9/24 Vectors!!!!
Pick up Problem Sets
Physics HW 11 & 12 2D Kinematics
AP Vectors More Practice
Tests are not graded. But I did sleep in a Holiday Inn last night.
Bonus Opportunity
9/25 Vectors!!!! Pick up a piece of
graph paper
Yesterday you picked up
Physics HW 11 & 12 2D Kinematics
AP Vectors More Practice
Tests are graded, but not entered.
Bonus Opportunity
9/26 Vectors!!!!
Yesterday we worked thorough ex 3 and 5.
Today you will complete the pirate vector lab
You will work in pairs.
Supplies: Book cover
Metric ruler
Protractor
Colored Pencils or markers
Lab directions
There are 12 clues
Test Corrections if below a 70
AP Bonus opportunities
9/29 Vectors!!!!
Friday we did the pirate lab. See me if you were absent.
Today we will look at the Girl Scout problem (this is
typically referred to as the river boat problem) and we
will consider Force as a vector (push or pull)
Test Corrections if below a 70
AP Bonus opportunities
9/30 Vectors!!!!
Get the calculator that matches your blue tag on your desk
Monday we did the Girl Scout problem (ex 6) and two
Forces at right angles (ex 7)
Today we will look at components, multiple vectors and
learn the component method.
If time we will look at problems that require correction of
motion (ie 15,16, & 19 on HW #12)
Test Corrections if below a 70 (This mon-thur)
AP Bonus opportunities
Pirate Vector Lab
Solve graphically (using a metric ruler and a protractor)
Read the story and follow the directions exactly.
Additional instructions: you must
Include arrows and units for each vector
Draw your resultant in a different color
Determine resultant magnitude and direction graphically
(using a metric ruler and a protractor)
Record answer (with units) and secret rule on blank area of
your map. Box your answer.
Rubric:
Name-Scale 1
Pirate Vector Lab Rubric:
1 Name-Scale
Vectors and Scalars
AP Physics B
Scalar
A SCALAR is ANY
quantity in physics that
has MAGNITUDE, but
NOT a direction
associated with it.
Magnitude – A numerical
value with units.
Scalar
Example
Magnitude
Speed
20 m/s
Distance
10 m
Age
15 years
Heat
1000
calories
Vector
A VECTOR is ANY
quantity in physics that
has BOTH
MAGNITUDE and
DIRECTION.
   
v , x, a, F
Vector
Velocity
Magnitude
& Direction
20 m/s, N
Acceleration 10 m/s/s, E
Force
5 N, West
Vectors are typically illustrated by
drawing an ARROW above the symbol.
The arrow is used to convey direction
and magnitude.
Applications of Vectors
VECTOR ADDITION – If 2 similar vectors point in the SAME
direction, add them.

Example: A man walks 54.5 meters east, then another 30
meters east. Calculate his displacement relative to where he
started?
54.5 m, E
+
84.5 m, E
30 m, E
Notice that the SIZE of
the arrow conveys
MAGNITUDE and the
way it was drawn
conveys DIRECTION.
Applications of Vectors
VECTOR SUBTRACTION - If 2 vectors are going in
opposite directions, you SUBTRACT.

Example: A man walks 54.5 meters east, then 30
meters west. Calculate his displacement relative to
where he started?
54.5 m, E
30 m, W
24.5 m, E
-
Non-Collinear Vectors
When 2 vectors are perpendicular, you must use
the Pythagorean theorem.
The hypotenuse in Physics
is called the RESULTANT.
A man walks 95 km, East then 55
km, north. Calculate his
RESULTANT DISPLACEMENT.
Finish
c2  a2  b2  c  a2  b2
55 km, N
Horizontal Component
Vertical
Component
c  Resultant  952  552
c  12050  109.8 km
95 km,E
Start
The LEGS of the triangle are called the COMPONENTS
BUT……what about the direction?
In the previous example, DISPLACEMENT was asked for
and since it is a VECTOR we should include a
DIRECTION on our final answer.
N
W of N
E of N
N of E
N of W
N of E
E
W
NOTE: When drawing a right triangle that
conveys some type of motion, you MUST
draw your components HEAD TO TOE.
S of W
S of E
W of S
E of S
S
BUT…..what about the VALUE of the
angle???
Just putting North of East on the answer is NOT specific enough
for the direction. We MUST find the VALUE of the angle.
To find the value of the
angle we use a Trig
function called TANGENT.
109.8 km
55 km, N
 N of E
95 km,E
opposite side 55
Tan 

