Physics for Students of Science and Engineering

By A. L. Stanford and J. M. Tanner

Physics for Students of Science and Engineering is a calculus-based textbook of introductory physics. The book reviews standards and nomenclature such as units, vectors, and particle kinetics including rectilinear motion, motion in a plane, relative motion. The text also explains particle dynamics, Newton's three laws, weight, mass, and the application of Newton's laws. The text reviews the principle of conservation of energy, the conservative forces (momentum), the nonconservative forces (friction), and the fundamental quantities of momentum (mass and velocity). The book examines changes in momentum known as impulse, as well as the laws in momentum conservation in relation to explosions, collisions, or other interactions within systems involving more than one particle. The book considers the mechanics of fluids, particularly fluid statics, fluid dynamics, the characteristics of fluid flow, and applications of fluid mechanics. The text also reviews the wave-particle duality, the uncertainty principle, the probabilistic interpretation of microscopic particles (such as electrons), and quantum theory. The book is an ideal source of reference for students and professors of physics, calculus, or related courses in science or engineering.

• Language: English
• Publisher: Academic Press
• Release date: Jun 28, 2014
• ISBN: 9781483220291

Book preview

1

Introduction

Publisher Summary

Physics is a natural science. It is one of humankind’s responses to its curiosity about how nature works and about how the universe is ordered. Like other modem natural sciences, physics has evolved to become a logical process based on the scientific method. This method is rooted in a philosophy that recognizes no truths and embraces no dogma but seeks to be completely objective and practical. Hypotheses proposed according to the scientific method are retained only if they enjoy continued and unfailing success. A single instance in which a hypothesis fails to predict successfully the outcome of a pertinent, repeatable experiment requires either rejection of the hypothesis or its modification to rectify that failure. Throughout the history of science, many hypotheses have been discarded and many have been changed. Those that have enjoyed some measure of success but are without extensive experimental verification over a long period of time are referred to as theories. Scientists do not believe the laws of physics; they merely use them in very practical ways, maintaining a healthy skepticism that permits continual checking of current laws and theories and encourages speculation about new hypotheses. In this way, the scientific method provides a rational approach to an intellectual and logical comprehension of natural phenomena.

1.1 Physics and the Scientific Method

Physics is a natural science. It is one of humankind’s responses to its curiosity about how nature works, about how the universe is ordered.

Like other modern natural sciences, physics has evolved to become a logical process based on the scientific method. This method is rooted in a philosophy that recognizes no truths and embraces no dogma but seeks to be completely objective and practical. The scientific method may be considered an investigative process composed of three parts:

1. Physical processes are observed and measured both quantitatively and qualitatively. This step necessarily includes the conception and definition of appropriate quantities by which measurements may be made.

2. A hypothesis is offered, usually in the form of a general principle or a mathematical statement of relationships between physical quantities (time and distance, for example). These principles or relationships can be used to predict the results of other similar physical processes.

3. The hypothesis is subjected to experimental tests of its validity. Its predictions are compared to actual measured values.

Hypotheses proposed according to the scientific method are retained only if they enjoy continued and unfailing success. A single instance in which a hypothesis fails to predict successfully the outcome of a pertinent, repeatable experiment requires either rejection of the hypothesis or its modification to rectify that failure. Throughout the history of science many hypotheses have been discarded, and many have been changed. Those that have enjoyed some measure of success but are without extensive experimental verification over a long period of time are referred to as theories (those that have not had some success are not referred to at all). Hypotheses that have withstood successfully the repeated and diverse trials of experiment are accorded the title law, but even the most venerated laws of physics are not considered true by scientists. Laws are, along with all the tenets of science, acceptable only as long as they continue to coincide with measurements of physical processes. Scientists do not believe the laws of physics; they merely use them in very practical ways, maintaining a healthy skepticism that permits continual checking of current laws and theories and encourages speculation about new hypotheses. In this way the scientific method provides a rational approach to an intellectual and logical comprehension of natural phenomena.

1.2 Units

Physics is a science of relationships and measurements of the physical world, and understanding physics requires an understanding of the measurement process. The measurements of physical quantities are determined quantitatively, in terms of some units like feet, meters, miles per hour, or kilograms.

Standards and Nomenclature

Measurements in terms of units require standards. For example, in the metric system of units the standard unit of length is the meter; a table that is 2.7 meters long, for example, is 2.7 times the length of the standard meter.

The standard unit of time is the second (s). Originally defined in terms of a fraction of a mean solar day on earth, the second is now defined in terms of certain electromagnetic emissions from the element cesium. Another basic standard in the metric system is the kilogram (kg), a unit of mass defined to be equal to the mass of a particular body of metal kept in France. The concept of mass will be considered later, but for now it is sufficient to note that the mass of a given object is an expression of the quantitity of matter contained in that body. The third basic unit of the metric system is the meter (m). Over the years the meter has been defined successively in terms of a quadrant of the surface of the earth, in terms of the distance between the marks on a metal bar, and in terms of a specified number of wavelengths of certain electromagnetic emissions from a particular species of atom. In 1983 the standard meter was redefined by international agreement to be the distance that light travels through a vacuum in 1/299,792,458 of a second. Thus the basic unit of length is defined in terms of our best measured value of the speed of light c. In 1983 the accepted value of the speed of light in vacuum was taken to be

The wide range of magnitudes of measurements encountered in physics makes it convenient to use multiples and submultiples of the standard units. The metric system is a decimal system, that is, it is based on powers of 10. This system is particularly amenable to using prefixes to specify multiplying factors that can be associated with units. Table 1.1 lists some of the common prefixes and those that will be used throughout this book. The prefixes or their abbreviations may be used with any metric unit or its abbreviation. For example, the kilowatt, or kW, is 10³ watts, and the nanosecond, or ns, is 10−9 second.

