Physics 40 Chapter 3: Vectors

Cartesian Coordinate System

• Also called rectangular coordinate system

• x- and y- axes intersect at the origin

• Points are labeled (x,y)

Polar Coordinate System

– Origin and reference line are noted

– Point is distance r from the origin in the direction of angle θ, ccw from reference line

– Points are labeled (r,θ)

Polar to Cartesian Coordinates

• Based on forming a right triangle from r and θ

• x = r cos θ• y = r sin θ

Cartesian to Polar Coordinates

• r is the hypotenuse and θan angle

θ must be ccw from positive x axis for these equations to be valid

2 2

tan yx

r x y

θ =

= +

Example 3.1

• The Cartesian coordinates of a point in the xy plane are (x,y) = (-3.50, -2.50) m, as shown in the figure. Find the polar coordinates of this point.

Solution:2 2 2 2( 3.50 m) ( 2.50 m) 4.30 mr x y= + = − + − =

1 2.50 mtan ( ) 35.53.50 m

− −= °

−

180 35.5 216θ = °+ ° = °

Vector quantities have both magnitude and direction

AThe length of the vector represents the magnitude of the vector. The orientation represents the direction or angle of the vector.

Example: velocity of 2 km/hr, 30 degrees north of east.2 km/hr is the magnitude, 30 degrees north or east, the direction.

Scalars only have magnitude: T = 82 degrees Celsius.

• The vector is written with an arrow over head and includes both magnitude and direction.

( )A= 2km, 30 NEo

A 2A magnitude km= = =30

2km

Writing Vectors• Text books use BOLD. Hard to write bold!

• Magnitude (length of vector) is written with no arrow or as an absolute value:

( )A= ,A θv

Vector Example• A particle travels from A

to B along the path shown by the dotted red line– This is the distance traveled

and is a scalar• The displacement is the

solid line from A to B– The displacement is

independent of the path taken between the two points

– Displacement is a vector

Equality of Two Vectors

• Two vectors are equalif they have the same magnitude and the same direction

• A = B if A = B and they point along parallel lines

• All of the vectors shown are equal

WARNING!Vector Algebra is Weird!!

Vectors have a different rules of operation for addition, subtraction, multiplication and division than ordinary real numbers!!!! Since vectors have magnitude and direction you can not always just simply add them!!!!

A & B co-linear A & B NOT colinear

• When you have many vectors, just keep repeating the process until all are included

• The resultant is still drawn from the origin of the first vector to the end of the last vector

Adding VectorsThe Graphical Method

Adding VectorsHead to Tail

R=A + B = B + A

When two vectors are added, the sum is independent of the order of the addition.– This is the commutative law

of addition– A + B = B + A– The sum forms the diagonal

of a Parallelogram!

Subtraction of Vectors

( )A C B= + −

Subtracting Vectors• If A – B, then use

A+(-B)• The negative of the

vector will have the same magnitude, but point in the opposite direction

• Two ways to draw subtraction:

How do you go from A to -B ?

The distance between two vectors is equal to the magnitude of the

difference between them! More on this later…

Parallelogram MethodAddition & Subtraction

R X Y= + R X Y= −

A

BR

1. Draw 1st vector tail A at origin2. Draw 2nd vector B with tail at the head

of the 1st vector, A. The angle of B is measured relative to an imaginary axis attached to the tail of B.

3. The resultant is drawn from the tail of the first vector to the head of the last vector.

R A B= +

R A B= −

A BR

AB

ROR

•Draw vectors to scale.•Draw the vectors “Tip to Tail.”•IMPORTANT: the angle of a vector is relative its own tail!•The resultant, R, is drawn from the tail of the first to the head of the last vector.•Use a ruler to MEASURE the resultant length.•Use a protractor to MEASURE the resultant angle.

