Problem 53 - University Physics with Modern Physics

Question

A swimming pool is 5.0 \(\mathrm{m}\) long, 4.0 \(\mathrm{m}\) wide, and 3.0 \(\mathrm{m}\) deep. Compute the force exerted by the water against (a) the bottom and (b) either end. (Hint: Calculate the force on a thin, horizontal strip at a depth \(h,\) and integrate this over the end of the pool.) Do not include the force due to air pressure.

Step-by-Step Solution

Answer

Force on the bottom: 588,000 N; either end: 176,400 N.

Step 1: Determine the Volume of the Pool

The volume of the pool is given by the formula for the volume of a rectangular prism: \(V = ext{length} \times ext{width} \times ext{depth}\). Substitute the values: \(V = 5.0 \, \mathrm{m} \times 4.0 \, \mathrm{m} \times 3.0 \, \mathrm{m} = 60.0 \, \mathrm{m}^3\).

Step 2: Calculate the Force on the Bottom of the Pool

The force exerted by the water on the bottom of the pool is given by the product of the pressure at the bottom and the area of the bottom. Pressure at a depth is given by \(P = \rho gh\), where \(\rho = 1000 \, \mathrm{kg/m}^3\) is the density of water, \(g = 9.8 \, \mathrm{m/s}^2\) is the acceleration due to gravity, and \(h = 3.0\, \mathrm{m}\) is the depth. Calculate \(P = 1000 \times 9.8 \times 3.0 = 29400 \, \mathrm{Pa}\). The area of the bottom is \(5.0 \, \mathrm{m} \times 4.0 \, \mathrm{m} = 20 \, \mathrm{m}^2\). Thus, the force on the bottom is \(F = P \times \text{Area} = 29400 \, \mathrm{Pa} \times 20 \, \mathrm{m}^2 = 588000 \, \mathrm{N}\).

Step 3: Understand the Force on Either End of the Pool

For the force on either end, calculate the pressure using a strip method. Consider a strip at a depth \(h\) with a thickness \(dh\). Pressure \(P = \rho gh\) acts on each of these horizontal strips. The area \(A\) of each strip is the width of the pool \(4.0 \, \mathrm{m}\) times the thickness \(dh\), so \(dA = 4.0 \, \mathrm{m} \times dh\).

Step 4: Integrate the Pressure Across the End

The force on a strip at depth \(h\) is \(dF = P \times dA = (\rho gh)(4.0 \, dh)\). To find the total force on one end, integrate from the top to the bottom of the pool (0 to 3.0 m): \[F = \int_0^3 (1000 \times 9.8 \times h \times 4.0) \, dh = 39200 \int_0^3 h \, dh = 39200 \left[\frac{h^2}{2}\right]_0^3 = 39200 \times \frac{3^2}{2} = 176400 \, \mathrm{N}.\]

Step 5: Summarize the Results

The force exerted by the water against the bottom of the pool is \(588000 \, \mathrm{N}\) and the force against either end of the pool is \(176400 \, \mathrm{N}\).

Key Concepts

Understanding Hydrostatic Pressure

Hydrostatic pressure is a fundamental concept in fluid mechanics that describes the pressure exerted by a fluid at rest. This pressure increases with depth due to the weight of the fluid above. When you dive deeper into a pool, you notice your ears feeling more pressure—that's hydrostatic pressure at work! Here’s how it works:
- **Definition:** Hydrostatic pressure at any given depth is given by the formula \( P = \rho gh \).
- \( \rho \) represents the density of the fluid (for water, it’s \( 1000 \, \text{kg/m}^3 \)).
- \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)).
- \( h \) is the depth in the fluid.
- **Example:** In our pool scenario, at the bottom (3 meters deep), the hydrostatic pressure is \( 29400 \, \text{Pa} \).
This concept is essential to determine forces exerted by water on submerged surfaces, which engineers and designers consider in many applications, from designing dams to calculating loads on marine vehicles.

Force Calculation in Fluid Mechanics

Calculating force in fluid mechanics often involves understanding the relationship between pressure and area. If you know the pressure at a surface and the area of that surface, you're halfway there! Here’s how it applies to the pool example:
- **Force on the Bottom:**
- The pressure at the bottom is already computed as \( 29400 \, \text{Pa} \).
- The area of the pool's bottom is \( 20 \, \text{m}^2 \) (calculated as length times width).
- Therefore, the force exerted by the water on the bottom is \( F = P \times \text{Area} = 29400 \, \text{Pa} \times 20 \, \text{m}^2 = 588000 \, \text{N} \).
- **Force on the Ends:** Unlike the bottom, the pressure against the vertical ends varies with depth. This requires a bit more complex calculation—the integration, which ties into the next section.

Integral Calculus in Fluid Mechanics

Integral calculus is a powerful mathematical tool used extensively in fluid mechanics to calculate varying forces, like those on vertical surfaces submerged in liquids.
- **Why Integration?** In the context of the pool, pressure increases with depth. To determine the total force on the ends of the pool, which are vertical, we divide the end into small strips and calculate the force on each strip separately. This is where the integral comes in handy—summing up the forces on all these strips.
- **Integration Process:**
- For a strip at depth \( h \), the force \( dF \) is \( P \times dA = (\rho gh)(4.0 \, dh) \).
- Conducting the integration, we have: \[F = \int_0^3 (1000 \times 9.8 \times h \times 4.0) \, dh = 39200 \int_0^3 h \, dh = 39200 \left[\frac{h^2}{2}\right]_0^3 = 176400 \, \text{N}.\] - This calculation gives the total force on one vertical end of the pool.
By understanding and applying integral calculus, you can account for changing pressures and compute forces in complex systems. It's a crucial tool in analyzing fluid-related phenomena.