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

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

3. 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.

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

5. Expectations (Continued):
College algebra and simple formula manipulation are assumed.
Example:
Solve for vo

6. Expectations (Continued)
You must be able to work in scientific notation.
Evaluate the following:
(6.67 x 10-11)(4 x 10-3)(2)
(8.77 x 10-3)2
F = =
Gmm’ r2
F = 6.94 x 10-9 N = 6.94 nN

7. Expectations (Continued)
You must be familiar with SI prefixes
The meter (m) m = 1 x 100 m
1 Gm = 1 x 109 m nm = 1 x 10-9 m
1 Mm = 1 x 106 m mm = 1 x 10-6 m
1 km = 1 x 103 m mm = 1 x 10-3 m

8. Expectations (Continued)
You must have mastered right-triangle trigonometry. y x R q
y = R sin q
x = R cos q
R2 = x2 + y2

9. 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

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

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

12. Displacement—A Vector Quantity
Displacement is the straight-line separation of two points in a specified direction.
A vector quantity:
Contains magnitude AND direction, a number, unit & angle.
(12 m, 300; 8 km/h, N) A B
D = 12 m, 20o q

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

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

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

16. 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. 0o
180o
270o
90o q 0o
180o
270o
90o R
40 m
50o
R is the magnitude and q is the direction.

17. Vectors and Polar Coordinates
Polar coordinates (R,q) are given for each of four possible quadrants: 0o
180o
270o
90o
(R,q) = 40 m, 50o
120o
60o
210o
50o
(R,q) = 40 m, 120o
3000
60o
60o
(R,q) = 40 m, 210o
(R,q) = 40 m, 300o

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

19. Trigonometry Review Application of Trigonometry to Vectors y = R sin qx R q
x = R cos q
R2 = x2 + y2

20. Example 1: Find the height of a building if it casts a shadow 90 m long and the indicated angle is 30o.
90 m
300
The height h is opposite 300 and the known adjacent side is 90 m. h
h = (90 m) tan 30o
h = 57.7 m

21. 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
y = R sin q x y R q
Finding components:
Polar to Rectangular Conversions

22. Example 2: A person walks 400 m in a direction of 30o N of E
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?
x = ?
y = ?
400 m
30o E N x y R q N E
The x-component (E) is ADJ:
x = R cos q
The y-component (N) is OPP:
y = R sin q

23. Note: x is the side adjacent to angle 300
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?
x = ?
y = ?
400 m
30o E N
Note: x is the side adjacent to angle 300
ADJ = HYP x Cos 300
x = R cos q
The x-component is:
Rx = +346 m
x = (400 m) cos 30o
= +346 m, E

24. Note: y is the side opposite to angle 300
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?
x = ?
y = ?
400 m
30o E N
Note: y is the side opposite to angle 300
OPP = HYP x Sin 300
y = R sin q
The y-component is:
Ry = +200 m
y = (400 m) sin 30o
= m, N

25. The x- and y- components are each + in the first quadrant
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?
Rx = +346 m
Ry = +200 m
400 m
30o E N
The x- and y- components are each + in the first quadrant
Solution: The person is displaced 346 m east and 200 m north of the original position.

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

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

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

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

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

31. 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 direction just like length does. We can treat force vectors just as we have length vectors to find the resultant force. The procedure is the same!
30 lb
40 lb
40 lb
30 lb
4 cm = 40 lb
3 cm = 30 lb

32. Finding Resultant: (Cont.)
Finding (R,q) from given (x,y) = (+40, -30) Rx
40 lb
30 lb f q
40 lb
30 lb Ry R
R = (40)2 + (30)2 = 50 lb
R = x2 + y2
tan f =
-30 40
q = 323.1o
f = -36.9o

33. Four Quadrants: (Cont.)
40 lb
30 lb R q Ry Rx
40 lb
30 lb R q Ry Rx
R = 50 lb
40 lb
30 lb R f q Ry Rx
40 lb
30 lb R f q Ry Rx
f = 36.9o; q = 36.9o; 143.1o; 216.9o; 323.1o

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

35. Displacement is 30 m west and 40 m north of the starting position.
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
R = Rxi + Ry j
Rx = - 30 m
Ry = + 40 m
-30 m
R = -30 i + 40 j
Displacement is 30 m west and 40 m north of the starting position.

36. Example 4 (Cont.): Next we find her displacement in R,q notation.
q = 126.9o
R = 50 m
(R,q) = (50 m, 126.9o)

37. Example 6: Town A is 35 km south and 46 km west of Town B
Example 6: Town A is 35 km south and 46 km west of Town B. Find length and direction of highway between towns. B
46 km
R = -46 i – 35 j
35 km
R = ?
f=?
R = 57.8 km A
q =
q =
f = S. of W.

38. 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.
280
F = 240 N F Fy Fx
Fx = -|(240 N) cos 280| = -212 N
Or in i,j notation:
F = -(212 N)i + (113 N)j
Fy = +|(240 N) sin 280| = +113 N

39. Example 8. Find the components of a 300-N force acting along the handle of a lawn-mower. The angle with the ground is 320.
320
F = 300 N F Fy Fx
32o
32o
Or in i,j notation:
F = -(254 N)i - (159 N)j
Fx = -|(300 N) cos 320| = -254 N
Fy = -|(300 N) sin 320| = -159 N

40. 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).

