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.