Problem 11 - The Calculus with Analytic Geometry

Question

The face of a dam adjacent to the water is inclined at an angle of \(30^{\circ}\) from the vertical. The shape of the face is a rectangle of width \(50 \mathrm{ft}\) and slant height \(30 \mathrm{ft}\). If the dam is full of water, find the total force due to liquid pressure on the face.

Step-by-Step Solution

Answer

The total force is 378,000 lb.

Step 1 - Understand the Problem

Determine the force on the face of a dam inclined at an angle of \(30^{\circ}\) from the vertical. The rectangular face has a width of \(50 \text{ft}\) and a slant height of \(30 \text{ft}\).

Step 2 - Calculate the Vertical Height

Using trigonometry, determine the vertical height \(h\) of the dam face. Since the slant height is \(30 \text{ft}\) and the angle is \(30^{\circ}\), calculate the height as follows: \( h = 30 \text{ft} \times \sin(30^{\circ}) = 30 \text{ft} \times 0.5 = 15 \text{ft} \).

Step 3 - Calculate the Depth to Area Centroid

The centroid of a rectangular area is located at half of its height. Therefore, depth to the centroid from the water surface is: \( d = 15 \text{ft} \/ 2 = 7.5 \text{ft} \).

Step 4 - Calculate the Hydrostatic Pressure Force

Use the hydrostatic pressure force formula on a plane surface: \( F = \rho g A h_c \), where \( \rho \) is the density of water \( = 62.4 \text{lb/ft}^3 \), \( g \) is gravitational acceleration \( = 32.2 \text{ft/s}^2 \), \( A \) is the area of the rectangle \( = 50 \text{ft} \times 30 \text{ft} = 1500 \text{ft}^2 \), and \( h_c \) is the depth to the centroid \( = 7.5 \text{ft} \). Plug in the values to get: \( F = (62.4 \text{lb/ft}^3) (32.2 \text{ft/s}^2) (1500 \text{ft}^2) (7.5 \text{ft}) = 33.6 \text{lb/ft} (1500 \text{ft}^2) (7.5 \text{ft}) \).

Step 5 - Final Calculation

Compute the final force: \( F = 33.6 \text{lb/ft} \times 1500 \text{ft}^2 \times 7.5 \text{ft} = 378,000 \text{lb} \).

Key Concepts

Trigonometry

Trigonometry plays a vital role in solving problems involving angles and lengths. In our exercise, we use trigonometry to determine the vertical height of the dam, which is critical for calculating hydrostatic pressure. Given that the slant height of the dam face is 30 ft and the angle from the vertical is 30°, we can calculate the vertical height using the sine function.
The formula is: \(h = 30 \text{ft} \times \text{sin}(30^\text{°})\). According to trigonometric values, \(\text{sin}(30^\text{°}) = 0.5\). Hence, \(h = 30 \text{ft} \times 0.5 = 15 \text{ft}\). This calculation helps in determining the depth of water above the centroid, which leads us to accurately compute the hydrostatic pressure.

Centroid Calculation

The centroid is the geometric center of an area, and it is essential when determining hydrostatic pressure on a submerged surface. For a rectangle, the centroid is located at the midpoint of its height. In this scenario, the height of the rectangular face of the dam is calculated as 15 ft.
The formula used is: \(d = \frac{h}{2}\). This gives us \(d = \frac{15 \text{ft}}{2} = 7.5 \text{ft}\). The depth to the centroid from the water surface is crucial for the next steps, as it determines the effective depth that contributes to the pressure exerted by the water. This midpoint helps in evaluating how much pressure the water exerts at different points on the dam face.

Hydrostatic Pressure

Hydrostatic pressure is the force exerted by a fluid due to its weight. It increases with depth and is fundamental in understanding forces on submerged surfaces. To determine the hydrostatic pressure force on the dam face, we use the formula: \(F = \rho g A h_c\), where:

• \(\rho\) is the density of water, which is 62.4 lb/ft³,
• \(g\) is the gravitational acceleration, 32.2 ft/s²,
• \(A\) is the area of the dam face, calculated as 50 ft × 30 ft = 1500 ft²,
• \(h_c\) is the depth to the centroid, 7.5 ft.

Plugging in these values, the calculation is: \(F = (62.4 \text{lb/ft}^3) (32.2 \text{ft/s}^2) (1500 \text{ft}^2) (7.5 \text{ft})\). When solved, it simplifies to \(F = 33.6 \text{lb/ft} \times 1500 \text{ft}^2 \times 7.5 \text{ft} = 378,000 \text{lb}\). This force is what needs to be countered by the structure of the dam to ensure it remains stable and functional under the pressure of the water.