CHAPTER 2: VECTOR COMPOETS DESCRIBE MOTIO I TWO DIMESIOS 2.1 Vector Methods in One Dimension Vectors may be pictured with sketches in which arrows represent quantities such as displacement, force and velocity. Such arrow diagrams can be used to show how vectors can be added, or combined in such a way as to find a single vector (called a resultant vector or just a resultant) that can do the same task as the ones that were combined. Vectors in sketches may be labeled with letters and/or numbers. For collinear vectors (along a single straight line called vectors in one dimension), positive and negative signs are sufficient to give directions. Vectors are added by positioning them tip-to-tail; the resultant then starts at the tail of the first vector added and goes to the tip of the last vector added. d 1 EXAMPLES 1. A car drives 12 km from Fort Saskatchewan to Sherwood Park, then continues 85 km further to Camrose. Sketch the displacements; show and calculate the total displacement. The total displacement the resultant can be thought of as the sum of the two displacements 12 km and 85 km. ote that vector starts at the start of the 12 km vector and ends at the end of the 85 km vector, and that the two added vectors are drawn tip-to-tail. Because these displacements are collinear (along a straight line), arithmetic is sufficient to find the magnitude and direction of : d 2 = d 1 + d 2 = 12 85 = 97 km 2. A cyclist rides 500 m E, then turns around and rides 200 m W. Sketch the displacements; show the total displacement. d 1 = +500 m d 2 = -200 m The total displacement the resultant can be thought of as the sum of the two displacements +500 m and 200 m. Again, the resultant vector starts at the start of the +500 m vector and ends at the end of the 200 m vector, and the two added vectors are drawn tip-to-tail. EXERCISE 2.1 = d 1 + d 2 = +500 + ( 200) = +300 m 1. Sketch each vector sum showing the total displacement or resultant for each of the following; calculate the magnitude and direction of the resultant. a) moving left 15 m, then right 8 m b) moving north 22 m, then south 30 m PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 1
c) moving southeast 15 km, then moving northwest 10 km d) moving up 15 m, then moving down 5.0 m, then moving up 8.0 m. 2.2 Motion in Two Dimensions Vectors that combine north-south directions with east-west directions are described as non-collinear vectors, or vectors in two dimensions. Two-dimensional vectors are added just as collinear vectors, by positioning them tip-to-tail; the resultant then starts at the tail of the first vector added and goes to the tip of the last vector added. EXAMPLE A boy walks east a distance of 40 m, then turns left and walks a further distance of 28 m. Sketch the resultant vector. Although it is still possible to write the vector equation as = d 1 + d 2, YOU CAOT FID BY SUBSTITUTIG VALUES ITO THE EQUATIO. For twodimensional vectors, either an accurate scale d 1 = 40 m E drawing (using a ruler and protractor), or vector components using trigonometry and the theorem of Pythagoras (used only for right-angled triangles) are needed to actually find the magnitude and direction of the resultant. d 2 = 28 m EXERCISE 2.2 1. Using a scale of 1 cm = 4.0 m, carefully add the two vectors from the example above. (Follow the steps on page 81 of your text.) Use your scale to determine the magnitude of the resultant; give the direction of the resultant as a number of degrees north of east. PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 2
2. Choose a suitable scale, then add the following vectors to find the magnitude and direction of the resultant. d 1 = 20 m d 2 = 15 m W d 3 = 10 m S The resultant for question #2 is about 18 m, with a direction of W 35 (35 north of west.) This method of giving a direction is called the navigator method. An alternate way of giving directions is called angles in standard position or the Cartesian (graph-based) method. In this method, the direction of the resultant for question #2 would be given as 145, as shown to the right. To state the direction of a vector using standard position angles, note that: The vector is placed on an x-y axis, with the tail at the origin The positive x-axis is defined as ; note the directions of the other axes at right The angle giving the vector s direction is measured counterclockwise from, and is defined as positive Your calculators are designed to operate using this method; for this and other reasons, standard position angles (the Cartesian method) are more common in applied science, and are your teacher s preference. EXAMPLES 145 35 18 27 y y 9 x x y Direction is 25 y Direction is 315 y Direction is 7 25 315 7 x 0 x 0 x 0 2 45 PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 3
