[{"Name":"Basic Explanation","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"1.1 Intro","Duration":"3m 6s","ChapterTopicVideoID":9113,"CourseChapterTopicPlaylistID":5411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.320 ","End":"00:02.925","Text":"Hello. In this unit,"},{"Start":"00:02.925 ","End":"00:05.025","Text":"we\u0027re going to be speaking about torque."},{"Start":"00:05.025 ","End":"00:08.790","Text":"This is the introduction to that term. What does this mean?"},{"Start":"00:08.790 ","End":"00:11.160","Text":"Now, in America it\u0027s called torque,"},{"Start":"00:11.160 ","End":"00:15.720","Text":"but if you are coming from England or other English-speaking countries,"},{"Start":"00:15.720 ","End":"00:20.235","Text":"they might call torque the moment of force or the moment."},{"Start":"00:20.235 ","End":"00:23.160","Text":"Notice, if we\u0027re speaking about the moment of inertia,"},{"Start":"00:23.160 ","End":"00:26.730","Text":"we will say the entire phrase, moment of inertia."},{"Start":"00:26.730 ","End":"00:29.490","Text":"If you ever hear just the term moments alone,"},{"Start":"00:29.490 ","End":"00:33.670","Text":"it\u0027s referring to torque or the moment of force."},{"Start":"00:33.670 ","End":"00:37.790","Text":"Let\u0027s give a brief description of what this is."},{"Start":"00:37.790 ","End":"00:42.185","Text":"Say you go to a door and you can see where it\u0027s hinges are,"},{"Start":"00:42.185 ","End":"00:46.305","Text":"this is its axis of rotation."},{"Start":"00:46.305 ","End":"00:52.940","Text":"It\u0027s clear that if you push it very close to the axis of rotation,"},{"Start":"00:52.940 ","End":"00:56.780","Text":"or if you push it as far away as you can from the axis of"},{"Start":"00:56.780 ","End":"01:00.980","Text":"rotation when you\u0027re further away from the axis of rotation,"},{"Start":"01:00.980 ","End":"01:07.785","Text":"you\u0027ll have to apply much less force in order to close it."},{"Start":"01:07.785 ","End":"01:13.535","Text":"Here, this is r_2 and this distance over here will be r_1."},{"Start":"01:13.535 ","End":"01:18.575","Text":"Torque is speaking about the distance from the axis of rotation,"},{"Start":"01:18.575 ","End":"01:21.140","Text":"and therefore how much force you\u0027re going to have to"},{"Start":"01:21.140 ","End":"01:26.100","Text":"apply to rotate the object given the distance away."},{"Start":"01:26.680 ","End":"01:31.620","Text":"Let\u0027s take a look at the equation for torque."},{"Start":"01:32.390 ","End":"01:38.345","Text":"Over here we can see that this is the equation for torque and its symbol is Tau,"},{"Start":"01:38.345 ","End":"01:39.875","Text":"and it\u0027s a vector quantity,"},{"Start":"01:39.875 ","End":"01:42.980","Text":"because you can see that there\u0027s the arrow on top of the Tau."},{"Start":"01:42.980 ","End":"01:46.155","Text":"It equals to the r,"},{"Start":"01:46.155 ","End":"01:48.240","Text":"which is also a vector quantity."},{"Start":"01:48.240 ","End":"01:52.595","Text":"The r denotes the distance from the axis of rotation."},{"Start":"01:52.595 ","End":"01:58.880","Text":"Then we have the cross-product of the force."},{"Start":"01:58.880 ","End":"02:04.140","Text":"The force is also a vector quantity and it\u0027s the force being applied."},{"Start":"02:04.810 ","End":"02:13.620","Text":"Now another example for torque is if we\u0027re trying to screw in a screw."},{"Start":"02:13.620 ","End":"02:18.005","Text":"If you notice, if you\u0027re going to try and twist the screw with your hand,"},{"Start":"02:18.005 ","End":"02:21.860","Text":"it\u0027s going to be very difficult to screw whatever you need to do."},{"Start":"02:21.860 ","End":"02:24.874","Text":"However, if you add in a wrench,"},{"Start":"02:24.874 ","End":"02:32.965","Text":"because it increases the radius from the axis of rotation."},{"Start":"02:32.965 ","End":"02:36.245","Text":"Because before when you were trying to spin the screw,"},{"Start":"02:36.245 ","End":"02:42.365","Text":"you were over here and you can see that the blue arrow is much further away."},{"Start":"02:42.365 ","End":"02:47.135","Text":"You\u0027ll notice that when you are using a wrench or something like that,"},{"Start":"02:47.135 ","End":"02:51.460","Text":"then it\u0027s much easier to screw in the screw."},{"Start":"02:51.460 ","End":"02:55.670","Text":"That is to give you a bit of an idea of what torque means."},{"Start":"02:55.670 ","End":"02:59.105","Text":"Now we\u0027re going to see how we use this equation,"},{"Start":"02:59.105 ","End":"03:04.055","Text":"which is going to be our central equation for working out all questions."},{"Start":"03:04.055 ","End":"03:06.720","Text":"That\u0027s the end of this lesson."}],"ID":9386},{"Watched":false,"Name":"1.2 Right Hand Rule","Duration":"2m 30s","ChapterTopicVideoID":9114,"CourseChapterTopicPlaylistID":5411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.150","Text":"Hello. In this lesson we\u0027re going to be speaking about"},{"Start":"00:03.150 ","End":"00:06.540","Text":"positioning the z-axis. What does that mean?"},{"Start":"00:06.540 ","End":"00:11.970","Text":"That means positioning the z-axis relative to our x and y-axis."},{"Start":"00:11.970 ","End":"00:17.805","Text":"That also includes the direction that the z-axis will be in;"},{"Start":"00:17.805 ","End":"00:20.535","Text":"be it pointing upwards or downwards."},{"Start":"00:20.535 ","End":"00:23.475","Text":"Let\u0027s see what this actually means."},{"Start":"00:23.475 ","End":"00:33.255","Text":"If I have my x going in this direction and my y going in this direction,"},{"Start":"00:33.255 ","End":"00:36.020","Text":"so the way I decide the direction of my z-axis,"},{"Start":"00:36.020 ","End":"00:40.890","Text":"I have to go from my x until my y, like so."},{"Start":"00:40.890 ","End":"00:45.135","Text":"Then through using my right-hand rule,"},{"Start":"00:45.135 ","End":"00:53.145","Text":"I will see that my z-axis will be going in this upwards direction."},{"Start":"00:53.145 ","End":"00:55.265","Text":"Aside from the right-hand rule,"},{"Start":"00:55.265 ","End":"00:57.410","Text":"how did I know to do this?"},{"Start":"00:57.410 ","End":"00:59.390","Text":"There\u0027s a simple equation,"},{"Start":"00:59.390 ","End":"01:01.160","Text":"which is this right over here."},{"Start":"01:01.160 ","End":"01:07.140","Text":"It says that my z is equal to my x cross my y."},{"Start":"01:07.420 ","End":"01:11.285","Text":"By doing this cross-product,"},{"Start":"01:11.285 ","End":"01:14.500","Text":"I can see that I\u0027ll get my z-axis."},{"Start":"01:14.500 ","End":"01:17.820","Text":"Now over here I\u0027ve put a picture of the right-hand rule."},{"Start":"01:17.820 ","End":"01:23.195","Text":"You can see over here that the index finger is the x-axis,"},{"Start":"01:23.195 ","End":"01:25.955","Text":"the middle finger represents the y-axis,"},{"Start":"01:25.955 ","End":"01:32.780","Text":"and then the thumb represents the direction that the z-axis will be in."},{"Start":"01:32.780 ","End":"01:37.055","Text":"Similarly, if I would draw, for instance, again,"},{"Start":"01:37.055 ","End":"01:45.485","Text":"that my x-axis is going like this and then if I have my y-axis going in this direction,"},{"Start":"01:45.485 ","End":"01:52.085","Text":"so you can see that the shortest way to rotate from my x and to my y is going like this."},{"Start":"01:52.085 ","End":"01:55.595","Text":"You can see it\u0027s the anticlockwise direction."},{"Start":"01:55.595 ","End":"02:00.875","Text":"If you all put up your right hand and look using your hand,"},{"Start":"02:00.875 ","End":"02:02.450","Text":"the right-hand rule to see,"},{"Start":"02:02.450 ","End":"02:09.750","Text":"you\u0027ll see very easily that the z-axis will be pointing in the downwards direction."},{"Start":"02:10.580 ","End":"02:12.945","Text":"That is the end of the lesson."},{"Start":"02:12.945 ","End":"02:15.274","Text":"Remember this rule in this equation."},{"Start":"02:15.274 ","End":"02:18.815","Text":"Also, the right-hand rule,"},{"Start":"02:18.815 ","End":"02:21.470","Text":"which I\u0027ll square over here, which is important."},{"Start":"02:21.470 ","End":"02:29.000","Text":"Also, the equation for the z equals x cross y."},{"Start":"02:29.000 ","End":"02:31.800","Text":"That\u0027s the end of this lesson."}],"ID":9387},{"Watched":false,"Name":"1.3 Torque of Rod","Duration":"8m 37s","ChapterTopicVideoID":9115,"CourseChapterTopicPlaylistID":5411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.320 ","End":"00:02.670","Text":"Hello. In this lesson,"},{"Start":"00:02.670 ","End":"00:06.875","Text":"we\u0027re going to see an example of finding the torque of a rod."},{"Start":"00:06.875 ","End":"00:10.065","Text":"This is a very simple introductory example,"},{"Start":"00:10.065 ","End":"00:16.350","Text":"so this pink arrow here is going to be the axis of rotation and we have a yellow rod."},{"Start":"00:16.350 ","End":"00:18.630","Text":"Because this pink arrow is the axis of rotation,"},{"Start":"00:18.630 ","End":"00:24.090","Text":"we can see that it will rotate along like this."},{"Start":"00:24.560 ","End":"00:32.100","Text":"Now suppose that I were to apply some force to a random point,"},{"Start":"00:32.100 ","End":"00:33.885","Text":"let\u0027s say over here."},{"Start":"00:33.885 ","End":"00:37.335","Text":"This is my force and remember that it\u0027s a vector,"},{"Start":"00:37.335 ","End":"00:40.090","Text":"so we put an arrow on top."},{"Start":"00:40.900 ","End":"00:46.250","Text":"Now we can see that we take our distance,"},{"Start":"00:46.250 ","End":"00:51.490","Text":"the distance that the force is being applied to from the axis of rotation is r,"},{"Start":"00:51.490 ","End":"00:54.905","Text":"so we have r until this point,"},{"Start":"00:54.905 ","End":"00:58.400","Text":"and I remember is also a vector quantity."},{"Start":"00:58.400 ","End":"01:03.260","Text":"Now we remember that we have our equation for our torque,"},{"Start":"01:03.260 ","End":"01:07.375","Text":"which is Tau is equal to"},{"Start":"01:07.375 ","End":"01:14.475","Text":"our our r vector cross our F. Now this triangle is cross,"},{"Start":"01:14.475 ","End":"01:18.130","Text":"so this also equals this."},{"Start":"01:18.530 ","End":"01:22.030","Text":"Let\u0027s see how we do this."},{"Start":"01:22.390 ","End":"01:25.910","Text":"Now, notice that the angle over here between"},{"Start":"01:25.910 ","End":"01:30.190","Text":"the radius and the force is going to be 90 degrees."},{"Start":"01:30.190 ","End":"01:32.915","Text":"Now let\u0027s see how we work this out."},{"Start":"01:32.915 ","End":"01:39.560","Text":"Now, another important thing to remember is that whenever you have r cross F,"},{"Start":"01:39.560 ","End":"01:43.560","Text":"that the size of this."},{"Start":"01:43.560 ","End":"01:49.430","Text":"If we\u0027re going to see the size because over here the direction is less interesting,"},{"Start":"01:49.430 ","End":"01:53.105","Text":"but the size of the torque is most interesting,"},{"Start":"01:53.105 ","End":"02:02.734","Text":"so the size of the torque is equal to the size of our r multiplied by the size of our f,"},{"Start":"02:02.734 ","End":"02:06.845","Text":"and then sine of the angle."},{"Start":"02:06.845 ","End":"02:09.320","Text":"Here, in this case,"},{"Start":"02:09.320 ","End":"02:15.630","Text":"our Theta because we saw our Theta is equal to 90 degrees."},{"Start":"02:15.630 ","End":"02:18.960","Text":"Because we see it\u0027s at 90 degrees to the radius,"},{"Start":"02:18.960 ","End":"02:24.065","Text":"therefore our Tau will equal to,"},{"Start":"02:24.065 ","End":"02:25.790","Text":"let\u0027s write it over here."},{"Start":"02:25.790 ","End":"02:32.745","Text":"The size of our Tau will equal to the size of our r and the size"},{"Start":"02:32.745 ","End":"02:40.310","Text":"of F because sine Theta will therefore be equal to 1 because of this."},{"Start":"02:41.990 ","End":"02:45.195","Text":"Now what we\u0027ve figured out is"},{"Start":"02:45.195 ","End":"02:50.070","Text":"the size of our torque"},{"Start":"02:50.070 ","End":"02:55.200","Text":"but now we want to figure out which direction the z-axis will be going in,"},{"Start":"02:55.200 ","End":"02:57.470","Text":"in which direction our torque is."},{"Start":"02:57.470 ","End":"03:03.270","Text":"How are we going to do this? We\u0027re going to be doing this through our right-hand rule."},{"Start":"03:03.650 ","End":"03:06.305","Text":"Now, because of force as a vector,"},{"Start":"03:06.305 ","End":"03:10.640","Text":"we learned earlier on in this course that we can move"},{"Start":"03:10.640 ","End":"03:16.100","Text":"the arrow anywhere as long as we keep its size and the direction that it\u0027s pointing in."},{"Start":"03:16.100 ","End":"03:20.240","Text":"So what I\u0027m going to do is I\u0027m going to move"},{"Start":"03:20.240 ","End":"03:26.415","Text":"this arrow over to here, keeping the direction."},{"Start":"03:26.415 ","End":"03:32.585","Text":"This is going to be the force and I\u0027ve kept the size and the direction of the arrow,"},{"Start":"03:32.585 ","End":"03:35.260","Text":"so I can rub this out now."},{"Start":"03:35.260 ","End":"03:37.445","Text":"Now I can say,"},{"Start":"03:37.445 ","End":"03:44.105","Text":"because I can look into my equation and I see that my Tau is equal to my r cross F."},{"Start":"03:44.105 ","End":"03:53.270","Text":"here my r can represent my x in the equation and my F can represent my y,"},{"Start":"03:53.270 ","End":"03:57.840","Text":"if we remember, if we look back to our original equation."},{"Start":"03:57.890 ","End":"04:04.580","Text":"We can see that our equation is z is equal to x cross y,"},{"Start":"04:04.580 ","End":"04:07.925","Text":"so now we\u0027re going to be looking our r,"},{"Start":"04:07.925 ","End":"04:10.285","Text":"because it\u0027s our first term,"},{"Start":"04:10.285 ","End":"04:12.875","Text":"so we can see like this,"},{"Start":"04:12.875 ","End":"04:15.570","Text":"this is like this,"},{"Start":"04:15.570 ","End":"04:19.340","Text":"so we can say that this is our x-axis and"},{"Start":"04:19.340 ","End":"04:23.765","Text":"that the direction that our force is going in is our y-axis,"},{"Start":"04:23.765 ","End":"04:26.830","Text":"so I\u0027ll just write x hat and y hat."},{"Start":"04:26.830 ","End":"04:29.975","Text":"Now if we use our right-hand rule,"},{"Start":"04:29.975 ","End":"04:33.230","Text":"everyone lift up your right hand and we can"},{"Start":"04:33.230 ","End":"04:41.345","Text":"see that when we\u0027re going from our x until our y,"},{"Start":"04:41.345 ","End":"04:46.410","Text":"the shortest route is to go in this clockwise direction."},{"Start":"04:46.410 ","End":"04:50.885","Text":"If you all do that, if you raise your right hand and you move your index finger"},{"Start":"04:50.885 ","End":"04:54.890","Text":"in the clockwise direction to the y-axis,"},{"Start":"04:54.890 ","End":"04:56.270","Text":"to your middle finger,"},{"Start":"04:56.270 ","End":"04:59.899","Text":"you\u0027ll see that your thumb is pointing in the upwards direction,"},{"Start":"04:59.899 ","End":"05:08.750","Text":"which means that the z-axis will also be pointing in the upward direction."},{"Start":"05:10.190 ","End":"05:17.165","Text":"now we have the size of the torque and also the direction that it\u0027s in,"},{"Start":"05:17.165 ","End":"05:21.995","Text":"we\u0027ve done this using more intuition and using the right-hand rule."},{"Start":"05:21.995 ","End":"05:28.775","Text":"Now let\u0027s see mathematically how we do this cross-product over here,"},{"Start":"05:28.775 ","End":"05:33.600","Text":"where we actually do the calculation."},{"Start":"05:34.160 ","End":"05:39.900","Text":"We\u0027re going to work out this cross-product,"},{"Start":"05:39.900 ","End":"05:43.085","Text":"so what we\u0027re going to do is we can see that our force"},{"Start":"05:43.085 ","End":"05:47.870","Text":"F is going in the direction that we\u0027ve said is our y-axis."},{"Start":"05:47.870 ","End":"05:52.290","Text":"Let\u0027s just give our F the name F,"},{"Start":"05:52.290 ","End":"05:58.605","Text":"so our F vector is equal to F_0 in the y-direction."},{"Start":"05:58.605 ","End":"06:04.760","Text":"Now, we can also see that our r is also only going in the x-direction so"},{"Start":"06:04.760 ","End":"06:11.280","Text":"we can say that our r vector is equal to x in the x-direction."},{"Start":"06:11.280 ","End":"06:16.050","Text":"The F has no values in the x or z-direction,"},{"Start":"06:16.050 ","End":"06:22.650","Text":"and the r has no values in the y or z-direction which we can see from our sketch above."},{"Start":"06:22.760 ","End":"06:24.915","Text":"Now let\u0027s see,"},{"Start":"06:24.915 ","End":"06:28.740","Text":"so our Tau our torque is equal to,"},{"Start":"06:28.740 ","End":"06:31.995","Text":"and now we\u0027re going to work out the determinant,"},{"Start":"06:31.995 ","End":"06:39.315","Text":"so the cross product of these 2 variables."},{"Start":"06:39.315 ","End":"06:43.373","Text":"We\u0027re going to have our x-direction, our y-direction,"},{"Start":"06:43.373 ","End":"06:46.890","Text":"and our z-direction then we\u0027ll fit in,"},{"Start":"06:46.890 ","End":"06:48.495","Text":"so we see that our r,"},{"Start":"06:48.495 ","End":"06:51.895","Text":"we go r cross F. First we\u0027ll add in our r,"},{"Start":"06:51.895 ","End":"06:54.065","Text":"so we can see in the x-direction,"},{"Start":"06:54.065 ","End":"06:58.225","Text":"we have x and then y and z-direction we have 0,"},{"Start":"06:58.225 ","End":"07:00.755","Text":"because we have no terms for y and z."},{"Start":"07:00.755 ","End":"07:03.200","Text":"Then we\u0027ll put in our F variable,"},{"Start":"07:03.200 ","End":"07:04.970","Text":"so in the x-direction,"},{"Start":"07:04.970 ","End":"07:09.570","Text":"we have 0, in the y-direction here we have F_0."},{"Start":"07:10.220 ","End":"07:16.330","Text":"In the z-direction again we have no value, so 0 again."},{"Start":"07:16.370 ","End":"07:19.880","Text":"Now if we work out the determinant,"},{"Start":"07:19.880 ","End":"07:25.780","Text":"we\u0027ll see that our values for the x-axis will equal 0,"},{"Start":"07:25.780 ","End":"07:29.760","Text":"and then our values for the y-axis will have negative,"},{"Start":"07:29.760 ","End":"07:33.485","Text":"our values for the y-axis will also be 0 in"},{"Start":"07:33.485 ","End":"07:37.805","Text":"the y-direction and then in our z-direction, we\u0027ll have positive."},{"Start":"07:37.805 ","End":"07:40.230","Text":"Then we\u0027ll have F_0,"},{"Start":"07:40.610 ","End":"07:48.020","Text":"x going in the z-direction because we can see here that if we cross this,"},{"Start":"07:48.020 ","End":"07:53.625","Text":"so we have x times F-0 minus 0 times 0."},{"Start":"07:53.625 ","End":"07:57.930","Text":"Now we can really see that we have 0 in the x-direction and"},{"Start":"07:57.930 ","End":"08:02.300","Text":"0 in the y-direction and we have this over here,"},{"Start":"08:02.300 ","End":"08:08.175","Text":"a positive in the z-direction."},{"Start":"08:08.175 ","End":"08:10.545","Text":"Then we have the size,"},{"Start":"08:10.545 ","End":"08:16.325","Text":"so the size of our force is F_0 and the size of our radius is x."},{"Start":"08:16.325 ","End":"08:20.130","Text":"The size means just without the direction,"},{"Start":"08:20.800 ","End":"08:24.740","Text":"so this is the size of our force and this is the size of"},{"Start":"08:24.740 ","End":"08:28.175","Text":"our radius and they\u0027re both going in the positive z-direction."},{"Start":"08:28.175 ","End":"08:32.515","Text":"We can see that the answer that we got is the correct answer."},{"Start":"08:32.515 ","End":"08:37.980","Text":"That\u0027s the end of this example for the torque of a rod."}],"ID":9388},{"Watched":false,"Name":"1.4 R Effective","Duration":"5m 19s","ChapterTopicVideoID":9116,"CourseChapterTopicPlaylistID":5411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.320 ","End":"00:05.835","Text":"Hello. In this lesson we\u0027re going to be speaking about the effective distance,"},{"Start":"00:05.835 ","End":"00:11.685","Text":"which is written as r vector F for effective."},{"Start":"00:11.685 ","End":"00:17.680","Text":"This is a trick for shortening the calculation for working out the torque."},{"Start":"00:18.500 ","End":"00:23.085","Text":"We know that we have this equation over here for the torque"},{"Start":"00:23.085 ","End":"00:27.495","Text":"is equal to r vector cross our force vector."},{"Start":"00:27.495 ","End":"00:30.014","Text":"Where our r, if you remember,"},{"Start":"00:30.014 ","End":"00:33.295","Text":"is the distance from the axis."},{"Start":"00:33.295 ","End":"00:38.345","Text":"From the axis of rotation and our F is the force being applied."},{"Start":"00:38.345 ","End":"00:43.498","Text":"We saw that there are 2 ways to solve this already."},{"Start":"00:43.498 ","End":"00:47.844","Text":"We\u0027ve seen the one that does the determinant way."},{"Start":"00:47.844 ","End":"00:54.800","Text":"Then the second way is by saying that the size of"},{"Start":"00:54.800 ","End":"01:01.535","Text":"the vector is equal to the size of our r cross our F,"},{"Start":"01:01.535 ","End":"01:07.910","Text":"which we said was equal to the size of our r multiplied by the size of our F,"},{"Start":"01:07.910 ","End":"01:13.000","Text":"multiplied by sine of the angle between them."},{"Start":"01:13.000 ","End":"01:17.948","Text":"Now in our lesson with the r effective,"},{"Start":"01:17.948 ","End":"01:21.950","Text":"our method that we\u0027re going to speak about now is similar to this method,"},{"Start":"01:21.950 ","End":"01:24.715","Text":"so similar to method number 2."},{"Start":"01:24.715 ","End":"01:29.360","Text":"The trick here is to take the component of one of"},{"Start":"01:29.360 ","End":"01:34.250","Text":"the vectors which is perpendicular to the other vector."},{"Start":"01:34.250 ","End":"01:38.540","Text":"Usually we take one of the components from the r vector,"},{"Start":"01:38.540 ","End":"01:41.835","Text":"which is perpendicular to the F vector."},{"Start":"01:41.835 ","End":"01:45.105","Text":"That\u0027s why we call this r effective."},{"Start":"01:45.105 ","End":"01:47.450","Text":"When we\u0027re speaking about r effective,"},{"Start":"01:47.450 ","End":"01:51.950","Text":"we\u0027re speaking about taking the component of either the r or F vector,"},{"Start":"01:51.950 ","End":"01:55.235","Text":"which is perpendicular to the other vector."},{"Start":"01:55.235 ","End":"02:00.205","Text":"This trick is useful when it\u0027s hard to work out the angle Alpha."},{"Start":"02:00.205 ","End":"02:03.920","Text":"Then we use this trick, usually we\u0027ll use a component from"},{"Start":"02:03.920 ","End":"02:08.670","Text":"the r vector and that is why it is called r effective."},{"Start":"02:09.320 ","End":"02:12.645","Text":"Let\u0027s see an example."},{"Start":"02:12.645 ","End":"02:18.245","Text":"Imagine that I have my r vector going in this direction,"},{"Start":"02:18.245 ","End":"02:24.380","Text":"and that I have my force vector going in this direction."},{"Start":"02:24.380 ","End":"02:30.365","Text":"Then I have my angle Alpha over here and it\u0027s given to me in the question."},{"Start":"02:30.365 ","End":"02:33.630","Text":"If I\u0027m going to use method number 2."},{"Start":"02:33.630 ","End":"02:35.585","Text":"Let\u0027s take a look."},{"Start":"02:35.585 ","End":"02:37.610","Text":"If I\u0027m going to say,"},{"Start":"02:37.610 ","End":"02:41.785","Text":"see what my r sine of Alpha is going to be."},{"Start":"02:41.785 ","End":"02:45.545","Text":"What this will mean is if I break this up,"},{"Start":"02:45.545 ","End":"02:48.155","Text":"it will mean I break it up into my components."},{"Start":"02:48.155 ","End":"02:51.440","Text":"I\u0027ll have my component going in this direction,"},{"Start":"02:51.440 ","End":"02:53.990","Text":"which will be my r cosine Alpha,"},{"Start":"02:53.990 ","End":"02:56.615","Text":"which is given from the dot product."},{"Start":"02:56.615 ","End":"02:59.585","Text":"But here because I\u0027m using the cross product."},{"Start":"02:59.585 ","End":"03:02.740","Text":"I will get this over here."},{"Start":"03:02.740 ","End":"03:06.075","Text":"Which is my r sine of Alpha."},{"Start":"03:06.075 ","End":"03:15.660","Text":"This is r sine of Alpha and it is at 90 degrees to my force vector."},{"Start":"03:16.370 ","End":"03:22.700","Text":"Then we\u0027ll also notice that my r sine of Alpha is the component of my r which is"},{"Start":"03:22.700 ","End":"03:30.090","Text":"perpendicular to my F. This connects to over here."},{"Start":"03:30.090 ","End":"03:33.830","Text":"This is exactly what my statement and Number 3 means."},{"Start":"03:33.830 ","End":"03:36.770","Text":"We\u0027re taking the components of the r vector."},{"Start":"03:36.770 ","End":"03:38.210","Text":"That\u0027s what we did over here,"},{"Start":"03:38.210 ","End":"03:41.780","Text":"which is perpendicular to my other vector,"},{"Start":"03:41.780 ","End":"03:43.450","Text":"which is my F vector."},{"Start":"03:43.450 ","End":"03:47.115","Text":"The component is r sine of Alpha."},{"Start":"03:47.115 ","End":"03:50.700","Text":"I\u0027m taking the size of my r vector,"},{"Start":"03:50.700 ","End":"03:55.200","Text":"which is just r and I\u0027m multiplying it by sine Alpha."},{"Start":"03:55.200 ","End":"03:59.505","Text":"I\u0027m getting the perpendicular component to the F vector."},{"Start":"03:59.505 ","End":"04:01.885","Text":"This is what this means."},{"Start":"04:01.885 ","End":"04:06.140","Text":"Now, of course this can go also the other direction."},{"Start":"04:06.140 ","End":"04:10.385","Text":"If I again draw that this is the direction of my r vector."},{"Start":"04:10.385 ","End":"04:15.365","Text":"That this is the direction of my F vector."},{"Start":"04:15.365 ","End":"04:22.505","Text":"Instead of taking the size of my r I can take the size of my F and sine Alpha."},{"Start":"04:22.505 ","End":"04:27.260","Text":"Then I will get the component which is perpendicular to"},{"Start":"04:27.260 ","End":"04:31.625","Text":"my r. If I break up my F into its separate components,"},{"Start":"04:31.625 ","End":"04:35.585","Text":"so I\u0027ll have the component going in this direction,"},{"Start":"04:35.585 ","End":"04:37.775","Text":"which is also the same as this."},{"Start":"04:37.775 ","End":"04:39.740","Text":"Notice it\u0027s at 90 degrees."},{"Start":"04:39.740 ","End":"04:43.620","Text":"Then I\u0027ll also have this component."},{"Start":"04:44.390 ","End":"04:51.015","Text":"You can see that this will be my F sine of Alpha."