Question

How deep must the center of a vertically oriented square plate with a side length of \(2 \mathrm{ft}\) be submerged in water, with a weight density of \(62.4 \mathrm{lb} / \mathrm{ft}^{3},\) for the fluid force on the plate to reach 1,000 lb?

Short answer

The plate's center must be submerged approximately 4.0064 ft deep.

Step-by-step solution

Understand the Problem

To find the depth at which the center of a square plate needs to be submerged so that the fluid force on it is 1,000 lb, we first consider that the force exerted by a fluid on a submerged surface is given by the integral of pressure over the area of the surface. In this case, a square plate of side length 2 ft is submerged.

Set Up the Force Equation

The force exerted by the fluid when a plate of width, parallel to the surface is submerged is obtained by the formula \( F = P \cdot A \), where \( P \) is the average fluid pressure at the depth of the center, and \( A \) is the area. For a vertically oriented square plate submerged, the area \( A = 2 \times 2 = 4 \; \text{ft}^2 \).

Calculate the Pressure at Center Depth

The fluid pressure at depth \( d \), where \( d \) is the depth of the center of the plate, is given by \( P = \text{weight density} \times \text{depth} = 62.4 \times d \). This yields an average pressure when considering the symmetry of a square plate submerged vertically.

Equate Fluid Force to Desired Force and Solve

We want the fluid force \( F = P \times A = 1,000 \; \text{lb} \). Substituting for \( P \) and \( A \): \( 62.4d \times 4 = 1,000 \). Solve for \( d \): \( 62.4d = \frac{1,000}{4} \), thus \( d = \frac{250}{62.4} \approx 4.0064 \; \text{ft} \).

Verify Reasonableness

Since calculations show that \( d \approx 4.0064 \; \text{ft} \) aligns with reasonable physical expectations for this scenario, verify that each calculation step checks out with given formulas. Confirm result integrity with unit consistency: weight density and depth are in \( \text{ft} \) and lbs.

Key concepts

Pressure Calculation

Understanding pressure in fluid mechanics is crucial, especially when dealing with submerged surfaces. Pressure is defined as the force exerted per unit area. In the context of a submerged object, fluid pressure increases with depth.
This is because pressure results from the weight of the fluid above the object.

For example, consider a plate submerged in water. The weight density of water is often given in units such as pounds per cubic foot (lb/ft³). To calculate the pressure exerted by the water at a specific depth, we use the formula: \[ P = \text{weight density} \times \text{depth} \] This formula helps predict how pressure changes as the depth increases. In our problem, the weight density is 62.4 lb/ft³. This tells us how much weight the water supplies over each cubic foot of area.
By multiplying by the depth, you find the pressure exerted on the submerged surface.

Integration Techniques

Integration is a powerful tool in fluid mechanics to calculate forces exerted by fluids. In cases of submerged surfaces, such as plates or walls, integrating pressure over the entire area gives the total force.
Pressure is not uniform across a submerged surface, so understanding how to apply integration in this context is vital.

When solving for the force exerted by water on a submerged plate, as in our example, the formula actually integrates pressure over the surface area. Here’s how it generally works: - Define the pressure distribution across the area. This varies with depth in our exercise given by \( P = 62.4 \, d \). - Compute the force by integrating pressure across the surface.While our example simplifies the calculation by averaging pressure at the center, knowing how integration applies helps in complex shapes or non-uniform pressure scenarios.

Force on Submerged Surfaces

When dealing with submerged surfaces, determining the force exerted by a fluid is key for engineers and designers.This is especially relevant for objects like dams or underwater structures.
The force a fluid exerts is influenced by both the area of the submerged surface, and the pressure acting upon it.

Let’s break it down further: - **Area Calculation**: For a square plate, the area is simple to calculate. If the plate's side is 2 ft, the area would be \( 2 \times 2 = 4 \, \text{ft}^2 \). - **Force Calculation**: Multiply the pressure by the area to find the total force. In our case, with average pressure calculated at depth, the force becomes \( F = P \times A \). - **Result Validation**: It's critical to verify that the solution aligns with physical expectations, such as ensuring that calculated quantities like force match real-world experience.

Through understanding these principles, we can apply these calculations to real-world problems, ensuring safety and stability in fluid-based scenarios.