Problem 43 - Thomas Calculus

Question

Tilted plate Calculate the fluid force on one side of a 5 \(\mathrm{ft}\) by 5 \(\mathrm{ft}\) 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 \text { ft 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) 12480 lb, (b) 8580 lb, (c) 8580 lb.

Step 1: Understanding Fluid Pressure

Fluid pressure is calculated using the formula \( P = ho g h \), where \( \rho \) is the density of the fluid (water, in this case, with \( \rho = 62.4 \text{ lb/ft}^3 \)), \( g \) is the gravitational acceleration (32.2 ft/s², but often absorbed in \( \rho \) for simplicity in imperial units), and \( h \) is the depth of water above the point.

Step 2: Calculating Fluid Force when the Plate is Flat

When the plate lies flat at the bottom, the depth above is constant at 8 ft. The pressure at this depth is \( P = 62.4 \times 8 \). The force on the plate is the pressure multiplied by the area of the plate, \( F = P imes A \), where \( A = 5 \times 5 = 25 \text{ ft}^2 \). Thus, \( F = 62.4 \times 8 \times 25 \).

Step 3: Calculating Fluid Force when the Plate is Vertical

When the plate is vertical, the depth varies from 3 to 8 ft. The average depth is \( \frac{3+8}{2} = 5.5 \text{ ft} \). Use this average depth for pressure: \( P = 62.4 \times 5.5 \). The force is \( F = P \times A \) with the same area \( A = 25 \text{ ft}^2 \), so \( F = 62.4 \times 5.5 \times 25 \).

Step 4: Calculating Fluid Force when the Plate is Tilted

When the plate is tilted at 45°, the effective depth changes linearly from 3 ft (top edge) to 8 ft (bottom edge). The average depth can be calculated as \( \frac{3 + 8}{2} = 5.5 \text{ ft} \), similar to the vertical position. Thus, \( P = 62.4 \times 5.5 \), and \( F = P \times A = 62.4 \times 5.5 \times 25 \).

Step 5: Final Calculation and Comparing Forces

The fluid force calculations for each part are as follows: - (a) Flat: \( F = 62.4 \times 8 \times 25 = 12480 \text{ lb} \).- (b) Vertical: \( F = 62.4 \times 5.5 \times 25 = 8580 \text{ lb} \).- (c) Tilted: \( F = 8580 \text{ lb} \) (same as vertical since the average depth remains 5.5 ft).

Key Concepts

Fluid Pressure

Fluid pressure is a crucial concept when working with forces applied by fluids like water. Pressure in a fluid is calculated using the formula: \[ P = \rho gh \]Here, \( \rho \) represents the density of the fluid, \( g \) is the gravitational acceleration, and \( h \) is the depth of the fluid above the point where the pressure is being measured. This formula indicates that pressure increases with depth due to the weight of the fluid above. In our specific example, water density \( \rho \) is given as 62.4 lb/ft³. Gravitational acceleration, \( g \), is often integrated into the density value in imperial units to simplify calculations. Therefore, pressure changes as the depth \( h \) of the water over the plate changes.

Density of Water

The density of water is usually a constant known value in physics and engineering problems. In the imperial unit system, the density of water is given as 62.4 lb/ft³. Density is an essential part of calculating pressure because it determines how much fluid mass is present over a given area.

• Importance: Density affects the weight of water above the plate, influencing fluid pressure.
• Consistency: In most calculations within a single body of water, density remains constant.

Understanding water density helps in calculating the fluid pressure exerted on submerged objects such as plates, determining the force necessary to maintain or move them.

Tilted Plate

A tilted plate introduces a bit of complexity into force calculations because the depth of water above each portion of the plate changes. The tilt angle, 45° in our exercise, affects how water's depth over the plate transitions from the top edge to the bottom edge.When a plate is tilted, the depth varies linearly from the shallowest point to the deepest. In this example:- Top edge: 3 ft depth- Bottom edge: 8 ft depthThe average depth, calculated as \((3 + 8)/2 = 5.5\) ft, is useful for a simplified approach to pressure calculation, particularly with uniform dimensions like our 5 by 5 ft plate.

Calculation of Pressure

Calculating pressure exerted by water involves multiplying the depth-average pressure from water with the area of the submerged object. The steps are as follows:

• Calculate depth-dependent pressure using \( P = \rho gh \).
• Identify the area \( A \) of the submerged plate, which in this case is always 25 ft².
• Compute the force as \( F = P \times A \).

For a flat laying plate, where water depth is constant, pressure is assessed directly using the constant depth. For vertical or tilted plates, the average depth simplifies calculations, particularly when the fluid's density remains constant. This method effectively allows us to determine forces exerted by fluid on objects submerged at varying angles and positions.