Transcription
Chapter 3 - VectorsI.DefinitionII. Arithmetic operations involving vectorsA) Addition and subtraction- Graphical method- Analytical method Vector componentsB) Multiplication
Review of angle reference system90º0º θ1 90º90º θ2 180ºθ2θ1180ºθ3180º θ3 270º0ºOrigin of angle reference systemθ4270º θ4 360º270ºAngle originΘ4 300º -60º
I. DefinitionVector quantity: quantity with a magnitude and a direction. It can berepresented by a vector.Examples: displacement, velocity, acceleration.Same displacementDisplacement does not describe the object’s path.Scalar quantity: quantity with magnitude, no direction.Examples: temperature, pressure
II.Arithmetic operations involving vectorsVector addition: s a b- Geometrical method a b s a bRules: a b b a(commutative law) (a b ) c a (b c )(3.1)(associative law)(3.2)
Vector subtraction: d a b a ( b )(3.3)Vector component: projection of the vector on an axis.a x a cos (3.4)a y a sin a a x2 a 2y Scalar components of aVector magnitude(3.5)tan ayaxVector direction
Unit vector:Vector with magnitude 1.No dimensions, no units.iˆ, ˆj , kˆ unit vectors in positive direction of x, y, z axes a a x iˆ a y ˆj(3.6)Vector componentVector addition:- Analytical method: adding vectors by components. r a b (a x bx )iˆ (a y by ) ˆj(3.7)
Vectors & Physics:-The relationships among vectors do not depend on the location of the origin ofthe coordinate system or on the orientation of the axes.- The laws of physics are independent of the choice of coordinate system.a a x2 a 2y a'2x a '2y(3.8) ' Multiplying vectors:- Vector by a scalar: f s a- Vector by a vector:Scalar product scalar quantity(dot product) a b ab cos a x bx a y by a z bz(3.9)
Rule: a b b a a b ab cos 1 ( 0 ) a b 0 cos 0 ( 90 )(3.10) i i j j k k 1 1 cos 0 1 i j j i i k k i j k k j 1 1 cos 90 0 a bcos a bAngle between two vectors:Multiplying vectors:- Vector by a vectorVector product vector(cross product) a b c (a y bz by a z )iˆ (bz a x a z bx ) ˆj (a x by bx a y )kˆc ab sin Magnitude
a b 0 sin 0 ( 0 ) a b ab sin 1 ( 90 )Vector productDirection right hand ruleRule: b a ( a b )(3.12) c perpendicular to plane containing a , b1) Place a and b tail to tail without altering their orientations.2) c will be along a line perpendicular to the plane that contains a and bwhere they meet.3) Sweep a into b through the smallest angle between them.
Right-handed coordinate systemzkijyxLeft-handed coordinate systemzkjyix
i i j j k k 1 1 sin 0 0 i i j j k k 0 i j ( j i ) k j k ( k j ) i k i (i k ) j
ˆ the result is a vector in the positive direction of the y axis, with aP1: If B is added to C 3iˆ 4j,magnitude equal to that of C. What is the magnitude of B?Method 2Method 1Isosceles triangle B C B (3iˆ 4 ˆj ) D DˆjC D 32 4 2 5 B (3iˆ 4 ˆj ) 5 ˆj B 3iˆ ˆj B 9 1 3.2θCDtan (3 / 4) 36.9 B / 2sin B 2 D sin 3.2D 2 2 BP2: A fire ant goes through three displacements along level ground: d1 for 0.4m SW, d2 0.5m E, d3 0.6m at60º North of East. Let the positive x direction be East and the positive y direction be North. (a) What are thex and y components of d1, d2 and d3? (b) What are the x and the y components, the magnitude and the directionof the ant’s net displacement? (c) If the ant is to return directly to the starting point, how far and in what directionshould it move?(a)d1x 0.4 cos 45 0.28mND45ºd4 d3d1Ed1 y 0.4 sin 45 0.28md 2 x 0.5md2 y 0d 3 x 0.6 cos 60 0.30md2d 3 y 0.6 sin 60 0.52m(b) d 4 d1 d 2 ( 0.28iˆ 0.28 ˆj ) 0.5iˆ (0.22iˆ 0.28 ˆj )m D d 4 d 3 (0.22iˆ 0.28 ˆj ) (0.3iˆ 0.52 ˆj ) (0.52iˆ 0.24 ˆj )mD 0.52 2 0.24 2 0.57m 0.24 24.8 North of East 0.52 tan 1 (c) Return vector negative of net displacement,D 0.57m, directed 25º South of West
