Module 1 Course Title Course Number Course Description : : : : Total Learning Time : Pre-requisites : Physics and Measurement Physics for Engineers Physics for Engineers 121 This course provides students with a clear and logical presentation of the basic concepts and principles of physics through a broad range of interesting real-world application. 5 units (4 units lecture, 1 unit laboratory) Calculus 1 Overview: In this module, it will discuss the different kinds of physical quantities and some preliminary concepts of measurements, units, modelling and estimation. Learning Outcomes: At the end of this module, the student should be able to: 1. Have a working knowledge to identify fundamental quantities in mechanics. 2. Have a working knowledge to express measurements in standard form, such as SI -Standard International/Metric or English System. 3. Convert one units from one measurement system to another or convert within a system. Indicative Content: This module discusses the following topics: Standards of Length, Mass and Time; Modelling and Alternative Representations; Dimensional Analysis; Conversion of Units; Estimates and Order-of-Magnitude Calculations; Significant Figures; and Vectors. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 1 College of Engineering, Architecture and Technology Name: ___________________________________ Date Submitted: __________ Course & Year: ____________________________ Rating: _________________ Pre-Assessment 1. A proton, which is the nucleus of a hydrogen atom, can be modelled as a sphere with a diameter of 2.4 m and a mass of 1.67 x 10 -27 kg. Determine the density of the proton. 2. Two spheres are cut from a certain uniform rock. One has radius 4.5 cm. The mass of the others is five times greater. Find its radius. 3. A solid piece of lead has a mass of 23. 94 g and a volume of 2.10 cm 3. From these data, calculate the density of lead in SI units (kg/m 3) 4. One gallon of paint (volume = 3.78 x 10 -3 m3) covers an area of 25.0 m3. What is the thickness of the fresh paint on the wall? 5. a) Compute the order of magnitude of the mass of a bathtub half full of water. b) Compute the order of magnitude of the mass of a bathtub half full of copper coins. 6. To an order of magnitude, how many piano tuners reside in NY City? The physicist Enrico Fermi was famous for asking this question like this one on oral PhD qualifying examination. 7. How many significant figures are in the following numbers? a) 78.9 ± 0.2 b) 3.788 x 109 c) 2.46 x 10-6 d) 0.0053 8. Prove that one solution of the equation 2x4 – 3x3 + 5x = 70 if x = - 2.22 9. The consumption of natural gas by a company satisfies the empirical equation V= 1.5t + 0.00800t2, where V is the volume of gas in millions of cubic feet and t is the time in months. Express this equation in units of cubic feet and seconds. Assume a moth is 30 days. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 2 10. A woman wishing to know the height of a mountain measures the angle of elevation of the mountaintop as 120. After walking 1 km closer to the mountain on level ground, she finds the angle to be 140. a) Draw a picture of the problem, neglecting the height of the woman’s eyes above the ground. b) Using the symbol y to represent the mountain height and the symbol x to represent the woman’s original distance from the mountain, label the picture. c) Using the labelled picture, write two trigonometric equations relating the two selected variables. d) Find the height y. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 3 Discussion: Physics, the most fundamental physical science, is concerned with the fundamental principles of the Universe. It is the foundation upon which the other sciences- astronomy, biology, chemistry, and geology-are based. It is also the basis of a large number of engineering applications. The beauty of physics lies in the simplicity of its fundamental principles and in the manner in which just a small number of concepts and models can alter and expand our view of the world around us. Standards of Length, Mass and Time To describe natural phenomena, we must make measurements of various aspects of nature. Each measurements is associated with a physical quantity, such as the length of an object. The laws of physics are expressed as mathematical relationships among physical quantities. In mechanics, the three dimensional quantities are length, mass and time. All other quantities in mechanics can be expressed in terms of these three. In 1960, an international committee established a set of standards for the fundamental quantities in science. It is called the SI (System International), and its fundamental units of length, mass and time are the meters, kilogram, and second, respectively. Other standards for SI fundamental units established by committee are those for temperature (kelvin), electric current (ampere), luminous intensity (candela), and the amount of substance (mole). Length –is the distance between two points in space. In 1120, the king of England decreed that the standard of length in his country would be named the yard and would be precisely equal to distance from the tip of his nose to the end of his outstretched arm. Similarly, the original standard for the foot adopted by the French was the length of the royal foot of King Louis XIV. Neither of these standards is constant in time; when a new king took the throne, length measurements changed! The French standard prevailed until 1799, when the legal standard of length in France became the meter (m), defined as one ten-millionth of the distance from the equator to the North Pole along one particular longitudinal line that passes through Paris. Notice that this value is an Earth –based standard that does not satisfy the requirement that it can be used throughout the Universe. As recently as 1960, the length of the meter was defined as the distance between two lines on a specific platinum-iridium bar stored under controlled conditions in France. Current requirements of science and technology, however, necessitate more accuracy than that with which the separation between the lines on the bar can be determined. In the 1960s and 1970s, the meter was defined to be equal to 1650763.73 wavelength of orange-red emitted from a krypton-86 lamp. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 4 In October 1983, however, the meter was redefined as the distance travelled by light in a vacuum during a time interval of 1/299 792 458 second. In effect, this latest definition establishes that the speed of the light in vacuum is precisely 299 792 458 meters per second. This definition of the meter is valid throughout the Universe based on the assumption that the light is the same everywhere. The speed of light also allows us to define the light-year- the distance that light travels through empty space in one year. Mass – is related to the amount of materials that is present in the object, or to how much that object resist changes in its motion. Mass is an inherent property of an object and is dependent of the object’s surroundings and of the method used to measure it. The SI fundamental unit of mass, the kilogram (kg), is defined as the mass of a specific platinum-iridium alloy cylinder kept at the International Bureau of Weights and Measures at Service, France. This mass standard was established in 1887 and has not been changed since that time because platinumiridium is an unusually stable alloy. Time – the standard of time was defined in terms of the mean solar day. (A solar day is the time interval between successive appearances of the Sun at the highest point it reaches in the sky each day). The fundamental unit of a second (s) was defined as (1/60)(1/60)(1/24) of a mean solar day. This definition is based on the rotation of one planet, the Earth. Therefore, this motion does not provide a time standard that is universal. In 1967, the second was redefined to take advantage of the high precision attainable in a device known as an atomic clock which measures vibrations of cesium atoms. One second is now defined as 9 192 631 770 times the period of vibration of radiation from the cesium -133 atom. You should note that we will use the notations time and time interval differently. A time is a description of an instant relative to a reference time. For example, t = 10 s, refer to an instant 10 s after the instants we have identified as t = 0. As another example, a time of 11:30 a.m., means an instant 11.5 hours after our reference time of midnight. On the other hand, a time interval refers to duration. He required 30 minutes to finish to finish the task. It is common to hear a “time of 30 minutes” in this latter example, but we will be careful to refer to measurements of duration as time intervals. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 5 Units and Quantities In addition to SI, another system of units, the US customary system, is still used in the United States despite acceptance of SI by the rest of the world. In this system, the units of length, mass and time are the foot, slug and second respectively. We shall use SI Units because they are almost universally accepted in science and industry. We shall make some limited use of US customary units in the study of classical mechanics. In addition to the fundamental SI units of meter, kilogram, and second, we can also use other units, such as millimetres and nanoseconds, where the prefixes milli and nano denote multipliers of the basic units based on various power of ten. The variable length, mass and time are examples of fundamental quantities. Most other variables are derived quantities, those that can be expressed as mathematical combination of fundamental quantities. Common examples are area (a product of two length) and speed (a ratio of a length to a time interval). Another example of a derived quantity is density. The density 𝜌 (Greek letterrho) of any substance is defined as its mass per unit volume. 𝜌 = m/v In terms of fundamental quantities, density is a ratio of mass to a product of three lengths. Aluminum, for example, has a density of 2.70 x 103 kg/m3, and iron has a density of 77.86 x 103 kg/m3. An extreme difference in density can be imagined by thinking about a 10-cm cube of Styrofoam in one hand and a 10-cm cube of lead in the other. Modelling and Alternative Representations One of the primary problems-solving methods in physics is to form an appropriate model of the problem. A model is a simplified substitute for the real problems that allows us to solve the problem in a relatively simple way. As long as the predictions of the model agree to our satisfaction with the actual behaviour of the real system, the model is valid. If the predictions do not agree, the model must be refined or replaced with another model. The power of modelling is in its ability to reduce a wide variety of very complex problems to a limited number of classes of problems that can be approached in similar ways. