Vector Quantities
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We will concern ourselves with two
measurable quantities:
Scalar quantities: physical
quantities expressed in terms of a
magnitude only. Scalar quantities
consist of a number and a unit.
Example: 100 m/s (no direction).
Vector Quantities

Vector quantities: quantities that are
described by a magnitude and a
direction. Vector quantities consist of a
number, a unit, and a direction
(basically, a scalar quantity with the
direction indicated). Example: if east is
considered to be positive, then west is
negative, so the vector can described as
100 m/s, east or +100 m/s; 80 m/s,
west or -80 m/s.
Vector Quantities
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
Vectors are represented by an arrow ().
The tail of the arrow will always be
placed at the point of origin for the
measurement of the desired quantity.
The length of the arrow indicates the
magnitude of the vector.
The orientation of the arrow indicates the
direction.
Vector Resolution
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
Resultant: is the single vector that
produces the same result as the
combination of the separate vectors.
Each separate vector is called a
component vector.
We will examine the impact of all vector
quantities acting upon an object as if all
the vectors originate at a single point.
Vector Addition
1.
2.
When two or more vectors act at the
same point in the same direction (in
other words, the angle between each of
the vectors is 0), the resultant is
determined by adding all the component
vectors together.
The direction of the resultant vector is
in the direction of the component
vectors.
Vector Addition


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
If you walk 5 m to the
right, stop, and then
walk 3 m to the right,
the total displacement
is: 5 m + 3 m = 8 m.
The 5 m vector and the
3 m vector are the
component vectors.
Resultant displacement
= 8 m to the right
Properties of Vectors
(HRW)
Vector Subtraction
1.
2.
When two vectors act at the same point,
but in opposite directions (the angle
between each vector is 180), the
resultant is equal to the difference
between the two vector quantities
(subtract).
The direction of the resultant is in the
direction of the component vector with
the largest magnitude.
Vector Subtraction
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



If you walk 5 m to the
right, stop, and then walk
3 m to the left, the total
displacement is:
5m–3m=2m
The 5 m vector and the 3
m vector are the
component vectors.
Resultant displacement = 2
m to the right.
Subtraction of Vectors
(HRW)
Head to Tail Method


When two or more vectors
act at the same time
(concurrently) on the same
point, the resultant can be
determined by placing the
vectors head to tail.
The tail of the first vector
(A) will begin at the origin
and the tail of the next
component vector (B) is
placed at the head of the
vector A.
Head to Tail Method
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

The tail of vector C is placed
at the head of vector B.
Each component vector is
drawn with the correct
orientation.
The order in which the
vectors are drawn does not
matter as long as the
magnitude and direction of
each vector is maintained
when drawn.
Head to Tail Method

The resultant vector
R would be the vector
beginning at the
origin and extending
in a straight line to
the head of the last
vector. The
determination of the
magnitude will
involve use of the
Pythagorean theorem
or the law of cosines.
Head to Tail Method
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

The direction of the resultant can be
determined by resolving the resultant
vector into x (horizontal) and y (vertical)
components and using a trig function
(sin, cos, or tan).
This will be described later.
Graphical Addition of Vectors & Adding
Vectors Algebraically (HRW)
For Vector Problems Involving
Right Angles
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
Example: move 5 m
along the x-axis and
then move 8 m up
the y-axis.
Place the origin of a
coordinate system at
the point where the
motion begins or
where the force is
applied; you will start
at (0,0) on the
coordinate system.
For Vector Problems Involving
Right Angles
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

Tail of the 5 m vector goes at the origin and
the head points along the x-axis and would
lie on 5 on the axis.
At the head of the 5 m vector, turn up and
the tail of the 8 m vector will begin and the
head points up the y-axis and would lie on 8
on the axis.
The head of the 8 m vector would lie on the
point (5,8) on an x-y coordinate plane.
For Vector Problems Involving
Right Angles
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
Resultant vector R
begins at the origin
(0,0) and ends at
the head of the 8 m
vector [at the point
(8,5)].
The 5 m vector and
the 8 m vector are
the component
vectors.
For Vector Problems Involving
Right Angles
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
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The resultant R is the
hypotenuse of a right
triangle formed by the
5 m vector and the 8 m
vector.
Use the Pythagorean
theorem to determine
the magnitude of the
resultant R:
R2 = A2 + B2
R2 = (5 m)2 + (8 m)2
R2 = 89 m2; R = 9.43 m
For Vector Problems Involving
Right Angles

