Lesson 1 Chapter 1: Units, Physical Quantities, and Vectors Course Plan Week Date Chapter Topics Units, Physical Quantities, and Vectors 1 20 February 2023 Chapter 1 2 27 February 2023 Chapter 2 and 3 Kinematics 3 6 March 2023 Chapter 4 Newton's Laws of Motion 4 13 March 2023 Chapter 5 Applying Newton's Laws 5 20 March 2023 Chapter 6 Work and Kinetic Energy 6 27 March 2023 Chapter 7 Potential Energy and Energy Conservation 7 3 April 2023 Chapter 8 Momentum, Impulse, and Collisions 8 10 April 2023 Chapter 8 Momentum, Impulse, and Collisions 9 17 April 2023 Chapter 9 Kinematics of Rotational Motion 10 24 April 2023 Chapter 10 Dynamics of Rotational Motion 11 1 May 2023 Chapter 10 Dynamics of Rotational Motion 12 8 May 2023 Chapter 12 Gravitation 13 15 May 2023 Chapter 13 Periodic Motion 14 22 May 2023 Chapter 13 Periodic Motion 2 Units, Physical Quantities, and Vectors Nature of Physics Physics wants to understand the nature Understanding the nature must relay on mathematical base Physics interprets the nature on the mathematical base It is a basic science that reveals the causes of the consequences of natural phenomena as a mathematical model. Examines the results of the observations, reveals theories about the reasons, derives mathematical expressions. Checks the validity of these statements and makes corrections or additions if necessary. 3 Units, Physical Quantities, and Vectors Measurements Measurements are important for consistency of observations. Physical quantities should be measurable. Examples of quantities that can be measured: length, temperature, time, etc. We must express the magnitudes of these quantities, and we use units for this. Different units can be used for a physical quantity. We must also define the conversions between these units. Meter (m) is the basic unit for length in the SI unit system (International System of Units - Le Système International d'Unités). Miles (mi) is the unit of length for the Royal Unit System (Imperial units). 4 Idealized Model What we call a model in physics is a simplified version of a very complex system to be fully understood. The figure shows a tennis ball thrown into the air. When we try to examine the system in all its details, we should consider each physical event during the movement of the ball, such as the rotation of the ball around its own axis, air resistance, and the shape of the ball due to not being a solid object. In such a case, the solution to the problem will be very complex. If we start from the simplest case instead, we can proceed. 5 Idealized Model 1. We can neglect air friction in the initial approximation, assuming that air resistance is not dominant over other effects on the case. 2. We can assume the object is a point object. We can make this assumption if the distance the object moves is too great for the dimensions of the object. 3. The point-mas acceptance will also eliminate the rotation of the object around its own axis, since the point-mass will be one-dimensional, the rotation is undefined and the change of its shape is also undefined. Under these assumptions, the motion of the tennis ball was modeled and turned into a solvable problem with the initial approximation. When analyzing the results, we should not overlook under which approaches we obtained these results. 