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 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
1h
Expectations (Continued):
• College algebra and simple formula
manipulation are assumed.
Example:
 v0  v f
x
 2
v0 

t

v f t  2x
t
Solve for vo
Expectations (Continued)
• You must be able to work in scientific
notation.
Evaluate the following:
(6.67 x 10-11)(4 x 10-3)(2)
F = -------- = ------------3)2
2
(8.77
x
10
r
Gmm’
F = 6.94 x 10-9 N = 6.94 nN
Expectations (Continued)
• You must be familiar with SI prefixes
The meter (m)
1 m = 1 x 100 m
1 Gm = 1 x 109 m
1 nm = 1 x 10-9 m
1 Mm = 1 x 106 m
1 mm = 1 x 10-6 m
1 km = 1 x 103 m
1 mm = 1 x 10-3 m
Expectations (Continued)
• You must have mastered right-triangle
trigonometry.
R
y
q
x
y
y = R sin q
sin q 
R
x
x = R cos q
cos q 
R
y 2
tan q 
R = x2 + y2
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 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 the length of the actual path
taken by an object.
s = 20 m
A
B
A scalar quantity:
Contains magnitude
only and consists of a
number and a unit.
(20 m, 40 mi/h, 10 gal)
Displacement—A Vector Quantity
• Displacement is the straight-line
separation of two points in a specified
direction.
D = 12 m, 20o
A
q
B
A vector quantity:
Contains magnitude
AND direction, a
number, unit & angle.
(12 m, 300; 8 km/h, N)
Distance and Displacement
• Displacement is the x or y coordinate of
position. Consider a car that travels 4
m, E then 6 m, W.
D
Net displacement:
4 m,E
x = -2
x = +4
6 m,W
D = 2 m, W
What is the distance
traveled?
10 m !!
Identifying Direction
A common way of identifying direction
is by reference to East, North, West,
and South. (Locate points below.)
Length = 40 m
N
40 m, 50o N of E
W
60o
60o
50o
60o
E
40 m, 60o N of W
40 m, 60o W of S
S
40 m, 60o S of E
Identifying Direction
Write the angles shown below by using
references to east, south, west, north.
N
W
45o
E
50o
S
N
W
E
S
0 S of
50
Click
to Esee the Answers
450 W. of
. .N
Vectors and Polar Coordinates
Polar coordinates (R,q) are an excellent
way to express vectors. Consider the
vector 40 m, 500 N of E, for example.
90o
180o
270o
90o
40 m
180o
50o
0o
270o
R
q
0o
R is the magnitude and q is the direction.
Vectors and Polar Coordinates
Polar coordinates (R,q) are given for each
of four possible quadrants:
90o
(R,q) = 40 m, 50o
120o
210o
180o
60o
60o
50o
60o
3000
270o
0o
(R,q) = 40 m, 120o
(R,q) = 40 m, 210o
(R,q) = 40 m, 300o
Rectangular Coordinates
y
(-2, +3)
(+3, +2)
+
(-1, -3)
+
x
Reference is made to
x and y axes, with +
and - numbers to
indicate position in
space.
Right, up = (+,+)
-
Left, down = (-,-)
(+4, -3)
(x,y) = (?, ?)
Trigonometry Review
• Application of Trigonometry to Vectors
Trigonometry
R
y
q
x
y
sin q 
R
x
cos q 
R
y
tan q 
x
y = R sin q
x = R cos q
R2 = x2 + y2
Example 1: Find the height of a building
if it casts a shadow 90 m long and the
indicated angle is 30o.
The height h is opposite 300 and
the known adjacent side is 90 m.
opp
h
tan 30 

adj 90 m
0
h
300
h = (90 m) tan 30o
90 m
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,q) are illustrated below.
x = R cos q
R
q
x
y
y = R sin q
Finding components:
Polar to Rectangular Conversions
Example 2: A person walks 400 m in a
direction of 30o N of E. How far is the
displacement east and how far north?
N
N
R
q
x
400 m
y
30o
E
y=?
x=?
The x-component (E) is ADJ:
x = R cos q
The y-component (N) is OPP:
y = R sin q
E
Example 2 (Cont.): A 400-m walk in a
direction of 30o N of E. How far is the
displacement east and how far north?
N
Note: x is the side
400 m
30o
y=?
x=?
E
x = (400 m) cos 30o
= +346 m, E
adjacent to angle 300
ADJ = HYP x Cos 300
x = R cos q
The x-component is:
Rx = +346 m
Example 2 (Cont.): A 400-m walk in a
direction of 30o N of E. How far is the
displacement east and how far north?
N
Note: y is the side
400 m
30o
y=?
x=?
E
opposite to angle 300
OPP = HYP x Sin 300
y = R sin q
y = (400 m) sin 30o
The y-component is:
= + 200 m, N
Ry = +200 m
Example 2 (Cont.): A 400-m walk in a
direction of 30o N of E. How far is the
displacement east and how far north?
N
400 m
30o
Rx =
Ry =
+200 m
E
The x- and ycomponents are
each + in the
first quadrant
+346 m
Solution: The person is displaced 346 m east
and 200 m north of the original position.
