Vector Procedures (PowerPoint)

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Transcript Vector Procedures (PowerPoint)

Scalars and Vectors
A scalar quantity is one that can be described
by a single number:
temperature, speed, mass
A vector quantity deals inherently with both
magnitude and direction:
velocity, force, displacement, acceleration
Scalars and Vectors
Arrows are used to represent vectors. The
direction of the arrow gives the direction of
the vector.
By convention, the length of a vector
arrow is proportional to the magnitude
of the vector.
8 lb
4 lb
Vectors
The plane travels with a velocity relative to the
ground which is vector sum of the plane’s velocity
(relative to the air) plus the wind velocity
The resultant is a combination of both motions
Vector Procedures
These are the rules you must play by to add vectors and find the
resultant:
1.
Draw each vector to scale and label
2.
Vectors should be drawn in order as they occurred and labeled (I.e.v1, v2…..)
3.
Draw each successive vector starting from the tip of the preceding vector. (This is called
the “tip to tail” method
4.
The vector sum is found by drawing the resultant vector arrow from the tail of the first
vector to the tip of the last vector
5.
For two-vector operations in two dimensions the resultant forms the diagonal of a
parallelogram with the component vectors forming the sides.
v1
Vector v1 is 1 unit east or +1 unit
A1
v2
A2
Vector v2 is 3 units west or - 3 units
Vector (v1 + v2) is 2 units west or - 2 units
v1 + v2 = R
v2

v1
R = Runits @ 0
R = A 1 + A2
( measured from East counterclockwise)
Use a ruler and vector rules to find the resultant of the following vector additions. Record your
answer using vector convention.
F1
R = F1 + F2
F2
F1
F2
P1
P1
P2
R = P 1 + P2
P2
R = P 2 + P1
P1
Vector Addition and Subtraction
When a vector is multiplied
by -1, the magnitude of the
vector remains the same, but
the direction of the vector is
reversed.
Vector Addition and Subtraction

B
 
AB

A
What is A  B ?

A
 
AB

B
Finding the components of a vector
To find the magnitude of a vector quantity in a particular direction you can use the parallelogram
rule
1.
Draw the vector and treat it as the resultant
2.
Form a parallelogram around the resultant in the two directions of interest. In physics
problems these are usually perpendicular directions to simplify the problem
3.
Draw the component vectors along the sides of the parallelogram
ah
a
E
a
Pythagorean Theorem:
a2 = ah2 + av2
av
Eparalel
E
Eperpendicular
Pythagorean Theorem:
E2 = Eparalel2 + Eperpendicular2
Using Trig to solve 2D-Vector Problems
Right-Angled Trig Relationships
Trig functions for right-angled triangles are just geometric ratios. You probably memorized
these ratios in your math classes using So/h..Ca/h..To/a. Here they are once more!
sin = o / h
cos = a / h
h
o
tan = sin / cos = o / a

a
Follow these rules!
1.
Sketch vectors on a coordinate system. (Two perpendicular axes)
2.
Find (resolve) the components of all vectors in the two directions of interest using the parallelogram
rule. Make sure that you use the normal conventions of + and – arithmetic quantities.
3.
Sum the components in the two directions of interest.
4.
Find the magnitude of R using Pythagorean theorem
5.
Use the tangent rule to determine the direction R as measured from the East by convention or a
specified direction
Practice, Practice…..Practice
Use trig relationships to determine i) the other component and ii) the resultant R
i)
vv
R
600
vh = 35 m/s
tan = vv / vh
vv = vh tan

vv = (35 m/s) tan 600
vv = 60.6 m/s
ii) R =  (vv2 + vh2)
R =  ((60.6 m/s)2 + (35 m/s)2)
R = 70 m/s
 = 1800 - 600 = 1200
R = 70 m/s @ 1200
Determine the resultant R of the following vectors by using the horizontal and vertical components
of A and B
H
R = 100 Newtons
A
+
A = 80 Newtons
A cos 600
-80 N cos 600
B = 60 Newtons
- 40 N
V
A sin 600
+80 N sin 600
+ 69.3 N
Av
600
-

AH
300
Bv
+
B cos 300
B
BH
R =  (RH2 + Rv2)
+60 N cos 300
B sin 300
60 N sin 300
+ 52 N
+ 30 N
+ 12 N
+ 99.3 N
R =  ((12 N)2 + (99.3 N)2)
R =A+B
R = 100 N
 = tan-1 (Rv / RH)
 = tan-1 (99.3 N / 12 N)
 = 830
R = 100 N @ 830