WEBVTT
00:00:03.439 --> 00:00:07.630
Suppose you get a text message. Your friend
tells you to go to Lobby 7 at MIT to find
00:00:07.630 --> 00:00:12.400
the gift they left you 7 meters from the center
of the lobby. Is that enough information to
00:00:12.400 --> 00:00:19.400
find the gift right away? As you can see,
there are many locations 7 meters from the
00:00:20.000 --> 00:00:24.010
center of the room. Don't forget that we live
in 3 dimensions, so there are actually even
00:00:24.010 --> 00:00:28.460
more points 7 meters away from the center
of the room. Fortunately, in this problem,
00:00:28.460 --> 00:00:33.280
you can ignore most of them since we don't
expect our gift to be hanging in mid air.
00:00:33.280 --> 00:00:36.960
Distance alone wasn't enough information.
It would have been helpful to have both the
00:00:36.960 --> 00:00:41.699
distance and the direction.
00:00:41.699 --> 00:00:45.829
This video is part of the Representations
video series. Information can be represented
00:00:45.829 --> 00:00:51.909
in words, through mathematical symbols, graphically,
or in 3-D models. Representations are used
00:00:51.909 --> 00:00:57.309
to develop a deeper and more flexible understanding
of objects, systems, and processes.
00:00:57.309 --> 00:01:03.089
Hi, my name is Dan Hastings and I am Dean
of Undergraduate Education and a Professor
00:01:03.089 --> 00:01:08.470
of Engineering Systems, and Aeronautics and
Astronautics here at MIT. Today, I'd like
00:01:08.470 --> 00:01:13.050
to talk to you about the utility of thinking
about displacements as vectors when trying
00:01:13.050 --> 00:01:18.750
to recall vector properties, and how you determine
if a physical quantity can be represented
00:01:18.750 --> 00:01:20.160
using vectors.
00:01:20.160 --> 00:01:25.789
Before watching this video, you should know
how to add and scale vectors. You should also
00:01:25.789 --> 00:01:31.730
understand how to decompose vectors, and how
to find perpendicular basis vectors.
00:01:31.730 --> 00:01:35.640
After you watch this video, you will be able
to understand the properties of vectors by
00:01:35.640 --> 00:01:39.750
using displacement as an example, and you
will be able to determine whether a physical
00:01:39.750 --> 00:01:43.990
quantity can be represented using vectors.
00:01:43.990 --> 00:01:50.990
Meet the vector. The vector is an object that
has both magnitude and direction. One way
00:01:51.140 --> 00:01:56.830
to represent a vector is with an arrow. You
have seen other algebraic representations
00:01:56.830 --> 00:02:02.640
of vectors as well. There are many physical
quantities that have both magnitude and direction.
00:02:02.640 --> 00:02:08.199
Can you think of some? Make a list of quantities
that can be described by a magnitude and direction.
00:02:08.199 --> 00:02:12.690
Feel free to discuss your list with other
people. We'll come back to this list at the
00:02:12.690 --> 00:02:13.090
end of the video.
00:02:13.090 --> 00:02:16.090
Pause the video here.
00:02:16.090 --> 00:02:23.090
In engineering, there are many physical quantities
of interest that have both magnitude and direction.
00:02:25.260 --> 00:02:31.120
Consider the following example: Here you see
a video of airflow over the wing of an F16
00:02:31.120 --> 00:02:35.760
fighter jet model in the Wright Brothers wind
tunnel at MIT. The air that flows over the
00:02:35.760 --> 00:02:40.940
wing has both speed and direction. The direction
is always tangent to the path of the airflow.
00:02:40.940 --> 00:02:42.390
We can represent the air velocity with an
arrow at each point around the wing. The length
00:02:42.390 --> 00:02:47.079
of the arrow represents speed, and the direction
represents the direction of motion. Such a
00:02:47.079 --> 00:02:51.780
collection of vectors is called a vector field.
The vector field of airflow over the wing
00:02:51.780 --> 00:02:57.020
creates a lift force via the Bernoulli effect.
This effect suggests that because the horizontal
00:02:57.020 --> 00:03:01.200
component of the airflow velocity is the same
throughout the flow field, the air flowing
00:03:01.200 --> 00:03:05.940
over the wing is moving faster than the air
flowing beneath the wing. This creates a difference
00:03:05.940 --> 00:03:11.410
in air pressure, which provides the lift force,
another physical quantity that we can represent
00:03:11.410 --> 00:03:16.770
with a vector. Depending on the angle of the
wing, the magnitude and direction of the lift
00:03:16.770 --> 00:03:21.530
force changes. Lift is just one example of
a vector quantity that is very important in
00:03:21.530 --> 00:03:27.550
designing aircraft. We are quite used to thinking
of forces as vectors, but do forces exhibit
00:03:27.550 --> 00:03:33.380
the properties necessary to be aptly represented
by vectors? Let's review the properties of
00:03:33.380 --> 00:03:38.569
the vector.