 0.5789
adjacent side 95
  Tan 1 (0.5789)  30
So the COMPLETE final answer is : 109.8 km, 30 degrees North of East
What if you are missing a component?
Suppose a person walked 65 m, 25 degrees East of North. What
were his horizontal and vertical components?
H.C. = ?
V.C = ?
25
65 m
The goal: ALWAYS MAKE A RIGHT
TRIANGLE!
To solve for components, we often use
the trig functions sine and cosine.
adjacent side
hypotenuse
adj  hyp cos 
cosine  
opposite side
hypotenuse
opp  hyp sin 
sine  
adj  V .C.  65 cos 25  58.91m, N
opp  H .C.  65 sin 25  27.47m, E
Example 1
A bear, searching for food wanders 35 meters east then 20 meters north.
Frustrated, he wanders another 12 meters west then 6 meters south. Calculate
the bear's displacement.
-
12 m, W
-
=
6 m, S
20 m, N
35 m, E

=
14 m, N
R  14 2  232  26.93m
14 m, N
R
23 m, E
14
 .6087
23
  Tan 1 (0.6087)  31.3
Tan 
23 m, E
The Final Answer: 26.93 m, 31.3 degrees NORTH of EAST
Example 2
A boat moves with a velocity of 15 m/s, N in a river which
flows with a velocity of 8.0 m/s, west. Calculate the
boat's resultant velocity with respect to due north.
Rv  82  152  17 m / s
8.0 m/s, W
15 m/s, N
Rv

8
Tan 
 0.5333
15
  Tan 1 (0.5333)  28.1
The Final Answer : 17 m/s, @ 28.1 degrees West of North
Example 3
A plane moves with a velocity of 63.5 m/s at 32 degrees South of East. Calculate
the plane's horizontal and vertical velocity components.
adjacent side
cosine  
hypotenuse
adj  hyp cos 
H.C. =?
32
63.5 m/s
opposite side
sine  
hypotenuse
opp  hyp sin 
V.C. = ?
adj  H .C.  63.5 cos 32  53.85 m / s, E
opp  V .C.  63.5 sin 32  33.64 m / s, S
Example 4
A storm system moves 5000 km due east, then shifts course at 40
degrees North of East for 1500 km. Calculate the storm's
resultant displacement.
1500 km
adjacent side
hypotenuse
V.C.
adj  hyp cos 
cosine  
opposite side
hypotenuse
opp  hyp sin 
sine  
40
5000 km, E
H.C.
adj  H .C.  1500 cos 40  1149.1 km, E
opp  V .C.  1500 sin 40  964.2 km, N
5000 km + 1149.1 km = 6149.1 km
R  6149.12  964.2 2  6224.14 km
964.2
 0.157
6149.1
  Tan 1 (0.364)  8.91
Tan 
R
964.2 km

6149.1 km
The Final Answer: 6224.14 km @ 8.91
degrees, North of East
Practice adding vectors Ex 5

Draw the following to scale:








Indicate scale.
Make sure you draw tail to tip for the two vectors.
Indicate the resultant with a dotted line. The tip of
the resultant meets the tip of the second vector.
You walk 5 m to the north and then 8 m east
Determine your resultant displacement graphically.
What did you get? I got 9.7m.
Determine your resultant displacement using the
Pythagorean formula.
Did you get 9.43 m?
Practice adding vectors Ex 5

Draw the following to scale:








Indicate scale.
Make sure you draw tail to tip for the two vectors.
Indicate the resultant with a dotted line. The tip of
the resultant meets the tip of the second vector.
You walk 5 m to the north and then 8 m east
Determine your resultant displacement graphically.
What did you get? I got 9.7m.
Determine your resultant displacement using the
Pythagorean formula.
Did you get 9.43 m?
Ex 5
8m
5m
Ex 5

“θ” or Theta, is any unknown angle but in this case it
is the angle between the two vectors



