Transcript Vectors & Scalars

Vectors and Scalars

Many quantities in geometry and physics,
such as area, volume, temperature, mass, and
time, can be characterized as a single real
number scaled to appropriate units of
measure. These are called scalar quantities,
and the real number associated with each is
called a scalar.

Other quantities, such as force, velocity, and
acceleration, involve both magnitude and direction and
cannot be characterized completely by a single real
number. A directed line segment is used to represent
such a quantity. A directed line segment has an initial
point and terminal point for example PQ would have
initial point P and terminal point Q, and its length or
(magnitude) is denoted as ||PQ|| or sometimes as |PQ|.
The set of all directed line segments that are equivalent
to a given directed line segment are vectors in the
plane. Vectors are denoted in typesets as boldface
letters such as u, v, and w. Written by hand they are
usually denoted as v, u,and w

A scalar quantity is a quantity that has
magnitude only and has no direction in space
Examples of Scalar Quantities:
 Length
 Area
 Volume
 Time
 Mass

A vector quantity is a quantity that has both
magnitude and a direction in space
Examples of Vector Quantities:
 Displacement
 Velocity (speed in a given direction)
 Acceleration (rate of which something speeds up
or slows down in a direction)
 Force (gravity for example, or pushing an object
with a force of 10 Newtons (N)).
Because these two vectors are the
same length but are in opposite
directions we have u and –u.



Vector diagrams are
shown using an
arrow
The length of the
arrow represents its
magnitude
The direction of the
arrow shows its
direction
u
-u
v
Because these two vectors (u and v) are
going in the same direction and have the
same magnitude (length) then the vectors
are the same u=v.

The resultant is the sum or the combined effect of
two vector quantities
Vectors in the same direction:
6N
4N
=
10 N
=
10 m
6m
4m
Vectors in opposite directions:
6 m s-1
10 m s-1
=
4 m s-1
6N
10 N
=
4N

The sum of two vectors can be represented geometrically
by positioning the vectors (without changing their
magnitudes or directions) so that the initial point of one
coincides with the terminal point of the other. The vector
u+v is called the resultant vector.
u
u+v
v

If you have a vector v and then –v is a vector of the same
magnitude, however it is directed in the opposite
direction. This is a property what we will use to subtract
two vectors such as u-v, we will change to addition of u
+ (-v) so you need to locate vector v, change its direction
and then move it to tip-tail.
-v
u-v
u
v
Animation done by
Taylor Cox (the guy)
When two vectors are joined
tail to tail

Complete the parallelogram

The resultant is found by
drawing the diagonal



When two vectors are joined
head to tail
Draw the resultant vector by
completing the triangle

Suppose that we have the following two vectors
Graphically represent u+v.
v
v
u
u
Resultant vector
3v
2u
v
u

