Download SPH 3U Grade 11 U Physics

SPH 3U Grade 11 U Physics
• This lesson will extend your knowledge of kinematics to two
dimensions.
• You will be able to solve problems involving displacement in
two dimensions using two, more accurate methods
SPH 3U Grade 11 U Physics
What are we going to cover today?
By the end of this lesson,
you will be able to:
Apply the Trig
Method
Apply the Comp
Method
Homework
Math Review
 Right angle triangles (90°)… what
math “tools” can we use?
 Pythagorean Theorem
 Trig ratio’s
 S-OH
 C-AH
 T-OA
Math Review
 Non-right angle triangle's…
what options exist?
 Sine Law
 Cosine Law
2
C
=
2
A
+
2
B
- 2ABcosc
Allows us to solve problems by looking at parts as right angle triangles
Also allows us to solve problems that involve non-right angle triangles
Let’s solve yesterdays hiker question using the
trigonometric method.
SPH 3U Grade 11 U Physics
Ex 1: A hiker walks 15 km [N], then 24 km [N60oW],
then 21 km [S]. What is his final
displacement?
Sol’n:
This is the diagram
24 km
21
km
after we apply vector
o
60.
addition
A
Next we break the vector
diagram into Triangles
What can we do from here?
15 km
B
Solve for length A and then
solve for length B
SPH 3U Grade 11 U Physics
We apply cosine law
to solve for the
unknown side
60.o
24 km
A
21 km
C2 = A2 + B2 - 2ABcosc
a
a   21   24   2  21 24  cos 60.
2
2
a  513.02
 a  22.65
2
2
60.o
o
SPH 3U Grade 11 U Physics
We apply sine law
to solve for the
angles
60.o
24 km
A
21 km
22.65 km
sin q sin60.

21
22.65
o
 q  53.4
o
60.o
q
SPH 3U Grade 11 U Physics
We apply cosine law
o
60.
again to solve for the
final answer
21 km
A
24 km
22.65 km
60.o
53.4o
66.6o
B
a  15   22.65  2 15 22.65  cos 66.6
2
2
a  468.16
 a  21.6
2
2
15 km
o
SPH 3U Grade 11 U Physics
We apply sine law
o
60.
again to solve for the
final angle
21 km
A
24 km
22.65 km
60.o
53.4o
66.6o
B
sin q sin66.6o

22.65
21.6
 q  74.2o
21.6 km
15 km
q
 Dd = 22 km [N74oW]
Break the VECTOR into its component pieces
(a y value and an x value)
Remember:
vector quantities can
be represented by
directed line segments
Dd
“All y, no x”
v av
“No y, all x”
r
“Some y, some x”
If I gave you 5 different “vectors” or movements…
how would you go about finding the TOTAL
DISPLACEMENT of all 5?
 Ie.
 10m N30ºE =
 13m E15ºN =
 21m E83ºN =
 20m W45ºN =
 11m N72ºE =
V1
V2
V3
V4
V5
Perhaps with a graph?
 But what the heck
does that tell us?
 If only we knew the total vertical
displacement & the total horizontal
displacement
 We could use Pythagorean Theorem to
solve for the total displacement
 What we can do is figure out the
INDIVIDUAL vertical and horizontal
displacement of EACH VECTOR  and
then add them up!
 If only we knew the total vertical
displacement & the total horizontal
displacement
 We could use Pythagorean Theorem to
solve for the total displacement
 What we can do is figure out the
INDIVIDUAL vertical and horizontal
displacement of EACH VECTOR  and
then add them up!
 By breaking up each individual vector (ex. 20 km
N45ºE) into a y value (ex. Vertical displacement) and
x value (horizontal displacement) we eliminate the
need to look at its direction – focusing instead on
the magnitude only
 This effectively turns any set of complicated vectors
that might be going in various directions, into a
streamlined set of values only going vertical or
horizontal (never both)
 This allows us to use our kinematics in 1D rules to
determine the overall vertical and horizontal
displacements – which will always form 90º triangles
– Pythagorean Theory - EASY!
We will let r represent our RESULTANT vector (this is the
same as our total displacement)
Δdtotal = r
r
ry
= Δdy
or the Δ
in
vertical
position
q
rx
= Δdx
Or the horizontal
displacement
What is the only
segment of the
triangle we
automatically know
from the start?
What determines the names
of the rest (think location)?
hyp
r
ryopp
?
Ө or 15º
adj
?
rx
Based on this information (hypotenuse and the Ө) how
can we figure out the adjacent and opposite lengths?
Adjacent  Cos Ө = adj / hyp Opposite  Sin Ө = opp / hyp
cos 15º = adj / 13m
Adj = 12.6m
sin 15º = opp / 13m
Opp = 3.4m
Use Pythagorean theory to check ans.
Now that we know how we
can use the trig formula’s
to solve for the adj or opp
lengths we can substitute
the variables to represent
the vertical and horizontal
displacement
Unknown Value
opp
opp ry ry
sin
sin
qq 
hyp
hyp r r
hyp
r
ropp
y
Ө or 15º
rx adj
Vertical Component
 ry  r sinq
hyp
ryopp
r
q
Unknown Value
adj rx
cosq 

