Question

How deep must the center of a vertically oriented circular plate with a radius of \(1 \mathrm{ft}\) 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 \(1,000 \mathrm{lb}\) ?

Short answer

The center of the plate must be submerged approximately 5.10 feet deep in water.

Step-by-step solution

Understand the problem

We need to determine the depth at which a circular plate must be submerged in water so that the force exerted by the water (fluid force) on the plate equals 1000 lb.

Recall the formula for fluid force on vertical submerged surfaces

The formula for fluid force on a vertically submerged surface is given by \( F = ho imes g imes A imes h \), where \( F \) is the fluid force, \( \rho \) is the density of the liquid, \( g \) is acceleration due to gravity, \( A \) is the area of the submerged surface, and \( h \) is the depth at which the center of gravity of the surface is submerged.

Simplify parameters

Since we are dealing with water, the weight density \( \rho \) is given as 62.4 lb/ft³. The acceleration due to gravity \( g \) is considered 1 in such calculations when using weight density.

Calculate the area of the circular plate

The area \( A \) of a circle is given by \( A = \pi r^2 \). Since the radius \( r \) is 1 ft, the area \( A = \pi \times (1)^2 = \pi \) ft².

Insert known values into the formula

We need the fluid force \( F \) to be 1000 lb. Using the formula \( F = \rho \cdot g \cdot A \cdot h \), we set up the equation:\[ 1000 = 62.4 \times \pi \times h \]

Solve for h

Isolate \( h \) by dividing both sides by \( 62.4 \times \pi \):\[ h = \frac{1000}{62.4 \times \pi} \]Calculate this using a calculator to find:\[ h \approx 5.10 \] ft.

Key concepts

Circular Plate

A circular plate is essentially a flat, round surface with a consistent radius from its center point. Understanding the shape and size of a circular plate is key to calculating the area, which is crucial for determining the fluid force exerted when submerged. In our problem, the circular plate has a radius of 1 ft.
Let's break down how the area is calculated for a circle. The formula for the area \( A \) of a circle is given by \( A = \pi r^2 \), where \( r \) is the radius.

Given:

• Radius \( r = 1 \) ft

By substituting the radius into the formula, we find that \( A = \pi \times (1)^2 = \pi \) square feet.
This area is fundamental to understanding how much surface will be in contact with the fluid, which directly impacts the force calculation.

Density of Water

The density of a substance is a measure of its mass per unit volume. For water, this is a widely known and consistent value, usually given as \(62.4 \text{ lb/ft}^3\) in terms of weight density. This measurement indicates how much water weighs per cubic foot.
In fluid force calculations, density is a crucial factor because it helps determine the total force exerted by the fluid on a submerged object. Here, it is the density of water that aids in calculating how much force is applied to our circular plate.

Several key points to note:

• The weight density of water is necessarily included in the fluid force formula \( F = \rho \cdot g \cdot A \cdot h \).
• In calculations involving weight density, the gravitational constant \( g \) is taken as 1.

This simplification allows the problem to focus purely on the interaction between the area of the submerged plate and the depth, without needing to factor in gravitational variations.

Depth of Submersion

Depth of submersion refers to how deep the center of an object is immersed beneath the surface of a fluid. It's crucial when determining the force exerted by the fluid on that object since fluid pressure increases with depth.

In the exercise, we were tasked to find the depth at which the circular plate should be submerged to experience a fluid force of \( 1000 \text{ lb} \).

• The fluid force formula is \( F = \rho \cdot g \cdot A \cdot h \), where \( h \) is the depth of submersion.
• By substituting into the equation, the focus was to isolate and solve for \( h \).

Thus, by dividing \( 1000 \text{ lb} \) by the product of the water's density and the plate's area, we effectively determined the depth at which the desired fluid force is achieved:
\[ h = \frac{1000}{62.4 \times \pi} \approx 5.10 \text{ ft} \]
This calculation indicates how deeply the center must be submerged for the circular plate to meet the specified fluid force condition.