3. Give the standard position angles (Cartesian method) for each of the vectors sketched at right. 3 15. y 52. 5 4. Convert to the opposite notation: navigator method (eg. 15 W) or Cartesian method (a standard position angle eg. 105.) avigator Cartesian avigator Cartesian 15. E S 52. W S 22. E 15 W 67. 1 4 65. 25 88.5 322 55. 2 x 0 In question #1 above (page 2), the resultant is about 49 m with a direction of E 35 (35 north of straight east, or 35 as a standard position angle.) ote that could be equally well described as the sum of the two perpendicular vectors, d 1 = 40 m E and d 2 = 28 m. These two vectors are then called the perpendicular components of, because their sum equals the vector. AY TWO-DIMESIOAL VECTOR CA BE WRITTE AS THE SUM OF TWO PERPEDICULAR VECTOR COMPOETS; that is, the two components can be used instead of the original vector. EXAMPLES Original Vector Vector Components d 2 d 1 ote that d 1, the first component drawn (starting at the tail of ), should always be horizontal. d 1 d 2 PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 4
5. Sketch perpendicular vector components for the following vectors; label the components as d 1 and d 2. Remember that d 1, the first component drawn (starting from the tail of ), should be horizontal. (ote that two of these vectors are already drawn as components.) a) b) c) d) e) f) To calculate component magnitudes, trigonometry is used. Recall that for right-angled triangles, the ratios of sine, cosine and tangent can be used to solve for unknown sides and angles. opposite sine θ = hypotenuse adjacent cosine θ = hypotenuse tangent θ = opposite adjacent Right triangle solutions may also require the Theorem of Pythagoras: SOH CAH TOA c 2 = a 2 + b 2 c is longest side hypotenuse θ adjacent opposite Finding angles in any triangle may be done with the sum of interior angles theorem: angle sum = 18 For right triangles, this implies that the two acute angles have a sum of 9. EXAMPLES PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 5
1. Find the perpendicular components of a displacement of 20 m at an angle of 6. sin 60 = d y 20 d y = 20 sin 60 d y = +17 m or 17 m cos 60 = d x 20 d x = 20 cos 60 d x = +10 m or 10 m E 20 m 6 d y d x ote that in general, for any vector A with a standard position angle of θ: A x = A cos θ A y = A sin θ For this to work dependably, A x must be the first vector added; i.e., A x must be drawn with its tail starting at the tail of the given vector A. 2. Find the perpendicular components of a displacement of 35.0 km at an angle of 21. 21 sin 210 = d y 35.0 d y = 35.0 sin 210 d y = 17.5 km or 17.5 km S cos 210 = d x 35.0 d x = 3 5. 0 cos 210 d x = - 3 0. 3 km or 30.3 km W d y d x 35.0 km OTE THAT USIG THE STADARD POSITIO AGLE OF 21 CORRECTLY GIVES THE EGATIVE SIG FOR d y. Using 3 would give the correct magnitude value, but the negative sign showing that the component is directed to the south would have to be attached separately. 6. Find perpendicular x and y vector components for each of the following vectors. Include a sketch for each question. a) A displacement of 220 cm in a direction of 65. b) A displacement of 900 km in a direction of 12 (93.0 cm; 199 cm) PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 6
(-450 km; 779 km) c) A velocity of 4.50 m/s in a direction of 25. (Don t worry that this is a velocity vector the same methods work equally well for all vectors.) (-1.54 m/s; -4.23 m/s) d) A velocity of 48.0 km/h in a direction of 33 (41.6 km/h; -24.0 km/h) If two component vectors are known, the preceding techniques can be used to find an actual vector that is the sum of given components. See text p. 85. EXAMPLES 1. A displacement vector has d x = 22.0 m and d y = +15.0 m. Find the magnitude and direction of the actual vector d. d d y = 15.0 m This is a right-angled triangle, so the Pythagorean Theorem can be used to find the magnitude of d; inverse tan is used to find the standard position angle θ: θ d x = 22.0 m d = ( 15.0) 2 + ( 22.0) 2 d = 26.627 m d = 26.6 m tan θ = 15.0 22.0 θ = 34.287 θ = 34.3 The absolute value signs around d show that only the magnitude is being found the direction has to be found separately. PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 7