00:03:38.569 --> 00:03:44.900
To add vector b to vector a, we connect the
tail of b to the tip of a and the sum is the
00:03:44.900 --> 00:03:49.819
vector that connects the tail of a to the
tip of b. An important property of vector
00:03:49.819 --> 00:03:56.400
addition is that it is commutative. That is
a + b = b + a. You can see this visually from
00:03:56.400 --> 00:04:03.300
the parallelogram whose diagonal represents
both sums simultaneously. Another important
00:04:03.300 --> 00:04:08.030
property is that vectors can be multiplied
by real numbers, which are called scalars,
00:04:08.030 --> 00:04:12.150
because they have the effect of scaling the
length of the vector. Multiplying by positive
00:04:12.150 --> 00:04:17.699
scalars increases the length for large scalars,
and shrinks the vector for scalars less than
00:04:17.699 --> 00:04:22.970
one. Multiplying a vector by -1 has the effect
of making the vector point in the opposite
00:04:22.970 --> 00:04:29.220
direction. Another important property of vectors
is that the initial point doesn't matter.
00:04:29.220 --> 00:04:33.380
Any vector pointing in the same direction
with the same magnitude represents the same
00:04:33.380 --> 00:04:38.770
vector. To make this seem less abstract, we
can think of vector properties in terms of
00:04:38.770 --> 00:04:45.750
displacement.
00:04:45.750 --> 00:04:52.110
Suppose you walk from a point P to a point
Q. The displacement, or change is position
00:04:52.110 --> 00:04:57.639
from P to Q, is aptly represented by an arrow
that starts at the point P and ends at the
00:04:57.639 --> 00:05:03.590
point Q. Let's see how displacement motivates
the correct form of vector addition. Consider
00:05:03.590 --> 00:05:07.560
the following example: you start at home,
which is represented by a star on the map.
00:05:07.560 --> 00:05:14.050
You walk 300 meters east to get a cup of tea
before you walk southeast 500 meters to school.
00:05:14.050 --> 00:05:19.300
After class you walk 400 meters southwest
of your school to play tennis. Your friend,
00:05:19.300 --> 00:05:23.990
who lives in your apartment complex, is going
to meet you there. What vector would represent
00:05:23.990 --> 00:05:28.690
the displacement vector for your friend who
leaves home directly and meets you to play
00:05:28.690 --> 00:05:35.690
tennis? Pause the video here and discuss your
answer with someone. Answer: The vector that
00:05:41.430 --> 00:05:46.199
starts at your home and moves down to the
tennis court. This is interesting because
00:05:46.199 --> 00:05:50.620
the arrow that connects your starting location
to your ending location represents the total
00:05:50.620 --> 00:05:55.389
displacement from your starting point. In
other words, this vector is the sum of the
00:05:55.389 --> 00:06:02.080
other 3 displacement vectors. Displacement
also helps you understand vector decomposition.
00:06:02.080 --> 00:06:06.910
Suppose you have walked a few blocks away,
represented by the following displacement.
00:06:06.910 --> 00:06:11.210
To get there, you probably didn't walk through
other people's houses and yards. Your path
00:06:11.210 --> 00:06:17.389
more likely looked something like this. This
process of breaking a vector down into component
00:06:17.389 --> 00:06:22.780
parts pointing along particular directions
is completely analogous to decomposing a vector
00:06:22.780 --> 00:06:28.639
into components that point along perpendicular
basis vectors. When in doubt about the mathematics
00:06:28.639 --> 00:06:33.240
of the vector, take a moment to rephrase your
problem in terms of displacements, and see
00:06:33.240 --> 00:06:39.110
if your intuition can guide the mathematics.
00:06:39.110 --> 00:06:46.110
Now, let's go back to forces -- do they have
the vector properties that we expect them
00:06:46.540 --> 00:06:51.830
to? When representing physical quantities
with vectors, the quantity must have both
00:06:51.830 --> 00:06:57.340
magnitude and direction. But it must also
scale and add commutatively. Let's see if
00:06:57.340 --> 00:06:59.320
force has these properties. Force seems to
have magnitude and direction. Force also scales
00:06:59.320 --> 00:07:03.770
appropriately. We think of forces as being
small or large, we can increase them and decrease
00:07:03.770 --> 00:07:10.770
them. When we draw a free body diagram, we
are implicitly assuming that forces are vectors,
00:07:14.150 --> 00:07:20.210
and that they add like vectors. But how do
we know this? We do an experiment. In this
00:07:20.210 --> 00:07:27.160
next segment we'll see a demonstration of
how forces, do indeed, add like vectors. [Pause]
00:07:27.160 --> 00:07:31.979
Here you see 3 Newton Scales connected by
strings. We'll call the two strings on top
00:07:31.979 --> 00:07:34.100
String A and String B.