TABLE 1.1

Common Prefixes and Their Multiplying Factors Associated with Physical Units

The metric system of units, known as the SI (for Système Internationale), will be used in this book along with the British Engineering system, often called the English system. The metric system uses the kilogram as the standard unit of mass, the meter for length, and the second for time. In the English system the standard force (the choice of force or mass as fundamental is arbitrary) is the pound (lb), defined to be that force with which the earth pulls on a mass of 0.45359237 kilogram at a certain location on the surface of the earth. The standard length in the English system is the yard (yd), which is specified in terms of the meter such that

It follows that

The unit of time, the second, is the same in the English and SI systems.

Conversion of Units

A basic skill necessary to the successful solution of many physics problems is the conversion of units between the metric and English systems. It may be necessary to determine, for example, the number of inches in a half mile (mi) or the number of meters in six feet. In any case, confusion and error can be avoided by using a simple procedure based on the principle that a given measure (including units) is not changed when multiplied by unity, that is, by the number 1. Of course, unity can be represented by any fraction in which the numerator and the denominator are equal or equivalent. The fractions 7/7, 3 ft/1 yd, and 2.54 cm/1 in., for example, are all equal to unity. A half mile can be converted to inches without changing its measure (that is, without changing the magnitude of its given length) by starting with the given value and multiplying it by appropriate fractions, each of which is equal to unity, as many times as needed:

The key to this procedure is finding the appropriate fractions equal to unity that should be used. Most people probably do not know offhand the number of inches in a mile, but many people know that 5280 feet are equal to a mile. Thus, the fraction that converts miles to feet is used, anticipating the next step, which uses 12 in. in a foot. Notice that in constructing the first fraction, 1 mi is placed in the denominator so that its unit cancels algebraically with that of the given 0.50 mi. At this stage, 0.50 mi has been converted to feet. Similarly, the second fraction is formed so that the unit feet cancels the previously obtained feet, leaving the final product expressed in inches. This process accomplishes the desired conversion. For a given conversion, an arbitrary number of fractions, each equal to unity, may be used as necessary to convert units step by step to provide the given measure in the desired units.

Here we have been treating units as if they were algebraic quantities, and indeed they may be manipulated algebraically. As a further example, the area of a rectangle with sides of lengths 3 m and 2 m is obtained by multiplying its height h by its width w.

in which multiplication of the unit of height by the unit of width gives the unit of area, m². A similar algebraic treatment of units applies to division. For example, an automobile traveling a distance of 120 mi on 3.0 gallons (gal) of fuel achieves a mileage rating of

Addition and subtraction of physical quantities require particular care with respect to units. Any terms to be added algebraically must be expressed in identical units before the usual rules of algebraic addition may be applied. The procedure for adding 3 ft and 2 in., for example, is

Note to the Student

The following exercises and those throughout the book are integral parts of the text. At times, new and essential material will be introduced only in the exercises. These exercises are, in general, one-step problems designed to ensure that the student understands the preceding textual material. Answers to the exercises usually accompany each problem or question. Any difficulty with an exercise should prompt the student to review the appropriate section of the text before proceeding with new material.

E 1.1

A person’s height is measured at 5 ft, 8 in. Express this height in centimeters.

Answer: 1.7 × 10² cm

E 1.2

A world-class sprinter can run the 100-m dash in about 10 s, which corresponds to an average speed of 10 m/s. Express this sprinter’s speed in miles per hour.

Answer: 22 mi/h

E 1.3

An acre is the area of a square plot of land measuring approximately 209 ft along each side. How many square meters correspond to an acre?

Answer: 4.1 × 10³ m²

E 1.4

The density of an object is equal to its mass divided by its volume. A cube of lead 4.2 cm along an edge has a mass of 865 grams (g). Calculate the density of lead in kg/m³.

Answer: 1.1 × 10⁴ kg/m³

E 1.5

The mass of the earth is 6.0 × 10²⁴ kg, and its radius is 4.0 × 10³ mi. Determine the average density of the earth in g/cm³.

Answer: 5.5 g/cm³

1.3 Vectors

Mathematics is the working language of physics. The material in this text requires facility with algebra and elementary plane geometry, a basic knowledge of analytic geometry, and a growing familiarity with the fundamental concepts of the calculus. Certain aspects of trigonometry are so frequently required that considerable facility with the manipulation of trigonometric relationships is needed. A review of trigonometry is provided in the Appendix. Similarly, necessary to the mastery of physics is a comfortable working knowledge of vector analysis. Some familiarity with vector notations and operations is necessary for handling the physical principles and their applications presented in this book. For these reasons, the more commonly used definitions and operations associated with vectors are presented here for review or study, as required. This section has been made compact intentionally. A student studying vectors for the first time may wish to consult a mathematics textbook.

Vector and Scalar Quantities

Certain physical quantities have a direction associated with them. One cannot, for example, exert a push, that is, a force, on an object without pushing in a specific direction. The push must be in some definite direction, such as vertically upward, eastward, or in a direction 30° north of westward. Such quantities also have a magnitude, of course. The push in our example must have associated with it some number of pounds of force applied in the specified direction. Such physical quantities, those that have both magnitude and direction, are called vector quantities. In contrast, physical quantities that have no direction associated with them, like volume, are called scalar quantities, and such quantities have associated magnitudes (such as a volume of 2.0 liters) but no associated directions.

Here we shall consider some of the basic operations involving vectors, the graphical or analytical representations of vector quantities. All the operations considered here will be used extensively throughout this book.

A vector quantity is represented graphically by an arrow, the length of which represents the magnitude of the quantity and the direction of which represents the direction of the quantity. Vectors having identical magnitudes and identical directions are mathematically equal. Thus, in Figure 1.1, the vectors A, B, and C are equal, or A = B = C. (Note that vector quantities are represented by boldface letters.) The magnitude of a vector A is represented by either A or |A|. In Figure 1.2 the vectors A, 2A, and –A are shown graphically. The vector 2A is twice the length of A and is in the same direction, and the vector –A is the same length as A but is in the opposite, or antiparallel, direction.

