﻿1 00:00:01,230 --> 00:00:08,230 Let’s look at another problem here. Let me just explain the problem first and then 2 00:00:53,940 --> 00:00:59,890 we will first draw the free body and seek to solve the problem. I will just explain 3 00:00:59,890 --> 00:01:06,890 to you, how you will go about solving the entire problem. There is a particular rod 4 00:01:10,540 --> 00:01:17,540 here A B, let me call this as E. So A E B that is pinned at A to the fixed frame, as 5 00:01:21,660 --> 00:01:28,660 you can see here it is hashed. This is a fixed frame. It is pinned to a body which is L in 6 00:01:30,150 --> 00:01:37,150 shape, B C D pinned to A B at B and it is pinned to a wheel which is resting on A E 7 00:01:43,180 --> 00:01:50,180 B at E and this wheel is just standing at E. Now this B C D is prevented from moving at C and this is the problem. 8 00:02:06,650 --> 00:02:13,650 We wished to find out all the unknowns related to this. When we say we have to solve the 9 00:02:14,430 --> 00:02:19,239 problem, we need to solve for the reactions from the fixed supports and the reactions 10 00:02:19,239 --> 00:02:26,239 that appear as an interaction between the rigid bodies. How do I go about doing this? 11 00:02:27,730 --> 00:02:34,489 The first exercise is to understand that there are three rigid bodies in this. 12 00:02:34,489 --> 00:02:41,489 One rigid body is this A E B. The other rigid body is B C D and the third rigid body is 13 00:02:42,780 --> 00:02:49,780 this particular wheel which is hinged about D. There are three rigid bodies, this will 14 00:02:52,870 --> 00:02:59,870 help me draw the free body diagrams appropriately. One is A B, you can either call it as A E 15 00:03:04,189 --> 00:03:11,189 B or A B. The other rigid body is B C D and the wheel D, the three rigid bodies that we 16 00:03:21,079 --> 00:03:28,079 have. What is the first exercise that we need to carry out? We need to remove the restrains 17 00:03:28,709 --> 00:03:34,010 that appear due to fixed frame of reference. So let’s do that particular exercise to 18 00:03:34,010 --> 00:03:41,010 start with. Now I am going to just shift to simple line diagram. I am just going to draw 19 00:03:41,180 --> 00:03:48,180 this A B as a line, B C D as an L and a circle for a wheel. This is a point A B, we have 20 00:03:54,639 --> 00:04:01,639 like this which is BCD and you have a wheel. Now what all have we removed here? We have 21 00:04:14,549 --> 00:04:20,350 removed the support that restrains motion in the horizontal direction which means I 22 00:04:20,350 --> 00:04:27,350 should have a reaction from the fixed frame on to this system of planar rigid bodies. 23 00:04:27,950 --> 00:04:33,810 So let’s call this as C or Cx along the x direction. So we are going to take this 24 00:04:33,810 --> 00:04:40,660 as x direction and this as y direction for our own convenience. 25 00:04:40,660 --> 00:04:47,660 At K it is pinned here which means the fixed frame of reference doesn’t let it move with 26 00:04:47,840 --> 00:04:52,700 respect to the fixed frame of reference at A, both in the vertical and the horizontal 27 00:04:52,700 --> 00:04:58,100 direction which means there will be two forces that will be needed by the fixed frame of 28 00:04:58,100 --> 00:05:05,100 reference in order to restrain the motion. Let’s call this as Ay and Ax. So we have 29 00:05:05,420 --> 00:05:12,420 3 unknowns Ax, Ay and Cx which are to be found out from equilibrium equations. In addition 30 00:05:16,910 --> 00:05:23,350 in this particular problem it is given that there is a moment acting here. Think of some 31 00:05:23,350 --> 00:05:30,350 motor that is applying a torque of 120 Newton meter. I need to transfer that over here. 32 00:05:32,580 --> 00:05:39,580 There is a 120 Newton meter that is acting external to this. Assume all the rigid bodies 33 00:05:40,650 --> 00:05:47,650 are negligible mass compared to the 120 Newton meter that is applied. The first free body 34 00:05:50,640 --> 00:05:56,710 concerns this entire system without the fixed frame of reference and it contains all the 35 00:05:56,710 --> 00:06:03,710 three. Then we seek to draw the free body of each one of them separately. 