Transcription of Chapter 3 - Vectors
1 Chapter 3 - Vectors I. Definition II. Arithmetic operations involving Vectors A) Addition and subtraction - Graphical method - Analytical method vector components B) Multiplication Review of angle reference system 90 . 0 < 1<90 . 90 < 2<180 . 2. 1 0 . 180 Origin of angle reference system 3. 4. 180 < 3<270 . 270 < 4<360 . 270 . Angle origin 4=300 =-60 . I. Definition vector quantity: quantity with a magnitude and a direction. It can be represented by a vector . Examples: displacement, velocity, acceleration. Same displacement Displacement does not describe the object's path. Scalar quantity: quantity with magnitude, no direction. Examples: temperature, pressure II. Arithmetic operations involving Vectors .
2 vector addition: s a b . b . a - Geometrical method . s a b Rules: . a b b a (commutative law) ( ).. (a b ) c a (b c ) (associative law) ( ).. vector subtraction: d a b a ( b ) ( ). vector component: projection of the vector on an axis. a x a cos . ( ) . Scalar components of a a y a sin . a a x2 a 2y vector magnitude ( ). ay tan vector direction ax Unit vector : vector with magnitude 1. No dimensions, no units. i , j , k unit Vectors in positive direction of x, y, z axes . a a x i a y j ( ). vector component vector addition: - Analytical method: adding Vectors by components.. r a b (a x bx )i (a y by ) j ( ). Vectors & Physics: -The relationships among Vectors do not depend on the location of the origin of the coordinate system or on the orientation of the axes.
3 - The laws of physics are independent of the choice of coordinate system. a a x2 a 2y a'2x a '2y ( ). ' . 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 ( ).. Rule: a b b a ( ) a b ab cos 1 ( 0 ).. a b 0 cos 0 ( 90 ).. 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 b Angle between two Vectors : cos . a b Multiplying Vectors : - vector by a vector vector product = vector (cross product).. a b c (a y bz by a z )i (bz a x a z bx ) j (a x by bx a y )k . c ab sin Magnitude vector product a b 0 sin 0 ( 0 ). Direction right hand rule a b ab sin 1 ( 90 ).. Rule: b a ( a b ) ( ).
4 C perpendicular to plane containing a , b 1) 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 b where they meet. 3) Sweep a into b through the smallest angle between them. Right-handed coordinate system z k j i y x Left-handed coordinate system z k i j x y . i i j j k k 0.. i i j j k k 1 1 sin 0 0 i j ( j i ) k . j k ( k j ) i . k i (i k ) j P1: If B is added to C = 3i + 4j, the result is a vector in the positive direction of the y axis, with a magnitude equal to that of C. What is the magnitude of B? Method 2. Method 1. Isosceles triangle . B C B (3i 4 j ) D D j . C D 32 4 2 5 C tan (3 / 4) . D. B (3i 4 j ) 5 j B 3i j B 9 1 B / 2.
5 Sin B 2 D sin 2 D 2 . B. P2: A fire ant goes through three displacements along level ground: d1 for SW, d2 E, d3= at 60 North of East. Let the positive x direction be East and the positive y direction be North. (a) What are the x and y components of d1, d2 and d3? (b) What are the x and the y components, the magnitude and the direction of the ant's net displacement? (c) If the ant is to return directly to the starting point, how far and in what direction should it move? (b).. (a) d 4 d1 d 2 ( j ) ( j )m N d1x cos 45 . D d 4 d 3 ( j ) ( j ) ( j )m D d1 y sin 45 E D 2 2 d 2 x 45 . d4 d3 d2 y 0 tan 1 North of East . d1 . d 3 x cos 60 d2 d 3 y sin 60 (c) Return vector negative of net displacement, D= , directed 25 South of West.
6 P2 (a) r d1 d 2 d 3 ? d1 4i 5 j 6k . (b) Angle between r and z ? d 2 i 2 j 3k . (c) Component of d1 along d 2 ? d 4i 3 j 2k . 3 . (d ) Component of d1 perpendicular to d 2 and in plane of d1 , d 2 ? . (a ) r d1 d 2 d 3 (4i 5 j 6k ) ( i 2 j 3k ) (4i 3 j 2k ) 9i 6 j 7k . 7 . (b) r k r 1 cos 7 cos 1 123.. r 9 2 6 2 7 2 d1perp d1.. d1 d 2. (c) d1 d 2 4 10 18 12 d1d 2 cos cos . d1d 2 d1//.. d d 12 d2. d1// d1 cos d1 1 2 d1d 2 d 2 12 2 2 32 (d ) d1 d12// d12perp d1 perp 2 2 d1 4 2 52 6 2 m P3 . If d1 3i 2 j 4k (d1 d 2 ) a contained in (d1 , d 2 ) plane (d1 d 2 ) (d1 4d 2 ) ? .. d 2 5i 2 j k (d1 4d 2 ) 4(d1 d 2 ) 4b perpendicular to (d1 , d 2 ) plane . a perpendicular to b cos 90 0 4a b 0. Tip: Think before calculate !
7 !! P4: Vectors A and B lie in an xy plane. A has a magnitude and angle 130 ; B has components Bx= , By= What are the angles between the negative direction of . the y axis and (a) the direction of A, (b) the direction of AxB, (c) the direction of Ax(B+3k)? y A.. 130 (a) Angle between y and A 90 50 140 .. x (b) Angle y , ( A B) C angle j , k because C perpendicular B . plane ( A, B) ( xy ) 90 .. (c) Direction A ( B 3k ) D.. E B 3k j 3k . i j k .. D A E 0 j . 3. D 2 2 . j D j ( j ) . j D . cos . 99.. 1 D . P5: A wheel with a radius of 45 cm rolls without sleeping along a horizontal floor. At time t1 the dot P painted on the rim of the wheel is at the point of contact between the wheel and the floor.
8 At a later time t2, the wheel has rolled through one-half of a revolution. What are (a) the magnitude and (b) the angle (relative to the floor) of the displacement P during this interval? y Vertical displacement: 2 R 1. Horizontal displacement: (2 R) d 2.. r ( )i ( ) j . r 2 x 2R . tan . R . P6: vector a has a magnitude of m and is directed East. vector b has a magnitude of m and is directed 35 West of North. What are (a) the magnitude and direction of (a+b)?. (b) What are the magnitude and direction of (b-a)?. (c) Draw a vector diagram for each combination.. a 5i . N . b b 4 sin 35 i 4 cos 35 j j -a 125 . a+b . b-a (a) a b j (b) b a b ( a ) j . a a b b a 2 8m W E.. tan tan .. or 180 ( ).
9 S. 180 North of West