2. Two forces act perpendicularly on an object. One force pushes due west with a magnitude of 6.00, and the second force pushes due south with a magnitude of 8.00. Find the magnitude and direction of the resultant force. 6.00 θ This question doesn t talk about components, but the approach to adding two perpendicular vectors is the same as adding two perpendicular components. Because the resultant force has a direction greater than 18 (as seen from the sketch), a reference angle β inside the right-angled triangle must be found before the standard position angle θ is determined. 8.00 β F F = ( 6.00) 2 + ( 8.00) 2 F = 10.00 F = 10.0 8.00 tan β = θ = 180 + 53.13 6.00 β = 53.13 θ = 233.1 θ = 180 + β θ = 233 7. Use the x and y components given to find the magnitude and direction of the resultant vector. a) F x = +120 ; F y = -75.0 b) d x = -340 km; d y = +225 km (142 at 328 ) c) v x = -55.0 m/s; v y = -38.0 m/s (408 km at 147 ) (66.9 m/s at 215 ) PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 8
8. List the four steps for the algebraic addition of non-collinear vectors. (Text p. 86.) 9. Complete the question below to find the resultant of two displacement vectors: d 1 = 510 km at 55. and d 2 = 380 km at 165. (Sketches are only approximately to scale.) x - components y - components d 1 55. d 1y d 1x = 510 cos 55. = d 1y = 510 sin 55. = d 1x d 2y d 2 165 d 2x = 380 cos 165 = d 2y = 380 sin 165 = d 2x Σd x =d 1x + d 2x = Σd y =d 1y + d 2y = Σ d y β d Σ d x θ d = β = θ = The resultant displacement is at an angle of. β is the Greek letter beta; θ is the Greek letter theta. Greek letters are often used to refer to angles. Σ is the Greek letter sigma; it means sum of. Σ d x means sum of the x-components; Σ d y means sum of the y-components. PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 9
10. Use the techniques and examples in the text on pages 84-89 to find the magnitude and direction of the resultant of each pair of vectors below. ote on example 2.5 on page 88: the sketches at the bottom of the page would be better with the horizontal components drawn on the x-axis. d 1 is shown correctly at right; redraw the sketch for d 2 correctly. d 1x d 1y d 1 4 20 m y 22 x a) A force of 12.0 at 75. and a force of 28.0 at 315. (Your text would give these forces as 120 [75.] and 28.0 [315 ]. Either notation is fine for the direction.) (24.3 at 34) b) A velocity of 75.0 km/h in a direction of 80. and a velocity of 75.0 km/h in a direction of 355 (111 km/h at 37.5 ) PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 10
c) A displacement of 55.0 m in a direction of 13 and a displacement of 75.0 m in a direction of 25. 2.3 Relative Motion (67.3 m at 205 ) All velocities are measured relative to something; usually, the reference is a stationary point attached to the earth. Relative motion problems usually involve two velocities that when combined determine an actual velocity with respect to the earth, or involve determining the velocity of one moving object from the point of view of a second also-moving object. Examples of the second type include airplanes flying through moving air (a wind), and canoeists or swimmers traveling across moving streams. These are examples of relative motion - motion measured with respect to an observer, who may also be moving. See text pp. 91-96. Omit examples 2.7-2.9; we will only consider relative velocity problems in two dimensions where right triangles are involved. Collinear relative motion is illustrated by questions #1-6 below. EXERCISE 2.3 1. From the point of view of a stationary observer, what is the velocity of each of the following? a) an unsecured lawnmower rolling in the box of a passing truck; the truck is traveling north at 20 m/s (ground velocity), and the mower is rolling south at 2.0 m/s with respect to the truck b) a pop can thrown by an inline skater moving east at 5.5 m/s (ground velocity); the can is thrown over the skater s shoulder and moves west at 8.0 m/s with respect to the skater c) a passenger walks toward the front of a bus at 1.0 m/s with respect to the bus; the bus is moving west at 30 m/s (ground velocity) d) the same passenger as in (c) if he runs towards the rear of the bus at 2.5 m/s with respect to the bus e) a ball thrown forward by a man running north at 4.0 m/s (ground velocity); the ball leaves his hand at 25 m/s, with respect to his hand f) the ball in question (e) if the man drops the ball so that it lands at his (moving) feet g) a passenger in an airplane that is taxiing south at 1.1 m/s (ground velocity); the passenger walks toward the rear of the airplane at 1.1 m/s with respect to the plane PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 11