Use a protractor to determine the angle “θ” of your
resultant.
Place the protractor along the axis of the initial
vector. Take the reading in reference to the
resultant.
Refer to diagram on next slide
Ex 5
8m
5m
θ
Ex 5







What value did you get with your protractor?
I got 59°
This would be described as 59° east of north
since the resultant is east of the north axis.
How could you do this mathematically?
Use SOH CAH TOA
Label the sides in reference to θ.
I would suggest using tan since you calculated the
hypotenuse.
Ex 5
Opp 8m
Adj
5m
θ
Ex 5








Using Tan:
Tan θ = opp/adj
Tan θ = 8m/5m
Tan θ = 1.6 THIS IS NOT THE ANGLE!!!
Find inverse tan.
Enter 2nd tan (1.6) enter
You should get 58.0° The direction is still East
of North
This is in the same range as what we got with
the protractor.
Example 6

A girl scout elects to swim across the river. The
river is 37.5 meters wide. A current flows
downstream at a rate of 0.66 m/s. If she initially
swims towards the boy scout camp (directly cross
the river) at a rate of 1.73 m/s, how long will it take
her to reach the far shore?
] 37.5m
1.73 m/s




Remember what the question asks.
How long does it take her to swim across?
To solve for time, what do we need to know?
Use velocity and displacement but only in
reference to crossing the river.
] 37.5m
1.73 m/s
v = d/t
 t = d/v
 t = 37.5m÷1.73m/s
 t = 21.7 s

Example 6
] 37.5m
1.73 m/s
Where exactly does the girl scout
end up on the far shore?
 What do we need to know?
 To determine displacement we need
velocity and time but only in reference
to downstream.

0.66m/s






To find where she ends up, what is the
downstream velocity?
0.66m/s
What is the time?
21.7 sec
Time is the same for both cross stream and
downstream.
14.3 m downstream
When working with multiple vectors
remember they are independent of
one another although they have a net
effect.
 In the case of the girl scout, her
overall (think resultant) velocity and
direction changed.
 Do you know how to solve for the
apparent resultant velocity and
direction?

0.66m/s
1.73 m/s
c
θ







Resultant velocity? Make sure you only use velocity vectors!
c2 = 1.732 + 0.662
c = 1.85 m/s
Which angle for direction?
tan θ = 1.73/0.66 = 2.62…
θ = 69.1°
vr = 1.85 m/s at 69.1° downstream in respect to shore
Fix next slide
0.66m/s
1.73 m/s
c
θ


What angle should the girl
scout enter the water
upstream to end up at the
boy scout camp??
Expand this
How are force vectors drawn?

Tail to tail
Resultant Force


The resultant force of concurrent forces can
be calculated using the parallelogram method
or using the component method.
We will use the parallelogram method first.







F1 and F2 are:
concurrent forces.
FR is the
Resultant force.
FR’ is the
equilibrant force.
The mathematical method used with such
diagram is called the parallelogram method.
EX 7 One force of 7.75 N is exerted on point R in the
direction of direct south. A second force of 15.50 N is
exerted on R in the direction of 90.0 west of south.
What is the magnitude and direction of the resultant
force in reference to the larger force?
15.50 N
θ
tan θ = 7.75 N / 15.50 N
θ = 26.6° south of west
7.75 N
(15.50 N)2 + (7.75 N)2 = FR2
17.3 N = FR
EX 8 A sled is being pulled up and to the right at an
angle of 27.00 to the horizontal. If the person is pulling
with a force of 165.0 N, what are the horizontal and
vertical components of the force?
Determine Y using sin
O= SH = sin(27)(165 N)
A = 74.91 N
165.0 N
27º
Determine X using cos
A= CH = cos(27)(165 N)
A = 147.02 N
What happens if the concurrent vectors
are not at right angles?


