Problem 41 - Thomas Calculus in SI Units

Question

Calculate the fluid force on one side of a \(1 \mathrm{m}\) by 1 \(\mathrm{m}\) square plate if the plate is at the bottom of a pool filled with water to a depth of \(2 \mathrm{m}\) and a. lying flat on its \(1 \mathrm{m}\) by \(1 \mathrm{m}\) face. b. resting vertically on a \(1 \mathrm{m}\) edge. c. resting on a \(1 \mathrm{m}\) edge and tilted at \(45^{\circ}\) to the bottom of the pool.

Step-by-Step Solution

Answer

a) 19600 N, b) 14700 N, c) 14700 N.

Step 1: Understand Fluid Force

Fluid force on an area submerged in a liquid is the pressure exerted by the fluid multiplied by the area of the surface. The pressure at a depth is given by \( P = \rho \cdot g \cdot h \), where \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the depth.

Step 2: Calculate Fluid Force when Plate is Lying Flat

For case (a), the plate is at a uniform depth of \(2 \; \mathrm{m}\). Given \( \rho = 1000 \, \mathrm{kg/m^3} \) (density of water) and \( g = 9.8 \, \mathrm{m/s^2} \), the pressure at \(2 \, \mathrm{m}\) depth is \( P = 1000 \times 9.8 \times 2 = 19600 \, \mathrm{N/m^2} \). The fluid force is then the pressure times the area: \( 19600 \, \mathrm{N/m^2} \times 1 \, \mathrm{m^2} = 19600 \, \mathrm{N} \).

Step 3: Calculate Fluid Force when Plate is Vertical

For case (b), the plate's depth varies linearly from \(1 \, \mathrm{m}\) to \(2 \, \mathrm{m}\). The average depth is \(1.5 \; \mathrm{m}\). Thus, the fluid pressure at this average depth is \( P = 1000 \times 9.8 \times 1.5 = 14700 \; \mathrm{N/m^2} \). Therefore, the force is \( 14700 \times 1 = 14700 \, \mathrm{N} \).

Step 4: Calculate Fluid Force when Plate is Tilted

For case (c), when tilted at \(45^\circ\), calculate the effective depth as the average. The plate spans vertically from \( rac{\sqrt{2}}{2} \, \mathrm{m} \) depth above \(2 \, \mathrm{m}\) to \(1 \, \mathrm{m}\) (as diagonal length \( \sqrt{2} \) will have components of \([1, 1])\). The effective depth is \( 1.5 \, \mathrm{m} \). Pressure is \( 14700 \times 1 = 14700 \, \mathrm{N} \).

Step 5: Conclusion

The total fluid forces for each condition: (a) is \(19600 \, \mathrm{N}\), (b) is \(14700 \, \mathrm{N}\), and (c) is \(14700 \, \mathrm{N}\). These forces are due to how depth variations are averaged through these positions.

Key Concepts

Hydrostatics

Hydrostatics is the branch of physics that deals with fluids at rest. It focuses on understanding how forces are exerted by a static fluid. In the context of our exercise, hydrostatics allows us to determine how water pressure varies with depth, influencing the fluid force acting on submerged objects like our square plate.

• Consider a pool filled with a fluid, such as water, that is not moving.
• The pressure at any point can be calculated using hydrostatic principles, and it increases linearly with depth due to the weight of the water above.
• We use this concept to calculate the pressure on one side of the submerged plate.

Understanding hydrostatics is crucial for solving problems involving submerged surfaces by quantifying how pressure at a certain depth affects that surface.

Pressure in Fluids

Pressure in a fluid is the force exerted per unit area. It is a fundamental concept in fluid mechanics and crucial for calculating the fluid force in our exercise. The formula used for calculating pressure in fluids is given by: \[ P = \rho \cdot g \cdot h \]Where:

• \( P \) is the pressure.
• \( \rho \) represents the density of the fluid.
• \( g \) is the acceleration due to gravity.
• \( h \) is the depth of the fluid above the point in question.

The pressure on the plate varies depending on its position in the water. For example, when the plate is flat at the pool's bottom, the pressure calculation is straightforward. However, when the plate is at an angle, the depth is calculated as an average to account for the varying pressure along the tilted surface.

Density of Fluids

Density is a measure of how much mass a substance has per unit volume, and it plays a vital role in determining fluid pressure. In our exercise, water density is given as \( 1000 \, \mathrm{kg/m^3} \), a common approximation for water.

• Higher density fluids exert more pressure at a given depth than lower density ones, as they weigh more over a given area.
• This directly affects the force perceived on submerged objects.
• In calculations, density is multiplied by gravitational acceleration and depth to give the pressure acting at a point in a fluid.

Understanding the role of fluid density is essential, as it helps us understand how different substances exert varying levels of pressure in similar situations.

Geometry in Calculus

Geometry and calculus interplay in analyzing how fluid force interacts with differently oriented surfaces. Accurate depth and pressure calculations depend on understanding the structure of the submerged object.

• When the plate is flat, its geometry makes calculations direct, as the depth is uniform.
• If the plate is vertical, the depth changes linearly from top to bottom, and the mean depth is used for calculations.
• For scenarios where the plate is tilted, calculus methods assist in determining the effective pressure by integrating over the surface.

By combining geometry with calculus, we determine average depths and use them to calculate the pressure distribution, illustrating the importance of geometry in fluid force calculations.