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Vectors
Turn in Freefall LabsDay 1: Read pp. 119 – 125 in
textbook
Day 2: Vector Sheet A
Day 3: Vector Sheet B
Note: Some additional reading sources have been placed online. These .pdf files introduce vectors, two dimensional kinematics, and projectile motion (the topics covered in this unit.) These are for your reference only, and not required reading. If you need a little extra help for vectors, please read through these packets.
Introduction to Vectors {2010}:
I. Introduction:
In physics, some physical quantities measure only an __________.
Other physical quantities must also include a ______________
with the measured amount. The quantities that measure only an
amount are called ___________. Those quantities that include a
direction are called ____________.
amount
direction
scalars
vectors
Examples of scalars:
mass, time, distance, speed, energy, temperature
Examples of vectors:
displacement, velocity, acceleration, force, momentum, fields (gravitional, electric and magnetic)
II. Representing Vectors:
When representing a vector in mathematical equations, some kind of notation must be used to distinguish the vector from the scalar. Scalars will be represented with ordinary letters, whereas vectors will be represented by a letter with an arrow over the top of it. For example:
TEtm ,,, = scalars
xpFav ,,,, = vectors
When drawing a picture detailing a given problem, vectors arerepresented by arrows. The length of the arrow represents the amount of the vector, and the direction of the arrow gives the direction of the vector.
velocity of an airplane
sum of two displacements.
boat traveling across a stream, flow of river carries boat downstream.
Note that the boat is not pointing in the direction it is traveling! We will see an example of this later…
III. Equality of Vectors:
If two vectors have the same length and the same direction, then
these two vectors are considered ___________. Any vector can be
moved from one place to another, and so long as the magnitude
(length) and direction are not changed, it will still be considered
the ____________ vector.
equal
same
Even though each of these vectors start and finish in a different location, each has the same length and direction as the others. These vectors are all considered equal to one another.
The picture can also be interpreted as the same vector drawn in 4 different locations.
Even though each of these vectors start and finish in a different location, each has the same length and direction as the others. These vectors are all considered equal to one another.
IV. Scalar Multiplication of Vectors:
A vector may be multiplied by a scalar. Multiplying a vector by a positive scalar will only change the length (and possibly the unit) of the vector quantity, but it will not change the direction of the vector.
Example #1: Given vector measures 3.00 cm towards the right, what is ?
A
A
3
rightcmA 00.3
IV. Scalar Multiplication of Vectors:
A vector may be multiplied by a scalar. Multiplying a vector by a positive scalar will only change the length (and possibly the unit) of the vector quantity, but it will not change the direction of the vector.
Example #1: Given vector measures 3.00 cm towards the right, what is ?
A
A
3
rightcmA 00.3
is a vector 3 times as long as vector , in the same direction.A
3 A
rightcmA 00.93
Just as a little extra, here is an example of a scalar multiplication that changes units: momentum is mass times velocity
vmp
kg s
ms
mkg
Since mass is a positive quantity, the momentum and the velocity vector always point in the same direction.
IV. Scalar Multiplication of Vectors: {continued…}
A vector may also be multiplied by a negative scalar. The negative sign will reverse the direction of the vector. The size of the scalar will change the length of the vector.
Example #2: Given vector measures 3.00 cm towards the right, what are and ?
A
A
rightcmA 00.3
A
2
The negative of a vector is a vector with the same length that points in the opposite direction.
rightcmA 00.3
leftcmA 00.3
leftcmA 00.322
leftcm00.6
V. Adding and Subtracting Vectors:
Two vectors may be added together provided that the values of the vectors have the same units. The diagrams below show the addition of two vectors, and .1V
2V
1V
2V
The addition of two vectors can be viewed as a series of operations: first do vector #1 then do vector #2.
The tip – to – tail method of adding vectors places the tail (starting point) of one vector at the tip (ending point) of another vector.
21 VV
1V
2VRV
The resultant vector extends from the start of the first vector to the end of the second vector.