P2 d1 4iˆ 5 ˆj 6kˆ d 2 iˆ 2 ˆj 3kˆ d 4iˆ 3 ˆj 2kˆ3 (a) r d1 d 2 d 3 ? (b) Angle between r and z ? (c) Component of d1 along d 2 ? (d ) Component of d1 perpendicular to d 2 and in plane of d1 , d 2 ? (a ) r d1 d 2 d 3 (4iˆ 5 ˆj 6kˆ) ( iˆ 2 ˆj 3kˆ) (4iˆ 3 ˆj 2kˆ) 9iˆ 6 ˆj 7kˆ 7 (b) r kˆ r 1 cos 7 cos 1 123 12.88 d1perpr 9 2 6 2 7 2 12.88m d1 d 2(c) d1 d 2 4 10 18 12 d1d 2 cos cos d1d 2 d d 12d1// d1 cos d1 1 2 3.2md1d 23.74d1θ d1//d2d 2 12 2 2 32 3.74m(d ) d1 d12// d12perp d1 perp 8.77 2 3.2 2 8.16md1 4 2 52 6 2 8.77 mP3 If d1 3iˆ 2 ˆj 4kˆ d 2 5iˆ 2 ˆj kˆTip: (d1 d 2 ) (d1 4d 2 ) ?Think before calculate !!! (d1 d 2 ) a contained in (d1 , d 2 ) plane (d1 4d 2 ) 4(d1 d 2 ) 4b perpendicular to (d1 , d 2 ) plane a perpendicular to b cos 90 0 4a b 0
P4:Vectors A and B lie in an xy plane. A has a magnitude 8.00 and angle 130º; B hascomponents Bx -7.72, By -9.20. What are the angles between the negative direction ofˆthe y axis and (a) the direction of A, (b) the direction of AxB, (c) the direction of Ax(B 3k)?yA (a) Angle between y and A 90 50 140 130ºxB (b) Angle y , ( A B) C angle ˆj , kˆ because C perpendicular plane ( A, B) ( xy ) 90 (c) Direction A ( B 3kˆ) D E B 3kˆ 7.72iˆ 9.2 ˆj 3kˆiˆ D A E 5.14ˆj6.13kˆ0 18.39iˆ 15.42 ˆj 94.61kˆ 7.72 9.20 3D 18.39 2 15.42 2 94.612 97.61 ˆj D ˆj (18.39iˆ 15.42 ˆj 94.61kˆ) 15.42 ˆj D 15.42 cos 97.61 991D
P5: A wheel with a radius of 45 cm rolls without sleeping along a horizontal floor. At time t1 the dot P paintedon the rim of the wheel is at the point of contact between the wheel and the floor. At a later time t2, thewheel has rolled through one-half of a revolution. What are (a) the magnitude and (b) the angle (relativeto the floor) of the displacement P during this interval?yVertical displacement: 2 R 0.9mHorizontal displacement:1(2 R) 1.41m2d r (1.41m)iˆ (0.9m) ˆj r 1.412 0.9 2 1.68mx 2R tan 32.5 R P6: Vector a has a magnitude of 5.0 m and is directed East. Vector b has a magnitude of 4.0 m and is directed35º West of North. What are (a) the magnitude and direction of (a b)?. (b) What are the magnitude anddirection of (b-a)?. (c) Draw a vector diagram for each combination.-abN125ºb-aaWa bE a 5iˆ b 4 sin 35 iˆ 4 cos 35 ˆj 2.29iˆ 3.28 ˆj (b) b a b ( a ) 7.29iˆ 3.28 ˆj(a) a b 2.71iˆ 3.28 ˆj b a 7.29 2 3.282 8ma b 2.712 3.282 4.25m 3.28 tan 50.43 2.71 3.28 tan 24.27.29 or 180 ( 24.2 ) 155.8 S180 155.8 24.2 North of West
2 2 B i j j B i j B C D B C B i j D Dj Method 1 Method 2 3.2 2 2 sin /2 2 sin tan (3/4) 36.9 B D D B D C B θ Isosceles triangle P2: A fire ant goes through three displacements along level ground: d 1 for 0.4m SW, d 2 0.5m E, d 3 0.6m at 60º North of East. Let the positive x direction be East and the positive y direction be North. (a) What are .