Adding VectorsThe Graphical Method

Each of the displacement vectors A and B shown has a magnitude of 3.00 m.  Find graphically  (a) A + B,  (b) A − B,  (c) B − AReport all angles counterclockwise from the positive xaxis. Use a scale of:  1 unit 0.5 m=

5.2m,60o 3m,330o 3m,150o

Adding Vectors: Graphical Method

xA cosθ= AA

yA sinθ= AA

( )x y= A , AA

2 2

-1tan

x y

yA

x

A A A

AA

θ

= +

⎛ ⎞= ⎜ ⎟

⎝ ⎠ ( ), AA A θ=

Vector Components

If the magnitude of r is 175m, find the magnitude of components x and y.

cos

θ= =x rx r r

175 cos50m= o

113=x m

sin

θ= =y rx r r

134=y m175 sin 50m= o

Example

C A B= +Vector AdditionComponents Method

C A B= +Vector Components

x x xC =A +B

y y yC =A +B

2 2

-1tan

x y

yC

x

C C C

CC

θ

= +

⎛ ⎞= ⎜ ⎟

⎝ ⎠

C A B= +Vector Components

x x xC =A +B

y y yC =A +B

2 2

-1tan

x y

yC

x

C C C

CC

θ

= +

⎛ ⎞= ⎜ ⎟

⎝ ⎠

Unit Vectors• A unit vector is a dimensionless vector

with a magnitude of exactly 1.• Unit vectors are used to specify a

direction and have no other physical significance

• The symbolsrepresent unit vectors• They form a set of mutually

perpendicular vectors (called a basis) • Any vector with component Ax and Ay

can be written in terms of unit vectors:

kand,j,i

ˆ ˆ ˆx y zA A A= + +A i j k

Right Handed Coordinate System

ˆ ˆ ˆi j k× =

Vectors in 3-D

r

The orientation of ijk doesn’t matter as long as they form a right-handed system!

Quiz Fun

Which statement is true about the unit vectors i, j and k?

a. Their directions are defined by a left-handed coordinate system.

b. The angle between any two is 90 degrees.c. Each has a length of 1 m.d. If i is directed east and j is directed south, k points up

out of the surface.e. All of the above.

Adding Vectors Using Unit Vectors

• R = A + B

• Rx = Ax + Bx and Ry = Ay + By

( ) ( )( ) ( )

ˆ ˆ ˆ ˆ

ˆ ˆ

ˆ ˆ

= + + +

= + + +

= +

x y x y

x x y y

x y

A A B B

A B A B

R R

R i j i j

R i j

R i j

2 2 1tan yx y

x

RR R R

Rθ −= + =

Each of the displacement vectors A and B shown has a magnitude of 3.00 m.  Find using components            (a) A + B,  (b) A − B,  (c) B − A,  (d) A − 2B.  Report all angles counterclockwise from the positive xaxis. Use a scale of:  1 unit 0.5 m=

( ) ( )ˆ ˆ3.00m cos 30.0 3.00msin 30.0ˆ ˆ2.60m 1.50m

= ° + °

= +

i j

i j

A

A ˆ3.00mB = j

ˆ ˆ(2.60 0)m (1.50 3.00)mA B+ = + + +i jˆ ˆ2.60m 4.50mA B+ = +i j

( )2 22.60m (4.50m) 5.20mA B+ = + =

1 4.5mtan 602.6m

θ − ⎛ ⎞= =⎜ ⎟⎝ ⎠

o

Adding Vectors: Component Method

Each of the displacement vectors A and B shown has a magnitude of 3.00 m.  Find using components            (a) A + B,  (b) A − B,  (c) B − A,  (d) A − 2B.  Report all angles counterclockwise from the positive xaxis. Use a scale of: 

ˆ ˆ(2.60 0)m (1.50 3.00)mA B− = − + −i j

ˆ ˆ2.60m 1.50mA B− = −i j

( )2 22.60m ( 1.50m) 3.00mA B− = + − =

1 1.5mtan 302.6m

− −⎛ ⎞ = −⎜ ⎟⎝ ⎠

o, 360 30 330θ = − =o o o

Adding Vectors: Component Method

1 unit 0.5 m=

( ) ( )ˆ ˆ3.00m cos 30.0 3.00m sin 30.0ˆ ˆ ˆ2.60m 1.50m 3.00m

= ° + °

= + =

i j

i j j

A

A B

Each of the displacement vectors A and B shown has a magnitude of 3.00 m.  Find using components      (a) A + B,  (b) A − B,  (c) B − A,  (d) A − 2B.  Report all angles counterclockwise from the positive xaxis. Use a scale of: 

