WEBVTT 00:00:00.000 --> 00:00:02.840 This is Giancoli Answers with Mr. Dychko. 00:00:03.920 --> 00:00:08.020 Since this tree sapling is in equilibrium it means all the forces balance out. 00:00:08.300 --> 00:00:10.420 So I have chosen a coordinate system 00:00:10.480 --> 00:00:13.320 so that one axis is along one of the forces 00:00:13.320 --> 00:00:16.480 we'll choose x-axis to be along force A 00:00:16.540 --> 00:00:20.740 and that will reduce our work a little bit because we don't have to worry 00:00:20.740 --> 00:00:22.740 about any trigonometry when we are dealing with force A— 00:00:22.740 --> 00:00:24.560 it's entirely in the x-direction. 00:00:25.140 --> 00:00:29.360 Now the force's balancing out means they balance out separately 00:00:29.360 --> 00:00:32.360 in the x-axis and along the y-axis 00:00:32.440 --> 00:00:34.400 so we'll have two different equations 00:00:34.600 --> 00:00:39.020 and so we can consider this x-component for force B up here 00:00:39.580 --> 00:00:43.740 pointing straight to the left that's F B x 00:00:43.800 --> 00:00:50.040 and that added to the x-component of force C, this is F C x 00:00:50.120 --> 00:00:57.300 those two have to equal the total x-forces going to the right which is just F A. 00:00:57.840 --> 00:01:01.800 So that's where this equation comes from and then we can solve for 00:01:01.800 --> 00:01:08.620 the x-component of force C by subtracting F B x from both sides. 00:01:10.160 --> 00:01:15.460 And then we'll replace F B x with F BsinΘ B 00:01:15.460 --> 00:01:18.680 where I have taken Θ B to be this angle between 00:01:18.760 --> 00:01:21.520 the vertical y-axis and the force 00:01:21.740 --> 00:01:29.940 and so we are gonna calculate that as being 105 minus 90, just 15 degrees. 00:01:32.560 --> 00:01:37.400 And F B x is the opposite leg of this triangle so we take 00:01:37.400 --> 00:01:42.580 sin of Θ B to get the opposite leg multiplied by the hypotenuse F B. 00:01:43.700 --> 00:01:44.720 Okay. 00:01:44.760 --> 00:01:47.160 And then the same idea for the y-direction 00:01:47.220 --> 00:01:49.080 and in the y-direction, there are only 00:01:49.080 --> 00:01:52.680 two forces to consider since force A is entirely in the x-direction. 00:01:52.880 --> 00:02:00.040 So the component of force B that's vertical, upwards, this is F B y 00:02:00.040 --> 00:02:05.040 has to equal the component of force C which is downwards, that's F C y. 00:02:05.620 --> 00:02:11.020 So F C y is F B times cos Θ B— 00:02:11.240 --> 00:02:17.200 cos is what we use to find the adjacent leg of the triangle— 00:02:17.380 --> 00:02:25.040 and here I mention again that Θ B is 105 degrees from force A 00:02:25.120 --> 00:02:28.080 minus the 90 degrees here 00:02:28.180 --> 00:02:32.480 and that gives us just a little sliver that's in the corner that slice 00:02:32.480 --> 00:02:34.680 in this triangle here 00:02:36.960 --> 00:02:41.520 and now we can calculate each of these components for force C. 00:02:41.600 --> 00:02:46.440 So force C in the x-direction is force A, 385 newtons, 00:02:46.540 --> 00:02:49.320 minus 475 times sin 15 00:02:49.420 --> 00:02:53.660 so I just substituted into this formula just now with numbers here 00:02:54.620 --> 00:02:57.700 and that works out to 262.06 newtons. 00:02:57.820 --> 00:03:00.040 And then force C in the y-direction 00:03:00.040 --> 00:03:05.780 is force B, 475 newtons, times cos 15 which is 458.81. 00:03:06.100 --> 00:03:08.700 So those are the two components for force C 00:03:08.780 --> 00:03:11.680 and then defined the hypotenuse... 00:03:11.920 --> 00:03:16.680 see what we figured out so far is, we found out that force C 00:03:16.940 --> 00:03:22.040 goes downwards this much, 458.81 newtons, 00:03:22.200 --> 00:03:28.800 and it goes to the right this much, F C x, 262.06 newtons 00:03:28.900 --> 00:03:33.700 that makes a right triangle with this resultant being the hypotenuse 00:03:34.180 --> 00:03:36.600 and we use Pythagoras to calculate it. 00:03:36.600 --> 00:03:38.120 So force C is the square root of 00:03:38.120 --> 00:03:40.560 F C x squared plus F C y squared 00:03:40.940 --> 00:03:44.320 so that's the square root of 262.06 newton squared 00:03:44.320 --> 00:03:46.840 plus 458.81 newton squared 00:03:46.840 --> 00:03:49.760 that gives 528.38. 00:03:49.840 --> 00:03:55.000 And then to figure out the angle and that is just the sliver in here, Θ C, 00:03:55.100 --> 00:04:00.380 that's the part between the vertical axis and the force 00:04:01.700 --> 00:04:04.020 so we are gonna have to add 90 to our answer here 00:04:04.260 --> 00:04:09.540 is gonna be the inverse tangent of the opposite which is F C x 00:04:09.640 --> 00:04:12.280 divided by the adjacent F C y. 00:04:13.080 --> 00:04:18.280 And that's inverse tangent of 262.06 divided by 458.81 newtons 00:04:18.280 --> 00:04:20.280 this gives 29.73. 00:04:20.540 --> 00:04:26.360 And we'll call that Θ or the answer that the question is looking for 00:04:26.360 --> 00:04:31.580 is the angle all the way from force A to force C so we'll just call that Θ 00:04:31.580 --> 00:04:35.120 and that's gonna be Θ C plus 90. 00:04:37.740 --> 00:04:43.860 So that's 90 plus 29.73, 129.73 should have three significant figures 00:04:43.860 --> 00:04:47.360 because you know the forces we were given are 00:04:47.360 --> 00:04:51.060 385 and 475 and they had three significant figures so 00:04:51.160 --> 00:04:54.240 we'll have to round this to three significant figures which makes it into 130. 00:04:54.440 --> 00:04:58.700 So our final answer is the force C is 528 newtons, 00:04:58.700 --> 00:05:01.060 130 degrees and then we have to 00:05:01.060 --> 00:05:04.320 say how this angle is measured because 00:05:04.320 --> 00:05:06.000 it's not in standard position. 00:05:06.000 --> 00:05:08.740 If you just said 130 degrees and that's all you said 00:05:08.740 --> 00:05:11.760 then the assumption would have to be that it's in standard position 00:05:11.760 --> 00:05:14.300 going counter-clockwise from the positive x-axis 00:05:14.300 --> 00:05:15.660 which is not the case 00:05:15.700 --> 00:05:18.100 so we have to be specific and say that our angle is 00:05:18.100 --> 00:05:21.680 measured clockwise from force A.