The parallelogram method of adding vectors places the tails (starting points) of both vectors at the origin. Recall that a parallelogram has two sets of parallel sides, with each member of the same set having the same length.
1V
2V
Next place copies of each vector at the tip of the other vector to finish off the parallelogram.
2V
1V
The resultant vector will extend from the origin out as a body diagonal.
RV
Note that when this is compared to the tip – to – tail method, the order of addition does not make any difference.
1221 VVVVVR
Why is this diagram incorrect?
1V
2V
C
Vector ‘C’ does not represent the sum of the two vectors. This vector is the wrong body diagonal on the parallelogram method. This is a common mistake, though.
Vector ‘C’ actually represents a difference, or subtraction, between the two vectors. From the tip – to – tail method, get:
CVV
12
12 VVC
This diagram shows again that the order of addition of vectors is not important. Note the parallelogram method for vector addition.
This diagram also shows that the order of addition of vectors is not important. Here, three vectors are added in various orders. The result is the same each time.
This diagram shows tip – to – tail method for adding three vectors.
Another diagram detailing order of addition…
This diagram details how to subtract two vectors. Just view the subtracted vector as adding the negative of the vector.
Another diagram showing how to interpret the subtraction of two vectors…
VI. Reporting Directions: Map Coordinates, Bearing, Azimuth
N
EW
S
Say a vector is given pointing as shown. How do we write the angle for this vector to show direction?A
30
The first system will write the numeric value of the angle to a one side of a given axis. This example would be written as:
30 degrees to the north side of the east axis.
NEeastofnorth 3030
Vector Directions: Map Coordinates
This angle is measured as NE for North of the East axis
This a
ngle
is m
easu
red
as
EN for E
ast o
f the
Nor
th a
xis
This angle is measured as NW for North of the West axis
WN
SW
WS ES
SE
N
W E
S
Vector Directions: Map Bearings
The angle is measured clockwise from the North axis. North is 0°, east is 90°, south is 180°, and west is 270°.
N
E
S
W
Vector Directions: Azimuth
The angle is measured as degrees East or West of the North/South axis.
The angle is written in the form N35°E, or S27°W.
A
Vector A is measured as N35°E, or 35° East of the North axis.
35°
27°
B
Vector B is measured as S27°W, or 27° West of the South axis.
N
E
S
W
EXAMPLES: 3. Convert 17° SW in to a WS direction , a bearing, and an azimuth.
17° SW
90° – 17° = 73° WS
This vector also measures as S73°W as an azimuth.
This vector also measures as 180° + 73° = 253° as a bearing.
N
E
S
W
4. Convert 34° ES in to a SE direction , a bearing, and an azimuth
34° ES
90° – 34° = 56° SE
Bearing = 90° + 56° = 146°
Azimuth = S34° E
N
E
S
W
5. Convert a bearing of 52° in to two different directions and an azimuth
Bearing 52° is
also 52° EN and azim
uth N52°E
Finally 90° – 52° = 38° NE
N
E
S
W
6. Convert a bearing of 306° in to two different directions and an azimuth
Bearing = 306°
Direction = 306° – 270° = 36° NW
Direction = 360° – 306° = 54° WN
Azimuth = N54° W
N
E
S
W
Angles measured from vertical and horizontal…….
Vertical
Horizontal
From vertical
Above horizontal
Below horizontal
Vectors, Day #2
Tonight: Vector Sheet A
Day 3: Vector Sheet B
VII: Combining vectors
____________________ : the combination of 2 or more vectors.
(also called... _______, _________, ______)
Resultant
sum total net
Example #6: A swimmer averages 3.00 km/h in still water. A river flows due east at a rate of 2.00 km/hr. Calculate the velocity of the swimmer if she swims East:
River Swimmer
Resultant is 3.00 km/hr + 2.00 km/hr = 5.00 km/hr East
Example #7: A swimmer averages 3.00 km/h in still water. A river flows due east at a rate of 2.00 km/hr. Calculate the velocity of the swimmer if she swims West:
River
Swimmer
Resultant is 3.00 km/hr West + 2.00 km/hr East = 1.00 km/hr West
Alternative: Let west be negative and east positive. The sum becomes:
(-3.00 km/hr) + (+2.00 km/hr) = -1.00 km/hr = 1.00 km/hr West
Example #8: A swimmer averages 3.00 km/h in still water. A river flows due east at a rate of 2.00 km/hr. Calculate the velocity of the swimmer if she swims North. Calculate the answer both using (a) trig and (b) graphically.