Figure 1.1 Three equal vectors. The vectors A, B, and C all have the same magnitude and direction.

Figure 1.2 Graphical representation of the effect of multiplying a vector by a scalar value. The vector 2A is twice the length of A and is in the same direction as A. The vector –A is the same length as A but is in an opposite direction.

Vector Addition and Subtraction in Polar Form

A vector may be represented in polar form by giving its magnitude and direction, such as A = 12 m ∠212°. Two vectors A and B may be added graphically, as shown in Figure 1.3. By placing the tail of B at the head of A, while maintaining the length and direction of both arrows, the sum C is formed by the vector from the tail of A to the head of B. Then,

Figure 1.3 The graphical addition of two vectors. The vector B is added to A by placing the tail of B to the head of A without changing the magnitude (length) or direction of either vector. The sum of A and B is the vector C, formed by drawing a vector from the tail of A to the head of B.

(1-1)

symbolically represents the addition operation. The magnitude and direction of C is determined by applying trigonometry to the triangle thus formed, that is, by using the laws of sines and cosines.

Example 1.1

PROBLEM

In Figure 1.3 let A = 4, B = 6, and θ = 60°. Calculate the magnitude of C and the angle between A and C.

SOLUTION

Figure 1.4 shows the triangle relating the known quantities A, B, and θ to the unknown quantities C, α,β, and γ. (Notice that the figure shows the known values for A, B, and θ and that the unknown quantities specifically asked for in the problem are labeled with question marks. Other unknown quantities in the triangle, like the angles α and γ, are labeled in case they are needed.)

Figure 1.4 Example 1.1

The law of cosines relates C to A, B, and γ, giving

Because γ and θ are supplementary angles, or γ + θ = 180°, it follows that

Thus, C is now determined by

The angle β is determined using the law of sines:

A simple check for this solution can be accomplished by calculating the angle α using the law of sines and then adding α, β, and γ to make sure they add to 180°. Thus, the law of sines gives

and the sum of the interior angles of the triangle is

E 1.6

If F = C + D, determine the magnitude of F and the angle between F and C.

Answer: 9.0, 39°

Exercise 1.6

E 1.7

Convince yourself that A + B = B + A, that is, that vector addition is commutative. (Hint: The sum of two vectors can be represented graphically as the diagonal of a parallelogram.)

Three or more vectors may be added graphically by extending the procedure of placing the tail of a vector to be added to the head of the last one added. Thus, in Figure 1.5, the sum A + B + C is drawn from the tail of the first to the head of the last. Because vector addition is commutative, the order in which the vectors A, B, and C are added does not affect the result. It follows then, for example, that

Figure 1.5 The graphical addition of three vectors. The vector sum of A, B, and C is obtained by placing the tail of B to the head of A, the tail of C to the head of B, and the sum A + B + C is drawn from the tail of A to the head of C.

(1-2)

Vector subtraction is defined by

(1-3)

The vector B is subtracted from the vector A by adding the negative of B to A. Figure 1.6 illustrates vector subtraction.

Figure 1.6 Graphical representation of vector subtraction. (a) A – B is formed by adding the vector –B to A. (b) To subtract D from C, the negative of D is formed and added to C.

E 1.8

If C = A – B, determine the magnitude of C and the angle between C and A.

Answer: 6.1, 25°

Exercise 1.8

E 1.9

If F = E – D, determine the magnitude of F and the angle between F and E.

Answer: 11, 75°

Exercise 1.9

Vector Multiplication

We have already seen that a vector may be multiplied by a scalar; in Figure 1.2 the vector 2A has twice the magnitude of A and the same direction as A. There are two distinct ways to multiply a vector by a vector. One way yields a product that is a scalar; the other way yields a product that is a vector.

The scalar product of two vectors A and B (written A · B and sometimes called the dot product) is defined by

(1-4)

where A and B are the magnitudes of A and B, and ϕ is the smaller angle between A and B when they are drawn with their tails at a common point. Notice that the result of this multiplication operation is not a vector but a real number with an algebraic sign. The following example and exercises will consider this aspect of the scalar product.

Example 1.2

PROBLEM

Two vectors A and B have magnitudes that are given by A = 6 and B = 5. Calculate the angle between the two vectors if (a) A · B = 18, and (b) A · B = −21.

SOLUTION

The unknown angle ϕ between A and B is shown in Figure 1.7. Using Equation (1-4) gives

Figure 1.7 Example 1.2

(a) If A · B = 18, then it follows that

(b) But if A · B = −21, then the calculation for ϕ is

E 1.10

Calculate the scalar product for each of the illustrated pairs of vectors.

Answers: (a) 41; (b) 0; (c) −20; (d) −12

Exercise 1.10

E 1.11

In the previous exercise you saw examples of pairs of vectors with scalar products that are positive, zero, or negative. See if you can generalize these results by completing the following sentences:

(a) The scalar product of two vectors is positive if the angle between the two is _____.

Answer: ≥0 and <90°

(b) The scalar product of two nonzero vectors is zero if the angle between the two is_____.

Answer: 90°. i.e., the vectors are perpendicular

(c) The scalar product of two vectors is negative if the angle between the two is _____.

Answer: 90° and ≤180°

(d) If a · b = ab, then the two vectors are _____.

Answer: parallel, i.e., in the same direction

(e) If a · b = −ab, then the two vectors are _____.

Answer: in opposite directions

E 1.12

You know that real number multiplication is commutative, that is, if r1 and r2 are real numbers, then r1r2 = r2r1. Is this also true for the scalar product of two vectors? In other words, does A · B = B · A?

