WEBVTT 00:00:00.000 --> 00:00:03.240 This is College Physics Answers with Shaun Dychko. 00:00:03.780 --> 00:00:07.700 This sandwich board is held together by this chain 00:00:07.780 --> 00:00:10.720 and it has a total mass of 8.00 kilograms 00:00:10.780 --> 00:00:14.440 and our job is to figure out what is the tension force on this chain. 00:00:15.080 --> 00:00:20.380 So we are going to figure out the normal force on the feet 00:00:20.420 --> 00:00:25.180 of this sandwich board and that can be found by saying that 00:00:25.220 --> 00:00:27.840 2 times the normal force on each foot 00:00:27.920 --> 00:00:29.840 equals the total weight downwards 00:00:29.840 --> 00:00:31.100 and when we look at it that way, 00:00:31.100 --> 00:00:33.100 we don't have to worry about this hinge. 00:00:33.100 --> 00:00:38.240 Now the force exerted on a side by the hinge, 00:00:38.300 --> 00:00:41.320 there might be some angle to this hinge force— 00:00:41.320 --> 00:00:42.500 we don't really know— 00:00:42.500 --> 00:00:46.300 but we can ignore any vertical component to the hinge force because that's internal 00:00:46.300 --> 00:00:51.400 to both sides and it's not something we need to be concerned with 00:00:51.400 --> 00:00:55.760 when we talk about the normal force and gravity. 00:00:56.920 --> 00:00:58.000 Okay! 00:00:58.240 --> 00:01:01.560 So the normal force is half the weight, in other words. 00:01:02.120 --> 00:01:06.600 Now we can also talk about the horizontal forces— 00:01:06.660 --> 00:01:09.920 this is the first condition of equilibrium that we are talking about— 00:01:09.920 --> 00:01:13.760 and the horizontal forces have to balance out to total zero 00:01:14.040 --> 00:01:17.180 and so the x-component of this hinge force 00:01:17.180 --> 00:01:20.160 which is the hinge force times cosine of Θ 00:01:20.240 --> 00:01:23.940 has to equal the tension force to the left. 00:01:26.780 --> 00:01:31.880 And thirdly, we need a third equation since we have a hinge force that we don't know; 00:01:31.880 --> 00:01:34.080 a tension force that we don't know; 00:01:34.080 --> 00:01:36.940 and an angle that we don't know 00:01:36.940 --> 00:01:39.300 so we need a third equation to deal with three unknowns. 00:01:39.480 --> 00:01:44.620 So we have the second condition of equilibrium which is that 00:01:44.620 --> 00:01:50.600 the total clockwise torques have to equal the total counter-clockwise torques. 00:01:51.840 --> 00:02:01.580 So the counter-clockwise torques are due to the tension force in the chain 00:02:01.580 --> 00:02:05.600 and it has a distance r subscript T, this is the lever arm 00:02:05.600 --> 00:02:11.060 for the tension force of 1.30 meters total height minus this 00:02:11.120 --> 00:02:15.220 0.50 meters from the hinge is where the chain's attached 00:02:15.300 --> 00:02:20.440 so that's 0.80 meters is the lever arm for the chain taking the pivot to be 00:02:20.440 --> 00:02:23.080 the foot of one side. 00:02:24.300 --> 00:02:30.080 And then we also have the weight exerting a counter-clockwise torque 00:02:30.080 --> 00:02:33.980 and it has a lever arm r W for weight 00:02:34.020 --> 00:02:37.740 of one-quarter this total width because 00:02:37.800 --> 00:02:40.340 we are told that this sandwich board is 00:02:40.340 --> 00:02:43.320 has a mass distribution that is uniform 00:02:43.380 --> 00:02:47.120 and so this hinge must be at the half-way point 00:02:47.120 --> 00:02:49.560 between the two feet 00:02:49.620 --> 00:02:56.040 and then this center of gravity has to be half-way point 00:02:56.080 --> 00:02:58.180 between the middle and the foot 00:02:58.180 --> 00:03:01.660 which makes it one-quarter of the way between the feet. 00:03:01.920 --> 00:03:06.180 So we have 1.10 meters—total distance