1 AP Physics Chapter 3 Vector. 2 AP Physics Turn in Chapter 2 Homework, Worksheet, & Lab Take quiz Lecture Q&A.

1

AP Physics Chapter 3

Vector

2

AP Physics

Turn in Chapter 2 Homework, Worksheet, & Lab

Take quiz Lecture Q&A

3

Vector and Scalar

Vector: – Magnitude: How large, how fast, … – Direction: In what direction (moving or pointing)– Representation depends on frame of reference

Scalar: – Magnitude only– No direction– Representation does not depend on frame of

reference

4

Examples of vector and scalar

Vectors:– Position, displacement, velocity, acceleration, force,

momentum, …

Scalars:– Mass, temperature, distance, speed, energy,

charge, …

5

Vector symbol

Vector: bold or an arrow on top and v V

– Typed: v and V or

Scalar: regular– v or V

v stands for the magnitude of vector v.

– Handwritten: and v V

6

Adding Vectors

Graphical– Head-to-Tail (Triangular)– Parallelogram

Analytical (by components)

7

Graphical representation of vector: Arrow

An arrow is used to graphically represent a vector.

ab

– The direction of the arrow represents the direction of the vector.– When comparing the magnitudes of vectors, we ignore

directions.

head

tail

– The length of the arrow represents the magnitude of the vector.

Vector a is smaller than vector b because a is shorter than b.

8

Equivalent Vectors

Two vectors are identical and equivalent if they both have the same magnitude and are in the same direction.

AB C

– A, B and C are all equivalent vectors.

– They do not have to start from the same point. (Their tails don’t have to be at the same point.)

9

Negative of Vector

Vector -A has the same magnitude as vector A but points in the opposite direction.

If vector A and B have the same magnitude but point in opposite directions, then A = -B, and B = -A

A -A

-A

10

Adding Vectors: Head-to-Tail

Head-to-Tail method:A

B

A

BA+B

Make sure arrows are parallel and of same length.

– Draw vector A– Draw vector B starting

from the head of A– The vector drawn from the

tail of A to the head of B is the sum of A + B.

Example: A + B

11

A+B=B+A

A

A

A

B

B

BA+BB+A

A + B

How about B + A?

What can we conclude?

12

A+B+C

A

B

A+B+C

C

A B C

Resultant vector:

from tail of first to head of last.

13

A-B=A+(-B)

A B

-B

A

A-B

– Draw vector A– Draw vector -B from head

of A.– The vector drawn from the

tail of A to head of –B is then A – B.

14

What Are the Relationships?

cb

a a

bc

b c a

0a b c

15

Magnitude of sum A + B = C

|A – B| C A + B

A and B in same direction

A and B in opposite direction

A and B at some angle

max. c min. c

16

Adding Vectors: Parallelogram

A+B

A+B

A

B

– Draw the two vectors from the same point

Advantage:No need to measure length.

– Construct a parallelogram with these two vectors as two adjacent sides

– The sum is the diagonal vector starting from the same tail point.

A B

17

Example

Vector a has a magnitude of 5.0 units and is directed east. Vector b is directed 35o west of north and has a magnitude of 4.0 units. Construct vector diagrams for calculating a + b and b – a. Estimate the magnitudes and directions of a + b and b – a from your diagram.

E

N

W

S

35o

18

Solution

35o

a

ba+

b

-a

b-a

a+b: 4.3 unit, 50o North of East

b-a: 8.0 unit, 66o West of North

E

S

W

N

Using ruler and protractor, we find:

50o 66o

19

Vector Components

Drop perpendicular lines from the head of vector a to the coordinate axes, the components of vector a can be found:

cos

sinx

y

a a

a a

x

y

ax

aya

is the angle between the vector and the +x axis.

ax and ay are scalars.

20

Finding components of a vector

Resolving the vector Decomposing the vector

21

Vector magnitude and direction

The magnitude and direction of a vector can be found if the components (ax and ay) are given:

2 2magnitude: x ya a a

is the angle from the +x axis to the vector.

