1 00:00:03,630 --> 00:00:06,510 When we analyze the problem about putting 2 00:00:06,510 --> 00:00:09,320 two blocks along a frictionless surface 3 00:00:09,320 --> 00:00:13,050 and pushing one and asking, what is the maximum force such 4 00:00:13,050 --> 00:00:15,290 that block 2 does not slip? 5 00:00:15,290 --> 00:00:16,910 We have three different systems. 6 00:00:16,910 --> 00:00:18,700 And now I want to focus a little bit 7 00:00:18,700 --> 00:00:23,010 on what would happen if we just naively chose our systems 8 00:00:23,010 --> 00:00:25,230 as both blocks together. 9 00:00:25,230 --> 00:00:29,050 So let's try to look at the types of issues 10 00:00:29,050 --> 00:00:31,190 that come up when we do that. 11 00:00:31,190 --> 00:00:34,890 So once again, let's draw free body diagrams. 12 00:00:34,890 --> 00:00:36,270 Now, we begin. 13 00:00:36,270 --> 00:00:40,290 We're pushing block 1 with a force. 14 00:00:40,290 --> 00:00:45,160 Gravitation is the sum of these two forces, 15 00:00:45,160 --> 00:00:52,560 because our system here is block 1 and block 2. 16 00:00:52,560 --> 00:00:54,620 What about normal forces? 17 00:00:54,620 --> 00:00:59,580 Well, the ground is acting on block 1. 18 00:00:59,580 --> 00:01:02,230 And now, here's the significant thing. 19 00:01:02,230 --> 00:01:06,980 What about all of those forces between blocks 1 and 2? 20 00:01:06,980 --> 00:01:17,803 Well, the forces between blocks 1 and 2 are internal forces. 21 00:01:22,490 --> 00:01:28,900 And we saw that they were friction forces, f12 and f21. 22 00:01:28,900 --> 00:01:32,200 This was the friction force between the two blocks, 23 00:01:32,200 --> 00:01:35,800 on block 2 due to 1 and the friction force on block 1 24 00:01:35,800 --> 00:01:36,960 due to 2. 25 00:01:36,960 --> 00:01:40,800 There were normal forces between these two blocks, 26 00:01:40,800 --> 00:01:43,235 but these are interaction pairs. 27 00:01:47,900 --> 00:01:51,930 And the sum of them are 0. 28 00:01:51,930 --> 00:01:58,210 And so we see that all internal forces form Newton's third law 29 00:01:58,210 --> 00:02:01,450 interaction pairs and the vector sum of them are 0. 30 00:02:01,450 --> 00:02:05,260 And that's why I don't need those internal forces 31 00:02:05,260 --> 00:02:06,990 on my free body diagram. 32 00:02:06,990 --> 00:02:09,840 If I were to draw them, I would have different arrows. 33 00:02:09,840 --> 00:02:14,210 For instance, I would have that arrow f21 34 00:02:14,210 --> 00:02:16,920 and I would have the arrow f12. 35 00:02:16,920 --> 00:02:20,570 And you can see that the sum of those cancel. 36 00:02:20,570 --> 00:02:28,680 I would have the arrow n21 and I would have the arrow n12. 37 00:02:28,680 --> 00:02:31,500 Arrows in opposite directions. 38 00:02:31,500 --> 00:02:36,360 The interaction pairs sum to 0, because they 39 00:02:36,360 --> 00:02:38,530 are internal forces. 40 00:02:38,530 --> 00:02:43,340 And again, this enables us now to just draw 41 00:02:43,340 --> 00:02:50,390 f equals m1 plus m2 times the acceleration of the system. 42 00:02:50,390 --> 00:02:54,310 And so we have our i hat and our j hat directions. 43 00:02:54,310 --> 00:02:57,352 Let's pick i hat and j hat. 44 00:02:57,352 --> 00:03:03,180 And now, we didn't include the kinetic friction force 45 00:03:03,180 --> 00:03:05,550 of the ground in that system. 46 00:03:05,550 --> 00:03:07,980 So let's make sure that that's there. 47 00:03:07,980 --> 00:03:12,330 And what we have is f minus f kinetic 48 00:03:12,330 --> 00:03:17,565 equals m1 plus m2 times a. 49 00:03:17,565 --> 00:03:18,940 And in the vertical direction, we 50 00:03:18,940 --> 00:03:25,200 have n ground 1 minus m1 plus m2 g equals 0. 51 00:03:25,200 --> 00:03:27,870 That gave us our same result before. 52 00:03:27,870 --> 00:03:34,650 Notice that f equals fk plus m1 plus m2 g. 53 00:03:34,650 --> 00:03:47,690 We know this is mu km1 plus m2 g plus m1 plus m2 g a. 54 00:03:47,690 --> 00:03:54,490 So we have the acceleration of the system 55 00:03:54,490 --> 00:04:03,890 depends on the force, mu k m1 plus m2g divided by m1 m2. 56 00:04:03,890 --> 00:04:08,300 But notice because our static friction 57 00:04:08,300 --> 00:04:10,480 is an internal force in this system, 58 00:04:10,480 --> 00:04:14,030 it never shows up in Newton's second law, 59 00:04:14,030 --> 00:04:17,790 so we were never able to apply the condition 60 00:04:17,790 --> 00:04:23,790 that f static max was mu static n, 61 00:04:23,790 --> 00:04:26,310 what we call the normal force between the blocks. 62 00:04:26,310 --> 00:04:30,900 And so we were unable to figure out what is the maximum force. 63 00:04:30,900 --> 00:04:35,690 All we can say is if I push f, that's the acceleration. 64 00:04:35,690 --> 00:04:40,020 But I cannot determine what maximum force will cause block 65 00:04:40,020 --> 00:04:43,570 2 to slip with respect to block 1. 66 00:04:43,570 --> 00:04:46,100 So when you pick your system like this, 67 00:04:46,100 --> 00:04:48,730 it's very quick to calculate a. 68 00:04:48,730 --> 00:04:49,990 No problem about that. 69 00:04:49,990 --> 00:04:53,930 But I am not able to answer any questions that 70 00:04:53,930 --> 00:04:58,760 require some type of information about the internal forces. 71 00:04:58,760 --> 00:05:03,160 So the art to choosing systems and free body diagrams 72 00:05:03,160 --> 00:05:06,980 is to think about the types of questions you're asking. 73 00:05:06,980 --> 00:05:09,120 If you have a question that involves something 74 00:05:09,120 --> 00:05:11,950 about a maximum condition on static friction, 75 00:05:11,950 --> 00:05:15,160 then you want to make sure that static friction is 76 00:05:15,160 --> 00:05:18,270 an external force to your system. 77 00:05:18,270 --> 00:05:20,800 If it's an internal force, like in this case, 78 00:05:20,800 --> 00:05:24,065 you will not be able to apply that condition.