1 00:00:03,510 --> 00:00:06,080 Today, we'd like to discuss tension in a rope. 2 00:00:06,080 --> 00:00:08,850 Under what conditions is the tension uniform? 3 00:00:08,850 --> 00:00:14,170 Recall that if we had a rope and say we were pulling-- here's 4 00:00:14,170 --> 00:00:15,170 our one end. 5 00:00:15,170 --> 00:00:18,570 We'll call that we're pulling at this end with the force a. 6 00:00:18,570 --> 00:00:23,770 And we pull on this end b with a force b. 7 00:00:23,770 --> 00:00:27,100 Then what we meant by tension at any point in the rope 8 00:00:27,100 --> 00:00:31,350 was that we took an imaginary slice of the rope p 9 00:00:31,350 --> 00:00:34,120 and we looked at the two sides. 10 00:00:34,120 --> 00:00:37,460 And here-- if we call this the left side and this is 11 00:00:37,460 --> 00:00:41,320 the right side-- then on the left side, 12 00:00:41,320 --> 00:00:43,970 we have that the right side of the rope 13 00:00:43,970 --> 00:00:49,190 is pulling the left side of the rope by some force f. 14 00:00:49,190 --> 00:00:51,590 And this is the left side. 15 00:00:51,590 --> 00:00:57,000 And the right side is being pulled by the left side. 16 00:00:57,000 --> 00:01:00,450 So that's the right side being pulled by the left. 17 00:01:00,450 --> 00:01:03,990 And what we define the tension at the point p 18 00:01:03,990 --> 00:01:09,840 was to be the magnitude of this action-reaction pair. 19 00:01:09,840 --> 00:01:15,100 And that's how we define tension at a point inside a rope. 20 00:01:15,100 --> 00:01:19,680 Now we'd like to address the question, under what conditions 21 00:01:19,680 --> 00:01:22,830 can we say that the tension is uniform? 22 00:01:22,830 --> 00:01:26,250 Which means that at all points p, the tension is the same. 23 00:01:26,250 --> 00:01:29,430 In order to analyze this, we'll apply Newton's second law. 24 00:01:29,430 --> 00:01:32,640 So let's begin with the same situation as before. 25 00:01:32,640 --> 00:01:36,690 Let's consider a rope where we're pulling this side 26 00:01:36,690 --> 00:01:37,880 with a force at the end. 27 00:01:37,880 --> 00:01:43,755 By the way, we can call the tension at that end T of a. 28 00:01:43,755 --> 00:01:46,900 And on this side, we can refer to this 29 00:01:46,900 --> 00:01:50,900 as the tension on side b. 30 00:01:50,900 --> 00:01:57,870 Now suppose that our mass of the rope is non-zero. 31 00:01:57,870 --> 00:02:00,920 And we now want to apply Newton's second law. 32 00:02:00,920 --> 00:02:03,450 But let's also apply the condition 33 00:02:03,450 --> 00:02:06,500 that the acceleration of the rope is non-zero. 34 00:02:06,500 --> 00:02:09,889 So you're holding the rope under some tension 35 00:02:09,889 --> 00:02:11,700 by pulling both sides. 36 00:02:11,700 --> 00:02:14,720 But the rope is not accelerating. 37 00:02:14,720 --> 00:02:17,600 And so what does that mean when we apply Newton's 38 00:02:17,600 --> 00:02:20,130 second law to the rope? 39 00:02:20,130 --> 00:02:22,800 Well, let's just arbitrarily call that 40 00:02:22,800 --> 00:02:24,350 our positive direction. 41 00:02:24,350 --> 00:02:29,370 We see by vector decomposition that we have Tb minus Ta 42 00:02:29,370 --> 00:02:31,210 equals mass of the rope. 43 00:02:31,210 --> 00:02:35,180 But because the acceleration of the rope is zero, 44 00:02:35,180 --> 00:02:36,930 this side is zero. 45 00:02:36,930 --> 00:02:39,820 And we can use Newton's second law 46 00:02:39,820 --> 00:02:42,910 to conclude that the tension in the rope 47 00:02:42,910 --> 00:02:46,180 is uniform when the rope is not accelerating. 48 00:02:46,180 --> 00:02:50,030 And this is what we'll call our case one. 49 00:02:50,030 --> 00:02:54,079 A zero tension in the rope is uniform. 50 00:02:54,079 --> 00:03:01,080 Now let's suppose case two, that a is not zero. 51 00:03:01,080 --> 00:03:07,510 And so now we have the rope. 52 00:03:07,510 --> 00:03:09,900 This is point b. 53 00:03:09,900 --> 00:03:12,250 Here's point a. 54 00:03:12,250 --> 00:03:14,213 We'll choose a uniform direction a. 55 00:03:14,213 --> 00:03:18,890 And now we can see when we apply F equals 56 00:03:18,890 --> 00:03:25,480 Ma, that Tb minus Ta equals mass of the rope 57 00:03:25,480 --> 00:03:27,710 times the acceleration of the rope. 58 00:03:27,710 --> 00:03:32,960 So you're pulling such that Tb is greater 59 00:03:32,960 --> 00:03:38,160 than Ta by precisely the quantity Mr times a. 60 00:03:38,160 --> 00:03:41,160 So the tension at the ends are not the same. 61 00:03:41,160 --> 00:03:44,120 And therefore, the tension in the rope is not uniform. 62 00:03:44,120 --> 00:03:48,960 Now there's a special case here that we want to consider. 63 00:03:48,960 --> 00:03:51,990 So even though a is nonzero, let's 64 00:03:51,990 --> 00:03:56,510 assume that the mass of the rope is very small. 65 00:03:56,510 --> 00:03:58,670 And this is often-- people call this 66 00:03:58,670 --> 00:04:00,851 assume the rope is massless. 67 00:04:00,851 --> 00:04:02,370 Well, that's a little strange. 68 00:04:02,370 --> 00:04:06,870 We can say that the very, very light string is often 69 00:04:06,870 --> 00:04:08,590 the same type of modeling. 70 00:04:08,590 --> 00:04:12,500 And under those circumstances with that assumption, 71 00:04:12,500 --> 00:04:14,990 you see that there is no contribution here. 72 00:04:14,990 --> 00:04:18,060 And with that assumption, that the tension 73 00:04:18,060 --> 00:04:21,829 is approximately uniform. 74 00:04:21,829 --> 00:04:23,210 Ropes are not massless. 75 00:04:23,210 --> 00:04:24,790 There's a small difference. 76 00:04:24,790 --> 00:04:30,980 But under those circumstances when we say a very light rope 77 00:04:30,980 --> 00:04:33,680 that-- even if it is accelerating-- implies 78 00:04:33,680 --> 00:04:40,610 that T is uniform in the rope. 79 00:04:40,610 --> 00:04:43,810 And we can say the rope is under uniform tension. 80 00:04:43,810 --> 00:04:47,790 And we will often refer to this case 81 00:04:47,790 --> 00:04:49,430 when we analyze problems where we're 82 00:04:49,430 --> 00:04:54,040 pulling objects, for instance, in a typical pulley problem.