Is that the same thing as v+(-u)?
v
u
-u
Sketch u+v, 2(u+v), u-v
u
v
Vector Algebra
-component form
-vector addition
-vector subtraction
-scalar multiplication
• Component form of a vector is essentially working backwards to
identify the two perpendicular vectors that would add (tip to tail) to
provide you with the current vector. These two perpendicular
vectors are essentially the x-direction and y-direction distances.
These values are referred to as the COMPONENTS of your particular
vector.
The x component of v is 3
The y component of v is 2
Thus we write v = (3,2) or
v  (3, 2)
This is referred to as COMPONENT FORM of v
v
3
2
• You can find the component form of a vector
by subtracting the ordered pairs of tip and tail.
General
Example
(x2,y2)
v
(x1,y1)
x2-x1
y2-y1
(4,3)
v
(1,1)
y2-y1 = 3-1 =2
x2-x1 =4-1=3
If we needed to find the length of v we would use the Pythagorean Theorem, and
then the Pythagorean Theorem would lead us directly to the distance formula.
Because we are talking about length it will be positive so we use notation that
implies absolute value. This is the equation for the size or MAGNITUDE of the vector.
| AB | ( x2  x1 ) 2  ( y2  y1 ) 2
Find the magnitude of the following
• Vector addition (algebraically) is quite simple if your
vectors are both expressed in component form.
• Suppose we have two vectors v and w. In component
form v=(a,b) and w=(c,d) then vector addition states
that v+w=(a,b)+(c,d), which means v+w = (a+c,b+d)
Example
• If v=(1,2) and w=(5,1) find v+w.
• v+w = (1,2)+(5,1) = (1+5, 2+1) = (6,3)
• v+w = (6,3).
• Think of it graphically/geometrically. The vectors can start anywhere in the
Cartesian Plane, lets assume that one starts at the origin, and the other starts at
the tip of the first (you can assume that they both start at the origin and then use
parallelogram technique to verify your answer)
w
1
5
v
2
1
Resultant vector v+w is purple
Identify its components.
Are they (6,3)?
Vector Subtraction
• Vector subtraction is the same concept if
v=(a,b) and w=(c,d) then v-w=(a-c, b-d)
Remember geometrically vector subtraction is the
same as addition however you add the opposite
direction of the vector. If I want to subtract v-w,
geometrically we would take v + (-w). And the (-w)
would be the same exact vector as w but going in the
opposite direction.
v
w
Scalar Multiplication
• Refers to the idea that we can take any vector
in component form and multiply it by any
constant that we want to. In light of doing so,
the components are both changed by the
same scalar.
• If v=(3,2) and we want 3v then we multiply
both components by 3, thus 3v=(9,6)
• The constant 3 is our SCALAR
Summary of Algebraic Operations on
Vectors
if u  x1 , y1 and v  x2 , y2 then,
u  v  x1  x2 , y1  y2
u  v  x1  x2 , y1  y2
cu  cx1 , cy1 , c 
Properties of Vectors
•
•
•
•
u+v = v+u (commutative)
u + (v + w) = (u + v) + w (associative)
u + 0 = u (identity of addition)
u + (-u) = 0 (subtraction)
Find the magnitude of the
resultant vector
39o
14 lb
10.3 Algebraic Interpretation of Vectors
• A vector with its initial point at the origin is called
a position vector.
• A position vector u with endpoint (a,b) is written
as u = a, b, where a is called the horizontal
component and b is called the vertical component
of u.
Copyright © 2011 Pearson Education, Inc.
Slide 10.3-27
10.3 Finding Horizontal and Vertical
Components
Horizontal and Vertical Components
The horizontal and vertical components, respectively,
of a vector u having magnitude |u| and direction angle
 are given by
a = |u| cos 
and b = |u| sin .
That is, u = a, b =  |u| cos  , |u| sin  .
.
Copyright © 2011 Pearson Education, Inc.
Slide 10.3-28


When resolving a vector into
components we are doing the
opposite to finding the resultant
We usually resolve a vector into
components that are
perpendicular to each other
Here a vector v is resolved into
an x component and a y
component
y

x

If a vector of magnitude v and makes an angle θ
with the horizontal then the magnitude of the
components are:
y=v Sin θ
x = v Cos θ
y
θ
y = v Sin θ

Proof:


x
Cos  
v
x  vCos 
x=vx Cos θ
y
Sin 
v
y  vSin
10.3 Finding Horizontal and Vertical
Components
Example From the figure,
the horizontal component is
a = 25.0 cos 41.7°  18.7.
The vertical component is
b = 25.0 sin 41.7°  16.6.
Copyright © 2011 Pearson Education, Inc.
Slide 10.3-31
2002 HL Sample Paper Section B Q5 (a)
A force of 15 N acts on a box as shown. What is the horizontal
component of the force?
HorizontalComponent x  15Cos60  7.5 N
VerticalComponent y  15Sin60  12.99 N
Vertical
12.99 N
Component
Solution:
60º
Horizontal
7.5 N
Component
Two forces are applied to a body, as shown. What is the magnitude
and direction of the resultant force acting on the body? A forces of
12N due East and 5N due South
Solution:


Complete the parallelogram (rectangle)
The diagonal of the parallelogram ac
represents the resultant force
The magnitude of the resultant is found using
Pythagorean Theorem on the triangle abc a
Magnitude ac  12  5
2
ac  13 N
5
Direction of ac : tan  
12
12
   tan 1  22.6
5
2
θ
12 N
d
5N

