Problem 58 - College Algebra with Corequisite Support

Question

For the following exercises, use the given information to answer the questions. The force exerted by the wind on a plane surface varies jointly with the square of the velocity of the wind and with the area of the plane surface. If the area of the surface is 40 square feet surface and the wind velocity is 20 miles per hour, the resulting force is 15 pounds. Find the force on a surface of 65 square feet with a velocity of 30 miles per hour.

Step-by-Step Solution

Answer

The force is approximately 54.84 pounds.

Step 1: Understand the Joint Variation Formula

The force exerted by the wind, denoted as \( F \), varies jointly with the square of the velocity \( v \) and the area \( A \). This means we can express \( F \) as \( F = k \cdot v^2 \cdot A \), where \( k \) is a constant.

Step 2: Calculate the Constant of Variation

Use the given values (Area = 40 square feet, Velocity = 20 miles per hour, Force = 15 pounds) to find \( k \). Substitute into the formula: \( 15 = k \cdot 20^2 \cdot 40 \). Solving for \( k \), we get:\[ 15 = k \cdot 400 \cdot 40 \] \[ 15 = 16000k \] \[ k = \frac{15}{16000} \] \[ k = 0.0009375 \]

Step 3: Use the Constant to Find the New Force

Now that we have \( k = 0.0009375 \), use it to find the force when the area is 65 square feet and the velocity is 30 miles per hour. Substitute these values into the formula:\[ F = 0.0009375 \cdot 30^2 \cdot 65 \] \[ F = 0.0009375 \cdot 900 \cdot 65 \] \[ F = 0.0009375 \cdot 58500 \] \[ F = 54.84375 \] Thus, the force on the new surface is approximately 54.84 pounds.

Key Concepts

Force Calculation

The concept of force calculation is fundamental in the realm of physics. It helps us understand how various factors can influence the force exerted on objects. In this context, the force exerted by the wind is calculated based on both the surface area and the speed at which the wind travels.

To calculate this force, we begin with the formula for joint variation: \[ F = k \cdot v^2 \cdot A \] where:

• \( F \) stands for the force exerted by the wind.
• \( v \) is the velocity of the wind.
• \( A \) represents the area of the surface.
• \( k \) is a constant particular to the conditions.

The equation clearly shows that the force increases quadratically with the velocity and linearly with the area. An increase in either factor results in a corresponding increase in force, demonstrating the power of the wind over larger or more exposed surfaces. This calculation is especially useful for designing buildings, structures, or any object exposed to strong winds. Understanding these principles ensures safety and efficiency.

Mathematical Modeling

Mathematical modeling is a powerful technique where real-world situations are translated into mathematical expressions. It allows us to predict and analyze outcomes based on given conditions. In this exercise, we use a model to describe how wind force behaves when involving variables like speed and area.

By setting up the equation \( F = k \cdot v^2 \cdot A \), we're modeling how changes in wind speed (\( v \)) and surface area (\( A \)) affect the force (\( F \)). This formula provides a framework to simulate different scenarios:

• It helps engineers design wind-resistant buildings.
• Allows planners to anticipate how different structures respond to variable wind conditions.
• Can be adapted to other contexts where joint variation applies, like fluid dynamics or aerodynamics.

The process of deriving such equations from observed reality is an example of mathematical modeling in action. It offers a blend of theoretical and practical insights, making it a cornerstone of scientific inquiry.

Constant of Variation

The constant of variation, denoted as \( k \), is a key component in the joint variation equation. It acts as a proportionality factor linking the wind's forces to its variables.

In the formula \( F = k \cdot v^2 \cdot A \), once you substitute known values for force, velocity, and area, you can solve for \( k \). Determining \( k \) is crucial because:

• It is specific to the conditions of the problem, such as surface texture or air density.
• Its value ensures that units on both sides of the equation match, keeping the calculation valid.
• Once obtained, it allows for predictive calculations under new conditions.

In the given exercise, by substituting the initial conditions (velocity 20 miles/hour, area 40 square feet, force 15 pounds), we computed \( k = 0.0009375 \). With this constant, one can assess how much force would be exerted under different conditions, as shown when determining the force for an area of 65 square feet and a velocity of 30 miles/hour. The constant thus provides both a specific measurement and a versatile tool for various predictive scenarios.