/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 Lift on an airplane. Air streams... [FREE SOLUTION] | 91Ó°ÊÓ

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Lift on an airplane. Air streams horizontally past a small airplane's wings such that the speed is \(70.0 \mathrm{~m} / \mathrm{s}\) over the top surface and \(60.0 \mathrm{~m} / \mathrm{s}\) past the bottom surface. If the plane has a mass of \(1340 \mathrm{~kg}\) and a wing area of \(16.2 \mathrm{~m}^{2},\) what is the net vertical force (including the effects of gravity) on the airplane? The density of the air is \(1.20 \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The net vertical force is -509.4 N.

Step by step solution

01

Calculate the Dynamic Pressure Difference

To find the dynamic pressure difference, use Bernoulli's equation for each surface:\[\Delta P = \frac{1}{2} \rho (v_{top}^2 - v_{bottom}^2)\]where \(\rho = 1.20 \mathrm{~kg/m}^3\), \(v_{top} = 70.0 \mathrm{~m/s}\), and \(v_{bottom} = 60.0 \mathrm{~m/s}\). Calculating the difference:\[\Delta P = \frac{1}{2} \times 1.20 \times (70^2 - 60^2) = \frac{1}{2} \times 1.20 \times (4900 - 3600) = \frac{1}{2} \times 1.20 \times 1300\]\[\Delta P = 780 \mathrm{~N/m}^2\]
02

Calculate the Lift Force

The lift force \(F_{lift}\) can be found by multiplying the pressure difference by the wing area \(A\):\[F_{lift} = \Delta P \times A\]Given \(\Delta P = 780 \mathrm{~N/m}^2\) and \(A = 16.2 \mathrm{~m}^2\), calculate the lift force:\[F_{lift} = 780 \times 16.2 = 12636 \mathrm{~N}\]
03

Calculate Gravitational Force

Gravitational force \(F_{gravity}\) on the airplane is calculated using:\[F_{gravity} = m \times g\]where \(m = 1340 \mathrm{~kg}\) and \(g = 9.81 \mathrm{~m/s}^2\). Thus,\[F_{gravity} = 1340 \times 9.81 = 13145.4 \mathrm{~N}\]
04

Determine the Net Vertical Force

The net vertical force \(F_{net}\) is the difference between the lift force and the gravitational force:\[F_{net} = F_{lift} - F_{gravity}\]Substitute the known values:\[F_{net} = 12636 - 13145.4 = -509.4 \mathrm{~N}\]
05

Conclusion

The net vertical force acting on the airplane is -509.4 N, which indicates that the gravitational force is larger than the lift force.

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Key Concepts

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

Bernoulli's Equation
The fundamental principle behind how airplanes achieve lift is largely derived from Bernoulli's Equation. This powerful equation relates the pressure, velocity, and height of a flowing fluid. In the context of an airplane, the equation helps explain how air pressure varies above and below the wings. Bernoulli's Equation can be written as:\[ P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant} \]where:
  • \( P \) is the pressure exerted by the fluid.
  • \( \rho \) is the fluid density.
  • \( v \) is the flow velocity of the fluid.
  • \( g \) is acceleration due to gravity.
  • \( h \) is the height of the fluid above a reference point.
For a wing, air flows faster over the top surface than the bottom, reducing the pressure on top and creating lift. This difference is a direct application of Bernoulli's principles.
Dynamic Pressure Difference
The dynamic pressure difference is crucial for generating lift on an airplane. It measures how much pressure changes due to different airspeeds over the aircraft surfaces. This difference is calculated by:\[ \Delta P = \frac{1}{2} \rho (v_{\text{top}}^2 - v_{\text{bottom}}^2) \]Here, \( \Delta P \) represents the pressure difference, \( \rho \) is the air density, \( v_{\text{top}} \) is the speed of airflow over the top of the wing, and \( v_{\text{bottom}} \) is the speed underneath.In our example, the speeds are 70 m/s over the top and 60 m/s below, with a density of 1.20 kg/m³. Substituting in these values gives the pressure difference:\[ \Delta P = \frac{1}{2} \times 1.20 \times (70^2 - 60^2) = 780 \text{ N/m}^2 \]
Lift Force Calculation
Lift force is the key element keeping an airplane in the air, counteracting its weight. To calculate the lift force, we multiply the dynamic pressure difference by the wing area.The equation for lift force \( F_{\text{lift}} \) is:\[ F_{\text{lift}} = \Delta P \times A \]where:
  • \( \Delta P \) is the dynamic pressure difference.
  • \( A \) is the wing area.
For the situation at hand, with \( \Delta P = 780 \text{ N/m}^2 \) and a wing area of 16.2 \( \text{m}^2 \), the lift force becomes:\[ F_{\text{lift}} = 780 \times 16.2 = 12,636 \text{ N} \]
Gravitational Force
Gravitational force is the pulling force exerted by the Earth that needs to be overcome by lift for an airplane to ascend. This force is straightforward to calculate using the mass of the airplane and gravitational acceleration.The formula is:\[ F_{\text{gravity}} = m \times g \]where:
  • \( m \) is the mass of the airplane.
  • \( g \) is the acceleration due to gravity, approximately \( 9.81 \text{ m/s}^2 \).
Given a plane mass of 1340 kg, the gravitational force becomes:\[ F_{\text{gravity}} = 1340 \times 9.81 = 13,145.4 \text{ N} \]
Net Vertical Force
This concept determines if an aircraft climbs, descends, or maintains level flight. The net vertical force is the resultant force combining both lift and gravity.It can be calculated by:\[ F_{\text{net}} = F_{\text{lift}} - F_{\text{gravity}} \]where:
  • \( F_{\text{lift}} \) is the upward lift force.
  • \( F_{\text{gravity}} \) is the downward gravitational force.
In the problem at hand, substituting the computed forces:\[ F_{\text{net}} = 12,636 - 13,145.4 = -509.4 \text{ N} \]The negative sign indicates that the gravitational force is stronger, resulting in a net downward force, meaning the plane would descend unless adjustments are made.