The vector product of two vectors A and B is written A × B and is sometimes called the cross product. This vector product is defined such that the magnitude of the product is given by

(1-5)

where, again, A and B are magnitudes of A and B, and ϕ is the smaller angle between the vectors A and B when they are drawn with their tails at a common point. The direction of the product A × B is perpendicular to both A and B, that is, it is perpendicular to the plane containing A and B. The direction of A × B is further specified by the so-called right-hand rule, in which the fingers of the right hand are curved in a direction from A toward B through the smaller angle between A and B. Then the extended thumb of the right hand is perpendicular to the plane of A and B and is in the direction of A × B. The right-hand rule determination of the direction of A × B is shown in Figure 1.8. B × A is a vector of the same magnitude as A × B but is in the opposite direction. Thus we conclude that B × A = −A × B; so the vector product is not commutative.

Figure 1.8 The right-hand rule for specifying the direction of A × B. When two vectors A and B are positioned with their tails at a common point, the vector A × B is perpendicular to the plane containing A and B in the direction of the thumb of the right hand when the fingers of that hand curve in the direction from A toward B through θ, the smaller of the two angles between A and B.

E 1.13

Calculate a × b for each of the vector pairs shown.

Answers: (a) 8.5, out of the page; (b) 8.5, into the page; (c) 8.5, out of the page; (d) 12, into the page; (e) zero

Exercise 1.13

E 1.14

Complete the following sentences:

(a) The vector product of two nonzero vectors is zero if _____.

Answer: (a) the two are parallel or antiparallel

(b) The magnitude of the vector product of two vectors of specified magnitudes is a maximum if _____.

Answer: the two are perpendicular

(c) If |a × b| = ab, then _____.

Answer: the two are perpendicular

(d) If |a × b| = 0 and a ≠ 0, b ≠ 0, then _____.

Answer: the two are parallel or antiparallel

E 1.15

Explain why a · (a × b) = 0 for any two vectors a and b.

Rectangular Components of Vectors

A vector can always be specified by its magnitude and direction, but it is often necessary or convenient to specify one by its rectangular components Ax and Ay (and Az if the vector is three dimensional). Ax, the x component of the vector A, is the x coordinate of the head of A when the vector A is drawn with its tail at the origin of a coordinate system. The y component of A, Ay, is similarly defined. Ax and Ay are illustrated in Figure 1.9(a). Note that the rectangular components of a vector are real numbers that may be positive, zero, or negative. Figure 1.9(b) illustrates a vector B having a negative x component and a positive y component.

Figure 1.9 Rectangular components of vectors. (a) The rectangular components Ax and Ay of a vector A in the first quadrant are both positive numbers. (b) For a vector B lying in the second quadrant, its rectangular components are Bx, a negative number, and By, a positive number.

A vector may be represented by giving its magnitude and direction (polar representation) or by giving its x and y components (rectangular representation). The equivalence of these two representations and the process of changing from one representation to the other may be seen by considering the vector C of Figure 1.10(a). The magnitude of C is 10, and its direction is 30° (measured counterclockwise from the positive x axis, according to convention). In the polar representation, C may be written as C = 10∠30°. The rectangular components of C are obtained by applying right-triangle trigonometry to Figure 1.10(a), giving

Figure 1.10 (a) Determination of rectangular components of a vector C that is specified in the polar representation. (b) Determination of the polar representation of a vector D with rectangular components that are known.

(1-6)

where C = 10 and θ = 30°. Then the components are given by

(1-7)

Alternatively, the magnitude and direction of a vector may be determined if its rectangular components are known. In Figure 1.10(b) the vector D has components Dx = 3 and Dy = 4. Using the Pythagorean theorem and trigonometry, we see that

(1-8)

so that

E 1.16

Calculate the rectangular components for each of the vectors shown.

Answers: ax = −1.7, ay = 4.7; bx = −3.0, by = −5.2; cx = −2.7, cy = 7.5

Exercise 1.16

E 1.17

Calculate the magnitude and direction of each of the vectors for which rectangular components are given. Answer in the form a = a∠θ, where θ is measured counterclockwise from the positive x direction.

(a) ax = 3.7 ay = 2.9

Answer: 4.7∠38°

(b) bx = −2.2 by = −1.3 (Be careful with the direction!)

Answer: 2.5∠210°

(c) cx = 1.9 cy = −2.2

Answer: 2.9∠310°

Vector Operations in Component Notation

So far vectors have been referred to as representations of quantities having both magnitude and direction. A more frequently useful representation in physical problems and in three-dimensional situations is component notation. In this form a vector is written in terms of its components

(1-9)

in which Ax, Ay, and Az are the x, y, and z components of A. The symbols î, ĵ, and are unit vectors in the positive x, y, and z directions, respectively, as shown in Figure 1.11. A unit vector has a length of one unit and is used simply to indicate direction. A vector A with components Ax = 3, Ay = 4, and Az = 2 could, for example, be represented by

Figure 1.11 The unit vectors. î, ĵ, and of a rectangular coordinate system.

(1-10)

The vector A is represented in Equation (1-10) in component notation and is shown in Figure 1.12. The component vectors of A, namely 3î, 4ĵ, and are seen in Figure 1.12(a) to be the projections of A onto the x, y, and z axes. Figure 1.12(b) illustrates how the vector A is the sum of its component vectors.

Figure 1.12 The rectangular component vectors of a vector A = 3î + 4ĵ + 2k. (a) The component vectors are the projections of A onto the rectangular axes. (b) The vector A is the sum of its component vectors.

E 1.18

Use the unit vectors î and ĵ to write in rectangular form each of the vectors shown here.

Answers: A = 4.0î −6.9ĵ; B = −3.7î + 4.7ĵ

Exercise 1.18

E 1.19

Write in polar form each of the vectors that are specified here:

Answers: a = 4.2∠311°; b = 5.0∠260°

Component notation makes it particularly easy to add or subtract vectors. The sum of and is obtained by algebraically adding the individual components:

(1-11)

Vector subtraction is similarly accomplished, subtracting algebraically the individual components:

(1-12)

E 1.20

For the vectors A and B of Exercise 1.18, calculate A – B and express the result in polar form.

