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# Ch vertical direction

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```									 Chapter 1. Vectors and Coordinate Systems
Our universe has three
dimensions, so some
quantities also need a
direction for a full
description. For example,
wind has both a speed and a
direction; hence the motion
of the wind is described by a
vector.
Chapter Goal: To learn how
vectors are represented and
used.
Student Learning Objectives – Ch. 1

• To understand the basic properties of vectors.
• To add and subtract vectors both graphically and
using components.
• To be able to resolve a vector into its components
and to reassemble vector components into a
magnitude and a direction.
• To work with tilted coordinate systems.
•A scalar quantity is one that can be completely described by
its magnitude (size).
•Examples include temperature, energy, and speed
•Some scalars can be described with negative numbers (1st
two examples, above, but not the third). In that case the
negative sign means the quantity is less than a positive
quantity.

+30°C                        -30° C
•A vector quantity has both magnitude and direction.
•Examples include velocity, force, and displacement
•Vectors are never negative. A vector has a magnitude, which
is an absolute value and a direction, which is usually given as
an angle.
•Vector components may be described with negative signs. In
that case the negative sign never means less than. It means
“left” or “down” on a graph, and west or south on a map.

The y component of                     The y component of
velocity is – 40 m/s.                  velocity is + 40 m/s.

Both cars are going at the exact same speed!
Arrows are used to represent vectors.
The direction of the arrow relative to
some reference point (north or + x axis)
gives the direction of the vector. The
length of a vector arrow is proportional to
the magnitude of the vector.

8m
4m
This is an example of a displacement vector with magnitude of 2
km and direction 30° north of east
Graphical Addition: Put the tail of one vector after
the head of the other. The resultant vector is an
arrow that starts where the first vector starts and
ends where the second vector ends.
This is easy if the two vectors
are co-linear, especially if they
point in the same direction.
scalars.

5m                  3m

8m
If the two vectors are not in
the same direction, then the
graphical process is the same,
but the magnitudes of the
vectors can no longer be
In this case, vector A has a
magnitude of 27.5 cm due
east and vector B has a
magnitude of 12.5 cm in a
direction 55° north of west.
In order to get a value for the
resultant vector, one needs a
ruler and a protractor.
Multiplication by a scalar

Multiplication by
a scalar serves to
stretch or shrink
the vector by a
factor equal to
that of the scalar.
1.6 Multiplication by -1 allows subtraction

When a vector is multiplied
by -1, the magnitude of the
vector remains the same, but
the direction of the vector is
reversed (i.e. the angle is
increased by 180°).

Note that A + B is not the
                                 reverse of A – B.
AB                   B


A                                   
A           B
 
AB
Multiplication by a negative scalar
Vector A has a magnitude of +2 and a direction of 30°
from the positive x axis. Vector -3A has:
A. Magnitude -6; direction - 30°
B. Magnitude -6, direction 210°
C. Magnitude 6, direction 210°
D. Magnitude 6, direction 30°
1.4 Trigonometry Review

ho
sin  
h

ha
cos 
h

ho
tan  
ha
“SOH-CAH-TOA”
1.4 Trigonometry Review

h sin   ho

h cos  ha

h  h  h (Pythagorean theorem)
2        2
o
2
a
 ho 
  tan  1
h 
 a
1.7 The Components of a Vector

     
x and y are called the x component

and the y component of r.
1.7 The Components of a Vector


The vector componentsof A are two perpendicular
        
vectorsA x and A y that are parallel to the x and y axes, and
      
add together using the Pythagorea Theoremso that A  A x  A y .
n

The scalar componentsAx and A y have the same magnitude as their

,
vector components but are positive if they point in the positive direction
and negative if they point in the negative direction.
1.7 The Components of a Vector
Help! My calculator won’t tell me when a
scalar component of a vector is negative if I
use these acute angles as given. And
inverse tangent doesn’t work right in
quadrants other than the 1st . What do I do?
1. You can convert all angles from acute to
their proper values. However, this doesn’t
solve the problem for inverse tangent.
2. Better, use absolute values for all
components. Draw the picture and                S   A
determine which components should be
negative and then manually insert the
negative sign when needed.
3 Remember: “All Students Take
Calculus” – whether they want to or not!
Use this in conjunction with inverse tangent
to determine the appropriate quadrant for θ.
T       C
Finding the components of an acceleration
vector
Finding the components of an acceleration
vector
If the angle is given with respect to the y-axis,

C1x= - C1sin φ (x-comp. is opposite); negative by
inspection
C1y = - C1 cos φ (y-comp. is adjacent); negative by
inspection
φ = tan-1 (|C1x |/|C1y|).
In what quadrant is vector C , for purposes
of finding components and θ?

