Chapter Three
Vectors
Vector Fundamentals
A vector quantity has two or more variables which define it.
A scalar quantity only has size (i.e. temperature, time,
energy, etc.)
Vectors in physics have magnitude and direction:
length – represents the magnitude
tail
head
Equivalent vectors are parallel with the same
magnitude.
Opposite/Negative vector are parallel
vectors with the same magnitude but in the
opposite direction.
 v1 
v 
 2
. 
 
. 
 vn 
Vector Addition
• Vector Addition is the combination of two
or more vectors into a single vector which
has the same effect/results.
• Resultant: Results of vector addition
R  F1  F2  F3  ...  Fn
Net Velocity Upstream
Upstream: Place vectors head to tail,
net result, 5 km/hr upstream
R  vb  vc
vc
R
vb
Note: Subtraction is adding the opposite
Net Velocity Downstream
Downstream: Place vectors head to tail
R  vb  vc
vc
R
vb
Can you explain why in one
case, addition results in the
difference between
magnitudes and the other it
is the sum of the
magnitudes?
What if nonparallel?
Triangle Method:
b
a
R
b
a
Parallelogram Method:
Commutative Property:
a b b a
Components
Any vector can be considered to be the resultant of an
infinite number of vectors.
X- and Y-Components
A  Ax  Ay
Ax  A Cos
y

A
Ay

Ay  A Sin
x
A
2
Ax  Ay
Ax
  tan
1
Ay
Ax
2
Unit Vectors
•Scalar Multiplication: The magnitude of any vector can
be multiplied by a scalar.
•Unit Vector: A vector with a magnitude of one: xˆ , yˆ , zˆ, etc
•Unit Vector Form: Scalar times a unit vector eg
Ax  Ax xˆ
Ax  Ax xˆ
A  Ax xˆ  Ay yˆ
A  Ax xˆ  Ay yˆ

 

A  B  Ax  Bx xˆ  Ay  By yˆ
A  B   Ax  Bx  xˆ   Ay  B y  yˆ
Unit vectors are often written without the absolute
value marks and vector symbols. Contextual
understanding.
Using Components to
Combine Vectors
Given two vectors with the following values:
A  Ax xˆ  Ay yˆ  3 xˆ  4 yˆ
B  Bx xˆ  By yˆ  5 xˆ  8 yˆ
 


A  B   Ax  Bx  x  Ay  By  y


 3  5x  4  8 y


 2 x  12 y
Comparison with the graphical
method!

y
 
A B 
B
A

x
 2  12
2
2
 4  144  148  12.2
Finding the angle the vector makes
with the y-axis?
A  B  2 xˆ  12 yˆ
opp = 2
y

adj = 12
x
opp 2 1
1 1
tan  


   tan
 9.50
adj 12 6
6
3 - Way Tug-o-War
Bugs Bunny, Yosemite Sam, and
the Tweety Bird are fighting over a
giant 450 g Acme super ball. If
their forces remain constant, how
far, and in what direction, will the
ball move in 3 s, assuming the
super ball is initially at rest ?
Sam:
111 N
Tweety:
64 N
38°
a, so we
But in order to find a, we
43°
To answer this question, we must find
can do kinematics.
must first find Fnet.
Bugs:
95 N
continued on next slide
3 - Way Tug-o-War…
Sam:
111 N
68.3384 N
38°
87.4692 N
43.6479 N
Tweety: 64 N
43°
46.8066 N
Bugs:
95 N
First, all vectors are split into horiz. & vert. comps. Sam’s are purple,
Tweety’s orange. Bugs is already done since he’s purely vertical.
The vector sum of all components is the same as the sum of the
original three vectors. Avoid much rounding until the end.
continued on next slide
3 - Way Tug-o-War…
68.3384 N
43.6479 N
87.4692 N
46.8066 N
95 N
16.9863 N
Next we combine all parallel vectors by
adding or subtracting:
68.3384 + 43.6479 - 95 = 16.9863, and
87.4692 - 46.8066 = 40.6626. A new
picture shows the net vertical and
horizontal forces on the super ball.
Interpretation: Sam & Tweety together
slightly overpower Bugs vertically by
about 17 N. But Sam & Tweety oppose
each other horizontally, where Sam
overpowers Tweety by about 41 N.
40.6626 N
continued on next slide
3 - Way Tug-o-War…
Fnet = 44.0679 N
16.9863 N

