WEBVTT 00:00:00.000 --> 00:00:03.240 This is College Physics Answers with Shaun Dychko. 00:00:03.540 --> 00:00:08.960 We are going to calculate the force on each of the feet of this horse 00:00:09.200 --> 00:00:12.400 and it's going to be a force that is somewhat upwards 00:00:12.400 --> 00:00:15.280 to balance the weight of the horse and rider 00:00:15.420 --> 00:00:17.640 and it's also going to be somewhat to the left 00:00:17.680 --> 00:00:21.860 to balance the force exerted by the wall to the right 00:00:22.320 --> 00:00:27.200 and so it's gonna have some angle Θ here above horizontal towards the wall. 00:00:27.340 --> 00:00:30.920 So the first condition of equilibrium is to say that the total 00:00:30.920 --> 00:00:33.820 forces in the x-direction and in the y-direction equal zero 00:00:34.640 --> 00:00:40.340 or you could say that the forces to the left equal the forces to the right. 00:00:40.560 --> 00:00:44.060 So this is 2 times because there are two feet 00:00:44.120 --> 00:00:46.380 2 times the force on a single foot 00:00:46.440 --> 00:00:52.800 multiplied by cosine Θ to give us this component which is horizontal here 00:00:53.340 --> 00:00:57.240 that equals the force exerted by the wall to the right. 00:00:59.160 --> 00:01:01.720 And then we can talk about the y-direction 00:01:01.720 --> 00:01:08.220 and say that the vertical component of the force exerted by a foot 00:01:08.340 --> 00:01:12.580 multiplied by 2 equals the total weight downwards, mg. 00:01:13.180 --> 00:01:15.460 And there are three unknowns here: 00:01:15.460 --> 00:01:17.440 we don't know the force exerted by the wall 00:01:17.440 --> 00:01:22.200 nor Θ nor do we know the force on a single foot 00:01:22.320 --> 00:01:25.100 and so we need to have three equations to solve three unknowns 00:01:25.100 --> 00:01:26.660 and so far we have two equations 00:01:26.700 --> 00:01:29.800 and then we'll use the second condition of equilibrium 00:01:29.800 --> 00:01:32.120 which says that the total clockwise torque 00:01:32.120 --> 00:01:36.500 has to balance the total counter clockwise torque to find equation (3). 00:01:36.760 --> 00:01:39.300 So we have the force exerted by the wall 00:01:39.500 --> 00:01:44.180 multiplied by its perpendicular distance to the pivot at the foot—1.20 meters— 00:01:44.240 --> 00:01:51.400 equals the counter-clockwise torque due to the weight 00:01:52.020 --> 00:01:56.620 which is 0.35 meters perpendicular distance to the pivot. 00:01:57.560 --> 00:02:01.180 So we divide both sides by 1.20 meters here 00:02:01.240 --> 00:02:05.000 and this gives us the force exerted by the wall 00:02:05.060 --> 00:02:08.040 mg times 0.35 divided by 1.20 00:02:08.840 --> 00:02:12.420 and then we can do some substituting and rearranging 00:02:12.420 --> 00:02:14.820 to eventually solve for some of the unknowns. 00:02:14.960 --> 00:02:17.100 So returning back to equation (1), 00:02:17.140 --> 00:02:21.700 we can solve it for the force exerted on a foot 00:02:21.980 --> 00:02:26.740 and it's going to be the force exerted by the wall divided by 2cos Θ 00:02:26.740 --> 00:02:29.820 when you divide both sides by 2cos Θ here 00:02:31.940 --> 00:02:38.220 and then substitute in the force exerted by the wall in place of F wall here. 00:02:38.360 --> 00:02:44.380 So this is substituting equation 3 version b in place of equation [1b] 00:02:44.440 --> 00:02:47.220 and then we'll call this result equation [1c] 00:02:47.280 --> 00:02:52.640 so we have mg times 0.35 meters divided by 2cos Θ times 1.20 meters. 00:02:53.820 --> 00:03:00.840 Then we'll look at equation 2 version b and substitute in what we have for 00:03:00.840 --> 00:03:02.840 the force on a foot 00:03:02.840 --> 00:03:09.580 equation [1c] in place of F in equation 2 version b. 00:03:09.680 --> 00:03:15.260 So solving... rewriting equation 2 but instead of the force on a foot, 00:03:15.260 --> 00:03:21.200 we are gonna write mg times 0.35 meters divided by 2cos Θ times 1.20 meters 00:03:21.360 --> 00:03:23.220 and so we do that here. 00:03:23.940 --> 00:03:26.360 Now whole bunch of things cancel which is convenient: 00:03:26.360 --> 00:03:30.780 the mg's cancel, these 2's cancel and sin Θ divided by cos Θ 00:03:30.900 --> 00:03:35.120 can be replaced with tangent Θ— that's a trigonometric identity— 00:03:35.260 --> 00:03:38.540 and that's gonna equal 1.20 meters divided by 0.35 00:03:38.540 --> 00:03:41.700 because the other thing we'll do here is multiply both sides by 1.20 00:03:41.700 --> 00:03:44.960 and divide both sides by 0.35 00:03:48.600 --> 00:03:53.420 and Θ then is gonna be inverse tangent of 1.20 divided by 0.35 00:03:53.420 --> 00:03:55.640 which is 73.7 degrees. 00:03:56.620 --> 00:04:00.880 Then returning back to equation 2, we'll call it version c, 00:04:00.880 --> 00:04:04.260 where we solve for the force on a foot 00:04:04.500 --> 00:04:08.480 that's going to be mg divided by 2sin Θ. 00:04:09.640 --> 00:04:13.360 We have 500 kilograms—total mass of the horse and rider— 00:04:13.360 --> 00:04:15.080 times 9.8 newtons per kilogram 00:04:15.080 --> 00:04:17.760 divided by 2 times sin of this angle that we just found 00:04:17.760 --> 00:04:20.420 and that is 2550 newtons. 00:04:20.560 --> 00:04:24.220 So the force on a single foot is 2550 newtons, 00:04:24.220 --> 00:04:28.040 73.77 degrees above horizontal towards the wall. 00:04:28.980 --> 00:04:31.000 Then the next question in part (b) asks 00:04:31.060 --> 00:04:34.260 what is the minimum coefficient of static friction 00:04:34.320 --> 00:04:37.280 that's necessary to keep these feet from slipping. 00:04:37.700 --> 00:04:39.580 So the total friction force is 00:04:39.580 --> 00:04:43.940 2 times the horizontal component of a force on a foot 00:04:44.100 --> 00:04:45.760 and that also is going to be 00:04:45.760 --> 00:04:48.400 the coefficient of static friction multiplied by 00:04:48.420 --> 00:04:51.920 the normal force which will in turn equal the weight 00:04:51.960 --> 00:04:53.580 of the horse and rider. 00:04:53.960 --> 00:04:58.080 So we'll solve for μ s by dividing both sides by mg 00:04:58.340 --> 00:05:01.700 so that's 2 times the force on a foot times cos Θ divided by mg. 00:05:01.700 --> 00:05:04.860 So that's 2 times 2552.1 newtons 00:05:04.860 --> 00:05:08.000 times cos of 73.7398 degrees 00:05:08.000 --> 00:05:11.200 divided by 500 kilograms times 9.8 newtons per kilogram 00:05:11.200 --> 00:05:17.280 which is 0.292 and that is the minimum coefficient of static friction.