41. Example 9. A boat moves 2. 0 km east then 4. 0 km north, then 3
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. E N
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 km, S D
4 km, N B
3 km, W C
2 km, E A
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.

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

43. Example 9 (Cont.) Find resultant displacement.
Resultant Sum is:
R = -1 i + 2 j E N
2 km, E A
4 km, N B
3 km, W C
2 km, S D
Now, We Find R, 
R = 2.24 km
Ry= +2 km
Rx = -1 km R f
 = N or W

44. Reminder of Significant Units:
2 km A
4 km B
3 km C D
For convenience, we follow the practice of assuming three (3) significant figures for all data in problems.
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, N of W

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

46. Example 10: Find R,q for the three vector displacements below:
B = 2.1 m, 200
C = 0.5 m, 900
C = m R q
B = 2.1 m
200 B
A = 5 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 ...)

47. Example 10: Find R,q for the three vector displacements below: (A table may help.)
B = 2.1 m
200 B
C = m R q
For i,j notation find x,y compo-nents of each vector A, B, C.
Vector f
X-component (i)
Y-component (j)
A=5 m 00
+ 5 m
B=2.1m
200
+(2.1 m) cos 200
+(2.1 m) sin 200
C=.5 m
900
+ 0.5 m
Rx = Ax+Bx+Cx
Ry = Ay+By+Cy

48. Example 10 (Cont. ): Find i,j for three vectors: A = 5 m,00; B = 2
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 = m
Ay = 0
Bx = m
By = m
Cx = 0
Cy = m
6.97 i j
A = 5.00 i j
B = 1.97 i j
C = i j
4. Add vectors to get resultant R in i,j notation.
R =

49. Example 10 (Cont. ): Find i,j for three vectors: A = 5 m,00; B = 2
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 j
Rx= 6.97 m R q
Ry m
Diagram for finding R,q:
5. Determine R,q from x,y:
R = 7.08 m
q = N. of E.

50. 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
Graphically, we use ruler and protractor to draw components, then measure the Resultant R,q
B = 40 m
30o R q
60o f
A = 20 m, E
R = (32.6 m, 143.0o)
Let 1 cm = 10 m

51. A Graphical Understanding of the Components and of the Resultant is given below:q
Note: Rx = Ax + Bx + Cx Cy By B
Ry = Ay + By + Cy C Ry A Rx Ax Cx Bx

52. Write each vector in i,j notation.
Example 11 (Cont.) Using the Component Method to solve for the Resultant. 60
30o R f q Ax B Bx Rx A C Cx Ry By Cy
Write each vector in i,j notation.
Ax = 20 m, Ay = 0
A = 20 i
Bx = -40 cos 60o = -20 m
By = 40 sin 60o = m
B = -20 i j
Cx = -30 cos 30o = -26 m
C = -26 i - 15 j
Cy = -30 sin 60o = -15 m

53. Example 11 (Cont.) The Component Method
Add algebraically: 60
30o R f q Ax B Bx Rx A C Cx Ry By Cy
A = 20 i
B = -20 i j
C = -26 i - 15 j
R = -26 i j
R = (-26)2 + (19.6)2 = 32.6 m R
-26
+19.6 f
tan f =
19.6
-26
q = 143o

54. Example 11 (Cont.) Find the Resultant.60
30o R f q Ax B Bx Rx A C Cx Ry By Cy
R = -26 i j R
-26
+19.6 f
The Resultant Displacement of the bike is best given by its polar coordinates R and q.
R = 32.6 m; q = 1430

55. Example 12. Find A + B + C for Vectors Shown below.Cx
A = 5 m, 900
B = 12 m, 00
C = 20 m, -350 A C
350 Cy R q
Ax = 0; Ay = +5 m
28.4 i j
A = i j
B = i j
C = 16.4 i – 11.5 j
Bx = +12 m; By = 0
Cx = (20 m) cos 350
Cy = -(20 m) sin -350
R =

56. Example 12 (Continued). Find A + B + C
350 R q
Rx = 28.4 m R q
Ry = m
R = 29.1 m
q = S. of E.

57. First Consider A + B Graphically:
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:
R = A + B B A R B A

58. 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 A -B A R’ -B

59. 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
R’ = A - B A R B R’ -B A

60. Example 13. Given A = 2. 4 km, N and B = 7
Example 13. Given A = 2.4 km, N and B = 7.8 km, N: find A – B and B – A.
A N
B N
A – B; B - A
A - B
B - A +B +A -A -B R R
(2.43 N – 7.74 S)
(7.74 N – 2.43 S)
5.31 km, S
5.31 km, N

61. 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) Rx Ry R q
Components of R:
Rx = R cos q
Ry = R sin q

62. 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: Rx Ry R q

63. 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).

64. 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 A -B A R’ -B

65. Conclusion of Chapter 3B - Vectors