1 00:00:03,570 --> 00:00:05,560 Let's look at a typical application 2 00:00:05,560 --> 00:00:08,570 of Newton's second law for a system of objects. 3 00:00:08,570 --> 00:00:13,000 So what I want to consider is a system of pulleys and masses. 4 00:00:13,000 --> 00:00:17,670 So I'll have a fixed surface here, a ceiling. 5 00:00:17,670 --> 00:00:20,770 And from the ceiling, we'll hang a pulley, which 6 00:00:20,770 --> 00:00:23,970 I'm going to call pulley A. And this pulley will 7 00:00:23,970 --> 00:00:28,350 have a rope attached to it, wrapped around it. 8 00:00:28,350 --> 00:00:31,090 And here we have object 1. 9 00:00:31,090 --> 00:00:34,220 And the rope goes around the pulley. 10 00:00:34,220 --> 00:00:40,880 And now it's going to go around another pulley B 11 00:00:40,880 --> 00:00:44,550 and fixed to the ceiling. 12 00:00:44,550 --> 00:00:46,880 So that's fixed. 13 00:00:46,880 --> 00:00:52,200 And hanging from pulley B is another mass 2. 14 00:00:52,200 --> 00:00:55,060 And our goal in applying Newton's second law 15 00:00:55,060 --> 00:00:59,180 is to find the accelerations of objects 1 and 2. 16 00:00:59,180 --> 00:01:01,390 Now, how do we approach this? 17 00:01:01,390 --> 00:01:03,260 Well, the first thing we have to do 18 00:01:03,260 --> 00:01:07,000 is decide if we're going to apply Newton's second law, what 19 00:01:07,000 --> 00:01:09,490 is the system that we'll apply it to? 20 00:01:09,490 --> 00:01:12,090 And there's many different ways to choose a system. 21 00:01:12,090 --> 00:01:14,240 When we look at this problem, we'll 22 00:01:14,240 --> 00:01:16,850 have several different systems. 23 00:01:16,850 --> 00:01:19,710 So let's consider the ones that we're going to look at. 24 00:01:19,710 --> 00:01:21,360 And the first one is very simple. 25 00:01:21,360 --> 00:01:24,640 It will be block 1. 26 00:01:24,640 --> 00:01:26,440 And the second system that we look at, 27 00:01:26,440 --> 00:01:31,470 we'll call that AB is pulley A. 28 00:01:31,470 --> 00:01:33,860 Now that brings us to an interesting question 29 00:01:33,860 --> 00:01:38,810 about pulley B and block 2, because we could separately 30 00:01:38,810 --> 00:01:43,210 look at pulley B, and we could separately consider block 2, 31 00:01:43,210 --> 00:01:44,979 or we consider them together. 32 00:01:44,979 --> 00:01:52,840 And I want to first consider separately pulley B 33 00:01:52,840 --> 00:01:55,680 and block 2. 34 00:01:55,680 --> 00:02:00,480 Now in some ways when you're looking at a compound system, 35 00:02:00,480 --> 00:02:02,600 and it has four objects, it makes sense 36 00:02:02,600 --> 00:02:05,720 to apply Newton's second law to each object separately. 37 00:02:05,720 --> 00:02:08,538 But we'll pay careful attention to the fact 38 00:02:08,538 --> 00:02:11,760 that object 2 is connected to pulley B. 39 00:02:11,760 --> 00:02:14,030 And eventually, we'll see that we 40 00:02:14,030 --> 00:02:16,560 can combine these two things. 41 00:02:16,560 --> 00:02:19,390 So the next step is once we have identified 42 00:02:19,390 --> 00:02:22,430 our object is to draw a free body force diagrams for each 43 00:02:22,430 --> 00:02:23,530 of the objects. 44 00:02:23,530 --> 00:02:27,760 So in order to do that, let's start with object 1. 