5
b
12
c
Find the magnitude (correct to two decimal places) and direction of the
resultant of the three forces shown below.
Solution:
Find the resultant of the two 5 N forces first (do right angles first)
ac  52  52  50  7.07 N
5
tan    1    45
5



Now find the resultant of the 10 N and
7.07 N forces
The 2 forces are in a straight line (45º +
135º = 180º) and in opposite directions
So, Resultant = 10 N – 7.07 N = 2.93 N
in the direction of the 10 N force
5
d
c
5
5N

a
90º
θ
45º
135º
5N
b


Here we see a table
being pulled by a force
of 50 N at a 30º angle
to the horizontal
When resolved we see
that this is the same as
pulling the table up
with a force of 25 N
and pulling it
horizontally with a
force of 43.3 N
y=25 N
30º
x=43.3 N

We can see that it
would be more
efficient to pull the
table with a diagonal
force of 50 N
Inclined Planes

A person in a wheelchair is moving up a ramp at constant speed. Their
total weight is 900 N. The ramp makes an angle of 10º with the
horizontal. Calculate the force required to keep the wheelchair moving
at constant speed up the ramp. (You may ignore the effects of friction).
Solution:
If the wheelchair is moving at constant speed (no acceleration), then
the force that moves it up the ramp must be the same as the
component of it’s weight parallel to the ramp.
Complete the parallelogram.
Component of weight
10º
parallel to ramp:
 900 Sin10  156 .28 N
80º
10º
Component of weight
perpendicular to ramp:
 900 Cos10  886 .33 N
900 N
886.33 N
10.3 Finding the Magnitude of a Resultant
Example Two forces of 15 and 22 newtons act on a point in
the plane. If the angle between the forces is 100°, find the
magnitude of the resultant force.
Solution
From the figure, the angles
of the parallelogram adjacent to angle P
each measure 80º, since they are
supplementary to angle P. The resultant
force divides the parallelogram into two
triangles. Use the law of cosines on
either triangle.
|v|2 = 152 + 222 –2(15)(22) cos 80º
Copyright © 2011 Pearson Education, Inc.

|v|  24 newtons
Slide 10.3-37
10.3 The Unit Vector
• A unit vector is a vector that has magnitude 1.
• Two very useful unit vectors
are defined as
i = 1, 0 and j = 0, 1.
i, j Forms for Unit Vectors
If v = a, b, then v = ai + bj.
Copyright © 2011 Pearson Education, Inc.
Slide 10.3-38
10.3 Dot Product
Dot Product
The dot product of two vectors u = a, b and
v = c, d  is denoted u · v, read “u dot v,” and given
by
u · v = ac + bd.
Example Find the dot product 2, 3 · 4, –1.
Solution
Copyright © 2011 Pearson Education, Inc.
2, 3 · 4, –1 = 2(4) + 3(–1)
=8–3
=5
Slide 10.3-39
10.3 Applying Vectors to a
Navigation Problem
Example
A plane with an airspeed of 192 mph is headed on a
bearing of 121º. A north wind is blowing (from north
to south) at 15.9 mph. Find the groundspeed and the
actual bearing of the plane.
Solution
Let |x| be groundspeed. We
must find angle . Angle AOC = 121º.
Find |x| using the law of cosines .
x  192  15.9  2(192)(15.9) cos121
2
2

2
 40,261 
x  200.7 mph
sin  sin 121

15.9
200.7
sin   .06792320
Copyright © 2011 Pearson Education, Inc.
   3.89
Slide 10.3-40
10.3 Finding a Required Force
Example
Find the force required to pull a wagon weighing
50 lbs up a ramp inclined at 20º to the horizontal.
(Assume no friction.)
Solution
The vertical 50 lb force BA
represents the force of gravity. BA is the
sum of the vectors BC and –AC. Vector
BC represents the force with which the
weight pushes against the ramp. Vector
BF represents the force required to pull
the weight up the ramp. Since BF and AC are equal, | AC | gives
the magnitude of the required force.
AC
sin20 
50
Copyright © 2011 Pearson Education, Inc.

AC  50sin20  17.1 lbs
Slide 10.3-41