The center of pressure is the point on a submerged surface where the resultant pressure force acts. It is not necessarily at the centroid of the area, especially for vertical surfaces, and is influenced by the distribution of pressure across the surface. Understanding the center of pressure is important for analyzing stability and structural integrity in fluid mechanics, particularly in applications like aquariums.

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