COMPONENTS OF VECTORS

Transcription

1 COMPONENTS OF VECTORS To describe motion in two dimensions we need a coordinate sstem with two perpendicular aes, and. In such a coordinate sstem, an vector A can be uniquel decomposed into a sum of two perpendicular vectors A = A + A where A is parallel to the ais while A is parallel to the ais, for eample A A A A (1) A Since the A vector is alwas parallel to the ais, we ma describe it b a single signed number A, which is positive when A points right but negative when A points left. Likewise, the A vectors ma be described b a single signed number A positive when A points up but negative when A points down. The two signed numbers A and A are called the componentsof the vector A. In two dimensions, an vector V can be completel specified b its components (V,V ). Describing motion in all 3 dimensions of space requires a coordinate sstem with 3 perpendicular aes (,,z). Consequentl, a 3D vector V has three components (V,V,V z ), and we need to know all 3 components to completel specif the vector. 1

2 Adding vectors in components. In component notations, adding vectors is ver eas: The components of a vector sum C = A+ B are simpl the algebraic sums C = A + B, C = A + B. (2) Here is the geometric eplanation of this rule: C B C B B (3) C A A A In the same wa, we ma sum up several vectors: To get the components of a vector sum C = A 1 + A 2 + A A n, (4) we separatel sum up the components of all the vectors and the components of all the vectors: C = A 1 + A 2 + A A n, (5) C = A 1 + A 2 + A A n. Note: the sums here are algebraic, so please mind the ± signs of the components. 2

3 For the 3D vectors, there are similar formulae, but there is one more algebraic sum for the z components: C = A 1 + A 2 + A A n, C = A 1 + A 2 + A A n, (6) C z = A 1z + A 2z + A 3z + + A nz. Vector subtraction in components works similar to vector addition. To get B = C A (7) in components, subtract the components of A from the components of C, B = C A, B = C A, (8) in 3D also B z = C z A z. Conversion from magnitude and direction to components. A vector quantit V has magnitude and direction. On a graph the magnitude is shown b the length of the arrowed line; algebraicall, the magnitude is a non-negative number V. In a 2D plane, the direction of a vector can be specified b the angle φ v it makes with the ais, for eample V φ v (9) Now let us draw a similar diagram which also includes the components (V,V ) of the 3

4 vector V: V V (10) φ v V Note the right triangle made b the three red lines; taking the ratios of this triangle s sides and appling basic trigonometr, we immediatel obtain V V = cosφ v, V V = sinφ v, (11) and therefore V = V cosφ v, V = V sinφ v. (12) The triangle on the diagram (10) is drawn for direction of Vin the first quadrant of the coordinate sstem (between the positive and positive direction, φ v < 90 ), but the formulae (12) for the components work for all possible directions, provided we alwas measure the angle φ v counterclockwise from the positive ais. For eample, for V in the second quadrant 90 < φ v < 180, V = V cosφ v < 0, V V = V sinφ v > 0, V φ v V (13) 4

5 Likewise, for V in the third quadrant 180 < φ v < 270, V = V cosφ v < 0, V = V sinφ v < 0, φ v V V V (14) or for V in the fourth quadrant 270 < φ v < 360, V = V cosφ v > 0, V = V sinφ v < 0, φ v V V V (15) 5

6 Conversion from components to magnitude and direction. Now supposed we know the (V,V ) components of a vector V; how do we find the vector s magnitude and direction? The magnitude V follows from the Pthagoras theorem for the right triangles on an of the diagrams on the last two pages: V 2 = V 2 + V 2 (16) regardless of the signs of the V and V and therefore V = V 2 + V 2. (17) To find the direction of V, we need a bit of trigonometr. Let s take the ratio of the two equations (12) for the components (V,V ): V V = V sinφ v V cosφ v = sinφ v cosφ v = tanφ v. (18) Thus the ratio V /V c gives us the tangent of the angle φ v, so naivel we ma calculate the angle φ v itself as the arc-tangent (the inverse tangent) of this ratio, φ v?? = arctan V V. (19) However, the formula ma be off b 180, so it might give us precisel the opposite direction. Indeed, the vectors V and V have opposite directions but similar ratios This ambiguit is related to the trigonometric identit V V = V V. (20) for an angle ϕ, tan(ϕ) = tan(ϕ±180 ). (21) 6

7 Therefore, given the components of a vector, its direction is φ v = either arctan V V or arctan V V ± 180, depending on the signs of the V and V. (22) Using conversion to add vectors. Consider a simple problem: A person rides a bike for 10.0 km in the direction 30 (counterclockwise from the ais), then changes direction to 150 (also counterclockwise from the ais) and rides for 20.0 km. Find his net displacement vector. D 2 D net = D 1 + D 2 D net (23) D 1 To solve this problem, we start b converting the displacement vectors D 1 and D 2 into components: D 1 = 10.0 km cos(30 ) = km, D 1 = (10.0 km;30 ) = D 1 = 10.0 km sin(30 ) = km, (24) D 2 = 20.0 km cos(150 ) = km, D 2 = (20.0 km;150 ) = D 2 = 20.0 km sin(150 ) = km. Net, we add the two vectors in components: D net D net = D 1 + D 2 = km km = 8.66 km, = D 1 + D 2 = km km = km. (25) Finall, we convert the components of the net displacement vector into its direction and 7