We will the component method
We will solve for resultant in respect to the X
axis in Quadrant I
What are the values for X & Y in each
quadrant ?
X is
X is
Y is
Y is
X is
X is
Y is
Y is
What are the values for X & Y in each
quadrant ?
X is +
X is -
Y is +
Y is +
X is +
X is -
Y is -
Y is -
Ex 8 Two forces act on point Q. The first, of 435 N, heads 18.0° north
of east. The second, of 220 N, is directed due west. Determine
magnitude and direction of the resultant force using the component
method.
Ex 9 Two forces act on point Q. The first, of 435 N, heads 18.0° north
of east. The second, of 220 N, is directed due west. Determine
magnitude and direction of the resultant force using the component
method.
435 N
18º
220 N
West
Ex 9 Two forces act on point Q. The first, of 435 N, heads 18.0° north
of east. The second, of 220 N, is directed due west. Determine
magnitude and direction of the resultant force using the component
method.
Describe any
angles outside
of quad 1 in
reference to x
axis of quad 1
180º
220 N
West
435 N
18º
Ex 9 Describe any angles outside of
quad 1 in reference to x axis of quad 1
435 N
18° N of E
Direction in
relation to
QI
18° N of E
220 N
W
Use 180°
magnitude
direction
EX 9 Determine the components and
total
Vector
435 N at 18° N of E
220 N at 180°CCW to E
Total
X component Y component
Cos H
Sin H
EX 9 Determine the components
Vector
435 N at 18° N of E
X component Y component
Cos H
Sin H
413.71 N
134.42 N
220 N at 180°CCW to E -220.00N
0N
Total
134.42 N
193.71N



Draw components on graph, using signs (+ or -) to
determine quadrant
Calculate FR with total X and Y
Calculate Angle in reference to X using tan. (absolute
values)
134.42 N
FR

193.71 N
Ex 9 Two forces act on point Q. The first, of 435 N, heads 18.0° north
of east. The second, of 220 N, is directed due west. Determine
magnitude and direction of the resultant force using the component
method.
435 N
18º
220 N
West
Example 9






FR = √(134.42 N)2 + (193.71N)2
FR = 235.78 N
Tan  = 134.42 N/193.71
 = 34.76°N of E
FINAL ANSWER
The FR is 235.78 N at 34.76°N of E
Ex. 10 What is the net force on the
rope?
Socks
Blue
Patches
Ex 10 A true story
Once upon a time 3 dogs (Socks, Blue, and
Patches) found an old smelly rope with a mass of
0.65 kg. They all began to tug on it at the same
time. Socks pulled on the rope with a magnificent
force of 185N at 15° S of W. Blue, thinking he
didn’t have to apply as much force as the other
smaller dogs, pulled on the rope with a force of
111N at 35° W of N. Little Patches, eager to
make up for his diminutive size, pulled on the
rope with 65 N at 60° N of E. In what direction
did the rope accelerate across the room? What
is the moral of this story?
How do we solve this?







What do we need to know to determine direction of the
rope?
Net Force on the rope.
There’s too many!!!!
Diagram the forces on an X-Y coordinate system
Describe Angles over 90 in reference to Quad I
List knowns and solve for X and Y components
Still use cos and sin (respectively) but use angles in
reference to Quad I
111 N: 35 + 90°= 125°
185N: 15°+ 180° = 195°
65 N
60° N of E
Direction in
relation to
QI
60° N of E
111 N
35° W of N
Use 125°
185 N
15° S of W
Use 195
magnitude
direction
Determine the components
Force Vector
65 N at 60° N of E
111 N at 125°CCW to E
185 N at 195°CCW to E
Total
X component Y component
Cos H
Sin H
Determine the components
Force Vector
65 N at 60° N of E
X component Y component
Cos H
Sin H
32.5 N
56.3 N
111 N at 125°CCW to E -63.7 N
90.9 N
185 N at 195°CCW to E -179 N
-47.9 N
Total
Determine the components
Force Vector
65 N at 60° N of E
X component Y component
Cos H
Sin H
32.5 N
56.3 N
111 N at 125°CCW to E -63.7 N
90.9 N
185 N at 195°CCW to E -179 N
-47.9 N
Total
99.3 N
-210 N



Draw components on graph, using signs (+ or -) to
determine quadrant
Calculate FR with total X and Y
Calculate Angle in reference to X using tan. (absolute
values)
FR
99 N

210 N
Example 10






FR = √2102 + 99.32
FR = 232 N
Tan  = 99.3/210
 = 25.3°N of W
FINAL ANSWER
The rope moves with a force of 232 N at
25.3° N of W