46–50. Force on dams The following figures show the shapes and dimensions of small dams. Assuming the water level is at the top of the dam, find the total force on the face of the dam.

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Identify the shape of the dam face, which is a rectangle with width 40 m and height 10 m, submerged in water up to the top.

Recall that the pressure exerted by a fluid at depth \(y\) is given by \(P(y) = \rho g y\), where \(\rho\) is the density of water, \(g\) is the acceleration due to gravity, and \(y\) is the depth measured from the surface.

Set up the coordinate system with \(y=0\) at the water surface and \(y\) increasing downward to 10 m at the bottom of the dam.

Express the force on a thin horizontal strip of the dam at depth \(y\) with thickness \(dy\) as \(dF = P(y) \times \text{area of strip} = \rho g y \times (40 \times dy)\).

Integrate this expression from \(y=0\) to \(y=10\) to find the total force: \(F = \int_0^{10} \rho g y \times 40 \, dy\).

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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 given by the formula P = ρgh, where ρ is the fluid density, g is acceleration due to gravity, and h is the depth below the surface.

Force on a Surface Due to Fluid Pressure

The total force exerted by a fluid on a submerged surface is found by integrating the pressure over the area. Since pressure varies with depth, the force is the integral of pressure times the differential area, often simplified by using the average pressure at the centroid of the surface.

Calculus Integration for Variable Pressure

Calculus integration is used to sum the infinitesimal forces over the dam face where pressure changes with depth. Setting up an integral with pressure as a function of depth and integrating over the height of the dam yields the total force.

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