6 Standard Units Measurable Physical Quantities: Length can be measured by ruler or meter. Time can be measured with a stopwatch. Describable (Operational description) Physical quantities: Average Speed: It is the measure of how long a certain distance has been covered, size can be obtained from the recipe. For this, the distance and how long it takes to cover this distance must be measured. ððð ð¡ðððð ðð£ððððð ð ðððð = ðžðððð ðð ð¡ððð The unit system has been defined so that the measurements made by different people at different places and at different times are comparable. Turkey and many other countries use the SI units. (International System of Units - Le Système International d'Unités) 7 SI Units Time: seconds (s) Length: Meters (m) There are Standard definitions for mass, length and time. Over time, new definitions have been created with more precise precision. Mass: Kilogram (kg) 1000 gram (g) 8 Huge – Tiny We must describe physical quantities with appropriate scales. We express the length of a straight line in our notebook in centimeters or millimeters. We express our height in centimeters or meters. We express our mass in kilograms and our age in years. We indicate the distance between the Earth and the Moon in kilometers, the distance between galaxies in light years, the distance between atoms in nanometers, and the duration of fluorescence in nanoseconds. 9 Unit Prefixes We need to describe physical quantities with appropriate scales. Multiplier Name Symbol Mass Length Time Memory Units 1015 Peta P 1012 Tera T TB 109 Giga G GB 106 Mega M MB 103 Kilo k kg km 100 - - [gram] [metre] [second] 10−3 Milli m mg mm ms 10−6 Micro m mg mm ms 10−9 Nano n nm ns 10−12 Pico p 10−15 Femto f [byte] fs 10 Unit Prefixes In addition to these, there are also inter units that we use. Multiplier Name Symbol Length Volume 103 Kilo k km 102 Hecto H Hm 101 Deca D Dm 100 - - [metre] [Litre] 10−1 Deci d dm dL 10−2 Centi c cm cL 10−3 Mili m mm mL 11 Unit Prefixes Below are the approximate sizes of some objects: 12 Unit Consistency and Unit Conversions We use symbols in expressions to correlate physical quantities. For example: We use the symbols ð for distance, ð¡ for time, and ð£ for velocity. If ð; in meters, ð¡; in seconds: ð ð£= ð¡ ð£; will be in meter/seconds. An equation should be dimensionally consistent. The speed calculated here must be in the unit of speed. Precisely speed is a measure of displacement, that is, how much time has been traveled a given distance. Its unit must be in the distance / time unit. From the SI system, this is m / s. km / h (kilometer / hour) is also the unit of speed. We talked about the SI and Royal Unit systems. The Royal Unit system can use miles per hour for speed. Example 1.1 Converting speed units: The official world land speed record is 1228 km/h, set on October 15, 1997, by Andy Green in the jet engine car Thrust SSC (Thrust Super Sonic Car). Express this speed in meters per second. km km 1000 m 1 h ð£ = 1228 = 1228 h h 1 km 3600 s 1000 m ð£ = 1228 = 341.11 m/s 3600 s 14 Uncertainty and Significant Figures 15 Order of Magnitude Estimation In some cases, we can make a rough estimate of the situation without detailed calculation. It is a situation we use very often in daily life. By rounding the available data, we can calculate the magnitude we want with a good approximation. When we sit in a cafe, looking around, most of us think about how well this place has