Answer: 14∠300°

E 1.21

Extending vector addition and subtraction to include three or more vectors is accomplished simply using component notation. For example,

For the vectors a = 2.7î −3.2ĵ, b = −1.3î −4.8ĵ, and c = 2.2î + 1.8ĵ, calculate b – a – c and express the result in rectangular form.

Answer: 3.6î −6.2ĵ

Scalar multiplication and vector multiplication in component notation require that we take particular note of the unit vectors î, ĵ, and . Because these three unit vectors are mutually perpendicular, the definition of a scalar product (Equation [1-4], A · B = AB cosϕ) demands that

(1-13)

A right-handed coordinate system, which will be used exclusively in this book, is an arrangement of the x, y, and z axes such that according to the right-hand rule of the vector cross product. Therefore, the definition of the vector product requires that

(1-14)

Using the scalar products of unit vectors of Equation (1-13), the scalar product of and becomes

(1-15)

All other terms of the scalar product, like AxByî · ĵ, are zero because, according to Equation (1-13), only the î · î, ĵ · ĵ, and terms are not zero.

Example 1.3

PROBLEM

If and , calculate (a) A · B and (b) the angle between A and B.

SOLUTION

(a) Because A and B are given in component form, Equation (1-15) gives

(b) Because the definition of A · B is

the angle ϕ between A and B may be calculated according to

E 1.22

If , and , calculate: (a) a · b, (b) a · c, (c) b · c, (d) a · (b + c).

Answers: (a) 0; (b) −41; (c) −28; (d) −41

E 1.23

For a, b, and c of Exercise 1.22, which of the vectors b, c, or (b + c) is perpendicular to a? Why?

Answer: b; because a · b = 0 = ab cosϕ. cosϕ= 0 and ϕ= 90°

E 1.24

If a = 2î + 2ĵ – , b = 9î + 20ĵ + 12 , and c = 2î −2ĵ + , calculate the angle between (a) a and b, (b) a and c, and (c) b and c.

Answers: (a) 52°; (b) 96°; (c) 98°

The vector product of A and B in component form is somewhat more complicated.

(1-16)

may be remembered more easily in the mnemonic form of a determinant:

(1-17)

Example 1.4

PROBLEM

If the vectors A and B are those given in Example 1.3, namely, A = 4î −6ĵ + 5 and B = 3î + 2î −7 ,

(a) calculate A × B.

(b) verify that |A × B| = AB sinϕ, using ϕ as determined in Example 1.3.

(c) show that A × B is perpendicular to both A and B.

SOLUTION

(a) Equation (1-16) applied to A and B gives

Alternatively, Equation (1-17) may be used to obtain

(b) The magnitude of A × B is given by

In Example 1.3 the angle between A and B was found to be 120°. It follows, then, that

(c) To show that A × B is perpendicular to A and to B, we need only demonstrate that A · (A × B) and B · (A × B) are each equal to zero. This is accomplished by using the result of part (a) to get

E 1.25

If A = 4î −7ĵ, B = 5î + 3ĵ, and C = −2î −4ĵ, calculate: (a) A × B; (b) B × C; (c) A × C.

Answers: (a) 47 ; (b) −14 ; (c) −30

E 1.26

Using the vectors a, b, and c, as given in Exercise 1.24, calculate a × b, b × c, and c × a.

Answers: 44î −33ĵ + 22 ; 44î + 15ĵ −58 ; 4ĵ + 8

1.4 Problem-Solving: A Strategy

For beginning students, the study of physics often amounts to solving problems. Indeed, learning basic physics means learning to solve word problems, that is, problems stated in words. Of course, a knowledge of the principles, definitions, and relationships of physics is a necessary prerequisite to the successful study of physics. A considerable portion of this book is devoted to those fundamentals. This section, however, and part of each subsequent chapter, is concerned with the presentation and development of a strategy designed to assist the student in the successful solution of problems. Here, then, we present an overall strategy for solving problems, a process intended to be applicable to any word problem.

Word problems generally lend themselves to analysis in four steps, which may be characterized as follows:

Step 1. Understanding and visualizing the problem. Identifying known and unknown quantities.

Step 2. Relating unknown quantities to known quantities.

Step 3. Executing the mathematics to obtain the solution.

Step 4. Checking the solution.

Let us examine the steps individually. Step 1 may seem obvious, because it is necessary to understand what a problem is about in order to attempt its solution with any chance of success. But understanding a problem means understanding all the words used to state it, not only their literal meanings but also their implications for the problem at hand. For example, the word frictionless appears frequently in physics problems. Most people would say this word suggests smoothness, but that association is not enough. In a physics problem, a frictionless surface can push on objects only in specific directions, and this implication may be crucial in the analysis of a given problem. Therefore, the technical definitions and implications of the words in a problem should be clearly understood before tackling the problem.

A second aspect of understanding a word problem hinges on recognizing common assumptions and agreements between the writer of the problem and the problem solver. For example, few problems specifically state that the physical situation of the problem takes place on or near the surface of the earth. It is generally understood that if the problem involves an alien planet, that condition would be specified in the statement of the problem. A perfectly unambiguous problem that details every possible contingency would require unreasonable space and effort for its statement. Common sense is the appropriate guide in this respect.

Step 1 also suggests visualizing the problem. Picturing the problem in one’s mind is not usually sufficient; a graphic representation, such as a drawing or sketch, however crude, is an immense aid in understanding a word problem. The process of translating the literal meanings of the words of a problem into an accurate mental image of what is taking place is a crucial aspect of problem-solving. Making a drawing, if possible, is the surest means of making a successful transition from words to a useful and effective mental image. In particular, it is helpful to identify the known and unknown quantities in the problem (the last task in Step 1) by putting appropriate labels with quantities directly on the drawing. Thus, a known distance may be labeled x = 12 m and an unknown force indicated by F = ? This technique makes the issues of the problem clearer and more meaningful.

