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Chapter 13: Problem 2

Atmospheric pressure is about \(1.0 \times 10^{5} \mathrm{~Pa}\). How large a force does the still air in a room exert on the inside of a window pane that is \(40 \mathrm{~cm} \times 80 \mathrm{~cm}\) ? The atmosphere exerts a force normal to any surface placed in it. Consequently, the force on the window pane is perpendicular to the pane and is given by $$F=P A=\left(1.0 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.40 \times 0.80 \mathrm{~m}^{2}\right)=3.2 \times 10^{4} \mathrm{~N}$$ Of course, a nearly equal force due to the atmosphere on the outside keeps the window from breaking.

Short Answer

Expert verified
The force exerted by the air on the window pane is \(3.2 \times 10^4\, \text{N}.\)

Step by step solution

01

Identify Given Values

We are given that the atmospheric pressure is \( 1.0 \times 10^5 \, \text{Pa} \). The dimensions of the window pane are \( 40 \, \text{cm} \times 80 \, \text{cm} \).
02

Convert Units to Meters

Convert the dimensions of the window pane from centimeters to meters. \( 40 \, \text{cm} = 0.40 \, \text{m} \) and \( 80 \, \text{cm} = 0.80 \, \text{m} \).
03

Calculate the Area of the Window Pane

Use the formula for the area of a rectangle, \( A = \text{length} \times \text{width} \). Substitute the converted values: \( A = 0.40 \times 0.80 = 0.32 \, \text{m}^2 \).
04

Calculate the Force Exerted by the Air

Use the formula \( F = P \times A \), where \( P \) is the pressure and \( A \) is the area. Substitute the given values: \( F = (1.0 \times 10^5 \, \text{Pa}) \times (0.32 \, \text{m}^2) = 3.2 \times 10^4 \, \text{N} \).

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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 upon the surface of the Earth. It is an essential concept in physics as it explains how pressure from the air around us can affect physical objects. Typically measured in Pascals (Pa), atmospheric pressure at sea level is approximately \(1.0 \times 10^{5} \, \text{Pa}\). This indicates that every square meter of surface area experiences a force equivalent to 100,000 Newtons due to the atmosphere. This pressure acts perpendicularly on surfaces, like window panes, walls, and roofs, influencing the stability of these structures. Understanding atmospheric pressure helps in designing buildings and various mechanical systems to withstand these forces effectively.
Force Calculation
Calculating the force exerted by atmospheric pressure on a surface involves understanding the relationship between pressure, force, and area. The equation \( F = P \times A \) is used to determine this force, where \( F \) is the force, \( P \) is the pressure, and \( A \) is the area. This formula highlights that the force exerted increases with greater pressure or a larger area. In our exercise, by substituting the known values of atmospheric pressure and the area of the window pane, we can solve for the force. This understanding is particularly useful in applications such as determining material strength for structures that will face atmospheric forces regularly.
Unit Conversion
Unit conversion is a critical step in solving physics problems, as it ensures that calculated values are consistent and accurate. In our exercise, the window pane's dimensions were initially given in centimeters, which needed conversion to meters鈥攖he standard unit for these calculations.
  • To convert from centimeters to meters, one must divide by 100, since 1 meter equals 100 centimeters. Thus, \( 40 \, \text{cm} \) becomes \( 0.40 \, \text{m} \), and \( 80 \, \text{cm} \) is \( 0.80 \, \text{m} \).
This step of converting units is crucial to avoid errors and ensure that all values used in calculations are compatible with SI (International System of Units) standards, particularly when calculating areas and forces.
Area Calculation
Calculating area accurately is essential in determining the force exerted on surfaces. In our problem, the window pane is a rectangle, so we use the formula for the area of a rectangle: \( A = \text{length} \times \text{width} \).
  • After converting the dimensions to meters, the area calculation becomes straightforward: \( A = 0.40 \, \text{m} \times 0.80 \, \text{m} = 0.32 \, \text{m}^2 \).
This area is then used in the force calculation. Correctly determining area is pivotal because even slight miscalculations can lead to significant errors in further mechanical computations, affecting the material selection and structural integrity of designs.

