Post on 31-Dec-2019
Scalars & Vectors
Vectors are quantities that have both a direction
and a magnitude (size).
Ex. 2 km, 30ο north of east
Examples of Vectors used in Physics
Displacement
Velocity
Acceleration
Force
Scalars are quantities that have only a
magnitude(size) are called.
Scalar Example Magnitude
Speed 20 m/s
Distance 10 m
Age 15 years
Heat 1000 calories
Vectors can be represented by words “Take your team 2 ‘clicks’ (km) north”
“US Air 45, new course 30o at 500 mph.”
Vectors can be represented by symbols In the text, boldface indicates vectors.
Examples:
Vectors can be represented graphically using arrows The direction of the arrow is the direction of the
vector. The length of the arrow tells the magnitude
Vectors can be moved parallel to themselves and still be the same vector
Vectors only tell amount and direction, so a vector doesn’t care where it starts.
tΔx
VaF av
The sum of two vectors is called the
resultant.
To add vectors graphically, draw each vector
to scale.
Place the tail of the second vector at the tip
of the first vector.
Vectors can be added in any order.
To subtract a vector, add its opposite.
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 30 m, E +
84.5 m, E
Notice that the SIZE
of the arrow conveys
MAGNITUDE and the
way it was drawn
conveys DIRECTION.
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
When 2 vectors are perpendicular, you must use the Pythagorean Theorem.
kmc
c
bacbac
8.10912050
255295Resultant
22222
95 km,E
55 km, N
Start
Finish The hypotenuse in Physics is
called the RESULTANT.
The LEGS of the triangle are called the COMPONENTS
Horizontal Component
Vertical
Component
Example: A man walks 95 km, East then 55 km, north. Calculate his RESULTANT DISPLACEMENT
In the previous example, DISPLACEMENT was asked for
and since it is a VECTOR we should include a DIRECTION
on our final answer.
NOTE: When drawing a right triangle that
conveys some type of motion, you MUST
draw your components HEAD TO TOE.
N
S
E W
N of E
E of N
S of W
W of S
N of W
W of N
S of E
E of S
N of E
Just putting North of East on the answer is NOT specific enough for
the direction. We MUST find the VALUE of the angle.
o30)5789.0(1
5789.09555
Tan
sideadjacentsideoppositeTan
95 km, E
To find the value of the
angle we use a Trig
function called TANGENT.
So the COMPLETE final answer = 109.8 km, 30 degrees North of East
N of E
55 km, N
109.8 km
Resolve each vector into x and y
components, using sin and cos.
Add the x components together to get the
total x component. Add the y component
together to get the total y component.
Find the magnitude of the resultant using
Pythagorean theorem.
Find the direction of the resultant using the
inverse tan function.
Any vector can be resolved, that is, broken
up, into two vectors, one that lies on the x-
axis and one on the y-axis.
An arrow is shot from a bow at an angle of 25ο
above the horizontal, with an initial speed of 45
m/s. Find the horizontal and vertical components
of the arrow’s initial velocity.
25o
vx
vy
?
?
25
m/s 45
y
x
v
v
v
o
m/s 4178.40
)cos(25m/s) 45(
cos
cos
x
x
x
x
v
v
vv
v
v
m/s 1901.19
)(25sinm/s) 45(
sin
sin
y
y
y
y
v
v
vv
v
v
Suppose a person walked 65 m, 25 degrees East of North.
What were his horizontal and vertical components?
EmCHopp
NmCVadj
hypopphypadj
hypotenuse
sideopposite
hypotenuse
sideadjacent
,47.2725sin65..
,91.5825cos65..
sincos
sincos
65 m 25
H.C. = ?
V.C = ?
The goal: ALWAYS MAKE A RIGHT
TRIANGLE!
To solve for components, we often use
the trig functions since and cosine.
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.
3.31)6087.0(
6087.23
14
93.262314
1
22
Tan
Tan
mR
35 m, E
20 m, N
12 m, W
6 m, S
- = 23 m, E
- = 14 m, N
The Final Answer: 26.93 m, 31.3 degrees NORTH or EAST
23 m, E
14 m, N R
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.
