## Lesson 1: Introduction and Mathematical Concepts

### Purpose

This lesson is an introduction to the basic physical and mathematical tools used in this course.

### Learning Objectives

After completing this lesson, you should be able to:

• 1.1 identify the fundamental physical quantities used in physics and their units.
• 1.2 determine the dimension of a quantity and perform a dimensional analysis on any equation.
• 1.3 convert quantities from one set of units to another.
• 1.4 differentiate between scalars and vectors.
• 1.5 represent vectors both graphically and mathematically.
• 1.6 calculate the magnitude and direction of a vector, and the components of a vector.
• 1.7 add vectors both graphically and algebraically.

Physics, Volume One

Part A: Units, Dimensions and Mathematical Concepts

• Chapter 1, Sections 1.1–1.4, pages 1–8
• Appendices A and B, page A–1

Part B: Scalars and Vectors

• Chapter 1, Sections 1.5–1.9, pages 8–20

Make sure that you test your understanding of the concepts by taking Self Assessment Tests 1.1 and 1.2. Be sure to take these tests when prompted to do so in the textbook, which displays their prompts on pages 12 and 17, respectively, using green-shaded text and icons of a computer mouse and a mortarboard hat. The tests are available on the Student Companion Site. Just click on the chapter selector in the upper-left of the page, then on the "Self Assessment Tests" link. Also, below the "Self Assessments" link is the "Concept Simulations" link, which will lead you to graphics and animations that will help further your understanding of the material being studied.

### Commentary

#### Part A: Units, Dimensions and Mathematical Concepts

Physics, one of the most fundamental sciences, can be defined as the study of the basic nature of matter and the interactions that govern its behavior. Physics deals with the fundamental laws of nature and many of their applications.

The study of physics involves:

• Observation of events (including designing and carrying out experiments).
• Invention/creation of theories (noting that theories are not derived directly from observations and that they also evolve or may change over time).
• Testing of its theories to see if their predictions are verified by experiment.

In physics you will use:

• Models, which are relatively simple and provide a structural similarity to the phenomena being studied.
• Theories, which are broader, more detailed, and attempt to solve a set of problems, often with great precision.
• Laws, which are certain and concise, but general statements about how nature behaves. These statements can take the form of a relationship or be stated as an equation between quantities.

Various quantities were created in physics for the purpose of describing the behavior of physical systems. There are three quantities: length, mass, and time, that are considered fundamental quantities and are used to create other quantities. One must be able to quantify these quantities—that is, to specify how much length, mass, or time a certain quantity has. This quantification is accomplished with a system of units.

• The unit of length is the meter (m).
• The unit of mass is the kilogram (kg).
• The unit of time is the second (s).

The system of units based on these choices is called SI units, which stands for System International. The units are defined as follows:

• Meter (m) is defined as the distance traveled by light in vacuum during a time of 1/299,792,458 of a second.
• Kilogram (kg) is defined as the mass of a specific platinum-iridium cylinder kept at the International Bureau of Weights and Measures at Sevres, France.
• Second (s) is defined as 9,192,631,770 times the period of vibration of radiation from the cesium-133 atom.

Quantities such as length, mass, and time can also be expressed in several different units; therefore it is important to know how to convert from one unit to another. For example, length can be measured in feet (see example 1 on page 3 of your textbook). In everyday life in the United States, speed is measured in miles per hour, but for scientific purposes, in SI units, speed is measured in meters per second. Let’s convert a speed of 60 miles/hour into meters/second. Consider that 1 mile = 1.609 kilometers, 1 kilometer = 1000 meters, and 1 h = 3600 s.

Example 1.1

See also example 2 on page 4 of your textbook. All physical quantities are derived from the fundamental quantities discussed earlier. When a quantity is broken down into its fundamental quantities we call this breakdown its dimension. The dimension, therefore, represents the fundamental type of the quantity. When indicating the dimension of a quantity only, we use capital letters and enclose them in brackets. Thus, the dimension of length is represented by [L], mass by [M], and time [T], as shown in your textbook.

All equations used in physics must be dimensionally consistent; that is, the dimensions of both sides of the equation must match. Two simple rules must be taken into consideration when doing dimensional analysis:

• Two quantities can be added or subtracted only if they have the same dimension.
• Two quantities can be equal only if they have the same dimension.