},{"Start":"04:51.015 ","End":"04:54.960","Text":"Here is my Alpha, my F sine of Alpha."},{"Start":"04:54.960 ","End":"04:58.730","Text":"Again, so I\u0027ve taken my component from my F vector,"},{"Start":"04:58.730 ","End":"05:03.450","Text":"which is perpendicular to my r vector."},{"Start":"05:03.800 ","End":"05:10.330","Text":"That is all that this statement means and we just call this r effective."},{"Start":"05:10.340 ","End":"05:12.620","Text":"That\u0027s the end of this lesson."},{"Start":"05:12.620 ","End":"05:15.770","Text":"The next lesson, we\u0027re going to see an example of how"},{"Start":"05:15.770 ","End":"05:19.980","Text":"to use r effective in a calculation to work out the torque."}],"ID":9389},{"Watched":false,"Name":"1.5 R Effective Example","Duration":"5m 7s","ChapterTopicVideoID":9117,"CourseChapterTopicPlaylistID":5411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.320 ","End":"00:02.985","Text":"Hello. In the last lesson,"},{"Start":"00:02.985 ","End":"00:07.830","Text":"we learned about how to use r_effective when working out the torque."},{"Start":"00:07.830 ","End":"00:11.024","Text":"Now we\u0027re going to show an example."},{"Start":"00:11.024 ","End":"00:17.490","Text":"Here there\u0027s a man pushing a box and say that the height of the box is half a meter,"},{"Start":"00:17.490 ","End":"00:21.885","Text":"and man is pushing the box at a force of 10 newtons."},{"Start":"00:21.885 ","End":"00:24.910","Text":"Now imagine that there\u0027s no friction in this question,"},{"Start":"00:24.910 ","End":"00:27.780","Text":"so the box is freely moving on the ground."},{"Start":"00:27.780 ","End":"00:34.493","Text":"Now imagine that eventually the box hits some rock in the way,"},{"Start":"00:34.493 ","End":"00:40.180","Text":"which means that the box is now going to rotate."},{"Start":"00:40.180 ","End":"00:43.025","Text":"It\u0027s going to move in this type of direction."},{"Start":"00:43.025 ","End":"00:45.590","Text":"This block over here,"},{"Start":"00:45.590 ","End":"00:50.609","Text":"this rock in the middle of the way has turned into its axis of rotation."},{"Start":"00:50.950 ","End":"00:54.290","Text":"Now, it\u0027s important to note that this is"},{"Start":"00:54.290 ","End":"00:57.680","Text":"the axis of rotation and not a point of rotation."},{"Start":"00:57.680 ","End":"00:59.615","Text":"This isn\u0027t a point."},{"Start":"00:59.615 ","End":"01:03.215","Text":"It\u0027s going to be some axis going in or out of the page."},{"Start":"01:03.215 ","End":"01:04.665","Text":"It doesn\u0027t really matter right now,"},{"Start":"01:04.665 ","End":"01:09.200","Text":"say that it\u0027s coming out of the page and it\u0027s always an axis of rotation."},{"Start":"01:09.200 ","End":"01:11.750","Text":"The box is going to rotate about this axis."},{"Start":"01:11.750 ","End":"01:15.635","Text":"Now, when the axis is going either in or out of the page,"},{"Start":"01:15.635 ","End":"01:17.930","Text":"then I won\u0027t draw it."},{"Start":"01:17.930 ","End":"01:25.640","Text":"But just know that if there\u0027s ever some point which is being highlighted,"},{"Start":"01:25.640 ","End":"01:27.660","Text":"then it\u0027s not a point of rotation,"},{"Start":"01:27.660 ","End":"01:29.465","Text":"it\u0027s still an axis of rotation."},{"Start":"01:29.465 ","End":"01:36.810","Text":"But I\u0027m not going to draw the 3-dimensional axis coming in or out of the page. Moving on."},{"Start":"01:37.730 ","End":"01:40.785","Text":"What we\u0027re going to do now is,"},{"Start":"01:40.785 ","End":"01:47.030","Text":"we\u0027re going to see what our radius is."},{"Start":"01:47.030 ","End":"01:54.242","Text":"Our axis of rotation until the point where our force is being applied,"},{"Start":"01:54.242 ","End":"01:56.925","Text":"so that\u0027s going to be that direction."},{"Start":"01:56.925 ","End":"02:04.055","Text":"Over here, this is going to be r. Over here,"},{"Start":"02:04.055 ","End":"02:09.560","Text":"this is going to be our Alpha angle between the force and our radius."},{"Start":"02:09.560 ","End":"02:15.550","Text":"Now notice that we don\u0027t know what our r is,"},{"Start":"02:15.550 ","End":"02:20.130","Text":"and we also don\u0027t know what our Alpha is."},{"Start":"02:20.130 ","End":"02:28.305","Text":"If we look back to our equation for our torque,"},{"Start":"02:28.305 ","End":"02:32.925","Text":"it\u0027s a tau and not a T. It equals to the size of"},{"Start":"02:32.925 ","End":"02:39.465","Text":"r and the size of f multiplied by sine of the angle."},{"Start":"02:39.465 ","End":"02:41.755","Text":"This here is an unknown,"},{"Start":"02:41.755 ","End":"02:44.480","Text":"and this here is an unknown."},{"Start":"02:44.480 ","End":"02:46.080","Text":"What we\u0027re going to do is,"},{"Start":"02:46.080 ","End":"02:48.890","Text":"we\u0027re going to use our r_effective in order to"},{"Start":"02:48.890 ","End":"02:52.400","Text":"solve this because that\u0027s the way that we know how to use this."},{"Start":"02:52.400 ","End":"02:56.885","Text":"If we break up our r into its components,"},{"Start":"02:56.885 ","End":"03:06.645","Text":"so we\u0027ll see that we have a component going over here in this direction,"},{"Start":"03:06.645 ","End":"03:12.170","Text":"and we also have a component going in this direction."},{"Start":"03:12.170 ","End":"03:15.170","Text":"Now, currently ignore the direction that"},{"Start":"03:15.170 ","End":"03:18.635","Text":"the arrows are facing because I realized that they are incorrect."},{"Start":"03:18.635 ","End":"03:28.145","Text":"But you can see that here we have this one is r sine of Alpha."},{"Start":"03:28.145 ","End":"03:30.015","Text":"Because if this is Alpha,"},{"Start":"03:30.015 ","End":"03:31.945","Text":"because of alternating angles,"},{"Start":"03:31.945 ","End":"03:33.890","Text":"this is also going to be Alpha,"},{"Start":"03:33.890 ","End":"03:37.315","Text":"and then this side is r cosine of Alpha."},{"Start":"03:37.315 ","End":"03:40.900","Text":"Now notice that because we said that our height is half a meter,"},{"Start":"03:40.900 ","End":"03:45.880","Text":"so our height is equal to half a meter,"},{"Start":"03:45.880 ","End":"03:50.185","Text":"which is also equal to r sine of Alpha."},{"Start":"03:50.185 ","End":"03:54.755","Text":"Also notice that it is at 90 degrees."},{"Start":"03:54.755 ","End":"03:58.810","Text":"That means that we\u0027re taking the component of r,"},{"Start":"03:58.810 ","End":"04:04.780","Text":"which is perpendicular at 90 degrees to our component of the other vector,"},{"Start":"04:04.780 ","End":"04:07.895","Text":"which in this case is going to be our force vector."},{"Start":"04:07.895 ","End":"04:13.545","Text":"In that case, we can just write in,"},{"Start":"04:13.545 ","End":"04:17.550","Text":"so therefore that the size of"},{"Start":"04:17.550 ","End":"04:25.175","Text":"our torque is equal to the size of our force,"},{"Start":"04:25.175 ","End":"04:26.750","Text":"which we know what that is,"},{"Start":"04:26.750 ","End":"04:31.675","Text":"it\u0027s 10 newtons multiplied by our r_effective."},{"Start":"04:31.675 ","End":"04:38.100","Text":"That means, so we know that our force is 10 newtons,"},{"Start":"04:38.100 ","End":"04:43.050","Text":"and we know that r_effective is equal to half a meter."},{"Start":"04:43.050 ","End":"04:49.545","Text":"Then we\u0027re just going to have 10 multiplied by 0.5,"},{"Start":"04:49.545 ","End":"04:53.940","Text":"and then we\u0027re going to get that that equals to 5."},{"Start":"04:53.940 ","End":"04:58.800","Text":"Then the units, because we have 10 newtons multiplied by half a meters,"},{"Start":"04:58.800 ","End":"05:04.935","Text":"so the units for torque are newton meters, written like that."},{"Start":"05:04.935 ","End":"05:08.710","Text":"That\u0027s the end of this lesson."}],"ID":9390},{"Watched":false,"Name":"1.6 Choosing An Angle","Duration":"2m ","ChapterTopicVideoID":9118,"CourseChapterTopicPlaylistID":5411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.490","Text":"Hello. In this lesson,"},{"Start":"00:02.490 ","End":"00:04.890","Text":"we\u0027re going to be speaking about choosing an angle."},{"Start":"00:04.890 ","End":"00:06.960","Text":"In 1 of the previous lessons,"},{"Start":"00:06.960 ","End":"00:08.880","Text":"we saw that if we had,"},{"Start":"00:08.880 ","End":"00:10.485","Text":"say on the x-axis,"},{"Start":"00:10.485 ","End":"00:12.970","Text":"we had a rod."},{"Start":"00:13.040 ","End":"00:19.245","Text":"Then we saw that if our radius was up until here,"},{"Start":"00:19.245 ","End":"00:21.150","Text":"this was our i."},{"Start":"00:21.150 ","End":"00:27.300","Text":"Then we had a force hitting the rod at 90 degrees perpendicular."},{"Start":"00:27.300 ","End":"00:29.805","Text":"We saw how to do the calculation."},{"Start":"00:29.805 ","End":"00:32.355","Text":"But now what I want to speak about is,"},{"Start":"00:32.355 ","End":"00:36.390","Text":"let\u0027s say that our force isn\u0027t hitting the right at 90 degrees,"},{"Start":"00:36.390 ","End":"00:42.000","Text":"but rather is hitting the rod at some angle Alpha."},{"Start":"00:42.000 ","End":"00:44.330","Text":"For the sake of this example,"},{"Start":"00:44.330 ","End":"00:48.660","Text":"let\u0027s say that our Alpha is equal to 30 degrees."},{"Start":"00:48.860 ","End":"00:52.115","Text":"Because it\u0027s a vector,"},{"Start":"00:52.115 ","End":"00:54.155","Text":"so as long as I keep its size and direction,"},{"Start":"00:54.155 ","End":"00:59.435","Text":"I can move it to any point on the screen that I want."},{"Start":"00:59.435 ","End":"01:02.030","Text":"I\u0027m going to move it up until here."},{"Start":"01:02.030 ","End":"01:05.780","Text":"Then we can see because of alternate angles that this will"},{"Start":"01:05.780 ","End":"01:09.920","Text":"also be 30 degrees, which equals Alpha."},{"Start":"01:09.920 ","End":"01:16.130","Text":"They\u0027re the same. You don\u0027t have to worry if you\u0027re taking this angle,"},{"Start":"01:16.130 ","End":"01:20.870","Text":"this angle Alpha, or if you are taking this angle over here,"},{"Start":"01:20.870 ","End":"01:23.030","Text":"which will be a 180 minus Alpha,"},{"Start":"01:23.030 ","End":"01:25.775","Text":"so this angle will be a 150 degrees."},{"Start":"01:25.775 ","End":"01:28.100","Text":"It doesn\u0027t matter. You can choose any,"},{"Start":"01:28.100 ","End":"01:36.635","Text":"because if you look at the sin(180 minus Alpha),"},{"Start":"01:36.635 ","End":"01:41.640","Text":"it will always be the sin(Alpha)."},{"Start":"01:41.780 ","End":"01:44.190","Text":"You\u0027ll get the same answer."},{"Start":"01:44.190 ","End":"01:46.820","Text":"All you have to do is do the sine of"},{"Start":"01:46.820 ","End":"01:51.140","Text":"the angle to get your force in the perpendicular direction,"},{"Start":"01:51.140 ","End":"01:57.810","Text":"and then you can carry on with your calculation."},{"Start":"01:57.810 ","End":"02:00.700","Text":"That\u0027s the end of this lesson."}],"ID":9391},{"Watched":false,"Name":"1.7 Positive Direction Of Rotation","Duration":"4m 16s","ChapterTopicVideoID":9119,"CourseChapterTopicPlaylistID":5411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.380 ","End":"00:02.850","Text":"Hello. In this lesson,"},{"Start":"00:02.850 ","End":"00:05.640","Text":"we\u0027re going to be using the rotating right-hand rule"},{"Start":"00:05.640 ","End":"00:10.005","Text":"in order to find the positive direction of rotation."},{"Start":"00:10.005 ","End":"00:16.725","Text":"If you all right now lift up your right hand and give a thumbs up."},{"Start":"00:16.725 ","End":"00:21.210","Text":"You can see that the thumb on this drawing,"},{"Start":"00:21.210 ","End":"00:28.995","Text":"the thumb represents this arrow over here pointing in this direction."},{"Start":"00:28.995 ","End":"00:35.840","Text":"This is the thumb and that represents the axis of rotation."},{"Start":"00:35.840 ","End":"00:41.670","Text":"Then what your fingers represent is this."},{"Start":"00:41.670 ","End":"00:49.860","Text":"Your fingers are going in this direction and this means that this is this arrow."},{"Start":"00:49.860 ","End":"00:55.020","Text":"This is the positive direction of rotation."},{"Start":"00:55.020 ","End":"00:57.520","Text":"If your z is pointing upwards,"},{"Start":"00:57.520 ","End":"01:01.025","Text":"then your positive direction of rotation is"},{"Start":"01:01.025 ","End":"01:06.850","Text":"the direction that your fingers are curled in and rotate in."},{"Start":"01:07.790 ","End":"01:10.269","Text":"Now let\u0027s give an example."},{"Start":"01:10.269 ","End":"01:13.075","Text":"For instance, if I have a graph."},{"Start":"01:13.075 ","End":"01:18.925","Text":"This is my z-axis and this is my x-axis,"},{"Start":"01:18.925 ","End":"01:23.215","Text":"and this is my y-axis over here."},{"Start":"01:23.215 ","End":"01:28.360","Text":"If the z-axis in this direction is the positive direction,"},{"Start":"01:28.360 ","End":"01:33.160","Text":"then if we use our rotating right-hand rule,"},{"Start":"01:33.160 ","End":"01:38.770","Text":"we\u0027ll see that if we have some rod over here on the y-axis,"},{"Start":"01:38.770 ","End":"01:41.440","Text":"it\u0027s always going to be rotating on the xy-plane."},{"Start":"01:41.440 ","End":"01:43.960","Text":"Using our rotating right-hand rule,"},{"Start":"01:43.960 ","End":"01:46.750","Text":"we can see that it\u0027s positive direction of"},{"Start":"01:46.750 ","End":"01:51.465","Text":"rotation will be in this direction anticlockwise."},{"Start":"01:51.465 ","End":"01:56.035","Text":"Now, I\u0027m actually going to change the color of the rod,"},{"Start":"01:56.035 ","End":"01:58.180","Text":"so that it doesn\u0027t confuse you."},{"Start":"01:58.180 ","End":"02:02.080","Text":"Say that the rod is this green color because I don\u0027t"},{"Start":"02:02.080 ","End":"02:05.485","Text":"want to confuse you and the pink represents the thumb,"},{"Start":"02:05.485 ","End":"02:08.855","Text":"which represents the axis of rotation."},{"Start":"02:08.855 ","End":"02:11.260","Text":"I\u0027ve color-coordinated everything."},{"Start":"02:11.260 ","End":"02:15.040","Text":"The blue arrow represents the positive direction of rotation,"},{"Start":"02:15.040 ","End":"02:18.010","Text":"the pink arrow represents the axis of rotation,"},{"Start":"02:18.010 ","End":"02:22.945","Text":"and this green is just the rod which is rotating on the xy-plane."},{"Start":"02:22.945 ","End":"02:28.240","Text":"Now, if we go back to the example that we saw a few lessons ago,"},{"Start":"02:28.240 ","End":"02:32.440","Text":"where we said that this is i vector that\u0027s"},{"Start":"02:32.440 ","End":"02:38.670","Text":"perpendicular to it our I vector we had a force vector pushing the rod."},{"Start":"02:41.680 ","End":"02:51.235","Text":"Then we said that our momentum would be the cross product between these 2 vectors."},{"Start":"02:51.235 ","End":"03:02.920","Text":"We said that our momentum would be the vector representing this direction on the z-axis."},{"Start":"03:03.710 ","End":"03:07.750","Text":"This was our torque."},{"Start":"03:08.480 ","End":"03:13.190","Text":"A torque, because we can see that it\u0027s pointing in"},{"Start":"03:13.190 ","End":"03:17.865","Text":"the positive direction of the z-axis in the positive direction."},{"Start":"03:17.865 ","End":"03:21.170","Text":"We can say that our torque is positive."},{"Start":"03:21.170 ","End":"03:24.905","Text":"Another way that we can say this and it\u0027s the exact same thing,"},{"Start":"03:24.905 ","End":"03:31.735","Text":"is we can see that our force is pushing the rod in a positive direction of rotation."},{"Start":"03:31.735 ","End":"03:34.230","Text":"If we say that this is the positive."},{"Start":"03:34.230 ","End":"03:39.125","Text":"Right now, he said that the anticlockwise direction is the positive direction."},{"Start":"03:39.125 ","End":"03:44.675","Text":"Because we can see our force is pushing the rod in the positive direction of rotation,"},{"Start":"03:44.675 ","End":"03:47.435","Text":"so we can see that our vector for our torque is"},{"Start":"03:47.435 ","End":"03:50.625","Text":"also going to be in the positive direction."},{"Start":"03:50.625 ","End":"03:52.935","Text":"It\u0027s going to be a positive vector."},{"Start":"03:52.935 ","End":"03:56.860","Text":"Those things mean the exact same."},{"Start":"03:56.990 ","End":"04:04.460","Text":"I wrote down now that the torque is positive if its direction is in the positive,"},{"Start":"04:04.460 ","End":"04:07.160","Text":"z-direction or the force is in"},{"Start":"04:07.160 ","End":"04:11.905","Text":"the positive direction of rotation and that both statements are equivalent."},{"Start":"04:11.905 ","End":"04:14.415","Text":"They both mean the same thing."},{"Start":"04:14.415 ","End":"04:17.290","Text":"That\u0027s the end of this lesson."}],"ID":9392},{"Watched":false,"Name":"1.8 Direction Of Torque","Duration":"5m 28s","ChapterTopicVideoID":9120,"CourseChapterTopicPlaylistID":5411,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.560 ","End":"00:02.820","Text":"Hello. In this lesson,"},{"Start":"00:02.820 ","End":"00:07.815","Text":"we\u0027re going to be discussing how to find the direction of the torque intuitively."},{"Start":"00:07.815 ","End":"00:11.880","Text":"We\u0027ve spoken about the right-hand rule and the rotating right-hand rule,"},{"Start":"00:11.880 ","End":"00:13.830","Text":"which are basically the same thing."},{"Start":"00:13.830 ","End":"00:18.105","Text":"However, some people might get a bit confused so I\u0027m going to discuss"},{"Start":"00:18.105 ","End":"00:23.120","Text":"another way of finding this using a different method."},{"Start":"00:23.120 ","End":"00:30.045","Text":"We\u0027ve already seen the example of the person pushing this box"},{"Start":"00:30.045 ","End":"00:38.700","Text":"forwards with some vector force F and suddenly the box hits a rock."},{"Start":"00:39.080 ","End":"00:41.925","Text":"After the box,"},{"Start":"00:41.925 ","End":"00:45.630","Text":"the corner has hit this rock in the way,"},{"Start":"00:45.630 ","End":"00:49.537","Text":"so we want to know which direction the torque will be in."},{"Start":"00:49.537 ","End":"00:54.230","Text":"This point is representing the axis of rotation,"},{"Start":"00:54.230 ","End":"00:59.660","Text":"and the axis is going either in or out of the page depending on the positive direction."},{"Start":"00:59.660 ","End":"01:04.405","Text":"Right now we\u0027re not going to focus on which direction is the positive direction."},{"Start":"01:04.405 ","End":"01:07.090","Text":"A lot of the time in physics,"},{"Start":"01:07.090 ","End":"01:12.220","Text":"the default is to say that the anticlockwise direction is the positive direction."},{"Start":"01:12.220 ","End":"01:15.165","Text":"There\u0027s not really a reason for this,"},{"Start":"01:15.165 ","End":"01:17.055","Text":"so this is just the default."},{"Start":"01:17.055 ","End":"01:22.939","Text":"I\u0027m right now going to say that this direction is the positive direction."},{"Start":"01:23.400 ","End":"01:27.820","Text":"I just chose that randomly it doesn\u0027t mean anything."},{"Start":"01:27.820 ","End":"01:34.270","Text":"What we want to know is if this force F is going to push"},{"Start":"01:34.270 ","End":"01:42.610","Text":"this box in the positive direction of rotation or in the negative direction of rotation."},{"Start":"01:42.610 ","End":"01:46.890","Text":"The negative direction being this direction."},{"Start":"01:46.890 ","End":"01:53.885","Text":"We can see in this case that it\u0027s going to push it in the negative direction of rotation."},{"Start":"01:53.885 ","End":"01:57.480","Text":"Let\u0027s see exactly why."},{"Start":"01:57.620 ","End":"02:03.125","Text":"What we can do with this example over here is that we can just"},{"Start":"02:03.125 ","End":"02:09.050","Text":"imagine that this man is pushing the box with a force F and it hits this rock."},{"Start":"02:09.050 ","End":"02:15.965","Text":"What you can do is you can imagine your thumb going over here."},{"Start":"02:15.965 ","End":"02:18.050","Text":"You have a blue nail."},{"Start":"02:18.050 ","End":"02:26.555","Text":"This is your thumb and it\u0027s holding the corner of the box on where the rock is."},{"Start":"02:26.555 ","End":"02:29.135","Text":"The rock being the axis of rotation."},{"Start":"02:29.135 ","End":"02:35.030","Text":"Then, you can try and imagine in your head which direction this force will push it in."},{"Start":"02:35.030 ","End":"02:40.190","Text":"You can see very easily that your force will be going in"},{"Start":"02:40.190 ","End":"02:49.655","Text":"this clockwise direction so that will be the direction that this box will rotate in."},{"Start":"02:49.655 ","End":"02:53.540","Text":"Bearing in mind that the point of the box which is"},{"Start":"02:53.540 ","End":"02:57.865","Text":"touching the axis of rotation isn\u0027t meant to move,"},{"Start":"02:57.865 ","End":"03:02.700","Text":"so it\u0027s on the axis of rotation and it isn\u0027t moving."},{"Start":"03:03.020 ","End":"03:05.930","Text":"From our last lesson,"},{"Start":"03:05.930 ","End":"03:10.130","Text":"we saw that if the force is pushing the object in"},{"Start":"03:10.130 ","End":"03:16.970","Text":"the negative direction of rotation,"},{"Start":"03:16.970 ","End":"03:21.165","Text":"then that means that the torque is negative,"},{"Start":"03:21.165 ","End":"03:22.940","Text":"we\u0027ll have a negative value."},{"Start":"03:22.940 ","End":"03:28.130","Text":"Here, our torque is going to be"},{"Start":"03:28.130 ","End":"03:34.860","Text":"less than 0 because it\u0027s rotating in the negative direction."},{"Start":"03:34.900 ","End":"03:40.370","Text":"Another way of thinking of this is if again,"},{"Start":"03:40.370 ","End":"03:44.840","Text":"your thumb is still holding the corner of the box at the axis of rotation,"},{"Start":"03:44.840 ","End":"03:49.280","Text":"and you know that the normal force is acting in this direction."},{"Start":"03:49.280 ","End":"03:53.200","Text":"You can see that if you follow with a normal force,"},{"Start":"03:53.200 ","End":"03:55.595","Text":"when you\u0027re holding this corner,"},{"Start":"03:55.595 ","End":"04:03.660","Text":"it\u0027s still going to be pushing in this direction which is still the negative direction."},{"Start":"04:04.850 ","End":"04:11.255","Text":"Let\u0027s imagine for a moment that there is no force being pushed over here,"},{"Start":"04:11.255 ","End":"04:13.250","Text":"and this man isn\u0027t over here."},{"Start":"04:13.250 ","End":"04:21.530","Text":"We\u0027ll know that we have the center of mass and we have a force pushing downwards,"},{"Start":"04:21.530 ","End":"04:25.910","Text":"which is our mg due to gravity and the mass of the box."},{"Start":"04:25.910 ","End":"04:34.050","Text":"We can see that our mg if the floor isn\u0027t here and our thumb is still over here."},{"Start":"04:34.050 ","End":"04:38.205","Text":"Our thumb is still holding our box at the corner,"},{"Start":"04:38.205 ","End":"04:39.660","Text":"the axis of rotation,"},{"Start":"04:39.660 ","End":"04:47.265","Text":"so we can see that our force mg will be working in this anticlockwise direction."},{"Start":"04:47.265 ","End":"04:51.615","Text":"Which is going in the positive direction."},{"Start":"04:51.615 ","End":"04:56.045","Text":"That\u0027s the way that we can see that the torque will be in the positive direction."},{"Start":"04:56.045 ","End":"04:59.880","Text":"Here, torque will be bigger than"},{"Start":"04:59.880 ","End":"05:05.720","Text":"0 because we can see that it will be rotating in the positive direction."},{"Start":"05:06.970 ","End":"05:09.440","Text":"You can use the right-hand rule,"},{"Start":"05:09.440 ","End":"05:11.345","Text":"the rotating right-hand rule,"},{"Start":"05:11.345 ","End":"05:15.770","Text":"or this way of thinking by putting your thumb at the corner of"},{"Start":"05:15.770 ","End":"05:20.675","Text":"the object which is touching the axis of rotation and imagining it that way."},{"Start":"05:20.675 ","End":"05:22.580","Text":"Either way they all mean the same thing,"},{"Start":"05:22.580 ","End":"05:25.700","Text":"so it really is whichever is easier for you."},{"Start":"05:25.700 ","End":"05:28.560","Text":"That\u0027s the end of this lesson."}],"ID":9393}],"Thumbnail":null,"ID":5411},{"Name":"Vector Multiplication","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"2.1 Cross Product Using Determinant","Duration":"11m 26s","ChapterTopicVideoID":9237,"CourseChapterTopicPlaylistID":9394,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.140 ","End":"00:03.555","Text":"Hello. In this lesson we\u0027re going to be speaking"},{"Start":"00:03.555 ","End":"00:06.750","Text":"about how to work out the vector product,"},{"Start":"00:06.750 ","End":"00:10.830","Text":"or also called the cross products, using the determinant."},{"Start":"00:10.830 ","End":"00:17.610","Text":"Now, you can also do it by finding the direction and the size of the vectors."},{"Start":"00:17.610 ","End":"00:22.560","Text":"However, right now we\u0027re going to be speaking about using the determinant method."},{"Start":"00:22.560 ","End":"00:27.180","Text":"Now, this lesson is going to be speaking about the cross-products in"},{"Start":"00:27.180 ","End":"00:33.740","Text":"a purely mathematical sense so if you already know how to do this from math\u0027s courses,"},{"Start":"00:33.740 ","End":"00:36.870","Text":"then you can skip this video."},{"Start":"00:37.070 ","End":"00:44.160","Text":"Let\u0027s begin. Let\u0027s say that we have vectors a and vector b."},{"Start":"00:44.810 ","End":"00:52.660","Text":"This symbol over here is the same as the symbol for a cross product."},{"Start":"00:52.660 ","End":"00:55.565","Text":"How we set this up is,"},{"Start":"00:55.565 ","End":"01:00.500","Text":"let\u0027s say we\u0027re using Cartesian coordinates in 3-dimensions."},{"Start":"01:00.500 ","End":"01:04.295","Text":"The first thing that we do is in the first line,"},{"Start":"01:04.295 ","End":"01:07.520","Text":"we write down x-hat,"},{"Start":"01:07.520 ","End":"01:11.885","Text":"y-hat, and z-hat."},{"Start":"01:11.885 ","End":"01:15.715","Text":"Then we put another straight line."},{"Start":"01:15.715 ","End":"01:17.995","Text":"You\u0027ll remember that when dealing with vectors,"},{"Start":"01:17.995 ","End":"01:23.709","Text":"a lot of the time you can write the vectors like this with square brackets,"},{"Start":"01:23.709 ","End":"01:27.880","Text":"which is also the same as this, these round brackets."