VECTORS
1. Scalars
• Just a Value
• This Value is called a Magnitude
2. Vectors
VECTORS
• Quantities that have 
MAGNITUDE (size or value)
AND
DIRECTION
REPRESENTATION OF VECTOR QUANTITIES
• VECTORS ARE
REPRESENTED BY AN
ARROW
•tip
•tail
•THE ARROW:
LENGTH =
• THE MAGNITUDE OR SIZE OF THE
VECTOR
THE ARROW’S DIRECTION =
• IS THE DIRECTION OF THE
VECTOR
EXAMPLES OF VECTORS
•
•
•
•
•
•
•
FORCE (a push or a pull)
ELECTRIC/MAGNETIC FIELD STRENGTH
ACCELERATION
TORQUE – twist causing rotation
DISPLACEMENT – not distance
MOMENTUM – possessed by moving mass
VELOCITY – not speed
EXAMPLES OF SCALARS
•
•
•
•
•
Mass
Time
Distance
Energy
+ Everything else that’s not a vector…..
These quantities have “NO DIRECTION”
• Take care here
• You Can NOT Add them like
regular numbers (called
Scalars)
VECTOR ADDITION (THE TIP-TO-TAIL
METHOD) FINDING THE RESULTANT
A
B
–The SUM or RESULT of Adding 2
Vectors is called
VECTOR ADDITION (THE TIP-TO-TAIL
METHOD) FINDING THE RESULTANT
A
B
A
B
A
B
REVIEWING VECTOR ADDITION
• ADD VECTORS IN ANY ORDER (A+B =
B+A)
• IF VECTORS ARE POINTING IN THE SAME
DIRECTION  THIS IS REGULAR
ALGABRAIC ADDITION
REVIEWING VECTOR ADDITION
• POSITION THE TAIL OF ONE VECTOR TO
THE TIP OF THE OTHER
• CONNECT FROM THE TAIL OF THE 1ST
VECTOR TO THE TIP OF THE LAST
– THIS IS THE RESULTANT
ADDITION CONTINUED
A
B
A
B
MORE VECTOR ADDITION
• SUPPOSE THE VECTORS FORM A RIGHT
ANGLE
• GRAPHICAL SOLUTIONS CAN ALWAYS
BE USED BUT…
– HERE IS A MATHEMATICAL SOLUTION….
– THIS SOLUTION USES THE PYTHAGOREAN
THEORUM…
C2 = A2 + B2
B = 3 lbs
ADDING VECTORS THAT ARE AT RIGHT ANGLES TO EACH
OTHER…
R2 = 16 + 9 = 25
R = 5 lbs
A = 4 lbs
R2
= A2
+
B2
R2 = 42 + 32
BUT R = 5 lbs IS ONLY
HALF AN ANSWER!!
WHY?????
REMEMBER !!!
•VECTORS HAVE 2 PARTS
MAGNITUDE
AND
DIRECTION !!!
HERE’S HOW TO FIND THE
DIRECTION
TRIG FUNCTIONS TO REMEMBER

A = 4 lbs
B = 3 lbs
TRIG CALCULATIONS
COS() = 4/5=0.8
SIN() = 3/5= 0.6
•USE YOUR CALCULATOR TO FIND
THE ANGLE THAT HAS THESE VALUES
OF SIN OR COS. Could also use TAN
AT LAST THE ANGLE
(THE VECTOR’S DIRECTION)
SIN(X) = 0.6  ANGLE (X) = 37 DEGREES
COS(X) = 0.8 ANGLE (X) = 37 DEGREES
SO, THE OTHER HALF OF OUR
ANSWER IS…..
RESULTANT…….
= 37 deg
A = 4 lbs
B = 3 lbs
5 lbs 37 degrees NORTH OF EAST
•Not
NORTHEAST
i.e. NE is 45 deg
SUMMARY
•A VECTOR IS A DIRECTED QUANTITY
THAT HAS BOTH A MAGNITUDE AND
DIRECTION
• IF THE ANGLE BETWEEN THE VECTORS IS
• 0 deg: algebraic addition (MAXIMUM ANS.)
• 180 deg: algebraic “subtraction” (MINIMUM ANS.)
• 90 deg: use Pythagorean Theorem to find
magnitude and trig functions to find the angle
• COULD YOU PASS A QUIZ ON
THIS MATERIAL????
RETURN TO
BEGINNING
• NOW?
• LATER, WITH STUDY?
CONTINUE TO
VECTOR MATH
Vector Concepts used in Physics – Fancy Foot Work
• Imagine you were asked to mark your
starting place and walk 3 meters North,
followed by two meters East.
Could you answer the following:
– How far did you walk?
– Where are you relative to your original
spot?
Fancy Foot Work
• How far did you walk?
– This requires a MAGNITUDE ONLY
• SCALAR QUANTITY called DISTANCE
• 3m + 2m = 5m
• Where are you relative to your original
spot?
– This requires both a MAGNITUDE &
DIRECTION
• VECTOR QUANTITY called DISPLACEMENT
Fancy Foot Work
• Where are you relative to your original spot?
– This requires both a MAGNITUDE & DIRECTION
• VECTOR QUANTITY called DISPLACEMENT
2m, E
End
3m, N
Displacement: Needs Magnitude & Direction
Start
N
W
E
Fancy Foot Work
• Magnitude =?
S
2m, E
• The Length of the
Hypotenuse
s2=(3m)2 + (2m)2
3m, N
Displacement = s