(for 3-D)2 2 2x y za a a

1 directi : non ta y

x

a

a

a

ax

ay

x

y

tan y

x

a

a

22

Example

A ship sets out to sail to a point 120 km due north. Before the voyage, an unexpected storm blows the ship to a point 100 km due east of its starting point. How far, and in what direction, must it now sail to reach its original destination?

23

Solution

It must sail 156 km at 39.8o West of North to reach its original destination.

a =

120

b = 100

c

E

Nc

2 2 2 2120 100 156a b km

1 1 100tan tan 39.8

120ob

a

A ship sets out to sail to a point 120 km due north. Before the voyage, an unexpected storm blows the ship to a point 100 km due east of its starting point. How far, and in what direction, must it now sail to reach its original destination?

24

Practice: What are the magnitude and direction of vector ˆ ˆ4 5 ?a i j

x

y

5

4a 2 2 2 24 5 6.4x ya a

1 1 5tan tan 51

4y o

x

a

a

4, 5,

?, ?

x ya a

a

25

Practice: What are the components of a vector that has a magnitude of 12 units and makes an angle of 126o with the positive x direction?

x

y

126o

12, 126

?, ?

o

x y

a

a a

cos 12cos126 7.1oxa a

sin 12sin126 9.7oya a

26

Unit vectors

Unit vector: magnitude of exactly 1 and points in a particular direction.– x direction:– y direction:

– z direction:

ii or

jj or

kk or

27

Vector components and expression

Any vector can be written in its components and the unit vectors:

ˆˆ

ˆˆ ˆ

ˆ

x y

y z

z

xa i a j a ka

b b i b j b k

, ,x y za a a

28

Terminology

axi is the vector component of a.

ax is the (scalar) component of a.

29

Example

Express the following vector in component and unit vector form.

x

y

=30o

a=12.0 units

xa

ˆ ˆ ˆ ˆ10.4 6.00x ya a i a j i j ya

cos 12.0cos30 10.4oa

sin 12.0sin30 6.00oa

ay

ax

30

Adding Vectors by Components

When adding vectors by components, we add components in a direction separately from other components.

r a b

r a b

s a b

3-D

x

y

ax

ay bx

by

a

b

r

y y yr a b xr x xa b

Component form:

rx

ry

z z zr a b

2-D

ˆˆ ˆx x y y z za b i a b j a b k

ˆˆ ˆx x y y z za b i a b j a b k

31

Example

The minute hand of a wall clock measures 10 cm from axis to tip. What is the displacement vector of its tip (a) from a quarter after the hour to half past, (b) in the next half hour, and (c) in the next hour?

•

•

•

32

Solution

O A (10, 0)

C (0,10)

B (0,-10)

)

A B

a

r

rAB

rBC

x

y

)

(0,10) (0, 10)

(0,20)

B C C B

b

r r r

cm

)

0C C

c

r r r

B Ar r(0, ) (11 ,0)0 0

(0 10, ) ( 10, 1010 0 )cm

or magnitude and direction form.

Try b) and c)

33

Practice

)

ˆ ˆˆ ˆ ˆ3 4ˆ4

a

a b i j k i j k

ˆ ˆˆ ˆ ˆ ˆ4

ˆ ˆˆ ˆ ˆ ˆ(4 1)

3

3 1( ) ( ) 5 44 3

4

1

a b i j k i j k

i j k i j k

)

0

c

a b c

ˆ ˆˆ ˆ ˆ ˆ(4 1) ( ) ( )43 3 21 1 5i j k i j k

c ˆˆ ˆ( ) 5 4 3a b i j k

Two vectors are given by a = 4i – 3j + k and b = -i + j + 4k. Find:

a) a + b

b) a – b

c) a vector c such that a – b + c = 0

)b

b a

34

Practice: 54-19The two vectors a and b in Fig. 3-29 have equal magnitudes of 10.0 m and the angels are 1 = 30o and 2 = 105o. Find the (a) x and (b) y components of their vector sum r, (c) the magnitude of r, and (d) the angle r makes with the positive direction of the x axis.

