WEBVTT 00:00:00.060 --> 00:00:02.806 This is College Physics Answers with Shaun Dychko. 00:00:03.700 --> 00:00:06.833 We're going to re-do this question that we've done before in an example 00:00:07.360 --> 00:00:12.693 but with the angle of between the arm and the bicep going to be 120 degrees. 00:00:12.960 --> 00:00:16.913 So the hand is down here and the book is on the hand down here. 00:00:17.726 --> 00:00:18.320 Okay. 00:00:18.320 --> 00:00:23.153 So, the torques that are going counter-clockwise 00:00:23.153 --> 00:00:24.966 of which there is only one due to the bicep 00:00:25.460 --> 00:00:27.800 has to equal the total torque going clockwise. 00:00:28.040 --> 00:00:30.326 We're assuming the pivot is at the elbow here 00:00:31.173 --> 00:00:37.233 and so the force exerted on the elbow has no term in our torque formula 00:00:37.233 --> 00:00:38.933 because its lever arm is zero. 00:00:39.926 --> 00:00:41.780 We're multiplying each of these forces 00:00:41.780 --> 00:00:46.646 by the perpendicular component of this distance to the elbow. 00:00:47.300 --> 00:00:50.886 So, this angle theta which is between the horizontal 00:00:50.886 --> 00:00:56.960 and the arm is going to be 120 minus 90, and that gives 30 degrees. 00:00:57.320 --> 00:01:01.680 So the horizontal component of each of these lengths here 00:01:01.826 --> 00:01:07.326 is going to be the hypotenuse which is the length, multiplied by cosine of thirty. 00:01:07.653 --> 00:01:12.113 If we consider this triangle for the bicep first of all, 00:01:13.293 --> 00:01:15.713 it's this triangle here where this is r b 00:01:15.926 --> 00:01:20.633 and this is r b perpendicular that we want to know, perpendicular to the force. 00:01:22.186 --> 00:01:25.186 This angle here is 30, that's a right triangle, 00:01:25.400 --> 00:01:29.760 and so we multiply the hypotenuse by cosine of 30 to get the adjacent. 00:01:32.080 --> 00:01:35.780 So we substitute that into each of these terms here 00:01:35.780 --> 00:01:41.573 and we have F b times r b cos theta equals force of weight on the arm, 00:01:41.740 --> 00:01:46.593 multiplied by the distance from the center of mass of the arm to the elbow, 00:01:46.760 --> 00:01:49.873 times cos theta, plus force of weight on the book, 00:01:50.506 --> 00:01:54.146 times the distance from the book to the elbow, multiplied by cos theta. 00:01:54.460 --> 00:01:57.773 But this is cos theta is a factor on every term 00:01:57.773 --> 00:02:01.186 and so we can divide both sides by cos theta and it disappears. 00:02:01.520 --> 00:02:06.520 Then also divide both sides by r b and you solve for the force due to the bicep. 00:02:07.766 --> 00:02:08.800 So then we plug in numbers. 00:02:09.100 --> 00:02:12.673 We have two and a half kilograms times 9.81 newtons per kilogram, 00:02:12.946 --> 00:02:15.553 this is the force of the weight of the arm, 00:02:15.746 --> 00:02:18.966 multiplied by its 16 centimeter distance to the elbow. 00:02:19.146 --> 00:02:22.240 I can use centimeters here because since we have centimeters on top 00:02:22.240 --> 00:02:24.866 and centimeters on the bottom, those units are going to cancel. 00:02:25.093 --> 00:02:28.133 It doesn't matter what the units are so long as they are the same on top and bottom. 00:02:28.813 --> 00:02:30.653 You could convert them to meters if you prefer. 00:02:31.740 --> 00:02:34.153 Then add to this the four kilogram mass of the book 00:02:34.153 --> 00:02:37.380 times 9.81 times 38 centimeters from the hand to the elbow, 00:02:37.753 --> 00:02:42.506 divided by four centimeters lever arm of the bicep, 00:02:42.646 --> 00:02:45.360 or you know, not quite the lever arm technically 00:02:45.360 --> 00:02:51.960 but it's the distance from where the bicep attaches to the arm bone, 00:02:52.066 --> 00:02:53.220 the radius or the ulna. 00:02:54.506 --> 00:02:57.406 Anyway this works out to 471 newtons 00:02:58.793 --> 00:03:02.060 which is the same as we had in the example by the way. 00:03:02.206 --> 00:03:04.340 The angle didn't make any difference because it canceled.