hyp r
rx
adj
Horizontal Component
 rx  r cosq
hyp
r
r
ry opp
q q
rx  r cosq
rx
adj
ry  r sinq
“x-component”
of r
“y-component”
of r
i.e. horizontal Δd
i.e. vertical Δd
Or the adjacent length
of the hypotenuse
Or the opposite length of
the hypotenuse
x
y
r
r
hyp
opp
r
y
qq
rxadj
What happens if we use
(or are provided) the
complimentary angle?
rx  r cosq
ry  r sinq
X-Axis Rule
1. Always take your degree from the x axis!
I.
If you take it from the y-axis you will have effectively
flipped the formula upside, causing cos to represent y
(instead of x) and vice versa
2. If the problem involves vectors in multiple
directions (i.e. east & west) – determine the
degree by counting from the SAME x-axis
I.
This will calculate vectors with opposite directions (i.e.
east vs west) as a positive vs. negative – ensuring an
accurate displacement measurement
After determining the X & Y values of every vector simply add
them up to determine the TOTAL-X & the TOTAL-Y displacement
E 45º N
E 45º S
E 315º S
But if we want our
formula to do the
thinking for us we
should use E 315º S
This will ensure it’s a
negative vertical
(south) value but still
a positive horizontal
(east) value
Consider the following vectors:
A (rh1) = 5.0 km due East
B(rh2) = 3.0 km East 450 North
Vector
x-component
rx
y-component
ry
(rh1) Cos q = ___ km(rx1)
(rh1) Sin q = ___ km(ry1)
5.0 cos 0 = 5.0 km
5.0 sin 0 = 0
(rh2) Cos q = ___ km(rx2)
(rh2) Sin q = ___ km(ry2)
3.0 cos 45 = 2 .m
3.0 sin45 = 2 .1 km
A
rh1
B
rh2
Resultant
rh
rx1 + rx2 = rxtotal
ry1 + ry2 = rytotal
7.1 km [E]
2.1 km [N]
SPH 3U Grade 11 U Physics
Every vector consists of two components.
• Vector components
are
 to each other
• Vector components
add together to form
the vector
rh
q
rx = r cos q
SPH 3U Grade 11 U Physics
Now let’s solve use the component method.
o
Ex
1:
A
hiker
walks
15
km
[N],
then
24
km
[N60
W],
Sol’n:
then 21 km [S]. What is his final
displacement?
First, we break each vector down into its components
Dd2 x  20.78 km
Dd2 y
21 km [S]
?
 12 km
60o
24 km [N60oW]
rx  r cosq
ry  r sinq
15 km [N]
Dd1x  0
Dd3x  0
Dd1y
Dd3y  15 km
?
 21 km
?
SPH 3U Grade 11 U Physics
Then we add the x and y comp’s separately
DdRx  Dd1x  Dd2 x  Dd3x
 0  (20?.78)  0
 20.78?km
DdRy  Dd1y  Dd2 y  Dd3y
 (21)  ?
(12)  (15)
 6 km ?
SPH 3U Grade 11 U Physics
Next we draw the comp’s head to tail and use the
Pythagorean Theorem
DdR
q
-20.78 km
DdR 
6
2
 (?20.78)2
 21.63 km?
6
tan q  ?
20.78
o
 q  16
?
 Dd = 22 km [W16oN]
SPH 3U Grade 11 U Physics
Read pgs 68 – 74
Review tutorials before attempting practice!
Questions “Practice”
1 – 2 pg 71
1 – 2 pg 74