Transcript L2 PYX VECTORS

CONTENTS
 Scalars and Vectors.
 Bearings & Compass headings.
 Drawing Vectors.
 Vector addition & subtraction.
 Relative velocity.
 Change in velocity.
 Exercises from Rutter.
 Homework.
Scalars and Vectors
Every quantity can have two types of value. They can either be:
a. SCALAR or
b. VECTOR
But what does this mean?
SCALAR quantity just gives the
MAGNITUDE
MAGNITUDE
VECTOR quantity gives
&
DIRECTION
Distance
Displacement
Speed
Velocity
Mass
Acceleration
Energy
Force
Density
Momentum
N
R  4  3  5 km
2
2
R=?
+ 3 Km
North

E
4 Km East
 3
  tan    36.90
 4
1
= Bearing of 053.10
Answer = 5 km 053.1o or 5 km E53.1oN
BEARINGS:
These can be written in two forms as:
a. Bearing which is a proportion of a circle
-
Three sig.figs.
E.g. 53o written as 053o. Move from north 53o round clockwise.
b. As a compass heading
-
Start with direction pointing
-
Place number of degrees from 0 – 90o
-
Place compass direction you are moving in after the number of
degrees.
E.g. 53o west of north can be rewritten N53oW
What is the resultant force R ?
6N
We ADD vectors HEAD to TAIL
8N
6N

R
R  82  62 10 N
8
tan    1.333
6
  53.1
Answer = 10N N53.1oE or 10N 053.1o
0
Two vectors à & Ñ are added together = à + Ñ
Ã
+
Ã
Ã
Ñ
Ñ
Ã+Ñ
1. First draw
vector Ã
2. Then draw
vector Ñ so
that the tail of
vector Ñ
touches the
head of
vector Ã
3. The resultant
is the vector
that starts at
the tail of
vector à and
ends at the
head of
vector Ñ.
Ñ
Example 1:
A tourist couple visited several places of interest around Dunedin.
They began by travelling 5.0km east from their hotel, then 3.0km
south, then 4.0km west and finally 6.0km north. What was their final
displacement, Ỹ, from the hotel?
N
SOLUTION:
Displacement is a vector quantity so
the four displacements covered by the
tourists can be represented by vectors
and added using a vector diagram. The
resultant is their final displacement
from the hotel.
Ỹ

5.0km
6.0km
3.0km
4.0km
0
1
SCALE
5
In the north/south direction the couple
have moved
N
3.0km S & 6.0km N = 3.0km N
In the east/west direction they have
moved
5.0km E & 4.0km W = 1.0km E
Resultant found by adding two vectors
together.
Ỹ
3.0km N

E
1.0km E
Ỹ = (3.02 + 1.02)
= 10
= 3.2km
tan = 3.0/1.0
= 3.0
 = 72o
Ans = 3.2km, 72o N of E OR 018o
Vector Subtraction
If any negative sign is used then this means that the direction has
changed to the opposite direction.
Ã
Two vectors à & Ñ are subtracted = à - Ñ
-
Ñ
-Ñ
Ã-Ñ
Ã
-Ñ
Ã
1. Begin with Ã
2. Add -Ñ to give 3. ….the
resultant,
à - Ñ.
To find the change in velocity, v, from an initial velocity vi to a final
velocity vf, the vector subtraction vf – vi is done.
vi
vf
v
vf
-vi
v = final velocity – initial velocity
v = vf – vi
v = vf + (-vi)
RESOLUTION OF VECTORS COMPONENTS
F
What is the component of the
force F in the x - direction?

x
C
C is the component of F in
the x - direction
C
 cos 
F
Dropping a perpendicular
Hence C = F cos 
What is the component of the
force F in the y - direction?
F
E