) )

cos 10cos30 8.66

sin 10sin30 5

ox

oy

a b

a a

a a

y

x

b

a

rb=105o+30o=135o

a

07.7135sin10

07.7135cos10o

y

ox

b

b

x

y

r

r

1

2

sin

cos

aa

aa

y

x

8.66 7.07 1.59x xa b

5 7.07 12.07y ya b

35

54-19 (Continued)

)c

r

)

r

d

2 2 2 21.59 12.07 12.2x yr r

1 1 12.07tan tan 82.5

1.59y o

x

r

r

36

Vector Multiplication

More:

ˆˆ ˆ2 4a i j k

ˆ ˆˆ ˆ ˆ ˆ4 4 2 4 4 4 1 8 16 4a i j k i j k

Scalar (aka dot or inner) product: a b Vector (aka cross) product: a b

We cannot write ab if a and b are vectors. But we still can write 2a since 2 is a scalar.

37

Scalar product: ab

(phi) is the angle between vector a and b. is always between 0o and 180o. (0o 180o) The scalar product is a scalar It has no

direction. What if the two vectors are perpendicular to each

other?

cosa b ab a

b

90o 0a b

: angle between vector and +x axis

: angle between two vectors

38

Physical meaning of ab

ba is the projection of b onto a.

ba

ab

a

b

cosab b

cosba a a

b

ab is the project of a onto b.

Also

cos aba b a ab

cos bba b a a b

39

Properties of scalar product

1. ab = ba2. ii = jj = kk = 13. ij = ji = jk = kj = ki = ik = 04. aa = a2

5. ab = 0 if a b

ˆ ˆˆ ˆ ˆ ˆx y z x y za b a i a j a k b i b j b k

6. ab = axbx+ayby+azbz

40

Angle between two vectors

cosa b ab

When we know the components:

When we know magnitudes and :

x x y y z za b a b a b a b

cos x x y y z zab a b a b a b Put together:

cos x x y y z za b a b a b

ab

1cos x x y y z za b a b a b

ab

41

Example:a. Determine the components and magnitude of r = a – b + c if a = 5.0i + 4.0j – 6.0k, b = -2.0i + 2.0j + 3.0k, and c = 4.0i + 3.0 j + 2.0k.b. Calculate the angle between r and the positive z axis.

)

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ(5.0 ) ( 2.0 ) (4.0 )

ˆˆ ˆ(5.0

6.0 3.0 2.0

6

4

2.0 4.0) ( ) ( )

ˆˆ ˆ11.0 5

.0 2.0 3

.0 3.0

.0

4.0 2 2.0

.0 7.0

.0 3.0

a

r a b c

i j k i j k i j k

i j k

i j k

x

z

O

B

A

2 2

)

11.0 5.0 12.1

b

OA y

11

5

-7

7.0AB

90oOAB

tan AOB

1tan 0.579 30.0oAOB

90 120.0o oCOB AOB

7.00.579

12.1

AB

OA

C•

42

Another approach

ˆˆ ˆ11.0 5.0 7.0r i j k

We are looking for the angle between r and any vector in the z direction. Let’s choose the unit vector in the z direction, k

ˆ ˆ0ˆ ˆ, a 0nd 1k i j k

ˆr k 11 0 5 0 7 1 7x x y y z zr k r k r k

22 211 5 7 13.9r

1k

ˆ cosr k rk

ˆ 7cos 0.5

13.9 1

r k

rk

1cos 0.5 120o

cosa b ab

43

Practice: 55-43For the vectors in Fig. 3-35, with a = 4, b =3, and c =5, calculate

b

a

c5 3

4

)

0

a

a b

(a) , (b) , and (c) .a b a c b c

a b

1 1

)

3tan tan 36.87

4o

b

b

a

180 36.87 143.17o o o

cos 4 5cos143.17 16oa c ac

A

B

1 1

)