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 6 The particle model is an example of the second category of models, which we will call simplification models. In simplification model, details that are not significant in determining the outcome of the problem are ignored. The third category is that of analysis model, which are general types of problems that we have solved before. An important technique in problem solving is to cast a new problem into a form similar to one we have already solved and which can be used as model. The fourth category is structural models, generally used to understand the behaviour of a system that is that is far different in scale from our macroscopic world- either mush smaller or much larger, so that we cannot interact with it directly. Intimately related to the notion of modelling is that of forming alternative representation of the problem that you are solving. A representation is a method of viewing or presenting the information related to the problems. The modelling approach recognizes that none of these processes affects the motion of the Earth around the Sun to a measurable degree. Therefore these details are all ignored. In addition, the size of the Earth does not affect the gravitational force between the Earth and the Sun; only the masses of the Earth and Sun and the distance between their centers determine this force. In simplified model, the Earth is imagined to be particle, an object with mass but zero size. This replacement of an extended object by a particle is called particle model, which is used extensively in physics. The two primary conditions for using the particle model are as follows: The size of the actual object is of no consequence in the analysis of its motion. Any internal processes occurring in the object are of no consequence in the analysis of its motion. Both of these conditions are in action in modelling the Earth as a particle. Its radius is not a factor in determining its motion, and internal processes such as thunderstorm, earthquakes, and manufacturing process can be ignored. Four categories of models will help us understand and solve Physics problems. 1. Geometric model- we form a geometric construction that represents the real condition. We then set aside the real problem and perform an analysis of the geometric construction. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 7 Example #1: You wish to find the height of a tree but cannot measure it directly. You stand 50.0 m from the tree and determine that the line of sight from the ground to the top of the tree makes an angle of 25.00 with the ground. How tall is the tree? The height of a tree can be found by measuring the distance from the tree and the angle of sight to the top above the ground. This problem is a simple of geometrically modelling the actual problem. Solution: Figure shows the tree and a right triangle corresponding to the information in the problem superimposed over it. (We assume that the tree is exactly perpendicular to a perfectly flat ground). In the triangle, we know the length of the horizontal leg. We can find the height of the tree by calculating the length of the vertical leg. We do so with the tangent function: tan 𝜃 = opposite / adjacent = h/50.0 m h = (50m) tan 𝜃 = (50m) tan 250 h = 23.3 m 2. Simplification model – details that are not significant in determining the outcome of the problem are ignored. When we study rotation, objects will be modelled as rigid objects. All molecules in a rigid object maintain their exact positions with respect to forces to one another. We adopt this simplification model because a spinning rock is much easier to analyze than a spinning block of gelatin, which is not a rigid object. Other simplification models will assume that quantities such as friction forces are negligible, remain constant, or are proportional to some power of the object’s speed. We will assume uniform metal beams, laminar flow of fluids, massless spring, symmetric distributions of electric charge, resistance free wires, thin lenses and many more are simplification models. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 8 3. Analysis model - an important technique in problem solving is to cast a new problem into a form similar to one have already solved and which can be used as a model. As we shall see, there are about two dozen analysis models that can be used to solve most of the problems you will encounter. All of the analysis models in classical physics will be based on four simplification models: particles, system, rigid bodies and wave. 4. Structural model – these model are generally used to understand the behaviour of a system that is far different in scale from our macroscopic world- either much smaller or larger- so that we cannot interact with it directly. As an example, the notion of hydrogen from an atom as an electron in a circular orbit around a proton is a structural model of an atom. The ancient geocentric model of the Universe, in which the Earth is theorized to be at the center of the Universe, is an example of a structural model for something larger in scale than our macroscopic world. Types of representation can be of assistance to help you understand and solve problems in different ways: 1. Mental representation – Imagine a scene that describes what is happening in the word problem, then let time progress so that you understand the situation and can predict what changes will occur in the situation. This step is critical in approaching every problem. 2. Pictorial representation – Drawing a picture of the situation described in the word problem can be great assistance in understanding the problem. 3. Simplified pictorial representation- It is often useful to redraw the pictorial representation without completing details by applying a simplification model. This process is similar to the discussion of the particle model described earlier. 