Problem 41 - Thomas Calculus

Question

Tilted plate Calculate the fluid force on one side of a 5 ft by 5 ft square plate if the plate is at the bottom of a pool filled with water to a depth of 8 \(\mathrm{ft}\) and \begin{equation} \begin{array}{l}{\text { a. lying flat on its } 5 \mathrm{ft} \text { by } 5 \text { ft face. }} \\ {\text { b. resting vertically on a } 5 \text { -ft edge. }} \\ {\text { c. resting on a } 5 \text { -ft edge and tilted at } 45^{\circ} \text { to the bottom of the pool. }}\end{array} \end{equation}

Step-by-Step Solution

Answer

a. 12,480 lb; b. 16,250 lb; c. 15,591 lb.

Step 1: Calculate the Force for a Flat Plate

For a flat plate, the depth of the fluid above is constant at 8 ft. The force is the pressure times the area: \( F = P \times A \). Pressure \( P = \rho \times g \times h \), where \( \rho = 62.4 \, \text{lb/ft}^3 \) is the density of water, \( g = 32.2 \, \text{ft/s}^2 \) is the acceleration due to gravity, and \( h = 8 \, \text{ft} \). Calculate the area \( A = 25 \, \text{ft}^2 \). Therefore, \( P = 62.4 \times 8 \) and \( F = 62.4 \times 8 \times 25 = 12,480 \, \text{lb} \).

Step 2: Calculate the Force for a Vertical Plate

For a vertical plate, the pressure varies with depth. We integrate pressures across the plate height with: \( F = \int_{0}^{5} \rho \times g \times (8 + y) \times 5 \, dy \). Simplifying: \( F = 5 \times \int_{0}^{5} 62.4 \times (8 + y) \, dy \). This computes to \( F = 5 \times 62.4 \times [8y + \frac{y^2}{2}]_0^5 = 5 \times 62.4 \times (40 + 12.5) = 16,250 \, \text{lb} \).

Step 3: Calculate the Force for a Tilted Plate at 45°

For a plate tilted at 45°, we need to adjust the integration for the variable depth of water. The depth of each portion of the plate is \( 8 + y \sin(45^{\circ}) = 8 + \frac{y}{\sqrt{2}} \). The force is \( F = \int_{0}^{5} \rho \times g \times (8 + \frac{y}{\sqrt{2}}) \times 5 \, dy \). Simplified and integrated, calculate: \( F = 5 \times 62.4 \times \left[8y + \frac{y^2}{2 \sqrt{2}}\right]_0^5 \). After evaluating, \( F = 5 \times 62.4 \times (40 + \frac{12.5}{\sqrt{2}}) \approx 15,591 \, \text{lb} \).

Key Concepts

Pressure Calculation

Pressure in fluids results from the weight of the fluid above exerting force across a surface. This pressure at a specific depth in a fluid is calculated using the formula: \( P = \rho \times g \times h \). Here, \( \rho \) represents the fluid's density, \( g \) is the acceleration due to gravity, and \( h \) is the depth of the fluid.

• For instance, in the original exercise, the density of water is given as \( 62.4 \, \text{lb/ft}^3 \).
• The acceleration due to gravity is \( 32.2 \, \text{ft/s}^2 \).
• For a flat plate at a constant depth, this means the entire plate experiences pressure from an 8 ft depth of water.

Thus, pressure calculations are straightforward in cases where the depth is constant across the plate.

Integration in Calculus

Integration is a pivotal tool in calculus, providing a way to calculate total quantities that vary across a domain. In the context of the exercise, integration is used to compute the fluid force on a vertical or tilted plate.

When the plate stands upright, the pressure isn't constant over its height, but changes with each vertical point. By setting up an integral \( F = \int_{0}^{5} \rho \times g \times (8 + y) \times 5 \, dy \), we account for how the force varies with depth. Here, \( y \) represents the variable depth along the height of the plate.

• This approach allows accumulating all incremental forces from various depths into a single total force.
• For tilted plates, the integration becomes slightly more complex since the depth changes with the angle, requiring adjustments in the integral to reflect the different depth contributions.

Hence, integration proves essential for accurately determining forces in systems where conditions like depth aren't uniform.

Hydrostatic Pressure

Hydrostatic pressure refers to the pressure exerted by a fluid at rest, due to the force of gravity. It depends on the fluid's density, gravity, and depth below the fluid's surface. This type of pressure increases uniformly with depth.

In simple terms, imagine layers of water: each layer presses down on the one below it, creating pressure. The deeper you go, the more layers there are above you, and the greater the pressure.

• In the exercise, hydrostatic pressure is uniform across the horizontal plate due to its constant depth.
• For vertical and tilted plates, the pressure varies with depth, showcasing non-uniform hydrostatic pressure across the surface.

Understanding hydrostatic pressure helps in predicting how pressure increases within a body of water as depth increases.

Acceleration Due to Gravity

The acceleration due to gravity, denoted as \( g \), represents the rate at which objects accelerate towards the Earth. This value is essential in calculating forces involving gravity-effected bodies because it dictates how strongly bodies experience gravitational pull.

• On Earth, this value is approximately \( 32.2 \, \text{ft/s}^2 \), influencing how water applies pressure based on depth.
• In the exercise, it remains constant, which simplifies calculations as gravity affects all vertical forces consistently.

By understanding gravity's constant pull, students can better grasp how fluid forces and pressure calculations factor gravity into their equations.