Problem 45 - Calculus Early Transcendentals

Question

A large building shaped like a box is 50 \(\mathrm{m}\) high with a face that is \(80 \mathrm{m}\) wide. A strong wind blows directly at the face of the building, exerting a pressure of \(150 \mathrm{N} / \mathrm{m}^{2}\) at the ground and increasing with height according to \(P(y)=150+2 y,\) where \(y\) is the height above the ground. Calculate the total force on the building, which is a measure of the resistance that must be included in the design of the building.

Step-by-Step Solution

Answer

Answer: The total force exerted by the wind on the face of the building is 800,000 N.

Step 1: Identify the pressure function

The pressure function given is \(P(y) = 150 + 2y\), where y is the height above the ground. This function describes how the pressure changes with height on the face of the building.

Step 2: Find the differential force

To calculate the total force, we need to integrate the pressure function with respect to the height (y). First, we find the differential force exerted by the wind at height y. This differential force, \(dF\), is equal to the pressure at height y multiplied by the width (80 m) and the small height interval \(\mathrm{dy}\): \(dF = P(y) \cdot 80 \mathrm{dy}\) Plugging in the pressure function, we get: \(dF = (150 + 2y) \cdot 80 \mathrm{dy}\)

Step 3: Integrate the differential force

Now we integrate the differential force with respect to y over the interval of the building height (0 to 50 m) to find the total force F: \(F = \int_{0}^{50} (150 + 2y) \cdot 80 \mathrm{dy}\) To evaluate the integral, we can use the linearity property and split the integral into two separate integrals: \(F = 80\int_{0}^{50} (150 \mathrm{dy} + 2y \mathrm{dy})\) \(F = 80\int_{0}^{50}150 \mathrm{dy} + 80\int_{0}^{50}2y \mathrm{dy}\)

Step 4: Evaluate the integrals and find the total force

Now, we evaluate each integral: \(F = 80[\int_{0}^{50} 150 \mathrm{dy}] + 80[\int_{0}^{50} 2y \mathrm{dy}]\) \(F = 80[150y\Big|_0^{50}] + 80[y^2\Big|_0^{50}]\) Evaluate the integrals at the limits: \(F = 80(150 \cdot (50) - 150 \cdot (0)) + 80((50)^2 - (0)^2)\) \(F = 80(7500) + 80(2500)\) \(F = 600000 + 200000\) \(F = 800000 \ \mathrm{N}\) The total force exerted by the wind on the face of the building is 800,000 N. This is the resistance that must be included in the design of the building.

Key Concepts

Integration

Integration is a fundamental concept in calculus that helps us find the accumulation of quantities, such as areas under curves or total accumulated forces. In this context, we are using integration to calculate the total force exerted by the wind on the building. The pressure applied by the wind changes with height, and integration allows us to sum up these varying pressures from the ground to the top of the building.

When integrating, we look for the total effect of a function over an interval. Here, the interval goes from the base of the building at 0 meters to its top at 50 meters.

• The integration of a pressure function involves setting up an integral over the height.
• Breaking down the integral can simplify the calculation.

In our problem, we integrated the pressure over the building's face to find the total force. This method is particularly helpful when dealing with functions that vary, as it captures the cumulative effects at each point along the height.

Differential Force

The idea of differential force is central to understanding how forces can act differently depending on their position or context, like height in this case. Differential force (\(dF\)) represents a small, incremental force exerted at a specific point.

In our problem, we calculate the differential force by taking the pressure at a given height and multiplying it with a small height interval.

• The pressure on the building's face differs with height, so it needs to be considered in small sections.
• Differential force allows us to apply calculus to sum these small forces across the entire height.

This method translates the continuous pressure across a large area into simpler, manageable components, which we can then add up through integration to find the total force. Understanding this concept is key to solving integrals involving physical forces.

Pressure Function

A pressure function models how pressure changes with specific conditions, such as height in our exercise. For our building, the given pressure function \(P(y) = 150 + 2y\), expresses how the pressure on the building increases as you go higher.

Pressure is measured in newtons per square meter, and this function helps visualize the pressure distribution across the surface.

• The first term (150 \ \text{N/m}^2) represents the pressure at ground level.
• The second term (2y) shows how pressure increments with every meter.

This function is vital because it can inform structural engineers how much force a structure must withstand at varying heights. By integrating this pressure function over the height, we can determine the total force acting on the structure, which is essential for safe and effective design.