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Chapter 13: Problem 28

At a height of \(10 \mathrm{~km}\) ( \(33000 \mathrm{ft}\) ) above sea level, atmospheric pressure is about \(210 \mathrm{~mm}\) of mercury. What is the net resultant normal force on a \(600 \mathrm{~cm}^{2}\) window of an airplane flying at this height? Assume the pressure inside the plane is \(760 \mathrm{~mm}\) of mercury. The density of mercury is \(13600 \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The net force on the window is approximately 43,970.4 N.

Step by step solution

01

Understand the Problem

We need to find the net force on a window due to different atmospheric pressures outside and inside an airplane flying at a specific height. We will use the pressure difference to find the force.
02

Convert Pressures to Pascal

First, convert the given pressures from millimeters of mercury to pascals (Pa), the SI unit of pressure. Use the conversion formula: \\[\text{Pressure (Pa)} = \text{Pressure (mmHg)} \times \text{Density of mercury} \times g\] \where \(g = 9.81 \text{ m/s}^2\).\- External pressure is 210 mmHg: \\[P_{\text{outside}} = 210 \times 13600 \times 9.81 \= 280,728 \text{ Pa}\] \- Internal pressure is 760 mmHg: \\[P_{\text{inside}} = 760 \times 13600 \times 9.81 \= 1,013,568 \text{ Pa}\]
03

Calculate Pressure Difference

Find the difference in pressure between the inside and outside of the plane: \\[\Delta P = P_{\text{inside}} - P_{\text{outside}} = 1,013,568 - 280,728 = 732,840 \text{ Pa}\]
04

Calculate the Net Force

Use the pressure difference to find the net force on the window using the equation: \\[F = \Delta P \times A\] \where \(A\) is the area of the window in square meters. Convert the area from cm虏 to m虏: \\[A = 600 \text{ cm}^2 = 0.06 \text{ m}^2\] \Calculate the force: \\[F = 732,840 \times 0.06 = 43,970.4 \text{ N}\]
05

Conclusion

The net resultant normal force acting on the window is approximately 43,970.4 N in the direction from inside to outside the plane.

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Key Concepts

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

Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of the atmosphere above a particular point. It varies with altitude; the higher you go, the lower the atmospheric pressure. At sea level, the average atmospheric pressure is roughly 101,325 Pascals (Pa), or 760 mm of mercury (mmHg). As an airplane ascends to 10 km above sea level, the atmospheric pressure drops significantly because there is less air above that altitude to exert force.

In our exercise, the atmospheric pressure at 10 km is given as 210 mmHg. This lower pressure affects how we calculate forces on structures like airplane windows, as it changes the pressure difference between the outside and inside of the aircraft.
Pressure Conversion
Converting pressure from one unit to another is crucial in scientific calculations. In this problem, the atmospheric pressure and the pressure inside the airplane are given in mmHg units. However, the standard unit of pressure in scientific calculations is the Pascal (Pa).

To convert mmHg to Pascals, we use the equation: \[\text{Pressure (Pa)} = \text{Pressure (mmHg)} \times \text{Density of mercury} \times g\] where the density of mercury is 13,600 kg/m鲁 and the gravitational acceleration \(g\) is 9.81 m/s虏.
  • External Pressure: 210 mmHg converts to 280,728 Pa.
  • Internal Pressure: 760 mmHg converts to 1,013,568 Pa.
Converting these units allows us to accurately calculate the pressure difference and force on the window.
Pressure Difference
The difference in pressure between the inside and the outside of an object, like an airplane window, is a key factor in determining the net force on that object. It's this pressure difference that creates a force that could potentially push against structures like the aircraft windows.

Here, we calculate the pressure difference (\(\Delta P\)) as the difference between internal pressure \(P_{\text{inside}}\) and external pressure \(P_{\text{outside}}\). Using the converted pressures:\[\Delta P = 1,013,568 - 280,728 = 732,840 \text{ Pa}\]Understanding this difference is essential as it determines the magnitude of the force exerted on the window.
Force on Window
The net force on a window due to a pressure difference is calculated using the formula: \[ F = \Delta P \times A \] where \(\Delta P\) is the pressure difference and \(A\) is the area of the window.