between the feet—divided by 4 00:03:06.180 --> 00:03:10.620 that's 0.275 meters is this r W. 00:03:11.020 --> 00:03:15.940 So we multiply that by the weight to get the additional counter-clockwise torque 00:03:16.460 --> 00:03:21.560 and then substituting mg over 2 for this F W because this is 00:03:21.560 --> 00:03:25.020 half the total weight 00:03:26.140 --> 00:03:32.600 and that has to equal the torque that's clockwise due to the hinge 00:03:32.700 --> 00:03:36.080 and so we have the lever arm of the hinge is r H, 00:03:36.120 --> 00:03:39.240 which is the height of the sandwich board, 00:03:39.340 --> 00:03:44.420 and multiplied by the hinge force times cos Θ because we want to 00:03:44.420 --> 00:03:46.960 get the component of the hinge force 00:03:47.040 --> 00:03:50.500 that's perpendicular to this lever arm. 00:03:52.680 --> 00:04:01.720 And then we substitute in F T in place of the F H cos Θ because our 00:04:02.340 --> 00:04:08.560 'total force horizontally equaling zero' condition for equilibrium led us to that 00:04:08.600 --> 00:04:11.580 and so we can replace F H cos Θ with F T. 00:04:11.960 --> 00:04:16.600 And then we are gonna subtract 00:04:16.680 --> 00:04:19.640 F Tr T from both sides 00:04:21.140 --> 00:04:27.040 and then factor out the F T and we have F T times r H minus r T— 00:04:27.100 --> 00:04:29.660 of course, I have also switched the sides around too— 00:04:29.720 --> 00:04:33.600 and so all this equals mgr W over 2 00:04:33.760 --> 00:04:38.400 and then divide both sides by r H minus r T. 00:04:41.960 --> 00:04:44.300 And so we have then that the tension force in the chain 00:04:44.300 --> 00:04:48.620 is mg lever arm of the weight divided by 2 times 00:04:48.620 --> 00:04:51.800 lever arm of the hinge minus the lever arm of the chain. 00:04:52.260 --> 00:04:54.940 So that's 8.00 kilograms times 9.80 newtons per kilogram 00:04:54.940 --> 00:04:59.580 times 0.275 meters divided by 2 times 1.30 meters minus 0.80 meters 00:04:59.640 --> 00:05:02.100 which is 21.6 newtons. 00:05:03.040 --> 00:05:08.120 And then in part (b), we want to find the hinge force 00:05:09.280 --> 00:05:15.320 and we can say that F H sin Θ plus the normal force equals 00:05:15.320 --> 00:05:18.520 the weight but we are... 00:05:19.080 --> 00:05:23.480 you know, this Θ is going to be zero 00:05:23.700 --> 00:05:30.300 because the weight is half the total weight of the sandwich board, mg over 2, 00:05:30.300 --> 00:05:34.200 the normal force is mg over 2 as well and that's what we figured out here 00:05:34.480 --> 00:05:39.060 and so that leaves nothing left over for this F H sin Θ 00:05:39.060 --> 00:05:42.280 it has to equal zero and of course, F H is not zero which means 00:05:42.280 --> 00:05:47.180 sin Θ is zero and that's true only when you have the angle being 0 degrees. 00:05:47.660 --> 00:05:48.860 Okay! 00:05:49.760 --> 00:05:54.340 So then we can say that the hinge force times cos Θ 00:05:54.340 --> 00:05:57.060 which is the horizontal component of the hinge force to the right 00:05:57.100 --> 00:05:58.860 has to equal the tension force to the left 00:05:58.920 --> 00:06:02.900 and then divide both sides by cos Θ but we have already established that 00:06:02.900 --> 00:06:06.240 this Θ is 0 and the cos of which is 1 00:06:06.280 --> 00:06:08.860 and so the hinge force then is going to 00:06:08.860 --> 00:06:11.960 have the same magnitude as the tension force in the chain 00:06:11.960 --> 00:06:16.180 which is 21.6 newtons and that's horizontally 00:06:16.180 --> 00:06:17.600 and that will be to the right on 00:06:17.600 --> 00:06:19.780 the right hand side of the sandwich board 00:06:19.780 --> 00:06:23.340 and it will be to the left on the left hand side of the sandwich board.