45 00:02:27,760 --> 00:02:32,620 And we want to consider the forces on object 1. 46 00:02:32,620 --> 00:02:34,710 Now that brings us to our first issue 47 00:02:34,710 --> 00:02:38,490 about what types of assumptions we're making in our system. 48 00:02:38,490 --> 00:02:40,380 For instance, we have a rope that's 49 00:02:40,380 --> 00:02:41,730 wrapped around this pulley. 50 00:02:41,730 --> 00:02:45,610 And we have two pulleys that in principle could be rotating. 51 00:02:45,610 --> 00:02:48,920 But what we'd like to do to simplify our analysis-- so 52 00:02:48,920 --> 00:02:53,070 let's keep track of some assumptions here. 53 00:02:53,070 --> 00:03:01,615 Our first assumption will be that the mass, MP, of pulley A 54 00:03:01,615 --> 00:03:05,740 and the mass of pulley B are approximately zero. 55 00:03:05,740 --> 00:03:08,300 Now the reason for that is that we're not 56 00:03:08,300 --> 00:03:09,750 going to consider any of the fact 57 00:03:09,750 --> 00:03:13,410 that these objects have to be put into rotational motion. 58 00:03:13,410 --> 00:03:16,270 Later on in the course, we'll see that this will give us 59 00:03:16,270 --> 00:03:19,150 a more complicated analysis. 60 00:03:19,150 --> 00:03:22,420 We're also going to assume that our rope is not slipping. 61 00:03:22,420 --> 00:03:28,650 So the rope is actually is just slipping on the pulleys. 62 00:03:28,650 --> 00:03:31,050 So what that means is it's just the rope is 63 00:03:31,050 --> 00:03:34,320 sliding as the objects move. 64 00:03:34,320 --> 00:03:37,910 Now again, what this is going to imply 65 00:03:37,910 --> 00:03:41,430 is that the tension in the rope-- this rope 66 00:03:41,430 --> 00:03:43,329 is also slipping. 67 00:03:43,329 --> 00:03:48,000 And the rope is massless as well. 68 00:03:48,000 --> 00:03:49,380 It's very light rope. 69 00:03:49,380 --> 00:03:50,970 And all of these assumptions we've 70 00:03:50,970 --> 00:03:54,140 seen when we analyze ropes tell us 71 00:03:54,140 --> 00:03:59,850 that the tension T is uniform in the rope. 72 00:03:59,850 --> 00:04:02,000 So that's our first assumption. 73 00:04:02,000 --> 00:04:04,280 And we need to think about this before we even 74 00:04:04,280 --> 00:04:06,490 begin to think about the forces on the object. 75 00:04:06,490 --> 00:04:08,480 And now we can draw our forces. 76 00:04:08,480 --> 00:04:09,690 What do we have? 77 00:04:09,690 --> 00:04:12,370 We have the gravitational force on object 1. 78 00:04:12,370 --> 00:04:16,519 And now we can identify the tension pulling in the string, 79 00:04:16,519 --> 00:04:18,579 pulling object 1 up. 80 00:04:18,579 --> 00:04:22,430 Now for every time we introduce a free body diagram, 81 00:04:22,430 --> 00:04:25,040 recall that we have to choose what 82 00:04:25,040 --> 00:04:27,550 we mean by positive directions. 83 00:04:27,550 --> 00:04:31,800 And in this case, I'm going to pick a unit vector down, j hat 84 00:04:31,800 --> 00:04:33,310 1 down. 85 00:04:33,310 --> 00:04:36,612 So that's my positive direction for force. 86 00:04:36,612 --> 00:04:40,180 Now before I write down all of Newton's laws, 87 00:04:40,180 --> 00:04:44,540 I'll just write down our various force diagrams. 88 00:04:44,540 --> 00:04:50,300 So for pulley A, I have two strings 89 00:04:50,300 --> 00:04:52,370 that are pulling it downwards. 