Vectors
Thanks to PowerPoint from
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
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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.
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http://www.khanacademy.org/science/physic
s/one-dimensional-motion/displacementvelocity-time/v/introduction-to-vectorsand-scalars
Surveyors use accurate measures of
magnitudes and directions to create
scaled maps of large regions.
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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
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College algebra and simple formula
manipulation are assumed.
Example:
 v0  v f
x
 2

t

Solve for vo
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Let’s do Table talks—I will call on one person
at one table. They will tell me their groups
answer. I will then call on each table and ask
if they agree or disagree and why.
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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’
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Turn to your neighbor and explain what the
exponent would be in this equation.
Let’s see who agrees and disagrees before
you figure out the answer
-8
Now with your table figure out the answersee if you can do it without a calculator.
F = 6.94 x 10-9 N = 6.94 nN
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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
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You are familiar with right-triangle
trigonometry.
y
y = R sin q
sin q 
R
R
y
x = R cos q
q
x
y 2
tan q 
R = x2 + y2
x
Length
Weight
Time
We begin with the measurement of length:
its magnitude and its direction.
 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 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)
• 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 !!
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
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
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.
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
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) = (?, ?)
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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
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
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
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
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
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
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.
90o
First Quadrant:
R is positive (+)
R
q
+
+
0o > q < 90o
0o
x = +; y = +
x = R cos q
y = R sin q
90o
180o
+
R
Second Quadrant:
R is positive (+)
q
90o > q < 180o
x=-;
y=+
x = R cos q
y = R sin q
Third Quadrant:
R is positive (+)
180o
q
180o > q < 270o
x=-
-
R
270o
y=-
x = R cos q
y = R sin q
Fourth Quadrant:
R is positive (+)
q
+ 360o
R
270o
270o > q < 360o
x=+
y=-
x = R cos q
y = R sin q
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
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 (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
f is S of E
q = 323.1o
30 lb
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
R
R = 50 lb
40 lb
Ry
R
30 lb
f = 36.9o; q = 36.9o; 143.1o; 216.9o; 323.1o