36 00:06:08,130 --> 00:06:15,130 Let’s start with the wheel because that seems to be the simplest one. This is the 37 00:06:19,370 --> 00:06:26,370 wheel D that we seek to draw. What happens here as we have discussed earlier, in order 38 00:06:27,700 --> 00:06:34,700 to draw the free body of this wheel, I need to make sure that I fix A E B and B C D the 39 00:06:38,590 --> 00:06:45,230 two other rigid bodies fixed to the fixed frame of reference which means A E B and B 40 00:06:45,230 --> 00:06:52,230 C D are not moving at all. In which case this point D of B C D will restrain this wheel 41 00:06:55,680 --> 00:07:02,680 from moving in x as well as y direction which means I will have a reaction that appears 42 00:07:04,780 --> 00:07:11,780 on the wheel as Dy and Dx that is responsible for preventing motion from x and y. Apart 43 00:07:15,300 --> 00:07:22,300 from that you already notice here that the wheel is resting on A E B which means there 44 00:07:26,210 --> 00:07:33,210 is a normal reaction equal to, let’s just call this as N E or since it is in y direction, 45 00:07:34,860 --> 00:07:41,860 we will just call it as Ey. We have a question here. Will I have a horizontal reaction over 46 00:07:44,220 --> 00:07:51,220 here? I don’t know. Let’s say it is not frictionless, it has friction. Let’s just 47 00:07:51,630 --> 00:07:56,790 introduce Ex for now. 48 00:07:56,790 --> 00:08:03,790 There is a resistance in the x direction. we will come back to simplification as far 49 00:08:03,830 --> 00:08:10,830 as this wheel is concerned in a while. Have you missed out anything else? the mass of 50 00:08:11,050 --> 00:08:17,170 this particular wheel is negligible which means there are only two points that i have 51 00:08:17,170 --> 00:08:22,730 to consider. One point of restrain coming from B C D and the other was one is due to 52 00:08:22,730 --> 00:08:29,730 resting at B. so this is complete as far as wheel D is concerned. Let’s seek to draw 53 00:08:31,210 --> 00:08:37,050 the other two free bodies. I am going to use line diagrams in order to make it simple. 54 00:08:37,050 --> 00:08:44,050 Now let’s just look at this particular B C D. If you look at B C D, we have to fix 55 00:08:48,490 --> 00:08:55,490 A E B and the wheel to the fixed frame of reference and have a look at what will be 56 00:08:55,550 --> 00:09:02,550 the restrains that will be offered. There are three points at which restrains will be 57 00:09:02,720 --> 00:09:09,720 offered on this B C D, one is at B by A E B. The other is at C due to fixed frame of 58 00:09:12,089 --> 00:09:19,089 reference and the other one is at D due to this wheel remaining un moveable and therefore 59 00:09:25,709 --> 00:09:30,889 I will have a reaction. 60 00:09:30,889 --> 00:09:35,730 For now since we already know that we have drawn Dx and Dy in this direction. We will 61 00:09:35,730 --> 00:09:42,730 take equal and opposite directions for this point D, this is Dy Dx. We have finished this. 62 00:09:51,579 --> 00:09:58,579 Looking at this there is a restrain offered here. So we will just stick to that. At B 63 00:10:00,970 --> 00:10:07,970 we will have a restrain in the vertical direction and therefore a force appears. There is a 64 00:10:08,050 --> 00:10:13,290 restrain in the horizontal direction which means there is another force Bx that appears. 65 00:10:13,290 --> 00:10:20,290 Are we complete? The answer is yes. The mass we have here is negligible which means we 66 00:10:20,490 --> 00:10:24,839 have completed the free body of B C D. 67 00:10:24,839 --> 00:10:31,839 This is x direction, this is y direction. What remains now is A E B. Again I am going 68 00:10:37,480 --> 00:10:44,480 to resort with line diagram, it makes it simple for drawing. At B remember I am going to fix 69 00:10:50,540 --> 00:10:57,540 B C D as well as the wheel E D which means at this particular point, I will have an equal 70 00:10:59,670 --> 00:11:06,670 and opposite reaction which means I will have By here and Bx like this. Simple. At E there 71 00:11:10,480 --> 00:11:17,480 is an interaction with the wheel and therefore I should take the opposite signs of this Ex 72 00:11:18,839 --> 