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Most popular questions from this chapter

Blood pressure on the moon. A resident of a lunar colony needs to have her blood pressure checked in one of her legs. Assume that we express the systemic blood pressure as we do on earth and that the density of blood does not change. Suppose also that normal blood pressure on the moon is still \(\frac{120}{80}\) (which may not actually be true). If the lunar colonizer has her blood pressure taken at a point on her ankle that is \(1.3 \mathrm{~m}\) below her heart, what will be her systemic blood-pressure reading, expressed in the standard way, if she has normal blood pressure? The acceleration due to gravity on the moon is \(1.67 \mathrm{~m} / \mathrm{s}^{2}\).

A cube of compressible material (such as Styrofoam or balsa wood) has a density \(\rho\) and sides of length \(L\). (a) If you keep its mass the same, but compress each side to half its length, what will be its new density, in terms of \(\rho ?\) (b) If you keep the mass and shape the same, what would the length of each side have to be (in terms of \(L\) ) so that the density of the cube was three times its original value?

Elephants under pressure. An elephant can swim or walk with its chest several meters underwater while the animal breathes through its trunk, which remains above the water surface and acts like a snorkel. The elephant's tissues are at an increased pressure due to the surrounding water, but the lungs are at atmospheric pressure because they are connected to the air through the trunk. Figure 13.47 shows the gauge pressures in an elephant's lungs and abdomen when the elephant's chest is submerged to a particular depth in a lake. In this situation, the elephant's diaphragm, which separates the lungs from the abdomen, must maintain the difference in pressure between the lungs and the abdomen. The diaphragm of an elephant is typically \(3.0 \mathrm{~cm}\) thick and \(120 \mathrm{~cm}\) in diameter. The maximum force the muscles of the diaphragm can exert is \(24,000 \mathrm{~N}\). What maximum pressure difference can the diaphragm withstand? A. \(160 \mathrm{~mm} \mathrm{Hg}\) B. \(760 \mathrm{~mm} \mathrm{Hg}\) C. \(920 \mathrm{~mm} \mathrm{Hg}\) D. \(5000 \mathrm{~mm} \mathrm{Hg}\)

A piece of wood is \(0.600 \mathrm{~m}\) long, \(0.250 \mathrm{~m}\) wide, and \(0.080 \mathrm{~m}\) thick. Its density is \(600 \mathrm{~kg} / \mathrm{m}^{3}\). What volume of lead must be fastened underneath it to sink the wood in calm water so that its top is just even with the water level? What is the mass of this volume of lead?

What radius must a water drop have for the difference between the inside and outside pressures to be \(0.0200 \mathrm{~atm} ?\) The surface tension of water is \(0.0728 \mathrm{~N} / \mathrm{m}\)

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