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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.R.85
Chapter 6, Problem 6.R.85

Force on a dam Find the total force on the face of a semicircular dam with a radius of 20 m when its reservoir is full of water. The diameter of the semicircle is the top of the dam.

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Identify the physical setup: The dam face is a vertical semicircle with radius \(r = 20\) m, and the water exerts pressure on it. The pressure at a depth \(h\) in a fluid is given by \(p = \rho g h\), where \(\rho\) is the density of water and \(g\) is the acceleration due to gravity.
Set up a coordinate system: Let the vertical axis \(y\) measure depth from the water surface downward, with \(y=0\) at the top (diameter) of the semicircle and \(y\) increasing downward to \(y=20\) m at the bottom of the dam.
Express the width of the dam at depth \(y\): Since the dam face is a semicircle, the horizontal width at depth \(y\) corresponds to the length of a horizontal chord of the semicircle. Using the circle equation \(x^2 + y^2 = r^2\), the half-width at depth \(y\) is \(x = \sqrt{r^2 - y^2}\), so the full width is \(2\sqrt{r^2 - y^2}\).
Write the differential force \(dF\) on a thin horizontal strip of thickness \(dy\) at depth \(y\): The pressure at depth \(y\) is \(p = \rho g y\), and the area of the strip is \(dA = \text{width} \times dy = 2\sqrt{r^2 - y^2} \, dy\). Therefore, \(dF = p \, dA = \rho g y \cdot 2\sqrt{r^2 - y^2} \, dy\).
Integrate the differential force over the depth of the dam from \(y=0\) to \(y=r\) to find the total force: \(F = \int_0^r 2 \rho g y \sqrt{r^2 - y^2} \, dy\). Set up this integral and prepare to evaluate it to find the total force on the dam face.

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Key Concepts

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

Hydrostatic Pressure Distribution

Hydrostatic pressure increases linearly with depth in a fluid at rest, given by p = ρgh, where ρ is fluid density, g is gravity, and h is depth. Understanding this pressure variation is essential to calculate the force exerted by water on submerged surfaces like a dam.
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Force on a Curved Surface

The total force on a curved surface submerged in a fluid is found by integrating the pressure over the surface area. For a semicircular dam, this involves setting up an integral that accounts for the shape and depth-dependent pressure to find the resultant force.
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Geometry of a Semicircle and Coordinate Setup

Accurately describing the semicircular dam requires understanding its geometry, including radius and orientation. Setting up a coordinate system helps express depth as a function of position along the dam face, which is crucial for integrating pressure over the curved surface.
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