To determine the direction of the
resultant vector:
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
Requires an angular measurement and
a direction moved from a reference
axis. Example: 40 above the x-axis.
Choose one of the two angles at the
point of origin. The reference axis is
the adjacent side of the selected angle.
For Vector Problems Involving
Right Angles
Trig Functions
Any one of the trig
functions
(sin,
cos,
opposite
or tan) can be used
sin θ 
hypotenuse to find the direction
of the resultant
adjacent
cos θ 
vector R.
hypotenuse
 Be sure the
calculator is in
opposite
tan θ 
degree mode!
adjacent

  sin
1
  cos
1
  tan
1
opposite
hypotenuse
adjacent
hypotenuse
opposite
adjacent
For Vector Problems Involving
Right Angles
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
Report the angle determined with
the trig function and the direction
moved from the reference axis.
Example:
8m
tan θ 
 1.6
5m
Use the inverse tan function (tan-1)
to determine the angle .
For Vector Problems Involving
Right Angles
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

 = tan-1 1.6
 = 57.995o
To get to the resultant,
you must start at the xaxis and rotate 57.995o
above the x-axis.
The magnitude of the
resultant is 9.43 m and
the direction is 57.995o
above the x-axis.
X and Y Components of a Vector
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
Vectors can be resolved (broken down)
into a component that acts along the
x-axis and a component that acts along
the y-axis.
The tail of the vector to be resolved is
placed at the origin and drawn as
indicated in the problem.
X and Y Components of a Vector

Ex: the 50 m/s
vector located 30°
above the positive xaxis will be resolved
into an x-component
and a y-component.
X and Y Components of a Vector
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
From the origin, draw a line
along the x-axis to a point
below the tip of the head of
the vector (the arrow head).
This is the x-component of
the vector.
From the origin, draw a line
along the y-axis to a point
adjacent to the tip of the
head of the vector (the
arrow head). This is the ycomponent of the vector.
X and Y Components of a Vector
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
Construct a parallelogram (either
a square or a rectangle). A
parallelogram is used because the
opposite sides of a parallelogram
are equal in magnitude. You also
have two right triangles which
you can use to solve for the
components.
The 50 m/s the diagonal of the
parallelogram and will be the
hypotenuse for the right triangle
you will use to determine the xcomponent and the y-component.
X and Y Components of a Vector

The x-component is
the adjacent side of
the right triangle
and you will use the
cosine function to
determine its
magnitude.
X and Y Components of a Vector
adj
cos θ 
hyp
x
cos 30 
50 m / s
cross multiply
x  50 m / s  cos 30
x  43 .3 m / s
X and Y Components of a Vector

The y-component is
the opposite side of
the right triangle
and you will use the
sine function to
determine its
magnitude.
X and Y Components of a Vector
opp
sin 
hyp
y
sin 30 
50 m / s
cross multiply
y  50 m / s  sin 30
y  25 m / s
X and Y Components of a Vector
Summarized
For problems involving multiple
vectors: Rectangular Resolution
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Resolve each vector into an xcomponent and a y-component using
the trig functions sine or cosine,
whichever is appropriate. Be careful to
denote the negative values for the xand y-components, when appropriate.
Add all the x-components together to
get one resultant x-component vector.
Add all the y-components together to get
one resultant y-component vector.
For problems involving multiple
vectors: Rectangular Resolution
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

Use the Pythagorean theorem to
determine the resultant vector.
Use a trig function (sine, cosine, or
tangent) to determine the angle of
orientation for the resultant vector.
Visit Adding Vectors Algebraically
(HRW)
Example 3C, p. 95
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Given two vectors:
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25.5 km at 35° south of east
41 km at 65° north of east
Draw the two vectors on the coordinate
grid. The tail of the first vector goes at
the origin.
Example 3C, p. 95