Step 2 offers the opportunity to use the principles, definitions, and relationships that have been learned. Every problem asks the solver to find some unknown quantity. To find it, the solver must relate that unknown quantity to those that are known or given in the problem. Here is where the student calls on the resources acquired during the study of principles and relationships. This step, relating unknown quantities to known quantities, usually becomes easier with experience and practice. In physics problems, the process of relating unknowns to knowns is almost always accomplished in the form of equations in which symbols are used to represent physical quantities.

Step 3, solving the equations obtained in Step 2, is not usually a concern of students prepared to study physics. It is assumed that students using this text can carry out the appropriate mathematical operations necessary to obtain the solution.

Step 4 serves as a check on both the mathematical execution of Step 3 and the selection of appropriate relationships in Step 2. Usually it is not possible to check the solution of a physics problem by direct substitution, in the sense that the solution of most algebra problems may be checked, but some procedures are often available that provide confidence in the reasonableness of a solution. The most obvious check is the examination of the units associated with the solution. If an answer that is supposed to be a distance turns up with units of time, it is immediately clear that an error has been made. Less obvious procedures, like estimations, are often appropriate. This type of check is illustrated in the four-step-strategy example that follows. Finally, solutions to physics problems may frequently be checked by what are sometimes called limiting cases. The ability to make these checks effectively usually depends on the problem solver’s experience in recognizing what occurs in simple physical situations and relating or comparing a particular solution to the simple situation. As an example, suppose we have found a solution that describes the motion of an object moving along a horizontal surface. We could examine the solution to see if it predicts what we would expect if the surface were frictionless. Although such procedures do not guarantee that a solution is correct, they afford a measure of confidence in the answer and provide the further benefit of exercising and strengthening our physical insight into the problem.

Let us now look at the use of the four-step strategy with a simple problem:

How long after midnight does it take for the hands of a clock to coincide again?

Step 1.

The only word in the problem that may need clarifying is coincide. The hands of a clock coincide when they point in the same direction. What are the assumptions made in the statement of the problem? Without belaboring this issue, the problem obviously assumes that we know the functional principles of how a clock indicates the passage of time. We are supposed to know that both hands of a clock coincide at midnight, that the minute hand moves through an angle of 360° every hour, and that the hour hand moves only 1/12 as fast, that is, through an angle of 30° every hour. To visualize the pertinent aspects of the problem, construct a simple diagram (see Fig. 1.13) that shows the positions of both hands of a clock when they first coincide after midnight. The figure also shows the angle through which each hand has turned since midnight. Let θm be the angle through which the minute hand has turned and θh be the angle through which the hour hand has turned.

Figure 1.13 The hands of a clock coinciding for the first time after midnight. The figure shows the angles θh and θm through which the hour hand and minute hand have turned since midnight.

Step 2.

This problem requires that we understand the notion of rate, that the minute hand moves at an angular rate of 360°/h, and the hour hand turns at the rate of 30°/h. The mathematical relationship that we can establish is suggested by the units in the problem. An angular rate R has the unit (deg/h). If we multiply R by the time t with the unit (h), we obtain an angle θ, measured in (deg). Thus, the appropriate relationship here is

or

(1-18)

We now apply Equation (1-18) to each hand of the clock, recognizing certain facts from the figure. We do not know the angle θh through which the hour hand has moved in the time t that bas elapsed since midnight, so we write

(1-19)

Neither θh nor t is a known value. Obviously we cannot solve one equation with two unknown quantities. We proceed, therefore, by considering further information in the figure, the information associated with the minute hand. The minute hand has moved through an angle θm, which is clearly 360° + θh, in the same elapsed time t. Then for the minute hand, we may write

(1-20)

Because t is the time elapsed since midnight, both Equations (1-19) and (1-20) include the same unknown quantities, θh and t. We now have a sufficient number of relationships to determine t, the desired quantity representing the time elapsed between midnight and the instant the hands next coincide.

Step 3.

The execution of the mathematical solution is, in this case, a simple algebraic substitution and manipulation. Substituting Equation (1-19) for θh into Equation (1-20) gives

(1-21)

Because t is the elapsed time since midnight, the result of Equation (1-21) is the value required in the statement of the problem.

Step 4.

How may we check the result? We may estimate what we might have expected without working the problem, namely in somewhat more than 1.0 h. When the minute hand has completed one revolution from its position at midnight, the hour hand has moved to 1 on the clock. Five minutes later, when t = 1 1/12 h, the minute hand has reached 1 on the clock, but the hour hand has moved forward slightly. It should then not take much longer for the hands to coincide. Because 1 1/11 h is only slightly greater than 1 1/12 h, we conclude that our answer is probably correct.

A second example, in which the steps are not discussed as extensively, demonstrates further the utility of this problem-solving process.

Example 1.5

PROBLEM

How many grams of a powder A with a density of 6.0 g/cm³ must be added to 10 g of a powder B, the density of which is 2.0 g/cm³, to give a mixture of density 5.0 g/cm³?

SOLUTION

Let x be the number of grams in powder A. Figure 1.14 depicts the situation. If the volume of A is added to the volume of B, we obtain the volume of A + B, or

Figure 1.14 Example 1.5

(1-22)

The units suggest that the volumes of A, B, and A + B may be obtained by dividing the number of grams of each material by the density of that material. Then Equation (1-22) becomes

(1-23)

Equation (1-23) may be solved algebraically:

Thus, 90 g of A (6.0 g/cm³) must be added to 10 g of B (2.0 g/cm³) to produce 100 g of a mixture with density 5.0 g/cm³. This result may be checked by direct substitution into Equation (1-23):

This problem-solving approach will be utilized throughout this book in three ways, with Problems, Examples, and Problem-Solving Summaries:

1. Each chapter contains illustrative Problems within the text. In these problems, extended discussions relate the principles, definitions, and relationships being studied to the problem-solving process. These problems emphasize specialized techniques and hints that are applicable to similar problems. Problem-solving hazards are pointed out in these discussions, and appropriate safeguards are recommended.