s=
13 m  3.6m
• Direction =  East of North
• Pick your Trig function
D  3.6 m, 33.7o E of N
• =33.7o, E of N
Fancy Foot Work
NOW, measure the angle from the +X axis……
2m, E
3m, N
Displacement = s =
=33.7
 = 90 – 33.7 = 56.3o
3 .6 m

s  3.6m, 560
Multiplying a Vector by a Scalar
• When you multiply a vector by a scalar, it only
affects the MAGNITUDE of the vector
** Not the direction**
• Example:

o
A  20m,45

o
3A  60m,45

o
5 A  100m,45
Vector Components
• Component means “part”
• A vector can be composed of many parts known
as components
• It’s best to break a vector down into TWO
perpendicular components. WHY?
– To use Right Triangle Trig
Vector Components
• Introducing Vector V
• Vector V’s X-Component is its Projection onto the X-axis
• Vector V’s Y-Component is its Projection onto the Y-axis

V
Sub Scripts in Action
Now we have a
Right Triangle
Vy
Vx
Vector Components
• Given this diagram, find V’s X & Y Components

V
=38.66o
Vx= 5
Vy= 4
What’s the Magnitude of Vector V?

2
2
V  5 4

V  41
Vector Components
• Now, knowing the magnitude of vector V,
verify the V’s X & Y components using Trig
Vx=?
41
Vy=?
Vy
38.66o
Vx
Vx  41(Cos(38.66))
5
Vy  41(Sin(38.66))
4


Vector Components
• Golden Rules of Vector Components
– 1. If you know the magnitude and direction of vector V to be
(V,), then you can find Vx & Vy by
• Vx=Vcos
• Vy=Vsin
– 2. If Vx & Vy are known, the magnitude of the vector can be found
with Pythagorean Theorem:
V  Vx  Vy
2
2
WHAT ARE THE X & Y COMPONENTS
OF VECTOR ‘A’?
THESE ARE The
VECTOR
COMPONENTS OF A
Sin OPP/HYP
Sin Ay/A
Ay = Asin
Ax
Cos  ADJ/HYP
Cos   Ax/A
Ax = Acos
Ay
A
Ay =Asin

Ax =Acos
AN EXAMPLE
• Suppose the magnitude of A = 5 and 
 37 deg. Find the VALUES of the X & Y
components.
Ay
 =37
Ay =Asin
5sin337
5(0.6)
Ax =Acos
5(0.8)
5cos37
4
QUESTION #1
A couple on vacation are about to go sight-seeing in a city
where the city blocks are all squares. They start out at
their hotel and tour the city by walking as follows:
1 block East; 2 blocks North; 3 blocks East;
3 blocks South; 2 blocks West;1 block
South; 6 blocks East; 8 blocks North; 8
blocks West. WHAT IS THEIR
DISPLACEMENT? (i.e., WHERE ARE
THEY FROM THEIR HOTEL)?
ANSWER #1:
USING GRAPH PAPER
1 block East; 2 blocks North; 3 blocks East; 3 blocks South;
2 blocks West;1 block South; 6 blocks East; 8 blocks North;
8 blocks West.
WHERE ARE THEY FROM
THE HOTEL?
THEY ARE 6
BLOCKS
NORTH OF
THE HOTEL
H
QUESTION #2
RIVER
A river flows in the east-west direction with a current 6 mph
eastward. A kayaker (who can paddle in still water at a maximum
rate of 8 mph) wishes to cross the river in his boat to the North. If
he points the bow of his boat directly across the river and paddles
as hard as he can, what will be his resultant velocity?
ANSWER #2
6
8
USE
PYTHAGOREAN
THEOREM

 = ARCTAN 6/8
R2 =
R2
R2
62 +
82
= 36 + 64
= 100
R = 10
= 37 deg.
R = 10 mph, 37 deg East of North
or
10 mph, bearing 037 deg.
QUESTION #3
RIVER
Desired path to the South shore
The kayaker wants to go directly across the river from the North
shore to the South shore, again, paddling as fast as he can. At
what angle should the kayaker point the bow of his boat so that
he will travel directly across the river? What will be his
resultant velocity?
ANSWER #3...
RIVER
Desired path to the South shore

RESULTANT
R VELOCITY

= arctan = arctan(opp/adj)
=arc tan(6/5.3)
VEL.RIVER 2
= arctan 1.1 = 49 deg
8 = 62 + Vr2
6 mph
upstream
Vr2 = 64 - 36 = 5.3 mph
QUESTION #4:
Two forces A and B of 80 and
60 newtons respectively, act
concurrently(at the same
point, at the same time) on
point P.
A
Calculate the resultant force.
P
60 newtons
B
ANSWER #4:
Pythagorean Theorem for
3-4-5 right triangle
A


P
60 newtons
B
= arctan (80/60)
= 53 deg