1 00:00:03,980 --> 00:00:08,530 And now I'm in position to talk about what 2 00:00:08,530 --> 00:00:10,650 is the maximum force. 3 00:00:10,650 --> 00:00:15,810 If I push F harder, as if I push F and the blocks go together, 4 00:00:15,810 --> 00:00:18,480 the accelerations are the same. 5 00:00:18,480 --> 00:00:19,900 I push F harder. 6 00:00:19,900 --> 00:00:21,460 Accelerations are the same. 7 00:00:21,460 --> 00:00:25,650 I push F harder, and I push F so hard, 8 00:00:25,650 --> 00:00:30,580 that the static friction no longer reaches 9 00:00:30,580 --> 00:00:33,580 its maximum value. 10 00:00:33,580 --> 00:00:36,800 And if I push F harder than that, 11 00:00:36,800 --> 00:00:39,210 I will not have-- the static friction 12 00:00:39,210 --> 00:00:42,880 can't get bigger, and the blocks 1 and 2 will start 13 00:00:42,880 --> 00:00:45,290 to slip relative to each other. 14 00:00:45,290 --> 00:00:53,080 So my no slipping condition is that I want two things. 15 00:00:53,080 --> 00:00:55,020 a1 equals a2. 16 00:00:55,020 --> 00:00:56,456 We'll call that a. 17 00:00:56,456 --> 00:01:06,730 And the maximum force condition is that F static 18 00:01:06,730 --> 00:01:13,050 is equal to maximum value. 19 00:01:13,050 --> 00:01:19,380 Now, what is the normal force between that we refer to 20 00:01:19,380 --> 00:01:20,600 in our fiction law? 21 00:01:20,600 --> 00:01:23,480 There's two normal forces-- the ground 22 00:01:23,480 --> 00:01:25,740 and the normal force between the surfaces. 23 00:01:25,740 --> 00:01:28,250 But the static friction that we're talking about 24 00:01:28,250 --> 00:01:30,230 is between the surfaces. 25 00:01:30,230 --> 00:01:35,430 So that's why we use N here and not N ground. 26 00:01:35,430 --> 00:01:40,520 And now I can solve for this F max. 27 00:01:40,520 --> 00:01:45,690 By the way, we also have the condition that fk is mu k. 28 00:01:45,690 --> 00:01:48,280 What normal force are we talking about? 29 00:01:48,280 --> 00:01:53,830 N, which we've called N ground 1. 30 00:01:53,830 --> 00:01:57,020 And now I look at my equations, and my goal 31 00:01:57,020 --> 00:02:03,460 is to solve for F. I know Ng1 from this equation. 32 00:02:03,460 --> 00:02:09,169 It's just equal to M1g plus N. I Know N from that equation. 33 00:02:09,169 --> 00:02:12,600 So Ng1 is just the sum of the masses times g. 34 00:02:12,600 --> 00:02:14,570 So I know this. 35 00:02:14,570 --> 00:02:17,904 I have f, which is m2a. 36 00:02:17,904 --> 00:02:24,550 And I can now apply my result. 37 00:02:24,550 --> 00:02:31,740 So what we'll do is we'll solve for the a's, a1 equals a2, 38 00:02:31,740 --> 00:02:34,070 in terms of F max. 39 00:02:34,070 --> 00:02:39,990 So over here we have that a is f over m2. 40 00:02:39,990 --> 00:02:43,360 That's from this equation. 41 00:02:43,360 --> 00:02:47,650 And now I'll substitute a1 is equal to that. 42 00:02:47,650 --> 00:02:49,990 I'll substitute that there. 43 00:02:49,990 --> 00:02:55,579 And I get that F Max is going to be 44 00:02:55,579 --> 00:03:01,133 equal to fk plus f static max. 45 00:03:04,680 --> 00:03:06,900 That's that one. 46 00:03:06,900 --> 00:03:16,460 Plus m1 times a1, which is f static max divided by m2. 47 00:03:16,460 --> 00:03:21,970 And so I get that F max equals-- now 48 00:03:21,970 --> 00:03:25,230 I'm going to substitute in all of these values. 49 00:03:25,230 --> 00:03:27,590 It's going to look a little complicated. 50 00:03:27,590 --> 00:03:32,070 And so I'd like to have a little space here for that, 51 00:03:32,070 --> 00:03:33,520 to get everything in here. 52 00:03:38,980 --> 00:03:49,870 And we'll see that it's equal to mu k m1 plus m2g, fk, mu k, 53 00:03:49,870 --> 00:03:56,770 plus m1 2mg plus f static max times 1 plus and m1 over m2. 54 00:03:56,770 --> 00:04:01,860 But f static max is mu N, and N is m2g. 55 00:04:01,860 --> 00:04:08,980 So I get m2g times 1 plus m1 over m2. 56 00:04:08,980 --> 00:04:14,110 And there is, if I push any harder than that, 57 00:04:14,110 --> 00:04:17,350 block 2 will slip with respect to block 1. 58 00:04:17,350 --> 00:04:22,050 Again, all of our terms have the dimensions of acceleration. 59 00:04:22,050 --> 00:04:26,110 This is dimensionless-- 1, dimensions of acceleration. 60 00:04:26,110 --> 00:04:29,150 And we did miss one little thing. 61 00:04:29,150 --> 00:04:33,820 We missed the coefficient of static friction. 62 00:04:33,820 --> 00:04:38,680 And there we have it, a tricky problem.