A.   1st
B.   2nd
C.   3rd
D.   4th
E.   Both 2nd and 4th
What are the x- and y-components Cx and
Cy of vector C ?

A.   Cx = 1 cm, Cy = –1 cm
B.   Cx = –3 cm, Cy = 1 cm
C.   Cx = –2 cm, Cy = 1 cm
D.   Cx = –4 cm, Cy = 2 cm
E.   Cx = –3 cm, Cy = –1 cm

Angle φ that specifies the direction of F1
is given by

A.   tan–1(F1y /F1x)
φ                      B.   tan–1(|F1y |/|F1x|)
C.   tan–1(F1x /F1y)
D.   tan–1(|F1x |/|F1y|)
Tilted axes
• Often it is convenient to use a tilted
axis system (to represent an object on
an incline for example).
• The axes stay perpendicular to each
other.
• The components correspond to axes,
not to “horizontal and vertical” so they
are also tilted.
• The gravitational force always points
in the “true vertical” direction.
Tilted axes
• Often it is convenient to use a tilted
axis system (to represent an object on
an incline for example).
• The axes are always perpendicular to
each other.
• The components correspond to axes,
not to “horizontal and vertical” so they
are also tilted.
• The gravitational force always points
in the “true vertical” direction,
regardless of axis orientation.
To find components of a vector parallel and perpendicular to
an inclined surface, as shown in (a):

•Draw a set of coordinate axes so
that one axis (usually x) is parallel
and the other is perpendicular to the
surface.
•Place the tail of the vector at the
origin of the axes.
•Draw components parallel to each
axis as shown in (b).
Ax = A cos 
Ay = A sin 
Note that  is defined relative to the
tilted x-axis and not to “horizontal”
Finding the force perpendicular to a surface
A horizontal force, F, of 10 Newtons is applied to an object on a
surface inclined at a 20° angle (object not pictured). Draw the
component of the force vector which is perpendicular to the
surface and find its magnitude.

Inclined surface
Finding the force perpendicular to a surface

In this case the coordinate axes are drawn right on the
inclined surface. The y- axis is perpendicular to the inclined
surface.
Example: EOC #47

A foot ball player runs the
pattern shown by the 3
displacement vectors A, B
and C. The magnitudes
of these vectors are A =
5.00 m, B = 15.0 m and C
= 18.0 m. Find the
magnitude and direction
of the resultant vector.
Example: EOC #52
Two geological field teams are working
in a remote area. A global positioning
system (GPS) tracker at their base
camp shows the location of the first
team as 38 km away, 19° north of
Second team
west, and the second team as 29 km                                  N (y)

First team
C
away, 35° east of north. When the

first team uses its GPS to check the                        A       35°
B=A+C

position of the second team, what        W
19°
E (x)

does the GPS give for the distance                 Base camp

between the teams and direction,θ,
measured from due east?
Graph of a line
The vertical axis value comes first
when naming graphs: i.e. this is a
position vs time.
The standard format is:
y = mx + b,
where m is the slope and b is the y-
intercept.
Note that the normal variable for
position is x and the quantity on the
horizontal axis is time (t) in which
case the equation format is:
x = mt + b
Graph of a line
m = “rise over run”, in this case
Δx/Δt, where:
Δx = x2-x1 and Δt = t2-t1 for any
two points on the line.
Note that the slope has units, in
this case:
Δx/Δt = 8m/2s = +4 m/s.
In this case the slope has units
of velocity and represents the
velocity of the object being
graphed.
Graph of a line
Find the equation of a line
if you know 2 points:
• use Δx/Δt to find m
• substitute the values of
one of the points into the
equation to find b:
x1 = mt1 + b
b = x1 – mt1
or
b = x2 – mt2
Equation of a line

What is the equation of
this velocity vs time
graph? What are the
units of the slope?
v 12m / s
m            6m / s / s
t   2s
v = 6 (m/s2) t + 5m/s or
v = 6t + 5

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