},{"Start":"01:27.880 ","End":"01:30.535","Text":"But when we\u0027re working out the determinant,"},{"Start":"01:30.535 ","End":"01:36.605","Text":"we use these straight lines to indicate that we\u0027re working out the determinant."},{"Start":"01:36.605 ","End":"01:41.500","Text":"I\u0027ll rub that out. Then what we do is on the second row,"},{"Start":"01:41.500 ","End":"01:48.705","Text":"we write down the components for our a vector."},{"Start":"01:48.705 ","End":"01:51.780","Text":"I\u0027ll have a in the x-component,"},{"Start":"01:51.780 ","End":"01:55.180","Text":"a in the y-component and a in the z component."},{"Start":"01:55.180 ","End":"01:57.415","Text":"Then on the third and final row,"},{"Start":"01:57.415 ","End":"02:00.230","Text":"we\u0027ll write also so for b,"},{"Start":"02:00.230 ","End":"02:07.170","Text":"b is x-component, b is y component and b is z component."},{"Start":"02:07.170 ","End":"02:11.459","Text":"Now, how do we do the determinant?"},{"Start":"02:11.459 ","End":"02:15.605","Text":"The first thing we do is we circle and we say,"},{"Start":"02:15.605 ","End":"02:19.415","Text":"first we\u0027re going to work out in the x-direction."},{"Start":"02:19.415 ","End":"02:22.100","Text":"You write down x-hat,"},{"Start":"02:22.100 ","End":"02:25.250","Text":"then you have these brackets before."},{"Start":"02:25.250 ","End":"02:27.875","Text":"Now, when you circle the x-hat,"},{"Start":"02:27.875 ","End":"02:32.550","Text":"then you cross out everything in the same column and then the same row."},{"Start":"02:32.650 ","End":"02:42.795","Text":"Then what you do is you multiply this by this and you subtract this by this."},{"Start":"02:42.795 ","End":"02:50.055","Text":"What you will have is you will have a_y multiplied by b_z"},{"Start":"02:50.055 ","End":"03:00.030","Text":"minus a_z multiplied by b_y so that it\u0027s easier to see,"},{"Start":"03:00.030 ","End":"03:06.910","Text":"minus b_y multiplied by a_z, it\u0027s the same thing."},{"Start":"03:08.270 ","End":"03:11.985","Text":"Now we\u0027re going to move on to the y-axis,"},{"Start":"03:11.985 ","End":"03:16.710","Text":"so will rub all of this out."},{"Start":"03:16.710 ","End":"03:21.140","Text":"Now as we move on to the column of y,"},{"Start":"03:21.140 ","End":"03:23.510","Text":"we have a sign change."},{"Start":"03:23.510 ","End":"03:29.735","Text":"We\u0027ll write here that we have a negative because before the x-axis we have a positive,"},{"Start":"03:29.735 ","End":"03:32.066","Text":"then the y-axis is a negative,"},{"Start":"03:32.066 ","End":"03:34.055","Text":"then the z-axis is a positive,"},{"Start":"03:34.055 ","End":"03:36.710","Text":"and if we had a fourth and fifth dimension,"},{"Start":"03:36.710 ","End":"03:40.605","Text":"so the fourth dimension would be a negative and the fifth would be a positive,"},{"Start":"03:40.605 ","End":"03:42.435","Text":"so it alternates along."},{"Start":"03:42.435 ","End":"03:44.840","Text":"Here we have a negative and here we had a positive,"},{"Start":"03:44.840 ","End":"03:46.780","Text":"we just didn\u0027t have to write it."},{"Start":"03:46.780 ","End":"03:48.930","Text":"Then we do the same thing."},{"Start":"03:48.930 ","End":"03:51.660","Text":"We\u0027re working on the y-direction."},{"Start":"03:51.660 ","End":"03:58.155","Text":"We can write y-hat and then we put in our brackets."},{"Start":"03:58.155 ","End":"04:01.415","Text":"Again, because we\u0027re working on the y column,"},{"Start":"04:01.415 ","End":"04:07.265","Text":"so we cross out the row and the column on the y."},{"Start":"04:07.265 ","End":"04:10.295","Text":"Now, we can do the multiplication."},{"Start":"04:10.295 ","End":"04:18.115","Text":"Again, we\u0027ll have this multiplied by this minus this times this."},{"Start":"04:18.115 ","End":"04:27.270","Text":"That will mean that we have a_xb_z minus b_xa_z."},{"Start":"04:32.120 ","End":"04:35.490","Text":"Now we\u0027ll move on to our z."},{"Start":"04:35.490 ","End":"04:44.670","Text":"Again, I can rub out these crosses and we can resume."},{"Start":"04:44.670 ","End":"04:48.290","Text":"Now I\u0027m circling my z-axis because we\u0027re"},{"Start":"04:48.290 ","End":"04:52.070","Text":"working in the z-direction and we can see here that there\u0027s a positive on top."},{"Start":"04:52.070 ","End":"04:53.881","Text":"We have positive,"},{"Start":"04:53.881 ","End":"04:57.535","Text":"and then we\u0027ll write z-hat over here and our brackets,"},{"Start":"04:57.535 ","End":"05:02.600","Text":"and then again we cross out the column and the row corresponding to our z."},{"Start":"05:02.600 ","End":"05:06.050","Text":"Then we have this times this,"},{"Start":"05:06.050 ","End":"05:09.245","Text":"minus this times this."},{"Start":"05:09.245 ","End":"05:18.330","Text":"Then we have a_xb_y minus b_xa_y."},{"Start":"05:19.070 ","End":"05:26.610","Text":"This is how you work out the determinant when you have a 3 by 3 vector."},{"Start":"05:28.700 ","End":"05:32.315","Text":"In the exam you can either work this out,"},{"Start":"05:32.315 ","End":"05:35.090","Text":"how I\u0027ve shown you and if you can remember how to do it."},{"Start":"05:35.090 ","End":"05:37.445","Text":"If you can\u0027t and you have enough space"},{"Start":"05:37.445 ","End":"05:40.119","Text":"on your notes that you\u0027re allowed to take into the exam,"},{"Start":"05:40.119 ","End":"05:47.165","Text":"so you can also bring in the equation and just apply it in the exam."},{"Start":"05:47.165 ","End":"05:53.370","Text":"But it\u0027s very simple and you can just practice it and it\u0027s really easy to do in the exam."},{"Start":"05:54.140 ","End":"05:57.365","Text":"Now let\u0027s take a look at an example."},{"Start":"05:57.365 ","End":"06:01.295","Text":"Say our vector a is equal to 1,"},{"Start":"06:01.295 ","End":"06:02.765","Text":"2, 3,"},{"Start":"06:02.765 ","End":"06:08.869","Text":"and we\u0027ll say vector b is equal to 1,"},{"Start":"06:08.869 ","End":"06:11.245","Text":"minus 1, 3,"},{"Start":"06:11.245 ","End":"06:14.240","Text":"what will our c be equal to?"},{"Start":"06:14.240 ","End":"06:19.320","Text":"Our c will be equal to a cross b."},{"Start":"06:19.320 ","End":"06:23.090","Text":"If we look at our equation over here,"},{"Start":"06:23.090 ","End":"06:28.875","Text":"I have a_yb_z, so that\u0027s 2 times 3,"},{"Start":"06:28.875 ","End":"06:35.565","Text":"so we\u0027ll have 2 times 3 minus a_zb_y,"},{"Start":"06:35.565 ","End":"06:42.045","Text":"so that will be minus 3 times negative 1,"},{"Start":"06:42.045 ","End":"06:44.760","Text":"and all of this in the x-direction."},{"Start":"06:44.760 ","End":"06:52.290","Text":"Then we\u0027ll look in the y-direction, a_xb_z minus again,"},{"Start":"06:52.290 ","End":"06:57.525","Text":"so a_xb_z is 1 times 3 minus b_xa_z,"},{"Start":"06:57.525 ","End":"07:02.250","Text":"which will be 1 times 3 again,"},{"Start":"07:02.250 ","End":"07:05.220","Text":"and all of this is in the y-direction."},{"Start":"07:05.220 ","End":"07:09.620","Text":"Then we have a positive and now in the z-direction,"},{"Start":"07:09.620 ","End":"07:19.440","Text":"so we\u0027ll have a_xb_y which will be 1 times negative 1 minus b_xa_y,"},{"Start":"07:19.440 ","End":"07:27.670","Text":"which will be minus 1 times 2 in the z-direction."},{"Start":"07:27.670 ","End":"07:31.230","Text":"If we just simplify this,"},{"Start":"07:31.230 ","End":"07:35.925","Text":"so then c will be equal to, let\u0027s see,"},{"Start":"07:35.925 ","End":"07:41.380","Text":"9, 0, negative 3."},{"Start":"07:42.620 ","End":"07:45.970","Text":"That is what c is equal to."},{"Start":"07:46.160 ","End":"07:54.085","Text":"Now let\u0027s talk about"},{"Start":"07:54.085 ","End":"08:00.680","Text":"the attributes or the qualities of the cross-product."},{"Start":"08:00.680 ","End":"08:05.910","Text":"The first attributes is that"},{"Start":"08:05.910 ","End":"08:11.325","Text":"if we have that our c vector is equal to our a vector cross our b vector,"},{"Start":"08:11.325 ","End":"08:16.095","Text":"so our c vector is perpendicular"},{"Start":"08:16.095 ","End":"08:23.805","Text":"to both our a vector and our b vector."},{"Start":"08:23.805 ","End":"08:29.300","Text":"That\u0027s the first trait and it\u0027s very useful to know. What does that mean?"},{"Start":"08:29.300 ","End":"08:33.275","Text":"If here we have our a vector in this direction,"},{"Start":"08:33.275 ","End":"08:36.695","Text":"and here we have b vector in this direction,"},{"Start":"08:36.695 ","End":"08:40.850","Text":"so our c vector will be in"},{"Start":"08:40.850 ","End":"08:47.180","Text":"this direction and it will be perpendicular to both a and to b,"},{"Start":"08:47.180 ","End":"08:56.250","Text":"and therefore also perpendicular to the plane going on ab."},{"Start":"08:56.540 ","End":"09:05.630","Text":"Another trait, the second trait is that if we\u0027re working out instead of a cross b,"},{"Start":"09:05.630 ","End":"09:07.985","Text":"we do b cross a,"},{"Start":"09:07.985 ","End":"09:16.670","Text":"it does not equal to a cross b."},{"Start":"09:16.670 ","End":"09:26.935","Text":"In fact, b cross a will equal to negative a cross b."},{"Start":"09:26.935 ","End":"09:29.270","Text":"If you flip these around,"},{"Start":"09:29.270 ","End":"09:32.030","Text":"this isn\u0027t like 2 times 3 and 3 times 2,"},{"Start":"09:32.030 ","End":"09:33.260","Text":"it makes a difference."},{"Start":"09:33.260 ","End":"09:34.895","Text":"You can\u0027t just do that."},{"Start":"09:34.895 ","End":"09:37.385","Text":"If we\u0027re doing b cross a,"},{"Start":"09:37.385 ","End":"09:43.880","Text":"then we will get that our arrow for c will be pointing in this direction,"},{"Start":"09:43.880 ","End":"09:47.105","Text":"so if c=b cross a,"},{"Start":"09:47.105 ","End":"09:52.950","Text":"and this c is a cross b."},{"Start":"09:54.560 ","End":"09:57.840","Text":"The last useful thing to note,"},{"Start":"09:57.840 ","End":"10:04.650","Text":"point number 3 is that if we want to find out the size of the vector c,"},{"Start":"10:04.730 ","End":"10:11.695","Text":"we have this absolute value which denotes the size of the vector."},{"Start":"10:11.695 ","End":"10:17.220","Text":"It equals the size of the vector a cross b,"},{"Start":"10:17.220 ","End":"10:23.840","Text":"and this equals to the size of the vector a multiplied by the size of the vector b,"},{"Start":"10:23.840 ","End":"10:27.390","Text":"these 2 will be scalar quantities,"},{"Start":"10:27.590 ","End":"10:36.380","Text":"scalar multiplied by sine of the angle between them."},{"Start":"10:36.380 ","End":"10:41.300","Text":"If this we say is the angle Alpha, and if we know that,"},{"Start":"10:41.300 ","End":"10:46.220","Text":"then we can find the size of vector c,"},{"Start":"10:46.220 ","End":"10:53.770","Text":"which means that this will then also therefore be scalar,"},{"Start":"10:54.710 ","End":"11:02.280","Text":"and it\u0027s a size and we won\u0027t have the direction of the vector."},{"Start":"11:02.950 ","End":"11:09.050","Text":"These are the 3 important things to remember when dealing with the cross-product,"},{"Start":"11:09.050 ","End":"11:11.220","Text":"aka the vector product,"},{"Start":"11:11.220 ","End":"11:15.065","Text":"and it\u0027s useful to remember and useful to write down in your notes."},{"Start":"11:15.065 ","End":"11:20.675","Text":"Also, don\u0027t forget how to work out the cross products whether it\u0027d be"},{"Start":"11:20.675 ","End":"11:22.910","Text":"remembering the method for working it out or"},{"Start":"11:22.910 ","End":"11:27.220","Text":"remembering the equation. That\u0027s the end of this lesson."}],"ID":9526},{"Watched":false,"Name":"2.2 Cross Product Using Direction And Size","Duration":"5m 23s","ChapterTopicVideoID":9238,"CourseChapterTopicPlaylistID":9394,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:02.850","Text":"Hello. In the previous lesson,"},{"Start":"00:02.850 ","End":"00:08.925","Text":"we spoke about working out the cross-product using the determinant."},{"Start":"00:08.925 ","End":"00:12.390","Text":"In this lesson we\u0027re going to learn another method,"},{"Start":"00:12.390 ","End":"00:15.810","Text":"which is by using the size and direction."},{"Start":"00:15.810 ","End":"00:19.725","Text":"For working out the size, we\u0027ll see how you do it now."},{"Start":"00:19.725 ","End":"00:24.600","Text":"Obviously, working out the cross-product through using the size and"},{"Start":"00:24.600 ","End":"00:27.090","Text":"direction and working out"},{"Start":"00:27.090 ","End":"00:30.840","Text":"the cross-product using the determinant will come to the exact same onset."},{"Start":"00:30.840 ","End":"00:34.095","Text":"This is the 2nd method of doing it."},{"Start":"00:34.095 ","End":"00:36.765","Text":"In order to work out the size,"},{"Start":"00:36.765 ","End":"00:39.900","Text":"if we have our vector c,"},{"Start":"00:39.900 ","End":"00:48.645","Text":"so the size of our c will be the size of vector a, cross vector b."},{"Start":"00:48.645 ","End":"00:52.470","Text":"This is the size of the cross-product,"},{"Start":"00:52.470 ","End":"00:58.577","Text":"and this is equal to the size of our vector a multiplied by the size of a vector b,"},{"Start":"00:58.577 ","End":"01:03.910","Text":"multiplied by sine of the angle between the 2."},{"Start":"01:03.910 ","End":"01:07.685","Text":"If for instance here I have my vector b,"},{"Start":"01:07.685 ","End":"01:10.535","Text":"and here I have my vector a,"},{"Start":"01:10.535 ","End":"01:15.980","Text":"so this angle over here is going to be my Alpha."},{"Start":"01:15.980 ","End":"01:20.195","Text":"That\u0027s perfect. That gives us our size."},{"Start":"01:20.195 ","End":"01:22.570","Text":"But what about direction?"},{"Start":"01:22.570 ","End":"01:28.591","Text":"The way we do this is through the right-hand rule."},{"Start":"01:28.591 ","End":"01:32.825","Text":"We\u0027re going to go over how we use the right-hand rule."},{"Start":"01:32.825 ","End":"01:34.820","Text":"If you take your right-hand,"},{"Start":"01:34.820 ","End":"01:37.495","Text":"so this is the right-hand."},{"Start":"01:37.495 ","End":"01:43.955","Text":"Then you have your fingers in a straight line going in the direction of your vector a,"},{"Start":"01:43.955 ","End":"01:45.455","Text":"why the vector a?"},{"Start":"01:45.455 ","End":"01:48.020","Text":"Because it\u0027s the 1st vector."},{"Start":"01:48.020 ","End":"01:54.050","Text":"Your fingers are in the direction of the first vector in the cross-products."},{"Start":"01:54.050 ","End":"01:57.665","Text":"Here, it\u0027s a, because we have a cross b,"},{"Start":"01:57.665 ","End":"02:02.710","Text":"so our fingers are in the direction of a."},{"Start":"02:02.710 ","End":"02:07.250","Text":"Then what we want to do is we want a form with our thumb,"},{"Start":"02:07.250 ","End":"02:10.490","Text":"90 degree angle to our fingers."},{"Start":"02:10.490 ","End":"02:14.210","Text":"Some of us are double jointed."},{"Start":"02:14.210 ","End":"02:16.255","Text":"Some of us aren\u0027t double jointed,"},{"Start":"02:16.255 ","End":"02:20.390","Text":"so not everyone will be able to get to this 90 degree angle."},{"Start":"02:20.390 ","End":"02:24.485","Text":"But just remember that whatever angle your thumb is at in reality,"},{"Start":"02:24.485 ","End":"02:28.970","Text":"it\u0027s meant to represent an angle of 90 degrees to your fingers."},{"Start":"02:28.970 ","End":"02:35.340","Text":"I apologize ahead of time for the picture."},{"Start":"02:35.340 ","End":"02:37.630","Text":"What do we want to do,"},{"Start":"02:37.630 ","End":"02:45.215","Text":"is we want to curve our fingers in the direction of vector b."},{"Start":"02:45.215 ","End":"02:48.185","Text":"This will now be our pinky."},{"Start":"02:48.185 ","End":"02:56.445","Text":"We\u0027re curving our fingers around in the direction of b."},{"Start":"02:56.445 ","End":"03:01.070","Text":"If I draw an arrow in a slightly different color,"},{"Start":"03:01.070 ","End":"03:04.100","Text":"let\u0027s see in a green."},{"Start":"03:04.100 ","End":"03:13.010","Text":"We will see that they are pointing in the direction of a vector b."},{"Start":"03:13.010 ","End":"03:22.160","Text":"Then we will see that the direction of our thumb is going to be equal"},{"Start":"03:22.160 ","End":"03:30.950","Text":"to the direction of our vector c. We can actually look over here and we can see,"},{"Start":"03:30.950 ","End":"03:37.560","Text":"so we\u0027re rotating in this direction."},{"Start":"03:37.560 ","End":"03:40.954","Text":"Then we can see that over here,"},{"Start":"03:40.954 ","End":"03:45.785","Text":"perpendicular to b vector and perpendicular to our a vector,"},{"Start":"03:45.785 ","End":"03:50.135","Text":"we\u0027ll have our c vector pointing in this upward direction,"},{"Start":"03:50.135 ","End":"03:54.330","Text":"which is in the same direction as our thumb."},{"Start":"03:54.400 ","End":"03:58.790","Text":"It\u0027s important to remember that the vector c,"},{"Start":"03:58.790 ","End":"04:03.110","Text":"which is a product of the cross-product between a and b,"},{"Start":"04:03.110 ","End":"04:08.250","Text":"is going to be perpendicular to both a and b."},{"Start":"04:08.360 ","End":"04:17.705","Text":"Another point which is useful to remember is that if I were to flip over my a and b,"},{"Start":"04:17.705 ","End":"04:20.330","Text":"so having a over here,"},{"Start":"04:20.330 ","End":"04:22.100","Text":"and b over here."},{"Start":"04:22.100 ","End":"04:26.630","Text":"Because I need to put my right hand,"},{"Start":"04:26.630 ","End":"04:31.145","Text":"and have my fingers pointing in the direction of"},{"Start":"04:31.145 ","End":"04:37.605","Text":"my 1st vector in the cross-product."},{"Start":"04:37.605 ","End":"04:39.275","Text":"Here it would be a,"},{"Start":"04:39.275 ","End":"04:43.205","Text":"and then I would have to rotate,"},{"Start":"04:43.205 ","End":"04:47.855","Text":"bend my fingers in the direction of b."},{"Start":"04:47.855 ","End":"04:51.345","Text":"This would be my direction of rotation."},{"Start":"04:51.345 ","End":"04:54.485","Text":"Then if you try and do this with your right-hand,"},{"Start":"04:54.485 ","End":"04:58.865","Text":"you\u0027re going to have a very difficult time bending your fingers,"},{"Start":"04:58.865 ","End":"05:03.645","Text":"and rotating from a to b when your thumb is pointing upwards."},{"Start":"05:03.645 ","End":"05:08.869","Text":"In this case, your thumb would be pointing in this downwards direction,"},{"Start":"05:08.869 ","End":"05:16.105","Text":"and so your torque will be in this direction over here."},{"Start":"05:16.105 ","End":"05:20.719","Text":"That\u0027s another very useful thing to remember."},{"Start":"05:20.719 ","End":"05:24.270","Text":"That\u0027s it. That\u0027s the end of this lesson."}],"ID":9527}],"Thumbnail":null,"ID":9394},{"Name":"Torque Equation","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"The Equation","Duration":"8m 55s","ChapterTopicVideoID":9239,"CourseChapterTopicPlaylistID":9395,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9239.jpeg","UploadDate":"2017-03-26T18:04:26.0770000","DurationForVideoObject":"PT8M55S","Description":null,"MetaTitle":"The Equation: Video + Workbook | Proprep","MetaDescription":"Torque - Torque Equation. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-1-mechanics-waves-and-thermodynamics/torque/torque-equation/vid9522","VideoComments":[],"Subtitles":[{"Start":"00:00.260 ","End":"00:07.065","Text":"Hello. In this lesson we\u0027re going to be speaking about the equation for our moments."},{"Start":"00:07.065 ","End":"00:13.740","Text":"As we know, if we have some body over here with a line coming out,"},{"Start":"00:13.740 ","End":"00:17.070","Text":"if I\u0027m going to apply a force in this direction,"},{"Start":"00:17.070 ","End":"00:21.030","Text":"then we know that the body is going to go in this direction,"},{"Start":"00:21.030 ","End":"00:23.085","Text":"rotate in this direction."},{"Start":"00:23.085 ","End":"00:29.265","Text":"Let\u0027s talk about the equation to describe this type of movement."},{"Start":"00:29.265 ","End":"00:33.570","Text":"The equation is the sum of all of the moments,"},{"Start":"00:33.570 ","End":"00:35.160","Text":"notice there\u0027s an arrow on top,"},{"Start":"00:35.160 ","End":"00:36.795","Text":"which means that it is a vector,"},{"Start":"00:36.795 ","End":"00:40.560","Text":"is equal to I multiplied by D Omega,"},{"Start":"00:40.560 ","End":"00:43.420","Text":"which is also a vector quantity by dt."},{"Start":"00:43.420 ","End":"00:50.030","Text":"We know that d Omega by dt is the change in angular velocity,"},{"Start":"00:50.030 ","End":"00:52.010","Text":"because Omega is the angular velocity,"},{"Start":"00:52.010 ","End":"00:53.180","Text":"and the I is of course,"},{"Start":"00:53.180 ","End":"00:54.685","Text":"our moment of inertia."},{"Start":"00:54.685 ","End":"01:00.320","Text":"What this actually means is that when there\u0027s moments acting, so for instance,"},{"Start":"01:00.320 ","End":"01:03.215","Text":"when I apply a force on this rod over here,"},{"Start":"01:03.215 ","End":"01:07.885","Text":"so I know that my angular velocity is meant to change."},{"Start":"01:07.885 ","End":"01:10.175","Text":"That is what this equation means."},{"Start":"01:10.175 ","End":"01:16.595","Text":"You\u0027ll notice that this equation is very similar to Newton\u0027s 2nd law."},{"Start":"01:16.595 ","End":"01:23.930","Text":"If you remember, it\u0027s the sum of all of the forces is equal to mass times acceleration,"},{"Start":"01:23.930 ","End":"01:30.190","Text":"which is our dv by dt, if you remember that."},{"Start":"01:30.190 ","End":"01:33.177","Text":"Here instead of the acceleration,"},{"Start":"01:33.177 ","End":"01:36.755","Text":"but linear, so we have angular velocity,"},{"Start":"01:36.755 ","End":"01:39.000","Text":"but that acceleration,"},{"Start":"01:39.000 ","End":"01:41.160","Text":"so of m, we have the moment of inertia,"},{"Start":"01:41.160 ","End":"01:44.405","Text":"and instead of our moment,"},{"Start":"01:44.405 ","End":"01:46.495","Text":"we have the sum of the forces."},{"Start":"01:46.495 ","End":"01:49.340","Text":"We\u0027re going to use this equation over here,"},{"Start":"01:49.340 ","End":"01:55.720","Text":"equation for moments in a very similar way that we\u0027ve been using our Newton\u0027s 2nd law."},{"Start":"01:55.720 ","End":"02:00.570","Text":"Similarly that we called this our dv by dt,"},{"Start":"02:00.570 ","End":"02:02.910","Text":"we called it our a,"},{"Start":"02:02.910 ","End":"02:09.705","Text":"acceleration, so we can say that d Omega by dt,"},{"Start":"02:09.705 ","End":"02:12.360","Text":"we call it Alpha."},{"Start":"02:12.360 ","End":"02:17.775","Text":"What is our Alpha? It\u0027s our angular acceleration."},{"Start":"02:17.775 ","End":"02:25.130","Text":"Our a, acceleration is being about how fast the velocity changes,"},{"Start":"02:25.130 ","End":"02:27.305","Text":"the rate of change of velocity,"},{"Start":"02:27.305 ","End":"02:30.500","Text":"and here, alpha, or angular acceleration,"},{"Start":"02:30.500 ","End":"02:34.415","Text":"is how fast the rate of change of velocity,"},{"Start":"02:34.415 ","End":"02:37.195","Text":"but in angular velocity."},{"Start":"02:37.195 ","End":"02:41.645","Text":"Again, like if I push this rod which is standing and over here,"},{"Start":"02:41.645 ","End":"02:45.170","Text":"it\u0027s angular velocity is Omega 0."},{"Start":"02:45.170 ","End":"02:47.000","Text":"When it\u0027s falling down,"},{"Start":"02:47.000 ","End":"02:52.465","Text":"what will its rate of change of angular velocity be?"},{"Start":"02:52.465 ","End":"02:56.435","Text":"If you really compare these equations,"},{"Start":"02:56.435 ","End":"02:59.300","Text":"Newton\u0027s 2nd law and the equation for moments,"},{"Start":"02:59.300 ","End":"03:03.950","Text":"it\u0027s a lot easier to remember and also to understand and also this equation gives"},{"Start":"03:03.950 ","End":"03:09.275","Text":"you a lot of better intuition as to what our moment of inertia is,"},{"Start":"03:09.275 ","End":"03:11.345","Text":"what our I is."},{"Start":"03:11.345 ","End":"03:15.200","Text":"It\u0027s very similar to the mass,"},{"Start":"03:15.200 ","End":"03:21.090","Text":"but if you want to think about it as in the mass when we\u0027re in circular motion."},{"Start":"03:22.000 ","End":"03:24.695","Text":"When we\u0027re moving around in a circle,"},{"Start":"03:24.695 ","End":"03:28.100","Text":"so that is what you can liken it to."},{"Start":"03:28.100 ","End":"03:30.620","Text":"In other words, the I,"},{"Start":"03:30.620 ","End":"03:34.880","Text":"the moment of inertia is how much the mass is"},{"Start":"03:34.880 ","End":"03:39.760","Text":"opposed to this force causing it to change direction."},{"Start":"03:39.760 ","End":"03:41.870","Text":"That\u0027s what this moment of inertia is."},{"Start":"03:41.870 ","End":"03:43.900","Text":"That\u0027s how you can think about it."},{"Start":"03:43.900 ","End":"03:50.735","Text":"What we\u0027re going to do is we\u0027re going to discuss how to deal with certain questions,"},{"Start":"03:50.735 ","End":"03:53.600","Text":"to deal with the moments."},{"Start":"03:53.600 ","End":"03:59.680","Text":"Imagine we have again this type of stand with the rod."},{"Start":"03:59.680 ","End":"04:05.670","Text":"Imagine that we have 2 forces and 2 moments acting upon this rod."},{"Start":"04:05.670 ","End":"04:08.985","Text":"We have over here our F_1,"},{"Start":"04:08.985 ","End":"04:12.154","Text":"and then we have over here acting,"},{"Start":"04:12.154 ","End":"04:21.495","Text":"we have our F_2 and they correspond to Tau 1 and Tau 2."},{"Start":"04:21.495 ","End":"04:25.235","Text":"Let\u0027s see how we answer this type of question."},{"Start":"04:25.235 ","End":"04:27.515","Text":"Just like F equals MA,"},{"Start":"04:27.515 ","End":"04:32.425","Text":"so we have the sum of all of our moments is going to be equal to,"},{"Start":"04:32.425 ","End":"04:38.205","Text":"so we\u0027ll have Tau 1 plus Tau 2."},{"Start":"04:38.205 ","End":"04:40.445","Text":"This is equal to,"},{"Start":"04:40.445 ","End":"04:44.565","Text":"as we know, I multiplied by Alpha."},{"Start":"04:44.565 ","End":"04:49.280","Text":"If we say that this is remains stationary,"},{"Start":"04:49.280 ","End":"04:52.820","Text":"so these forces are equal and opposite."},{"Start":"04:52.820 ","End":"05:00.285","Text":"Alpha our d Omega by dt will be 0 because this will remain stationary."},{"Start":"05:00.285 ","End":"05:04.775","Text":"There\u0027s no angular velocity and definitely no change in angular velocity."