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Exploration 1: 2D Vectors
As an example of how to add together vector quantities which are not colinear, imagine that you rolled a marble 10 m across the floor in a direction 37 o
north of east. The marble then struck a wall, and rolled back 8 m in a direction
30o west of north. What is the resultant displacement of the marble?
To see how a graphical method can be used to add vectors, refer to the
figure below. We represent the first displacement by a vector pointing 37o north

of east, scaled to represent 10 m in length. This vector is labeled A in the

diagram. The second displacement of the marble, represented by vector B in the


diagram, is drawn in a similar manner. To add the vectors A and B (and thus the


displacements) we draw B extending from the tip of A . This method of vector
addition is known as the tip-to-tail method.



The sum of A and B is called the resultant and is shown by the vector R .
The magnitude of the resultant displacement R can be determined by measuring
with a ruler and applying whatever scaling factor was used to draw the original

vectors. The angle between R and east can be measured with a protractor. For

this example R , the final location of the marble, is found to be given by a vector
13.7 m at an angle of 73o north of east.



Now suppose you wished to add a third vector C to the A and B as given in

the previous example. Vector C has a magnitude of 6 m and points due west.


Vectors A and B are drawn tip-to-tail as before, but before drawing in the


resultant, C is drawn tip-to-tail to B . All three vectors are now joined tip-to-tail




on the diagram. We can find the new resultant R , the vector sum of A + B + C



by drawing a vector from the initial point of A (the tail of A ) to the tip of C . As


before, the magnitude of R is obtained by measuring the length of R on the

diagram and applying the scaling factor. The direction of R is obtained by
measuring the angle to the E-W axis with a protractor. These two
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
measurements provide all the data needed to define the vector R in terms of its

magnitude and direction. From the diagram we obtain the magnitude of R as
13.1 m and the direction as 81o north of west. This same procedure can be
applied to any number of vectors.
Subtraction of Vectors
You can find the difference between two vectors by using the same basic

procedure. To find the value of the resultant vector when vector B is subtracted
 



from vector A ( A – B ), you add the vector (- B ) to the vector A , and follow the

same procedure as was used for adding vectors. The vector (- B ) is a vector that

has the same magnitude as vector B , but is in the opposite direction. Using the


same vectors A and B as is the previous examples, the graphical subtraction of




B from A is illustrated below. Note that the “ B ,” shown below is actually (- B ).

In this example of vector subtraction R is about 7o south of east with a
magnitude of 12 m.
It is also possible to use the method used Module 1. Put the tails of the two
vectors together, and draw an arrow from the tip of the subtrahend vector to the
tip of the minuend vector.
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The Component Method
It is not always convenient to graphically add vectors, and, as with any
graphical method, it is only as accurate as the person doing the scaling and
drawing. A more accurate result is obtained by using a method based on
trigonometry. This method is called the component method, and is based on the
fact that for any vector, a right triangle can be formed when placing the vector at
the origin of a Cartesian coordinate system.

Consider again A from the previous examples. When drawn on the
Cartesian coordinate system, the vector makes an angle of 37 o to the positive x
axis.

If a line is drawn straight down from the tip of A , perpendicular to the x
axis, the length of that line is the amount of the vector that is in the y direction.
That is, it is the projection of the vector in the y direction. We call this the y


component of A , designated Ay. The line drawn from the tip of A goes to a
point on the x axis. The distance from the origin to the point where the vertical
line intersects the x axis is the amount of the vector that is in the x direction.
That is, it is the projection of the vector in the x direction. We call this the x

component of A , designated Ax.

The vector A and its components form a right triangle, so the basic
trigonometric relationships of sine, cosine and tangent will apply to the triangle.

From the diagram above, we note that the x component of A is the side

adjacent the angle that A makes with the x axis, and that the length, or

magnitude, of A forms the hypotenuse of the triangle. From trigonometry we
know that the cosine of an angle is defined as the side adjacent the angle
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divided by the hypotenuse, or in this case, the x component divided by the

magnitude of A .
a
h
A
cos   x
A
A x  A cos 
cos  

This gives a method of finding the x component of A . In this example,
A = 10 m and  = 37o.
A x  A cos   10 cos 37 o  8 m

The same concept can be applied to find the y component of A , this time
using the definition of the sine of an angle as the side opposite the angle divided
by the hypotenuse.
sin  
sin  
o
h
Ay
A
A y  A sin   10 sin 37 o  6 m
Any vector can be written in terms of its components. When adding 2 or
more vectors, the components of the resultant vector are given by the algebraic
sum of the corresponding components. In other words, the x component of the
resultant is determined by the sum of the x components of all the vectors being
added together. As an example of how this works, we’ll go back to the previous


example of adding together A and B .


The vector A was 10 m at an angle of 37o to the positive x axis, and B was
8 m at an angle of 120o from the positive x axis. The first step in applying the
component method is to draw a sketch showing both vectors on the coordinate
system.
The next step is to determine the components of the vectors. This is
summarized in the table below.
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Vector

A : 10 m, 37o

B : 8 m, 120o

Resultant (sum) R
x component
y component
37o
10 sin 37o = 6 m
8 sin 120o = 6.9 m
12.9 m
10 cos
=8m
o
8 cos 120 = -4 m
4m
The components of the resultant vector
are found by summing the respective components of
the individual vectors, as seen in the table. Once the

components of R are known, the resultant can be
sketched on a coordinate system. Since both

components are positive, R is in the first quadrant of
the coordinate system.

