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Chapter 11: Problem 73

The construction of a flat rectangular roof \((5.0 \mathrm{m} \times 6.3 \mathrm{m})\) allows it to withstand a maximum net outward force that is 22000 \(\mathrm{N}\) . The density of the air is 1.29 \(\mathrm{kg} / \mathrm{m}^{3}\) . At what wind speed will this roof blow outward?

Short Answer

Expert verified
The wind speed is approximately 32.89 m/s.

Step by step solution

01

Understand the Problem

The problem is asking for the wind speed at which the outward force on a flat rectangular roof exceeds the maximum limit of 22000 N. The force is due to a pressure difference caused by wind blowing over the top of the roof.
02

Calculate Roof Area

The area \( A \) of the roof can be calculated using the formula for the area of a rectangle: \( A = \text{length} \times \text{width} = 5.0 \, \text{m} \times 6.3 \, \text{m} = 31.5 \, \text{m}^2 \).
03

Understand Bernoulli's Equation

Bernoulli's principle relates the speed of a fluid (in this case, wind) to its pressure. The pressure difference \( \Delta P \) that causes the force on the roof can be calculated with \( \Delta P = \frac{1}{2} \rho v^2 \), where \( \rho \) is the density of air and \( v \) is the wind speed.
04

Express Force in Terms of Pressure Difference

The force due to the pressure difference \( \Delta P \) is given by the formula \( F = \Delta P \times A \). Here, \( A = 31.5 \, \text{m}^2 \) and the maximum force \( F = 22000 \, \text{N} \). So, \( \Delta P = \frac{F}{A} = \frac{22000}{31.5} \, \text{N/m}^2 \approx 698.41 \, \text{N/m}^2 \).
05

Apply Bernoulli's Equation to Solve for Wind Speed

Using \( \Delta P = \frac{1}{2} \rho v^2 \), substitute \( \Delta P = 698.41 \, \text{N/m}^2 \) and \( \rho = 1.29 \, \text{kg/m}^3 \). Solve for \( v \): \[ 698.41 = \frac{1}{2} \times 1.29 \times v^2 \]Calculate \( v \): \[ v^2 = \frac{698.41 \times 2}{1.29} \approx 1082.49 \]\[ v \approx \sqrt{1082.49} \approx 32.89 \, \text{m/s} \]
06

Conclusion

The wind speed needed to exert the maximum allowed net outward force on the roof is approximately 32.89 m/s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wind Speed Calculation
Calculating wind speed involves understanding how the force exerted by moving air affects a flat surface. In our case, the surface is a roof with dimensions, and we want to find the speed at which wind will create sufficient pressure to exert a given force. We start by knowing the force limit the roof can withstand, which is 22000 N.

By using Bernoulli's principle, we relate the pressure difference caused by wind to the wind speed. Once we have the required pressure difference (\(\Delta P\)), we can calculate the wind speed \(v\) by rearranging the equation \(\Delta P = \frac{1}{2} \rho v^2\). Here, \(\rho\) is the air density.

This approach tells us how different speeds can affect structures. When you plug in the values and solve for \(v\), you find that about 32.89 m/s of wind speed can push the roof outward.
Pressure Difference
Pressure difference is a crucial factor in understanding how wind affects surfaces like roofs. When wind moves over a roof, it creates a difference in pressure between the inside and outside. This difference is what generates a lifting force.

To find this pressure difference (\(\Delta P\)), we need the maximum force that the roof can handle divided by its area. Using \(F = \Delta P \times A\), and knowing both \(F\) and \(A\), you can find \(\Delta P\). This step reveals how pressure changes with speed, allowing us to calculate the wind speed required to reach this pressure difference using Bernoulli鈥檚 equation.
Fluid Dynamics
Fluid dynamics is the study of how liquids and gases move and interact. In the context of wind and roofs, it helps us understand how air flows and the effects of this flow. Bernoulli's principle, a key concept in fluid dynamics, tells us that an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or potential energy.

When applying these principles to calculate wind speed, we're looking at how the moving air's speed caused by pressure change translates to a physical force on objects. Remember: Faster air movement over the roof results in lower pressure on top. This pressure difference, comprehended through fluid dynamics and Bernoulli's principle, is what allows us to solve for wind speed using mathematical methods.

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Most popular questions from this chapter

A paperweight, when weighed in air, has a weight of \(W=6.9 \mathrm{N}\) . When completely immersed in water, however, it has a weight of \(W_{\text { in water }}=4.3 \mathrm{N}\) . Find the volume of the paperweight.

Water is circulating through a closed system of pipes in a two-floor apartment. On the first floor, the water has a gauge pressure of \(3.4 \times 10^{5} \mathrm{Pa}\) and a speed of 2.1 \(\mathrm{m} / \mathrm{s}\) . However, on the second floor, which is 4.0 \(\mathrm{m}\) higher, the speed of the water is 3.7 \(\mathrm{m} / \mathrm{s}\) . The speeds are different because the pipe diameters are different. What is the gauge pressure of the water on the second floor?

A siphon tube is useful for removing liquid from a tank. The siphon tube is first filled with liquid, and then one end is inserted into the tank. Liquid then drains out the other end, as the drawing illustrates. (a) Using reasoning similar to that employed in obtaining Torricelli鈥檚 theorem (see Example 16), derive an expression for the speed \(v\) of the fluid emerging from the tube. This expression should give \(v\) in terms of the vertical height \(y\) and the acceleration due to gravity \(g\) . (Note that this speed does not depend on the depth \(d\) of the tube below the surface of the liquid.) (b) At what value of the vertical distance y will the siphon stop working? (c) Derive an expression for the absolute pressure at the highest point in the siphon (point \(A )\) in terms of the atmospheric pressure \(P_{0},\) the fluid density \(\rho, g,\) and the heights \(h\) and \(y\) (Note that the fluid speed at point \(A\) is the same as the speed of the fluid emerging from the tube, because the cross-sectional area of the tube is the same everywhere.)

The main water line entrs a house on the first floor. The line has a gauge pressure of \(1.90 \times 10^{5} \mathrm{Pa}\) . \((\mathrm{a})\) A faucet on the second floor, 6.50 \(\mathrm{m}\) above the first floor, is turned off. What is the gauge pressure at this faucet? (b) How high could a faucet be before no water would flow from it, even if the faucet were open?

A ship is floating on a lake. Its hold is the interior space beneath its deck; the hold is empty and is open to the atmosphere. The hull has a hole in it, which is below the water line, so water leaks into the hold. The effective area of the hole is \(8.0 \times 10^{-3} \mathrm{m}^{2}\) and is located 2.0 \(\mathrm{m}\) beneath the surface of the lake. What volume of water per second leaks into the ship?
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