},{"Start":"05:04.775 ","End":"05:07.385","Text":"This will cross out and equal to 0."},{"Start":"05:07.385 ","End":"05:11.225","Text":"Similarly, if these forces are equal and opposite,"},{"Start":"05:11.225 ","End":"05:15.470","Text":"but this rod is moving at a constant angular velocity,"},{"Start":"05:15.470 ","End":"05:17.945","Text":"constant circles, then again,"},{"Start":"05:17.945 ","End":"05:21.335","Text":"our d Omega by dt is going to be equal to 0."},{"Start":"05:21.335 ","End":"05:30.830","Text":"If we\u0027re"},{"Start":"05:30.830 ","End":"05:33.020","Text":"moving at constant linear velocity,"},{"Start":"05:33.020 ","End":"05:36.140","Text":"we\u0027ll say that acceleration is equal to 0."},{"Start":"05:36.140 ","End":"05:40.220","Text":"Similar here, if we\u0027re traveling at a constant angular velocity,"},{"Start":"05:40.220 ","End":"05:43.600","Text":"our angular acceleration will be 0."},{"Start":"05:43.600 ","End":"05:45.440","Text":"We can see that in this question,"},{"Start":"05:45.440 ","End":"05:49.295","Text":"specifically the sum of our moments is going to be equal to 0."},{"Start":"05:49.295 ","End":"05:52.475","Text":"Then we just have a normal equation where we can just work out"},{"Start":"05:52.475 ","End":"05:58.710","Text":"what our Moment number 1 and our Moment number 2 is equal to."},{"Start":"05:58.710 ","End":"06:02.135","Text":"Ut\u0027s very similar to when we use that Newton\u0027s 2nd law."},{"Start":"06:02.135 ","End":"06:06.260","Text":"The sum of all of the forces is equal to mass times acceleration."},{"Start":"06:06.260 ","End":"06:08.270","Text":"We had a body that wasn\u0027t moving,"},{"Start":"06:08.270 ","End":"06:11.840","Text":"so we\u0027d get that the sum of all of the forces was equal to 0,"},{"Start":"06:11.840 ","End":"06:14.010","Text":"and then we\u0027d solve."},{"Start":"06:14.020 ","End":"06:18.385","Text":"I\u0027m just going to write 3 notes for you."},{"Start":"06:18.385 ","End":"06:21.450","Text":"These are the notes that I have."},{"Start":"06:21.450 ","End":"06:23.345","Text":"If the body is stationary,"},{"Start":"06:23.345 ","End":"06:26.045","Text":"then Sigma Tau is equal to 0."},{"Start":"06:26.045 ","End":"06:28.955","Text":"The sum of all of the moments will be equal to 0."},{"Start":"06:28.955 ","End":"06:31.460","Text":"Similarly, if the body is stationary,"},{"Start":"06:31.460 ","End":"06:37.640","Text":"then we can select the axis of rotation to be anywhere we want. What is that mean?"},{"Start":"06:37.640 ","End":"06:39.810","Text":"Let\u0027s say we\u0027re dealing with this rod."},{"Start":"06:39.810 ","End":"06:42.060","Text":"If we say that this is stationary,"},{"Start":"06:42.060 ","End":"06:44.855","Text":"it doesn\u0027t matter we can say that the axis of rotation is here,"},{"Start":"06:44.855 ","End":"06:46.730","Text":"or here, or here."},{"Start":"06:46.730 ","End":"06:49.880","Text":"Wherever it might be it doesn\u0027t matter."},{"Start":"06:49.880 ","End":"06:53.480","Text":"If the body is rotating,"},{"Start":"06:53.480 ","End":"06:55.760","Text":"then we will choose the axis of rotation in"},{"Start":"06:55.760 ","End":"06:59.525","Text":"our calculations to be the actual axes of rotation of the body."},{"Start":"06:59.525 ","End":"07:02.390","Text":"If this rod is actually rotating,"},{"Start":"07:02.390 ","End":"07:05.795","Text":"so of course it\u0027s clear that it will be rotating around this point,"},{"Start":"07:05.795 ","End":"07:09.905","Text":"so we will choose this to be our axis of rotation."},{"Start":"07:09.905 ","End":"07:15.790","Text":"No other point but the actual natural axis of rotation of our body."},{"Start":"07:15.790 ","End":"07:20.210","Text":"Similarly, I\u0027ve added Point number 3 that if the body is stationary,"},{"Start":"07:20.210 ","End":"07:24.710","Text":"then we can also say that the sum of the forces is also equal to 0."},{"Start":"07:24.710 ","End":"07:27.320","Text":"That\u0027s another equation that you could add in."},{"Start":"07:27.320 ","End":"07:28.940","Text":"If the body is stationary,"},{"Start":"07:28.940 ","End":"07:30.600","Text":"the sum of the moments is equal to 0,"},{"Start":"07:30.600 ","End":"07:32.900","Text":"the sum of the forces is equal to 0,"},{"Start":"07:32.900 ","End":"07:35.765","Text":"and our axes of rotation can be anywhere we want,"},{"Start":"07:35.765 ","End":"07:37.790","Text":"because the body isn\u0027t rotating."},{"Start":"07:37.790 ","End":"07:39.560","Text":"If the body is rotating,"},{"Start":"07:39.560 ","End":"07:41.600","Text":"then we choose the axis of rotation in"},{"Start":"07:41.600 ","End":"07:45.175","Text":"our calculations to be the actual axes of rotation of the body."},{"Start":"07:45.175 ","End":"07:48.185","Text":"To sum up what we\u0027ve learned this lesson,"},{"Start":"07:48.185 ","End":"07:51.920","Text":"we\u0027ve seen that the equation for a moment or photic,"},{"Start":"07:51.920 ","End":"07:58.520","Text":"is the sum of all of the moments is equal to I multiplied by d Omega by dt,"},{"Start":"07:58.520 ","End":"08:05.135","Text":"where I is the moment of inertia and d Omega by dt is the angular acceleration."},{"Start":"08:05.135 ","End":"08:11.000","Text":"We have 2 cases where either the body is stationary."},{"Start":"08:11.000 ","End":"08:17.060","Text":"In this type of case, then we know that the sum of all the moments will be equal to 0,"},{"Start":"08:17.060 ","End":"08:21.290","Text":"and also we can say that the sum of all of the forces is equal to 0."},{"Start":"08:21.290 ","End":"08:26.355","Text":"This can be another equation in order to solve our problem."},{"Start":"08:26.355 ","End":"08:30.490","Text":"We can select the axis of rotation to be anywhere we want in this case,"},{"Start":"08:30.490 ","End":"08:33.830","Text":"so it can be anywhere along the body."},{"Start":"08:33.830 ","End":"08:36.965","Text":"If the body is rotating, on the other hand,"},{"Start":"08:36.965 ","End":"08:39.560","Text":"then we can choose the axis of rotation in"},{"Start":"08:39.560 ","End":"08:42.535","Text":"our calculations to be the actual axis of rotation."},{"Start":"08:42.535 ","End":"08:48.215","Text":"We\u0027ll just have 1 possible axis of rotation when we\u0027re doing our calculations."},{"Start":"08:48.215 ","End":"08:49.835","Text":"In the next lesson,"},{"Start":"08:49.835 ","End":"08:53.720","Text":"we\u0027re going to look at a question which deals with this."},{"Start":"08:53.720 ","End":"08:56.910","Text":"That\u0027s the end of our lesson."}],"ID":9522},{"Watched":false,"Name":"Example- Moments on a Triangle","Duration":"33m 1s","ChapterTopicVideoID":10444,"CourseChapterTopicPlaylistID":9395,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.284","Text":"Hello. So in this lesson,"},{"Start":"00:03.284 ","End":"00:05.310","Text":"we\u0027re doing an exercise,"},{"Start":"00:05.310 ","End":"00:08.055","Text":"and we\u0027re given an equilateral triangle."},{"Start":"00:08.055 ","End":"00:10.050","Text":"Just to remind anyone who doesn\u0027t know,"},{"Start":"00:10.050 ","End":"00:15.480","Text":"an equilateral triangle says that all the sides are of equal length,"},{"Start":"00:15.480 ","End":"00:21.355","Text":"and that also means that all the angles in the corners of the triangle are also equal."},{"Start":"00:21.355 ","End":"00:27.720","Text":"We\u0027re given a triangle with each side of the triangle is of length a."},{"Start":"00:27.740 ","End":"00:30.120","Text":"We\u0027re also given forces."},{"Start":"00:30.120 ","End":"00:31.845","Text":"We have F_1, F_2,"},{"Start":"00:31.845 ","End":"00:33.775","Text":"F_3, and F_4."},{"Start":"00:33.775 ","End":"00:37.609","Text":"We\u0027re being told that the y-axis is going up here,"},{"Start":"00:37.609 ","End":"00:40.610","Text":"the x-axis is going to the right."},{"Start":"00:40.610 ","End":"00:42.289","Text":"Then in our first question,"},{"Start":"00:42.289 ","End":"00:45.830","Text":"we\u0027re being told to calculate the torque of the forces in the diagram."},{"Start":"00:45.830 ","End":"00:48.559","Text":"That\u0027s F_1 and through F_4."},{"Start":"00:48.559 ","End":"00:51.814","Text":"About the axes perpendicular to the page"},{"Start":"00:51.814 ","End":"00:55.445","Text":"which goes through the bottom left corner of the triangle."},{"Start":"00:55.445 ","End":"00:58.464","Text":"We\u0027re being told that over here,"},{"Start":"00:58.464 ","End":"01:02.370","Text":"we have our axes."},{"Start":"01:02.370 ","End":"01:06.830","Text":"I\u0027ve drawn it as a dot because it\u0027s perpendicular to the page."},{"Start":"01:06.830 ","End":"01:12.349","Text":"If you imagine an arrow pointing out of the screen and in your direction,"},{"Start":"01:12.349 ","End":"01:14.419","Text":"so that\u0027s the direction,"},{"Start":"01:14.419 ","End":"01:17.965","Text":"it\u0027s the z-axis out of the screen."},{"Start":"01:17.965 ","End":"01:21.620","Text":"We\u0027re being told that the mass of the triangle is capital M,"},{"Start":"01:21.620 ","End":"01:28.565","Text":"and that the center of mass is located at 1/2a, 1/2 root 3a."},{"Start":"01:28.565 ","End":"01:32.430","Text":"Let\u0027s say it\u0027s somewhere over here."},{"Start":"01:32.960 ","End":"01:36.140","Text":"Now, the next thing that we\u0027re going to do is we\u0027re going"},{"Start":"01:36.140 ","End":"01:40.395","Text":"to choose our positive direction of rotation."},{"Start":"01:40.395 ","End":"01:43.039","Text":"Now, because we\u0027ve already decided that"},{"Start":"01:43.039 ","End":"01:46.415","Text":"our z-axis is coming out of the screen and into our eye,"},{"Start":"01:46.415 ","End":"01:49.415","Text":"if you look at using the right-hand rule,"},{"Start":"01:49.415 ","End":"01:50.450","Text":"if you have your thumb,"},{"Start":"01:50.450 ","End":"01:53.550","Text":"which represents the z-axis and it\u0027s pointing towards you,"},{"Start":"01:53.550 ","End":"01:56.464","Text":"you\u0027ll see that the positive direction of rotation,"},{"Start":"01:56.464 ","End":"01:59.210","Text":"the direction that your fingers are curling towards,"},{"Start":"01:59.210 ","End":"02:02.585","Text":"is going to be in the anticlockwise direction."},{"Start":"02:02.585 ","End":"02:07.905","Text":"That also translates as from x to y."},{"Start":"02:07.905 ","End":"02:14.060","Text":"This is the positive direction of rotation through the right-hand rule."},{"Start":"02:14.270 ","End":"02:19.130","Text":"We\u0027re going to start by answering question Number 1,"},{"Start":"02:19.130 ","End":"02:23.090","Text":"and we\u0027re going to start by speaking about a force F_1."},{"Start":"02:23.090 ","End":"02:28.055","Text":"To remind you, the equation for torque is Tau,"},{"Start":"02:28.055 ","End":"02:30.270","Text":"representing the torque and it\u0027s a vector,"},{"Start":"02:30.270 ","End":"02:33.139","Text":"which is equal to your r vector,"},{"Start":"02:33.139 ","End":"02:35.960","Text":"which is the vector pointing from the angle of"},{"Start":"02:35.960 ","End":"02:40.170","Text":"rotation until the point where your force is acting,"},{"Start":"02:40.170 ","End":"02:47.210","Text":"and then it\u0027s cross multiplication multiplied by your vector for force."},{"Start":"02:47.210 ","End":"02:54.039","Text":"If you want to find out just the size of your torque,"},{"Start":"02:54.039 ","End":"02:58.699","Text":"so it\u0027s going to be equal to the size of your r vector,"},{"Start":"02:58.699 ","End":"03:01.519","Text":"the size of your F vector,"},{"Start":"03:01.519 ","End":"03:06.435","Text":"multiplied by sine of the angle."},{"Start":"03:06.435 ","End":"03:09.110","Text":"Currently, with our F_1,"},{"Start":"03:09.110 ","End":"03:14.420","Text":"we don\u0027t know the angle between it and the side of the triangle."},{"Start":"03:14.420 ","End":"03:17.094","Text":"Let\u0027s see how we solve this."},{"Start":"03:17.094 ","End":"03:20.270","Text":"Now, specifically in the case of F_1,"},{"Start":"03:20.270 ","End":"03:26.239","Text":"we can see that it\u0027s working exactly from the axis of rotation."},{"Start":"03:26.239 ","End":"03:28.895","Text":"It\u0027s coming out of the axes of rotation."},{"Start":"03:28.895 ","End":"03:32.180","Text":"As we said, our r vector is from the axis of"},{"Start":"03:32.180 ","End":"03:36.390","Text":"rotation until the point where the force is being applied."},{"Start":"03:36.390 ","End":"03:40.565","Text":"Here we see that the force is being applied at the axes of rotation."},{"Start":"03:40.565 ","End":"03:51.099","Text":"That means that r vector for Number 1 is going to be equal to 0."},{"Start":"03:52.370 ","End":"03:57.120","Text":"Now, because I know what my r_1 is and it\u0027s equal to 0,"},{"Start":"03:57.120 ","End":"04:00.245","Text":"it doesn\u0027t matter which equation I plug this into,"},{"Start":"04:00.245 ","End":"04:06.575","Text":"and it doesn\u0027t matter what my force is or what the angle is, it doesn\u0027t matter."},{"Start":"04:06.575 ","End":"04:09.065","Text":"Because I\u0027m multiplying by 0."},{"Start":"04:09.065 ","End":"04:17.525","Text":"This means that r torque at this point is also going to be equal to 0."},{"Start":"04:17.525 ","End":"04:20.809","Text":"As a bit of a generalization,"},{"Start":"04:20.809 ","End":"04:27.109","Text":"if you have a force acting at your axis of rotation,"},{"Start":"04:27.109 ","End":"04:31.200","Text":"then the torque is going to be equal to 0."},{"Start":"04:31.250 ","End":"04:34.444","Text":"It doesn\u0027t matter where your center of rotation is."},{"Start":"04:34.444 ","End":"04:37.279","Text":"If there\u0027s a force acting right there,"},{"Start":"04:37.279 ","End":"04:41.220","Text":"the torque is going to be equal to 0."},{"Start":"04:41.420 ","End":"04:45.904","Text":"If you\u0027re in the exam and you have from a certain point,"},{"Start":"04:45.904 ","End":"04:48.185","Text":"lots of voices coming out,"},{"Start":"04:48.185 ","End":"04:51.724","Text":"let\u0027s say if this question was given to you in the exam,"},{"Start":"04:51.724 ","End":"04:55.895","Text":"this corner, you can see that there\u0027s 2 forces acting over here."},{"Start":"04:55.895 ","End":"04:58.790","Text":"If you want to make the question a little bit easier,"},{"Start":"04:58.790 ","End":"05:01.770","Text":"then what you could do unless told otherwise,"},{"Start":"05:01.770 ","End":"05:05.690","Text":"if you are given the opportunity to choose your center of rotation,"},{"Start":"05:05.690 ","End":"05:09.145","Text":"then you could choose it to be this coin over here,"},{"Start":"05:09.145 ","End":"05:12.220","Text":"and then your torque for these 2 forces,"},{"Start":"05:12.220 ","End":"05:15.050","Text":"F_2 and F_3 will equal 0."},{"Start":"05:15.050 ","End":"05:17.630","Text":"That\u0027s a nice little trick that you can use."},{"Start":"05:17.630 ","End":"05:20.990","Text":"It doesn\u0027t really matter, but it could help you out in the exam."},{"Start":"05:21.590 ","End":"05:26.990","Text":"Now let\u0027s take a look at force Number 2."},{"Start":"05:27.660 ","End":"05:31.270","Text":"Let\u0027s take a look at what is happening."},{"Start":"05:31.270 ","End":"05:35.835","Text":"We\u0027re going to take the size."},{"Start":"05:35.835 ","End":"05:37.849","Text":"Then afterwards using the right-hand rule,"},{"Start":"05:37.849 ","End":"05:40.260","Text":"we\u0027re going to find the direction of the torque,"},{"Start":"05:40.260 ","End":"05:41.749","Text":"the direction of the vector."},{"Start":"05:41.749 ","End":"05:44.285","Text":"That means I must find the size of the vector."},{"Start":"05:44.285 ","End":"05:47.030","Text":"We\u0027re going to take the size of r,"},{"Start":"05:47.030 ","End":"05:52.684","Text":"now what is the size of r. We\u0027re going from axis of rotation"},{"Start":"05:52.684 ","End":"05:59.705","Text":"all the way until we get to the point where our F_2 is working."},{"Start":"05:59.705 ","End":"06:06.289","Text":"Obviously, we\u0027re going along this whole side,"},{"Start":"06:06.289 ","End":"06:10.040","Text":"which we know we\u0027re given in the question is of length a."},{"Start":"06:10.040 ","End":"06:15.525","Text":"The size of our vector r is going to just be a,"},{"Start":"06:15.525 ","End":"06:19.280","Text":"and then we\u0027re multiplying it by the size of the force."},{"Start":"06:19.280 ","End":"06:23.280","Text":"Our force is 10 in newtons."},{"Start":"06:23.650 ","End":"06:28.155","Text":"We\u0027re going to say that it\u0027s times 10."},{"Start":"06:28.155 ","End":"06:33.170","Text":"Then we\u0027re going to multiply it by sine of the angle."},{"Start":"06:33.170 ","End":"06:34.909","Text":"Now, what is the angle?"},{"Start":"06:34.909 ","End":"06:39.514","Text":"What we can see is that if we carry on in this direction,"},{"Start":"06:39.514 ","End":"06:46.790","Text":"so the angle between F_2 and here is going to be the same angle as over here."},{"Start":"06:46.790 ","End":"06:53.705","Text":"Because we know that the angles in an equilateral triangle are equal to 60 degrees,"},{"Start":"06:53.705 ","End":"06:56.509","Text":"so if this angle is 60 degrees,"},{"Start":"06:56.509 ","End":"06:59.795","Text":"then this angle over here is going to be 60 degrees,"},{"Start":"06:59.795 ","End":"07:03.215","Text":"so we\u0027re going to multiply by sine of 60."},{"Start":"07:03.215 ","End":"07:06.619","Text":"Now, another way that we could have done it instead of taking"},{"Start":"07:06.619 ","End":"07:11.400","Text":"this angle is we could have worked out what this angle is."},{"Start":"07:12.110 ","End":"07:19.775","Text":"Now, it doesn\u0027t matter because then we\u0027d have sine of 180 minus 60,"},{"Start":"07:19.775 ","End":"07:22.410","Text":"so that will give us this angle."},{"Start":"07:23.690 ","End":"07:32.270","Text":"As we know when we\u0027re doing sine of 180 minus an angle or sine of the angle,"},{"Start":"07:32.270 ","End":"07:34.990","Text":"the answer is going to be the same anyway."},{"Start":"07:34.990 ","End":"07:42.110","Text":"We would have gotten the same answer either way. Let\u0027s see."},{"Start":"07:42.110 ","End":"07:47.580","Text":"This is going to be equal to 5 root 3a."},{"Start":"07:47.720 ","End":"07:49.935","Text":"That\u0027s the size of our torque,"},{"Start":"07:49.935 ","End":"07:52.650","Text":"and now let\u0027s figure out the direction."},{"Start":"07:52.650 ","End":"07:56.825","Text":"First, let\u0027s speak about the more mathematical way."},{"Start":"07:56.825 ","End":"07:58.925","Text":"We have the right-hand rule,"},{"Start":"07:58.925 ","End":"08:03.200","Text":"which uses the index finger to represent"},{"Start":"08:03.200 ","End":"08:09.050","Text":"the first thing that you\u0027re multiplying by when you are doing the cross-product,"},{"Start":"08:09.050 ","End":"08:12.605","Text":"and then your middle finger to represent the second item,"},{"Start":"08:12.605 ","End":"08:20.025","Text":"and then your thumb represents the direction that the result is acting in."},{"Start":"08:20.025 ","End":"08:21.975","Text":"In this case, the torque."},{"Start":"08:21.975 ","End":"08:26.330","Text":"Here, your index finger will represent the r vector."},{"Start":"08:26.330 ","End":"08:29.539","Text":"Over here, we have the r vector."},{"Start":"08:29.539 ","End":"08:33.140","Text":"Then your middle finger represents the force."},{"Start":"08:33.140 ","End":"08:37.339","Text":"You can see that if you\u0027re going to apply the force,"},{"Start":"08:37.339 ","End":"08:41.245","Text":"the force is going in the rightwards direction."},{"Start":"08:41.245 ","End":"08:47.735","Text":"Then you have your index finger pointing upwards in the page."},{"Start":"08:47.735 ","End":"08:51.020","Text":"In this direction, this is your index finger."},{"Start":"08:51.020 ","End":"08:55.720","Text":"Then your middle finger is going to be pointing in this direction."},{"Start":"08:55.720 ","End":"09:00.020","Text":"Then you\u0027ll see that in order to get this configuration on your right hand,"},{"Start":"09:00.020 ","End":"09:03.840","Text":"you\u0027re going to have to have your thumb facing downwards."},{"Start":"09:03.840 ","End":"09:06.729","Text":"If your thumb is facing downwards,"},{"Start":"09:06.729 ","End":"09:11.719","Text":"then the torque is going in the negative z direction."},{"Start":"09:11.719 ","End":"09:19.765","Text":"Which means that the direction of the force is going in the clockwise direction."},{"Start":"09:19.765 ","End":"09:22.220","Text":"Then you know that that\u0027s"},{"Start":"09:22.220 ","End":"09:25.370","Text":"the negative direction to what we said is the positive direction."},{"Start":"09:25.370 ","End":"09:28.920","Text":"We have a negative over here."},{"Start":"09:29.510 ","End":"09:33.485","Text":"If we draw this onto the graph,"},{"Start":"09:33.485 ","End":"09:37.745","Text":"so we can say that if this is our r vector,"},{"Start":"09:37.745 ","End":"09:46.581","Text":"so we can see that our force vector is going in this direction."},{"Start":"09:46.581 ","End":"09:51.190","Text":"Our force vector is going in this direction."},{"Start":"09:51.190 ","End":"09:55.750","Text":"Then you can see that to get from your r to your F,"},{"Start":"09:55.750 ","End":"09:57.849","Text":"you have to go in this direction,"},{"Start":"09:57.849 ","End":"10:01.165","Text":"which is anticlockwise so there\u0027s a negative."},{"Start":"10:01.165 ","End":"10:07.239","Text":"Now, another way that you can look at it from a more intuitive perspective is if you,"},{"Start":"10:07.239 ","End":"10:12.715","Text":"for instance, put your thumb on this corner and you hold it down really, really hard."},{"Start":"10:12.715 ","End":"10:16.030","Text":"If you disregard all the other forces, F_1,"},{"Start":"10:16.030 ","End":"10:18.939","Text":"F_4, and F_3, then you just look,"},{"Start":"10:18.939 ","End":"10:23.369","Text":"if you\u0027re holding down this corner with"},{"Start":"10:23.369 ","End":"10:25.829","Text":"your thumb over here and you\u0027re pushing"},{"Start":"10:25.829 ","End":"10:29.190","Text":"this corner in the direction that the force is pointing in,"},{"Start":"10:29.190 ","End":"10:34.240","Text":"then you\u0027ll see that the whole triangle will move in this direction which"},{"Start":"10:34.240 ","End":"10:36.460","Text":"is the negative direction because"},{"Start":"10:36.460 ","End":"10:39.550","Text":"we said that the anticlockwise direction is the positive."},{"Start":"10:39.550 ","End":"10:41.965","Text":"Then you have a minus over here."},{"Start":"10:41.965 ","End":"10:45.370","Text":"Now, moving on to F_3."},{"Start":"10:45.370 ","End":"10:53.679","Text":"We\u0027re going to try and work out the torque of the force. Let\u0027s see."},{"Start":"10:53.679 ","End":"10:57.370","Text":"Now, we can notice that if this is 60 degrees and we just"},{"Start":"10:57.370 ","End":"11:01.345","Text":"said for T2 that this angle was the same,"},{"Start":"11:01.345 ","End":"11:06.264","Text":"so also 60 degrees we can see that the angle"},{"Start":"11:06.264 ","End":"11:14.830","Text":"between our force F_3 our vector and our vector for r,"},{"Start":"11:14.830 ","End":"11:20.725","Text":"I\u0027m reminding you that our r vector was going also in this direction."},{"Start":"11:20.725 ","End":"11:26.619","Text":"The angle between our force and our r is 0 and sine of 0,"},{"Start":"11:26.619 ","End":"11:28.630","Text":"as you know, is equal to 0."},{"Start":"11:28.630 ","End":"11:34.810","Text":"Then we\u0027ll get that I torque 3 is equal to 0."},{"Start":"11:34.810 ","End":"11:41.440","Text":"That\u0027s because the angle Alpha between r and F"},{"Start":"11:41.440 ","End":"11:49.160","Text":"is equal to 0 and sine of 0 is equal to 0."},{"Start":"11:49.500 ","End":"11:56.080","Text":"Now let\u0027s see what our torque number 4 is equal to."},{"Start":"11:56.080 ","End":"11:59.694","Text":"Before we move on, as a general rule,"},{"Start":"11:59.694 ","End":"12:03.534","Text":"if the angle between 2 vectors is equal to 0,"},{"Start":"12:03.534 ","End":"12:09.050","Text":"then its moment, its torque will also be equal to 0, always."},{"Start":"12:09.240 ","End":"12:12.354","Text":"Now, if we\u0027re going to take a look,"},{"Start":"12:12.354 ","End":"12:18.040","Text":"then we can see that our r vector is going in this direction."},{"Start":"12:18.040 ","End":"12:21.040","Text":"All along until it gets over here."},{"Start":"12:21.040 ","End":"12:24.490","Text":"This is r vector imagine that this is a straight line."},{"Start":"12:24.490 ","End":"12:27.744","Text":"Our r vector is,"},{"Start":"12:27.744 ","End":"12:31.569","Text":"as we can see along the x-axis and we"},{"Start":"12:31.569 ","End":"12:36.220","Text":"have our F_4 which we can see is 10 newtons in the y-direction."},{"Start":"12:36.220 ","End":"12:38.050","Text":"It\u0027s on the y-axis,"},{"Start":"12:38.050 ","End":"12:42.580","Text":"which means that there\u0027s 90 degrees between the 2."},{"Start":"12:42.580 ","End":"12:47.380","Text":"Let\u0027s write down the size of the torque so it\u0027s our r vector, the size of it."},{"Start":"12:47.380 ","End":"12:50.890","Text":"Again, we\u0027re going along this side,"},{"Start":"12:50.890 ","End":"12:52.209","Text":"which as we know is length a,"},{"Start":"12:52.209 ","End":"12:54.325","Text":"because it\u0027s an equilateral triangle."},{"Start":"12:54.325 ","End":"12:58.329","Text":"We have a multiplied by the size of the force,"},{"Start":"12:58.329 ","End":"13:01.359","Text":"which is 10 and then sine,"},{"Start":"13:01.359 ","End":"13:06.445","Text":"the angle between them and the angle between them is 90 degrees."},{"Start":"13:06.445 ","End":"13:09.310","Text":"Sine of 90 is equal to 1,"},{"Start":"13:09.310 ","End":"13:14.350","Text":"so we have that the torque is equal to 10a."},{"Start":"13:14.350 ","End":"13:17.605","Text":"Now, let\u0027s see the direction."},{"Start":"13:17.605 ","End":"13:21.295","Text":"What you can do is either with the right-hand rule,"},{"Start":"13:21.295 ","End":"13:26.154","Text":"if you take your fingers in the direction of the force,"},{"Start":"13:26.154 ","End":"13:28.434","Text":"you can see that it\u0027s going to move"},{"Start":"13:28.434 ","End":"13:33.295","Text":"this whole configuration is going to go in this direction,"},{"Start":"13:33.295 ","End":"13:36.579","Text":"which as we can see is the anticlockwise direction,"},{"Start":"13:36.579 ","End":"13:39.024","Text":"which we\u0027ve said as the positive direction."