1.28)5333.0(
5333.015
8
/17158
1
22
Tan
Tan
smRv
15 m/s, N
8.0 m/s, W
Rv
The Final Answer : 17 m/s, @ 28.1 degrees West of North
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.
SsmCVopp
EsmCHadj
hypopphypadj
hypotenuse
opposite
hypotenuse
adjacent
,/64.3332sin5.63..
,/85.5332cos5.63..
sincos
sinecos
63.5 m/s
32
H.C. =?
V.C. = ?
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.
NkmCVopp
EkmCHadj
hypopphypadj
hypotenuse
opp
hypotenuse
adj
,2.96440sin1500..
,1.114940cos1500..
sincos
sinecosine
0.20)364.0(
364.01.2649
2.964
1.28192.9641.2649
1
22
Tan
Tan
kmR
5000 km, E
40
1500 km
H.C.
V.C.
1500 km + 1149.1 km = 2649.1 km
2649.1 km
964.2 km R
The Final Answer: 2819.1 km @ 20
degrees, East of North
We use the term VECTOR RESOLUTION to suggest
that any vector which IS NOT on an axis MUST be
broken down into horizontal and vertical
components.
BUT --- the ultimate and
recurring themes in
physics is take any and all
vectors and turn
them all into ONE BIG
RIGHT TRIANGLE.
1. Make a drawing showing all the vectors, angles, and given directions.
2. Make a chart with all the horizontal components in one column and all the vertical components on the other.
3. Make sure you assign a negative sign to any vector which is moving WEST or SOUTH.
4. Add all the horizontal components to get ONE value for the horizontal. Do the same for the vertical.
5. Use the Pythagorean Theorem to find the resultant and Tangent to find the direction.
A search and rescue operation produced the
following search patterns in order:
1: 30 meters, west
2: 65 meters, 32 degrees East of South
3: 130 meters, east
4: 42 meters, 22 degrees West of North
1: 30 meters, west
2: 65 meters, 32
degrees East of
South
3: 130 meters, east
4: 42 meters, 22
degrees West of
North
130 m, E
30 m, W
65 m
32ο 42 m
22ο
1: 30 meters, west
2: 65 meters, 32
degrees East of South
3: 130 meters, east
4: 42 meters, 22
degrees West of
North
Leg Horizontal Vertical
1 30 m 0 m
2
3 130 m 0 m
4
65 meters, 32 degrees
East of South
Leg Horizontal Vertical
1 30 m 0 m
2
3 130 m 0 m
4
65 m 32
h.c.
v.c
65cos32 =
55.12 m
65sin32 =34.44 m
55.12 m 34.44 m
42 meters, 22 degrees
West of North
Leg Horizontal Vertical
1 30 m 0 m
2 34.44 m 55.12 m
3 130 m 0 m
4
42 m 22 v.c
h.c. 42sin22 =15.73 m
15.73 m
42cos22=38.94 m
38.94 m
1: 30 meters, west
2: 65 meters, 32
degrees East of
South
3: 130 meters, east
4: 42 meters, 22
degrees West of
North
Leg Horizontal Vertical
1 30 m 0 m
2 34.44 m 55.12 m
3 130 m 0 m
4 15.73 m 38.94 m
-
-
-
1: 30 meters, west
2: 65 meters, 32
degrees East of South
3: 130 meters, east
4: 42 meters, 22
degrees West of
North
Leg Horizontal Vertical
1 -30 m 0 m
2 34.44 m -55.12 m
3 130 m 0 m
4 -15.73 m 38.94 m
Total 118.71 m -16.18 m
What does this mean??? 118.71 m
-16.18 m
2 2118.71 ( 16.18) 119.81R m
1
16.180.136
118.71
(0.136) 7.76
Tan
Tan
65 m
32
130 m, E
42 m 22
30 m, W 118.71 m
-16.18 m
Final Answer: 119.81 m, 7.76 degrees, South of East