Signs for Rectangular Coordinates
90o
First Quadrant:
R is positive (+)
R
q
+
+
0o > q < 90o
0o
x = +; y = +
x = R cos q
y = R sin q
Signs for Rectangular Coordinates
90o
180o
+
R
Second Quadrant:
R is positive (+)
q
90o > q < 180o
x=-;
y=+
x = R cos q
y = R sin q
Signs for Rectangular Coordinates
Third Quadrant:
R is positive (+)
180o
q
180o > q < 270o
x=-
-
R
270o
y=-
x = R cos q
y = R sin q
Signs for Rectangular Coordinates
Fourth Quadrant:
R is positive (+)
q
+ 360o
R
270o
270o > q < 360o
x=+
y=-
x = R cos q
y = R sin q
Resultant of Perpendicular Vectors
Finding resultant of two perpendicular vectors is
like changing from rectangular to polar coord.
R
q
x
R x y
2
y
2
y
tan q 
x
R is always positive; q 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 has40direction
just like length
lb
does. We can treat force vectors just as40
welb
have length vectors to find the resultant
force. The procedure is the same!
4 cm = 40 lb
30 lb
30 lb
3 cm = 30 lb
Finding Resultant: (Cont.)
Finding (R,q) from given (x,y) = (+40, -30)
40 lb
Rx
q
f
R=
tan f =
Ry
R
30 lb
x2 + y 2
-30
40
R=
40 lb
30 lb
(40)2 + (30)2 = 50 lb
f = -36.9o
q = 323.1o
R
q
Ry
f
40 lb
R = 50 lb
Rx
40 lb Rx
q
q
30 lb
R
Ry
Rx 40 lb
Rx
f
f
Ry
30 lb
R
q
30 lb
Four Quadrants: (Cont.)
R = 50 lb
40 lb
Ry
R
30 lb
f = 36.9o; q = 36.9o; 143.1o; 216.9o; 323.1o
Unit vector notation (i,j,k)
y
j
k
z
Consider 3D axes (x, y, z)
i
x
Define unit vectors, i, j, k
Examples of Use:
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;
then 40 m, N. Write her displacement
in i,j notation and in R,q notation.
In i,j notation, we have:
+40 m
R
f
-30 m
R = Rxi + Ry j
Rx = - 30 m
Ry = + 40 m
R = -30 i + 40 j
Displacement is 30 m west and 40 m
north of the starting position.
Example 4 (Cont.): Next we find her
displacement in R,q notation.
+40
m
R
40
0
tan f 
; f = 59.1
30
f
q = 1800 – 59.10
-30 m
q = 126.9o
R  (30)  (40)
2
2
R = 50 m
(R,q) = (50 m, 126.9o)
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
f?
R  (46 km) 2  (35 km) 2 35
km
B
R=?
R = 57.8 km
A
46 km
tan f 
35 km
q = 1800 + 52.70
f = 52.70 S. of W.
q = 232.70
Example 7. Find the components of the 240-N
force exerted by the boy on the girl if his arm
makes an angle of 280 with the ground.
F = 240 N
Fy
F
280
Fy
Fx
Fx = -|(240 N) cos 280| = -212 N
Fy = +|(240 N) sin 280| = +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 lawnmower. The angle with the ground is 320.
F = 300 N
Fx
32o
32o
Fx = -|(300 N) cos 320| = -254 N
Fy = -|(300 N) sin 320| = -159 N
Fy
320
F
Fy
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,q).
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
N
3 km, W
C
B
4 km, N
A
2 km, E
E
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.
Example 9 (Cont.) Find resultant displacement.
3. Write each vector
in i,j notation:
A = +2 i
B=
+4j
C = -3 i
D=
-2j
D
2 km, S
N
3 km, W
C
B
4 km, N
A
2 km, E
E
R = -1 i + 2 j
4. Add vectors A,B,C,D
algebraically to get
resultant in i,j notation.
1 km, west and 2
km north of origin.
5. Convert to R,q notation
See next page.
Example 9 (Cont.) Find resultant displacement.
Resultant Sum is:
R = -1 i + 2 j
D
2 km, S
N
3 km, W
C
Now, We Find R, q
R  (1)  (2)  5
2
2
B
4 km, N
A
2 km, E
R = 2.24 km
2 km
tan f 
1 km
f = 63.40 N or W
R
f
Rx = -1 km
Ry= +2
km
E
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, 63.40 N of W
Significant Digits for Angles
Since a tenth of a
degree can often be
significant, sometimes a
fourth digit is needed.
q=
36.9o;
323.1o
q
Rule: Write angles to
the nearest tenth of a
degree. See the two
examples below:
30 lb
R
q
Ry
Rx
Rx
f
40 lb
40 lb
Ry
R
30 lb
Example 10: Find R,q for the three vector
displacements below:
A = 5 m, 00
B = 2.1 m, 200
C = 0.5 m, 900
R
q
A=5m
B
C=
0.5 m
200
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,q)
3. Write each vector in i,j notation. (Continued ...)
Example 10: Find R,q for the three vector
displacements below: (A table may help.)
