Chapter 3B -- Vectors

Chapter 3B - Vectors
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
© 2007

Vectors
Surveyors use accurate measures of
magnitudes and directions to create
scaled maps of large regions.

Objectives: After completing this
module, you should be able to:
• Demonstrate that you meet mathematics
expectations: : unit analysis, algebra, scientific
notation, and right-triangle triangle trigonometry.
• Define and give examples of scalar and vector
quantities.
• Determine the components of a given vector.
• Find the resultant of two or more vectors.

Expectations
• You must be able convert units of
measure for physical quantities.
Convert 40 m/s into kilometers per hour.
m 1 km 3600 s
40--- x ---------- x -------- = 144 km/h
s 1000 m 1 h

Expectations (Continued):
• College algebra and simple formula
manipulation are assumed.
Example:
x
v0
v f
2
t
Solve for v o
v
0
vt
f
t
2x

Expectations (Continued)
• You must be able to work in scientific
notation.
Evaluate the following:
Gmm’ (6.67 x 10 -11 )(4 x 10 -3 )(2)
F = -------- = ------------
(8.77 x 10 -3 ) 2
r 2
F = 6.94 x 10 -9 -9 N = 6.94 nN

Expectations (Continued)
• You must be familiar with SI prefixes
The meter (m) 1 m = 1 x 10 0 m
1 Gm = 1 x 10 9 m 1 nm = 1 x 10 -9 m
1 Mm = 1 x 10 6 m 1 m = 1 x 10 -6 m
1 km = 1 x 10 3 m 1 mm = 1 x 10 -3 m

y
Expectations (Continued)
• You must have mastered right-triangle
triangle
trigonometry.
x
R
sin
cos
y
R
x
R
y = R sin
x = R cos
y
tan R 2 = x 2 + y 2
x

Mathematics Review
If you feel you need to
brush up on your
mathematics skills, try
the tutorial from Chap.
2 on Mathematics. Trig
is reviewed along with
vectors in this module.
Select Chap. 2 from the On-Line Learning
Center in in Tippens—Student Edition

Physics is the Science of
Measurement
Length
Weight
Time
We begin with the measurement of length:
its magnitude and its direction.

Distance: A Scalar Quantity
• Distance is is the length of the actual path
taken by an object.
A
s = 20 m
B
A scalar quantity:
Contains magnitude
only and consists of a
number and a unit.
(20 m, 40 mi/h, 10 gal)

Displacement—A A Vector Quantity
• Displacement is is the straight-line
separation of two points in a specified
direction.
D = 12 m, 20 o
A
B
A vector quantity:
Contains magnitude
AND direction, a
number, unit & angle.
(12 m, 30 0 ; 8 km/h, N)

Distance and Displacement
• Displacement is is the x or y coordinate of
position. Consider a car that travels 4
m, E then 6 m, W.
D
4 m,E
Net displacement:
D = 2 m, W
x = -2
6 m,W
x = +4
What is the distance
traveled?
10 m !!

Identifying Direction
A common way of identifying direction
is is by reference to East, North, West,
and South. (Locate points below.)
N
Length = 40 m
40 m, 50 o N of E
W
50 o
60 o 60 o 60 o
E
40 m, 60 o N of W
40 m, 60 o W of S
S
40 m, 60 o S of E

Identifying Direction
Write the angles shown below by using
references to east, south, west, north.
W
N
E
45 o
N
S
50 o
W
S
E
50 Click 0 S of to Esee the Answers 45 0 W . of . .
N

Vectors and Polar Coordinates
Polar coordinates ((R,) ) are an excellent
way to to express vectors. Consider the
vector 40 m, 50 0 N of of E, EE, for example.
40 m
90 o R
90 o
180 o
50 o
180 o
270 o
0 o
270 o
0 o
R is the magnitude and is the direction.