earned, and we do a rough finger count to calculate how much that amount is. How many tables are around, how long does a customer stay at the table on average and what is the average bill they pay, how many days a week, how many hours a day? Here are the necessary information for the endorsement. Of course there are also expenses, ... The difference will give you a rough overview of the earnings of this cafe. Of course, you do not trust this rough calculation and open a cafe, you will make a detailed feasibility study for this. 16 Example 1.4 An order-of-magnitude estimate: You are writing an adventure novel in which the hero escapes across the border with a billion dollars' worth of gold in his suitcase. Is this possible? Would that amount of gold fit in a suitcase? Would it be too heavy to carry? 3 19.3 g/cm In real 1 ons Gold 200 ~ 600 1 ons ≈ 30 g assuming 1 Billion $ Gold ≈ 108 g We got if 100 Ton 10 m3 50 Ton ~ 200 Ton 1 g/cm3 108 cm3 10 g/cm3 100 m3 Do the calculation for 100000 $ worth 5 carat (1 gram) diamonds. Is the result more plausible? 17 Vectors A number and with a unit may be sufficient to express some physical quantities such as temperature, mass, density. Since the direction is also associated with some other quantities, only number and unit are not enough. The motion of the plane is an example for such a situation In such cases, we must express these quantities with vectors. Vectors in three dimensional physical space will also have three components. ðŽÔŠ 18 Vectors ðŽÔŠ For example, we must use a vector to express displacement. 2 1 ðÔŠ When the object is displaced from 1 to 2 over any path, the net displacement must be expressed with a vector. The magnitude of this vector ð ÔŠ shown in blue indicates the amount of displacement and the direction of the arrow indicates the direction of the displacement. As can be seen, the displacement was from 1 to 2. Changing the direction of the arrow is sufficient to express the displacement from 2 to 1. 19 Vectors ðŽÔŠ ðµ ðŽÔŠ −ðŽÔŠ ð¶ÔŠ Vectors showing the same direction are called parallel, and vectors pointing opposite directions are called anti-parallel. Two vectors of the same length but opposite direction are equal to each other with the (-) sign. ÔŠ indicated by ðŽÔŠ The magnitude of the vector is ðŽ| Displacement cannot be interpreted as ðŽÔŠ = 6 m. The statement should also contain the information for direction. 20 Vector Operations 1. Vector Sum (addition, subtraction) 2. Vector Product (Scalar, Vectorial) ðŽÔŠ ðµ 21 Vector addition and subtraction Addition of vector ðŽÔŠ and ðµ : ðµ ðŽÔŠ ðµ ð¶ÔŠ = ðŽÔŠ + ðµ ðŽÔŠ Subtraction vector ðµ from ðŽÔŠ : −ðµ ðŽÔŠ −ðµ ðµ ðŽÔŠ ð· = ðŽÔŠ − ðµ = ðŽÔŠ + −ðµ 22 Vector addition and subtraction Sum of more than one vector: ð¶ÔŠ ðž ðµ ð· ðºÔŠ = ðŽÔŠ + ðµ + ð¶ÔŠ + ð· + ðž + ð¹ÔŠ ð¹ÔŠ ðŽÔŠ ðºÔŠ 23 Example 1.5 A cross-country skier skis 1 km north and then 2 km east on a horizontal snow field. How far and in what direction is she from he starting point? Net displacement is in North-East direction. North ðµ ðŽÔŠ East ð¶ÔŠ = ðŽÔŠ + ðµ The magnitude of the displacement can be found from the right triangle: ð¶ÔŠ 2 ð¶ÔŠ 2 2 = ðŽÔŠ + ðµ 2 = (1 km)2 +(2 km)2 = 5 km2 ð¶ÔŠ = 5 km = 2.24 km For the direction, we can determine the angle the plane makes with the north: 2 km tan ð = =2 1 km ð = 63.4° 24 Vector Components The sum of two or