00:11:25,839 and Ey and insert those as the forces, so Ex and Ey. This way we can avoid adding more 73 00:11:30,490 --> 00:11:36,930 number of unknowns, we take equal and opposite reactions automatically in this. At A we have 74 00:11:36,930 --> 00:11:43,930 already drawn for the system of rigid body. I will have Ax and Ay appearing. We have drawn 75 00:11:47,639 --> 00:11:54,639 all the free body diagrams here that of wheel, that of B C D and that of A E B. 76 00:12:01,369 --> 00:12:07,410 The next exercise is to find out what all unknowns do we have to solve. Let’s look 77 00:12:07,410 --> 00:12:14,410 at that. The simplest one is this. I am going to start with this particular wheel. Let me 78 00:12:17,689 --> 00:12:24,689 draw that again here. Now we are looking at the unknowns. For the wheel we have Dx and 79 00:12:37,930 --> 00:12:44,930 Dy, Ex and Ey. We already accounted for this, we have Cx that needs to be solved and we 80 00:12:51,369 --> 00:12:55,389 have Bx and By. 81 00:12:55,389 --> 00:13:02,389 For this body we have to solve for Ax and Ay, we already accounted for Ex, Ey, Bx, By 82 00:13:06,329 --> 00:13:13,329 which means Ax and Ay. In all how many? 2 plus 2, 4 plus 4, 5 plus 4 is 9 unknowns. 83 00:13:23,550 --> 00:13:30,550 Can I write down 9 equations? The answer is yes. Because I have 1, 2, 3 rigid bodies. 84 00:13:33,740 --> 00:13:40,740 Each can generate 3 equations and therefore I can solve for the 9 unknowns. Now if I do 85 00:13:42,709 --> 00:13:45,189 it this way, it will be a long procedure. 86 00:13:45,189 --> 00:13:52,189 Let me look for certain simplifications that I can do. So simplification one. If we examine 87 00:14:01,509 --> 00:14:08,509 the wheel, let me draw it over here. You have Dx, Dy, Ex, Ey. These are the only forces 88 00:14:22,869 --> 00:14:29,869 that are acting. If I take moment about this particular point and immediate result is that 89 00:14:34,949 --> 00:14:41,949 Ey is along the vertical direction that passes through D. Dx and Dy don’t take part in 90 00:14:43,309 --> 00:14:50,040 that moment equilibrium which means only Ex takes part in the moment equilibrium. Therefore 91 00:14:50,040 --> 00:14:57,040 if I take moment about D lets say positive, there is only one force that results in that. 92 00:14:59,220 --> 00:15:06,220 Let’s say this radius is r, we already know that this is 0.2 meters. We get it is anticlockwise 93 00:15:08,629 --> 00:15:15,629 in its motion Ex times 0.2 should be equal to 0. It immediately gives me Ex equals zero. 94 00:15:17,749 --> 00:15:24,749 There is no horizontal force that appears on this. This is clear? Now if Ex is equal 95 00:15:31,730 --> 00:15:38,730 to 0, sigma Fx equals 0 will immediately tell me that Dx is equal to 0. You get this? I 96 00:15:44,249 --> 00:15:51,249 am just going to erase it to make it simple, this is not there, this is not there. Like 97 00:15:53,040 --> 00:15:59,910 what we did earlier, for a rigid body where there are only two forces two points at which 98 00:15:59,910 --> 00:16:06,910 forces are acting, immediate answer that we get here is Dy and Ey are opposite to each 99 00:16:07,819 --> 00:16:09,439 other. 100 00:16:09,439 --> 00:16:16,439 So sigma Fy equals 0 gives me Ey equals minus Dy. I need not now look at all these four. 101 00:16:27,989 --> 00:16:34,989 I just need one unknown, either Ey or Dy and that will reduce the unknowns in the other 102 00:16:36,149 --> 00:16:43,149 free body diagrams. You have Dx and Dy here, you have Ex and Ey over here. So let’s seek 103 00:16:54,160 --> 00:17:01,160 to now remove those so that we simplify. Please remember these simplifications will help us 104 00:17:01,329 --> 00:17:08,329 solve in a much better way. I am going to remove this, I am going to call this as Ey 105 00:17:18,899 --> 00:17:25,899 equal and opposite. So I am going to remove this and shall we just take this instead of 106 00:17:26,399 --> 00:17:32,760 Ey, we will just put it as normal reaction. That’s the reaction that we see from the 107 00:17:32,760 --> 00:17:39,760 ground for a wheel. There will be N acting on this. Here also I don’t have this Ex, 108 00:17:44,970 --> 00:17:51,059 