In this problem, \(\Delta P\) was calculated as 732,840 Pa and the area \(A\) is 0.06 m虏 (converted from 600 cm虏). By substituting these values into the formula, we find:\[ F = 732,840 \times 0.06 = 43,970.4 \text{ N} \] This force acts in the direction from inside to outside of the airplane. Understanding this concept helps in designing and testing aircraft windows to ensure they can withstand such forces.

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Most popular questions from this chapter

When a submarine dives to a depth of \(120 \mathrm{~m}\), to how large a total pressure is its exterior surface subjected? The density of seawater is about \(1.03 \mathrm{~g} / \mathrm{cm}^{3}\). $$ \begin{aligned} P &=\text { Atmospheric pressure }+\text { Pressure of water } \\ &=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+\rho g h=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+\left(1030 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(120 \mathrm{~m}) \\ &=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+12.1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=13.1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.31 \mathrm{MPa} \end{aligned} $$

A piece of pure gold \(\left(\rho=19.3 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is suspected to have a hollow center. It has a mass of \(38.25 \mathrm{~g}\) when measured in air and \(36.22 \mathrm{~g}\) in water. What is the volume of the central hole in the gold? Remember that you go from a density in \(\mathrm{g} / \mathrm{cm}^{3}\) to \(\mathrm{kg} / \mathrm{m}^{3}\) by multiplying by 1000 . From \(\rho=m / V\), Actual volume of \(38.25 \mathrm{~g}\) of gold \(=\frac{0.03825 \mathrm{~kg}}{19300 \mathrm{~kg} / \mathrm{m}^{3}}=1.982 \times 10^{-6} \mathrm{~m}^{3}\) Volume of displaced water \(=\frac{(38.25-36.22) \times 10^{-3} \mathrm{~kg}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}=2.030 \times 10^{-6} \mathrm{~m}^{3}\) $$ \text { Volume of hole }=(2.030-1.982) \mathrm{cm}^{3}=0.048 \mathrm{~cm}^{3} $$

A balloon and its gondola have a total (empty) mass of \(2.0 \times 10^{2} \mathrm{~kg}\). When filled, the balloon contains \(900 \mathrm{~m}^{3}\) of helium at a density of \(0.183 \mathrm{~kg} / \mathrm{m}^{3}\). Find the added load, in addition to its own weight, that the balloon can lift. The density of air is \(1.29 \mathrm{~kg} / \mathrm{m}^{3}\).

A metal cube, \(2.00 \mathrm{~cm}\) on each side, has a density of \(6600 \mathrm{~kg} / \mathrm{m}^{3}\). Find its apparent mass when it is totally submerged in water.

Find the density \(\rho\) of a fluid at a depth \(h\) in terms of its density \(\rho_{0}\) at the surface. If a mass \(m\) of fluid has volume \(V_{0}\) at the surface, then it will have volume \(V_{0}-\Delta V\) at a depth \(h\). The density at depth \(h\) is then $$ \rho=\frac{m}{V_{0}-\Delta V} \quad \text { while } \quad \rho_{0}=\frac{m}{V_{0}} $$ which gives $$ \frac{\rho}{\rho_{0}}=\frac{V_{0}}{V_{0}-\Delta V}=\frac{1}{1-\left(\Delta V / V_{0}\right)} $$ However, from Chapter 12, the bulk modulus is \(B=P /\left(\Delta V / V_{0}\right)\) and so \(\Delta V / V_{0}=P / B\). Making this substitution, we obtain $$ \frac{\rho}{\rho_{0}}=\frac{1}{1-P / B} $$ If we assume that \(\rho\) is close to \(\rho_{0}\), then the pressure at depth \(h\) is approximately \(\rho_{0} g h\), and so $$ \frac{\rho}{\rho_{0}}=\frac{1}{1-\left(\rho_{0} g h / B\right)} $$
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