Vectors and Polar Coordinates
Polar coordinates ((R,) ) are given for each
of four possible quadrants:
210 o
90 o
120 o
60 o 50 o
0 o
60
60 o o
(R,) ) = 40 m, 50 o
(R,) ) = 40 m, 120 o
180 o 270 o
300 0
(R,) ) = 40 m, 210 o
(R,) ) = 40 m, 300 o

Rectangular Coordinates
(-2, +3)
-
+
y
-
(+3, +2)
+
x
Reference is made to
x and y axes, with +
and - numbers to
indicate position in
space.
Right, up = (+,+)
Left, down = (-,-)
(-1,
-3)
(+4, -3)
(x,y) = (?, ?)

Trigonometry Review
• Application of Trigonometry to Vectors
y
Trigonometry
x
R
sin
cos
y
R
x
R
y = R sin
x = R cos
y
tan R 22 = x 22 + y 22
x

Example 1: Find the height of a building
if it casts a shadow 90 m long and the
indicated angle is 30 o .
The height h is opposite 30 0 and
the known adjacent side is 90 m.
h
30 0
90 m
0
tan 30
opp
adj
h
90 m
h = (90 m) tan 30 o
h = 57.7 m

Finding Components of Vectors
A component is the effect of a vector along
other directions. The x and y components of
the vector (R, are illustrated below.
R
x
x = R cos
y = R sin
y
Finding components:
Polar to Rectangular Conversions

Example 2: A person walks 400 m in a
direction of 30 o N of E. How far is the
displacement east and how far north?
N
N
R
400 m
y
y = ?
E
x E
x = ?
The x-component (E) is ADJ:
The y-component (N) is OPP:
x = R cos
y = R sin

Example 2 (Cont.): A 400-m walk in a
direction of 30 o N of E. How far is the
displacement east and how far north?
N
400 m
x = ?
y = ?
E
Note: x is the side
adjacent to angle 30 0
ADJ = HYP x Cos 30 0
x = R cos
x = (400 m) cos 30 o
= +346 m, E
The x-component x
is:
R x = +346 m

Example 2 (Cont.): A 400-m walk in a
direction of 30 o N of E. How far is the
displacement east and how far north?
N
400 m
x = ?
y = ?
E
Note: y is the side
opposite to angle 30 0
OPP = HYP x Sin 30 0
y = R sin
y = (400 m) sin 30 o
= + 200 m, N
The y-component y
is:
R y = +200 m

Example 2 (Cont.): A 400-m walk in a
direction of 30 o N of E. How far is the
displacement east and how far north?
N
400 m
R x =
+346 m
R y =
+200 m
E
The x- x and y- y
components are
each + in the
first quadrant
Solution: The person is displaced 346 m east
and 200 m north of the original position.

Signs for Rectangular Coordinates
90 o R
+
+
0 o
R is positive (+)
0 o > < 90 o
x = +; y = +
First Quadrant:
x = R cos
y = R sin

Signs for Rectangular Coordinates
90 o
R
+
180 o
Second Quadrant:
R is positive (+)
90 o > < 180 o
x = - ; y = +
x = R cos
y = R sin

Signs for Rectangular Coordinates
Third Quadrant:
R is positive (+)
180 o > < 270 o
180 o 270 o
-
R
x = - y = -
x = R cos
y = R sin

Signs for Rectangular Coordinates
Fourth Quadrant:
R is positive (+)
+ 360 o
270 o > < 360 o
x = + y = -
270 o
R
x = R cos
y = R sin

Resultant of Perpendicular Vectors
Finding resultant of two perpendicular vectors is
like changing from rectangular to polar coord.
R
x
y
R
x
2
y
2
tan
y
x
R is always positive; is from + x axis

Example 3: A 30-lb
southward force
and a 40-lb
eastward force act on a
donkey at the same time. What is the
NET or resultant force on the donkey?
Draw a rough sketch.
Choose rough scale:
Ex: 1 cm = 10 lb
Note: Force has 40 direction lb just like length
does.
40 lb
We can treat force vectors just as we
have length vectors to find the resultant
force. The procedure is is the same!
4 cm = 40 lb
30 lb
30 lb
3 cm = 30 lb