4. Graphical representation- In some problems, drawing a graph that describes the situation can be very helpful. A graphical representation is different from a pictorial representation, which is also a two –dimensional display of information but whose axes, if any, represent length coordinates. In a graphical representation, the axes may represent any two related variables. Therefore, a graphical representation is generally not something you would see when observing the situation in the problem with your eyes. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 9 5. Tabular representation – It is sometime helpful to organize the information in tabular form to help make it clearer. The periodic table of the elements is an extremely useful tabular representation of information in chemistry and physics. 6. Mathematical representation – The ultimate goal in solving a problem is often the mathematical representation. You want to move from the information contained that allow you to understand what is happening, to one or more equations that represent the situation in the problem and that can be solved mathematically for the desired result. Fig 1A. A pictorial representation of pop foul being hit by a baseball player. Fig. 1B. A simplified pictorial 1A Representation for the situation Fig. 1C. A graphical representation of the position as a function of time of a block shanging from a spring and oscillating. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 10 Dimension Analysis The word dimension denotes the physical nature quantity. The distance between two points, for example, can be measured in feet, meters, or furlongs, which are all different units for expressing the dimension of length. The symbol we use to specify the dimension of length, mass and time are L, M, and T. respectively. We shall often use brackets [ ] to denote the dimension of a physical quantity. For example, the symbol we use for speed is v, and in our notation, the dimensions of speed are written [v] = L/T. As another example, the dimensions of area A are[A ] = L2. The dimensions and units of area, volume, speed, and acceleration are listed below. In many situations, you may have to check a specific equation to see if it matches your expectations. A useful procedure for doing that, called dimensional analysis, can be used because dimensions can be treated as algebraic quantities. For example, quantities can be added or subtracted only if they have the same dimensions. Furthermore, the terms on both sides of an equation must have the same dimensions. By following these simple rules, you can use dimensional analysis to determine whether an expression has the correct form. Any relationship can be correct only if the dimensions on both sides of the equation are the same. To illustrate this procedure, suppose you are interested in an equation for the position x of a car at a time t if the car starts from rest at x = 0 and moves with constant acceleration a. The correct expression for this situation is x= 1/2at2 as we show in the next topic. The quantity x on the left side has the dimension of length. For the equation to be dimensionally correct, the quantity on the right side must also have the dimension of length. We can perform a dimensional check by substituting the dimensions for acceleration. L/T2 and time, T, into the equation. That is, the dimensional form of the equation x=1/2at2 is: L = L/ T2 . T2 = L The dimensions of time cancel shown, leaving the dimension of length on the righthand side to match that on left. A more general procedure using dimensional analysis is to set up an expression of the form x ∝ a nt n where n and m are exponents that must be determined and the symbol ∝ indicates a proportionality. This relationship is correct only if the dimensions of both sides are the same. Because the dimension of the left side is length, the dimension of the right side must also be length. That is, [ a n t n ] = L = L1 T0 ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 11 Table 1.1 Dimensions and Units of Four Derived Quantities Quantity Area Volume Speed Acceleration (A) (V) (s) (a) Dimensions L2 L3 L/T L/T2 SI Units m2 m3 m/s m/s2 US customary units ft2 ft3 ft/s ft/s2 Because the dimensions of acceleration are L/T2 and the dimension of time is T, we have (L/T2) n T m = L1 T 0 (Ln T m -2n ) = L1 T 0 The exponents of L, and T must be the same on both sides of the equation. From the exponents of L, we see immediately that n =1. From the exponents of T, we see that m - 2n = 0, which, once we substitute for n, gives us m = 2. Returning to our original expression x ∝ a n t m, we conclude that x ∝ at2. Example # 2. Analysis of an Equation Show that the expression v = at, where v represents speed, a acceleration, and t an instant of time, is dimensionally correct. Identify the dimensions of v from Table 1.1 [ 𝑣 ] = L/T Identify the dimensions of a from Table 1.1 and multiply by dimension of t [ 𝑎𝑡 ] = L/T2 T = L/T Therefore, v = at is dimensionally correct because we have the same dimensions on both sides. (If the expression were given a v = at2, it would be dimensionally in correct. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 12 Example # 3. Analysis of a Power Law Suppose we are told that the acceleration of a particle moving with uniform speed v in a circle of radius r is proportional to some power of r, say r n , and some power of v, say v m. Determine the values of n and m and write the simplest form of an equation for the acceleration. Write an expression for a with a dimensionless constant of proportionality k: a = k r nv m Substitute the dimensions of a, r, and v L/T2 = Ln (L/T) m = Ln + m / T m Equate the exponents of L and T so that the dimensional equation is balanced: n + m = 1 and m = 2 Solve the two equations for n: n=-1 Write the acceleration expression: a = k r -1 v 2 = k v 2/r Conversion of Units Sometimes, it is necessary to convert units from one measurement system to another or convert within a system. Conversion factors between SI and US customary units of length are follows. 