Question

A diving pool that is 4 m deep and full of water has a viewing window on one of its vertical walls. Find the force on the following windows. The window is a circle, with a radius of \(0.5 \mathrm{m}\), tangent to the bottom of the pool.

Short answer

Answer: The force acting on the circular viewing window is approximately 30,808 N.

Step-by-step solution

Identify Given Variables

We're given the following information: - The diving pool is 4 m deep, so the depth above the center of the window is 4 m. - The window is a circle with a radius of \(0.5 \mathrm{m}\). - The density of water, \(\rho\), is \(1000 \mathrm{\frac{kg}{m^3}}\). - Gravitational acceleration, \(g\), is \(9.81 \mathrm{\frac{m}{s^2}}\).

Calculate the Pressure Acting on the Window

We'll use the formula for pressure: \(P = \rho g h\) Plugging in the given variables: \(P = 1000 \mathrm{\frac{kg}{m^3}} * 9.81 \mathrm{\frac{m}{s^2}} * 4 \mathrm{m}\) Calculating the pressure: \(P = 39240 \mathrm{\frac{N}{m^2}}\)

Calculate the Area of the Circular Window

The area of a circle \(A\) is given by: \(A = \pi r^2\) Using the radius given (0.5 m), we can calculate the area of the viewing window: \(A = \pi (0.5 \mathrm{m})^2\) Calculating the area: \(A \approx 0.785 \mathrm{m^2}\)

Calculate the Force on the Window

Now that we have the pressure and area, we can find the force acting on the window using the following formula: \(F = P * A\) Plugging in the calculated pressure and area: \(F = 39240 \mathrm{\frac{N}{m^2}} * 0.785 \mathrm{m^2}\) Calculating the force: \(F \approx 30808 \mathrm{N}\) So, the force acting on the circular viewing window is approximately 30,808 N.

Key concepts

Circular Area

When solving problems related to circular areas, like the window described in the diving pool exercise, it's important to understand how to calculate the area of a circle. The area of a circle is calculated using the formula:
\( A = \pi r^2 \)

• \( \pi \) is a constant approximately equal to 3.14159.
• \( r \) represents the radius of the circle, which is the distance from the center of the circle to any point on its boundary.

For the circular window in this exercise, the radius is given as 0.5 meters. Substituting this into the area formula:
\( A = \pi (0.5 \ m)^2 \)
You calculate the area to be approximately 0.785 square meters. Understanding this concept helps in problems where you need to find how much space a circular object covers or when calculating forces distributed over circular areas.

Force Calculation

In problems involving submerged surfaces like the diving pool's window, calculating the force exerted by the fluid is key. The force can be determined once you know the pressure exerted by the fluid and the area over which this pressure acts. The formula used here is:
\( F = P \times A \)

• \( F \) represents the force in newtons (N).
• \( P \) is the pressure in pascals (Pa), or newtons per square meter (\( \frac{N}{m^2} \)).
• \( A \) is the area in square meters (\( m^2 \)).

From our exercise, the pressure calculated was 39240 pascals, and the previously calculated area of the circle was 0.785 square meters. Thus, the force is calculated as:
\( F = 39240 \ \frac{N}{m^2} \times 0.785 \ m^2 \)
This results in a force of approximately 30808 newtons. Understanding the relationship between pressure, force, and area is crucial in fluid mechanics when assessing the impact that fluids have on structures.

Fluid Mechanics

Fluid mechanics is the branch of physics concerned with fluids (liquids and gases) and the forces on them. A key concept within fluid mechanics is hydrostatic pressure, which describes the pressure exerted by a fluid at equilibrium due to the force of gravity. Hydrostatic pressure increases with depth, as more fluid weight rests above the point of measurement.
To calculate hydrostatic pressure, we use:

• \( P = \rho g h \)
• \( \rho \) is the fluid density, measured in kilograms per cubic meter (\( \frac{kg}{m^3} \)). For water, \( \rho \) is typically 1000 \( \frac{kg}{m^3} \).
• \( g \) represents gravitational acceleration, approximately 9.81 meters per second squared (\( \frac{m}{s^2} \)) on Earth.
• \( h \) is the height of the fluid column above the point of interest, measured in meters.

In our diving pool exercise, the window is subject to hydrostatic pressure from the water 4 meters above it. By understanding these principles, we can better analyze how fluids exert pressure on submerged surfaces.