Finding Resultant: (Cont.)
Finding ( (R, R,) ) from given ( (x,y)) = (+40, -30)
40 lb
R x
40 lb
R y
30 lb
R
30 lb
R = x 2 + y 2
R = (40) 2 + (30) 2 = 50 lb
tan = -30
40
= -36.9 o
= 323.1 o

R y
30 lb
Four Quadrants: (Cont.)
R
R x
40 lb
40 lb
R y
40 lb R x R x 40 lb
30 lb R
R = 50 lb
R = 50 lb
R
R x
R
30 lb
R y
R y
30 lb
= 36.9 o ; = 36.9 o ; 143.1 o ; 216.9 o ; 323.1 o

Unit vector notation (i,j,k(
i,j,k)
y Consider 3D axes (x, y, z)
k
j
i
x
Define unit vectors, i, j, k
Examples of Use:
z
40 m, E = 40 i 40 m, W = -40 i
30 m, N = 30 j 30 m, S = -30 j
20 m, out = 20 k 20 m, in = -20 k

Example 4: A woman walks 30 m, W; W
then 40 m, N. Write her displacement
in i,j notation and in R, notation.
+40 m R
In i,j notation, we have:
R = R x i + R y j
R x = - 30 m
R y = + 40 m
-30 m
R = -30 i i + 40 j
Displacement is is 30 m west and 40 m
north of the starting position.

Example 4 (Cont.): Next we find her
displacement in R, notation.
+40
m
R
-30 m
40
tan ; = 59.1
30
= 180 0 – 59.1 0
= 126.9 o
R ( 30) 2
(40)
2
R = 50 m
(R,) = (50 m, 126.9 o )
0

Example 6: Town A is 35 km south and 46 km
west of Town B. Find length and direction of
highway between towns.
46 km
R = -46
i – 35 j
R
(46 km) (35 km)
2 2
R = 57.8 km
46 km
tan
35 km
35
km
A
R = ?
B
= 180 0 + 52.7 0
= 52.7 0 S. of W.
= 232.7 0

Example 7. Find the components of the 240-N
force exerted by the boy on the girl if his arm
makes an angle of 28 0 with the ground.
F y
F = 240 N
F
28 0
F y
F x
F x = -|(240 N) cos 28 0 | = -212 N
F y = +|(240 N) sin 28 0 | = +113 N
Or in i,j notation:
F = -(212
N)i + (113 N)j

Example 8. Find the components of a 300-N
force acting along the handle of a lawn-
mower. The angle with the ground is 32 0 .
F = 300 N
32 o
F x
32 o
F y
32 0
F
F y
F x = -|(300 N) cos 32 0 | = -254 N
F y = -|(300 N) sin 32 0 | = -159 N
Or in i,j notation:
F = -(254
N)i - (159 N)j

Component Method
1. Start at origin. Draw each vector to scale
with tip of 1st to tail of 2nd, tip of 2nd to
tail 3rd, and so on for others.
2. Draw resultant from origin to tip of last
vector, noting the quadrant of the resultant.
3. Write each vector in i,j notation.
4. Add vectors algebraically to get resultant in
i,j notation. Then convert to (R,(
R,).

Example 9. A boat moves 2.0 km east then
4.0 km north, then 3.0 km west, and finally
2.0 km south. Find resultant displacement.
1. Start at origin.
Draw each vector to
scale with tip of 1st
to tail of 2nd, tip of
2nd to tail 3rd, and
so on for others.
D
2 km, S
2. Draw resultant from origin to tip of last
vector, noting the quadrant of the resultant.
Note: The scale is approximate, but it is still
clear that the resultant is in the fourth quadrant.
N
3 km, W
C
A
2 km, E
B
4 km, N
E

Example 9 (Cont.) Find resultant displacement.
3. Write each vector
in i,j notation:
A = +2 i
B = + 4 j
C = -3 i
D = - 2 j 4. Add vectors A,B,C,D
algebraically to get
R = -1 i + 2 j resultant in i,j notation.
1 km, west and 2
km north of origin.
D
2 km, S
N
3 km, W
C
A
2 km, E
B
4 km, N
E
5. Convert to R, notation
See next page.