more vectors is again a vector. We can write a vector as the sum of other vectors. Specifically, we can choose these two vectors to be perpendicular to each other. ðŽÔŠ ðŽÔŠ2 ðŽÔŠ ðŽÔŠ2 ðŽÔŠ1 ðŽÔŠ1 ðŽÔŠ = ðŽÔŠ1 + ðŽÔŠ2 ðŽÔŠ = ðŽÔŠ1 + ðŽÔŠ2 If vector ðŽÔŠ1 is the component of ðŽÔŠ in the ð¥-direction and vector ðŽÔŠ2 is the component of ðŽÔŠ in the ðŠ-direction, ; ðŽÔŠ = ðŽÔŠð¥ + ðŽÔŠðŠ ðŽÔŠ1 and ðŽÔŠ2 vectors are the components ÔŠ of the vector ðŽ. 25 Vector Components y ðŽÔŠ ðŽÔŠðŠ ð 0 We can define the magnitudes of the components of the vector ðŽÔŠ in the direction of ð¥ and ðŠ and interpret the vector ðŽÔŠ as its components: ðŽÔŠð¥ x ðð¥ = ðŽÔŠð¥ = ðŽÔŠ cos(ð) ðŽÔŠð¥ / ðŽÔŠð¥ Gives the unit vector ðÆž that is in the direction of ð¥. ðŽÔŠðŠ / ðŽÔŠðŠ Gives the unit vector ðÆž that is in the direction of ðŠ. In terms of the unit vector ðÆž and ðÆž vector a can be written as: ðŽÔŠ = ðð¥ ð Æž + ððŠ ð Æž ððŠ = ðŽÔŠðŠ = ðŽÔŠ sin(ð) With the additional component in the third direction in three dimensional space; ðŽÔŠ = ðð¥ ð Æž + ððŠ ð Æž + ðð§ ðá 26 Example 1.6 Finding components: (a) What are the ð¥ and ðŠ-components of vector ð· in Fig. 1.19 a? The magnitude of the vector is ð· = 3 m and the angle ðŒ = 45°. (b) What are the ð¥ and ðŠcomponents of vector ðž in Fig. 1.19 b? The magnitude of the vector is ð· = 4.5 m and the angle ðœ = 37°. ð·ð¥ = ð· cos ðŒ = 3 m cos −45° = +2.41 m ð·ðŠ = ð· sin ðŒ = 3 m sin −45° = −2.41 m ðžð¥ = ðž sin ðœ = 4.5 m sin 37° = +2.71 m ðžðŠ = ðž cos ðœ = 4.5 m cos 37° = +3.59 m 27 Vector Operations with Components Method y ÔŠ we can calculate the direction and From the components of ðŽ, magnitude ( ðŽÔŠ ) of the vector ðŽÔŠ ðŽÔŠ ðŽÔŠðŠ ð 0 ðŽÔŠð¥ ðŽÔŠðŠ tan ð = ðŽÔŠð¥ The direction of the vector ðŽÔŠ can be stated with respect to the ð¥-axis. x ðŽÔŠ and 2 = ðŽÔŠð¥ ð = ðððð¡ðð 2 + ðŽÔŠðŠ ðŽÔŠðŠ ðŽÔŠð¥ 2 = ð¡ðð−1 ðŽÔŠðŠ ðŽÔŠð¥ 28 Vector Product Vector multiplied by a scalar : ð· = ð ðŽÔŠ Multiplying vector ðŽÔŠ with a scalar like ð give a new vector like ð·. ðŽÔŠ ð· = ð ðŽÔŠ ÔŠ If ð is positive (ð ≥ 0) ð· is also in the same direction, ð· has the same alignment with ðŽ. ÔŠ and the magnitude of ð· is ð times the magnitude of ðŽ. Multiplying a vector by a scalar is equivalent to multiplying its components by that scalar. ð·ð¥ = ð ðŽÔŠð¥ ð· = ð ðŽÔŠ = ð ðð¥ ð Æž + ððŠ ð Æž = ððð¥ ð Æž + ðððŠ ð Æž = ð ðŽÔŠð¥ + ð ðŽÔŠðŠ ð·ðŠ = ð ðŽÔŠðŠ 29 Vector Sum by Components Method y ð¶ÔŠ = ðŽÔŠ + ðµ ð¶ÔŠ ðµðŠ ðµ ðŽÔŠ = ðð¥ ðÆž + ððŠ ðÆž ðŽÔŠð¥ = ðð¥ ðÆž ðŽÔŠðŠ = ððŠ ðÆž ðµ = ðð¥ ðÆž + ððŠ ðÆž ðµð¥ = ðð¥ ðÆž ðµðŠ = ððŠ ðÆž ððµ ðŽÔŠðŠ 0 ð ÔŠ ððŽ ðŽ ðŽÔŠð¥ x ðµð¥ The lengths of vectors in the same direction can be summed directly. ðð¥ ðÆž + ðð¥ ðÆž = ðð¥ ðÆž ððŠ ðÆž + ððŠ ðÆž = ððŠ ðÆž ð¶ÔŠ = ðð¥ ðÆž + ððŠ ðÆž The sum of ðŽÔŠ and ðµ is again a vector. The magnitude of this vector is The alignment of this vector with respect to the ð¥ axis is ð¶ÔŠ 2 = ð¶ÔŠð¥ 2 + ð¶ÔŠðŠ ð = ð¡ðð−1 2 ð¶ÔŠðŠ ð¶ÔŠð¥ 30 Vector Sum by Components Method Addition of more than one vector: ð = ðŽÔŠ + ðµ + ð¶ÔŠ + ð· + ⯠+ ðÔŠ ð ð¥ ðÆž = ðð¥ ðÆž + ðð¥ ðÆž + ðð¥ ðÆž + ðð¥ ðÆž + ⯠+ ð§ð¥ ðÆž and ð ðŠ ðÆž = ððŠ ðÆž + ððŠ ðÆž + ððŠ ðÆž + ððŠ ðÆž + ⯠+ ð§ðŠ ðÆž ð = ðð¥ ðÆž + ððŠ ðÆž If the vectors are three-dimensional: ð ð§ ðà· = ðð§ ðà· + ðð§ ðà· + ðð§ ðà· + ðð§ ðà· + ⯠+ ð§ð§ ðà· ð = ðð¥ ðÆž + ððŠ ðÆž + ðð§ ðà· ð A ð-dimensional vector ð = à· ðð ð Æž ð=1 ðð is the magnitude of the component of the vector in the direction of ð and ðÆž is the unit vector in this direction. 