For i,j notation
find x,y components of each
vector A, B, C.
Vector
f
A=5 m
00
R
q
A=5m
X-component (i)
+5m
B=2.1m 200 +(2.1 m) cos 200
C=.5 m 900
B
0
Rx = Ax+Bx+Cx
C=
0.5 m
200
B = 2.1 m
Y-component (j)
0
+(2.1 m) sin 200
+ 0.5 m
Ry = Ay+By+Cy
Example 10 (Cont.): Find i,j for three
vectors: A = 5 m,00; B = 2.1 m, 200; C = 0.5 m,
900.
X-component (i)
Y-component (j)
Ax = + 5.00 m
Ay = 0
Bx = +1.97 m
By = +0.718 m
Cx = 0
Cy = + 0.50 m
4. Add vectors to
get resultant R
in i,j notation.
A = 5.00 i +
0j
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:
A = 5 m,00; B = 2.1 m, 200; C = 0.5 m, 900.
R = 6.97 i + 1.22 j
5. Determine R,q from x,y:
R  (6.97 m)  (1.22 m)
2
R = 7.08 m
1.22 m
tan f 
6.97 m
Diagram for
finding R,q:
R
2
q
Ry
1.22 m
Rx= 6.97 m
q = 9.930 N. of E.
Example 11: A bike travels 20 m, E then 40 m
at 60o N of W, and finally 30 m at 210o. What
is the resultant displacement graphically?
C = 30 m
B=
40 m
30o
R
f
q
Graphically, we use
ruler and protractor
to draw components,
then measure the
Resultant R,q
60o
A = 20 m, E
Let 1 cm = 10 m
R = (32.6 m, 143.0o)
A Graphical Understanding of the Components
and of the Resultant is given below:
Cy
By
Note: Rx = Ax + Bx + Cx
30o
B
C
R
Ry
f
q
Ry = Ay + By + Cy
60o A
Rx
Cx
0
Ax
Bx
Example 11 (Cont.) Using the Component
Method to solve for the Resultant.
Cy B
y
Ry
Write each vector
in i,j notation.
30o
C
R
f
B
q
Rx
Cx
Bx
60
Ax = 20 m, Ay = 0
A
A = 20 i
A
Bx = -40 cos 60o = -20 m
x
By = 40 sin 60o = +34.6 m
Cx = -30 cos 30o = -26 m
Cy = -30 sin 60o = -15 m
B = -20 i + 34.6 j
C = -26 i - 15 j
Example 11 (Cont.) The Component Method
Cy B
y
Ry
+19.6
30o
C
R
f
Add algebraically:
A = 20 i
B
q
B = -20 i + 34.6 j
60
Rx
A
Cx
x
Bx
R
f
-26
C = -26 i - 15 j
A
R=
R = -26 i + 19.6 j
(-26)2 + (19.6)2 = 32.6 m
tan f =
19.6
-26
q = 143o
Example 11 (Cont.) Find the Resultant.
R = -26 i + 19.6 j
Cy B
y
30o
B
C
Ry
R
f
q
Rx
Cx
60
A
+19.6
A
Bx
R
f
-26
x
The Resultant Displacement of the bike is best
given by its polar coordinates R and q.
R = 32.6 m; q = 1430
Example 12. Find A + B + C for Vectors
Shown below.
B
Cx
A = 5 m, 900
350 C
y
A
B = 12 m, 00
C
q
0
C = 20 m, -35
R
Ax = 0; Ay = +5 m
Cx = (20 m) cos 350
A=
0 i + 5.00 j
B = 12 i +
0j
C = 16.4 i – 11.5 j
Cy = -(20 m) sin -350
R = 28.4 i - 6.47 j
Bx = +12 m; By = 0
Example 12 (Continued). Find A + B + C
B
Rx = 28.4 m
q
350
A
R
C
q
R
Ry = -6.47 m
R  (28.4 m)  (6.47 m)
2
6.47 m
tan f 
28.4 m
2
R = 29.1 m
q = 12.80 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
R=A+B
R
A
A
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
A
-B
R’
A
A
-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
R=A+B
R
A
A
B
R’ = A - B
A
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
A
B
2.43 N 7.74 N
B-A
+B
-A
R
(2.43 N – 7.74 S)
(7.74 N – 2.43 S)
5.31 km, S
5.31 km, N
Summary for Vectors
 A scalar quantity is completely specified
by its magnitude only. (40 m, 10 gal)
 A vector quantity is completely specified by
its magnitude and direction. (40 m, 300)
Components of R:
Rx = R cos q
Ry = R sin q
R
q
Rx
Ry
Summary Continued:
 Finding the resultant of two perpendicular
vectors is like converting from polar (R, q)
to the rectangular (Rx, Ry) coordinates.
Resultant of Vectors:
R x y
2
y
tan q 
x
2
R
q
Rx
Ry
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 (Rx,Ry).
 Add vectors algebraically to get resultant
in i,j notation. Then convert to (R,q).
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
A
-B
R’
A
A
-B
Conclusion of Chapter 3B - Vectors