1 mile = 1609 m = 1.609 km 1 ft. = 0.3048 m = 30.48 cm 1m= 1 in. = 0.0254 m = 2.54 cm 39.37 in. = 3.281 ft. A more complete list of conversion factors see Conversion Table. Like dimensions, units can be treated as algebraic quantities that can cancel each other. For example, suppose we wish to convert 15 in to cm. 15 in = (15 in) (2.54 cm / 1 in) = 38.1 cm ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 13 Where the ratio in parentheses is equal to 1. We express 1 as 2.54 cm/1 in. (rather than 1 in/2.54 cm) so that the unit “inch” in the denominator cancels with unit in the original quantity. The remaining unit is the centimeter, our desired result. Example # 4: a) On an interstate highway in a rural region of Wyoming, a car is traveling at a speed of 38 m/s. Is the driver exceeding the speed limit of 75 mi/h? Solution: Convert meters to miles and second to hours: (38 m/s) (1 mi/1609 m) (60 s/ min) (60 min / 1 h) = 85 mi/h The driver is indeed exceeding the speed limit and should slow down. b) What if the driver were from outside the United States and is familiar with speeds measured in kilometre per hour? What is the speed of the car in kph? We can convert our final answer to the appropriate units: (85 mi/h) (1.609km/1 mi) = 137 km/h Estimates and Order -of-Magnetic Calculations Suppose someone asks you the number of bits of data on a typical Blu-ray Disc. In response, it is not generally expected that you would provide the exact number but rather an estimate, which may be expressed in scientific notation. The estimate may be made even more approximate by expressing it as an order of magnitude, which is a power of 10 determined as follows: 1. Express the number in scientific notation, with the multiplier of the power of 10 between 1 and 10 and a unit. 2. If the multiplier is less than 3.162 (the square root of 10), the order of magnitude of the number is the power of 10 in the scientific notation. If the multiplier is greater than 3.162, the order of magnitude is one larger than the power of 10 in the scientific notation. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 14 We use the symbol – for “is on the order of”. Use the procedure above to verify the orders of magnitude for the following lengths: 0.0086 m ~ 10-2 m 0.0021 ~ 10-3 m 720 m ~ 103 m Usually, when an order-of-magnitude estimate is made, the results are reliable to within about a factor of 10. Inaccuracies caused by guessing too low for one number are often cancelled by other guesses that are too high. You will find that with practice your guesstimates become better and better. Estimation problems can be fun to work because you freely drop digits, venture reasonable approximations for unknown numbers, making simplifying assumptions, and turn the question around into something you can answer in your head or with minimal mathematical manipulation on paper. Because of the simplicity of these types of calculation, they can be performed on a small scrap of paper and are often called back-of-the-envelope calculations. Example # 5: Estimate the number of breaths taken during an average human lifetime. Solution: We start by guessing that the typical human lifetime is about 70 years. Think about the average number of breaths that a person takes 1 min. This number varies depending on whether the person is exercising, sleeping, angry, serene, and so forth. To the nearest order of magnitude, we shall choose 10 breaths per minute as our estimate. (This estimate is certainly closer to the true average value than an estimate of 1 breath per minute or 100 breaths per minute). Find the approximate number of minutes in a year: 1 yr. (400 days/1 yr.) (25 h/1 day) (60 min/1 h) = 6 x 105 min Find the approximate number of minutes in a 70-year lifetime: number of minutes = (70 yr.) (6 x 105 min/yr.) = 4 x 107 min Find the approximate number of breaths in a lifetime Number of breaths = (10 breaths/min) (4 x 107 min) = 4 x 108 breaths Therefore, a person takes on the order of 109 breaths in a lifetime. Notice how much simpler it is in the first calculation above to multiply 400 x 25 than it is to work with the more accurate 365 x 24. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 15 Significant Figures When certain quantities are measured, the measured values are known only to within the limits of the experimental uncertainty. The value of this uncertainty can depend on various factors, such as the quality of the apparatus, the skill of the experimenter, and the number of measurements performed. The number of significant figures in a measurement can be used to express something about the uncertainty. The number of significant figures related to the number of numerical digits used to express the measurement. As an example of significant figures, suppose we are asked to measure the radius of a Blu-ray Disc using a meter stick as a measuring instrument. Let assume the accuracy to which we can measure the radius of the disc is ± 0.1 cm. Because of the uncertainty of ± 0.1 cm, if the radius is measured to be 6.0 cm, we can claim only that its radius lies somewhere between 5.9 cm and 6.1 cm. In this case, we say that the measured value of 6.0 cm has two significant figures. Note that the significant figures include the first estimated digit. Therefore, we could write the radius as (6.0 ± 0.1) cm. Zeros may or may not be significant figures. Those used to position the decimal point in such numbers as 0.03 and 0.0075 are not significant. Therefore, there are one and two significant figures, respectively, in