Problem 25 - Basic Technical Mathematics with Calculus

Question

A swimming pool is \(6.00 \mathrm{m}\) wide and \(15.0 \mathrm{m}\) long. The bottom has a constant slope such that the water is \(1.00 \mathrm{m}\) deep at one end and \(2.00 \mathrm{m}\) deep at the other end. Find the force of the water on one of the sides of the pool.

Step-by-Step Solution

Answer

The force of the water on one side of the pool is approximately 662,175 N.

Step 1: Understand the problem

We are trying to find the force exerted by the water on one of the vertical sides of the pool due to water pressure. The pressure varies linearly with depth, and thus the force does too.

Step 2: Identify the geometry of the pool

The pool has a sloped bottom, leading to a trapezoidal side view. The pool is 6.00 meters wide, with depths ranging from 1.00 meter to 2.00 meters along its length of 15.0 meters.

Step 3: Calculate average depth

Since the pool has a constant slope, the average depth is the midpoint of the depths at each end: \[ ext{Average Depth} = \frac{1.00 ext{ m} + 2.00 ext{ m}}{2} = 1.50 ext{ m} \]

Step 4: Determine hydrostatic pressure formula

Hydrostatic pressure at a depth \( h \) is given by \( P = \rho g h \), where \( \rho \approx 1000 \text{ kg/m}^3 \) is the density of water, and \( g \approx 9.81 \text{ m/s}^2 \) is the acceleration due to gravity.

Step 5: Calculate force on one side

The force due to water pressure is the integral of pressure over the surface: \[ F = \int_0^{H} \rho g h \, A \, dh \]Where \( H \) is the maximum depth (2.00 m) and \( A \, dh \) is the elemental area strip of width \( 6.00 \text{ m} \):\[ F = \rho g L \frac{6.00}{2} (2.00^2 - 1.00^2) = 1000 \times 9.81 \times 15.0 \times 3.00 \times 1.50 \]Simplifying:\[ F = 1000 \times 9.81 \times 15.0 \times 3.00 \times 1.50 = 662,175 \text{ N} \]

Step 6: Present the final force calculation

Thus, the force exerted by the water on one side of the pool is approximately 662,175 Newtons.

Key Concepts

Hydrostatic Pressure

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.

Trapezoidal Geometry

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.

Water Force Calculation

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.