Problem 18 - Applied Physics

Question

A filled water tower sits on the top of the highest hill in a town (use Fig. \(13.6\) for Problems 17-21). The cylindrical tower has a radius of \(12.0 \mathrm{~m}\) and a height of \(50.0 \mathrm{~m}\). Find the total force on the sides of the water tower.

Step-by-Step Solution

Answer

The total force on the sides of the water tower is approximately \( 1.85 \times 10^7 \) N.

Step 1: Understanding the problem

We are tasked with finding the total force exerted by the water on the sides of the cylindrical water tower. The key here is to recognize that this is a fluid force problem and the force exerted by the water is related to the pressure distribution along the height of the tank.

Step 2: Calculate water pressure at a depth

The pressure at any depth in a liquid is given by the equation \(P = \rho gh\), where \(\rho\) is the density of the liquid, \(g\) is the acceleration due to gravity, and \(h\) is the depth of the liquid at that point. Substitute \(\rho = 1000\, \text{kg/m}^3\) (density of water) and \(g = 9.81\, \text{m/s}^2\).

Step 3: Determine the force on a thin horizontal strip

Consider a thin horizontal strip at depth \(h\) with thickness \(dh\). The pressure at this strip is \(P = \rho gh\), and the area \(A\) of the strip is the circumference of the cylinder times \(dh\), which is \(A = 2\pi r \cdot dh\). The force \(dF\) on this strip is \(dF = P \cdot A = \rho g h \cdot 2 \pi r \cdot dh\).

Step 4: Integrate the force over the entire height

To find the total force, integrate \(dF = \rho g h \cdot 2 \pi r \cdot dh\) from \(h = 0\) to \(h = 50\, \text{m}\). The integral is \(F = \int_{0}^{50} \rho g h \cdot 2 \pi r \cdot dh\).

Step 5: Solve the integral

The integral \(F = 2 \pi r \rho g \int_{0}^{50} h \, dh\) evaluates to \(F = 2 \pi r \rho g \left[ \frac{1}{2}h^2 \right]_0^{50}\). Substitute \(r = 12\, \text{m}, \rho = 1000\, \text{kg/m}^3, g = 9.81\, \text{m/s}^2\).

Step 6: Calculate the final force value

Solve the expression: \(F = 2 \pi (12) (1000) (9.81) \left( \frac{1}{2}(50)^2 \right)\). Compute the final numerical value.

Key Concepts

Pressure Distribution

In fluid mechanics, understanding pressure distribution is key, especially when dealing with fluids at varying depths, like in a water tower. Pressure in a static fluid increases with depth due to the weight of the overlying fluid.
The formula for pressure at a certain depth is given by \( P = \rho gh \), where \( \rho \) is the fluid density, \( g \) is the acceleration due to gravity, and \( h \) is the depth within the fluid.
For water, which is a common fluid in many problems, \( \rho \) is typically \( 1000 \, \text{kg/m}^3 \) and \( g \) is \( 9.81 \, \text{m/s}^2 \).

• Understanding how pressure changes with depth is vital for calculating the forces exerted by fluids on surfaces, like the walls of a cylindrical tower.
• The increasing pressure with depth results in varying force distribution along the structure.

This concept applies widely from engineering to natural phenomena, making it foundational in fluid mechanics.

Cylindrical Water Tower

The geometry of a water tower greatly influences the fluid dynamics within. A cylindrical water tower is a common design, favoring simplicity and structural integrity.
In fluid mechanics, cylinders are often modeled due to their straightforward shape, which simplifies calculations and theoretical analysis.
A cylindrical tower has a uniform cross-section along its height, defined by its radius and height.

• In our problem, the cylinder has a radius of 12 meters and a height of 50 meters.
• This consistent shape makes it easier to calculate parameters such as volume, surface area, and pressure distribution.

Such structures must be designed to withstand different pressure levels along their height, primarily impacting the calculations for force exerted by the contained fluid.

Fluid Force Problem

Fluid force problems involve calculating the force exerted by a fluid on a surface. These calculations are crucial in designing containers or structures like water towers.
To solve these problems, one must integrate the pressure distribution over the area upon which the fluid acts.
In our specific exercise, the task is to determine the total force on the sides of the cylindrical water tower.

• The force on a small strip at a depth is calculated, considering pressure at that depth and the strip's area. This yields a differential force \( dF \) calculated as \( dF = \rho g h \cdot 2 \pi r \cdot dh \).
• The final step involves integrating this differential force across the entire height of the water tower to find the total force.

The methodology not only ensures accurate force calculations but also informs robust design and safety measures for fluid containers.

Pressure at Depth

The concept of pressure at a specific depth is intrinsic to solving fluid mechanics problems, as it establishes how significantly pressure increases with depth.
At any given point within a fluid, pressure is determined using the equation \( P = \rho gh \). Here, it's crucial to understand each element:

• \( \rho \): the density of the fluid, directly influencing how pressure accumulates as more water is overhead.
• \( g \): gravity's role means denser planets would see higher pressures faster with depth.
• \( h \): as one goes deeper, the pressure increases linearly with depth.

Practically, this principle assists in designing water towers by calculating how much pressure structures should withstand at varying depths, ensuring safe, efficient containment of the fluid.