4tan tan 53.13

3o

b

aA

b 180 53.13 127o o oB

cos 3 5cos127 9ob c bc B

44

Approach 2

x

y

a

c bb

c

ˆ ˆ4 0a i j

ˆ0 3b i j

ˆ ˆ4 3c i j

a c

b c

-4

-3 4 4 0 3 16

0 4 3 3 9

a b

4 0 0 3 0

45

Approach 3

b

a

c5 3

4

)aa b

)b

c a b

a c a a b a a a b

2 16a )c

0a b c

b c b a b a b

2 9b b b

0a b

46

Approach 4

b

a

c5 3

4

)b

a c

)c

b c

a c 16

da

d = -c

ce

aad aa

b c

b b 9eec

e = -b

a d ��������������

e c

47

Vector Product: c = a b

Magnitude of c is:

– Place the vectors a and b so that their tails are at the same point.

– Extend your right arm and fingers in the direction of a.– Rotate your hands along your arm so that you can flap

your fingers toward b through the smaller angle between a and b. Then

– Your outstretched thumb points in the direction of c.

A

A B

c is a vector, and it has a direction given by the right-hand-rule (RHR):

sinc ab

48

Properties of cross product

b a = - (a b) a b is a, and a b is ba a = 0

jik

ikj

kji

ˆˆˆ

ˆˆˆ

ˆˆˆ

kbabajbabaibababa xyyxzxxzyzzyˆˆˆ

ˆ ˆˆ ˆ ˆ ˆ

x y z x y za b a i a j a k b i b j b k

49

Practice:Three vectors are given by a = 3.0i + 3.0j – 2.0k, b = -1.0i –4.0j + 2.0k, and c = 2.0i + 2.0j + 1.0k. Find (a) a • (b c), (b) a • (b + c), and (c) a (b + c).

)a

b c

( )a b c

ˆ( 4.0)(1.0) (2.0)(2.0)

ˆ (2.0)(2.0) ( 1.0)(1.0)

ˆ ˆˆ ˆ ( 1.0)(2.0) ( 4.0)(2.0) 8.0 5.0 6.0

i

j

k i j k

ˆ ˆˆ ˆ ˆ ˆ(3.0 3.0 2.0 ) ( 8.0 5.0 6.0 )i j k i j k

(3.0)( 8.0) (3.0)(5.0) ( 2.0)(6.0) 21

ˆ ˆˆ ˆ ˆ ˆ1.0 4.0 2.0 2.0 2.0 1.0i j k i j k

kbabajbabaibababa xyyxzxxzyzzyˆˆˆ

50

Pg52-53P (2)

)

ˆ ˆˆ ˆ ˆ ˆ1.0 4.0 2.0 2.0 2.0 1.0

b

b c i j k i j k

( )a b c

ˆˆ ˆ( 1.0 2.0) ( 4.0 2.0) (2.0 1.0)

ˆˆ ˆ1.0 2.0 3.0

i j k

i j k

ˆ ˆˆ ˆ ˆ ˆ(3.0 3.0 2.0 ) (1.0 2.0 3.0 )i j k i j k

(3.0)(1.0) (3.0)( 2.0) ( 2.0)(3.0) 9

51

Pg52-53P (3)

kjicb

c

ˆ0.3ˆ0.2ˆ0.1

)

( )a b c ˆ ˆˆ ˆ ˆ ˆ(3.0 3.0 2.0 ) (1.0 2.0 3.0 )i j k i j k

ˆ(3.0)(3.0) ( 2.0)( 2.0)

ˆ ( 2.0)(1.0) (3.0)(3.0)

ˆ (3.0)( 2.0) (3.0)(1.0)

ˆˆ ˆ5.0 11.0 9.0

i

j

k

i j k

52

Law of cosine

Cabbac cos2222

b

a

c

C

53

Law of Sine

C

c

B

b

A

a

sinsinsin

a

c

b

B A

C

sin sin sinA B C

a b c or