these two values. When the zeros come after other digits, however, there is a possibility of misinterpretation. For example, suppose the mass of an object is given as 1500 g. This value is ambiguous because we do not know whether the last two zeros are being used to locate the decimal point on whether they represent significant figures in the measurements. To remove this ambiguity, it is common to use scientific notation to indicate the number of significant. In this case, we would express the mass as 1.5 x 10 3 g if there are two significant figures in the measurement value, 1.5 x 10 3 g if there are three significant figures and 1.500 x 103 g if there are four. The same rules holds for number less tahn1, so 2.3 x 10-4 has two significant figures (and therefore could be written 0.00023) and 2.50 x 10-4 has three significant figures (and therefore written as 0.000230). s In problem solving, we often combine quantities mathematically through multiplication, division, addition, subtraction, and so forth. When doing so, you must make sure that the result has the appropriate number of significant figures. A good rule of thumb to use in determining the number of significant figures that can be claimed in a multiplication or a division is as follows: When multiplying several quantities, the number of significant figures in the final answer is the same as the number of significant figures in the quantity having the smallest number of significant figures. The same rule applies to division. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 16 Let’s apply the rule to find the area of the Blu-ray Disc whose radius we measured above. Using the equation for the area of a circle, A = 𝜋r2 = 𝜋(66666.0 cm) 2 = 1.1 x 102 cm2 If you perform this calculation on your calculator, you will likely see 113 097 335.5. It should be clear that you don’t want to keep all of these digits, but you might be tempted to report the result as 113 cm2. This result is not justified because it has three significant figures, whereas the radius only has two. Therefore, we must report the result with only two significant figures as shown above. For addition and subtraction, you must consider the number of decimal places when you are determining how many significant figures to report: When numbers are added or subtracted, the number of decimal places in the result should equal the smallest number of decimal places of any term in the sum or difference. As an example of this rule, consider the sum 23.2 + 5.174 = 28.4 Notice that we do not report the answer as 28.374 because the lowest number of decimal places is one, for 23.2. Therefore, our answer must have only one decimal place. The rule for addition and subtraction can often result in answers that have a different number of significant figures than the quantities with which you start. For example, consider these operations that satisfy the rule: 1.0001 + 0.0003 = 1.0004 1.0002 - 0.998 = 0.004 In the first example, the result has five significant figures even though one of the terms, 0.0003, has only one significant figure. Similarly, in the second calculation, the result has only one significant figure even though the numbers being subtracted have four and three, respectively. Example # 6. A carpet is to be installed in a rectangular room whose length is measured to be 12.71m and whose width is measured to be 3.36 m. Find the area of the room. Solution: If you multiply 12.71 by 3.46 m on your calculator, you will see an answer of 43.9766 m2. How many of these numbers should you claim? Our rule of thumb for multiplication tells us that you can claim only the number of significant figures in your answer as are present in the measured quantity having the lowest number of ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 17 significant figures. In this example, the lowest number significant figures is three in 3.46 m, so we should express our final answer as 44.0 m2. Vectors Physical quantities fall into two categories. One type, known is scalar quantity can be completely described by a single number (with appropriate units) giving its magnitude or size. Some common scalars are mass, temperature, volume and speed. For example, a car’s speed can be completely described by noting the number on its speedometer. A basketball’s mass can be specified by measuring a single number with a scale. The other type of quantity known as a vector quantity, has both magnitude and a direction. Velocity is a common vector quantity with magnitude specifying how fast an object is moving and a direction specifying the direction of travel. For example, a car’s velocity could specified by noting it was traveling 60 miles per hour in the direction due north. A car traveling 60 miles per hour in an eastward direction would have different velocity vector but the same scalar speed. In general, a vector quantity is characterized by having both a magnitude and direction. By contrast, a scalar quantity has magnitude, but no direction. Equality of Two Vectors Two vectors A and B are equal if they have the same magnitude and the same direction. They need not be located at the same point in space. The four vectors are all equal to each other. Moving a vector from one point in space to another doesn’t change its magnitude or its direction. A + B = C differ significantly from A + B = C The first is a vector sum, which must be handled graphically or with components, whereas the second is a simple arithmetic sum of numbers. Adding Vectors When two or more vectors are added, they must all have the same units. For example, it doesn’t make sense to add a velocity vector, carrying units of meters per second, to a displacement vector, carrying units of meters. Scalar obey the same rule: It would be similarly meaningless to add temperatures to volumes or masses to time intervals. Vectors can be added geometrically or algebraically. To add vector B to vector A geometrically, first draw A on a piece of graph paper to some scale, such as 1 cm = 1 m, so that its direction is specified relative to a coordinate system. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 18 Then draw vector B to the same scale with the tail of B starting at the tip of A, as shown in Fig.1.12a. Vector B must be drawn along the direction that makes the proper angle relative to vector A. The resultant vector R = A + B is the vector drawn from the tail of A to the tip of B. This procedure is known as the triangle method of addition When two vectors are added, their sum is independent of the order of the addition: A + B = B + A. This relationship can be seen from the geometric construction in Fig. 1.12b, and is called the commutative law of addition. This same general approach can also be used to add more than two vectors, as is done in Fig 1.13 for four vectors. The resultant vector sum R = A + B + C + D is the vector sum drawn from the tail of the first vector to the tip of the last. Again, the order in which the vectors are added is unimportant. Fig. 1.11 These four vectors are equal because they have equal lengths and point in the same direction (a) (b) Fig. 1.12 (a) When vector B is added to vector A, the vector sum R is the vector that runs from the tail of A to the tip of B. (b) Here the resultant runs from the tail of B to the tip of A. These constructions prove that A + B = B + A ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 19 Fig. 1.13 A geometric construction for summing four vectors. The resultant vector R is the vector that complete the polygon. Negative of a Vector The negative of the vector A is defined as the vector that gives zero when added to A. This means that A and –A have the same magnitude but opposite in directions. Subtracting Vectors Vector subtraction makes use of the definition of the negative of a vector. We define the operation A –B as the vector –B added to the vector A: A – B = A + (-B) Vector subtraction is really a special case of vector addition. The geometric construction for subtracting two vectors is shown in Fig. 1.14 Fig. 1.14 This construction shows how to subtract vector B from vector A. The vector –B has the same magnitude as vector B but points in the opposite direction. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 20 Multiplying or Dividing a Vector by a Scalar Multiplying or dividing a vector by a scalar gives a vector. For example, if vector A is multiplied by the scalar number 3, the result, written 3A, is a vector with a magnitude three times that of A and pointing in the same direction. If we multiply vector A by the scalar -3, the result is -3A, a vector with a magnitude three times that of A and pointing in the opposite direction (because of the negative sign). Example: A car travels 20 km due north and then 35 km in the direction 60 0 west of north, as in fig. 1.15. Using a graph, find the magnitude and direction of a single vector that gives the net effect of the car’s trip. This vector is called the car’s resultant displacement. Fig. 1.15 A graphical method for finding the resultant displacement vector R=A+B Solution: Let A the first displacement vector, 20 km north, and B the second displacement vector, extending west of north. Carefully graph two vectors, drawing a resultant vector R with its base touching the base of A and extending to the tip of B. Measure the length of this vector, which turns out to be about 48 km. The angle 𝛽, measured with a protractor, is about 390 west of north. Components of a Vector One method of adding vector make use of the projections of a vector along the axes of a rectangular coordinate system. These projections are called components. Any vectors can be completely described by its components. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 21 Consider a vector A in a rectangular coordinate system, as shown in fig. 1.16. A can be expressed as the sum of two vectors Ax parallel to the x-axis, and Ay, parallel to the y-axis. Mathematically, A = Ax + Ay Fig. 1.16 Any vector A lying in the xy-plane can be represented by its rectangular components Ax and Ay. where Ax and Ay are component vectors of A. The projection of A along the x-axis, Ax, is called the x-component of A, and the projection of A along the y-axis, Ay, is called the y-component of A. These components can be either positive or negative numbers with units. From the definition of sine and cosine, we see that cos 𝜃 = Ax/A and sin 𝜃 = Ay/A, so the components of A are: Ax = A cos 𝜽 (1.4a) Ay = A sin 𝜽 (1.4b) These components form two sides of a right triangle having a hypotenuse with magnitude A. It follows that A’s magnitude and direction are related to its components through the Pythagorean theorem and the definition of the tangent: A = √ Ax2 + Ay2 Tan 𝜃 = Ax/Ay ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE (1.5) (1.6) 22 Fig. 1.17 The angle 𝜽 need not always be defined from the positive x-axis. To solve for the angle 𝜃, which is measured counterclockwise from the positive xaxis by convention, the inverse tangent can be taken of both sides of equation 𝜃 = tan-1 (Ax /Ay) This formula gives the right answer for 𝜃 only half the time! The inverse tangent function returns values only from -900 to + 900, so the answer in your calculator window will only be correct if the vector happens to lie in the first or fourth quadrant. If it lies in the second or third quadrant, adding 180 0 to the number in the calculator