90 00:04:52,370 --> 00:04:56,380 So I have tension and tension. 91 00:04:56,380 --> 00:05:00,110 And this string, I'm going to call that T2, 92 00:05:00,110 --> 00:05:04,140 is holding that pulley up. 93 00:05:04,140 --> 00:05:07,160 So we have the force diagram. 94 00:05:07,160 --> 00:05:10,420 Now I could write MAG, but we've assumed 95 00:05:10,420 --> 00:05:12,920 that the pulley is massless. 96 00:05:12,920 --> 00:05:17,250 And again, I'll call j hat A down. 97 00:05:17,250 --> 00:05:22,480 For object 2, let's do pulley B first. 98 00:05:22,480 --> 00:05:24,596 Now what are the forces on pulley B? 99 00:05:24,596 --> 00:05:29,980 I have strings on both sides, T. Pulley B is massless, 100 00:05:29,980 --> 00:05:33,916 so I'm not putting gravitational force. 101 00:05:33,916 --> 00:05:39,100 And this string is pulling B downwards, so that's T3. 102 00:05:39,100 --> 00:05:44,110 And again, we'll write j hat B downwards. 103 00:05:44,110 --> 00:05:47,180 And finally, I have block 2. 104 00:05:47,180 --> 00:05:50,850 So I'll draw that over here. 105 00:05:50,850 --> 00:05:53,230 I'll write block 2. 106 00:05:53,230 --> 00:05:57,170 In fact, let's say a little space here. 107 00:05:57,170 --> 00:05:59,750 We'll have j hat B downwards. 108 00:05:59,750 --> 00:06:02,430 Now block 2, what do we have there? 109 00:06:02,430 --> 00:06:04,770 We have the string pulling up block 2, 110 00:06:04,770 --> 00:06:08,760 which we've identified as T3. 111 00:06:08,760 --> 00:06:11,670 And we have the gravitational force on block 2 112 00:06:11,670 --> 00:06:14,450 downward, M2 g. 113 00:06:14,450 --> 00:06:17,220 And there we have j had 2. 114 00:06:17,220 --> 00:06:21,400 So I've now drawn the free body diagram of the various objects. 115 00:06:21,400 --> 00:06:24,460 And that enables me to apply Newton's second law 116 00:06:24,460 --> 00:06:26,410 for each of these objects. 117 00:06:26,410 --> 00:06:27,520 So let's begin. 118 00:06:27,520 --> 00:06:29,720 We'll start with object 1. 119 00:06:29,720 --> 00:06:33,920 We have-- remember in all cases, we're going to apply F 120 00:06:33,920 --> 00:06:35,880 equals m a. 121 00:06:35,880 --> 00:06:41,860 So for object 1, we have m1g positive downward minus 122 00:06:41,860 --> 00:06:46,140 T is equal to m1 a1. 123 00:06:46,140 --> 00:06:51,110 And that's our F equals m a on object 1. 124 00:06:51,110 --> 00:06:54,330 So sometimes we'll distinguish that the 125 00:06:54,330 --> 00:06:57,140 forces we're getting from our free body diagram. 126 00:06:57,140 --> 00:07:01,080 And A is a mathematical description of the motion. 127 00:07:01,080 --> 00:07:09,740 For block 2, we have m2g minus T3 is equal to m2 a2. 128 00:07:09,740 --> 00:07:18,590 And now for pulley A, we have 2T pointing downwards 129 00:07:18,590 --> 00:07:21,770 minus T2 going upwards. 130 00:07:21,770 --> 00:07:25,350 And because pulley A is massless, 131 00:07:25,350 --> 00:07:29,670 this is zero even though pulley A may 132 00:07:29,670 --> 00:07:32,880 be-- it's actually fixed too. 133 00:07:32,880 --> 00:07:34,310 So it's not even accelerating. 