00:07:34.100 --> 00:07:38.120
String A is 135 degrees off of horizontal.
00:07:38.120 --> 00:07:44.770
String B is 45 degrees off of horizontal.
The scale reads out the magnitude of the tension
00:07:44.770 --> 00:07:46.250
force on each string.
00:07:46.250 --> 00:07:51.009
We first want to get a reading of the scales
while there is no mass added to the system.
00:07:51.009 --> 00:07:56.610
The scales do not have very precise measurement;
we can only guarantee the measurement to within
00:07:56.610 --> 00:07:58.120
.5 Newtons.
00:07:58.120 --> 00:08:05.120
When looking at the bottom scale, we see that
the reading is approximately -.3 Newtons.
00:08:05.180 --> 00:08:12.180
The tension on string A is approximately .5
Newtons, and the tension on string B is .3
00:08:13.259 --> 00:08:18.069
Newtons. These tension forces are due to the
weight of the bottom scale and the strings.
00:08:18.069 --> 00:08:21.940
So we will need to subtract these amounts
off of any reading when mass is added into
00:08:21.940 --> 00:08:26.440
the system to get the tension force of the
mass alone.
00:08:26.440 --> 00:08:32.639
Let's add a 1kg mass to the hook below the
bottom scale. The bottom scale now reads about
00:08:32.639 --> 00:08:39.559
9.6 Newtons. Now we look at the top two scales.
We see that the tension force on string A
00:08:39.559 --> 00:08:46.560
is 7.4 Newtons, and the tension on string
B is 7.5 Newtons. We want to decompose these
00:08:49.700 --> 00:08:54.640
forces to see if they do in fact add like
vectors. Note that Newton's second law says
00:08:54.640 --> 00:09:00.260
that the sum of these forces must be zero,
since our system of strings and scales is
00:09:00.260 --> 00:09:01.279
stationary.
00:09:01.279 --> 00:09:05.540
Let's start by subtracting off the readings
we got from our Newton scale system with no
00:09:05.540 --> 00:09:11.779
added mass to find the net tension force due
to the mass. The tension in the string A is
00:09:11.779 --> 00:09:18.779
7.4-.5 = 6.9 N. And the tension on string
B is 7.5-.3=7.2 N. We find that the net force
00:09:25.860 --> 00:09:32.860
down is 9.6-(-.3) = 9.9 Newtons. Using F=mg,
we would predict that the force due to a 1kg
00:09:35.339 --> 00:09:40.760
mass would be 9.8Newtons. So the fact that
we are measuring 9.9Newtons indicates that
00:09:40.760 --> 00:09:46.990
we have some experimental error in our measurements.
How would you use this setup to show whether
00:09:46.990 --> 00:09:53.990
or not forces add like vectors?
00:09:59.910 --> 00:10:04.580
We want to see that the forces sum to zero.
To do this, let's decompose the forces into
00:10:04.580 --> 00:10:11.080
horizontal and vertical components. We use
the fact that the string A is at a 135 degree
00:10:11.080 --> 00:10:16.540
angle. Because the magnitude of sin(135 )and
cos(135) are both one over square root of
00:10:16.540 --> 00:10:21.390
2, we simply need to divide by the square
root of two. We find that the horizontal and
00:10:21.390 --> 00:10:27.959
vertical components of this tension force
are approximately 4.9 Newtons. Because sin(45)
00:10:27.959 --> 00:10:34.269
and cos(45) are both one over the square root
of two, we divide by the square root of two
00:10:34.269 --> 00:10:41.269
and find that each component is approximately
5.1 N. Thus the horizontal forces subtract
00:10:43.019 --> 00:10:47.640
to give a net force of .2N in the positive
x direction.
00:10:47.640 --> 00:10:52.240
The three vertical components add to 10-9.9
= .1 in the positive y direction. Because
00:10:52.240 --> 00:10:58.130
.1 and .2 are small, and because we know that
there are errors associated with the limits
00:10:58.130 --> 00:11:03.810
of accuracy of out measurements, we can be
confident that these forces do, in fact, sum
00:11:03.810 --> 00:11:04.790
to 0.