31 Example 1.7 (1/2) Adding vectors with components: Three players on a reality TV show are brought to the center of a large, flat field. Each is given a meter stick, a compass, a calculator, a shovel, and (in a different order for each contestant) the following three displacements: 72.4 m, 32.0" east of north, 57.3 m, 36.0" south of west, 17.8 m straight south The three displacements lead to the point where the keys to a new Porsche are buried. Two players start measuring immediately, but the winner first calculates where to go. What does she calculate? ð = ðŽÔŠ + ðµ + ð¶ÔŠ North 36° ðµ ðŽÔŠ ð¶ÔŠ 32° West ð ðŽÔŠ = ðð¥ ðÆž + ððŠ ðÆž = 72.4 sin(32°) ðÆž + cos(32°) ðÆž 0 South ðµ = ðð¥ ðÆž + ððŠ ðÆž = 57.3 cos(−36°) ðÆž + sin(−36°) ðÆž East ð¶ÔŠ = ðð¥ ðÆž + ððŠ ðÆž = 72.4 cos(−90°) ðÆž + sin(−90°) ðÆž ð = ðð¥ + ðð¥ + ðð¥ ðÆž + ððŠ + ððŠ + ððŠ ðÆž 32 Example 1.7 (2/2) North 36° ðµ 32° ð ðµ = −46.36 m ðÆž − 33.68 m ðÆž ðŽÔŠ ð¶ÔŠ West ðŽÔŠ = 38.37m ðÆž + 61.40 m ðÆž 0 Eat ð¶ÔŠ = 0 m ðÆž − 17.80 m ðÆž ð = ðð¥ + ðð¥ + ðð¥ ðÆž + ððŠ + ððŠ + ððŠ ðÆž South ð = −7.99 m ðÆž + 9.92 m ðÆž ð = −7.99 m 2 + 9.92 m 2 = 12.7 m ð = ð¡ðð−1 9.92 m = 129° −7.99 m 33 Determination of Axes To determine the orthogonal coordinate axis set in three-dimensional space: ð§ ðŠ ð¥ We can choose the first axis in any direction. Let's call this axis the x-axis. The second axis should be perpendicular to the first axis. We can choose an infinite number of axes perpendicular to the first axis. All of these axes remain in the plane perpendicular to the x axis. We can specify any one and call it the y-axis. As the third axis, we must determine the axis that will be perpendicular to these first two axes. The axis perpendicular to both axes must be in a direction perpendicular to the ð¥ − ðŠ plane formed by these two axes. ð We will decide which of the two possible axes is the axis according to the right hand rule. When we stretch our four fingers of our right hand in the direction of the ð¥ axis and rotate them shortly towards ðŠ axis, our thumb will point in the direction of the ð§ axis. 34 Unit Vectors ðŽÔŠ = ðŽÔŠð¥ + ðŽÔŠðŠ + ðŽÔŠð§ z ÔŠ ðŠ, ð§) or ðŽ(ð¥, ÔŠ ð, ∅) ðŽ(ð, ðŽÔŠð¥ = ðŽÔŠ sin ð cos ∅ ðŽÔŠð§ ðŽÔŠð¥ 0 ð ðŽÔŠ ðŽÔŠðŠ ∅ x ðŽÔŠ = ðð¥ ðÆž + ððŠ ðÆž + ðð§ ðà· ðŽÔŠðŠ = ðŽÔŠ sin ð sin ∅ y ðŽÔŠð§ = ðŽÔŠ cos ð and ðŽÔŠ 2 = ðŽÔŠð¥ 2 + ðŽÔŠðŠ 2 + ðŽÔŠð§ 2 ðÆž = ðŽÔŠð¥ / ðŽÔŠð¥ ðÆž = ðŽÔŠðŠ / ðŽÔŠðŠ unit vectors in related directions. ðà· = ðŽÔŠð§ / ðŽÔŠð§ 35 Example 1.9 Given the two displacements ð· = 6ðÆž + 3ðÆž − ðà· m and ðž = 4ðÆž − 5ðÆž + 8ðà· m find the magnitude of the displacements 2ð· − ðž. ð¹ÔŠ = 2ð· − ðž ð¹ÔŠ = 2 6ðÆž + 3ðÆž − ðà· m − 4ðÆž − 5ðÆž + 8ðà· m ð¹ÔŠ = 12 − 4 m ðÆž + 6 + 5 m ðÆž + −2 − 8 m ðà· ð¹ÔŠ = 8 ðÆž + 11 ðÆž − 10 ðà· m ð¹ÔŠ = 8 m 2 + 11 m 2 + −10 m 2 ð¹ÔŠ = 17 m 36 Vector Multiplication Scalar Product The scalar (dot) product of ðŽÔŠ and ðµ is given with ðŽÔŠ â ðµ. The scalar (dot) product of two vectors gives a scalar quantity. This magnitude gives the projection of one vector on the other. ðŽÔŠ ð ðŽÔŠ ð· = ðµ cos ð ðµ ð¶ÔŠ = ðŽÔŠ cos ð ð ðµ ðŽÔŠ â ðµ = ðµ â ðŽÔŠ = ðŽÔŠ ðµ cos ð 37 Vector Multiplication Scalar Product The scalar (dot) product of a vector by itself gives the square of its magnitude. ðŽÔŠ â ðŽÔŠ = ðŽÔŠ ðŽÔŠ cos 0 = ðŽÔŠ 2 z The scalar (dot) product of ðŽÔŠ and ðµ ðà· ðŽÔŠ = ðð¥ ðÆž + ððŠ ðÆž + ðð§ ðà· ðµ = ðð¥ ðÆž + ððŠ ðÆž + ðð§ ðà· ðÆž 0 x y ðÆž This magnitude gives the projection of one vector on the other. ðŽÔŠ â ðµ = ðµ â ðŽÔŠ = ðŽÔŠ ðµ cos ð Remember that the unit vectors are perpendicular to each other. 