1 00:00:00,980 --> 00:00:04,160 Let's think about calculating torque. 2 00:00:04,160 --> 00:00:08,240 I have a force, F, being applied to an object. 3 00:00:08,240 --> 00:00:10,140 And that object is pivoted at a point, 4 00:00:10,140 --> 00:00:14,165 Q. I'll define R as the vector from the pivot point, Q, 5 00:00:14,165 --> 00:00:18,080 to the point of the application of the force. 6 00:00:18,080 --> 00:00:22,430 Torque is defined as the cross product, R cross F, 7 00:00:22,430 --> 00:00:25,040 which we can also write as R times F 8 00:00:25,040 --> 00:00:27,290 times the sine of theta. 9 00:00:27,290 --> 00:00:29,600 But what does that mean? 10 00:00:29,600 --> 00:00:32,270 I'll introduce two different intuitive ways of understanding 11 00:00:32,270 --> 00:00:34,340 this mathematical expression. 12 00:00:34,340 --> 00:00:38,000 First, we can think about this as the magnitude of F times 13 00:00:38,000 --> 00:00:40,940 the quantity, R sine theta. 14 00:00:40,940 --> 00:00:42,920 If we look at the direction of the force, 15 00:00:42,920 --> 00:00:46,010 then it is easy to see there R sine theta 16 00:00:46,010 --> 00:00:51,250 is just the part of R that is perpendicular to the force. 17 00:00:51,250 --> 00:00:53,940 So here the torque is equal to the magnitude 18 00:00:53,940 --> 00:00:56,070 of the force, times the magnitude 19 00:00:56,070 --> 00:00:59,250 of this perpendicular part of the vector R. 20 00:00:59,250 --> 00:01:02,070 This perpendicular portion of the distance vector R 21 00:01:02,070 --> 00:01:06,420 is often referred to as the moment arm. 22 00:01:06,420 --> 00:01:11,039 We can instead group F with sine theta. 23 00:01:11,039 --> 00:01:13,320 So this means that we're thinking of the cross-product 24 00:01:13,320 --> 00:01:17,400 as the distance R multiplied by F sine theta. 25 00:01:17,400 --> 00:01:19,830 The component of the force F, that's 26 00:01:19,830 --> 00:01:23,320 perpendicular to the position vector R. 27 00:01:23,320 --> 00:01:27,090 Here I'm calculating torque as the distance 28 00:01:27,090 --> 00:01:30,000 between the pivot point, and where the force is applied, 29 00:01:30,000 --> 00:01:32,490 times the perpendicular part of the force. 30 00:01:35,670 --> 00:01:39,000 Be very careful about the sine theta term. 31 00:01:39,000 --> 00:01:42,180 In many cases the angle theta that is given in the problem 32 00:01:42,180 --> 00:01:45,450 is not the angle that is used in the expression. 33 00:01:45,450 --> 00:01:49,280 In this example, F times the sine of theta, 34 00:01:49,280 --> 00:01:52,650 will actually give the parallel component of the force. 35 00:01:52,650 --> 00:01:54,330 Therefore it is much better to remember 36 00:01:54,330 --> 00:01:57,270 to take the perpendicular component, rather than just 37 00:01:57,270 --> 00:01:59,564 memorizing sine of theta.