1 00:00:03,410 --> 00:00:05,750 We'd like to consider torques on a body. 2 00:00:05,750 --> 00:00:10,930 Let's draw an arbitrary body, and let's consider a point s, 3 00:00:10,930 --> 00:00:13,050 that we're about to calculate the torque. 4 00:00:13,050 --> 00:00:14,750 Now, we know that forces on the body 5 00:00:14,750 --> 00:00:17,180 can be both internal and external. 6 00:00:17,180 --> 00:00:19,700 And we'd like to show that all internal torques will 7 00:00:19,700 --> 00:00:21,140 cancel in pairs. 8 00:00:21,140 --> 00:00:25,920 The way we'll do that is, suppose we pick an object. 9 00:00:25,920 --> 00:00:29,930 We'll label it with mass mi, and another object, 10 00:00:29,930 --> 00:00:31,580 we'll label that with mass mj. 11 00:00:31,580 --> 00:00:34,700 These are small mass elements in the body. 12 00:00:34,700 --> 00:00:39,410 And we'd like to know something about the internal forces. 13 00:00:39,410 --> 00:00:42,860 Now, let's make the assumption-- and this is the key property-- 14 00:00:42,860 --> 00:00:45,800 that the force due to this interaction between j 15 00:00:45,800 --> 00:00:49,355 and the i-th particle pointing that way-- and here's 16 00:00:49,355 --> 00:00:54,660 the Newton's third law pair-- that these forces lie-- 17 00:00:54,660 --> 00:00:58,190 are directed along the line connecting the two bodies. 18 00:00:58,190 --> 00:00:59,990 With this assumption, we'll now show 19 00:00:59,990 --> 00:01:03,890 that the torque due to these two internal forces, [INAUDIBLE] 20 00:01:03,890 --> 00:01:06,270 Newton's third law pairs will cancel. 21 00:01:06,270 --> 00:01:08,330 So let's calculate that out. 22 00:01:08,330 --> 00:01:18,080 So we draw our vector from rsi, and our other vector rsj. 23 00:01:18,080 --> 00:01:20,810 And now we're in position to add these two torques. 24 00:01:20,810 --> 00:01:25,400 So we have the torque on s due to this pair is 25 00:01:25,400 --> 00:01:33,650 equal to the sum of rsi cross fji plus 26 00:01:33,650 --> 00:01:40,700 rs-- that's an r-- make sure we get that right. 27 00:01:40,700 --> 00:01:47,479 We have rsj cross fij. 28 00:01:47,479 --> 00:01:55,460 Now, the third law pair says fji is equal to minus fiji. 29 00:01:55,460 --> 00:01:59,200 And so, if we substitute-- let's put the minus sign over here-- 30 00:01:59,200 --> 00:02:06,973 we get rsi minus rsj cross fji. 31 00:02:12,630 --> 00:02:16,060 Now, let's look at this vector in particular. 32 00:02:16,060 --> 00:02:18,890 We can draw it again over here, just to see it. 33 00:02:18,890 --> 00:02:20,760 Here s. 34 00:02:20,760 --> 00:02:24,474 Here is rsi. 35 00:02:24,474 --> 00:02:30,742 Here is the vector rs, that's rsj. 36 00:02:34,210 --> 00:02:42,250 And we want to now consider the vector rsi minus rsj. 37 00:02:42,250 --> 00:02:45,250 Notice that this vector is directed 38 00:02:45,250 --> 00:02:50,290 along the lines connecting the i-th and the j-th particle. 39 00:02:50,290 --> 00:02:54,730 And we've made an assumption that fji is also 40 00:02:54,730 --> 00:02:56,000 along that line. 41 00:02:56,000 --> 00:02:58,990 So these two vectors, in this particular case, 42 00:02:58,990 --> 00:03:04,960 are either parallel or anti-parallel. 43 00:03:04,960 --> 00:03:10,240 And hence, the torque due to the sum of these internal 44 00:03:10,240 --> 00:03:13,210 forces cancel in pairs. 45 00:03:13,210 --> 00:03:15,490 And this means we only need to address 46 00:03:15,490 --> 00:03:17,950 the torque due to external forces that 47 00:03:17,950 --> 00:03:20,866 are acting on individual elements in a body.