The following questions deal with relative motion through water or air. ote that air velocity (or water velocity) is the velocity an airplane (or boat or swimmer) would have through still air (across a still body of water.) Wind velocity (current or stream velocity) is the motion of the air with respect to the ground (the motion of a stream or river with respect to its banks), and ground velocity is the motion of an object like an airplane or boat with respect to a stationary observer on the ground. 2. An airplane with an air velocity of 160 km/h flies north. The plane encounters a headwind (wind blowing in the opposite direction to the plane s velocity) with a velocity of 45.0 km/h. What is the ground velocity of the airplane? (This headwind would be correctly described as a north wind - a wind blowing from north to south.) 3. What is the ground velocity of the airplane in question #2 if the wind it encounters is a tailwind (wind blowing in the same direction as the plane s velocity)? 4. How long would the plane in question #3 take to fly from Calgary to Edmonton (about 300 km)? 5. A canoeist who can paddle at 2.40 m/s in still water points his canoe downstream in a creek that flows east at 1.50 m/s. What is the canoeist s ground velocity? How far downstream will the canoeist be after 45.0 s? 6. A swimmer who is able to swim in still water at 1.00 m/s swims upstream against a current of 0.500 m/s in a stream that flows south. What is the swimmer s ground velocity? How long would the swimmer take to move 300 m north from her starting point? EXAMPLES O-COLLIEAR RELATIVE MOTIO 1. An airplane heads due north; the plane has an airspeed of 410 km/h. The plane encounters a wind from the west of 95.0 km/h. Find the ground velocity of the airplane. OTE: Heads due north means the body of the plane as seen by a stationary observer is pointing due north, as sketched at right. But because of the west wind, the plane s actual motion over the ground will be somewhere east of north as seen in the solution. Airplane points this way Due to wind, airplane travels this way In vector terms, the sum of the plane s air velocity (v a ) plus the wind velocity (v w ) will equal the plane s ground velocity (v g ). PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 12
Draw a vector sketch; use Pythagoras and tangent to find the magnitude and direction of the ground velocity. v g = + 410 2 2 410 95.0 tan θ = 95.0 v g θ v a v g = 420.86 km/h θ =76.95 v w The airplane s ground velocity is 421 km/h at 77.. 2. A canoeist who can paddle at 3.50 m/s in still water leaves the south bank dock at point A and aims his canoe straight across a river that flows west at 2.00 m/s. As shown in the sketch, the river is 35.0 m wide. The canoeist lands at point B on the opposite bank. OTE THAT THE CAOE IS POITIG STRAIGHT ACROSS THE STREAM BUT BECAUSE OF THE CURRET, ITS MOTIO DIRECTIO IS AT A SLIGHT AGLE DOWSTREAM B 3.50 m/s A C DO T MIX THIS DISTACE THE WIDTH OF THE RIVER ITO YOUR VECTOR ADDITO PROBLEM FOR VELOCITIES 2.00 m/s 35.0 m OT TO SCALE a) Find the ground velocity v g (with respect to the earth (bank of the river) magnitude and direction) of the canoeist. v g v g = + 3.50 2 2 3.50 2.00 tan β = 2.00 v w β θ v g = 4.031 m/s β =60.26 θ =119.7 v c The canoe s ground velocity is 4.03 m/s at 12. b) Find the time needed to cross the river. (ote that this time depends only on the velocity straight across and requires the distance straight across.) = v t t = 35.0 3.50 t = v t =10.0 s The canoe takes 10.0 s to cross the river. c) Find how far downstream the canoeist will land. During the time he is in the water, the canoeist will be carried downstream by the current; distance downstream depends only on time in water and velocity of current. = v t = (2.00)(10.0) = 20.0 m The canoe s landing point (B) is 20.0 m downstream. PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 13
d) In what direction must the canoe point to land at point C, directly across the river? How long would the crossing take? ote from the vector sketch that the ground velocity (the velocity sum that gives his actual motion) must now point straight across, and that the paddling direction is no longer specified logically, the canoeist will have to point his canoe a bit upstream to counteract the effect of the current and allow his ground velocity to be straight across the stream. The ground velocity is the vector sum of the canoe s water velocity and the stream velocity. v w v c θ v g The negative sign appears because you are using Pythagoras (c 2 = a 2 + b 2 ) solved for b (the ground velocity.) v v g g = 2.00 2 2 3.50 2.00 cos θ = 3.50 = 2.872 m/s θ =55.15 The triangle sides used to find θ are the adjacent side (v c ) and the hypotenuse (v w ); cosine is the trig function chosen. To find the new (longer) crossing time, use the new ground velocity together with the width of the stream. = v t t = 35.0 2.872 t = t = 12.20 s v The new ground velocity to land at C is 2.87 m/s at 55.2. The new time would be 12.2 s. 7. A bird flies with an airspeed of 12.5 m/s straight east. The bird encounters a wind from the south of 8.50 m/s. Find the bird s ground velocity (magnitude and direction.) (15.1 m/s at 34.2 ) 8. The bird in question #7 decides it wants to fly with a ground velocity directed straight east. What is the new direction of the bird s airspeed velocity? (317 ) 9. How long would the bird, flying as in question #8, need to cross a field 300 m wide? (32.7 s) PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 14
2.4 Projectile Motion Any non-powered object moving above the earth's surface is a projectile; examples include thrown or kicked balls, arrows, water drops from sprayers, bullets from guns, etc. Projectiles follow a path called a trajectory. A projectile can be launched horizontally (as shown to the right) or with an initial upward or downward velocity. Your text in this section considers both types of projectiles; we will only work with horizontal projections (zero initial vertical velocity), as covered on pages 102-108. Initial vertical velocities equal zero Vertical velocities increase due to acceleration of gravity; same for both objects. Vertical displacement is y (a negative value) Horizontal range is x Horizontal velocity for projected object remains constant. The motion of a projectile can MOTIO OF PROJECTED AD FREE-FALLIG OBJECTS be analyzed using the independence of horizontal and vertical motions. These perpendicular motions don't affect each other; they can be examined separately using the vector equations of linear motion. (ote that as a general rule, any two perpendicular vectors do not affect each other.) Generally, neglecting friction: the vertical motion of a projectile is accelerated at -9.81 m/s 2 (acceleration due to gravity, or g) for a horizontally projected object, the initial vertical velocity is zero horizontal velocity v x of a projectile is constant once it is launched finding and using time of flight is often the key to solving these problems Three motion equations can be used to solve projectile problems with horizontal projection: x = v x t and y = v i t + 1 2 a ( y t)2 for vertical displacement, and v yf = v yi + a y t to determine fine vertical velocity. Because v yi (initial vertical velocity) is zero, these become: x = v x t, y = 1 2 a ( y t)2 and v yf = a y t. Again, note that time of flight ( t) is a common variable in both equations; finding t is often the key to solving a projectile problem. Also, the value of a y will be the acceleration of gravity. EXAMPLE A ball is thrown horizontally at 6.00 m/s from the top of a 22.0 m high building. 1. Determine how long the ball is in the air. To find the time of flight t, use the vertical displacement equation and the acceleration due to gravity; solve for t. The time of flight is 2.12 s. 2. Find how far from the base of the building the ball lands. y = 1 2 a y( t) 2 t = 2(-22.0) 9.81 t = 2 y a y t = 2.118 s PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 15
During the time the ball is in the air, it is traveling horizontally at 6.00 m/s. Its horizontal range x is found using the time of flight from part (a), together with this horizontal velocity. x = v x t x = (6.00)(2.118) x =12.71 m The ball lands at 12.7 m from the base. 3. Determine the ball's vertical velocity component just before it strikes the ground. Time of flight is used with the acceleration of gravity to find v yf. v yf = a y t v yf = ( 9.81)(2.118) v yf = 20.78 m/s The vertical velocity component just above the ground is -20.8 m/s. 4. Find the ball s actual velocity (that is, the sum of its vertical and horizontal components magnitude and direction, as an angle measured from horizontal) just before it strikes the ground. 2 2 20.78 v = 6.00 + 20.78 tan θ = 6.00 v = 21.63 m/s θ =73.89 The actual velocity is 21.6 m/s, at an angle of 73.9 below horizontal. v v x θ v yf EXERCISE 2.4 1. A rock is thrown at 12.5 m/s horizontally from the top of an observation tower, and lands 3.40 s later. How high is the tower, and how far from the base of the tower does the rock land? (56.7 m high; 42.5 m from base) 2. What is the actual velocity of the rock in question #1 just before it lands? Give a direction measured from ground level. (35.6 m/s at 69.5 from ground) PHYSICS 20 OTES AD OUTLIE QUESTIOS CHAPTER 2 REVISED JAUARY 08 PAGE 16