(b) Graphical Solution :
Solve the vector sum by drawing the vectors on a piece of paper. Use a ruler and protractor to put a scale to the length and to make the angle measurements for the vectors. On your picture, draw the resultant vector and measure its length with the ruler and its direction with the protractor. Finally, convert that length with the scaling factor you used to draw the original vectors.
(a) Trig Solution:
Draw the vectors
River 2.00 km/hr East
Sw
imm
er
3.00
km
/hr
Nor
th
Draw the resultant
By Pythagorean Theorem the resultant is:
222 cba 22 bac
hrkmc 61.3
The angle is needed for direction:
NEhrkm
hrkm
3.5600.2
00.3tan 1
Ex. #9: A swimmer can swim at a speed of 3.00 km/h relative to still water. The swimmer wishes to cross the river from the south bank to the north bank, but the river flows eastward at a rate of 2.00 km/h. (a) Which direction should the swimmer aim so that she can reach the opposite bank directly? (b) How fast is she effectively moving across the river? (c) If the river is 255 meters wide, how long will it take her to cross the river?
Do not write all this, just think logically about it:First, the 3.00 km/hr is the speed of the swimmer relative to the water, independent of whether the water is moving relative to land! If the water is moving relative to land, then the velocity of the swimmer adds as a vector to the velocity of the water to make the overall velocity of the swimmer relative to the ground.
Always use logic to solve these problems. Where is the swimmer starting? Where does the swimmer want to end? What does the swimmer have to do to make this possible?
For example, this swimmer wants to go due north. The river flows to the east. Which way should the swimmer point to arrive due north? Directly across or at some angle? Why?
Since the swimmer wants to go north, but the current will move the boat to the east, the swimmer must turn to an angle opposite to the current. In other words, the boat must aim west of north! Set up the picture as follows:
Desired path of sw
imm
er.
Direction the swimm
er must aim
, also
the speed of the swimm
er relative to
water. 3.00 km/hr This is not the
actual path of the swimm
er!
Velocity of water = 2.00 km/hr
hrkm
hrkm
00.3
00.2sin 1
Direction of swimmer:
WN 8.41
As for the time to cross, this depends on how far across and how fast the swimmer actually crosses the river. Remember, the 3.00 km/hr is how fast the swimmer travels relative to the water, not the shore! Solve for the speed of the swimmer across the river with Pythagorean’s Theorem.
2.00 km/hr
3.00 km/hr
A =
actual speed.
222 bac
22 00.200.3 hrkm
hrkmA
hrkmA 24.2
hrkm
ttimehrkm
114.024.2
255.0
84.6t minutes
REVIEW PROBLEM: A boat that goes 8.00 km/h in still water is to cross a river 385 m wide. The river is flowing south with a velocity of 2.50 km/h. (a) If he starts from the west bank of the river, where should the river boat pilot aim the boat so as to go directly across the river? (b) How long will it take to cross the river?
Desired v
River vBoat v
, where
to aim
NEhrkm
hrkm
2.1800.8
50.2sin 1
22 50.200.8 hrkm
hrkmA
sm
hrkmA 11.260.7
04.318211.2
385 s
mt
sm
min
VECTORS DAY 2: Vectors not at right angles
Vectors not at right angles: can be solved by making _________ and
then applying ____________ (law of __________ and ____________).