Example 9 (Cont.) Find resultant displacement.
Resultant Sum is:
R = -1 i + 2 j
Now, We Find R,
2 2
R ( 1) (2) 5
R = 2.24 km
D
2 km, S
N
3 km, W
C
A
2 km, E
B
4 km, N
E
2 km
tan
1 km
= 63.4 0 N or W
R x = -1 km
R
R y = +2
km

Reminder of Significant Units:
For convenience, we
follow the practice of
assuming three (3)
significant figures for
all data in problems.
D
2 km
N
3 km
C
A
2 km
B
4 km
E
In the previous example, we assume that the
distances are 2.00 km, 4.00 km, and 3.00 km.
Thus, the answer must be reported as:
R = 2.24 km, 63.4 0 N of W

Significant Digits for Angles
Since a tenth of a
degree can often be
significant, sometimes a
fourth digit is needed.
Rule: Write angles to to
the nearest tenth of of
a degree. See the
two examples below:
R
R x
R x
40 lb
40 lb
30 lb
R y
R y
= 36.9 o ; 323.1 o
R
30 lb

Example 10: Find R, for the three vector
displacements below:
A = 5 m, 0 0
B = 2.1 m, 20 0
C = 0.5 m, 90 0
R
A = 5 m
B
20 0
C =
0.5 m
B = 2.1 m
1. First draw vectors A, B, and C to approximate
scale and indicate angles. (Rough drawing)
2. Draw resultant from origin to tip of last vector;
noting the quadrant of the resultant. (R,)
3. Write each vector in i,j notation. (Continued ...)

Example 10: Find R, for the three vector
displacements below: (A table may help.)
For i,j notation
find x,y components
of each
vector A, B, C.
A = 5 m
Vector X-component (i)(
Y-component (j)(
A=5 m 0 0 + 5 m 0
20 0
B = 2.1 m
B=2.1m 20 0 +(2.1 m) cos 20 0 +(2.1 m) sin 20 0
C=.5 m 90 0 0 + 0.5 m
R
R x
= A x
+B x
+C x
R y
= A y
+B y
+C y
B
C =
0.5 m

Example 10 (Cont.): Find i,j for three
vectors: A = 5 m,0 0 ; B = 2.1 m, 20
90 0 .
= 2.1 m, 20 0 ; C = 0.5 m,
X-component (i)(
Y-component (j)(
A x
= + 5.00 m A y
= 0
B x
= +1.97 m B y
= +0.718 m
C x
= 0 C y
= + 0.50 m
4. Add vectors to
get resultant R
in i,j notation.
A = 5.00 i + 0 j
B = 1.97 i + 0.718 j
C = 0 i + 0.50 j
R = 6.97 i + 1.22 j

Example 10 (Cont.): Find i,j for three vectors:
R
A = 5 m,0 0 ; B = 2.1 m, 20 0 ; C = 0.5 m, 90 0 .
R = 6.97 i + 1.22 j
5. Determine R, from x,y:
(6.97 m) (1.22 m)
R = 7.08 m
2 2
Diagram for
finding R,
R x = 6.97 m
1.22 m
tan = 9.93 0
6.97 m
N. of E.
R
R y
1.22 m

Example 11: A bike travels 20 m, E then 40 m
at 60 o N of W, and finally 30 m at 210 o . What
is the resultant displacement graphically?
C = 30 m
30 o R
B =
40 m
60 o
Graphically, we use
ruler and protractor
to draw components,
then measure the
Resultant R,
A = 20 m, E
Let 1 cm = 10 m
R = (32.6 m, 143.0 o )

A Graphical Understanding of the Components
and of the Resultant is given below:
Note: R x = A x + B x + C x
C y
B y
C
B
0
R y = A y + B y + C y
R y
30 o R
60 o
A
R x
A x
C x
B x