40.6626 N
Find Fnet using the Pythagorean theorem. Find  using
trig: tan = 16.9863 N / 40.6626 N. The newtons cancel out,
so  = tan-1(16.9863 / 40.6626) = 22.6689. (tan-1 is the same
as arctan.) Therefore, the superball experiences a net force
of about 44 N in the direction of about 23 north of west.
This is the combined effect of all three cartoon characters.
continued on next slide
3 - Way Tug-o-War…
a = Fnet / m = 44.0679 N / 0.45 kg = 97.9287 m/s2. Note the
conversion from grams to kilograms, which is necessary since 1
m/s2 = 1 N / kg. As always, a is in the same direction as Fnet.. a
is constant for the full 3 s, since the forces are constant.
97.9287 m/s2
22.6689
Now it’s kinematics time:
Using the fact
x = v0 t + 0.5 a t 2
= 0 + 0.5 (97.9287)(3)2
= 440.6792 m  441 m,
rounding at the end.
So the super ball will move about 441 m at about 23 N of W. To find out how
far north or west, use trig and find the components of the displacement
vector.
Practice Problem
The 3 Stooges are fighting over a 10 000 g (10 thousand gram) Snickers
Bar. The fight lasts 9.6 s, and their forces are constant. The floor on
which they’re standing has a huge coordinate system painted on it, and
the candy bar is at the origin. What are its final coordinates?
Hint: Find this
angle first.
Larry:
150 N
Curly: 1000
N
78
93
Moe:
500 N
Answer:
( -203.66 , 2246.22 )
in meters
How to move a stubborn mule
It would be pretty tough to budge this mule by pulling
directly on his collar. But it would be relatively easy to
budge him using this set-up.
(explanation on next slide)
Big Force
Little Force
How to move a stubborn mule…
little force
tree
mule
overhead view
Just before the mule budges, we have static equilibrium.
This means the tension forces in the rope segments must
cancel out the little applied force. But because of the small
angle, the tension is huge, enough to budge the mule!
(more explanation on next slide)
little force
tree
T
T
mule
How to budge a stubborn mule (final)
Because  is so small, the tensions must be large to have vertical
components (orange) big enough to team up and cancel the little
force. Since the tension is the same throughout the rope, the big
tension forces shown acting at the middle are the same as the
forces acting on the tree and mule. So the mule is pulled in the
direction of the rope with a force equal to the tension. This set-up
magnifies your force greatly.
little force

tree
T

T
mule
River Crossing
campsite
Current
river
0.3 m/s
boat
You’re directly across a 20 m wide river from your buddies’ campsite.
Your only means of crossing is your trusty rowboat, which you can
row at 0.5 m/s in still water. If you “aim” your boat directly at the
camp, you’ll end up to the right of it because of the current. At what
angle should you row in order to trying to land right at the campsite,
and how long will it take you to get there?
continued on next slide
River Crossing
campsite
0.3 m/s
0.5 m/s
river
 0.4 m/s
Current 0.3 m/s
boat
Because of the current, your boat points in the direction of red but
moves in the direction of green. The Pythagorean theorem tells us that
green’s magnitude is 0.4 m/s. This is the speed you’re moving with
respect to the campsite. Thus:
t = d / v = (20 m) / (0.4 m/s) = 50 s.  = tan-1(0.3 / 0.4)  36.9.
continued on next slide
Law of Sines
The river problem involved a right triangle. If it hadn’t we would have had to
use either component techniques or the two laws you’ll also do in trig class:
Law of Sines & Law of Cosines.
C
b
A
Law of Sines:
a
B
c
sin
a
sin B
=
b
sin C
=
c
Law of Cosines
C
b
A
c
a
B
a 2 = b 2 + c 2 - 2bc cosA
These two sides
are repeated.
This side is always opposite this angle.
It doesn’t matter which side is called a, b, and c, so long as the two rules
above are followed. This law is like the Pythagorean theorem with a built in
correction term of -2 b c cos A. This term allows us to work with non-right
triangles. Note if A = 90, this term drops out (cos 90 = 0), and we have the
normal Pythagorean theorem.
Wonder Woman Jet Problem
Suppose Wonder Woman is flying her invisible jet. Her onboard controls
display a velocity of 304 mph 10 E of N. A wind blows at 195 mph in the
direction of 32 N of E. What is her velocity with respect to Aqua Man, who is
resting poolside down on the ground?
WA = vel. of Wonder Woman w/ resp. to the air
v
vAG = vel. of the air w/ resp. to the ground (and Aqua Man)
vWG = vel. of Wonder Woman w/ resp. to the ground (Aqua Man)
We know the first two vectors; we need to find
the third. First we’ll find it using the laws of
sines & cosines, then we’ll check the result
using components. Either way, we need to make
a vector diagram.
continued on next slide
Wonder Woman Jet Problem
32
vWG
10
vWA + vAG = vWG
(cont.)
32
80 100

vWG
80
The 80 angle at the lower right is the complement of the 10 angle.
The two 80 angles are alternate interior. The 100 angle is the
supplement of the 80 angle. Now we know the angle between red
and blue is 132.
continued on next slide
Wonder Woman Jet Problem
By the law of cosines v 2 = (304)2 + (195)2 - 2 (304) (195) cos
132. So, v = 458 mph. Note that the last term above appears
negative, but it’s actually positive, since cos 132 < 0. The
law of sines says:
sin 132
sin
v
=
195
So, sin = 195 sin 132 / 458, and   18.45
132
v

80
This mean the angle between green and the
horizontal is 80 - 18.45  61.6
Therefore, from Aqua Man’s perspective,
Wonder Woman is flying at 458 mph at
61.6 N of E.
Using the Component Method
This time we’ll add vectors via components as we’ve done
before. Note that because of the angles given here, we use
cosine for the vertical comp. of red but sine for the vertical
comp. of blue. All units are mph.
103.3343
32
165.3694
299.3816
10
52.789
continued on next
Component Method…
165.3694
52.789
103.3343
Combine vertical & horiz. comps. separately and use Pythag.
theorem.  = tan-1(218.1584 / 402.7159) = 28.4452.  is
measured from the vertical, which is why it’s 10 more than
.
218.1584 mph
52.78
9
165.3694
103.3343

Comparison of Methods
We ended up with same result for Wonder Woman doing it
in two different ways. Each way requires some work. You
will only want to use the laws of sines & cosines if:
• the vectors form a triangle.
• you’re dealing with exactly 3 vectors.
(If you’re adding 3 vectors, the resultant makes
a total of 4, and this method would require using 2
separate triangles.)
Regardless of the method, draw a vector diagram! To
determine which two vectors add to the third, use the
subscript trick.