Problem 33 - Algebra and Trigonometry

Question

Drag Force on a Boat The drag force \(F\) on a boat is jointly proportional to the wetted surface area \(A\) on the hull and the square of the speed \(s\) of the boat. A boat experiences a drag force of 220 lb when traveling at 5 mi/h with a wetted surface area of 40 \(\mathrm{ft}^{2} .\) How fast must a boat be traveling if it has 28 \(\mathrm{ft}^{2}\) of wetted surface area and is experiencing a drag force of 175 lb?

Step-by-Step Solution

Answer

The boat must travel at approximately 5.33 mi/h.

Step 1: Understanding the Relationship

Drag force is given by the formula \( F = k imes A imes s^2 \), where \( F \) is the drag force, \( A \) is the wetted surface area, \( s \) is the speed of the boat, and \( k \) is a constant of proportionality.

Step 2: Finding the Constant of Proportionality

Use the initial condition to find \( k \). Substitute the known values into the formula: \( 220 = k imes 40 imes 5^2 \). Simplifying, we get \( 220 = k imes 40 imes 25 \). So, \( 220 = 1000k \). Solve for \( k \) to get \( k = \frac{220}{1000} = 0.22 \).

Step 3: Applying the Constant to New Conditions

Now that we know \( k = 0.22 \), substitute it into the equation for the new conditions: \( 175 = 0.22 imes 28 imes s^2 \). Simplify to \( 175 = 6.16 imes s^2 \).

Step 4: Solving for the Speed

Solve for \( s \) in the equation \( 175 = 6.16 imes s^2 \) by dividing both sides by 6.16: \( s^2 = \frac{175}{6.16} \approx 28.41 \). Take the square root to find \( s \): \( s = \sqrt{28.41} \approx 5.33 \). Thus, the speed \( s \approx 5.33 \) mi/h.

Key Concepts

Understanding Drag Force

Drag force is a crucial concept when studying how objects move through fluids, like air or water. It's the resistance an object faces when moving through such a medium. For boats, this force can substantially affect speed and fuel efficiency. The drag force

• Opposes the direction of movement
• Increases with speed
• Depends on the size and shape of the object's surface in contact with the fluid

For the boat in this problem, drag force, denoted as \( F \), is given by the equation \( F = k \times A \times s^2 \). Here,

• \( A \) represents the wetted surface area, the part of the boat immersed in water.
• \( s \) is the speed of the boat.
• \( k \) is a constant of proportionality.

Understanding how these factors interact is vital for finding how fast a boat can travel under different conditions.

Exploring Wetted Surface Area

The wetted surface area is the portion of the boat's hull that is submerged and in direct contact with water. It plays an important role in determining the drag force.

This area is a significant factor because:

• Larger wetted areas result in more contact with water, thus increasing drag.
• Minimizing it can lead to more efficient travel.
• Factors such as the boat's shape, weight distribution, and load affect it.

In the exercise, the wetted surface area changed from 40 \( \text{ft}^2 \) to 28 \( \text{ft}^2 \). Since a smaller wetted surface generally means less drag, the speed calculation must adjust to achieve the same force conditions. This demonstrates how modifying the boat’s geometry affects performance significantly.

Finding the Proportionality Constant

The proportionality constant, \( k \), is a key component that helps relate the drag force to the speed and wetted surface area. It represents some constant factor, specific to a particular configuration or set of conditions. In this exercise, by using known quantities:

• Drag force \( F = 220\,\text{lb} \)
• Wetted surface area \( A = 40\,\text{ft}^2 \)
• Speed \( s = 5\,\text{mi/h} \)

We were able to solve for \( k \) using \( 220 = k \times 40 \times 5^2 \). Solving, we found that \( k = 0.22 \).

This constant allows calculations under different conditions by retaining proportional relationships between force, area, and speed.

Performing Speed Calculation

Speed calculation is often an end goal when examining the effects of drag on moving objects. Once we know the drag force, wetted area, and the proportionality constant, we can find the speed under new conditions. For our example, substituting \( k = 0.22 \) into the formula with the new conditions:

• Drag force \( F = 175\,\text{lb} \)
• Wetted area \( A = 28\,\text{ft}^2 \)

The equation is \( 175 = 0.22 \times 28 \times s^2 \). Solving for \( s \) gives \( s^2 \approx 28.41 \), and the square root gives us \( s \approx 5.33\,\text{mi/h} \).

This calculation shows how different wetted surface areas affect the speed needed to experience a specified drag force. Understanding such relationships is fundamental in designing and operating efficient boats.