Example 11 (Cont.) Using the Component
Method to solve for the Resultant.
C y
B y
30 o R
C
B
Write each vector
in i,j notation.
A x = 20 m, A y = 0
R y
R x
60
A
A
A = 20 i
B x = -40 cos 60 o = -20 m
C x
B x
x
B y = 40 sin 60 o = +34.6 m
C x = -30 cos 30 o = -26 m
C y = -30 sin 60 o = -15 m
B = -20 i + 34.6 j
C = -26 i - 15 j

Example 11 (Cont.) The Component Method
C y
B y
30 o R
C
B
Add algebraically:
A = 20 i
B = -20 i + 34.6 j
R y
R x
C x
B x
60
A
A
x
C = -26 i - 15 j
R = -26 i + 19.6 j
+19.6
R
-26
R = (-26) 2 + (19.6) 2 = 32.6 m
tan = 19.6
-26
= 143 o

Example 11 (Cont.) Find the Resultant.
C y
B y
30 o C
R
B
R = -26 i + 19.6 j
R y
R x
C x
B x
60
A
A
x
+19.6
R
-26
The Resultant Displacement of the bike is best
given by its polar coordinates R and .
R = 32.6 m; = 143 0

Example 12. Find A + B + C for Vectors
Shown below.
B
A = 5 m, 90 0
B = 12 m, 0 0 A
C = 20 m, -35 0
R
C x
C y
C
A x = 0; A y = +5 m
B x = +12 m; B y = 0
C x = (20 m) cos 35 0
C y = -(20 m) sin -35
0
A = 0 i + 5.00 j
B = 12 i + 0 j
C = 16.4 i – 11.5 j
R = 28.4 i - 6.47 j

Example 12 (Continued). Find A + B + C
A
B
R
C
R x = 28.4 m
R
R y = -6.47 m
2 2
R (28.4 m) (6.47 m) R = 29.1 m
6.47 m
tan = 12.8 0
28.4 m
S. of E.

Vector Difference
For vectors, signs are indicators of direction.
Thus, when a vector is subtracted, the sign
(direction) must be changed before adding.
First Consider A + B Graphically:
B
A
R = A + B
A
R
B

Vector Difference
For vectors, signs are indicators of direction.
Thus, when a vector is subtracted, the sign
(direction) must be changed before adding.
Now A – B: First change sign (direction)
of B, then add the negative vector.
B -B
A
A
A
R’
-B

Addition and Subtraction
Subtraction results in a significant difference
both in the magnitude and the direction of
the resultant vector. |(A – B)| = |A| - |B|
Comparison of addition and subtraction of B
B
A
R = A + B
A
R
R’ = A - B
A
B R’ -B

Example 13. Given A = 2.4 km, N and
B = 7.8 km, N: find A – B and B – A.
A – B;
B - A
A - B
+A
-B
R
B - A
-A
+B
R
A
2.43 N
B
7.74 N
(2.43 N – 7.74 S)
5.31 km, S
(7.74 N – 2.43 S)
5.31 km, N

Summary for Vectors
• A scalar quantity is completely specified
by its magnitude only. (40(
m, m 10 gal)
• A vector quantity is completely specified by
its magnitude and direction. (40(
m, 30 0 )
Components of R:
R x = R cos
R y = R sin
R
R x
R y

Summary Continued:
• Finding the resultant of two perpendicular
vectors is like converting from polar (R, )
to the rectangular (R x , R y ) coordinates.
Resultant of Vectors:
R x y
tan
2 2
y
x
R
R x
R y

Component Method for Vectors
• Start at origin and draw each vector in
succession forming a labeled polygon.
• Draw resultant from origin to tip of last
vector, noting the quadrant of resultant.
• Write each vector in i,j notation ( (R x ,R y ).
• Add vectors algebraically to get resultant
in i,j notation. Then convert to ( (R

Vector Difference
For vectors, signs are indicators of direction.
Thus, when a vector is subtracted, the sign
(direction) must be changed before adding.
Now A – B: First change sign (direction)
of B, then add the negative vector.
B -B
A
A
A
R’
-B

Conclusion of Chapter 3B - Vectors