},{"Start":"13:39.024 ","End":"13:44.919","Text":"Another way you can do it is you can say that if you put your thumb here again,"},{"Start":"13:44.919 ","End":"13:47.290","Text":"holding down the axis of rotation,"},{"Start":"13:47.290 ","End":"13:49.509","Text":"and then you apply this upwards force,"},{"Start":"13:49.509 ","End":"13:52.269","Text":"you\u0027ll see that the entire triangle will"},{"Start":"13:52.269 ","End":"13:55.599","Text":"rotate in this direction which again is the anticlockwise direction,"},{"Start":"13:55.599 ","End":"13:57.160","Text":"which is the positive direction."},{"Start":"13:57.160 ","End":"14:00.759","Text":"Another way with the other right-hand rule is if you"},{"Start":"14:00.759 ","End":"14:05.365","Text":"have your index finger pointing in the direction of the radius,"},{"Start":"14:05.365 ","End":"14:07.839","Text":"so your index finger is going to be pointing in"},{"Start":"14:07.839 ","End":"14:11.994","Text":"this direction then you have your middle finger"},{"Start":"14:11.994 ","End":"14:18.309","Text":"pointing in the direction of the force that is going to be in this direction."},{"Start":"14:18.309 ","End":"14:20.900","Text":"Now remember with vectors"},{"Start":"14:22.430 ","End":"14:28.945","Text":"that you can move them around as long as they keep their size and direction."},{"Start":"14:28.945 ","End":"14:31.359","Text":"I could have also drawn the force up in"},{"Start":"14:31.359 ","End":"14:34.420","Text":"this direction or I could have drawn the force like this."},{"Start":"14:34.420 ","End":"14:39.165","Text":"It doesn\u0027t matter as long as it maintains the same size and direction."},{"Start":"14:39.165 ","End":"14:43.245","Text":"If I wanted to rotate my radius until my force,"},{"Start":"14:43.245 ","End":"14:46.185","Text":"I would have to go in the anticlockwise direction,"},{"Start":"14:46.185 ","End":"14:49.385","Text":"which is the positive direction."},{"Start":"14:49.385 ","End":"14:53.079","Text":"Again, we can see that and also we can see"},{"Start":"14:53.079 ","End":"14:56.170","Text":"that if we\u0027re doing the right-hand rule and our radius,"},{"Start":"14:56.170 ","End":"14:57.459","Text":"this is our index finger,"},{"Start":"14:57.459 ","End":"14:59.830","Text":"and that forces is the middle finger,"},{"Start":"14:59.830 ","End":"15:03.564","Text":"then we can see that our thumb will be pointing outwards of the page,"},{"Start":"15:03.564 ","End":"15:07.790","Text":"which is also the direction of the axis of rotation,"},{"Start":"15:08.730 ","End":"15:11.185","Text":"it\u0027s coming out of the page."},{"Start":"15:11.185 ","End":"15:13.285","Text":"That\u0027s the positive z direction."},{"Start":"15:13.285 ","End":"15:18.144","Text":"Again, we can see that this is in the positive direction."},{"Start":"15:18.144 ","End":"15:19.915","Text":"This was the end of question number 1."},{"Start":"15:19.915 ","End":"15:22.915","Text":"Now let\u0027s go into question number 2."},{"Start":"15:22.915 ","End":"15:27.565","Text":"Question number 2, we\u0027re told to work out the moment of force of gravity."},{"Start":"15:27.565 ","End":"15:31.690","Text":"Now remember that the moment of force is the same as the torque."},{"Start":"15:31.690 ","End":"15:35.830","Text":"We\u0027re being told that the triangle\u0027s mass is M,"},{"Start":"15:35.830 ","End":"15:43.030","Text":"and then its center of mass is located at half of a and 1 over 2 root 3a."},{"Start":"15:43.030 ","End":"15:46.060","Text":"That\u0027s over here. I labeled it already."},{"Start":"15:46.060 ","End":"15:51.760","Text":"Now, what we have to do is just like in our force diagrams,"},{"Start":"15:51.760 ","End":"15:58.030","Text":"we label the arrow of the force acting from gravity."},{"Start":"15:58.030 ","End":"16:06.115","Text":"As we know that that\u0027s Mg and our mass is capital M. We have capital Mg acting downwards."},{"Start":"16:06.115 ","End":"16:09.370","Text":"Now, this is a little bit of a trick question."},{"Start":"16:09.370 ","End":"16:14.170","Text":"Let\u0027s begin. We\u0027re going to have our torque due to gravity."},{"Start":"16:14.170 ","End":"16:18.970","Text":"I\u0027ll write here g for gravity is going to be equal to,"},{"Start":"16:18.970 ","End":"16:21.234","Text":"so what is our force?"},{"Start":"16:21.234 ","End":"16:24.115","Text":"Because we\u0027re using the equation for the size."},{"Start":"16:24.115 ","End":"16:30.235","Text":"Our force, of course, is Mg. We\u0027re writing over here Mg. Now,"},{"Start":"16:30.235 ","End":"16:34.584","Text":"we have to multiply it by the radius,"},{"Start":"16:34.584 ","End":"16:38.890","Text":"which is going to be this."},{"Start":"16:38.890 ","End":"16:46.490","Text":"This is our r vector and then we multiply it by sine of the angle."},{"Start":"16:47.310 ","End":"16:52.959","Text":"This can be very complicated because then we have to find out what this distance is,"},{"Start":"16:52.959 ","End":"16:54.520","Text":"what this distance is,"},{"Start":"16:54.520 ","End":"16:57.820","Text":"and then use Pythagoras in order to find this."},{"Start":"16:57.820 ","End":"17:01.179","Text":"But then we also have to know what this angle is and"},{"Start":"17:01.179 ","End":"17:04.539","Text":"this angle and everything and it\u0027s really difficult and it\u0027s really long."},{"Start":"17:04.539 ","End":"17:08.065","Text":"We can do it but why when there\u0027s an easier way?"},{"Start":"17:08.065 ","End":"17:13.839","Text":"The simpler way is instead of doing r multiplied by the sine of the angle between them,"},{"Start":"17:13.839 ","End":"17:18.160","Text":"you find the component of your r vector,"},{"Start":"17:18.160 ","End":"17:22.040","Text":"which is perpendicular to the force."},{"Start":"17:22.350 ","End":"17:26.080","Text":"I wrote it over here so that you can remember"},{"Start":"17:26.080 ","End":"17:29.800","Text":"this and maybe also write this into your notes."},{"Start":"17:29.800 ","End":"17:35.110","Text":"Now, what is the component which is perpendicular to the force?"},{"Start":"17:35.110 ","End":"17:38.270","Text":"That means this."},{"Start":"17:38.430 ","End":"17:41.815","Text":"Then we know that its length,"},{"Start":"17:41.815 ","End":"17:43.449","Text":"from what is given,"},{"Start":"17:43.449 ","End":"17:48.820","Text":"we know that the center of mass is located in the x-direction at half a,"},{"Start":"17:48.820 ","End":"17:50.440","Text":"from what\u0027s given over here."},{"Start":"17:50.440 ","End":"17:56.020","Text":"We know that this length over here is going to be half of a over 2."},{"Start":"17:56.020 ","End":"18:01.045","Text":"Then we just multiply this by a over 2, and that\u0027s it."},{"Start":"18:01.045 ","End":"18:06.759","Text":"Then that\u0027s the only calculation that you have to do and nothing else."},{"Start":"18:06.759 ","End":"18:08.965","Text":"Now, this by the way,"},{"Start":"18:08.965 ","End":"18:15.475","Text":"is also called the r effective because it\u0027s the effects of radius that we\u0027re using."},{"Start":"18:15.475 ","End":"18:18.145","Text":"This over here is r eff."},{"Start":"18:18.145 ","End":"18:23.620","Text":"It\u0027s a over 2 and that is also speaking about this."},{"Start":"18:23.620 ","End":"18:28.135","Text":"Now, let\u0027s go on to question number 3."},{"Start":"18:28.135 ","End":"18:30.085","Text":"In question number 3,"},{"Start":"18:30.085 ","End":"18:31.390","Text":"we\u0027re being told to, again,"},{"Start":"18:31.390 ","End":"18:35.395","Text":"work out the moments about the center of mass of the triangle."},{"Start":"18:35.395 ","End":"18:39.100","Text":"In our first question, we were asked to calculate"},{"Start":"18:39.100 ","End":"18:43.750","Text":"the moments about this axis of rotation, but now,"},{"Start":"18:43.750 ","End":"18:47.770","Text":"they\u0027re telling us that our center of mass is our new axis of rotation and"},{"Start":"18:47.770 ","End":"18:52.840","Text":"to just do the exact same thing as in question 1 but about a different axis of rotation."},{"Start":"18:52.840 ","End":"19:00.970","Text":"We\u0027re also told that the angle between F_1 and the side of the triangle is 60 degrees."},{"Start":"19:00.970 ","End":"19:03.880","Text":"Now we\u0027re being told that this is 60 degrees,"},{"Start":"19:03.880 ","End":"19:05.935","Text":"this angle over here."},{"Start":"19:05.935 ","End":"19:09.530","Text":"Let\u0027s see how we do this."},{"Start":"19:09.690 ","End":"19:17.875","Text":"What we\u0027re going to do is we\u0027re going to start with force F_1."},{"Start":"19:17.875 ","End":"19:22.765","Text":"What we want to do is we have to have our radius."},{"Start":"19:22.765 ","End":"19:29.500","Text":"Remember r vector going from the axis of rotation until where the force is being applied."},{"Start":"19:29.500 ","End":"19:32.260","Text":"That means like this."},{"Start":"19:32.260 ","End":"19:36.380","Text":"This is a vector going in this direction."},{"Start":"19:37.560 ","End":"19:41.990","Text":"Now we have to write down our equation for"},{"Start":"19:41.990 ","End":"19:46.999","Text":"torque for number 1 and we\u0027re going to do the size of the vector,"},{"Start":"19:46.999 ","End":"19:48.320","Text":"which we still don\u0027t know."},{"Start":"19:48.320 ","End":"19:51.335","Text":"We\u0027re just going to leave it as an unknown,"},{"Start":"19:51.335 ","End":"19:55.234","Text":"multiplied by the size of the F vector,"},{"Start":"19:55.234 ","End":"19:59.465","Text":"which is 10 and multiplied by the sine of the angle."},{"Start":"19:59.465 ","End":"20:01.819","Text":"Now, because this is an equilateral triangle,"},{"Start":"20:01.819 ","End":"20:04.729","Text":"so we know that the center of mass,"},{"Start":"20:04.729 ","End":"20:08.329","Text":"the line going from the center of mass to each corner is"},{"Start":"20:08.329 ","End":"20:12.310","Text":"an angle bisection and it goes right through the center of the angle."},{"Start":"20:12.310 ","End":"20:17.239","Text":"Which means that if each angle in the equilateral triangle is 60 degrees,"},{"Start":"20:17.239 ","End":"20:22.305","Text":"so half of 60 degrees is going to be 30 degrees."},{"Start":"20:22.305 ","End":"20:28.750","Text":"Then we have 30 degrees plus 60 degrees over here is going to be 90 degrees."},{"Start":"20:28.750 ","End":"20:32.389","Text":"Then we multiply by sine of that angle which is sine 90,"},{"Start":"20:32.389 ","End":"20:35.635","Text":"which is going to be equal to 1."},{"Start":"20:35.635 ","End":"20:44.015","Text":"Now, what we have to do is we have to find out the size of our r vector is."},{"Start":"20:44.015 ","End":"20:53.265","Text":"By using Pythagoras, we know that this height over here is a over 2 root 3,"},{"Start":"20:53.265 ","End":"20:57.620","Text":"and that this length over here is a over 2."},{"Start":"20:57.620 ","End":"21:02.990","Text":"By doing Pythagoras, the radius is the hypotenuse,"},{"Start":"21:02.990 ","End":"21:13.390","Text":"so we have half of a squared plus 1 over 2 root 3a."},{"Start":"21:13.390 ","End":"21:18.145","Text":"All of this squared and then with the square root."},{"Start":"21:18.145 ","End":"21:26.625","Text":"Then we\u0027ll get the result that it equals to 1 over root 3a."},{"Start":"21:26.625 ","End":"21:31.285","Text":"The length of the radius is 1 over root 3a,"},{"Start":"21:31.285 ","End":"21:37.360","Text":"so then we just substitute this in so we have 10 times 1 times 1 over root 3,"},{"Start":"21:37.360 ","End":"21:44.150","Text":"which will equal to 10a divided by root 3."},{"Start":"21:44.490 ","End":"21:53.665","Text":"That is our torque of force 1 and now all we have to do is find the direction."},{"Start":"21:53.665 ","End":"21:57.160","Text":"Again, doing exactly what we\u0027ve spoken about."},{"Start":"21:57.160 ","End":"22:01.450","Text":"If you hold your thumb at the center of mass,"},{"Start":"22:01.450 ","End":"22:07.209","Text":"because now this is the axis of rotation and then you"},{"Start":"22:07.209 ","End":"22:13.705","Text":"apply the force F_1 and you\u0027re pressing down at the center of mass,"},{"Start":"22:13.705 ","End":"22:22.150","Text":"then you\u0027ll see that the whole triangle is going to spin in the anticlockwise direction,"},{"Start":"22:22.150 ","End":"22:24.175","Text":"which is the negative direction."},{"Start":"22:24.175 ","End":"22:26.770","Text":"Similarly with the right-hand rule,"},{"Start":"22:26.770 ","End":"22:34.365","Text":"if you take your index finger and point it in the direction of your radius vector,"},{"Start":"22:34.365 ","End":"22:35.685","Text":"which is going to be this."},{"Start":"22:35.685 ","End":"22:42.210","Text":"This is your index finger and then you have your middle finger going in the force,"},{"Start":"22:42.210 ","End":"22:43.739","Text":"the direction of the force,"},{"Start":"22:43.739 ","End":"22:46.210","Text":"which is like this."},{"Start":"22:46.370 ","End":"22:50.490","Text":"Then you\u0027ll see in order for your radius to spin to"},{"Start":"22:50.490 ","End":"22:54.269","Text":"your force is going to have to go in this direction."},{"Start":"22:54.269 ","End":"22:57.479","Text":"Again, this is perpendicular and then you can see again that"},{"Start":"22:57.479 ","End":"23:01.780","Text":"it\u0027s anticlockwise and so it\u0027s in the negative direction."},{"Start":"23:02.670 ","End":"23:07.070","Text":"Then we\u0027ll put a negative over here."},{"Start":"23:07.830 ","End":"23:12.790","Text":"Now, we\u0027re going to take a look at Force Number 2 so"},{"Start":"23:12.790 ","End":"23:18.230","Text":"this is Tau 2 and I\u0027m going to rub this out."},{"Start":"23:19.260 ","End":"23:23.679","Text":"Now, the first thing we do is that we draw"},{"Start":"23:23.679 ","End":"23:29.979","Text":"our radius from the axis of rotation until the point where our force is being applied,"},{"Start":"23:29.979 ","End":"23:32.844","Text":"so this is straight up."},{"Start":"23:32.844 ","End":"23:38.299","Text":"We see that our r is going in the y direction."},{"Start":"23:38.460 ","End":"23:42.835","Text":"Then we\u0027re told here that F_2 is in the x direction,"},{"Start":"23:42.835 ","End":"23:47.350","Text":"which means that they\u0027re perpendicular one to another."},{"Start":"23:47.350 ","End":"23:57.804","Text":"We have the size of our r and it is going to be just the same as what we worked out,"},{"Start":"23:57.804 ","End":"24:00.250","Text":"because it\u0027s an equilateral triangle."},{"Start":"24:00.250 ","End":"24:05.965","Text":"The size of our r is going to be 1 over root 3a."},{"Start":"24:05.965 ","End":"24:10.179","Text":"Then we\u0027re going to do multiplied by the size of the force,"},{"Start":"24:10.179 ","End":"24:11.665","Text":"which is 10,"},{"Start":"24:11.665 ","End":"24:15.880","Text":"and multiplied by sine of the angle between which is again"},{"Start":"24:15.880 ","End":"24:20.545","Text":"90 degrees and sin of 90 is equal to 1, so times 1."},{"Start":"24:20.545 ","End":"24:27.595","Text":"Again, we\u0027re going to get 10a divided by root 3."},{"Start":"24:27.595 ","End":"24:33.145","Text":"Now, we just have to again decide the direction of the force."},{"Start":"24:33.145 ","End":"24:36.100","Text":"Again, if you put your thumb on the center of mass and"},{"Start":"24:36.100 ","End":"24:40.630","Text":"you push the triangle in the direction that the force is going,"},{"Start":"24:40.630 ","End":"24:47.545","Text":"then you\u0027ll see that the triangle is going to rotate again in the clockwise direction,"},{"Start":"24:47.545 ","End":"24:49.435","Text":"which is the negative direction."},{"Start":"24:49.435 ","End":"24:51.715","Text":"Similarly, if we draw this out,"},{"Start":"24:51.715 ","End":"24:55.600","Text":"we have our r vector going in this direction and"},{"Start":"24:55.600 ","End":"25:02.110","Text":"our force vector going in this direction."},{"Start":"25:02.110 ","End":"25:06.220","Text":"Then because our r is the first component in our equation,"},{"Start":"25:06.220 ","End":"25:10.284","Text":"so we have to have the force so it will be going like this,"},{"Start":"25:10.284 ","End":"25:13.059","Text":"which is the anticlockwise direction, again,"},{"Start":"25:13.059 ","End":"25:18.400","Text":"so that\u0027s our negative direction with how we defined it."},{"Start":"25:18.400 ","End":"25:24.500","Text":"Now let\u0027s speak about the torque of Force Number 3."},{"Start":"25:25.380 ","End":"25:27.580","Text":"Force Number 3,"},{"Start":"25:27.580 ","End":"25:29.620","Text":"it\u0027s being applied at the exact same points,"},{"Start":"25:29.620 ","End":"25:36.130","Text":"so my radius, my r vector is going to the exact same place and again,"},{"Start":"25:36.130 ","End":"25:40.075","Text":"as we know our r vector is going to be equal to 1 over"},{"Start":"25:40.075 ","End":"25:45.520","Text":"root 3a multiplied by the size of the force,"},{"Start":"25:45.520 ","End":"25:48.280","Text":"which again is 10 newtons."},{"Start":"25:48.280 ","End":"25:50.425","Text":"Now, as we know,"},{"Start":"25:50.425 ","End":"25:53.350","Text":"F_3 is pointing in the same direction."},{"Start":"25:53.350 ","End":"25:57.939","Text":"It\u0027s in the same line as this side of the triangle,"},{"Start":"25:57.939 ","End":"26:02.710","Text":"which means that the angle here is equal to 60 degrees as well."},{"Start":"26:02.710 ","End":"26:11.469","Text":"What is this angle is going to be 90 plus 60,"},{"Start":"26:11.469 ","End":"26:16.010","Text":"which is going to be equal to sine of 150."},{"Start":"26:18.390 ","End":"26:28.105","Text":"Or similarly, we could have said that it was sine of 180 minus 150,"},{"Start":"26:28.105 ","End":"26:35.289","Text":"which would have given us sine of 30 so you can say either a sign of 150 or sine of 30,"},{"Start":"26:35.289 ","End":"26:37.779","Text":"we will get the same answer."},{"Start":"26:37.779 ","End":"26:44.004","Text":"Sine of 150 is equal to sine of 130 and sine of 130 is equal to a 1/2"},{"Start":"26:44.004 ","End":"26:52.250","Text":"so then 10 divided by 2 is equal to 5 multiplied by a divided by root 3."},{"Start":"26:52.470 ","End":"26:56.185","Text":"Now we have to see what the sign is."},{"Start":"26:56.185 ","End":"27:01.585","Text":"Again, if I\u0027m holding my thumb down at the center of mass and I\u0027m applying this force,"},{"Start":"27:01.585 ","End":"27:08.170","Text":"we\u0027ll see that the triangle will rotate in the anticlockwise direction and also,"},{"Start":"27:08.170 ","End":"27:09.310","Text":"if I draw this,"},{"Start":"27:09.310 ","End":"27:13.825","Text":"so I\u0027ll have the force going in this direction and"},{"Start":"27:13.825 ","End":"27:22.000","Text":"my r vector going in this direction so to get my r vector to reach my force,"},{"Start":"27:22.000 ","End":"27:23.949","Text":"it\u0027s going to go in this direction,"},{"Start":"27:23.949 ","End":"27:28.060","Text":"which is the anticlockwise direction."},{"Start":"27:28.060 ","End":"27:31.600","Text":"Again, this is going to be a negative."},{"Start":"27:31.600 ","End":"27:36.434","Text":"Now let\u0027s see what our torque for 4 is."},{"Start":"27:36.434 ","End":"27:43.585","Text":"This time,"},{"Start":"27:43.585 ","End":"27:52.525","Text":"my r vector is going to be going from the center of mass,"},{"Start":"27:52.525 ","End":"27:55.390","Text":"from the axis of rotation,"},{"Start":"27:55.390 ","End":"28:01.780","Text":"until the point where my force is being applied, so that\u0027s that."},{"Start":"28:01.780 ","End":"28:03.940","Text":"Again, my r vector,"},{"Start":"28:03.940 ","End":"28:05.709","Text":"because it\u0027s an equilateral triangle,"},{"Start":"28:05.709 ","End":"28:09.235","Text":"is going to be 1 over root 3a,"},{"Start":"28:09.235 ","End":"28:11.440","Text":"like we previously worked out,"},{"Start":"28:11.440 ","End":"28:14.455","Text":"multiplied by the size of the force,"},{"Start":"28:14.455 ","End":"28:16.105","Text":"so multiplied by 10,"},{"Start":"28:16.105 ","End":"28:22.420","Text":"multiplied by sine of the angle between them."},{"Start":"28:22.420 ","End":"28:28.750","Text":"Now here what we could do is we know that this angle is at 90 degrees,"},{"Start":"28:28.750 ","End":"28:32.335","Text":"so what we could do is we could say that because it\u0027s having,"},{"Start":"28:32.335 ","End":"28:38.215","Text":"this angle over here is going to be 30 degrees and then work out with this angle"},{"Start":"28:38.215 ","End":"28:45.370","Text":"over here is and then we could find out sign of that."},{"Start":"28:45.370 ","End":"28:49.105","Text":"Or alternatively, we can again use our r effective,"},{"Start":"28:49.105 ","End":"28:54.130","Text":"remember that that means taking the perpendicular component of"},{"Start":"28:54.130 ","End":"29:00.265","Text":"our r vector to a force vector so that means this component,"},{"Start":"29:00.265 ","End":"29:03.565","Text":"because this is going along the x axis,"},{"Start":"29:03.565 ","End":"29:06.729","Text":"which is perpendicular to the y-axis and we\u0027re being told that"},{"Start":"29:06.729 ","End":"29:10.780","Text":"our F_4 is working in the y direction,"},{"Start":"29:10.780 ","End":"29:13.900","Text":"and as we know, because it\u0027s an equilateral triangle,"},{"Start":"29:13.900 ","End":"29:17.185","Text":"that this is going to be 1/2a,"},{"Start":"29:17.185 ","End":"29:19.360","Text":"this length over here."},{"Start":"29:19.360 ","End":"29:25.190","Text":"Then we just multiply it by 1/2(a)."},{"Start":"29:26.640 ","End":"29:29.485","Text":"But then because we\u0027re using the r effective,"},{"Start":"29:29.485 ","End":"29:34.300","Text":"then we don\u0027t need the size of this, we don\u0027t need that."},{"Start":"29:34.300 ","End":"29:37.449","Text":"The only time we would have needed the size of"},{"Start":"29:37.449 ","End":"29:43.105","Text":"our r angle is if we\u0027re then also using the sine of that angle."},{"Start":"29:43.105 ","End":"29:45.490","Text":"Because we\u0027re not using that,"},{"Start":"29:45.490 ","End":"29:48.805","Text":"we\u0027re just using our r effective so we just have this,"},{"Start":"29:48.805 ","End":"29:52.610","Text":"which is going to be equal to 5a."},{"Start":"29:53.790 ","End":"29:57.519","Text":"Again, let\u0027s go over what happened."},{"Start":"29:57.519 ","End":"30:01.045","Text":"If you\u0027re using the size of your r vector,"},{"Start":"30:01.045 ","End":"30:05.665","Text":"then you\u0027re also working out sine of the angle and then you have to work out the angle."},{"Start":"30:05.665 ","End":"30:08.920","Text":"If it\u0027s a little bit tricky to find the angle,"},{"Start":"30:08.920 ","End":"30:11.380","Text":"then you\u0027re not going to use the size of the r vector."},{"Start":"30:11.380 ","End":"30:14.830","Text":"You are going to use the size of your r effective."},{"Start":"30:14.830 ","End":"30:18.339","Text":"Then the size of your r effective is going to be"},{"Start":"30:18.339 ","End":"30:24.280","Text":"the size of the perpendicular component to your force vector."},{"Start":"30:24.280 ","End":"30:29.571","Text":"The perpendicular component of your r vector to your force vector."},{"Start":"30:29.571 ","End":"30:35.319","Text":"That is going to be 5a and then to work out the sign, so again,"},{"Start":"30:35.319 ","End":"30:39.909","Text":"if you hold your thumb here at the axis of rotation,"},{"Start":"30:39.909 ","End":"30:42.190","Text":"and then you apply the force,"},{"Start":"30:42.190 ","End":"30:46.209","Text":"you\u0027ll see that the triangle will spin in the anticlockwise direction."},{"Start":"30:46.209 ","End":"30:49.290","Text":"Similarly, if we draw our r vector,"},{"Start":"30:49.290 ","End":"30:53.850","Text":"it\u0027s going in this direction and our force vector is going"},{"Start":"30:53.850 ","End":"30:58.315","Text":"in this direction and then to get our r into our force,"},{"Start":"30:58.315 ","End":"31:00.850","Text":"we have to go in the anticlockwise direction,"},{"Start":"31:00.850 ","End":"31:02.319","Text":"which as we said before,"},{"Start":"31:02.319 ","End":"31:03.820","Text":"is our positive direction,"},{"Start":"31:03.820 ","End":"31:08.810","Text":"so it\u0027s in the positive direction, like so."},{"Start":"31:08.970 ","End":"31:17.800","Text":"Now finally, we can work out the torque of our gravity."},{"Start":"31:17.800 ","End":"31:23.350","Text":"As we know, we have our M_g axing downwards."},{"Start":"31:23.350 ","End":"31:26.125","Text":"Now, if I were to now draw"},{"Start":"31:26.125 ","End":"31:33.279","Text":"the radius vector from my axis of rotation to where my force is being applied."},{"Start":"31:33.279 ","End":"31:36.490","Text":"The angle is going to be 0,"},{"Start":"31:36.490 ","End":"31:40.719","Text":"which means that the torque is equal to 0."},{"Start":"31:40.719 ","End":"31:45.309","Text":"So that is the end of this lesson."},{"Start":"31:45.309 ","End":"31:48.460","Text":"Now, the important things to take from this is 1,"},{"Start":"31:48.460 ","End":"31:57.370","Text":"knowing that if the angle between your radius vector and your force vector is equal to 0,"},{"Start":"31:57.370 ","End":"32:01.429","Text":"then your torque is going to be equal to 0."},{"Start":"32:02.990 ","End":"32:09.824","Text":"Similarly, if your radius vector is coming out of your axis of rotation,"},{"Start":"32:09.824 ","End":"32:15.164","Text":"it means that the length of your r vector is going to be equal to 0,"},{"Start":"32:15.164 ","End":"32:19.115","Text":"which also means that your torque is going to be equal to 0."},{"Start":"32:19.115 ","End":"32:23.649","Text":"Then lastly, if you see that you\u0027re having"},{"Start":"32:23.649 ","End":"32:26.335","Text":"a complicated time trying to find out"},{"Start":"32:26.335 ","End":"32:30.310","Text":"the angle between your r vector and your force vector,"},{"Start":"32:30.310 ","End":"32:31.750","Text":"then if possible,"},{"Start":"32:31.750 ","End":"32:34.165","Text":"you can work out the r effective,"},{"Start":"32:34.165 ","End":"32:37.990","Text":"which is the component of your r vector,"},{"Start":"32:37.990 ","End":"32:42.115","Text":"which is perpendicular to your F vector."},{"Start":"32:42.115 ","End":"32:47.004","Text":"Then you can just use the r effective"},{"Start":"32:47.004 ","End":"32:51.730","Text":"multiplied by the size of your force and that will give you the torque."},{"Start":"32:51.730 ","End":"32:53.710","Text":"Then you don\u0027t have to include"},{"Start":"32:53.710 ","End":"32:58.704","Text":"your original r vector or sine of the angle into the equation."},{"Start":"32:58.704 ","End":"33:01.819","Text":"That\u0027s the end of this lesson."}],"ID":10792},{"Watched":false,"Name":"Proof - Torque Due to Gravity Acts on Center of Mass","Duration":"8m 39s","ChapterTopicVideoID":10445,"CourseChapterTopicPlaylistID":9395,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.980","Text":"Hello, in this lesson,"},{"Start":"00:01.980 ","End":"00:08.220","Text":"we\u0027re going to be speaking about why the gravitational force acts on the center of mass."