The magnitude of R is found using the Pythagorean theorem, where the two
components are the legs of the triangle and the magnitude of the vector is the
hypotenuse.
R  R 2x  R 2y  4 2  12.9 2  13.5 m

The direction of R is found by using the definition of the tangent of the

angle. The two components of R are the sides opposite and adjacent the angle

that R makes to the x axis. Since the tangent is defined as the side opposite
divided by the side adjacent and both sides are known, the inverse tangent can
be used to find the angle.
o Ry

a Rx
12.9
tan  
4
1  12.9 
o
  tan 
  73
 4 
This result matches that obtained using the graphical method.
tan  
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Extension 1: 2D Vectors
1. Define vector:
2. For the following questions, which of the following statements involve vector
quantities, and which involve scalar quantities?
a) I ran two miles at the lake. ______________
b) I ran two miles due north along the lake. ______________
c) The plane flew 90 knots on a heading of 600 east of north.
______________
d) The car drove at 60 mph._________________
e) The temperature was minus 20 degrees. ______________
f) The cart moved 20 meters in the negative x direction. ___________
Vector A points due west, while vector B points due south.
3. Does the direction (A + B) point north or south of due west?
4. Does the direction (A – B) point north or south of due west?
Make a drawing in both cases to show your reasoning.
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5. Vectors A and B are shown in the drawing below. Vector R (not shown) is the
resultant vector where R = A + B. Indicate the signs (negative or positive) of the
following scalar components:
Ax ____Ay ____Bx ____
By ____
Rx ____
Ry, ____
Refer to the vectors drawn above. Draw the following resultant vectors to scale.
Use a straightedge to transfer the vectors from place to place.
6. (A+B)
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7. (A-B)
8. (C+D)
9. (C-D)
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10. A pilot flies her route in 2 straight line segments. The displacement vector
A for the first segment has a magnitude of 244 km and a direction 30.0 0 north of
east. The displacement vector B for the second segment has a magnitude of
175 km and a direction due west. Draw the vectors on an appropriate coordinate
system and use the component method to find the resultant vector R, and the
direction relative to east.
11. Draw the components of vector B which are parallel to and perpendicular to
the inclined surface. Then determine the magnitude and sign of the
components. (HINT) Draw your coordinate axes so that the x-axis is parallel to
the incline (that is, 200 from true horizontal) and the y-axis is perpendicular to the
x-axis. Transfer the vector B so that its tail is at the origin. Now draw
components parallel and perpendicular to the axes .
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Exercises 21-23: Draw and label the vectors on the axes. Draw and label the
angle, θ. Then determine the magnitude of the vector and the angle:
Ax = -1, Ay = 2
Bx = 0, By = -2
Cx = 3, Cy = -2
A = ________
B = ________
C = ________
θ = ________
θ = ________
θ = ________
Exercises 24-26: Draw and label the vectors on the axes. Draw and label the
angle, θ.Then determine the magnitude of the vector and the angle:
Ax = 1, Ay = 2
Bx = -2, By = 2
Cx = 3, Cy = -1
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Exercises 35-37: The vector A is 5 units long and is directed 300 above
true horizontal. Determine the components Ax and Ay for the three coordinate
systems shown below. Show your work below the figure.
The next three exercises use graphical representation to remind us that
vector addition is done using the x and y components independently. For each
force, determine the x and y components by counting the number of squares.
Include the appropriate “+” and “-“ signs. Then, add the total number x and y
squares from all the forces and draw the net force vector on the graph provided.
Example
x
y
F1 (N)
0
2
F2 (N)
-2.5
0
F3 (N)
0
-3
F4 (N)
3.5
-3.5
Fnet (N)
1
-4.5
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x
y
x
y
x
y
F1 (N)
F2 (N)
F3 (N)
F4 (N)
Fnet (N)
F1 (N)
F2 (N)
F3 (N)
F4 (N)
Fnet (N)
F1 (N)
F2 (N)
F3 (N)
F4 (N)
Fnet (N)
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The next three exercises review the addition of forces using the
mathematical procedure of vector addition. The procedure for all of these
problems is shown below. Note that all angles are taken relative to the x-axis:
1.
In the space below the table, add the vectors using:
2.
Fx = F cos Θ
3.
Fy = F sin Θ
4.
Put in the appropriate “+” and “-“ signs based on quadrant.
Add components and determine the net force vector using:
5.
Fnet  x 2  y 2
6.
θ = tan-1 opp leg/adj leg
7.
Draw the net force vector on the graph provided.
Example
FORCE
x
component
F cos 
y
component
F sin 
F1 (N)
40 cos 45o
28.3
0
40 sin 45o
28.3
30
F2 (N)
F3 (N)
Fnet (N)
-50 cos 30o -50 sin 30o
-43.3
-25
-15
33.3
Fnet = 36.5 N
x axis
Θ = 66o above –
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FORCE
x
component
F cos 
y
component
F sin 
F1 (N)
F2 (N)
F3 (N)
Fnet (N)
Fnet =
FORCE
Θ=
x
component
F cos 
y
component
F sin 
F1 (N)
F2 (N)
F3 (N)
Fnet (N)
Fnet =
FORCE
Θ=
x
component
F cos 
y
component
F sin 
F1 (N)
F2 (N)
F3 (N)
Fnet (N)
Fnet =
Θ=
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