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Hydrostatic pressure is the pressure exerted by a fluid at equilibrium at a given depth due to the force of gravity. In a swimming pool, hydrostatic pressure increases with depth because the weight of the water above adds to the pressure exerted at any depth.
Understanding hydrostatic pressure helps to determine how forces distribute along surfaces submerged in a fluid. The formula to calculate this pressure at any depth is given by \[P = \rho g h\]where:

• \(P\) is the pressure at a given depth
• \(\rho\), which is approximately \(1000 \, \text{kg/m}^3\), represents the density of water
• \(g\), approximately \(9.81 \, \text{m/s}^2\), is the acceleration due to gravity
• \(h\) is the depth

This relationship shows that hydrostatic pressure directly depends on how deep a point is submerged in the fluid. Hence, in our exercise, it's essential to consider the varying depth to accurately calculate the force exerted by water.

The geometry of the pool can be visualized as a trapezoid when viewed from the side. This is because the pool's bottom slopes from one depth to another, creating a shape where the top and bottom sides are parallel but of different lengths, and the vertical sides are the non-parallel segments.
A trapezoidal shape simplifies calculating average values, such as the effective depth in our pool example. The depth varies from 1.00 meter at one end to 2.00 meters at the other end. Calculating the average depth involves finding the midpoint:\[\text{Average Depth} = \frac{1.00 \, \text{m} + 2.00 \, \text{m}}{2} = 1.50 \, \text{m}\]The trapezoidal view accounts for the continuous change in depth across the length of the pool, allowing for an effective simplification in force calculations by using this average depth in further computations.

To find the force exerted by water on the side of a pool, we integrate the hydrostatic pressure over the area of the side. This involves considering how the pressure varies with depth and summing these effects across the entire side. \[F = \int_0^{H} \rho g h \, A \, dh\]

• \( A \, dh \) denotes a small strip of area on the pool's side, perpendicular to the water's depth.
• The pool is 6 meters wide, so each strip is 6 meters long vertically.
• \( H \) represents the maximum depth, which is 2 meters.

By using the trapezoidal geometry and average depth, we simplify this to a direct calculation:\[F = \rho g L \frac{6.00}{2} (2.00^2 - 1.00^2) = 1000 \times 9.81 \times 15.0 \times 3.00 \times 1.50 = 662,175 \, \text{N}\]This force calculation gives a value of approximately 662,175 Newtons, describing the total force the water exerts against one of the pool's sides due to the uniform distribution of pressure along the depth.