Xem mẫu

20.7 A Brief Introduction to Vectors 671 20.7 A BRIEF INTRODUCTION TO VECTORS Questions of displacement and velocity have Þgured prominently in our studies. When we talk about displacement, we ask Òhow far?Ó and Òin what direction?Ó When we talk about velocity, we ask Òhow fast?Ó and Òin what direction?Ó WeÕve been answering the question Òin what direction?Ó by using a plus or a minus sign. This has restricted the type of motion we describe. Our objects are going up or down; our boatsareheadedeastorwest; ourcarstravelonlongstraighthighways; ourtrainsareeither inbound or outbound. If we wander along a meandering path, we can report our velocity only with reference to some benchmark point on the path; weÕre either heading toward it or away from it. We can free ourselves from this world of dichotomy by using an arrow to indicate direction. In fact, we can have the arrow answer both the amount and direction questions by having the length of the arrow indicate Òhow farÓ or Òhow fastÓ and the direction the arrow is pointing indicate the direction of the displacement or motion. An arrow used in this way is called a vector. DeÞning Vectors D e f i n i t i o n A vector has two deÞning characteristics: length (or magnitude) and direction. Avectorisoftendenotedbyv,aboldfaceletterwithanarrowaboveit,andrepresented by an arrow. The length of v is denoted by v. Vectors can be used in a large variety of situations; for example, vectors can be used to model velocity, force, and displacement. A velocity vector is a vector whose length represents speed and whose direction indicates the direction of motion. A force vector is a vector whose magnitude corresponds to the magnitude of the force and whose direction indicates the direction of application of the force. We can give a similar interpretation for a displacement vector. We refer to the head of the vector (the tip of the arrow) and the tail of the vector (the other end). A vector represented by an arrow whose tail is at 1, 2 and whose head is at 1, 2 can be denoted by . 672 CHAPTER 20 TrigonometryÑCircles and Triangles vector head tail B = (b1, b2) |b2 —a2| A = (a1, a2) (i) Figure 20.46 |b1 —a1| (ii) Using the Pythagorean Theorem we calculate the length of as 2 22 1 12. Twovectors,uandvareequivalentiftheyhavethesamemagnitudeandthesamedirection. For example, all the vectors drawn in Figure 20.47 are equivalent. y 2 1 —2 —1 1 2 x 1 2 Figure 20.47 The simplest way of determining that these vectors are all equivalent is to notice that each is constructed (from tail to head) by a displacement of 2 units in the horizontal direction followed by a displacement of 1 unit in the vertical direction. We say that each of these vectors has a horizontal component of 2 and a vertical component of 1. The horizontal and vertical components of a vector together completely determine the vector. They determine its direction, and its magnitude is given by horizontal component2 vertical component2. Horizontal and Vertical Components of Vectors Consider the vector , where 0, 0 and is a point on the unit circle. The length of is 1. LetÕs suppose that makes an angle of with the positive -axis, where [0, 2] and is swept out counterclockwise from the positive -axis. The coordinates of are (cos , sin ), where cos is the horizontal component of and sin is the vertical component of . 20.7 A Brief Introduction to Vectors 673 y 1 O 1 x P Figure 20.48 Given any vector v of length we can position v so that its tail is at the origin. Its head lies somewhere on a circle of radius centered at the origin, as shown in Figure 20.49. Let be the angle v makes with the positive -axis, where [0, 2]. The horizontal component of v cos , and the vertical component of v sin . y L Lsin x Lcos Figure 20.49 If the tail of a vector is at the origin, the horizontal component is the -coordinate of the head and the vertical component is the -coordinate of the head. EXERCISE 20.10 is the displacement vector for a turtle. Its horizontal component is 3 and its vertical component is 4. (a) How far has the turtle moved? (b) If the turtle started at 1, 3, where did it end up? ANSWERS (a) 5 units (b) 4, 1 674 CHAPTER 20 TrigonometryÑCircles and Triangles EXERCISE 20.11 v is the velocity vector at a certain instant for an object in motion. At this instant the object is traveling southwest at 50 miles per hour. Let due east correspond to the positive -axis. What are the horizontal and vertical components of the velocity? 50 50 45 v vertical component horizontal component Figure 20.50 EXAMPLE 20.15 SOLUTION A wheel 60 inches in diameter is oriented vertically and spinning steadily at a rate of 1 revolution every 2 minutes. ThereÕs a piece of gum stuck on the rim. Let be the position vector pointing from the center of the wheel to the wad of gum. At time 0, the vector is pointing in the same direction as the positive -axis. Find a formula for , the vertical component of . This is simply the Ferris wheel question formulated using different language. G(t) 30 30 sin t t Figure 20.51 Since 00, a formula of the form 30 sin can be used. One revolution is completed in 2 minutes, so the period is 2. 2 2 4. 30 sin4 The purpose of Example 20.15 is to show that we already know how to Þnd the vertical and horizontal components of a vector. Given a vector (such as a velocity vector, force vector, or displacement vector), physicists, engineers, and architects often Þnd it useful to resolve15 the vector into its vertical and horizontal components. The next example asks for the resolution of a force vector into its horizontal and vertical components. EXAMPLE 20.16 A suitcase is being pulled along a horizontal airport ßoor. A force of 50 pounds is being applied at an angle of 35 with the horizontal. (a) What is the component of the force in the direction of motion? (b) What is the component of the force perpendicular to the direction of motion? 15 To ÒresolveÓ a vector is to separate it into its constituent parts. 20.7 A Brief Introduction to Vectors 675 SOLUTION (a) Let the component of the force in the direction of motion. cos 35 50 50 cos 35 50 cos 35180 40.96 Approximately 40.96 pounds of force are being applied in the direction of motion. 50 lbs y28.68 lbs 35 x40.96 lbs Figure 20.52 (b) Let the component of the force perpendicular to the direction of motion. sin 35 50 50 sin 35 28.68 Approximately 28.68 pounds of force are being applied perpendicular to the direction of motion. What is the signiÞcance of what we have calculated? WeÕve resolved the force into two perpendicular components. If a force of (50 sin 35) pounds is applied upwards and simultaneously a force of (50 cos 35) pounds is applied in the direction of motion, the sum of these two forces is equivalent to the original force. EXAMPLE 20.17 Consider the trajectory made by some thrown or launched object. The object could be a tennis ball, a javelin, a basketball, or a stone. The exact nature of the trajectory is of critical importancetoathletesandmarksmenalike.Ifweneglectairresistanceandspinandconsider onlytheforceofgravity,thenthetrajectoryoftheobjectisdeterminedbytheinitialvelocity of the object, given by the velocity vector v0. LetÕs denote the length of v0 by 0 (with no arrow above it). 0 is the initial speed. WeÕll denote the angle of launch by , where is in theinterval(0,2).ForsimplicityÕssake, weÕllassumetheobjectislaunchedfromground level. The only force acting on the object is the force of gravity, resulting in a downward acceleration of meters per second squared. (a) Find the vertical position of the object at time . (b) When will the object hit the ground? (c) How far away will the object be from its launching spot when it hits the ground? ... - tailieumienphi.vn