1 00:00:03,630 --> 00:00:06,690 We would now like to consider a system of particles. 2 00:00:06,690 --> 00:00:11,940 And let's draw them as 1, 2, 3. 3 00:00:11,940 --> 00:00:17,100 And let's refer to this one as the j particle of mass mj. 4 00:00:17,100 --> 00:00:19,500 And these particles, system of particles 5 00:00:19,500 --> 00:00:21,180 is in a gravitational field. 6 00:00:21,180 --> 00:00:24,420 And I would like to consider the torque about some point 7 00:00:24,420 --> 00:00:26,420 s for this system of particles. 8 00:00:26,420 --> 00:00:28,140 And in particular, I'd like to know 9 00:00:28,140 --> 00:00:31,410 how do I treat the gravitational force that's acting on all 10 00:00:31,410 --> 00:00:33,540 of these individual particles. 11 00:00:33,540 --> 00:00:36,510 Well, the way we'll analyze this is 12 00:00:36,510 --> 00:00:41,460 recall that the definition of torque about some point. 13 00:00:41,460 --> 00:00:46,080 Now we have n particles labeled by index j. 14 00:00:46,080 --> 00:00:48,180 It's a vector from the place we're 15 00:00:48,180 --> 00:00:52,710 calculating the torque to where the j-th particle is located. 16 00:00:52,710 --> 00:00:56,280 So we draw that vector rsj. 17 00:00:56,280 --> 00:00:58,560 And we cross that with the force that's 18 00:00:58,560 --> 00:01:01,710 acting on the j-th particle, which in this case 19 00:01:01,710 --> 00:01:04,080 is the gravitational force mjg. 20 00:01:07,230 --> 00:01:11,690 And this is the torque about s, and it's this complicated sum. 21 00:01:11,690 --> 00:01:13,950 But the sum will simplify in the following way. 22 00:01:13,950 --> 00:01:15,570 Let's just write out a couple of terms 23 00:01:15,570 --> 00:01:17,380 to see what it actually looks like. 24 00:01:17,380 --> 00:01:27,240 So we have rs1 cross m1g plus rs2 cross m2g. 25 00:01:27,240 --> 00:01:30,060 And we just keep on going for the n terms. 26 00:01:30,060 --> 00:01:36,270 Now quantity m1 is the scalar units of kilograms in SI units. 27 00:01:36,270 --> 00:01:39,870 But the cross product is between this vector and that vector. 28 00:01:39,870 --> 00:01:45,570 So I can actually rewrite this as m1 rs1 cross g 29 00:01:45,570 --> 00:01:51,509 plus m2 rs2 cross g, et cetera. 30 00:01:51,509 --> 00:01:54,300 And you see that the g term is the same 31 00:01:54,300 --> 00:01:56,320 for every single object. 32 00:01:56,320 --> 00:02:03,900 So I can write this sum-- j goes from 1 to n as mj rsj 33 00:02:03,900 --> 00:02:08,190 and pull the g out of the sum, because every term has 34 00:02:08,190 --> 00:02:10,650 the same g in the cross product. 35 00:02:10,650 --> 00:02:13,440 Now let's focus on the meaning of this sum, 36 00:02:13,440 --> 00:02:16,470 where we're weighting position by mass. 37 00:02:16,470 --> 00:02:21,090 Recall that if you have the center of mass of an object, 38 00:02:21,090 --> 00:02:22,800 and we wanted to find out where is 39 00:02:22,800 --> 00:02:26,880 the location of the center of mass with respect to point s, 40 00:02:26,880 --> 00:02:32,910 then we can calculate this vector rs cm by our definition 41 00:02:32,910 --> 00:02:38,940 that the location of the center of mass with respect to s 42 00:02:38,940 --> 00:02:46,860 is given by the sum j goes from 1 to n of mj rsj. 43 00:02:46,860 --> 00:02:51,840 And we divide that by the total mass mj. 44 00:02:51,840 --> 00:02:56,460 We'll denote this term by m total for the total mass. 45 00:02:56,460 --> 00:02:59,250 So what we see here is the total mass times 46 00:02:59,250 --> 00:03:04,530 Rscm is equal to exactly the sum that we 47 00:03:04,530 --> 00:03:08,790 have in that expression mj rsj. 48 00:03:08,790 --> 00:03:11,730 That's precisely that term. 49 00:03:11,730 --> 00:03:16,140 And so we can conclude that the torque about the center of mass 50 00:03:16,140 --> 00:03:21,090 is m total Rscm cross g. 51 00:03:21,090 --> 00:03:23,760 Now remember again m total is a scalar. 52 00:03:23,760 --> 00:03:26,680 The cross product is between these two vectors. 53 00:03:26,680 --> 00:03:34,829 And so finally, we write this as Rscm cross m total g. 54 00:03:34,829 --> 00:03:38,220 And this is how we apply the gravitational force 55 00:03:38,220 --> 00:03:39,870 to a system of particles. 56 00:03:39,870 --> 00:03:41,890 Now what does that mean? 57 00:03:41,890 --> 00:03:45,210 Well, that's just denote our system like this, 58 00:03:45,210 --> 00:03:47,910 and let's say here is the center of mass, 59 00:03:47,910 --> 00:03:49,960 and here's the point s. 60 00:03:49,960 --> 00:03:57,680 So what we need to do is we need to draw the vector Rscm, 61 00:03:57,680 --> 00:04:03,260 and apply all the gravitational force at the center of mass. 62 00:04:03,260 --> 00:04:06,890 And then our torque about s is the vector Rs 63 00:04:06,890 --> 00:04:09,920 to where the force is acting cross m total g. 64 00:04:09,920 --> 00:04:12,980 So in conclusion, when you have a rigid body or a system 65 00:04:12,980 --> 00:04:14,930 of particles, and you want to calculate 66 00:04:14,930 --> 00:04:18,500 the torque due to a uniform gravitational field, 67 00:04:18,500 --> 00:04:23,540 then we place-- the total gravitational force is 68 00:04:23,540 --> 00:04:25,490 acting at the center of mass. 69 00:04:25,490 --> 00:04:27,710 And that gives us the torque about s 70 00:04:27,710 --> 00:04:31,930 to the individual torques of the system 71 00:04:31,930 --> 00:04:34,550 of the gravitational force about s. 72 00:04:34,550 --> 00:04:38,440 All the force is placed at the center of mass.