To draw 35° south of
east: place the
protractor on the origin
with the 90° mark on
the south axis and the
0° mark on the east
axis. Measure 35° from
the east axis toward the
south axis. Draw the
25.5 km vector from the
origin at 35°.
Example 3C, p. 95
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
Make the head of the 25.5 km vector the
origin. From the head of the 25.5 km vector,
draw the 41 km vector with its tail at the
origin and measure the 65° angle from the
horizontal axis.
To draw 65° north of east: place the
protractor on the origin with the 90° mark on
the north axis and the 0° mark on the east
axis. Measure 65° from the east axis toward
the north axis. Draw the 41 km vector from
the origin at 65°.
Example 3C, p. 95
Example 3C, p. 95
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
The resultant vector
R is the vector that
begins at the origin
and ends at the
head of the 41 km
vector.
Determine the xcomponent and the
y-component for the
two vectors.
Example 3C, p. 95
Example 3C, p. 95

25.5 km: x-component
x1
cos 35 
25 .5 km
x1  25 .5 km  cos 35
x1  20 .89 km
Example 3C, p. 95

25.5 km: y-component
y1
sin 35 
25 .5 km
y1  25 .5 km  sin 35
y1  14.63 km
Example 3C, p. 95
Example 3C, p. 95

41 km: x-component
x2
cos 65 
41km
x 2  41km  cos 65
x 2  17.33 km
Example 3C, p. 95

41 km: y-component
y2
sin 65 
41km
y 2  41 km  sin 65
y 2  37 .16 km
Example 3C, p. 95
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
Place the two x-component vectors and
the two y-component vectors at the
origin.
Examine each vector to determine if it
should have a positive or negative sign
based upon its direction on the x-y
coordinate plane.
Example 3C, p. 95
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Both x1 and x2 will
be positive.
Y2 will be positive.
Y1 will be negative.
Because x1 and x2
are in the same
direction, add them
to get a single xcomponent vector.
Example 3C, p. 95
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x1 + x2 = 20.89 km + 17.33 km =
38.22 km
The x-component of the resultant is
38.22 km.
y1 and y2 are in opposite directions,
make sure they have the correct signs
and add them to get a single ycomponent vector.
Example 3C, p. 95
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y1 + y2 = -14.63 km + 37.16 km = 22.53 km
The y-component of the resultant is 22.53
km.
Place the tail of the x-component of the
resultant at the origin along the positive xaxis.
Place the tail of the y-component of the
resultant at the origin along the positive yaxis.
Example 3C, p. 95
Example 3C, p. 95
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
Redraw the ycomponent vector
with its tail at the
head of the xcomponent vector.
The tail of the
resultant R begins at
the origin and goes
to the head of the ycomponent vector.
Example 3C, p. 95

Use a2 + b2 = R2 to determine R.
38.2 km
2
 22 .53 km  R
2
2
2
2
1459 .24 km  507 .6 km  R
2
1966 .84 km  R
2
2
2
R  1966 .84 km  44 .35 km
Example 3C, p. 95
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
The magnitude of the
resultant vector is 44.35
km.
Use sin, cos, or tan to
determine the direction
of the resultant. This
requires an angle
measurement and the
orientation of the
vector.
Example 3C, p. 95
22.53 km
tan θ 
 0.5898
38.2 km
1
θ  tan 0.5898
θ  30.53
o
Example 3C, p. 95
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The resultant vector is 30.53° north of the east axis. In
other words, to locate the resultant, you start at the east
axis and measure an angle of 30.53° towards the north
axis.
For determining the orientation of a vector, the angle is
given first, then the direction of rotation is given and
then the axis from which the rotation occurred is given.
An angle measurement of 59.47° east of the north axis is
also correct when determining the location of the
resultant vector.
Note: For orientations given as northwest, northeast,
southeast, or southwest, θ = 45°.
Casao’s Wheel of Directions