},{"Start":"00:08.220 ","End":"00:13.140","Text":"Let\u0027s see, now, this lesson is especially important for the physicists."},{"Start":"00:13.140 ","End":"00:16.365","Text":"If you\u0027re learning engineering or something like that,"},{"Start":"00:16.365 ","End":"00:18.780","Text":"then you can skip this lesson,"},{"Start":"00:18.780 ","End":"00:21.375","Text":"or if it interests you, then you can watch it."},{"Start":"00:21.375 ","End":"00:26.290","Text":"But for the physicists amongst us, this is important."},{"Start":"00:26.300 ","End":"00:30.600","Text":"Let\u0027s consider that we have some axis of rotation over"},{"Start":"00:30.600 ","End":"00:36.040","Text":"here and some kind of body over here."},{"Start":"00:37.220 ","End":"00:43.850","Text":"Now, we say that we want to find out the torque of this body and we\u0027re going"},{"Start":"00:43.850 ","End":"00:50.560","Text":"to call it torque mg because we\u0027re working out from the gravitational force."},{"Start":"00:50.560 ","End":"00:58.895","Text":"Now, this body can be split up into many small sections of mass such as this."},{"Start":"00:58.895 ","End":"01:02.675","Text":"This mass, this square that I just drew over here,"},{"Start":"01:02.675 ","End":"01:07.530","Text":"this will say that its mass is m_i."},{"Start":"01:07.540 ","End":"01:14.390","Text":"Now, if I draw the radius from the axis of rotation until my small mass,"},{"Start":"01:14.390 ","End":"01:18.030","Text":"so this is my radius vector."},{"Start":"01:18.040 ","End":"01:24.900","Text":"Now, I can say that the force going down in this direction,"},{"Start":"01:24.900 ","End":"01:31.920","Text":"in the direction of negative y is going to be m_ig,"},{"Start":"01:31.920 ","End":"01:33.680","Text":"in the negative y-direction."},{"Start":"01:33.680 ","End":"01:37.798","Text":"Now, I can say that my torque,"},{"Start":"01:37.798 ","End":"01:43.695","Text":"number 1 is going to be equal to,"},{"Start":"01:43.695 ","End":"01:46.410","Text":"so what was my equation?"},{"Start":"01:46.410 ","End":"01:49.050","Text":"It\u0027s going to be equal to r,"},{"Start":"01:49.050 ","End":"01:50.565","Text":"this will be r_1,"},{"Start":"01:50.565 ","End":"01:54.555","Text":"r_1 vector cross-product,"},{"Start":"01:54.555 ","End":"02:00.680","Text":"my force, so it\u0027s m_ig and"},{"Start":"02:00.680 ","End":"02:07.890","Text":"it\u0027s going to be in the negative y direction because it\u0027s pointing downwards."},{"Start":"02:09.080 ","End":"02:14.615","Text":"This is the torque for just this 1 square over here."},{"Start":"02:14.615 ","End":"02:16.865","Text":"However, if I want to find my total,"},{"Start":"02:16.865 ","End":"02:22.350","Text":"my torque of the whole entire shape, then I can say,"},{"Start":"02:22.350 ","End":"02:28.335","Text":"that it\u0027s the sum of all of my torques,"},{"Start":"02:28.335 ","End":"02:33.630","Text":"which is the sum of all of this."},{"Start":"02:33.630 ","End":"02:36.870","Text":"Now, how can I rearrange this equation?"},{"Start":"02:36.870 ","End":"02:40.280","Text":"Now, you\u0027ll notice that my g can exit the brackets,"},{"Start":"02:40.280 ","End":"02:46.430","Text":"because it\u0027s constant and it\u0027s never changing and right now,"},{"Start":"02:46.430 ","End":"02:50.420","Text":"it\u0027s just represented as a number and this is it\u0027s vector direction,"},{"Start":"02:50.420 ","End":"02:52.445","Text":"so you can take it out."},{"Start":"02:52.445 ","End":"02:54.950","Text":"We can write here and we can also take out the negative,"},{"Start":"02:54.950 ","End":"02:56.855","Text":"so we\u0027ll have negative g,"},{"Start":"02:56.855 ","End":"02:59.670","Text":"the sum on i\u0027s."},{"Start":"03:00.580 ","End":"03:05.640","Text":"Then we can also say that"},{"Start":"03:05.640 ","End":"03:13.055","Text":"my m_i is also just a number and we\u0027re not taking into account the vector direction,"},{"Start":"03:13.055 ","End":"03:17.730","Text":"so I can also move my m over here,"},{"Start":"03:17.730 ","End":"03:21.930","Text":"so I can say that it\u0027s my m_i multiplied by"},{"Start":"03:21.930 ","End":"03:29.525","Text":"my r_i vector cross-product with my y-direction."},{"Start":"03:29.525 ","End":"03:31.730","Text":"Again, we\u0027re summing on all of this."},{"Start":"03:31.730 ","End":"03:35.945","Text":"The reason I could move my m_i from here,"},{"Start":"03:35.945 ","End":"03:38.600","Text":"is because it\u0027s just a scalar number."},{"Start":"03:38.600 ","End":"03:40.384","Text":"It isn\u0027t dependent on anything,"},{"Start":"03:40.384 ","End":"03:42.110","Text":"it\u0027s a scalar number,"},{"Start":"03:42.110 ","End":"03:49.630","Text":"which means that I can put it next to my r. Now,"},{"Start":"03:49.630 ","End":"03:55.235","Text":"I\u0027m going to remind you of this identity."},{"Start":"03:55.235 ","End":"04:00.550","Text":"If we have a cross c,"},{"Start":"04:00.550 ","End":"04:06.037","Text":"and these are vectors plus b cross c,"},{"Start":"04:06.037 ","End":"04:15.540","Text":"this is going to be equal to a plus b cross c,"},{"Start":"04:15.540 ","End":"04:17.908","Text":"so this is equal to this."},{"Start":"04:17.908 ","End":"04:23.670","Text":"Then I\u0027m also going to remind you of our equation for r_cm."},{"Start":"04:24.440 ","End":"04:31.475","Text":"Our r_cm is equal to the sum of all of the masses"},{"Start":"04:31.475 ","End":"04:39.300","Text":"multiplied by their positions divided by the total mass."},{"Start":"04:40.480 ","End":"04:44.420","Text":"Let\u0027s get back to what we were doing."},{"Start":"04:44.420 ","End":"04:50.820","Text":"Now, we can rewrite this like we did over here."},{"Start":"04:50.820 ","End":"04:58.140","Text":"This is going to be equal to negative g multiplied by m_1,"},{"Start":"04:58.140 ","End":"05:04.120","Text":"r_1 plus m_2,"},{"Start":"05:04.120 ","End":"05:09.630","Text":"r_2 plus dot and all of this is going to"},{"Start":"05:09.630 ","End":"05:17.565","Text":"be cross the y-direction."},{"Start":"05:17.565 ","End":"05:26.620","Text":"I\u0027ve just opened up this summing and written cross y outside of the brackets,"},{"Start":"05:26.620 ","End":"05:30.335","Text":"according to this identity over here."},{"Start":"05:30.335 ","End":"05:32.660","Text":"Instead of doing m_i,"},{"Start":"05:32.660 ","End":"05:36.585","Text":"r_i cross y plus m_2,"},{"Start":"05:36.585 ","End":"05:38.505","Text":"r_2 cross y plus m_3,"},{"Start":"05:38.505 ","End":"05:43.375","Text":"r_3 cross y, I\u0027ve just skipped all of that and written it out like this."},{"Start":"05:43.375 ","End":"05:45.725","Text":"Now, from here,"},{"Start":"05:45.725 ","End":"05:55.990","Text":"I can write that it\u0027s equal to negative g multiplied by the sum of m_i, r_i."},{"Start":"05:55.990 ","End":"06:02.405","Text":"All I\u0027ve done is contracted this giant bracket over here into the sum and then again,"},{"Start":"06:02.405 ","End":"06:05.730","Text":"I write cross y hat."},{"Start":"06:05.900 ","End":"06:09.600","Text":"Now, notice, that this sum of m_i,"},{"Start":"06:09.600 ","End":"06:13.260","Text":"r_i is the same as our r_cm,"},{"Start":"06:13.260 ","End":"06:18.815","Text":"but it\u0027s multiplied by the total mass of the system."},{"Start":"06:18.815 ","End":"06:25.250","Text":"Then I can write this out as negative g multiplied by the mass,"},{"Start":"06:25.250 ","End":"06:27.185","Text":"the total mass of the system,"},{"Start":"06:27.185 ","End":"06:35.490","Text":"multiplied by r_cm and then cross y."},{"Start":"06:35.490 ","End":"06:39.640","Text":"Now, just like what I did on this step,"},{"Start":"06:39.640 ","End":"06:44.185","Text":"when I moved my m_ig from my y hats,"},{"Start":"06:44.185 ","End":"06:50.035","Text":"over to outside the brackets and to be the coefficient of my r. So now,"},{"Start":"06:50.035 ","End":"06:53.320","Text":"I can just do the same move that I did here,"},{"Start":"06:53.320 ","End":"06:56.155","Text":"but going in the backwards direction."},{"Start":"06:56.155 ","End":"07:01.300","Text":"Then I can say, that this is equal to my r center of"},{"Start":"07:01.300 ","End":"07:11.580","Text":"mass cross and then I\u0027ll move my m_t and my negative g,"},{"Start":"07:11.580 ","End":"07:18.575","Text":"so I\u0027ll have negative m_tg,"},{"Start":"07:18.575 ","End":"07:22.365","Text":"y hat, in the y-direction."},{"Start":"07:22.365 ","End":"07:26.785","Text":"I just did the same move that I did between steps 1 and 2,"},{"Start":"07:26.785 ","End":"07:29.905","Text":"in my last step and now we can see that"},{"Start":"07:29.905 ","End":"07:35.395","Text":"my m_tg is the total force acting on the entire body,"},{"Start":"07:35.395 ","End":"07:37.895","Text":"mass total times gravity,"},{"Start":"07:37.895 ","End":"07:42.135","Text":"in the negative y-direction which we know,"},{"Start":"07:42.135 ","End":"07:47.505","Text":"it\u0027s going to be m_tg in the negative y-direction,"},{"Start":"07:47.505 ","End":"07:52.561","Text":"which corresponds to this downwards arrow that we always know exists."},{"Start":"07:52.561 ","End":"07:55.000","Text":"Then, we have our r_cm,"},{"Start":"07:55.000 ","End":"07:58.300","Text":"which is the location of our center of mass."},{"Start":"07:58.300 ","End":"08:05.000","Text":"Now, we can see that instead of summing on all the small pieces of mass and"},{"Start":"08:05.000 ","End":"08:11.675","Text":"finding out what they are and what their radius is and doing some complicated equation,"},{"Start":"08:11.675 ","End":"08:15.830","Text":"we can just see that this is the normal equation for the torque."},{"Start":"08:15.830 ","End":"08:19.527","Text":"We\u0027re multiplying our radius,"},{"Start":"08:19.527 ","End":"08:21.065","Text":"our position vector,"},{"Start":"08:21.065 ","End":"08:24.225","Text":"cross product with our force."},{"Start":"08:24.225 ","End":"08:28.440","Text":"It\u0027s the exact same equation for our torque equation,"},{"Start":"08:28.440 ","End":"08:32.060","Text":"but now we can see that it\u0027s just acting on the center of mass."},{"Start":"08:32.060 ","End":"08:37.040","Text":"It makes the equation a lot easier and that\u0027s why we do this calculation."},{"Start":"08:37.040 ","End":"08:39.990","Text":"That\u0027s the end of this lesson."}],"ID":10793}],"Thumbnail":null,"ID":9395},{"Name":"Exercises","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise - Two People Holding a Plank","Duration":"13m 35s","ChapterTopicVideoID":10446,"CourseChapterTopicPlaylistID":9396,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.015","Text":"Hello. In this lesson,"},{"Start":"00:03.015 ","End":"00:05.130","Text":"we\u0027re given a question."},{"Start":"00:05.130 ","End":"00:08.250","Text":"We\u0027re being told 2 people hold a wooden plank of"},{"Start":"00:08.250 ","End":"00:13.140","Text":"mass 12 kilograms and of length 1.5 meters."},{"Start":"00:13.140 ","End":"00:19.470","Text":"We\u0027re going to assume that the mass of the plank is spread uniformly which means"},{"Start":"00:19.470 ","End":"00:25.800","Text":"that the center of mass is going to be somewhere around here, center of mass."},{"Start":"00:25.800 ","End":"00:30.130","Text":"It\u0027s going to be located at 0.75 meters."},{"Start":"00:30.130 ","End":"00:32.000","Text":"Then on the plank,"},{"Start":"00:32.000 ","End":"00:35.360","Text":"0.5 meters away from the person on the right."},{"Start":"00:35.360 ","End":"00:39.645","Text":"Over here, a box of mass 8 kilograms is placed."},{"Start":"00:39.645 ","End":"00:42.875","Text":"Again, we\u0027re going to be speaking about the center of mass of the box."},{"Start":"00:42.875 ","End":"00:46.415","Text":"The center of mass of the box is placed 0.5"},{"Start":"00:46.415 ","End":"00:50.440","Text":"meters away from person number 2, the person on the right."},{"Start":"00:50.440 ","End":"00:54.205","Text":"We\u0027re being told that the people are stationary."},{"Start":"00:54.205 ","End":"00:58.820","Text":"Now we\u0027re being asked what force does each person apply to the system."},{"Start":"00:58.820 ","End":"01:01.760","Text":"We\u0027re talking about F_1 and F_2."},{"Start":"01:01.760 ","End":"01:04.260","Text":"We\u0027re being asked to work that out."},{"Start":"01:05.000 ","End":"01:08.000","Text":"Let\u0027s begin answering this question."},{"Start":"01:08.000 ","End":"01:11.660","Text":"Now because we\u0027re being told that the people are stationary,"},{"Start":"01:11.660 ","End":"01:15.470","Text":"also the entire system is stationary in this case."},{"Start":"01:15.470 ","End":"01:19.370","Text":"We can say we\u0027ve been doing the whole time"},{"Start":"01:19.370 ","End":"01:25.030","Text":"that the sum of all of the forces in the system is equal to 0."},{"Start":"01:25.030 ","End":"01:30.095","Text":"If this was moving at constant velocity or that its stationary,"},{"Start":"01:30.095 ","End":"01:32.465","Text":"so the sum of the forces is equal to 0."},{"Start":"01:32.465 ","End":"01:35.765","Text":"Now, if the sum of all of the forces is equal to 0,"},{"Start":"01:35.765 ","End":"01:38.645","Text":"if you remember our equation for torque."},{"Start":"01:38.645 ","End":"01:42.335","Text":"Let\u0027s just write this up here in red."},{"Start":"01:42.335 ","End":"01:51.420","Text":"A torque is equal to our radius cross product with our force vector."},{"Start":"01:52.310 ","End":"01:56.825","Text":"Because the sum of all of our forces is equal to 0,"},{"Start":"01:56.825 ","End":"02:04.365","Text":"that means that the sum of all of our torques is also going to be equal to 0."},{"Start":"02:04.365 ","End":"02:06.515","Text":"When dealing with a question like this,"},{"Start":"02:06.515 ","End":"02:08.390","Text":"this is the first thing that you do."},{"Start":"02:08.390 ","End":"02:11.127","Text":"It\u0027s stationary or moving at a constant velocity,"},{"Start":"02:11.127 ","End":"02:13.055","Text":"sum of all of our forces is equal to 0."},{"Start":"02:13.055 ","End":"02:16.770","Text":"Therefore, sum of all the torques is equal to 0."},{"Start":"02:17.450 ","End":"02:21.620","Text":"Now, the next thing I have to do is I have to write out"},{"Start":"02:21.620 ","End":"02:25.805","Text":"which forces are acting on my plank."},{"Start":"02:25.805 ","End":"02:31.639","Text":"Obviously, I have my normal force for this."},{"Start":"02:31.639 ","End":"02:33.380","Text":"I have some normal force,"},{"Start":"02:33.380 ","End":"02:34.730","Text":"let\u0027s call it N_1,"},{"Start":"02:34.730 ","End":"02:37.070","Text":"which is acting on the plank, and also,"},{"Start":"02:37.070 ","End":"02:41.665","Text":"of course, it\u0027s also acting in this direction."},{"Start":"02:41.665 ","End":"02:44.615","Text":"Then coming from my center of mass,"},{"Start":"02:44.615 ","End":"02:48.220","Text":"I have the mass."},{"Start":"02:48.220 ","End":"02:55.575","Text":"We\u0027ll call it Mg from the center of mass of the plank."},{"Start":"02:55.575 ","End":"02:57.701","Text":"These were our forces."},{"Start":"02:57.701 ","End":"03:02.185","Text":"Now, let\u0027s go on to writing the sum of all of our forces."},{"Start":"03:02.185 ","End":"03:05.925","Text":"We\u0027ll start with our forces in the y-direction."},{"Start":"03:05.925 ","End":"03:12.065","Text":"We\u0027ll write that the sum of all of our forces in the y-direction is going to be equal to."},{"Start":"03:12.065 ","End":"03:17.780","Text":"Let\u0027s just label that our positive y-direction is going upwards."},{"Start":"03:17.780 ","End":"03:24.670","Text":"We have our F_1 plus our F_2,"},{"Start":"03:24.670 ","End":"03:29.270","Text":"which are pointing in the upward direction applied by the people, minus Mg minus N_1."},{"Start":"03:34.100 ","End":"03:38.140","Text":"As we know, the sum of all of the forces in the y-direction,"},{"Start":"03:38.140 ","End":"03:39.595","Text":"because the system is stationary,"},{"Start":"03:39.595 ","End":"03:41.635","Text":"is going to be equal to 0."},{"Start":"03:41.635 ","End":"03:44.170","Text":"Now let\u0027s work out what our N_1 is equal to."},{"Start":"03:44.170 ","End":"03:47.455","Text":"Now, if you remember, something is resting on a table,"},{"Start":"03:47.455 ","End":"03:51.370","Text":"then they\u0027ll have a normal force acting upwards."},{"Start":"03:51.370 ","End":"03:56.785","Text":"Then that will counter the mass of the object multiplied by gravity,"},{"Start":"03:56.785 ","End":"03:59.500","Text":"the force, pointing downwards."},{"Start":"03:59.500 ","End":"04:03.745","Text":"Then because it\u0027s stationary,"},{"Start":"04:03.745 ","End":"04:10.045","Text":"you\u0027ll have that the force is going to be N minus mg."},{"Start":"04:10.045 ","End":"04:13.175","Text":"The sum of all the forces is going to be equal to 0,"},{"Start":"04:13.175 ","End":"04:16.369","Text":"which means that you\u0027ll get that N is equal to"},{"Start":"04:16.369 ","End":"04:21.245","Text":"mg. Over here it\u0027s the exact same principle."},{"Start":"04:21.245 ","End":"04:24.425","Text":"You have N_1 pointing downwards,"},{"Start":"04:24.425 ","End":"04:27.300","Text":"which is going to be equal to our mg."},{"Start":"04:27.300 ","End":"04:29.795","Text":"Here, we\u0027ll just say that it\u0027s N_1."},{"Start":"04:29.795 ","End":"04:33.410","Text":"Now let\u0027s see what our N_1 is equal to."},{"Start":"04:33.410 ","End":"04:37.205","Text":"N_1 is equal to the mass of the box, which is 8,"},{"Start":"04:37.205 ","End":"04:42.750","Text":"multiplied by g. We\u0027ll just say that here that g"},{"Start":"04:42.750 ","End":"04:48.370","Text":"is equal to 10 m/s^2."},{"Start":"04:48.370 ","End":"04:54.630","Text":"This will equal to 8.10."},{"Start":"04:54.630 ","End":"04:57.695","Text":"Now, what we\u0027re going to do is we\u0027re going to write the equation for"},{"Start":"04:57.695 ","End":"05:01.390","Text":"the sum of all of the moments. Let\u0027s see."},{"Start":"05:01.390 ","End":"05:06.305","Text":"We have that the sum of all of the moments is equal to."},{"Start":"05:06.305 ","End":"05:09.035","Text":"Now, as you remember,"},{"Start":"05:09.035 ","End":"05:13.880","Text":"we have to find an axis of rotation in order to do this calculation."},{"Start":"05:13.880 ","End":"05:17.465","Text":"Now because we\u0027re not specified any axes in the question,"},{"Start":"05:17.465 ","End":"05:22.860","Text":"let\u0027s just say that the axes of rotation is here by person number 1."},{"Start":"05:23.480 ","End":"05:29.430","Text":"Now, let\u0027s work out the moment of F_1."},{"Start":"05:29.430 ","End":"05:31.750","Text":"As we can see, it\u0027s F_1,"},{"Start":"05:31.750 ","End":"05:39.050","Text":"but the radius vector is coming right from the axis of rotation."},{"Start":"05:39.050 ","End":"05:41.600","Text":"The radius is going to be equal to 0."},{"Start":"05:41.600 ","End":"05:44.430","Text":"So that means that the torque for F_1 is going to be equal to 0."},{"Start":"05:47.330 ","End":"05:52.680","Text":"Now we\u0027re going to find out the torque of our Mg."},{"Start":"05:52.680 ","End":"05:54.785","Text":"Soon we\u0027ll add in our numbers."},{"Start":"05:54.785 ","End":"05:56.735","Text":"Right now, just so you know,"},{"Start":"05:56.735 ","End":"06:00.050","Text":"I\u0027m using the other equation that we have for torque,"},{"Start":"06:00.050 ","End":"06:01.175","Text":"which is the size,"},{"Start":"06:01.175 ","End":"06:04.415","Text":"which means that afterwards we have to figure out the direction,"},{"Start":"06:04.415 ","End":"06:09.365","Text":"which is equal to the size of the radius vector multiplied by"},{"Start":"06:09.365 ","End":"06:16.650","Text":"the size of the force vector multiplied by the sine of the angle between the two."},{"Start":"06:17.600 ","End":"06:21.215","Text":"Now, let\u0027s take a look."},{"Start":"06:21.215 ","End":"06:27.525","Text":"Our next one is going to be the torque of our mg,"},{"Start":"06:27.525 ","End":"06:30.435","Text":"the center of mass of the plank."},{"Start":"06:30.435 ","End":"06:35.085","Text":"We know that it\u0027s located 0.75 meters away."},{"Start":"06:35.085 ","End":"06:45.715","Text":"We can write the size of the force is going to be Mg multiplied by 0.75 meters away."},{"Start":"06:45.715 ","End":"06:48.725","Text":"Then multiplied by sine of the angle."},{"Start":"06:48.725 ","End":"06:49.835","Text":"Now as we can see,"},{"Start":"06:49.835 ","End":"06:52.820","Text":"the angle between the plank and the Mg is 90 degrees."},{"Start":"06:52.820 ","End":"06:58.290","Text":"I\u0027ll write over here as well as a little note that sine of 90 degrees is equal to 1."},{"Start":"07:00.880 ","End":"07:03.275","Text":"That\u0027s something useful to note."},{"Start":"07:03.275 ","End":"07:05.990","Text":"Then we multiply by 1 over here."},{"Start":"07:05.990 ","End":"07:08.824","Text":"Now we have to look at the direction."},{"Start":"07:08.824 ","End":"07:11.660","Text":"Let\u0027s say which direction we want to"},{"Start":"07:11.660 ","End":"07:14.630","Text":"put in the positive direction and which in the negative."},{"Start":"07:14.630 ","End":"07:16.610","Text":"If this is the x-axis,"},{"Start":"07:16.610 ","End":"07:19.700","Text":"so let\u0027s say that if we\u0027re going anticlockwise,"},{"Start":"07:19.700 ","End":"07:22.410","Text":"that this is the positive direction."},{"Start":"07:22.410 ","End":"07:27.210","Text":"As you can see, if we draw out these vectors."},{"Start":"07:27.210 ","End":"07:32.870","Text":"If I have my radius vector going in this direction and I"},{"Start":"07:32.870 ","End":"07:38.565","Text":"have my force vector going in this direction,"},{"Start":"07:38.565 ","End":"07:41.315","Text":"we can see to get from my radius to my force,"},{"Start":"07:41.315 ","End":"07:43.430","Text":"I have to go in this direction,"},{"Start":"07:43.430 ","End":"07:46.205","Text":"which is the clockwise direction."},{"Start":"07:46.205 ","End":"07:49.805","Text":"Which means that it\u0027s in the negative direction."},{"Start":"07:49.805 ","End":"07:54.360","Text":"Here, I put a minus because it\u0027s in the negative direction."},{"Start":"07:55.250 ","End":"08:05.715","Text":"Then my next is going to be the torque of my box of N_1."},{"Start":"08:05.715 ","End":"08:09.345","Text":"The torque of the box. Let\u0027s see what this is going to be."},{"Start":"08:09.345 ","End":"08:12.755","Text":"It\u0027s going to be the size of the force,"},{"Start":"08:12.755 ","End":"08:14.300","Text":"which as we know,"},{"Start":"08:14.300 ","End":"08:17.930","Text":"we\u0027re going to say that it\u0027s mg,"},{"Start":"08:17.930 ","End":"08:21.960","Text":"soon we\u0027ll put in the actual values for N_1,"},{"Start":"08:21.960 ","End":"08:26.160","Text":"multiplied by the radius."},{"Start":"08:26.160 ","End":"08:28.980","Text":"We know that it\u0027s located 1.5 meters away,"},{"Start":"08:28.980 ","End":"08:33.280","Text":"so multiplied by 1.5."},{"Start":"08:35.120 ","End":"08:40.030","Text":"That\u0027s a number, not 1 times 5 but 1.5."},{"Start":"08:40.400 ","End":"08:44.605","Text":"Then again, it\u0027s going to be multiplied by sine of the angle."},{"Start":"08:44.605 ","End":"08:51.380","Text":"Again, the angle is 90 degrees so it\u0027s going to be multiplied by 1."},{"Start":"08:51.590 ","End":"08:56.960","Text":"Now, the sign, because it\u0027s pointing in the exact direction as what we just worked out,"},{"Start":"08:56.960 ","End":"09:01.340","Text":"is the negative direction so we\u0027ll put a minus over here for"},{"Start":"09:01.340 ","End":"09:07.020","Text":"the exact same reason that we did for the torque of the center of mass of the plank."},{"Start":"09:07.020 ","End":"09:10.405","Text":"Now, we\u0027re going to put of F_2,"},{"Start":"09:10.405 ","End":"09:15.650","Text":"so the torque of F_2. Let\u0027s see what this is."},{"Start":"09:15.650 ","End":"09:17.060","Text":"It\u0027s the size of the force,"},{"Start":"09:17.060 ","End":"09:18.335","Text":"which is F_2,"},{"Start":"09:18.335 ","End":"09:25.940","Text":"multiplied by the size of the radius from the axis of rotation until F_2,"},{"Start":"09:25.940 ","End":"09:31.140","Text":"which is going to be multiplied by the length of the plank,"},{"Start":"09:31.140 ","End":"09:33.750","Text":"which is 1.5 meters."},{"Start":"09:33.750 ","End":"09:36.320","Text":"Then multiplied by the sine of the angle between them,"},{"Start":"09:36.320 ","End":"09:37.819","Text":"which is also 90 degrees,"},{"Start":"09:37.819 ","End":"09:41.260","Text":"so sine of 90 is equal to 1."},{"Start":"09:41.260 ","End":"09:44.640","Text":"Now let\u0027s take a look at the sine."},{"Start":"09:44.640 ","End":"09:48.480","Text":"If I draw my radius vector going in"},{"Start":"09:48.480 ","End":"09:52.940","Text":"its direction and my force vector going in its direction,"},{"Start":"09:52.940 ","End":"09:56.600","Text":"to get from my radius vector until my force vector,"},{"Start":"09:56.600 ","End":"09:58.880","Text":"I\u0027m going in the anticlockwise direction,"},{"Start":"09:58.880 ","End":"10:01.820","Text":"which is what we said was the positive direction."},{"Start":"10:01.820 ","End":"10:04.445","Text":"I put in a plus over here."},{"Start":"10:04.445 ","End":"10:06.050","Text":"Then, like we said earlier,"},{"Start":"10:06.050 ","End":"10:08.810","Text":"that the sum of all of the torque is going to be equal to 0,"},{"Start":"10:08.810 ","End":"10:12.160","Text":"so this is equal to 0."},{"Start":"10:12.860 ","End":"10:16.790","Text":"Now I have two equations and two unknowns."},{"Start":"10:16.790 ","End":"10:22.425","Text":"My one unknown is my F_1 and my other unknown is my F_2."},{"Start":"10:22.425 ","End":"10:27.210","Text":"Then I can solve my torque equation to find out what my F_2 is,"},{"Start":"10:27.210 ","End":"10:30.305","Text":"and then I can substitute it into"},{"Start":"10:30.305 ","End":"10:34.675","Text":"my force equation to find out what F_1 is and then I\u0027m done."},{"Start":"10:34.675 ","End":"10:37.415","Text":"In 1 second, we\u0027re going to begin the algebra."},{"Start":"10:37.415 ","End":"10:41.645","Text":"Also another way that I could have done this instead of writing out a force equation"},{"Start":"10:41.645 ","End":"10:46.715","Text":"is just like I wrote out my torque equation from this axis of rotation,"},{"Start":"10:46.715 ","End":"10:49.910","Text":"I could have also given another axis of rotations such"},{"Start":"10:49.910 ","End":"10:54.605","Text":"as at person number 2 or at the center of mass, whatever it might be."