1 unit 0.5 m=

ˆ ˆ2.60m 1.50mA = +i j ˆ3.00mB = j

ˆ ˆ(0 2.60)m (3.00 1.50)mB A− = − + −i j

ˆ ˆ2.60m 1.50mB A− = − +i j

( )2 22.60m (1.50m) 3.00mB A− = − + =

1 1.5mtan 30-2.6m

− ⎛ ⎞ = −⎜ ⎟⎝ ⎠

o, 180 30 150θ = − =o o o

Adding Vectors: Component Method

Each of the displacement vectors A and B shown has a magnitude of 3.00 m.  Find using components      (a) A + B,  (b) A − B,  (c) B − A,  (d) A − 2B.  Report all angles counterclockwise from the positive xaxis. Use a scale of: 

1 unit 0.5 m=

ˆ ˆ2.60m 1.50mA = +i j ˆ2 6.00mB = j

ˆ ˆ2 (2.60 0)m (1.50 6.00)mA B− = − + −i j

ˆ ˆ2 2.60m 4.50mA B− = −i j

( )2 22 2.60m ( 4.50m) 5.2mA B− = + − =

1 4.5mtan 602.6m

− −⎛ ⎞ = −⎜ ⎟⎝ ⎠

o, 360 60 300θ = − =o o o

Adding Vectors: Component Method

The distance between two vectors is equal to the magnitude of the

difference between them!

Ch3. Two vectors in the xy plane have Cartesian coordinates (2.00, −4.00) m and (−3.00, 3.00) m. Draw the vectors, write them with ijk vectors and find the distance between the points.

Mule ProblemThe helicopter view shows two people pulling on a stubborn mule The system is in equilibrium. Write the force vectors in ijk form for both people and find the net force they pull with. Then find the force, as an ijk vector and also magnitude and direction that the mule pulls back with. Forces are measured in units of newtons (N).

( )3ˆ ˆ39.3 181 N= − = − −F F i j

( ) ( )1ˆ ˆ120cos 60.0 120sin 60.0

ˆ ˆ (60.0 104 )N

= ° + °

= +

N NF i j

i j

( )1 2ˆ ˆ39.3 181 N= + = +F F F i j 2 239.3 181 185 N= + =F

1 181tan 77.839.3

θ − ⎛ ⎞= = °⎜ ⎟⎝ ⎠

( ) ( )2ˆ ˆ80.0cos 75.0 80.0sin 75.0

ˆ ˆ ( 20.7 77.3 )N

= − ° + °

= − +

N NF i j

i j

1 181tan 77.839.3

− ⎛ ⎞= = °⎜ ⎟⎝ ⎠

SWθ

ProblemThree displacement vectors of a croquet ball are shown in Figure P3.49, where |A| = 20.0 units, |B| = 40.0 units, and |C| = 30.0 units. Find (a) the resultant in unit‐vector notation and (b) the magnitude and direction of the resultant displacement.

40.0cos 45.0 30.0cos 45.0 49.5= ° + ° =xR

( ) ( )2 249.5 27.1 56.4= + =R

40.0sin 45.0 30.0sin 45.0 20.0 27.1= ° − ° + =yR

ˆ ˆ49.5 27.1= +R i j

1 27.1tan 28.749.5

θ − ⎛ ⎞= = °⎜ ⎟⎝ ⎠

ProblemA person going for a walk follows the path shown. The total trip consists of four straight‐line paths. At the end of the walk, what is the personʹs resultant displacement measured from the starting point?

( ) ( )( ) ( )

1

2

3

4

ˆ100ˆ300

ˆ ˆ ˆ ˆ150cos 30.0 150sin 30.0 130 75.0ˆ ˆ ˆ ˆ200cos 60.0 200sin 60.0 100 173

=

= −

= − ° − ° = − −

= − ° + ° = − +

m

m

m m m m

m m m m

d i

d j

d i j i j

d i j i j

1 202tan 57.2130

180 237

φ

θ φ

− ⎛ ⎞= = °⎜ ⎟⎝ ⎠

= + = °

( )1 2 3 4ˆ ˆ130 202 m= + + + = − −R d d d d i j

( ) ( )2 2130 202 240 m= − + − =R

A

BR

-RThe Balancing Force!

R = A+ BNext Lab! Force Table!0 = A+ B+(- R )