2. Each chapter includes Examples, typical problems associated with the topics of that chapter. These examples, which use the problem-solving strategy without extended discussion, are intended as models for the student.

3. Each chapter contains a Problem-Solving Summary, which reiterates the principles and relationships of the chapter in terms of problem-solving. The summary offers suggestions on how problem categories may be identified: When is a problem about energy, or momentum, or rotation, for example. Each chapter summary suggests how the basic problem-solving strategy may be modified or specialized to apply most effectively to the specific topical material of that chapter. Finally, the problem-solving summary of each chapter attempts to identify common pitfalls of problems and suggests how these hazards may be recognized and avoided.

Problems

Note to the Student

The problems at the end of each chapter are separated into two groups, A and B. Group A problems are based on material contained in the preceding text. Group B problems are, in general, more challenging and frequently present new material.

GROUP A

1.1. The density of mercury is 13.6 g/cm³.

(a) Calculate the density of mercury in kg/m³.

(b) One kilogram of mass weighs approximately 2.2 pounds. Could you lift a suitcase filled with mercury if the dimensions of the suitcase are 15 cm × 50 cm × 60 cm?

1.2. A person who weighs 150 lb has a mass of approximately 70 kg. Using the fact that an average person floats in water with very little exposed volume and, therefore, has approximately the same density as water, namely 1.0 g/cm³, estimate the volume in m³ and ft³ of a person weighing 150 lb.

1.3. If A = 8î −9ĵ and B = 12∠300°,

(a) express A in polar form.

(b) express B in rectangular form.

(c) calculate A + B.

(d) calculate A · B.

1.4. If k and m are the vectors shown,

(a) express k in polar form.

(b) express m in rectangular form.

(c) calculate |2k + m|.

(d) calculate k · m.

Figure 1.15 Problem 1.4

1.5. Point B is 15 m east of point A, and point C is 10 m south of point B.

(a) How far is point C from point A?

(b) What is the direction from point A to point C?

(c) Let î point to the east and ĵ to the north. Express the vector d drawn from point A to point C in rectangular and polar forms.

1.6. If r = 6∠110° and s = 8∠340°, calculate

(a) r – s.

(b) 2r + 3s.

(c) r · s,

(d) r × s.

1.7. If K = 7î −3ĵ + 2 and L = 4î + 5ĵ −3 , calculate

(a) the magnitudes of K and L.

(b) |K + L|.

(c) K ·L.

(d) the cosine of the angle between K and L.

(e) the angle between K and L.

1.8. If D = −5î + 6ĵ −3 and E = 7î + 8ĵ + 4 , determine

(a) the magnitudes of D and E.

(b) D · E.

(c) D × E.

(d) the angle between D and E.

1.9. Using vectors R, S, and T (see Fig. 1.16), evaluate

Figure 1.16 Problem 1.9

(a) R · (S × T).

(b) R × (S × T).

(c) (S · T) R.

1.10. Given that r = 12∠70°, s = −8î + 6ĵ, t = 4î + 5ĵ, and u = 2r −3s + t, express u in rectangular form.

1.11. Points B and C are located relative to the origin, as shown in Figure 1.17.

Figure 1.17 Problem 1.11

(a) Calculate the length of the line segment connecting points B and C.

(b) Calculate the angle between the line segment connecting B and C and the line segment connecting O and B.

(c) If d is a vector starting at B and ending at C, write d in polar form and in component form.

1.12. For the vectors A, B, and C in Figure 1.18, calculate

Figure 1.18 Problem 1.12

(a) A · (B – C).

(b) (A ·B) C.

(c) |A – B|.

1.13. Given that A = 5∠40°, B = 8∠230°, C = 4∠340°, and D = 2A + B −3C, express D in polar form.

1.14. An airplane starting from airport A flies 300 km east, then 350 km 30° west of north, and then 150 km north to arrive finally at airport B. There is no wind on this day.

(a) In what direction should the pilot fly to travel directly from A to B?

(b) How far will the pilot travel in this direct flight?

1.15. A racetrack consists of two parallel sides with semicircular ends, as shown in Figure 1.19. The vector d connects track positions A and B. Express d in polar form.

Figure 1.19 Problem 1.15

1.16. Consider the triangle ABC with vectors b and c constructed, as shown in Figure 1.20. Show that 1/2|b × c| equals the area of the triangle.

Figure 1.20 Problem 1.16

1.17. Consider the 2 ft × 3 ft × 5 ft rectangular volume shown in Figure 1.21.

Figure 1.21 Problem 1.17

(a) Calculate the length of a vector drawn diagonally from corner A to corner B.

(b) Find the angle between the diagonal vector of part (a) and a vector drawn from point A to point C.

1.18. A person walks from A to B to C along the path shown in Figure 1.22. The distance R is 35 ft.

Figure 1.22 Problem 1.18

(a) Find the total distance walked.

(b) If D is a vector from A to C, express D in terms of î and ĵ.

(c) Express D in polar form.

GROUP B

1.19. Unit vectors are used to indicate direction. Given any vector v, a unit vector parallel to v, written , is given by v/v. If a = 2î – ĵ + 2 and b = 9î + 20ĵ + 12 , find

(a) a unit vector â in the same direction as a.

(b) a unit vector in the same direction as b.

(c) a unit vector perpendicular to both a and b.

(d) Is unique? Why?