38 Vector Multiplication Scalar Product ðŽÔŠ â ðŽÔŠ = ðŽÔŠ ðŽÔŠ cos 0 = ðŽÔŠ 2 The scalar (dot) product of unit vectors ðÆž z ðà· 0 x ðÆž ðà· ðÆž ðÆž y ðà· ðÆž ðÆž There is no projection onto each other. ðÆž ðÆž The projection onto itself other is one. ðÆž â ðÆž = ðÆž â ðà· = ðà· â ðÆž = 0 ðÆž ðÆž ðà· ðà· ðÆž â ðÆž = ðÆž â ðÆž = ðà· â ðà· = 1 39 Vector Multiplication Scalar Product ðŽÔŠ â ðµ = ðµ â ðŽÔŠ = ðŽÔŠ ðµ cos ð ðÆž â ðÆž = ðÆž ðÆž cos 0 = 1 ðŽÔŠ â ðµ = ðð¥ ðÆž + ððŠ ðÆž + ðð§ ðà· â ðð¥ ðÆž + ððŠ ðÆž + ðð§ ðà· ðÆž â ðÆž = ðÆž ðÆž cos 90 = 0 ðŽÔŠ â ðµ = ðð¥ ðÆž â ðð¥ ðÆž + ððŠ ðÆž â ðð¥ ðÆž + ðð§ ðà· â ðð¥ ðÆž ðÆž â ðà· = ðÆž ðà· cos 90 = 0 + ðð¥ ðÆž â ððŠ ðÆž + ððŠ ðÆž â ððŠ ðÆž + ðð§ ðà· â ððŠ ðÆž + ðð¥ ðÆž â ðð§ ðà· + ððŠ ðÆž â ðð§ ðà· + ðð§ ðà· â ðð§ ðà· ðŽÔŠ â ðµ = ðð¥ ðð¥ + ððŠ ððŠ + ðð§ ðð§ and ðŽÔŠ â ðŽÔŠ = ðð¥ ðð¥ + ððŠ ððŠ + ðð§ ðð§ = ðð¥ 2 + ððŠ 2 + ðð§ 2 Gives the 3-D Pythagorean theorem. 40 Example 1.10 Find the scalar product ðŽÔŠ â ðµ of the two vectors in figure. The magnitudes of the vectors are ðŽÔŠ = 4 and ðµ = 5. ðŽÔŠ â ðµ = ðŽÔŠ ðµ cos ð y ðŽÔŠ â ðµ = 4 â 5 cos 130° − 53° = 20 cos 77° = 4.5 ðµ 130° We can get the same result by finding the components. ðŽÔŠ ðÆž 53° ðÆž x ðð¥ = 4 cos 53° = 2.407 ðð¥ = 5 cos 130° = −3.214 ððŠ = 4 sin 53° = 3.195 ððŠ = 5 sin 130° = 3.830 ðð§ = 0 ðð§ = 0 ðŽÔŠ â ðµ = ðð¥ ðð¥ + ððŠ ððŠ + ðð§ ðð§ = 2.407 −3.214 + 3.195 3.830 + 0 = 4.5 41 3D Vectors z ðŽÔŠ = 2ðÆž + 3ðÆž + 3ðà· z ðŽÔŠð¥ = 2ðÆž ðµðŠ ðŽÔŠðŠ = 3ðÆž ðŽÔŠð§ = 3ðà· ðà· à· ðµ = −3ðÆž + 2ðÆž − ð. y ðÆž y ðÆž ðŽÔŠð¥ ðµðŠ = 2ðÆž ðµð§ = −ðà· ð ðÆž ðµð¥ = −3ðÆž ðµ ðà· ðÆž ð ðµð§ ðµð¥ ðŽÔŠð§ ðŽÔŠ x ðŽÔŠðŠ x 42 Example 1.11 à· Find the angle between the two vectors ðŽÔŠ = 2ðÆž + 3ðÆž + ðà· and ðµ = −4ðÆž + 2ðÆž − ð. ðŽÔŠ â ðµ = ðŽÔŠ ðµ cos ð ðŽÔŠ â ðµ = 2 −4 + 3 2 + (1)(−1) ðŽÔŠ â ðµ = ðð¥ ðð¥ + ððŠ ððŠ + ðð§ ðð§ ðŽÔŠ 2 = 22 + 32 + 12 ðŽÔŠ = 14 ðµ 2 = (−4)2 +22 + (−1)2 ðµ = 21 ðŽÔŠ â ðµ = −8 + 6 − 1 = −3 ðŽÔŠ â ðµ cos ð = ðŽÔŠ ðµ cos ð = −3 14 21 = −3 7 6 ð = 100° 43 Vector Multiplication Vector Product The vector (cross) product of ðŽÔŠ with ðµ : ðŽÔŠ × ðµ The cross product of two vectors gives a vector quantity. The direction of this vector is perpendicular to the plane formed by the vector ðŽÔŠ and ðµ, and its direction is determined by the right-hand rule. The magnitude of this vector ise given by ðŽÔŠ ðµ sin ð ð¶ÔŠ = ðŽÔŠ × ðµ = −ðµ × ðŽÔŠ ð¶ÔŠ = ðŽÔŠ ðµ sin ð ð¶ÔŠ = ðŽÔŠ × ðµ ðµ ð ðŽÔŠ 44 Vector Product ð¶ÔŠ = ðŽÔŠ ðµ sin ð The vector (cross) product of ðŽÔŠ with ðµ : ðŽÔŠ × ðµ ðµ sin ð ð¶ÔŠ = ðŽÔŠ × ðµ ð¶ÔŠ = ðŽÔŠ × ðµ ðµ ðµ ð ð ðŽÔŠ ðŽÔŠ ðŽÔŠ sin ð −ð¶ÔŠ = ðµ × ðŽÔŠ −ð¶ÔŠ = ðµ × ðŽÔŠ Magnitude of the shaded area: ðŽÔŠ ðµ sin ð Magnitude of the shaded area: ðŽÔŠ ðµ sin ð ð¶ÔŠ = ðŽÔŠ × ðµ = −ðµ × ðŽÔŠ The magnitude of the vector (cross) product ðŽÔŠ × ðµ is the area formed by this two vectors, and the direction of the resultant vector can be found by the right hand rule. 45 Vector Multiplication Vector Product ð¶ÔŠ = ðŽÔŠ ðµ sin ð ðÆž × ðÆž = 0 ðŽÔŠ × ðµ = ðð¥ ðÆž + ððŠ ðÆž + ðð§ ðà· × ðð¥ ðÆž + ððŠ ðÆž + ðð§ ðà· ðÆž × ðÆž = ðà· ðŽÔŠ × ðµ = ðð¥ ðÆž × ðð¥ ðÆž + ððŠ ðÆž × ðð¥ ðÆž + ðð§ ðà· × ðð¥ ðÆž ðÆž × ðà· = −ðÆž + ðð¥ ðÆž × ððŠ ðÆž + ððŠ ðÆž × ððŠ ðÆž + ðð§ ðà· × ððŠ ðÆž + ðð¥ ðÆž × ðð§ ðà· + ððŠ ðÆž × ðð§ ðà· + ðð§ ðà· × ðð§ ðà· ðŽÔŠ × ðµ = ððŠ ðð§ − ðð§ ððŠ ðÆž + ðð§ ðð¥ − ðð¥ ðð§ ðÆž + ðð¥ ððŠ − ððŠ ðð¥ ðà· = ðð¥ ðÆž + ððŠ ðÆž + ðð§ ðà· ð¶ÔŠ = ðŽÔŠ × ðµ 46 Vector Multiplication Vector Product The vector (cross) product of ðŽÔŠ and ðµ is given by ðÆž ðŽÔŠ × ðµ = ðð¥ ðð¥ ðÆž ððŠ ððŠ ðà· ððŠ 1+1 ðð§ = −1 ððŠ ðð§ ðÆž ðŽÔŠ × ðµ = ðð¥ ðð¥ ðð§ ðð¥ 1+2 ðð§ ðÆž + −1 ðð¥ ðÆž ððŠ ððŠ ðà· ðð§ ðð§ ðð¥ ðð§ 1+3 ðð§ ðÆž + −1 ðð¥ ððŠ à· ð ððŠ ðŽÔŠ × ðµ = ððŠ ðð§ − ðð§ ððŠ ðÆž + ðð§ ðð¥ − ðð¥ ðð§ ðÆž + ðð¥ ððŠ − ððŠ ðð¥ ðà· 47 Example 1.12 Vector ðŽÔŠ has magnitude 6 units and is in the direction of the +ð¥ − axis. Vector ðµ has magnitude 4 units and lies in the ð¥ðŠ − plane, making an angle of 30° with the +ð¥ − axis. Find the vector product ðŽÔŠ × ðµ. ðŠ ðŽÔŠ × ðµ = 6ðÆž × 4 cos 30° ðÆž + 4 sin(30°)ðÆž ðµ ðŽÔŠ × ðµ = 24 sin(30°)ðà· ð = 30° ðŽÔŠ ð§ ð¥ ðŽÔŠ × ðµ = 12ðà· ð¶ÔŠ = ðŽÔŠ × ðµ 48 Problems Q 1.49 (a) Write each vector in Fig. 1.37 in terms of the unit vectors ðÆž and ðÆž . (b) Use unit ÔŠ where ð¶ÔŠ = 3ðŽÔŠ − 4ðµ. (c) Find the magnitude and vectors to express the vectorð¶, ÔŠ direction of ð¶. ðŽÔŠ = ðð¥ ðÆž + ððŠ ðÆž + ðð§ ðà· ðŽÔŠ = 3.6 cos 70° m ðÆž + 3.6 sin 70° m ðÆž ðµ = ðð¥ ðÆž + ððŠ ðÆž + ðð§ ðà· ðµ = 2.4 cos 210° m ðÆž + 2.4 sin 210° m ðÆž ð¶ÔŠ = 3ðŽÔŠ − 4ðµ ð¶ÔŠ = 3 3.6 cos 70° m ðÆž + 3.6 sin 70° m ðÆž − 4 2.4 cos 210° m ðÆž + 2.4 sin 210° m ðÆž ð¶ÔŠ = 3 1.23 m ðÆž + 3.38 m ðÆž − 4 −2.08 m ðÆž − 1.2 m ðÆž ð¶ÔŠ = 12.01 m ðÆž + 14.94 m ðÆž and ð = ð¡ðð−1 ð¶ÔŠ â ð¶ÔŠ = ð¶ÔŠ 2 ð¶ÔŠ = 12.01 m 2 + 14.94 2 = 19.17 m 14.94 m = 51.2° 12.01 m 50 Q 1.51 à· a unit vector? Justify your answer. (b) Can a unit vector have any (a) Is the vector (ðÆž + ðÆž + ð) components with magnitude greater than unity? Can it have any negative components? In each case, justify your answer. (c) If ðŽÔŠ = ð(3ðÆž + 4ð), Æž where ð is a constant, determine the value of ð that makes ðŽÔŠ a unit vector. ðŽÔŠ = ðÆž + ðÆž + ðà· ðŽÔŠ â ðŽÔŠ = ðŽÔŠ 2 ðŽÔŠ = 12 + 12 + 12 = 3 ≠ 1 Is not a unit vector Any component of a unit vector cannot be greater than 1 or less than −1 ðŽÔŠ = ð(3ðÆž + 4ð)Æž ðŽÔŠ = ðð¥ ðð¥ + ððŠ ððŠ + ðð§ ðð§ > 1 ðŽÔŠ = 3ð 2 + 4ð 2 = ðŽÔŠ = ±5ð = 1 25ð2 ð = ±1/5 51 Q 1.86 (1/2) For the two vectors ðŽÔŠ and ðµ in Fig. 1.37, (a) find the scalar product ðŽÔŠ â ðµ, and (b) find the magnitude and direction of the vector product ðŽÔŠ × ðµ. ðŽÔŠ = 3.6 cos 70° m ðÆž + 3.6 sin 70° m ðÆž = 1.23 m ðÆž + 3.38 m ðÆž ðµ = 2.4 cos 210° m ðÆž + 2.4 sin 210° m ðÆž = −2.08 m ðÆž − 1.2 m ðÆž ðŽÔŠ â ðµ = ðŽÔŠ ðµ cos ð ðŽÔŠ = 1.23 m 2 + 3.38 m 2 = 3.6 m ðµ = −2.08 m 2 + −1.2 m 2 = 2.4 m ðŽÔŠ â ðµ = 3.6 m 2.4 m cos 210° − 70° = 6.62 m2 ðŽÔŠ â ðµ = ðð¥ ðð¥ + ððŠ ððŠ + ðð§ ðð§ = 1.23 m −2.08 m + 3.38 m −1.2 m + 0 = 6.62 m2 52 Q 1.86 (2/2) For the two vectors ðŽÔŠ and ðµ in Fig. 1.37, (a) find the scalar product ðŽÔŠ â ðµ, and (b) find the magnitude and direction of the vector product ðŽÔŠ × ðµ. ðŽÔŠ = 1.23 m ðÆž + 3.38 m ðÆž + 0 m ðà· ðŽÔŠ = 3.6 m ðµ = −2.08 m ðÆž − 1.2 m ðÆž + 0 m ðà· ðµ = 2.4 m ðÆž ðŽÔŠ × ðµ = ðð¥ ðð¥ ðÆž ððŠ ððŠ ðà· ðð§ ðð§ ð¶ÔŠ = ðŽÔŠ ðµ sin ð = 3.6 m 2.4 m sin 140° = 5.5 m ðŽÔŠ × ðµ = ððŠ ðð§ − ðð§ ððŠ ðÆž + ðð§ ðð¥ − ðð¥ ðð§ ðÆž + ðð¥ ððŠ − ððŠ ðð¥ ðà· = ðð¥ ððŠ − ððŠ ðð¥ ðà· = = 1.23 m −1.2 m − 3.38 m −2.08 m ðà· 1.23 m −1.2 m − 3.38 m −2.08 m ðà· = −1.48 + 7.03 ðà· = 5.5ðà· 53 Q 1.89 à· do the following. (a) Find the magnitude Given two vectors ðŽÔŠ = −2ðÆž + 3ðÆž + 4ðà· and ðµ = 3ðÆž + 1ðÆž − 3ð, of each vector. (b) Write an expression for the vector difference ðŽÔŠ − ðµ, using unit vectors. (c) Find ÔŠ Explain. the magnitude of the vector difference ðŽÔŠ − ðµ. Is this the same as the magnitude of ðµ − ðŽ? ðŽÔŠ = −2ðÆž + 3ðÆž + 4ðà· ðŽÔŠ â ðŽÔŠ = ðŽÔŠ ðµ = 3ðÆž + 1ðÆž − 3ðà· 2 ðŽÔŠ = −2 2 + 32 + 42 = 29 ≅ 5.38 ðµ = 32 + 12 + −3 2 = 19 ≅ 4.36 ðŽÔŠ − ðµ = −2ðÆž + 3ðÆž + 4ðà· − 3ðÆž + 1ðÆž − 3ðà· = −5ðÆž + 2ðÆž + 7ðà· ðŽÔŠ − ðµ = −5 2 + 22 + 72 = 25 + 4 + 49 = 78 ≅ 8.83 ðŽÔŠ − ðµ = ðµ − ðŽÔŠ = 78 ≅ 8.83 and 54 Q 1.93 A cube is placed so that one comer is at the origin and three edges are along the ð¥, y, and ð§-axes of a coordinate system (Fig. 1.43). Use vectors to compute (a) the angle between the edge along the z-axis (line ab) and the diagonal from the origin to the opposite comer (line ad), and (b) the angle between line ac (the diagonal of a face) and line ad. ðŽÔŠ = 02 + 02 + 12 = 1 ðµ = 12 + 12 + 12 = 3 Vector from ð to ð ð¶ÔŠ = 02 + 12 + 12 = 2 ðµ = 1ðÆž + 1ðÆž + 1ðà· ðŽÔŠ â ðµ = ðŽÔŠ ðµ cos ð Vector from ð to ð ðŽÔŠ â ðµ = ðð¥ ðð¥ + ððŠ ððŠ + ðð§ ðð§ ð¶ÔŠ = 0ðÆž + 1ðÆž + 1ðà· ðŽÔŠ â ðµ = 0 1 + 0 1 + 1 1 = 1 Vector from ð to ð ðŽÔŠ = 0ðÆž + 0ðÆž + 1ðà· cos ð = 1 3 ð = 54.7° ðµ â ð¶ÔŠ = 1 0 + 1 1 + 1 1 = 2 ðµ â ð¶ÔŠ = ðµ ð¶ÔŠ cos ðœ cos ðœ = 2 3 2 ðœ = 35.3° Q 1.95 You are given vectors ðŽÔŠ = 5ðÆž − 6.5ðÆž and ðµ = −3.5 + 7ð.Æž A third vector ð¶ÔŠ lies in the xy-pIane. Vector ÔŠ and the scalar product of ð¶ÔŠ with ðµ is 15. From this information, ð¶ÔŠ is perpendicular to vector ðŽ, ÔŠ find the components of vector ð¶. ðŽÔŠ = 5ðÆž − 6.5ðÆž ðµ = −3.5ðÆž + 7ðÆž If vector ð¶ÔŠ is in the ð¥ðŠ plane; ð¶ÔŠ = ðð¥ ðÆž + ððŠ ðÆž If ðŽÔŠ is perpendicular to ð¶ÔŠ 5 ððŠ = ð 6.5 ð¥ ðŽÔŠ â ð¶ÔŠ = ðð¥ ðð¥ + ððŠ ððŠ + ðð§ ðð§ = 0 5 −3.5ðð¥ + 7 ð = 15 6.5 ð¥ ðŽÔŠ â ð¶ÔŠ = 5ðð¥ − 6.5ððŠ = 0 70 −3.5 + ð = 15 13 ð¥ ðµ â ð¶ÔŠ = ðð¥ ðð¥ + ððŠ ððŠ + ðð§ ðð§ = 15 ðµ â ð¶ÔŠ = −3.5ðð¥ + 7ððŠ = 15 15 ðð¥ = ≅8 1.89 5 5 ððŠ = ðð¥ ≅ 8 = 6.1 6.5 6.5 56 Q 1.96 Two vectors A and B have magnitude ðŽ = 3 and ðµ = 3. Their vector product is ðŽÔŠ × ðµ = −5ðà· + 2ð.Æž What is the angle between ðŽÔŠ and ðµ? ð¶ÔŠ = ðŽÔŠ × ðµ = −5ðà· + 2ð.Æž ð¶ÔŠ â ð¶ÔŠ = ð¶ÔŠ ð¶ÔŠ = ðŽÔŠ ðµ sin ð 2 ð¶ÔŠ = −5 2 + 22 = 29 ð¶ÔŠ 29 sin ð = = ÔŠ 3 â3 ðŽ ðµ ð = 36.8° 57 Q 1.102 à· called the position vector, points from the origin (0,0,0) to an arbitrary The vector ðÔŠ = ð¥ ðÆž + ðŠðÆž + ð§ð, point in space with coordinates (ð¥, ðŠ, ð§). Use what you know about vectors to prove the following: All points (ð¥, ðŠ, ð§) that satisfy the equation ðŽð¥ + ðµðŠ + ð¶ð§ = 0, where ðŽ,ðµ, and ð¶ are constants, lie in à· Sketch a plane that passes through the origin and that is perpendicular to the vector ðŽðÆž + ðµ ðÆž + ð¶ ð. this vector and the plane. ðÔŠ â ðÔŠ = ðŽð¥ + ðµðŠ + ð¶ð§ = 0 ðÔŠ = ð¥ðÆž + ðŠðÆž + ð§ðà· ðÔŠ = ðŽðÆž + ðµðÆž + ð¶ ðà· If vector ðÔŠ is perpendicular to vector ðÔŠ ðÔŠ â ðÔŠ = ðð¥ ðð¥ + ððŠ ððŠ + ðð§ ðð§ = 0 so; ðŽð¥ + ðµðŠ + ð¶ð§ = 0 than ðÔŠ â ðÔŠ = 0 The projection of vector ðÔŠ onto vector ðÔŠ is zero so, ðÔŠ ⊥ ðÔŠ 58
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