E is the component of F in
the y - direction
x
y
E
 sin 
F
Dropping a perpendicular
Hence E = F sin 
Example 2:
The vertical/horizontal reference frame (a) shows a vector
representing a 10.0N force.
The direction of the vector is angled up from the horizontal at 30o.
The two force vectors which represent the components of this 10N
force are found by drawing two construction lines, one parallel to
the vertical axis and the other parallel to the horizontal axis. The
components are labelled (b).
The 10N vector and its components forma right angle triangle (c).
Use trigonometry to calculate the vertical & horizontal components.
b.
a.
vertical
c.
10N
30o
horizontal
10N
vertical
10N
30o
30o
x
horizontal
y
SOLUTION:
Vertical component
Sin 30o = opposite/hypotenuse
Sin 30o = y/10
y = 10sin30o
y = 5.00N
Horizontal component
Cos 30o = adjacent/hypotenuse
Cos 30o = x/10
x = 10cos30o
x = 8.66N
This means that the 10N
force at 30o is made up of
a horizontal component of
8.66N and a vertical
component of 5.00N
From this the 10N force at
30o has a horizontal effect
of 8.66N and a vertical
effect of 5.00N
READ INFORMATION PAGE 37 - 39
COMPLETE QUESTIONS
FROM
RUTTER
RELATIVE VELOCITY
This is the velocity of one object in relation to another object. The
velocity of an object may appear to be different on where it is
measured from.
What do you notice about
the motion of each of the
planes relative to a person
watching from the
ground?
Example 3:
A motorbike moves forward at 100kmh-1 and beside it is a jeep
moving at 50kmh-1 in the same direction.
How does fast does the bike appear to be moving observed by the
jeep driver?
50kmh-1 
100kmh-1 
SOLUTION:
Bike moves forward fro the jeep at 100-50 = 50kmh-1.
To the driver of the jeep it is though the jeep is stationary and the bike
is moving at 50kmh-1.
Example 4:
A tank rides horizontally at 50kmh-1 through the rain. The steady
rain falls vertically at 15kmh-1. What is the velocity of the rain
relative to the tank?
Tank
50kmh-1
Rain 15kmh-1
SOLUTION:
The two velocities are both given relative to the ground. Yet
experience tells us that relative to the tank the rain would appear to
be coming towards their face than straight down.
The velocity of the rain relative to the tank is found by subtracting
the tanks velocity from the velocity of the rain.
Using Pythagorus (where v is the magnitude):
v = 152 + 502
= 52kmh-1
Using trigonometry
=
tan-1(15/50)
 = 17o
Vrain rel tank
15kmh-1 -vrain

50kmh-1 -vtank
52kmh-1 at an angle of 17o to the horizontal
Observe the motion of the river boats below. In each case, the boat is
heading across the river. In the top case, the presence of a current causes
the boat to travel downstream as it simultaneously moves across the river;
the result is that it reaches the opposite shore at a point downstream from
where it started. In the bottom case, the absence of a current means that the
boat moves straight across the river; it ultimately reaches a point on the
opposite shore directly across from where it started. It is important to note
that the current is what carries the boat down the stream; it is not the motor
of the boat that provides its downstream motion.
Example 5:
A boat starts to cross a river which is 2.0km wide. The boat travels
at 20kmh-1 relative to the surface of the water and points directly
across the river. The river is moving 6.0kmh-1 downstream relative
to the river bank.
a. What is the boat’s velocity relative to the river bank?
b. How long does the boat take to cross the river?
c. At what point does the boat land on the other side?
6.0kmh-1 relative to the bank
20kmh-1
20kmh-1 relative to
the water’s surface
SOLUTION:
The current is relative to the river bank and the river bank (and any
observer standing on the bank) is at rest.
6.0kmh-1
a. The boats velocity relative to the
river bank is the combined result
of its velocity relative to the river
and the velocity of the river
relative to the river bank:
vboat rel river + vriver rel bank = vboat rel bank

20kmh-
v boat rel bank
1

Using Pythagorus and trigonometry, the boat moves with a
speed of:
v = (202 + 6.02)
tan  = 20/6.0
v = 21kmh-1
 = 73o
Vboat rel bank = 21kmh-1 at 73o
b. The boat continues to travel across the river at 20kmh-1, despite the
fact that it is also being carried downstream. The river’s motion
does not affect the boat’s movement across the river because the
motion of the river is at right angles to the direction in which the
boat is pointed.
time taken to cross the river is t = d/v
= 2.0/20
= 0.10hours (6minutes)
c. As well as travelling across the river, the boat is acrried at 6.0kmh-1
downstream for 0.10 hours (the time taken to cross the river). The
distance travelled downstream is:
distance d = vt
= 6.0 x 0.10
= 0.60km
The boat touches the opposite river bank 0.60km downstream from
a point directly across from the river.
READ INFORMATION PAGE 40 - 42
COMPLETE QUESTIONS
FROM
RUTTER
COMPLETE HOMEWORK
WORKSHEET ON
VECTORS