00:11:04.790 --> 00:11:10.260
So this demo does in fact suggest that forces
add like vectors. But we want to make sure
00:11:10.260 --> 00:11:15.110
that this wasn't an artifact of having so
much symmetry in the system. To do this, we
00:11:15.110 --> 00:11:20.459
move string B 60 degrees off of horizontal.
00:11:20.459 --> 00:11:26.690
As you can see, the tension force on the bottom
string did not change. It still reads 9.6N.
00:11:26.690 --> 00:11:32.790
But the force on each upper Newton scale has
changed. The tension force on String A is
00:11:32.790 --> 00:11:38.589
5.5N, and the tension force on the string
B is 7.5N
00:11:38.589 --> 00:11:45.120
We leave it as an exercise to you to decompose
the tension forces into horizontal and vertical
00:11:45.120 --> 00:11:52.120
components and verify that within the expected
measurement error the forces sum to zero.
00:11:59.589 --> 00:12:02.380
Forces really do add like vectors!!
00:12:02.380 --> 00:12:08.040
Okay, so forces can indeed be represented
with vectors. Let's look back at the list
00:12:08.040 --> 00:12:13.000
you generated of physical quantities with
both magnitude and direction. If force was
00:12:13.000 --> 00:12:20.000
on your list, we now know that force is indeed
a vector quantity. Maybe you also listed rotation.
00:12:20.300 --> 00:12:25.279
Let's see if rotations have vector properties.
Rotation seems like a physical quantity that
00:12:25.279 --> 00:12:31.089
has magnitude and direction. The direction
could be determined by the axis of rotation.
00:12:31.089 --> 00:12:36.670
We choose which way the arrow points based
on the right hand rule. The magnitude determines
00:12:36.670 --> 00:12:43.640
how many radians through which the object
rotates. Consider the following two rotations:
00:12:43.640 --> 00:12:50.640
Rz rotates an object by 90 degrees about the
z-axis: which rotates an object as such. Ry
00:12:51.940 --> 00:12:58.940
rotates an object by 90 degrees or π/2 radians
about the y-axis, which rotates the same object
00:12:59.529 --> 00:13:06.529
in this manner. Do rotations scale like vectors?
Let's see what happens if we take the vector
00:13:06.750 --> 00:13:13.680
that represents a rotation of π/2 radians
about the z-axis and add it to itself -- it
00:13:13.680 --> 00:13:20.680
seems that it should be a rotation of π radians,
or 180 degrees. And this agrees with what
00:13:21.220 --> 00:13:28.220
we get by rotating 90 degrees about the z-axis
twice. So scaling rotations makes sense. Question:
00:13:29.750 --> 00:13:36.750
Do rotations add like vectors? If we rotate
the object a quarter turn about the z-axis,
00:13:41.420 --> 00:13:47.230
followed by a quarter turn about the y-axis,
the object ends up in the following position.
00:13:47.230 --> 00:13:52.860
If instead we rotate a quarter of a turn about
the y-axis, followed by a quarter turn about
00:13:52.860 --> 00:13:59.860
the z-axis, the object ends up in this position.
Are the ending positions the same for the
00:13:59.950 --> 00:14:06.519
two different permutations of rotations? [Pause]
No, they are not. This means that rotations
00:14:06.519 --> 00:14:12.260
don't add commutatively, but vector addition
must be commutative. So this tells us that
00:14:12.260 --> 00:14:17.709
we CANNOT use vectors to represent rotations.
You'll learn that it is better to use matrices
00:14:17.709 --> 00:14:23.950
and matrix multiplication to represent combinations
of rotations. The tricky thing is that a vector
00:14:23.950 --> 00:14:29.670
can be used to represent the rotation rate,
the time derivative of rotation, quite well.
00:14:29.670 --> 00:14:35.149
During this video, you came up with several
physical quantities that you theorized behave
00:14:35.149 --> 00:14:42.149
like vectors. Consider the following list
of quantities. Compare our list to your own
00:14:42.170 --> 00:14:48.300
list, and determine whether each one is best
represented by a vector, a scalar, or neither.
00:14:48.300 --> 00:14:55.300
You may need to design an experiment or thought
experiment in order to verify your hypothesis.
00:14:55.990 --> 00:14:59.639
To review, you have learned that:
00:14:59.639 --> 00:15:03.220
Displacements help guide our intuition for
vector algebra.
00:15:03.220 --> 00:15:07.130
Physical quantities can be represented with
vectors only when they have magnitude, have
00:15:07.130 --> 00:15:13.050
direction, scale, and add commutatively.
Forces can be represented by vectors, while
00:15:13.050 --> 00:15:19.459
rotations cannot.