},{"Start":"10:54.605 ","End":"10:56.960","Text":"Then from that axis of rotation,"},{"Start":"10:56.960 ","End":"10:59.510","Text":"I could have written another equation for the torque,"},{"Start":"10:59.510 ","End":"11:04.080","Text":"and then just solved with 2 torque equations."},{"Start":"11:04.080 ","End":"11:06.270","Text":"That\u0027s another way of solving this."},{"Start":"11:06.270 ","End":"11:11.075","Text":"Let\u0027s begin. Let\u0027s begin by working out what our F_2 is."},{"Start":"11:11.075 ","End":"11:13.700","Text":"What we\u0027re going to do is we\u0027re going to go to"},{"Start":"11:13.700 ","End":"11:16.805","Text":"our torque equation and isolate out the F_2,"},{"Start":"11:16.805 ","End":"11:20.850","Text":"which means moving our torque for our Mg"},{"Start":"11:20.850 ","End":"11:25.175","Text":"and our torque for our N_1 over to the other side of the equation."},{"Start":"11:25.175 ","End":"11:34.865","Text":"We\u0027re going to have that 1.5 of F_2 is going to be equal to Mg. Now,"},{"Start":"11:34.865 ","End":"11:37.640","Text":"this is our M, which is the mass of the plank,"},{"Start":"11:37.640 ","End":"11:38.990","Text":"which is 12 kilograms,"},{"Start":"11:38.990 ","End":"11:44.305","Text":"and we said that our g is going to be equal to 10 meters per second."},{"Start":"11:44.305 ","End":"11:49.980","Text":"We said that above. We\u0027re going to have 120 multiplied by 0.75,"},{"Start":"11:49.980 ","End":"11:57.450","Text":"and then plus our mg,"},{"Start":"11:57.450 ","End":"12:01.770","Text":"which here is 80 multiplied by 1."},{"Start":"12:01.770 ","End":"12:06.830","Text":"Now, if we divide both sides by 1.5,"},{"Start":"12:06.830 ","End":"12:17.665","Text":"we\u0027ll get that our F_2 is equal to 113.3 recurring newtons."},{"Start":"12:17.665 ","End":"12:22.355","Text":"Now, if we substitute this into our force equation,"},{"Start":"12:22.355 ","End":"12:27.455","Text":"we can get what our F_1 is equal to. Let\u0027s see."},{"Start":"12:27.455 ","End":"12:36.089","Text":"We have that our F_1 plus our 113.3 recurring,"},{"Start":"12:36.089 ","End":"12:37.740","Text":"that\u0027s our F_2,"},{"Start":"12:37.740 ","End":"12:41.340","Text":"minus our Mg,"},{"Start":"12:41.340 ","End":"12:46.450","Text":"which we said our M is equal to 12 kilograms."},{"Start":"12:46.550 ","End":"12:49.500","Text":"Our energy is equal to 10, so it\u0027s 120,"},{"Start":"12:49.500 ","End":"12:52.358","Text":"and minus our N_1,"},{"Start":"12:52.358 ","End":"12:54.285","Text":"and our N_1 is 80,"},{"Start":"12:54.285 ","End":"12:57.305","Text":"is going to be equal to 0."},{"Start":"12:57.305 ","End":"13:01.195","Text":"Then if we isolate out our F_1,"},{"Start":"13:01.195 ","End":"13:11.530","Text":"we will get that our F_1 is equal to 86.6 recurring newtons."},{"Start":"13:11.990 ","End":"13:15.530","Text":"That is the end of this exercise."},{"Start":"13:15.530 ","End":"13:24.065","Text":"But as you can see with our people that F_2 is going to be greater than our F_1."},{"Start":"13:24.065 ","End":"13:30.685","Text":"Person number 2 is applying more force because the box is closer to him."},{"Start":"13:30.685 ","End":"13:35.460","Text":"That\u0027s the end of the lesson on to the next one."}],"ID":10794},{"Watched":false,"Name":"Exercise - Ladder on Wall","Duration":"16m 32s","ChapterTopicVideoID":10447,"CourseChapterTopicPlaylistID":9396,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.500","Text":"Hello. This is a very basic question where we\u0027re being told that"},{"Start":"00:04.500 ","End":"00:09.435","Text":"we have a ladder over here of mass m and of length L,"},{"Start":"00:09.435 ","End":"00:14.640","Text":"which is leaning on a smooth wall but a rough floor,"},{"Start":"00:14.640 ","End":"00:18.960","Text":"which means that we have some coefficient of friction over here."},{"Start":"00:18.960 ","End":"00:24.580","Text":"Then we\u0027re being asked what forces are acting on the ladder."},{"Start":"00:25.490 ","End":"00:30.135","Text":"The first thing that we\u0027re going to do is we\u0027re going to"},{"Start":"00:30.135 ","End":"00:35.085","Text":"draw on all of the forces acting on the ladder."},{"Start":"00:35.085 ","End":"00:39.375","Text":"First of all, we have perpendicular to the wall."},{"Start":"00:39.375 ","End":"00:44.790","Text":"The force always acts perpendicular to the surface."},{"Start":"00:44.790 ","End":"00:48.350","Text":"Here\u0027s the wall, we have a normal force,"},{"Start":"00:48.350 ","End":"00:49.520","Text":"let\u0027s call it N_1."},{"Start":"00:49.520 ","End":"00:52.625","Text":"This is from the wall until the ladder."},{"Start":"00:52.625 ","End":"00:54.815","Text":"Then over here, again,"},{"Start":"00:54.815 ","End":"00:57.545","Text":"perpendicular from the floor."},{"Start":"00:57.545 ","End":"00:59.450","Text":"Here\u0027s a right angle."},{"Start":"00:59.450 ","End":"01:01.670","Text":"We have N_2,"},{"Start":"01:01.670 ","End":"01:06.680","Text":"the normal force of the floor acting on the ladder."},{"Start":"01:06.680 ","End":"01:09.305","Text":"Then in this direction over here,"},{"Start":"01:09.305 ","End":"01:10.955","Text":"we have another force,"},{"Start":"01:10.955 ","End":"01:13.400","Text":"which is our frictional force,"},{"Start":"01:13.400 ","End":"01:17.150","Text":"because we know that the floor is rough."},{"Start":"01:17.150 ","End":"01:21.289","Text":"Then from the center of mass,"},{"Start":"01:21.289 ","End":"01:25.010","Text":"we have our force due to gravity,"},{"Start":"01:25.010 ","End":"01:28.250","Text":"Mg, our weight."},{"Start":"01:28.250 ","End":"01:33.720","Text":"This, as we know, always acts from the center of mass."},{"Start":"01:34.430 ","End":"01:37.415","Text":"Now, if you\u0027re asking ourselves,"},{"Start":"01:37.415 ","End":"01:41.075","Text":"if this frictional force is equal to Mu N,"},{"Start":"01:41.075 ","End":"01:45.635","Text":"so the answer is no. Why not?"},{"Start":"01:45.635 ","End":"01:50.705","Text":"Now the reason is because the frictional force isn\u0027t at a maximum here,"},{"Start":"01:50.705 ","End":"01:53.975","Text":"or we don\u0027t know if it\u0027s a maximum here. Why is that?"},{"Start":"01:53.975 ","End":"01:58.950","Text":"We could lean the ladder at a larger angle Alpha"},{"Start":"01:58.950 ","End":"02:04.700","Text":"and it could be that our ladder will still be balanced and still remain stationary,"},{"Start":"02:04.700 ","End":"02:09.400","Text":"which means that we could have a larger frictional force."},{"Start":"02:09.400 ","End":"02:16.235","Text":"That\u0027s why we don\u0027t know and can\u0027t say for sure that this f is equal to"},{"Start":"02:16.235 ","End":"02:22.490","Text":"Mu N. What we can say is that f is smaller,"},{"Start":"02:22.490 ","End":"02:24.905","Text":"or equal to mu N,"},{"Start":"02:24.905 ","End":"02:26.750","Text":"so it could be equal to mu N,"},{"Start":"02:26.750 ","End":"02:31.230","Text":"but it also could be anything below that value."},{"Start":"02:32.510 ","End":"02:36.225","Text":"We know what our value for m is,"},{"Start":"02:36.225 ","End":"02:38.300","Text":"because it\u0027s given to us in the question,"},{"Start":"02:38.300 ","End":"02:40.910","Text":"but we don\u0027t know what our values for N_1,"},{"Start":"02:40.910 ","End":"02:44.515","Text":"N_2, or f are equal to."},{"Start":"02:44.515 ","End":"02:48.020","Text":"We\u0027re trying to work out what these forces are,"},{"Start":"02:48.020 ","End":"02:49.820","Text":"so we have 3 unknowns,"},{"Start":"02:49.820 ","End":"02:51.740","Text":"which means that I\u0027m going to have to have"},{"Start":"02:51.740 ","End":"02:55.300","Text":"3 equations in order to find out what they are."},{"Start":"02:55.300 ","End":"02:57.885","Text":"What equations am I going to use?"},{"Start":"02:57.885 ","End":"03:05.315","Text":"First of all, I\u0027m going to work out the sum of all of the forces in the x-direction."},{"Start":"03:05.315 ","End":"03:07.789","Text":"That will be 1 equation."},{"Start":"03:07.789 ","End":"03:14.465","Text":"My second equation will be the sum of all of the forces in the y-direction,"},{"Start":"03:14.465 ","End":"03:21.480","Text":"and then my third equation is going to be the sum of all my torques,"},{"Start":"03:21.480 ","End":"03:23.980","Text":"or all of my moments."},{"Start":"03:24.220 ","End":"03:27.770","Text":"Now, a quick note when am I allowed to"},{"Start":"03:27.770 ","End":"03:30.410","Text":"use this equation for the sum of all of my torques,"},{"Start":"03:30.410 ","End":"03:33.095","Text":"when can I use it and when is it useful?"},{"Start":"03:33.095 ","End":"03:36.475","Text":"Every time I don\u0027t have some kind of point mass,"},{"Start":"03:36.475 ","End":"03:43.885","Text":"so every time I have a rigid body where the mass is spread out along some shape."},{"Start":"03:43.885 ","End":"03:48.064","Text":"Then my body can move, it can rotate,"},{"Start":"03:48.064 ","End":"03:50.840","Text":"which means that using this equation,"},{"Start":"03:50.840 ","End":"03:54.660","Text":"the sum of all of the torques comes in very handy."},{"Start":"03:54.980 ","End":"04:01.355","Text":"Before we start plugging in our values for the sum of the forces or torques,"},{"Start":"04:01.355 ","End":"04:05.380","Text":"let\u0027s first define which are the positive directions,"},{"Start":"04:05.380 ","End":"04:08.510","Text":"so the direction of the x-axis,"},{"Start":"04:08.510 ","End":"04:11.425","Text":"let\u0027s say that this is the positive direction."},{"Start":"04:11.425 ","End":"04:16.838","Text":"Then let\u0027s say that this is the positive direction of the y-axis,"},{"Start":"04:16.838 ","End":"04:19.460","Text":"and let\u0027s say that"},{"Start":"04:19.460 ","End":"04:28.140","Text":"a rotation anticlockwise is the positive direction of rotation."},{"Start":"04:28.940 ","End":"04:33.505","Text":"Let\u0027s begin with a sum of all of the forces in the x-direction."},{"Start":"04:33.505 ","End":"04:37.105","Text":"We can see that I have 2 forces in the x-direction."},{"Start":"04:37.105 ","End":"04:43.230","Text":"I have N_1 and I have f. Now,"},{"Start":"04:43.230 ","End":"04:48.100","Text":"I could have drawn my N_1 or my f pointing in the opposite direction,"},{"Start":"04:48.100 ","End":"04:49.885","Text":"in the negative x-direction,"},{"Start":"04:49.885 ","End":"04:53.035","Text":"and it doesn\u0027t make any difference because then I\u0027ll just solve"},{"Start":"04:53.035 ","End":"04:56.590","Text":"my 3 equations for 3 unknowns, and in the end,"},{"Start":"04:56.590 ","End":"05:03.510","Text":"I\u0027ll get a sine before my N_1 or my f be at a positive or a negative."},{"Start":"05:03.510 ","End":"05:07.420","Text":"That would tell me that the way I drew my arrow is either in"},{"Start":"05:07.420 ","End":"05:12.290","Text":"the correct direction or I have to change it to point in the opposite direction."},{"Start":"05:12.290 ","End":"05:14.917","Text":"With my answer that I\u0027ll get at the end,"},{"Start":"05:14.917 ","End":"05:19.640","Text":"I\u0027ll know if I drew these 2 arrows pointing in the right direction, or not."},{"Start":"05:19.640 ","End":"05:21.935","Text":"That\u0027s also correct for my N_2,"},{"Start":"05:21.935 ","End":"05:28.400","Text":"the only force here that I always have to draw pointing downwards is"},{"Start":"05:28.400 ","End":"05:31.520","Text":"my mg because I know that my mg is always going to"},{"Start":"05:31.520 ","End":"05:35.600","Text":"be pointing downwards towards the center of the Earth."},{"Start":"05:35.600 ","End":"05:38.690","Text":"That\u0027s the only force I rarely have to draw in the right direction."},{"Start":"05:38.690 ","End":"05:42.420","Text":"Everything else I can draw either in the positive,"},{"Start":"05:42.420 ","End":"05:44.570","Text":"or negative x, or y directions,"},{"Start":"05:44.570 ","End":"05:48.445","Text":"and they will sort themselves out as I solve the question."},{"Start":"05:48.445 ","End":"05:53.585","Text":"The sum of all of my forces in the x-direction is equal to N_1 plus f,"},{"Start":"05:53.585 ","End":"05:55.340","Text":"that\u0027s just how I drew it,"},{"Start":"05:55.340 ","End":"06:00.320","Text":"and because I know that my ladder is leaning on the wall and the floor,"},{"Start":"06:00.320 ","End":"06:03.065","Text":"so I know that my ladder is static."},{"Start":"06:03.065 ","End":"06:06.535","Text":"It isn\u0027t moving. Nowhere in the question is that being mentioned."},{"Start":"06:06.535 ","End":"06:09.530","Text":"That means that if there is no movement,"},{"Start":"06:09.530 ","End":"06:11.105","Text":"there was no velocity."},{"Start":"06:11.105 ","End":"06:13.640","Text":"Which means that if there\u0027s 0 velocity,"},{"Start":"06:13.640 ","End":"06:15.575","Text":"there\u0027s also 0 acceleration,"},{"Start":"06:15.575 ","End":"06:22.230","Text":"which means the sum of all of my forces is equal to 0 in this direction."},{"Start":"06:22.880 ","End":"06:28.093","Text":"Now let\u0027s look at the sum of all my forces in the y-direction."},{"Start":"06:28.093 ","End":"06:29.900","Text":"In the positive y-direction,"},{"Start":"06:29.900 ","End":"06:31.460","Text":"as we defined it,"},{"Start":"06:31.460 ","End":"06:34.706","Text":"I have my N_2 pointing in that direction.,"},{"Start":"06:34.706 ","End":"06:37.385","Text":"and then in the negative y-direction,"},{"Start":"06:37.385 ","End":"06:39.060","Text":"I have my mg."},{"Start":"06:39.060 ","End":"06:42.335","Text":"I\u0027ll write negative mg. Again,"},{"Start":"06:42.335 ","End":"06:45.200","Text":"in the y-direction, I have no movement."},{"Start":"06:45.200 ","End":"06:46.955","Text":"My ladder is stationary,"},{"Start":"06:46.955 ","End":"06:50.330","Text":"which means that again, 0 velocity means 0 acceleration."},{"Start":"06:50.330 ","End":"06:53.865","Text":"We know f is equal to mass times acceleration,"},{"Start":"06:53.865 ","End":"06:55.925","Text":"so we could say that,"},{"Start":"06:55.925 ","End":"06:59.310","Text":"again, this is equal to 0."},{"Start":"07:00.320 ","End":"07:04.270","Text":"Now let\u0027s go on to the sum of all of our torques."},{"Start":"07:04.270 ","End":"07:07.165","Text":"Before I can start writing the equation,"},{"Start":"07:07.165 ","End":"07:10.690","Text":"I have to define some axis of rotation."},{"Start":"07:10.690 ","End":"07:13.975","Text":"Now, in this question specifically,"},{"Start":"07:13.975 ","End":"07:16.705","Text":"we know that our ladder is stationary,"},{"Start":"07:16.705 ","End":"07:22.085","Text":"which means that I can choose my axis of rotation to be anywhere that I want."},{"Start":"07:22.085 ","End":"07:27.505","Text":"If I was told that my ladder is rotating about some point,"},{"Start":"07:27.505 ","End":"07:30.100","Text":"then that point has to be my axis of"},{"Start":"07:30.100 ","End":"07:34.160","Text":"rotation and I have to solve the question according to that point."},{"Start":"07:34.160 ","End":"07:37.450","Text":"But here, it\u0027s stationary so it doesn\u0027t really matter,"},{"Start":"07:37.450 ","End":"07:39.235","Text":"so I can choose whatever I want."},{"Start":"07:39.235 ","End":"07:44.440","Text":"Let\u0027s see how I\u0027m going to choose my axis of rotation now."},{"Start":"07:44.440 ","End":"07:47.455","Text":"What am I going to do?"},{"Start":"07:47.455 ","End":"07:52.360","Text":"What I want to do is I want to make this question as easy to solve as possible."},{"Start":"07:52.360 ","End":"08:00.705","Text":"As we know, if I have an axis of rotation and a force acting at the axis of rotation,"},{"Start":"08:00.705 ","End":"08:02.680","Text":"because of the equation for torque,"},{"Start":"08:02.680 ","End":"08:09.235","Text":"I know that a force acting at the axis of rotation is going to give me 0 torque."},{"Start":"08:09.235 ","End":"08:12.205","Text":"It\u0027s not going to contribute to the total torque."},{"Start":"08:12.205 ","End":"08:15.220","Text":"I want to make this as easy to solve as possible."},{"Start":"08:15.220 ","End":"08:18.220","Text":"I see at this point over here in the ladder,"},{"Start":"08:18.220 ","End":"08:24.915","Text":"I have both N_2 and f acting over here at this point."},{"Start":"08:24.915 ","End":"08:28.625","Text":"Therefore, if I say that my axis of rotation is over here,"},{"Start":"08:28.625 ","End":"08:33.625","Text":"that means that N_2 and f contribute 0 torques to my equation,"},{"Start":"08:33.625 ","End":"08:36.820","Text":"which would make my equation a lot easier to solve."},{"Start":"08:36.820 ","End":"08:41.185","Text":"If I put my axis of rotation anywhere else, let\u0027s say here,"},{"Start":"08:41.185 ","End":"08:47.210","Text":"then only my N_1 would contribute nothing and it\u0027s still have 3 variables in there."},{"Start":"08:47.490 ","End":"08:50.515","Text":"My axis of rotation is here,"},{"Start":"08:50.515 ","End":"08:54.580","Text":"and now we can start writing this out. Let\u0027s see."},{"Start":"08:54.580 ","End":"08:59.770","Text":"I\u0027ll have my N_2 multiplied by its distance away from the axis of rotation,"},{"Start":"08:59.770 ","End":"09:02.050","Text":"which is 0."},{"Start":"09:02.050 ","End":"09:04.270","Text":"That\u0027s great."},{"Start":"09:04.270 ","End":"09:09.535","Text":"That contributes 0 to the torque, plus my force,"},{"Start":"09:09.535 ","End":"09:11.260","Text":"my frictional force f,"},{"Start":"09:11.260 ","End":"09:16.030","Text":"multiplied by its distance away from the axis of rotation, which is again 0."},{"Start":"09:16.030 ","End":"09:18.175","Text":"It\u0027s acting at the axis of rotation."},{"Start":"09:18.175 ","End":"09:20.515","Text":"Again, multiplied by 0."},{"Start":"09:20.515 ","End":"09:25.090","Text":"Again, this contributes 0 to my equation."},{"Start":"09:25.090 ","End":"09:34.360","Text":"Then the next force is my mg. Before we start working out our mg,"},{"Start":"09:34.360 ","End":"09:36.580","Text":"let\u0027s work out our angles."},{"Start":"09:36.580 ","End":"09:40.285","Text":"Here, of course, between the wall and the floor,"},{"Start":"09:40.285 ","End":"09:43.770","Text":"we know that there\u0027s always going to be a 90-degree angle."},{"Start":"09:43.770 ","End":"09:47.110","Text":"Here, we\u0027re being told that our angle over here is Alpha,"},{"Start":"09:47.110 ","End":"09:56.680","Text":"which means that this angle over here is going to be 90 minus Alpha."},{"Start":"09:56.680 ","End":"10:01.330","Text":"If you work this out, we know that the sum of all of the angles in a triangle is 180."},{"Start":"10:01.330 ","End":"10:05.830","Text":"180 minus 90 is 90 minus Alpha,"},{"Start":"10:05.830 ","End":"10:08.215","Text":"and then we get this equation over here."},{"Start":"10:08.215 ","End":"10:10.540","Text":"That\u0027s that angle over here."},{"Start":"10:10.540 ","End":"10:14.665","Text":"Then we can say that in that case,"},{"Start":"10:14.665 ","End":"10:22.495","Text":"that the angle over here between the ladder and our mg is going to be,"},{"Start":"10:22.495 ","End":"10:25.985","Text":"because this is parallel to this,"},{"Start":"10:25.985 ","End":"10:28.660","Text":"I know I didn\u0027t draw it exactly,"},{"Start":"10:28.660 ","End":"10:34.480","Text":"this line over here representing the wall is parallel to this arrow over"},{"Start":"10:34.480 ","End":"10:40.165","Text":"here for mg. Then this side is obviously the same ladder."},{"Start":"10:40.165 ","End":"10:48.775","Text":"We can say that this angle over here is also equal to 90 minus Alpha."},{"Start":"10:48.775 ","End":"10:52.735","Text":"This angle right over here."},{"Start":"10:52.735 ","End":"10:58.540","Text":"Now let\u0027s work out our torque for mg. First, of course,"},{"Start":"10:58.540 ","End":"11:01.150","Text":"we write in our force, which is mg,"},{"Start":"11:01.150 ","End":"11:06.400","Text":"and then we multiply it by its distance away from the axis of rotation."},{"Start":"11:06.400 ","End":"11:13.090","Text":"We know that mg always acts from the center of mass and we\u0027re"},{"Start":"11:13.090 ","End":"11:16.210","Text":"assuming because we weren\u0027t told otherwise that the mass"},{"Start":"11:16.210 ","End":"11:19.900","Text":"is evenly distributed along the length of the ladder,"},{"Start":"11:19.900 ","End":"11:24.400","Text":"which means that the center of mass is going to be right in the center of the ladder."},{"Start":"11:24.400 ","End":"11:26.200","Text":"The ladder length is L,"},{"Start":"11:26.200 ","End":"11:29.515","Text":"which means that its center is in its center,"},{"Start":"11:29.515 ","End":"11:31.855","Text":"which is at L divided by 2."},{"Start":"11:31.855 ","End":"11:37.630","Text":"Then what we want to do is we want to multiply this by sine of"},{"Start":"11:37.630 ","End":"11:43.450","Text":"the angle between our force and our radius,"},{"Start":"11:43.450 ","End":"11:47.140","Text":"where this over here is our radius."},{"Start":"11:47.140 ","End":"11:51.445","Text":"I\u0027ll actually draw that for you in red."},{"Start":"11:51.445 ","End":"11:58.375","Text":"This is our radius R over here, specifically."},{"Start":"11:58.375 ","End":"12:00.490","Text":"The angle over here,"},{"Start":"12:00.490 ","End":"12:04.465","Text":"as we said, is sine of 90 minus Alpha."},{"Start":"12:04.465 ","End":"12:13.360","Text":"I\u0027m going to multiply this by sine of that angle, 90 minus Alpha."},{"Start":"12:13.360 ","End":"12:16.440","Text":"As we know, sine of 90"},{"Start":"12:16.440 ","End":"12:20.940","Text":"minus Alpha is the same as cosine of Alpha, but it doesn\u0027t really matter."},{"Start":"12:20.940 ","End":"12:23.290","Text":"You could also leave it like this."},{"Start":"12:23.940 ","End":"12:28.510","Text":"Now, let\u0027s look at the sine so we can see that it\u0027s a positive because"},{"Start":"12:28.510 ","End":"12:34.050","Text":"this force is acting to move the ladder downwards,"},{"Start":"12:34.050 ","End":"12:37.185","Text":"in this direction, anticlockwise,"},{"Start":"12:37.185 ","End":"12:41.715","Text":"which as we can see, is what we defined as the positive direction of rotation."},{"Start":"12:41.715 ","End":"12:44.135","Text":"We leave that as a positive."},{"Start":"12:44.135 ","End":"12:51.310","Text":"Now, our final force to deal with in the torque equation is our N_1."},{"Start":"12:51.310 ","End":"12:56.275","Text":"First of all, we can see that our N_1 is trying to move"},{"Start":"12:56.275 ","End":"13:03.415","Text":"our ladder such that it rotates in this direction, which is clockwise."},{"Start":"13:03.415 ","End":"13:05.574","Text":"That, as we defined,"},{"Start":"13:05.574 ","End":"13:08.695","Text":"clockwise is the negative direction."},{"Start":"13:08.695 ","End":"13:14.215","Text":"We\u0027ll put a minus over here and then we\u0027ll write in our force N_1"},{"Start":"13:14.215 ","End":"13:21.175","Text":"multiplied by its distance from the axis of rotation, so the radius."},{"Start":"13:21.175 ","End":"13:28.435","Text":"Now our radius is from the axis of rotation all the way to the tip-top of the ladder."},{"Start":"13:28.435 ","End":"13:37.030","Text":"That is length L. Then we want to multiply it by sine of the angle between these two."},{"Start":"13:37.030 ","End":"13:39.775","Text":"Between this and this."},{"Start":"13:39.775 ","End":"13:42.655","Text":"What is this angle over here?"},{"Start":"13:42.655 ","End":"13:46.195","Text":"We can see that this is alternate angles."},{"Start":"13:46.195 ","End":"13:54.925","Text":"We can see that our N_1 arrow is parallel to the line over here representing the floor."},{"Start":"13:54.925 ","End":"13:59.290","Text":"If this angle, therefore over here is Alpha,"},{"Start":"13:59.290 ","End":"14:01.930","Text":"so this angle on this side,"},{"Start":"14:01.930 ","End":"14:03.790","Text":"we can see a Z over here,"},{"Start":"14:03.790 ","End":"14:05.710","Text":"this is also going to be Alpha."},{"Start":"14:05.710 ","End":"14:10.880","Text":"We\u0027re just going to multiply this by sine of Alpha."},{"Start":"14:11.910 ","End":"14:17.210","Text":"This is of course all equal to 0."},{"Start":"14:18.510 ","End":"14:24.115","Text":"Now we have 3 equations with 3 unknowns."},{"Start":"14:24.115 ","End":"14:26.905","Text":"All I have to do is rearrange these equations,"},{"Start":"14:26.905 ","End":"14:30.445","Text":"substitute in whatever I need,"},{"Start":"14:30.445 ","End":"14:35.230","Text":"and then I will be able to solve this and find out what my N_1 is equal to,"},{"Start":"14:35.230 ","End":"14:36.910","Text":"what my N_2 is equal to,"},{"Start":"14:36.910 ","End":"14:40.060","Text":"and what my frictional force F is equal to"},{"Start":"14:40.060 ","End":"14:44.140","Text":"with relation to my known quantities such as mg,"},{"Start":"14:44.140 ","End":"14:46.640","Text":"L, and so on."},{"Start":"14:47.430 ","End":"14:52.030","Text":"I won\u0027t solve this right now because you can do this,"},{"Start":"14:52.030 ","End":"14:53.380","Text":"this is just algebra."},{"Start":"14:53.380 ","End":"14:57.190","Text":"Now, just a possible question that they could ask is they could ask"},{"Start":"14:57.190 ","End":"15:03.355","Text":"what maximum value for Alpha can I obtain"},{"Start":"15:03.355 ","End":"15:08.500","Text":"given some coefficient of friction such that my ladder will"},{"Start":"15:08.500 ","End":"15:14.440","Text":"remain stationary and won\u0027t slide and fall onto the ground, for instance."},{"Start":"15:14.440 ","End":"15:17.380","Text":"If they asked me that type of question,"},{"Start":"15:17.380 ","End":"15:20.350","Text":"as in how far away can this corner of the ladder"},{"Start":"15:20.350 ","End":"15:24.925","Text":"be along the floor such that my ladder remains stationary,"},{"Start":"15:24.925 ","End":"15:27.879","Text":"if they ever asked me that question,"},{"Start":"15:27.879 ","End":"15:31.270","Text":"then that means that my angle over here, Alpha,"},{"Start":"15:31.270 ","End":"15:32.740","Text":"is also an unknown,"},{"Start":"15:32.740 ","End":"15:34.840","Text":"which would mean that I have 4 unknowns,"},{"Start":"15:34.840 ","End":"15:39.475","Text":"which means that I would need 4 equations in order to solve this."},{"Start":"15:39.475 ","End":"15:42.715","Text":"Then and only in that type of question where they\u0027re"},{"Start":"15:42.715 ","End":"15:46.075","Text":"asking me specifically for the maximum,"},{"Start":"15:46.075 ","End":"15:49.060","Text":"or here it would actually be the minimum value of Alpha"},{"Start":"15:49.060 ","End":"15:52.495","Text":"because the angle would decrease as I move this along here."},{"Start":"15:52.495 ","End":"15:56.310","Text":"Then and only then can I use the equation"},{"Start":"15:56.310 ","End":"16:00.245","Text":"of f for my fourth equation in order to solve this."