window will always give the right answer. The angle must be measured from positive x –axis. Other choices of reference line are possible, but certain adjustments must then be made. If a coordinate system other than the one shown in Fig. 1.16 is chosen, the components of the vector must be modified accordingly. In many applications it’s more convenient to express the components of a vector in a coordinate system having axes that are not horizontal and vertical but are still perpendicular to each other. Suppose a vector B makes an angle 𝜃’ with the x’-axis defined in Fig. 1.18. The rectangular components of B along the axes of the figure are given by Bx’ = B cos 𝜃’ and By’ = B sin 𝜃’. The magnitude and direction of B are then obtained from expressions equivalent to equations 1.5 & 1.6. Fig. 1.18 The components of vector B in a tilted coordinate system. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 23 Sample Problem: a) Find horizontal and vertical components of the d = 1 x 102 m displacement of a superhero who flies from the top of a tall building along the path shown in the fig.1.20a b) Suppose instead the superhero leaps in the other direction along a displacement vector B to the top of a flagpole where the displacement components given by Bx = - 25 m and By = 10 m. Find the magnitude and direction of the displacement vector. Fig. 1.20 a) Find the vector components of A from its magnitude and direction Ax = A cos 𝜃 = (1 x 10 2 m) cos (300) = 86.6 m Ay = A sin 𝜃 = (1 x 10 2 m) sin (-300) = -50.0 m b) Find the magnitude and direction of the displacement vector B from its components. B = √ Bx2 + By2 = √(−25 m)2 + (10m)2 = 26.9 m 𝜃 = tan-1 (By/Bx) = tan-1 (10/-25) 𝜃 = -21.80 ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 24 Adding Vectors Algebraically The graphical method of adding vectors is valuable in understanding how vectors can be manipulated, but most of the time vectors are added algebraically in terms of their components. Suppose R = A + B. Then the components of the resultant vector R are given by: Rx = Ax + Bx Ry = Ay + By So x- components are added only to x-components, and y-components only to y- components. This magnitude and direction of R can be consequently be found with equations 1.5 and 1.6. Subtracting two vectors works the same way because it’s a matter of adding the negative of one vector to another vector. You should make a rough sketch when adding or subtracting vectors, in order to get an approximate geometric solution as a check. Sample Problem: A hiker begins a trip by first walking 25 km 450 south of east from her base camp. On the second day she walks 40 km in a direction 60 0 north of east, at which point she discovers a forest ranger’s tower. a) Determine the components of the hiker’s displacement in the first and second days. b) Determine the components of the hiker’s total displacement for the trip. c) Find the magnitude and direction of the displacement from base camp. (a) (b) Fig. 1.21 a) Hiker’s path and the resultant vector. b) Components of the hiker’s total displacement from camp. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 25 Solution: a) Find the components of A Ax = A cos (450) = (25 km) (0.0707) = 17.7 km Ay = A sin (-450) = (25km) (0.0707) = - 17.7 km Find the components of B Bx = B cos 600 = (40 km) (0.50) = 20 km By = B sin 600 = (40 km) (0.806) = 34.6 km b) Find the components of the resultant vector, R = A + B To find Rx add the x-components of A & B: Rx = Ax + Bx = 17.7 km + 20 km = 37.7 km To find Ry , add the y –components of A & B: Ry = Ay + By = - 17.7 km + 34.6 km = 16.9 km c) Find the magnitude and direction of R Use the Pythagorean theorem to get the magnitude: R = √ Rx2 + Ry2 = √ (37.7 km)2 + (16.9 km)2 = 41.3 km Calculate the direction of R using the inverse tangent function: 𝜃 = tan-1 (16.9 km/37.7 km) = 21.40 NE ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 26 College of Engineering, Architecture and Technology Name: ___________________________________Date Submitted: __________ Course & Year: ____________________________ Rating: _________________ Evaluation: 1. a) Estimate the number of times your heart beats in a month. b) Estimate the number of human heartbeats in an average lifetime. 2. Answer each question yes or no. Must two quantities have the same dimensions a) b) c) d) e) If you are adding them? If you are multiplying them? If you are subtracting them? If you are dividing them? If you are equating them? 3. Under what circumstances would a vector have components that are equal? 4. How many significant figures are there in a) b) c) d) 78.9 ± 0.2 3.788 x 109 2.46 x 1026 0.0032 5. A firkin is an old British unit of volume equal to 9 gallons. How many cubic meters are there in 6 firkins? 6. The speed of light is about 3 x 108 m/s. Convert this figure to miles per hour. 7. A ladder 9 m long leans against the side of a building. If the ladder is inclined at an angle of 750 to the horizontal, what is the horizontal distance from the bottom of the ladder to the building? ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 27 8. In fig. P1.49, find (a) the side opposite 𝜃 (b) the side adjacent to ∅ (c) cos 𝜃, and (d) tan ∅. 9. A person walks 250 north of east for 3.10 km. How much far due to north and how far due east would she have to walk to arrive at the same location? 10. A commuter plane starts from an airport and takes the route shown in fig. P 1.71. The plane first flies to city A, located 175 km 20 0 west of north, to city B. Finally, the plane flies 190 km due to west, to city C. Find the location of city C relative to the location of the starting point. ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE 28