Example 1: A boy walks 1.55 km East and then 2.45 km 37.0° NE.Find his displacement (which includes both magnitude and direction!).
triangle
trigonometry sines cosines
N
E
W
S
37.0°143.0°
1.55 km
2.45 kmResultant
Solve for the resultant using the law of cosines:
CabbaR cos2222
143cos45.255.1245.255.1 22 kmkmkmkmR
kmR 80.3
Solve for the direction angle, , using the law of sines:
kmkm 80.3
143sin
45.2
sin
NE 8.22
N
EW
S
Example 2: Find the displacement: 122 km 37.0° NE, 175 km 17.5° WS
A =122 km
B=
175
km
R
37.0°
17.5
°
37.0°
35.5°
Upper right corner:90 - 37.0 - 17.5 = 35.5°
First, solve for R with law of cosines.
CABBAR cos2222
5.35cos1751222175122 22 kmkmkmkmR
kmR 104664.103 Now solve for . Since this may be an ambiguous case, solve using the law of cosines.
cos2222 ARRAB
222cos2 BRAAR
AR
BRA
2cos
222
AR
BRA
2cos
2221
664.1031222
175664.103122cos
2221
4.101
Now subtract 37.0° from . This will give R from the east axis.
SEDirection 4.640.374.101
Example #3: An airplane is flying with an airspeed of 255 mph, and the compass indicates the plane is flying at an angle of 35.0° NW. If the wind is blowing 65.0 mph from the west, what is the actual velocity of the airplane relative to the ground?
N
S
EW
255 mph
35.0°
65.0 mph
R
35.0°
from west means towards east
First, solve for R with law of cosines.
CABBAR cos2222
0.35cos0.6525520.65255 22 mphmphmphmphR
mphR 20517.205 Now solve for .
mphmph 17.205
0.35sin
0.65
sin
5.10
0.35direction NW 5.45
N
EW
S
Example #4: To be on schedule a pilot needs to arrive at airport in 5.00 hours. The airport is located 1500 km 40.0° NW of her present location. If the wind averages 70.0 km/h from the southwest, where should she aim and how fast should she fly?
R = Desired velocity = 300 km/h
40.0°
Wind is from the southwest, so towards the northeast. The angle is 45.0°.
45.0°wind = 70.0 km/h
P = pilot should fly
45.0° 40.0°
95.0°
First, solve for P with law of cosines. P is where the pilot should fly, R is the resultant of 300 km/h, and W is the wind at 70.0 km/h.
0.95cos2222 RWWRP
0.95cos70300270300 22 kmkmkmkmP
hkmP 314944.313
Use law of sines to solve for .
hkm
hkm 944.313
0.95sin
0.70
sin
8.12
Subtract from 40.0° to get the direction.
NWdirection 2.278.120.40
The End
N
EW
S
Example 5: The pilot of a river boat (that goes 10.0 km/h in still water) wants to go to a dock that is located 300 m upstream from his present location on the southern bank of a1200m wide river. If the river is flowing 3.00 km/h eastward (a) where should he point the boat to go directly to the dock and (b) how long will it take to cross the river
120
0 m
300 m
R = desired
W = 3.00 km/h
P = pilot aim = 10.0 km/h
This would be a good bonus question!
Solve first for angle .
m
m
1200
300tan 0.14
Next solve for angle . 0.760.140.9090
Next solve for angle . 0.1040.760.180180
Next use the law of sines to solve for angle .
hkm
hkm 0.10
0.104sin
00.3
sin
9.16
Finally add angle onto angle for the direction.
WNdirection 9.309.160.14
(b) For the time to cross, we need the speed along the actual path the boat takes, which is the desired path. Because of the current, the boat’s actual speed is slower than 10.0 km/h. First solve for angle then use law of sines to solve for R.
0.599.160.1040.180180
0.104sin
0.10
sinhkmR
sm
hkmR 455.284.8
Finally find the distance of travel with Pythagorean theorem.
mmmd 12373001200 22
Solve for the time with d = v t
min40.8504455.2
1237 s
m
v
dt
sm