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An airplane wing is designed so that the speed of the air across the top of the wing is \(251 \mathrm{~m} / \mathrm{s}\) when the speed of the air below the wing is \(225 \mathrm{~m} / \mathrm{s}\). The density of the air is \(1.29 \mathrm{~kg} / \mathrm{m}^{3} .\) What is the lifting force on a wing of area \(24.0 \mathrm{~m}^{2} ?\)

Short Answer

Expert verified
The lifting force on the wing is approximately 191,373 N upwards.

Step by step solution

01

Understand the Problem

We need to calculate the lifting force on the airplane wing caused by the difference in airspeed across the top and bottom of the wing. We will use Bernoulli's principle and the given wing area to find the solution.
02

Apply Bernoulli's Equation

According to Bernoulli's principle, the pressure difference \( \Delta P \) between the top and bottom of the wing is given by: \[ \Delta P = \frac{1}{2} \rho (v_{bottom}^2 - v_{top}^2) \]where \( v_{bottom} = 225 \) m/s, \( v_{top} = 251 \) m/s, and \( \rho = 1.29 \) kg/m³.
03

Calculate Pressure Difference

Substitute the values into the Bernoulli's equation:\[ \Delta P = \frac{1}{2} \times 1.29 \times (225^2 - 251^2) \]Solving the expression in the parenthesis first gives:\[ 225^2 = 50625; \quad 251^2 = 63001 \]\[ 225^2 - 251^2 = 50625 - 63001 = -12376 \]Then calculating the pressure difference:\[ \Delta P = \frac{1}{2} \times 1.29 \times (-12376) \approx -7973.88 \, \text{Pa} \]
04

Calculate Lifting Force

Now, use the pressure difference to find the lifting force \( F_L \):\[ F_L = \Delta P \times A \]Substitute in the area and pressure difference:\[ F_L = -7973.88 \, \text{Pa} \times 24.0 \, \text{m}^2 \]\[ F_L \approx -191373.12 \, \text{N} \]The negative sign indicates that the direction of lift is upward, opposite to the defined positive pressure direction.

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Key Concepts

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

Aerodynamics
Aerodynamics is a branch of physics concerned with the study of air movement around objects, often used in the context of vehicles like cars and airplanes. It explores how air interacts with solid surfaces, such as an airplane wing, and how these interactions can be used to influence motion.
In our problem, aerodynamics plays a crucial role. The wing of the airplane is designed specially to manage air flow differently over the top and the bottom surfaces. This difference creates variations in air speed, which are fundamental to generating lift.
  • The faster-moving air over the top of the wing reduces pressure compared to the slower-moving air underneath.
  • This imbalance in air speed and pressure causes the airplane to lift.
Overall, aerodynamics helps experts design wings that use these principles to improve lift efficiency, minimizing fuel consumption and increasing flight stability.
Pressure Difference
The concept of pressure difference is central to Bernoulli's principle and to understanding how lift is created on a wing. When air flows over a wing, a pressure differential arises between the upper and lower surfaces. Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure. Let's put it into easier words:
  • If the air moves faster across the top than below the wing, the pressure above is lower than the pressure below.
  • This lower pressure on top helps to create a pressure difference known as \( \Delta P \).
Mathematically, the pressure difference can be calculated using the equation:\[\Delta P = \frac{1}{2} \rho (v_{bottom}^2 - v_{top}^2)\]Where:
  • \( \rho \) is the air density, \( v_{bottom} \) is the speed of the air below the wing, and \( v_{top} \) is the speed over the wing.
  • The negative result is interpreted as the direction of the lift being upward due to lower pressure on the top.
This concept is fundamental not only in aviation but also in many engineering applications involving fluid dynamics.
Lifting Force
Lifting force is the push that keeps an airplane in the sky, counteracting gravity. It emerges from the difference in air pressure created by the speed of airflow over the wing, essentially utilizing Bernoulli’s principle. In the given exercise, once the pressure difference \( \Delta P \) is calculated using the speeds and densities provided, we can find the lifting force \( F_L \). The formula used is:\[F_L = \Delta P \times A\]Where:
  • \( F_L \) is the lifting force.
  • \( \Delta P \) is the pressure difference.
  • \( A \) is the area of the wing's surface.
The resulting force provides the upward push needed to counteract the weight of the plane. Importantly, the negative sign in the pressure calculation indicates the lift's upward direction, contrary to the assumed pressure flow.
Through the combination of this pressure difference over a sufficient wing area, airplanes can fly and maneuver effectively by maximizing lifting force.

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Most popular questions from this chapter

An antifreeze solution is made by mixing ethylene glycol \(\left(\rho=1116 \mathrm{~kg} / \mathrm{m}^{3}\right)\) with water. Suppose the specific gravity of such a solution is \(1.0730 .\) Assuming that the total volume of the solution is the sum of its parts, determine the volume percentage of ethylene glycol in the solution.

Interactive LearningWare 11.2 at reviews the approach taken in problems such as this one. A small crack occurs at the base of a 15.0 -m-high dam. The effective crack area through which water leaves is \(1.30 \times 10^{-3} \mathrm{~m}^{2}\). (a) Ignoring viscous losses, what is the speed of water flowing through the crack? (b) How many cubic meters of water per second leave the dam?

The aorta carries blood away from the heart at a speed of about \(40 \mathrm{~cm} / \mathrm{s}\) and has a radius of approximately \(1.1 \mathrm{~cm}\). The aorta branches eventually into a large number of tiny capillaries that distribute the blood to the various body organs. In a capillary, the blood speed is approximately \(0.07 \mathrm{~cm} / \mathrm{s},\) and the radius is about \(6 \times 10^{-4} \mathrm{~cm} .\) Treat the blood as an incompressible fluid, and use these data to determine the approximate number of capillaries in the human body.

A 1.3-m length of horizontal pipe has a radius of \(6.4 \times 10^{-3} \mathrm{~m}\). Water within the pipe flows with a volume flow rate of \(9.0 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\) out of the right end of the pipe and into the air. What is the pressure in the flowing water at the left end of the pipe if the water behaves as (a) an ideal fluid and (b) a viscous fluid \(\left(\eta=1.00 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\right)\) ?

A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at \(11.0 \mathrm{~m} / \mathrm{s}\) over the top of the plate only, it is possible to keep the plate in a horizontal position, as illustrated in part \(a\) of the drawing. To what value should the air speed be reduced so that the plate is kept at a \(30.0^{\circ}\) angle with respect to the vertical, as in part \(b\) of the drawing? (Hint: Apply Bernoulli's equation in the form of Equation \(11.12 .)\)

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