this is nothing but the N acting from the wheel. Now we have reduced it. 109 00:17:51,059 --> 00:17:57,870 We don’t have this anymore, this anymore, instead it is replaced by a single unknown 110 00:17:57,870 --> 00:18:04,870 N. Now instead of 9 equations, we have 1, 2, 3, 4, 5, 6 equations to be solved. Mind 111 00:18:10,500 --> 00:18:17,500 you in doing that we have already used up all the three equations necessary for the 112 00:18:17,519 --> 00:18:24,519 wheel. Going back to this, there are 1, 2, 3, 4 unknowns present in this. 1, 2, 3, 4, 113 00:18:32,640 --> 00:18:39,640 5 unknowns present in this. In order to solve for this, what would be the points we take 114 00:18:41,669 --> 00:18:48,669 in order to write down the moment equations. 115 00:18:49,419 --> 00:18:55,149 Is there a simplification possible in this? There are simplifications that we can think 116 00:18:55,149 --> 00:19:02,149 of but mind you, I don’t have to look at the wheel anymore. That’s already taken 117 00:19:02,750 --> 00:19:09,750 care of. How do I solve for the other unknowns? We already saw 5 here, 5 there. If I take 118 00:19:15,640 --> 00:19:22,640 moment about this point, what all unknowns can I avoid? I can avoid Ax, Ay as well as 119 00:19:24,059 --> 00:19:31,059 Bx which means I will have an equation that is in terms of N and By. Is that clear? So 120 00:19:32,220 --> 00:19:39,220 N and By will take part, if I have equal zero. This will have N1 and By taking part. Go to 121 00:19:46,460 --> 00:19:53,460 the other free body, what is the equation that gives me a relationship between N and 122 00:19:54,330 --> 00:20:00,309 By only? That’s not very difficult, it is the vertical force equilibrium. 123 00:20:00,309 --> 00:20:07,309 So far this, I will do this. I will do sigma Fy equal zero that it will not involve Bx 124 00:20:12,179 --> 00:20:19,179 as well as Cx, it will involve only these two N and By. This involved N and By, this 125 00:20:22,179 --> 00:20:29,179 involved N and By, I can solve using these two equations N and By. Once I have solved 126 00:20:30,370 --> 00:20:37,370 for N and By, I am now left with Ax, Ay, Bx. in this case Bx and Cx. If I take the vertical 127 00:20:43,200 --> 00:20:49,139 equilibrium here, Ay can be solved for because N and By are already known. So I can solve 128 00:20:49,139 --> 00:20:54,799 for that by simply taking sigma Fy equals zero here. 129 00:20:54,799 --> 00:21:01,799 I am left with the horizontal equilibrium there that will relate Ax and Bx. If I take 130 00:21:03,309 --> 00:21:10,309 the horizontal equilibrium there, it will relate Bx and Cx. I am sorry, I should have 131 00:21:14,419 --> 00:21:21,419 inserted 120 Newton meter here. Taking moment about A, I should involve this 120 Newton 132 00:21:24,260 --> 00:21:31,260 meter. So Ax equals Bx, Cx equals Bx are the two equations I get. I need three equations 133 00:21:34,690 --> 00:21:41,690 because I have Ax, Bx and Cx. How do I get the third one? I will go back to this guide 134 00:21:44,010 --> 00:21:50,210 and if I take moment about this, I avoid Ax and Ay. There is only Cx that needs to be 135 00:21:50,210 --> 00:21:52,250 solved here. 136 00:21:52,250 --> 00:21:59,250 Remember this system of planar bodies free body will help me solve for Cx directly. This 137 00:22:01,889 --> 00:22:08,850 is an important point you have to note. If I do sigma MA equals zero here, I will get 138 00:22:08,850 --> 00:22:15,850 this directly. Having got on this directly I can find out Bx, having found out Bx I can 139 00:22:17,190 --> 00:22:23,909 found out Ax which means I have solved for all the unknowns in this particular problem. 140 00:22:23,909 --> 00:22:28,519 Please use this kind of an approach in order to solve the problem, else you will end up 141 00:22:28,519 --> 00:22:33,720 with 9 equations with 9 unknowns that need to be solved and it will be too cumbersome 142 00:22:33,720 --> 00:22:35,080 for you to solve. 