Notice that only the dimension needs to be the same, not the units. It is perfectly valid to write "12 inches = 1 foot," because both of them are lengths, [L] = [L], even though their units are different. However, it is not valid to write "x inches = t seconds," because they have different dimensions, [L] and [T]. In dimensional analysis, purely numerical factors are ignored because they are dimensionless. Since there are dimensionless quantities, dimensional consistency does not guarantee that the equation is physically correct, but it makes for a quick and easy first check.

Example 1.2

The period T (dimension [T]) of a swinging pendulum depends only on the length of the pendulum d (dimension [L]), and the acceleration of gravity g (dimension [L]/[T]2). Which of the formulas for T could be correct?

1. T = 2π(dg)2
$\left[T\right]=2\pi {\left(\left[L\right]\frac{\left[L\right]}{{\left[T\right]}^{2}}\right)}^{2}⇒\left[T\right]=2\pi {\left(\frac{{\left[L\right]}^{2}}{{\left[T\right]}^{2}}\right)}^{2}⇒\left[T\right]=2\pi \frac{{\left[L\right]}^{4}}{{\left[T\right]}^{4}}$

This is not correct because the dimensions on the left do not match the dimensions on the right.

2. $T=2\pi \frac{d}{g}$

$\left[T\right]=2\pi \frac{\left[L\right]}{\left[L\right]/{\left[T\right]}^{2}}⇒\left[T\right]=2\pi {\left[T\right]}^{2}$

This is not correct because the dimensions on the left do not match the dimensions on the right.

3. $T=2\pi \sqrt{\frac{d}{g}}$

$\left[T\right]=2\pi \sqrt{\frac{\left[L\right]}{\left[L\right]/{\left[T\right]}^{2}}}⇒\left[T\right]=2\pi \sqrt{{\left[T\right]}^{2}}⇒\left[T\right]=2\pi \left[T\right]$

This could be the correct formula for T: the dimensions on the left match the dimensions on the right. The constant 2π is ignored because it is dimensionless.

#### Part B: Scalars and Vectors

First, we need to be clear on the difference between a scalar and a vector:

• Scalars are numbers with units; these units may be positive, negative, or zero.
• Vectors are mathematical quantities with both a magnitude and a direction.
• A quantity that is a vector is denoted with an arrow above the symbol for that quantity, or with a bolded symbol.
• A vector is represented graphically by an arrow whose direction is the same as the direction of the vector and whose length is proportional to the magnitude of the vector.
Note: Throughout the commentaries and in all Worked Problems, vectors will be denoted with an arrow on top. In your textbook vectors are denoted by bolded letters. When solving problems that involve vectors you should never forget to add the arrow on top of the letter representing the vector. Forgetting it leads to confusion between scalar quantities and vector quantities.

Vectors have the following two properties:

• Equality: Two or more vectors are equal if their magnitudes and directions are the same.
• The negative of a vector is a vector having the same magnitude but opposite direction.
 Figure 1.1. The air traffic controller in this picture is viewing vectors, which show both the direction and the speed of each aircraft being monitored.

There are two methods for adding two vectors:

1. The Triangle Method: Without changing the direction or the magnitude of the vectors, place the tail of the second vector in the tip of the first vector. The arrow drawn from the tail of the first vector to the tip of the second vector represents the sum, or resultant, of the two vectors, $\stackrel{\to }{R}=\stackrel{\to }{A}+\stackrel{\to }{B}$ .

2. The Parallelogram Method: Without changing the direction or the magnitude of the vectors, place the tail of the second vector at the tail of the first vector. Build a parallelogram that has the two vectors as two adjacent sides. The arrow drawn along the diagonal of the parallelogram starting from the common tail of the two vectors represents the sum, or resultant, of the two vectors, $\stackrel{\to }{R}=\stackrel{\to }{A}+\stackrel{\to }{B}$ .

Note: Make sure you do not mix the two methods, or the result will be wrong in that case. Also, make sure that you never add a vector and a scalar, because this is not possible.
Example 1.3

Using the Triangle Method and the Parallelogram Method, find the resultant of vector $\stackrel{\to }{A}$  and $\stackrel{\to }{B}$  for the following cases:

Using the Triangle Method:

Using the Parallelogram Method:

Another simple way of adding two vectors is by using vector components. Any vector can be expressed as the sum of two other vectors (representing the projection of the vector along two perpendicular directions), called the vector components. The scalar component (or simply component) of a vector along a line in space is the length of the projection of the vector on that line.

To determine the components of a vector we need a coordinate system. A coordinate system is an artificial grid that you place on a problem in order to make quantitative measurements. When using a coordinate system you choose:

• where to place the origin.
• how to orient the axes.

