Download VECTORS comp box method addition 2015-16

WHAT’S THE SHAPE?
LOOK AT THE VECTOR,AND THE AXES. WHAT SHAPE CAN
YOU MAKE?
Original Vector
Y component Vector
VY
X component Vector
VX
Component – Means to be a piece, or apart, of something bigger
A Component Vector is a smaller vector that is apart of a larger one
ANALYTICAL METHOD
• Use trigonometry to break vectors into
smaller vectors called Component Vectors.
– The X component Vector is along the X
axis
– The Y component Vector is along the Y
axis
• Trigonometry deals with angles, and
triangles.
• This makes addition easier.
• Please remember Trig. is your friend.
• We can add these component vectors to find
the resultant.
TRIGONOMETRY
Opposite

•
•
•
•
Adjacent
SINE  = opposite/hypotenuse
COS  = adjacent/hypotenuse
TAN  = opposite/adjacent
SOH - CAH - TOA
BREAKING INTO COMPONENTS
Use Sine and Cosine:
Vector (V)

x component (VX)
y component (VY)
Sine  = Y/V
VY = V*Sine 
Cos  = X/V
VX = V*Cos 
1) Cosine closes the angle
2) Sine skies, Cosine coasts
BREAK INTO COMPONENTS
First
Nowuse
thatsine,
the original
and cosine
vector
to break
are broken
the vectors
into into
their
smaller
components
pieces the original two vectors are gone
Adding the components
Resultant
When
we can
draw
thethese
resultant
weeasily
get a right
Now, we
add
vectors
by thetriangle,
tip tail method
we get a right triangle EVERY TIME WE DO THIS.
FINDING THE RESULTANT
 Means
Angle
Y Total (YT)
R
X Total (XT)
Use Pythagorean Theorem to find the magnitude
VR2 = XT2 + YT2
VR = XT2 + YT2
To find the direction use the tangent
Tan R = YT/XT
R = Tan-1(YT/XT)
Solving a problem
20 meters at 30 degrees + 10 meters at 60 degrees
VY = 10m *Sin 60
VY = 10m *.866
VY = 8.66 m
10 m
60O
20 m
30O
Vx = 20m *Cos 30
Vx = 20m *.866
Vx = 17.32 m
Vx = 10m *Cos 60
Vx = 10m *.5
Vx = 5 m
VY = 20m *Sin 30
VY = 20m *.5
VY = 10 m
Nowuse
that
the and
original
vector
are broken
into into
First
sine,
cosine
to break
the vectors
smaller
pieces the original two are gone
their
components
VY = 8.66 m
VY = 8.66 m
Resultant
10 m + 8.66 m = 18.66 m
VY V
= x10
= 5mm
V17.32m
x = 17.34 m+
Vx = m
5m
5m = 22.34
VY = 10 m
Now
that
weahave
these
component
vectors
we lengths
can
addofthem
Once
we
have
the
right
triangle
we can
find
the
Now
we
have
regular
right
triangle
and
by using
trigonometry
Tip
Tail
and
form
a right
Triangle
each
side
by
simply
adding
the vector length’s together
we
can
find
the
resultant
a = 22.34m
b = 18.66 m
C= resultant
Resultant
18.66 m

C2
a2
+b2
22.32 m
=
C2 = (22.32m)2 + (18.66m)2
C2 = (498.18m2) + (348.2m2)
C2 = 846.38m2
C = 846.38m2 = 29.09m
 = tan-1 (b/a)
 = tan-1(18.66m/22.32m )
 = tan-1 (.836)
 = 39.9O
The final answer is 29.09 meters at 39.9O
WOW, That a lot of work!!
(Is there an easier way, to do this?)
Yes, there is a much faster and easier way, called
the “box method”
The box method is a technique that organizes all
the work into a few simple steps
that anyone can do
Lets look at the same problem again using the box method,
on the side will be the longer method
In the end chose the method that works best for you.
The box method
Step 1: Make a box like this one the top, and bottom rows
are always done this way
Vectors
You will need 1
row for each
Vector.
This problem
has 2 vectors
so we need
2 columns
Total
Vx = Vcos () Vy = Vsin ()
Step 2: Fill in the vector columns
(this sets up the problem)
20 meters at 30 degrees + 10 meters at 60 degrees
Then use
Write
The the
Vector’s
Cosine
Length
And sine
and
Equations
Angle
To fill in
In
thethe
next
Vector
two
column
columns
Vectors
Vx = Vcos () Vy = Vsin ()
20 m at
30O
20m*cos(30O)
17.32 m
20m*sin(30O)
10 m
10 m at
60O
10m*cos(60O)
5m
10m*sin(60O)
8.66 m
22.32 m
18.66 m
Total
Finally
add up
the two
Columns
Using the total columns gets us the
right triangle again
Vx total = a = 22.34m
Vy total = b = 18.66 m
C= resultant
Resultant
18.66 m

C2
a2
+b2
22.32 m
=
C2 = (22.32m)2 + (18.66m)2
C2 = (498.18m2) + (348.2m2)
C2 = 846.38m2
C = 846.38m2 = 29.09m
 = tan-1 (b/a)
 = tan-1(18.66m/22.32m )
 = tan-1 (.836)
 = 39.9O
The final answer is 29.09 meters at 39.9O