143 00:22:35,080 --> 00:22:40,059 It is important to look at simplifications that you can do, you will always have to look 144 00:22:40,059 --> 00:22:46,429 at, is there a simple way to solve for one unknown or the other. If you do use this approach, 145 00:22:46,429 --> 00:22:53,429 it will be possible for you to solve many problems in a very simple way. You can draw, 146 00:22:54,909 --> 00:23:00,450 you can make these strips in order to understand the problem that we discussed. This is that 147 00:23:00,450 --> 00:23:07,450 A B, let me draw this point of interaction as E. Now this is a point fixed to the fixed frame, 148 00:23:15,399 --> 00:23:22,269 this is the point A fixed to the fixed frame and remember this is not fixed to the fixed 149 00:23:22,269 --> 00:23:29,269 frame but it fixes B C D and A E B. Now we will remove each restrain and insert a force. 150 00:23:36,090 --> 00:23:43,090 As before this has x direction this has y direction. 151 00:23:43,509 --> 00:23:49,330 This restrain essentially does not allow the points C from moving in this direction. I 152 00:23:49,330 --> 00:23:56,330 am going to remove this and insert a force like this. Nice to do this kind of an exercise, 153 00:24:01,090 --> 00:24:08,090 this gives us understanding. In order to make this body free from the fixed frame, I need 154 00:24:08,529 --> 00:24:15,070 to remove A. What does A do? It doesn’t allow the point A to move either in vertical 155 00:24:15,070 --> 00:24:22,070 or in horizontal direction. I am going to insert the two restrains Ax and Ay. Now this 156 00:24:25,620 --> 00:24:32,440 is a free body that you have for the entire system. Supposing I have to remove this particular 157 00:24:32,440 --> 00:24:39,440 planar body from the system, I need to fix A B and fix B C D. The pin at point D does 158 00:24:44,539 --> 00:24:51,509 not allow it to move. So what I am going to do is I am going to remove this and insert 159 00:24:51,509 --> 00:24:58,509 two forces that prevented movement at D. In addition at this point E, there is horizontal 160 00:25:06,659 --> 00:25:13,659 and vertical restrains so I can add those two. An equal and opposite reaction has to 161 00:25:17,090 --> 00:25:19,490 be inserted over here. 162 00:25:19,490 --> 00:25:25,909 I am just going to remove this and insert it over here. There is a restrain this way 163 00:25:25,909 --> 00:25:30,980 equal and opposite restrain as inserted over here, let me just move it over here like this. 164 00:25:30,980 --> 00:25:37,940 There is one more like this that I have to insert, an equal and opposite reaction at 165 00:25:37,940 --> 00:25:44,940 D. Here we have made a simplification to find out that these two are zero. We might as well 166 00:25:47,730 --> 00:25:53,440 remove them which mean this is also zero. You can do this kind of simplification in 167 00:25:53,440 --> 00:25:58,480 order to understand, how to draw the free body diagram. 168 00:25:58,480 --> 00:26:04,970 Now if you look at it, I have a force over here let me retain that force. There is another 169 00:26:04,970 --> 00:26:11,970 force on this. If I have to remove A E B now, I have to remove the restrain here. When I 170 00:26:13,190 --> 00:26:20,190 remove the restrain here, I need to add the two restrains related to that and this is 171 00:26:24,000 --> 00:26:27,899 the free body diagram of B C D. 172 00:26:27,899 --> 00:26:33,580 In a very similar way you can do the same thing. This point is restrained by B C D and 173 00:26:33,580 --> 00:26:40,580 therefore let me just take this out. It will have a reaction like this. A is already restrained 174 00:26:44,120 --> 00:26:51,080 So let me just take those two. These are the two other restraints. At E also I have one 175 00:26:51,080 --> 00:26:58,080 restraint that comes from the wheel. I have used something like this, so this is like 176 00:26:58,470 --> 00:27:05,470 this. This is E upward equal and opposite reaction occurs here and this is the free 177 00:27:05,500 --> 00:27:12,500 body apart from the moment that occurs at A. This should be very clear to you. Try this 178 00:27:12,730 --> 00:27:19,730 exercise a few times to get a physical feel of how to draw the free body. Once you become 179 00:27:20,000 --> 00:27:27,000 a master in free body diagrams then lot of things can be accomplished in a very simple 180 00:27:41,470 --> 00:27:48,470 way. Thank you. 181