A Conventional Cartesian coordinate system and the quadrants of the xy plane are shown at right. We will use the following convention for measuring angles: an angle made by a vector will be measured counterclockwise from the positive x-axis.

Consider the two-dimensional coordinate system, xy.

• $\stackrel{\to }{{A}_{x}}$  is the x-vector component of vector $\stackrel{\to }{A}$ .
• $\stackrel{\to }{{A}_{y}}$  is the y-vector component of vector $\stackrel{\to }{A}$ .
• Ax is the x-scalar component of vector $\stackrel{\to }{A}$  and the magnitude of vector component $\stackrel{\to }{{A}_{x}}$ . It can have a positive, negative, or zero value.
• Ay is the y-scalar component of vector $\stackrel{\to }{A}$  and the magnitude of vector component $\stackrel{\to }{{A}_{y}}$ . It can have a positive, negative, or zero value.
• Θ is the angle formed by vector $\stackrel{\to }{A}$  with the positive x-axis, measured counterclockwise.

Vector $\stackrel{\to }{A}$  is split into two components, one pointing in the x-direction and the other in the y-direction. These directions are perpendicular to each other and form two sides of a right angle triangle having a hypotenuse with magnitude A. The magnitude of vector components Ax and Ay will be given by:

Ax = A cos θ
Ay = A sin θ

The magnitude and direction of vector $\stackrel{\to }{A}$  are related to its components through the Pythagorean theorem and the definition of the tangent:

From the expression for the magnitude of a vector, one can see that the magnitude of a vector is always positive (because taking the square root can never result in a negative number).

A vector can be written in component form by using unit vectors. A unit vector is a vector of magnitude 1 that is usually oriented along the axes of the coordinate system used to decompose a vector in components.

In this case, a vector can be written as:

$\stackrel{\to }{A}={A}_{x}\stackrel{^}{x}+{A}_{y}\stackrel{^}{y}$  where the “hat” symbol above the unit vectors $\stackrel{^}{x}$ and $\stackrel{^}{y}$ indicates that these are unit vectors (each with a magnitude of 1).

If $\stackrel{\to }{R}=\stackrel{\to }{A}+\stackrel{\to }{B}$ , then the components of the resultant vector are Rx = Ax + Bx and
Ry = Ay + By .

Example 1.4

Vector $\stackrel{\to }{A}$  has a magnitude of 16 and points due east. Vector $\stackrel{\to }{B}$  has a magnitude of 5 and points 30° north of west.

1. Find the x and y components of vectors $\stackrel{\to }{A}$ and $\stackrel{\to }{B}$ .
2. The x and y components of vectors $\stackrel{\to }{A}$  and $\stackrel{\to }{B}$  are found using the equations for vector components. First draw a coordinate system on which you represent vectors $\stackrel{\to }{A}$  and $\stackrel{\to }{B}$ .

Then, write the x and y components of vectors $\stackrel{\to }{A}$  and $\stackrel{\to }{B}$

where ΘA and ΘB are the angles made by vectors $\stackrel{\to }{A}$  and $\stackrel{\to }{B}$  with the positive x-axis, measured counterclockwise. Using the data provided by the problem, and making sure that our calculators are set to "degree" mode, as we are using degrees, we now see that:

$\begin{array}{l}\left\{\begin{array}{l}{A}_{x}=\left(16\right)\mathrm{cos}{0}^{0}=16\\ {A}_{y}=\left(16\right)\mathrm{sin}{0}^{0}=0\end{array}\\ \left\{\begin{array}{l}{B}_{x}=\left(5\right)\mathrm{cos}\left({180}^{0}-{30}^{0}\right)=5\mathrm{cos}{150}^{0}=-4.3\\ {B}_{y}=\left(5\right)\mathrm{sin}\left({180}^{0}-{30}^{0}\right)=5\mathrm{sin}{150}^{0}=2.5\end{array}\end{array}$

As seen on the drawing, Bx, the x component of vector $\stackrel{\to }{B}$ is negative.