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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 glass stopper has a mass of \(2.50 \mathrm{~g}\) when measured in air, \(1.50 \mathrm{~g}\) in water, and \(0.70 \mathrm{~g}\) in sulfuric acid. What is the density of the acid? What is its specific gravity? The \(F_{B}\) on the stopper in water is \((0.00250-0.00150)(9.81) \mathrm{N}\). This is the weight of the displaced water. Since \(\rho=m / V\), or \(\rho g=F_{W} / V\), $$\begin{aligned} \text { Volume of stopper } &=\text { Volume of displaced water }=\frac{\text { weight }}{\rho \mathrm{g}} \\ V &=\frac{(0.00100)(9.81) \mathrm{N}}{\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=1.00 \times 10^{-6} \mathrm{~m}^{3} \end{aligned}$$ The buoyant force in acid is $$\left[(2.50-0.70) \times 10^{-3}\right](9.81) \mathrm{N}=(0.00180)(9.81) \mathrm{N}$$ But this is equal to the weight of displaced acid, \(m g\). Since \(\rho=m / V\), and since \(m=0.00180 \mathrm{~kg}\) and \(V=1.00 \times 10^{-6} \mathrm{~m}^{3}\) $$\rho \text { of acid }=\frac{0.00180 \mathrm{~kg}}{1.00 \times 10^{-6} \mathrm{~m}^{3}}=1.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$$ Then, for the acid, $$\text { sp } \mathrm{gr}=\frac{\rho \text { of acid }}{\rho \text { of water }}=\frac{1800}{1000}=1.8$$ Alternative Method $$\begin{array}{l} \text { Weight of displaced water }=\left[(2.50-1.50) \times 10^{-3}\right](9.81) \mathrm{N} \\ \text { Weight of displaced acid }=\left[(2.50-0.70) \times 10^{-3}\right](9.81) \mathrm{N} \end{array}$$ so sp gr of acid \(=\frac{\text { Weight of displaced acid }}{\text { Weight of equal volume of displaced water }}=\frac{1.80}{1.00}=1.8\) Then, since sp gr of acid \(=(\rho\) of acid \() /(\rho\) of water \()\), $$ \rho \text { of acid }=(\text { sp gr of acid })(\rho \text { of water })=(1.8)\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)=1.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} $$

A cube of wood floating in water supports a 200-g mass resting on the center of its top face. When the mass is removed, the cube rises \(2.00 \mathrm{~cm}\). Determine the volume of the cube.

A partly filled beaker of water sits on a scale, and its weight is \(2.30 \mathrm{~N}\). When a piece of metal suspended from a thread is totally immersed in the beaker (but not touching bottom), the scale reads \(2.75 \mathrm{~N}\). What is the volume of the metal? The water exerts an upward buoyant force on the metal. According to Newton's Third Law of action and reaction, the metal exerts an equal downward force on the water. It is this force that increases the scale reading from \(2.30 \mathrm{~N}\) to \(2.75 \mathrm{~N}\). Hence the buoyant force is \(2.75-2.30=0.45 \mathrm{~N}\). Then, because $$F_{B}=\text { weight of displaced water }=\rho_{w} g V=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(V)$$ we have the volume of the displaced water, and of the piece of metal, namely, $$ V=\frac{0.45 \mathrm{~N}}{9810 \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}^{2}}=46 \times 10^{-6} \mathrm{~m}^{3}=46 \mathrm{~cm}^{3} $$

Downward forces of \(45.0 \mathrm{~N}\) and \(15.0 \mathrm{~N}\), respectively, are required to keep a plastic block totally immersed in water and in oil, respectively. If the volume of the block is \(8000 \mathrm{~cm}^{3}\), find the density of the oil.
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