134 00:07:34,310 --> 00:07:36,930 And what we see here is this equation-- 135 00:07:36,930 --> 00:07:39,510 I'm going to quickly note that it tells us 136 00:07:39,510 --> 00:07:47,270 that the string holding pulley 2 up, T2, is equal to 2T. 137 00:07:47,270 --> 00:07:49,890 So we can think of if, we want to know what 138 00:07:49,890 --> 00:07:52,330 T2 is, we need to calculate T. 139 00:07:52,330 --> 00:07:58,495 And finally we have B. And what is the forces on B? 140 00:07:58,495 --> 00:08:03,150 We have T3 minus 2T. 141 00:08:03,150 --> 00:08:06,260 And again pulley B is 0. 142 00:08:06,260 --> 00:08:12,400 And so we see that T3 is equal to 2T. 143 00:08:12,400 --> 00:08:17,880 Now if you think about what I said before about combining 144 00:08:17,880 --> 00:08:22,610 systems, if we combine pulley B in block 2, 145 00:08:22,610 --> 00:08:25,420 visually what we're doing is we're just adding these 146 00:08:25,420 --> 00:08:27,640 to free body diagram together. 147 00:08:27,640 --> 00:08:33,400 When we have a system B and block 2. 148 00:08:33,400 --> 00:08:36,640 Let's call this j hat downwards. 149 00:08:36,640 --> 00:08:39,570 And when we add these free body diagram together, 150 00:08:39,570 --> 00:08:43,940 you see that the T3 is now internal force to the system. 151 00:08:43,940 --> 00:08:47,080 It cancels in pair by Newton's second law. 152 00:08:47,080 --> 00:08:50,760 And all we have is the two strings going up, 153 00:08:50,760 --> 00:08:55,740 so we have T and T. And we have the gravitational force 154 00:08:55,740 --> 00:08:57,180 downward. 155 00:08:57,180 --> 00:09:02,950 And separately, when we saw that T3 equals 2T 156 00:09:02,950 --> 00:09:10,420 and we apply it there, then if we consider a system B2, 157 00:09:10,420 --> 00:09:16,680 and look at our free body diagram, we have m2g minus 2T-- 158 00:09:16,680 --> 00:09:23,330 and notice we have the same result their 2T equals m2 a2. 159 00:09:23,330 --> 00:09:26,820 So in principle now-- and I'll outline our equations. 160 00:09:26,820 --> 00:09:30,080 We have equation 1. 161 00:09:30,080 --> 00:09:33,030 We have equation 2. 162 00:09:33,030 --> 00:09:38,410 And in these two equations, we have three unknowns, T, A1, 163 00:09:38,410 --> 00:09:41,314 and A2, but only two equations. 164 00:09:41,314 --> 00:09:42,730 And so you might think, what about 165 00:09:42,730 --> 00:09:45,300 this missing third equation here? 166 00:09:45,300 --> 00:09:49,260 However, in this equation, we have a fourth unknown, T3. 167 00:09:49,260 --> 00:09:53,760 And this equation is just relating to T and T3. 168 00:09:53,760 --> 00:09:55,800 So in principle, we would have four unknowns 169 00:09:55,800 --> 00:09:56,980 and three equations. 170 00:09:56,980 --> 00:10:00,540 Or if we restrict our attention to these two equations, 171 00:10:00,540 --> 00:10:05,080 we have three unknowns and two equations. 172 00:10:05,080 --> 00:10:11,040 Are unknowns T, A1 and A2. 173 00:10:11,040 --> 00:10:12,910 These are our unknowns. 174 00:10:12,910 --> 00:10:16,030 And now our next step is to try to figure out 175 00:10:16,030 --> 00:10:19,800 what is the missing condition that's relating 176 00:10:19,800 --> 00:10:21,600 the sum of these unknowns. 177 00:10:21,600 --> 00:10:23,730 And that will be a constraint condition 178 00:10:23,730 --> 00:10:26,090 that we'll analyze next.