},{"Start":"16:00.245 ","End":"16:04.610","Text":"That f is equal to Mu N. Only in"},{"Start":"16:04.610 ","End":"16:08.780","Text":"that type of format of questions can I use this equation because then I know"},{"Start":"16:08.780 ","End":"16:13.130","Text":"that my maximum value for my frictional force and if"},{"Start":"16:13.130 ","End":"16:18.950","Text":"I reduce this or take this corner of the ladder further away,"},{"Start":"16:18.950 ","End":"16:22.820","Text":"then my frictional force of static friction won\u0027t"},{"Start":"16:22.820 ","End":"16:27.750","Text":"hold through and my ladder will just slide and completely fall."},{"Start":"16:28.920 ","End":"16:32.570","Text":"That is the end of our lesson."}],"ID":10795},{"Watched":false,"Name":"Exercise - Rod Resting on Paper","Duration":"18m 47s","ChapterTopicVideoID":10448,"CourseChapterTopicPlaylistID":9396,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:08.325","Text":"Hello. A rod of length L and of mass M is attached to the ceiling via an axis."},{"Start":"00:08.325 ","End":"00:12.195","Text":"The other end of the rod is resting on a sheet of paper,"},{"Start":"00:12.195 ","End":"00:15.000","Text":"which is itself resting on the floor."},{"Start":"00:15.000 ","End":"00:18.420","Text":"We can imagine that the paper is massless."},{"Start":"00:18.420 ","End":"00:23.670","Text":"The angle between the rod and the axis is Beta and the coefficient of"},{"Start":"00:23.670 ","End":"00:30.610","Text":"static friction between the rod and the paper and also the paper on the floor is Mu_S."},{"Start":"00:30.610 ","End":"00:32.945","Text":"Question number 1 is,"},{"Start":"00:32.945 ","End":"00:37.220","Text":"the paper is pulled rightwards with a force F. What is"},{"Start":"00:37.220 ","End":"00:42.620","Text":"the minimum force required in order to pull the paper from underneath the rod?"},{"Start":"00:42.620 ","End":"00:46.830","Text":"We can assume that the rod remains stationary."},{"Start":"00:48.050 ","End":"00:51.500","Text":"The first thing that we\u0027re going to do is we\u0027re going"},{"Start":"00:51.500 ","End":"00:54.785","Text":"to start drawing our free body diagram."},{"Start":"00:54.785 ","End":"00:56.410","Text":"First things first,"},{"Start":"00:56.410 ","End":"00:59.675","Text":"we have this force over here,"},{"Start":"00:59.675 ","End":"01:04.890","Text":"which is our F. I\u0027ll just go over this in blue."},{"Start":"01:04.890 ","End":"01:07.775","Text":"We have a force which is pulling the piece of paper."},{"Start":"01:07.775 ","End":"01:09.920","Text":"Then on the piece of paper,"},{"Start":"01:09.920 ","End":"01:16.499","Text":"we have the normal force from the floor pointing upwards,"},{"Start":"01:16.499 ","End":"01:18.930","Text":"so normal force from the floor,"},{"Start":"01:18.930 ","End":"01:23.450","Text":"and then going downwards still, on the piece of paper,"},{"Start":"01:23.450 ","End":"01:33.725","Text":"so we have the normal force applied onto the paper by the rod."},{"Start":"01:33.725 ","End":"01:38.630","Text":"Then on the piece of paper between the rod and the piece of paper,"},{"Start":"01:38.630 ","End":"01:40.805","Text":"so over here,"},{"Start":"01:40.805 ","End":"01:44.045","Text":"we have static frictional force."},{"Start":"01:44.045 ","End":"01:48.830","Text":"Then between the paper and the floor also in this direction,"},{"Start":"01:48.830 ","End":"01:52.978","Text":"pointing in the opposite direction to the direction of motion,"},{"Start":"01:52.978 ","End":"01:59.520","Text":"we have static frictional force also between the paper and the floor."},{"Start":"02:00.500 ","End":"02:03.710","Text":"Now if we look at the question, we\u0027re being asked,"},{"Start":"02:03.710 ","End":"02:09.155","Text":"what is the minimum force required in order to pull the paper from underneath the rod?"},{"Start":"02:09.155 ","End":"02:12.965","Text":"Now, when we look at this type of phrasing,"},{"Start":"02:12.965 ","End":"02:16.625","Text":"we haven\u0027t really seen a question like this, or have we?"},{"Start":"02:16.625 ","End":"02:22.280","Text":"Another word for the minimum force required in order to move the paper from underneath"},{"Start":"02:22.280 ","End":"02:26.750","Text":"the rod is the maximal force"},{"Start":"02:26.750 ","End":"02:32.645","Text":"required such that the paper isn\u0027t moved from underneath the rod."},{"Start":"02:32.645 ","End":"02:36.110","Text":"What we\u0027re trying to find is"},{"Start":"02:36.110 ","End":"02:46.770","Text":"the max force such that paper doesn\u0027t move."},{"Start":"02:47.080 ","End":"02:52.290","Text":"This type of phrasing we\u0027ve come across in the previous lessons,"},{"Start":"02:52.290 ","End":"02:54.530","Text":"so we know how to solve this question."},{"Start":"02:54.530 ","End":"02:57.780","Text":"Let\u0027s carry on and see how we do this."},{"Start":"02:58.370 ","End":"03:04.685","Text":"Question number 1 is we\u0027re actually trying to instead answer this question."},{"Start":"03:04.685 ","End":"03:07.420","Text":"It means the exact same thing as what\u0027s written here,"},{"Start":"03:07.420 ","End":"03:09.970","Text":"just phrased in a way that we know how to solve it."},{"Start":"03:09.970 ","End":"03:12.095","Text":"Now, as we know,"},{"Start":"03:12.095 ","End":"03:16.015","Text":"because we\u0027re being told that the rod remains stationary,"},{"Start":"03:16.015 ","End":"03:18.280","Text":"and we\u0027re dealing with static friction,"},{"Start":"03:18.280 ","End":"03:25.330","Text":"then we know that if we\u0027re trying to find our F_max,"},{"Start":"03:25.330 ","End":"03:30.244","Text":"that means that we\u0027re trying to find our f_s max."},{"Start":"03:30.244 ","End":"03:31.929","Text":"Because when it\u0027s stationary,"},{"Start":"03:31.929 ","End":"03:34.870","Text":"we know that the sum of all of the forces is equal to 0,"},{"Start":"03:34.870 ","End":"03:37.195","Text":"which means that if we know our F_max,"},{"Start":"03:37.195 ","End":"03:41.615","Text":"then we know our f_s max because they are equal."},{"Start":"03:41.615 ","End":"03:46.680","Text":"Then when we reach this F_max or f_s max,"},{"Start":"03:46.680 ","End":"03:53.285","Text":"that\u0027s when we reach the boundary of where the rod remains stationary and then moves."},{"Start":"03:53.285 ","End":"03:56.480","Text":"What we\u0027re trying to find is f_s max,"},{"Start":"03:56.480 ","End":"03:57.755","Text":"which as we know,"},{"Start":"03:57.755 ","End":"04:01.020","Text":"is equal to Mu_s,"},{"Start":"04:01.020 ","End":"04:03.245","Text":"the coefficient of static friction,"},{"Start":"04:03.245 ","End":"04:07.500","Text":"multiplied by N, the normal force."},{"Start":"04:08.390 ","End":"04:13.745","Text":"Now what we\u0027re going to do is we\u0027re going to work out the sum of all of the forces."},{"Start":"04:13.745 ","End":"04:24.285","Text":"First, we\u0027re going to do the sum of all of the forces on the sheet of paper."},{"Start":"04:24.285 ","End":"04:30.410","Text":"Let\u0027s work out the sum of all of the forces in the y direction is going to be"},{"Start":"04:30.410 ","End":"04:36.880","Text":"equal to the normal force which is acting on the piece of paper from the floor."},{"Start":"04:36.880 ","End":"04:41.330","Text":"N, I\u0027ll just write F for floor, minus,"},{"Start":"04:41.330 ","End":"04:43.460","Text":"because the arrow is in the opposite direction,"},{"Start":"04:43.460 ","End":"04:49.355","Text":"the normal force being applied onto the sheet of paper by the rod."},{"Start":"04:49.355 ","End":"04:51.140","Text":"N rod, N_r."},{"Start":"04:51.140 ","End":"04:54.935","Text":"As we know, because there\u0027s no motion in the y direction,"},{"Start":"04:54.935 ","End":"04:59.000","Text":"we know that the sum of all of the forces in the y direction is equal to 0."},{"Start":"04:59.000 ","End":"05:05.840","Text":"Therefore, we know that the normal force on sheet of paper coming from"},{"Start":"05:05.840 ","End":"05:13.040","Text":"the floor is equal to the normal force on the sheet of paper being applied by the rod."},{"Start":"05:13.040 ","End":"05:19.090","Text":"Now let\u0027s work out the sum of all of the forces in the x direction."},{"Start":"05:19.090 ","End":"05:24.020","Text":"We know if we take this way to be the positive x direction, it doesn\u0027t really matter."},{"Start":"05:24.020 ","End":"05:27.580","Text":"We have a pulling force which is pulling the sheet of paper,"},{"Start":"05:27.580 ","End":"05:29.885","Text":"so F, and then we have,"},{"Start":"05:29.885 ","End":"05:33.175","Text":"in the opposite direction, so minus,"},{"Start":"05:33.175 ","End":"05:39.485","Text":"our frictional force due to the rod being in contact with the paper,"},{"Start":"05:39.485 ","End":"05:45.395","Text":"and the frictional force due to the paper being in contact with the floor."},{"Start":"05:45.395 ","End":"05:49.955","Text":"Now as we saw, the frictional force is our f_s max,"},{"Start":"05:49.955 ","End":"05:55.939","Text":"which is equal to the coefficient of friction multiplied by the various normal forces."},{"Start":"05:55.939 ","End":"05:58.160","Text":"Now, we can see that because our N_f,"},{"Start":"05:58.160 ","End":"05:59.975","Text":"our N floor, and our N_r,"},{"Start":"05:59.975 ","End":"06:01.340","Text":"N rod, are equal,"},{"Start":"06:01.340 ","End":"06:06.920","Text":"so we can just use N to represent both of them because they\u0027re equal to the same."},{"Start":"06:06.920 ","End":"06:15.045","Text":"That means that we have minus 2 times our static frictional force."},{"Start":"06:15.045 ","End":"06:20.870","Text":"Then, because we\u0027re trying to find the maximum force such that the paper doesn\u0027t move,"},{"Start":"06:20.870 ","End":"06:23.270","Text":"that means that there\u0027s also no motion,"},{"Start":"06:23.270 ","End":"06:25.865","Text":"we\u0027re trying to find the boundary of where there still"},{"Start":"06:25.865 ","End":"06:28.930","Text":"isn\u0027t any motion in this system over here."},{"Start":"06:28.930 ","End":"06:30.395","Text":"If there\u0027s no motion,"},{"Start":"06:30.395 ","End":"06:36.125","Text":"that means that the sum of all of the forces in the x direction is also equal to 0."},{"Start":"06:36.125 ","End":"06:39.650","Text":"Therefore, we get that our force, F,"},{"Start":"06:39.650 ","End":"06:45.025","Text":"which is pulling the sheet of paper must be equal to 2F_s."},{"Start":"06:45.025 ","End":"06:48.335","Text":"As we know, because we\u0027re trying to find the maximum,"},{"Start":"06:48.335 ","End":"06:50.330","Text":"so this is also going to be the maximum,"},{"Start":"06:50.330 ","End":"06:51.800","Text":"so we can substitute that in."},{"Start":"06:51.800 ","End":"07:00.710","Text":"We can see that our maximum force is going to be equal to 2Mu_s,"},{"Start":"07:00.710 ","End":"07:02.165","Text":"which is given in the question,"},{"Start":"07:02.165 ","End":"07:07.272","Text":"multiplied by our normal force."},{"Start":"07:07.272 ","End":"07:10.300","Text":"We\u0027re almost there."},{"Start":"07:10.300 ","End":"07:16.120","Text":"All we have to do is we have to find out what this normal force is equal to."},{"Start":"07:16.120 ","End":"07:17.575","Text":"In order to do that,"},{"Start":"07:17.575 ","End":"07:23.750","Text":"we\u0027re now going to draw a free body diagram for the forces acting on the rod."},{"Start":"07:24.390 ","End":"07:28.540","Text":"Now, I\u0027m going to move over to the red color."},{"Start":"07:28.540 ","End":"07:33.490","Text":"All the blue arrows over here represent the forces on"},{"Start":"07:33.490 ","End":"07:38.875","Text":"the paper and now everything in red represents the forces on the rod."},{"Start":"07:38.875 ","End":"07:42.910","Text":"First of all, we have the normal force due to"},{"Start":"07:42.910 ","End":"07:47.050","Text":"the paper applying a normal force onto the rod."},{"Start":"07:47.050 ","End":"07:52.360","Text":"This is normal force from the paper and then we\u0027re going to"},{"Start":"07:52.360 ","End":"07:58.435","Text":"have the frictional force applied by the paper onto the rod."},{"Start":"07:58.435 ","End":"08:02.620","Text":"If we can see here that the frictional force applied"},{"Start":"08:02.620 ","End":"08:07.765","Text":"to the paper by the rod is going in the leftward direction,"},{"Start":"08:07.765 ","End":"08:12.010","Text":"so that means that the frictional force applied to"},{"Start":"08:12.010 ","End":"08:17.260","Text":"the rod by the paper is going to be in the opposite direction to this."},{"Start":"08:17.260 ","End":"08:20.560","Text":"It\u0027s going to be in this direction,"},{"Start":"08:20.560 ","End":"08:26.600","Text":"F_s, applied by the paper onto the rod."},{"Start":"08:26.790 ","End":"08:30.040","Text":"What I just said now it might be a bit confusing."},{"Start":"08:30.040 ","End":"08:34.090","Text":"Just go over that section of the video if you\u0027re getting confused."},{"Start":"08:34.090 ","End":"08:37.825","Text":"Now, there\u0027s no normal force between the floor and the rod"},{"Start":"08:37.825 ","End":"08:43.250","Text":"because the floor is in indirect contact with the rod but rather the papers."},{"Start":"08:44.280 ","End":"08:52.420","Text":"The next force that we have is our Mg acting from the center of mass of the rod,"},{"Start":"08:52.420 ","End":"08:56.635","Text":"Mg, and then we also have"},{"Start":"08:56.635 ","End":"09:01.285","Text":"a force acting from the axis connecting the rod to the ceiling."},{"Start":"09:01.285 ","End":"09:05.230","Text":"However, we\u0027re not going to deal with this with regards to forces,"},{"Start":"09:05.230 ","End":"09:09.170","Text":"but we\u0027re going to deal with this with regards to torque."},{"Start":"09:10.380 ","End":"09:16.405","Text":"Now we\u0027re going to work on the rod."},{"Start":"09:16.405 ","End":"09:22.870","Text":"On rod, I\u0027m going to scroll down a little bit to give us a little bit more space."},{"Start":"09:22.870 ","End":"09:25.330","Text":"Now what we\u0027re going to do because we\u0027re dealing with"},{"Start":"09:25.330 ","End":"09:28.840","Text":"this axis of rotation on the ceiling."},{"Start":"09:28.840 ","End":"09:35.690","Text":"We\u0027re going to work out the sum of all of the moments or of all of the torques."},{"Start":"09:36.150 ","End":"09:40.225","Text":"Let\u0027s look at all of the forces that we have acting."},{"Start":"09:40.225 ","End":"09:42.415","Text":"What we\u0027re going to do, first of all,"},{"Start":"09:42.415 ","End":"09:48.625","Text":"is we\u0027re going to define some direction of rotation to be the positive direction."},{"Start":"09:48.625 ","End":"09:52.915","Text":"Let\u0027s say that this is the positive direction of rotation."},{"Start":"09:52.915 ","End":"09:57.775","Text":"Now what we can do before we start with the calculations,"},{"Start":"09:57.775 ","End":"10:00.880","Text":"we can also see that if this angle over here is"},{"Start":"10:00.880 ","End":"10:04.614","Text":"Beta and these 2 lines over here are parallel,"},{"Start":"10:04.614 ","End":"10:10.315","Text":"then that means that this angle over here is also Beta."},{"Start":"10:10.315 ","End":"10:13.960","Text":"Now, let\u0027s begin by writing all the torques."},{"Start":"10:13.960 ","End":"10:17.950","Text":"The torque due to the normal from the paper acting on"},{"Start":"10:17.950 ","End":"10:23.530","Text":"the rod is going to be equal to N multiplied by L,"},{"Start":"10:23.530 ","End":"10:25.390","Text":"the length of the rod,"},{"Start":"10:25.390 ","End":"10:30.340","Text":"multiplied by sine of the angle."},{"Start":"10:30.340 ","End":"10:33.730","Text":"Now, of course, this is in the positive direction."},{"Start":"10:33.730 ","End":"10:36.970","Text":"As we can see, this torque is aiming to rotate"},{"Start":"10:36.970 ","End":"10:40.960","Text":"the rod in the direction that we defined as the positive direction."},{"Start":"10:40.960 ","End":"10:44.890","Text":"The next torque that we have is the torque due to"},{"Start":"10:44.890 ","End":"10:49.405","Text":"the static friction of the paper on the rod."},{"Start":"10:49.405 ","End":"10:52.510","Text":"That again is pointing in a direction"},{"Start":"10:52.510 ","End":"10:55.540","Text":"as to turn the rod in the positive direction of rotation,"},{"Start":"10:55.540 ","End":"10:59.545","Text":"so this is again going to be a positive and then it\u0027s going to be"},{"Start":"10:59.545 ","End":"11:07.780","Text":"F_s multiplied by L sine of Beta."},{"Start":"11:07.780 ","End":"11:10.510","Text":"This is not sine of Beta."},{"Start":"11:10.510 ","End":"11:14.110","Text":"It\u0027s in fact, in this case, cosine of Beta."},{"Start":"11:14.110 ","End":"11:15.280","Text":"If you can\u0027t remember why,"},{"Start":"11:15.280 ","End":"11:17.994","Text":"you please go back to one of the previous lessons."},{"Start":"11:17.994 ","End":"11:26.365","Text":"Then we have to include the torque acting due to gravitational force."},{"Start":"11:26.365 ","End":"11:28.045","Text":"I\u0027ll write it over here."},{"Start":"11:28.045 ","End":"11:31.120","Text":"Now we can see that this torque is aiming to"},{"Start":"11:31.120 ","End":"11:34.630","Text":"rotate the rod in this anticlockwise direction,"},{"Start":"11:34.630 ","End":"11:37.690","Text":"which is what we deemed as the negative direction of rotation,"},{"Start":"11:37.690 ","End":"11:40.300","Text":"so we\u0027ll have a minus over here."},{"Start":"11:40.300 ","End":"11:44.575","Text":"Then the force is Mg,"},{"Start":"11:44.575 ","End":"11:51.115","Text":"and it\u0027s acting from the center of mass of the rod."},{"Start":"11:51.115 ","End":"11:53.035","Text":"The center of mass,"},{"Start":"11:53.035 ","End":"11:56.710","Text":"if a mass is distributed evenly along the rod,"},{"Start":"11:56.710 ","End":"12:00.445","Text":"is going to be at the point L over 2,"},{"Start":"12:00.445 ","End":"12:10.195","Text":"so halfway through and then multiplied by sine of Beta."},{"Start":"12:10.195 ","End":"12:14.485","Text":"All of this is going to be equal to 0."},{"Start":"12:14.485 ","End":"12:17.395","Text":"Now we know that our F_s over here,"},{"Start":"12:17.395 ","End":"12:20.515","Text":"because we\u0027re trying to find our F_s max,"},{"Start":"12:20.515 ","End":"12:30.160","Text":"so we know that this is equal to Mu_s multiplied by N. Now let\u0027s write this out."},{"Start":"12:30.160 ","End":"12:32.890","Text":"I\u0027m going to scroll down a little bit again."},{"Start":"12:32.890 ","End":"12:35.995","Text":"We can say that this is equal to,"},{"Start":"12:35.995 ","End":"12:43.689","Text":"we can take out the common factor N and here we can move this over to the other side,"},{"Start":"12:43.689 ","End":"12:49.840","Text":"so we\u0027ll have that this is N multiplied by L,"},{"Start":"12:49.840 ","End":"12:56.500","Text":"sine of Beta plus Mu_s,"},{"Start":"12:56.500 ","End":"13:02.365","Text":"L cosine of Beta which is equal to,"},{"Start":"13:02.365 ","End":"13:04.870","Text":"when we move this over to the other side,"},{"Start":"13:04.870 ","End":"13:12.160","Text":"Mg, L over 2 sine of Beta."},{"Start":"13:12.160 ","End":"13:15.760","Text":"Now we can see that we have L on both sides,"},{"Start":"13:15.760 ","End":"13:22.600","Text":"so we can divide both sides by L. Then we can isolate out our N,"},{"Start":"13:22.600 ","End":"13:25.630","Text":"because if you remember, this is what we\u0027re trying to find."},{"Start":"13:25.630 ","End":"13:27.685","Text":"We want to work out what our N is."},{"Start":"13:27.685 ","End":"13:31.885","Text":"We\u0027ll just divide both sides by what is inside these brackets."},{"Start":"13:31.885 ","End":"13:37.735","Text":"Therefore, I get that N is equal to Mg,"},{"Start":"13:37.735 ","End":"13:43.790","Text":"sine of Beta divided by"},{"Start":"13:46.830 ","End":"13:52.555","Text":"sine of"},{"Start":"13:52.555 ","End":"13:58.915","Text":"Beta plus Mu_s,"},{"Start":"13:58.915 ","End":"14:05.215","Text":"cosine of Beta and then we also have this 2 over here."},{"Start":"14:05.215 ","End":"14:10.460","Text":"I\u0027ll put this as a common denominator."},{"Start":"14:12.090 ","End":"14:15.615","Text":"This is what N is equal to."},{"Start":"14:15.615 ","End":"14:21.050","Text":"Now, because what we\u0027re actually trying to find is our F max,"},{"Start":"14:21.050 ","End":"14:22.625","Text":"which is also our F min."},{"Start":"14:22.625 ","End":"14:24.350","Text":"I can\u0027t we remember what we did."},{"Start":"14:24.350 ","End":"14:34.055","Text":"Therefore, we can say that our F min is equal to just what our F max is which,"},{"Start":"14:34.055 ","End":"14:38.735","Text":"as we said, is equal to this 2F_s,"},{"Start":"14:38.735 ","End":"14:49.020","Text":"which was 2Mu_s multiplied by N. Then we will have Mu_s multiplied by N,"},{"Start":"14:49.020 ","End":"14:55.685","Text":"which is Mg sine of Beta divided by."},{"Start":"14:55.685 ","End":"14:58.880","Text":"Here, we\u0027re multiplying by 2,"},{"Start":"14:58.880 ","End":"15:00.470","Text":"and here we\u0027re dividing by 2,"},{"Start":"15:00.470 ","End":"15:02.375","Text":"so the 2\u0027s will cancel out."},{"Start":"15:02.375 ","End":"15:12.330","Text":"It will just be divided by sine of Beta plus Mu_s cosine of Beta."},{"Start":"15:13.110 ","End":"15:17.800","Text":"This is the final answer to question number 1."},{"Start":"15:17.800 ","End":"15:20.185","Text":"Now, let\u0027s move on to question number 2."},{"Start":"15:20.185 ","End":"15:23.260","Text":"Question number 2 is to redo question number 1,"},{"Start":"15:23.260 ","End":"15:30.565","Text":"but this time when the force is acting leftwards."},{"Start":"15:30.565 ","End":"15:35.215","Text":"If a force F is acting in the leftwards direction,"},{"Start":"15:35.215 ","End":"15:38.709","Text":"that means that this F just simply needs to change directions,"},{"Start":"15:38.709 ","End":"15:45.160","Text":"so the arrow will be pointing in this direction instead."},{"Start":"15:45.160 ","End":"15:49.480","Text":"Then all we have to do is we have to go over here"},{"Start":"15:49.480 ","End":"15:53.680","Text":"and we just change the sign of this F_max."},{"Start":"15:53.680 ","End":"15:59.810","Text":"This F_max will become minus because it\u0027s pointing in the negative x-direction."},{"Start":"15:59.810 ","End":"16:04.280","Text":"Then we know that our static frictional force,"},{"Start":"16:04.280 ","End":"16:05.900","Text":"or frictional force in general,"},{"Start":"16:05.900 ","End":"16:09.830","Text":"always acts in the direction opposite to the direction of motion."},{"Start":"16:09.830 ","End":"16:12.755","Text":"Which means that if this is the direction of motion,"},{"Start":"16:12.755 ","End":"16:16.210","Text":"so our F_s will be working in the opposite direction."},{"Start":"16:16.210 ","End":"16:18.055","Text":"Now, it will be pointing rightwards."},{"Start":"16:18.055 ","End":"16:22.820","Text":"That would mean that this would also be a minus over here."},{"Start":"16:22.820 ","End":"16:25.220","Text":"In fact, nothing here changes,"},{"Start":"16:25.220 ","End":"16:27.395","Text":"if this side is a minus and this side is a minus,"},{"Start":"16:27.395 ","End":"16:30.290","Text":"it\u0027s the exact same thing as saying that this side"},{"Start":"16:30.290 ","End":"16:33.360","Text":"is a positive and this side is a positive."},{"Start":"16:33.630 ","End":"16:37.065","Text":"What we see is that our F_max,"},{"Start":"16:37.065 ","End":"16:39.109","Text":"when we substitute in our values,"},{"Start":"16:39.109 ","End":"16:45.615","Text":"it\u0027s still going to be equal to 2Mu_s multiplied by our normal."},{"Start":"16:45.615 ","End":"16:49.005","Text":"Great. We don\u0027t have to change anything here."},{"Start":"16:49.005 ","End":"16:52.230","Text":"Now we still have to work out what our normal is."},{"Start":"16:52.690 ","End":"16:55.520","Text":"To work out our normal, just like before,"},{"Start":"16:55.520 ","End":"16:58.415","Text":"we\u0027re going to find the torques on the rod."},{"Start":"16:58.415 ","End":"17:00.860","Text":"Now, the only thing that we can see,"},{"Start":"17:00.860 ","End":"17:10.070","Text":"so our normal forces are still working in the y-direction but our F_s,"},{"Start":"17:10.070 ","End":"17:16.410","Text":"our static friction, we said it\u0027s now going to be pointing in this direction."},{"Start":"17:16.410 ","End":"17:18.305","Text":"In the opposite direction."},{"Start":"17:18.305 ","End":"17:23.500","Text":"This will now become a minus over here and"},{"Start":"17:23.500 ","End":"17:31.335","Text":"then that means that this over here will be a minus as well."},{"Start":"17:31.335 ","End":"17:37.480","Text":"This is a minus because we said that our F_s is equal to Mu_s N,"},{"Start":"17:37.480 ","End":"17:40.480","Text":"so here we have the Mu_s N. Here,"},{"Start":"17:40.480 ","End":"17:43.630","Text":"it\u0027s also a minus and then in order to get our N,"},{"Start":"17:43.630 ","End":"17:46.405","Text":"we simply divided over here."},{"Start":"17:46.405 ","End":"17:48.965","Text":"We\u0027ll add our minus over here."},{"Start":"17:48.965 ","End":"17:52.340","Text":"We divided both sides by what\u0027s inside the brackets"},{"Start":"17:52.340 ","End":"17:55.640","Text":"over here and that means that our final answer for"},{"Start":"17:55.640 ","End":"18:02.210","Text":"question number 2 is going to be the exact same answer that we got for question number 1."},{"Start":"18:02.210 ","End":"18:06.035","Text":"The only difference being that here in the denominator,"},{"Start":"18:06.035 ","End":"18:09.170","Text":"we\u0027ll have a minus instead of a plus."},{"Start":"18:09.170 ","End":"18:12.960","Text":"But it\u0027s the exact same answer aside from that."},{"Start":"18:13.770 ","End":"18:17.150","Text":"This is the final answer for question number 2."},{"Start":"18:17.150 ","End":"18:20.595","Text":"The only difference is this minus over here in the denominator."},{"Start":"18:20.595 ","End":"18:22.370","Text":"As we can see,"},{"Start":"18:22.370 ","End":"18:25.925","Text":"we didn\u0027t have to do any of the calculations again,"},{"Start":"18:25.925 ","End":"18:34.150","Text":"all we had to know is that if the direction of our force F is changed,"},{"Start":"18:34.150 ","End":"18:37.699","Text":"so our F_s direction always has to be opposite,"},{"Start":"18:37.699 ","End":"18:40.400","Text":"so we had to change that and then just follow"},{"Start":"18:40.400 ","End":"18:44.985","Text":"that through in all the working out and we\u0027ll get the answer."},{"Start":"18:44.985 ","End":"18:47.930","Text":"That\u0027s the end of this lesson."}],"ID":10796}],"Thumbnail":null,"ID":9396}]

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