8 magnitude. For the magnitude we have and hence As to the direction, D net 2 = ( D net ) 2 ( ) + D net 2 = ( 8.66 km) 2 + ( km) 2 = 75 km km 2 = 300 km 2 (26) D net = 300 km km. (27) D net D net = km 8.66 km Dnet = 1.73 = arctan D net = 60. (28) The 60 angle is the same as = +300, which correspond to the direction of D net being in the fourth quadrant. However, the signs of the components D net < 0, D net > 0 show that the direction of D net is in the second quadrant. This means that the arc-tangent is off b 180, so the correct direction of the net displacement vector is φ net = = +120 (counterclockwise from the ais). (29) The above eample illustrate a general rule for calculating sums of several vectors, A net = A 1 + A A n. (30) (1) First, convert all the vectors into components, A i, = A i cosφ i, A i, = A i sinφ i, for i = 1,2,...,n. (31) (2) Second, add the vectors in components, A net = A 1, + A 2, + + A n,, A net = A 1, + A 2, + + A n,. (32) (3) Finall, convert the components of the A net into its magnitude and direction. 8

9 Navigation Convention In navigation, the directions are given b angles counted clockwise from North instead of counterclockwise from the ais (whatever that might be). For eample, an airplane fling in the Southwest direction which is 225 clockwise from North is said to have heading 225. N 225 W E (33) v plane S To avoid confusion when ou work a navigation-related problem, it is best to avoid the and aes altogether and use the North and East aes instead of them. In particular, for the vectors ou should use the North and East components instead of the and components. In this convention, V N = V cosφ v, V E = V sinφ v, for φ v counted clockwise from North. (34) Foreample, foraplaneflingatspeedv = 220MPHintheSouthwest direction(φ v = 225 ), the velocit vector v has components v N = 220 MPH cos225 = 155 MPH, v E = 220 MPH sin225 = 155 MPH. (35) Note negative signs of the components since the plane is fling South and West instead of North and East. 9

10 Multipling vectors b scalars. The product of a scalar s and a vector V is a vector s V. Its magnitude is the absolute value of s times the magnitude of V, s V = s V. (36) The direction of the product sv is the same as direction of V for positive s but opposite from the direction of V for negative s. (For s = 0 the product sv = 0 the zero vector and its direction is undefined.) In components, B = sa has B = s A, B = s A, and in 3D also B z = s A z. (37) The product of a vector and a scalar obes the usual algebraic rules for opening parentheses: s ( V1 + V ) 2 = sv1 + sv 2, s ( V1 V ) 2 = sv1 sv 2, (s 1 +s 2 ) V = s 1 V + s2 V, (s 1 s 2 ) V = s 1 V s2 V, (38) ( ) s 1 s2v = (s1 s 2 ) V,... Phsical Eample: For a motion at constant velocit vector v i.e., motion at constant speed in a constant direction the displacement vector after time t is given b D = t v. (39) Interms ofthe time-dependent positionvector (AKAradius-vector) R(t) ofthemoving bod whose components R (t) and R (t) are simpl the time-dependent coordinates (,) of 10

11 the bod the displacement vector is the vector difference D = R = R(t) R 0, (40) so eq. (39) becomes R(t) = R 0 + t v. (41) In components, this formula means (t) = 0 + t v and (t) = 0 + t v (42) uniform motion in both and directions. Division: You cannot divide a scalar b vector, or a vector b another vector. However, ou can divide a vector b a scalar simpl multipl the vector b the inverse scalar, V s def = 1 s V. (43) For eample, to find the average velocit vector of some bod, divide its displacement vector b the time this displacement took, v avg = R t. (44) In the limit of a ver short time interval, this formula gives ou the instantaneous velocit vector v(t) = lim t 0 R(t+ t) R(t). (45) t In general, the velocit vector changes with time, which leads to the acceleration vector v(t+ t) v(t) a(t) = lim. (46) t 0 t 11

12 Motion at constant acceleration. Consider motion of some bod having a constant acceleration vector a. (Neither direction nor magnitude of a changes with time.) The velocit vector of such a bod changes with time according to v(t) = v 0 + t a (47) where v 0 is the initial velocit vector at time t 0 = 0. Note: eq. (46) looks simple in vector notations, or in components v (t) = v 0 + t a, v (t) = v 0 + t a, (48) but it leads to rather complicated formulae for the time dependence of the speed v(t) and the direction of motion. The time-dependent position vector R(t) of a uniforml accelerating bod is given b R(t) = R 0 + t v t2 a, (49) or in components (t) = 0 + t v t2 a, (t) = 0 + t v t2 a. (50) Projectile Motion A projectile is an object ou shoot, kick, throw, or otherwise send fling towards a target, for eample a basketball, a bullet, or a grenade. In phsics, projectile motion is a motion of a bod that has been released with some initial velocit (which generall has both horizontal and vertical components) and then flies free from forces other than gravit and air resistance. When the air resistance ma be neglected which is the onl case we shall stud in this class the projectile has a constant acceleration vector a = g due to gravit. Consequentl, the projectile s velocit vector v(t) and position vector R(t) evolve with time according to eqs. (47) through (50). 12

13 A projectile moves in a vertical plane, so its motion can be described using 2D vectors and their components. Let s use a coordinate sstem where the ais points verticall up while the ais is horizontal. (In 3D, the ais points along the horizontal components of the initial velocit.) In these coordinates, the downward acceleration vector a = g has components a = 0, a = g. (51) Consequentl, eqs. (48) and (50) become v (t) = v 0 = const (time independent), (t) = 0 + v 0 t, v (t) = v 0 g t, (52) (t) = 0 + v 0 t g 2 t2. Note that the first two of these equations which describe the horizontal motion are completel independent from the last two equations describing the vertical motion. Thus, the projectile moves horizontall at constant velocit as if there were no vertical motion, and at the same time it moves verticall up and down at constant acceleration as if there were no horizontal motion! 13