3. Find the magnitude and direction of vector $\stackrel{\to }{C}=\stackrel{\to }{A}+2\stackrel{\to }{B}$ .
4. Write vectors $\stackrel{\to }{A}$  and $\stackrel{\to }{B}$  in component form:

$\begin{array}{l}\left\{\begin{array}{l}\stackrel{\to }{A}={A}_{x}\stackrel{^}{x}+{A}_{y}\stackrel{^}{y}⇒\stackrel{\to }{A}=\left(16\right)\stackrel{^}{x}+\left(0\right)\stackrel{^}{y}=16\stackrel{^}{x}\\ \stackrel{\to }{B}={B}_{x}\stackrel{^}{x}+{B}_{y}\stackrel{^}{y}⇒\stackrel{\to }{B}=\left(-4.3\right)\stackrel{^}{x}+\left(2.5\right)\stackrel{^}{y}=-4.3\stackrel{^}{x}+2.5\stackrel{^}{y}\end{array}\\ \stackrel{\to }{C}=\stackrel{\to }{A}+2\stackrel{\to }{B}=\left({A}_{x}\stackrel{^}{x}+{A}_{y}\stackrel{^}{y}\right)+2\left({B}_{x}\stackrel{^}{x}+{B}_{y}\stackrel{^}{y}\right)\\ \stackrel{\to }{C}=\left(16\stackrel{^}{x}\right)+2\left(-4.3\stackrel{^}{x}+2.5\stackrel{^}{y}\right)\\ \stackrel{\to }{C}=16\stackrel{^}{x}+\left(-8.6\right)\stackrel{^}{x}+5\stackrel{^}{y}=7.4\stackrel{^}{x}+5\stackrel{^}{y}\end{array}$

The numbers in front of the unit vectors $\stackrel{^}{x}$  and $\stackrel{^}{y}$  represent the x and y components of vector $\stackrel{\to }{C}$ .

$\begin{array}{l}\stackrel{\to }{C}=7.4\stackrel{^}{x}+5\stackrel{^}{y}⇒\left\{\begin{array}{l}{C}_{x}=7.4\\ {C}_{y}=5\end{array}\\ C=\sqrt{{C}_{x}^{2}+{C}_{y}^{2}}⇒C=\sqrt{{7.4}^{2}+{5}^{2}}=8.9\end{array}$

The angle made by vector $\stackrel{\to }{C}$  with the positive x axis is:

$\theta ={\mathrm{tan}}^{-1}\left(\frac{{C}_{y}}{{C}_{x}}\right)⇒\theta ={\mathrm{tan}}^{-1}\left(\frac{5}{7.4}\right)⇒\theta ={34}^{0}$ .

### Worked Problems

The Worked Problems represent typical physics problems that cover the material you should have mastered after studying this lesson. These problems are also representative of the type of problems you will find on your exams.

Full and detailed solutions for these problems are provided in PDF format by clicking on the link at the bottom of this section. You are strongly encouraged to try to solve these problems on your own before viewing the answers.

• Begin each problem by recording the physical quantities given in the problem.
• Draw a schematic representation of the problem. Include all the given physical quantities on this diagram.
• Using the knowledge you gained by studying Lesson 1, try to solve the problems. Use your textbook readings, especially the solved problems in your textbook, as sources of help.
• After solving the problems, check your answer against the solutions provided. Even if you have found the correct answer, please pay close attention to the method used by the instructor to solve the problems. A systematic way of approaching and solving these problems is presented in the full solutions. You are strongly advised to try to master the technique of solving physics problems in a systematic way.

Part A: Units, Dimensions and Mathematical Concepts

• Chapter 1, Problem 8, textbook page 22

Part B: Scalars and Vectors

• Chapter 1, Problems 28, 38, 46, and 50, beginning on textbook page 23.

View the Worked Problems

### Study Questions

Make sure that you have tested your understanding of the concepts by taking Self-Assessment Tests 1.1 and 1.2 (prompted on pages 12 and 17, respectively, marked with a computer mouse and mortarboard hat). The tests are provided on the Student Companion Site by clicking on the "Self Assessment Tests" link in the left column. The "Concept Simulations" provided on the same site will also help to further your understanding of the material. Next, work the following problems:

Part A: Units, Dimensions and Mathematical Concepts

• Chapter 1, Problems 1, 7, 9, 13, and 17, beginning on textbook page 21.

Part B: Scalars and Vectors

• Chapter 1, Problems 23, 25, 29, 35, 37, and 57, beginning on textbook page 23.

#### Answers to the Study Questions

• The final answers to all these problems are located at the back of your textbook, beginning on page A–10.
• You may view fully detailed explanations of the solutions to problems 1, 9, 13, 23, 25, 35, and 57 by consulting your Student Solutions Manual beginning on page 3.
• You may view interactive solutions to problems 17 and 29 by visiting the Student Companion Site and selecting Chapter 1 under "Browse by Chapter." On the page that comes up, click on the "Interactive Solutions" link, then choose "Ch 1" on the menu page that comes up. Finally, click on the button for the question you wish to check interactively.