WEBVTT 00:00:00.000 --> 00:00:02.840 This is Giancoli Answers with Mr. Dychko. 00:00:04.040 --> 00:00:07.240 With the pivot being at the elbow joint here, 00:00:07.320 --> 00:00:10.740 there are only three forces that enter into our torque equation. 00:00:10.800 --> 00:00:11.780 So we have the force 00:00:11.780 --> 00:00:14.060 due to the bicep muscle upwards here 00:00:14.320 --> 00:00:17.720 and its insertion point is here, 00:00:17.720 --> 00:00:19.860 6 centimeters away from the pivot, 00:00:20.520 --> 00:00:25.360 and we have the weight of the arm with its center of gravity here 00:00:25.360 --> 00:00:29.760 at some distance l a for lever arm of the arm 00:00:30.160 --> 00:00:33.660 and then this ball is in the hand here at the end of the arm 00:00:33.660 --> 00:00:41.060 with a mass M times g at some distance lever arm b, l b, from the elbow. 00:00:41.660 --> 00:00:44.320 So the clockwise torque equals the counter-clockwise torque 00:00:44.380 --> 00:00:47.740 so we have the torque due to the ball mg times l b 00:00:47.820 --> 00:00:52.120 plus the torque due to the arm, mass of the arm times g times 00:00:52.120 --> 00:00:55.180 distance from the center of mass of the arm to the elbow 00:00:55.380 --> 00:00:59.680 equals the force exerted by the muscle 00:00:59.740 --> 00:01:02.420 times the distance from the muscle to the elbow. 00:01:03.660 --> 00:01:06.860 And so M, the mass that can be held here, 00:01:06.920 --> 00:01:09.580 is gonna be, after we take this term to the right hand side 00:01:09.580 --> 00:01:13.340 by subtracting it from both sides and then dividing both sides by gl b, 00:01:13.660 --> 00:01:18.720 it's gonna be F Ml M minus mgl a divided by gl b. 00:01:18.960 --> 00:01:22.480 So we have to look at this picture [9-13a] 00:01:22.480 --> 00:01:24.600 to figure out all the distances here. 00:01:24.600 --> 00:01:29.020 So we have 450 newtons is the force applied by the bicep muscle 00:01:29.240 --> 00:01:33.160 times 6 centimeters converted into meters, 0.060 meters, 00:01:33.360 --> 00:01:37.380 minus 2 kilogram—mass of the arm— times 9.8 newtons per kilogram 00:01:37.500 --> 00:01:43.940 times 0.15 meters—distance from the center of gravity of the arm to the elbow joint— 00:01:44.180 --> 00:01:50.100 all divided by 9.8 newtons per kilogram times 0.35 meters—length of the arm— 00:01:50.320 --> 00:01:52.820 and that gives about 7.0 kilograms 00:01:52.820 --> 00:01:56.080 is the maximum ball mass that could be held.