1.20. Any vector b can be written as the sum of two mutually perpendicular vectors b|| and b that are parallel (or antiparallel) to and perpendicular to any other vector a, as depicted in Figure 1.23.

Figure 1.23 Problem 1.20

(a) Show that b|| = (â · b)â where â = a/a is a unit vector parallel to a.

(b) Show that b = b – (â · b)â.

(c) Verify directly that b|| and b are perpendicular.

(d) If a = 12î −16ĵ + 15 and b= 10î + 10ĵ −5 , calculate b|| and b .

1.21. If A = 6î −8ĵ, B = −8î + 3ĵ, and C = 26ĵ + 19ĵ, determine a and b so that

2

Particle Kinematics

Publisher Summary

This chapter provides an overview of particle kinematics. Kinematics is the branch of mechanics that describes the motion of physical bodies. It is distinguished from dynamics, which considers relationships between the motions of physical bodies and the forces (pushes or pulls) that are associated with those motions. The chapter also discusses models of physical bodies that may be considered mathematical points for the purpose of mathematical and physical analyses. A physical body may be regarded as a particle when only its translational motion is being considered, that is, when any vibrations or rotations of the body are of no consequence to the problem at hand. Large or small bodies may be treated as particles. The earth in its orbit or an automobile moving along a highway may each be thought of as a particle, provided the interest lies in its translational motion only. Perhaps the simplest example of kinematics is the motion of a particle constrained to move along a straight line. The speed of the particle at any instant is how fast the particle is moving at that instant. Its speed at a given instant is equal to the distance the particle would traverse each second if it proceeded at the rate it is moving at the given instant.

Kinematics is the branch of mechanics that describes the motion of physical bodies. It is distinguished from dynamics, which considers relationships between the motions of physical bodies and the forces (pushes or pulls) that are associated with those motions.

For the present we shall be interested in the kinematics of particles, models of physical bodies that may be considered mathematical points for the purpose of mathematical and physical analyses. A physical body may be regarded as a particle when only its translational motion is being considered, that is, when any vibrations or rotations of the body are of no consequence to the problem at hand. Large or small bodies may be treated as particles. The earth in its orbit or an automobile moving along a highway may each be thought of as a particle provided we are interested in its translational motion only. Thus, in describing translational motion of physical bodies, we may use the model of a point particle. In this chapter, we shall consider the kinematics of particles.

To be physically meaningful a description of motion must be measurable relative to some frame of reference, such as a coordinate system that could be established within a laboratory. With this in mind, a description of the motion of a particle at a given instant should answer the following questions:

1. Where is the particle relative to a chosen reference frame?

2. How fast is the particle moving, and in what direction is it moving?

3. To what extent and in which direction is its motion changing? In this chapter we shall consider how these questions may be answered quantitatively and how the answers are interpreted and applied.

2.1 Motion Along a Straight Line (Rectilinear Motion)

Position, Velocity, and Acceleration

Perhaps the simplest example of kinematics is the motion of a particle constrained to move along a straight line. Suppose such a particle is moving along the x axis, as shown in Figure 2.1. In Figure 2.1(a) the position of the particle is specified by a position vector r, a vector that originates at the coordinate origin and terminates at the particle P. As the particle moves, the vector r changes, so r is a function of time and may be written as r(t) to emphasize its dependence on time. Then for this one-dimensional case, the position vector r is specified by

Figure 2.1 (a) The position vector r locating a particle P lying on the axis. (b) The instantaneous velocity vector v for a particle P moving along the x axis.

(2-1)

The average velocity of a particle as it moves from a position r1 at time t1 to a position r2 at time t2 is defined to be

(2-2)

where Δxî = (x2 –x1)î is the displacement of the particle during the time interval Δt = t2 – t1. Both displacement and average velocity are vector quantities.

The instantaneous velocity v of the particle at point P in Figure 2.1 is defined to be the limit of the average velocity as the time interval Δt approaches zero while the displacement always includes the point P. In the notation of the calculus, instantaneous velocity is

(2-3)

Thus, the instantaneous velocity v(t), a vector quantity illustrated in Figure 2.1(b), is the rate at which the position of a particle is changing with respect to time. The magnitude of v at a given instant is |dx/dt| = |vx|, the speed of the particle; the motion of the particle is in the positive x direction if dx/dt is positive and in the negative x direction if dx/dt is negative. The speed of the particle at any instant is how fast the particle is moving at that instant. Its speed at a given instant is equal to the distance the particle would traverse each second if it proceeded at the rate it is moving at the given instant.

The rate at which the velocity of a particle is changing at any instant is the instantaneous acceleration a of that particle. Then the acceleration

(2-4)

is a vector quantity. Here the magnitude of the instantaneous acceleration is |dvx/dt|, which specifies how much the speed is changing per unit time at a given instant. If the rate dvx/dt is positive, the acceleration is in the positive x direction; if dvx/dt is negative, the acceleration of the particle is in the negative x direction.

The unit of position is the meter or the foot, depending on the system of units being used. Similarly, the unit of velocity or speed is m/s (or ft/s), and the unit of acceleration is m/s² (or ft/s²). The unit of acceleration may become more meaningful if it is rewritten (m/s)/s, which suggests that a constant acceleration of 3 m/s² means that the velocity is changing 3 m/s every second.

Problem-solving and analysis of rectilinear motion can be simplified by using the position, velocity, and acceleration vectors rather than the unit vector notation, because x, vx, and ax contain equivalent information to r, v, and a. For example, suppose that at a given instant the position, velocity, and acceleration of a particle are r = 3 î m, v = 4 î m/s, and a = −2 î m/s². It follows that x = +3 m (the particle is 3 m from the origin in the positive x direction), vx = +4 m/s (the particle is moving in the positive x direction with a speed of 4 m/s), and ax = −2 m/s² (the acceleration of the particle is 2 m/s²