Newton's First and Third Laws
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The Static Friction
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The Kinetic Friction
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Newton's Second Law
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Additional Exercises
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[{"Name":"Newton\u0027s First and Third Laws","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Forces And The Laws Of Motion","Duration":"6m 38s","ChapterTopicVideoID":10271,"CourseChapterTopicPlaylistID":8960,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/10271.jpeg","UploadDate":"2017-09-04T11:32:45.0070000","DurationForVideoObject":"PT6M38S","Description":null,"MetaTitle":"Forces And The Laws Of Motion: Video + Workbook | Proprep","MetaDescription":"Dynamics - Newton\u0027s First and Third Laws. 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/dynamics/newton%27s-first-and-third-laws/vid10601","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.085","Text":"Hello. In this lesson,"},{"Start":"00:02.085 ","End":"00:06.750","Text":"we\u0027re going to be talking about forces and the laws of motion."},{"Start":"00:06.750 ","End":"00:11.130","Text":"Right now we\u0027re in our chapter of dynamics."},{"Start":"00:11.130 ","End":"00:12.555","Text":"Now, what is dynamics?"},{"Start":"00:12.555 ","End":"00:15.840","Text":"Dynamics is a part if classical mechanics which deals with"},{"Start":"00:15.840 ","End":"00:20.490","Text":"explaining how forces and torques affect a body\u0027s motion."},{"Start":"00:20.490 ","End":"00:23.775","Text":"Now, torques, we\u0027re going to speak about a little bit later."},{"Start":"00:23.775 ","End":"00:25.950","Text":"But in the meantime, let\u0027s speak about the basics,"},{"Start":"00:25.950 ","End":"00:30.165","Text":"which is a force. What is a force?"},{"Start":"00:30.165 ","End":"00:35.100","Text":"Now, intuitively, a force can be described as a push or a pull."},{"Start":"00:35.100 ","End":"00:37.740","Text":"But if we want to go a little bit more in-depth,"},{"Start":"00:37.740 ","End":"00:43.640","Text":"a force is any interaction with the body which can alter the motion of the body."},{"Start":"00:43.640 ","End":"00:46.265","Text":"Force is a vector quantity,"},{"Start":"00:46.265 ","End":"00:50.630","Text":"and that means that it has both size and direction."},{"Start":"00:50.630 ","End":"00:53.900","Text":"Force is denoted by the letter capital F, okay,"},{"Start":"00:53.900 ","End":"00:55.975","Text":"capital F for force,"},{"Start":"00:55.975 ","End":"01:00.590","Text":"and it is measured with the SI units in Newtons,"},{"Start":"01:00.590 ","End":"01:06.455","Text":"which is denoted by a capital N. When we\u0027re talking about forces,"},{"Start":"01:06.455 ","End":"01:12.320","Text":"it\u0027s customary to talk about contact forces and non-contact forces."},{"Start":"01:12.320 ","End":"01:15.485","Text":"Let\u0027s first speak about contact forces."},{"Start":"01:15.485 ","End":"01:21.020","Text":"Now, contact forces is anything that you have to physically touch in order"},{"Start":"01:21.020 ","End":"01:27.553","Text":"to exert a force and in order to cause some kind of change in the motion."},{"Start":"01:27.553 ","End":"01:31.550","Text":"A contact force examples are a horse pulling a carriage,"},{"Start":"01:31.550 ","End":"01:33.650","Text":"a hammer hitting a nail."},{"Start":"01:33.650 ","End":"01:36.065","Text":"Things that have to touch."},{"Start":"01:36.065 ","End":"01:43.040","Text":"Non-contact forces are forces that are exerted without physically touching one another."},{"Start":"01:43.040 ","End":"01:47.285","Text":"For instance, a magnet exerting its fields,"},{"Start":"01:47.285 ","End":"01:49.535","Text":"which then attracts the paperclip,"},{"Start":"01:49.535 ","End":"01:53.660","Text":"magnetic attraction, or the Earth\u0027s pulling the moon."},{"Start":"01:53.660 ","End":"01:55.340","Text":"That\u0027s for instance, gravity."},{"Start":"01:55.340 ","End":"02:01.700","Text":"Another example is the attractive force between an electron and its atom\u0027s core."},{"Start":"02:01.700 ","End":"02:04.370","Text":"An interesting little side point or"},{"Start":"02:04.370 ","End":"02:09.170","Text":"side fact is that the force responsible for contact forces,"},{"Start":"02:09.170 ","End":"02:13.610","Text":"so all the forces that there has to be to touch between the 2 objects."},{"Start":"02:13.610 ","End":"02:19.715","Text":"In fact, that force that\u0027s responsible for that is an electromagnetic force."},{"Start":"02:19.715 ","End":"02:22.505","Text":"It\u0027s in fact a non-contact force and"},{"Start":"02:22.505 ","End":"02:28.400","Text":"this electromagnetic force acts between the electrons on the surfaces of"},{"Start":"02:28.400 ","End":"02:32.930","Text":"the objects such that they\u0027re never actually touching and the force that you"},{"Start":"02:32.930 ","End":"02:37.909","Text":"feel is due to the electromagnetic force between those electrons."},{"Start":"02:37.909 ","End":"02:41.165","Text":"That is something that we\u0027re not going to talk about now,"},{"Start":"02:41.165 ","End":"02:45.260","Text":"you\u0027ll learn a little bit about that in your next physics course,"},{"Start":"02:45.260 ","End":"02:49.340","Text":"Physics 2, which is electricity and you\u0027ll"},{"Start":"02:49.340 ","End":"02:54.665","Text":"learn probably a lot more about that in your quantum mechanics courses."},{"Start":"02:54.665 ","End":"02:57.920","Text":"If multiple forces are acting on a body,"},{"Start":"02:57.920 ","End":"03:05.200","Text":"then the total force or the net force on the body is simply the sum of all of the forces."},{"Start":"03:05.200 ","End":"03:10.160","Text":"The net force, which is the total force acting on a body,"},{"Start":"03:10.160 ","End":"03:15.910","Text":"is the vector sum of all of the forces acting on the body."},{"Start":"03:15.910 ","End":"03:18.105","Text":"Let\u0027s go back. Now,"},{"Start":"03:18.105 ","End":"03:21.695","Text":"we said that a force is a vector quantity."},{"Start":"03:21.695 ","End":"03:24.230","Text":"Let\u0027s speak about how it\u0027s denoted."},{"Start":"03:24.230 ","End":"03:32.180","Text":"It\u0027s denoted by the letter capital F. Sometimes because force is a vector quantity,"},{"Start":"03:32.180 ","End":"03:37.880","Text":"so sometimes it\u0027s also denoted as capital F with an arrow on top."},{"Start":"03:37.880 ","End":"03:43.030","Text":"This arrow on top means that this is a vector quantity."},{"Start":"03:43.030 ","End":"03:47.165","Text":"Now when we were dealing with the net force,"},{"Start":"03:47.165 ","End":"03:50.975","Text":"it\u0027s the vector sum of all of the forces acting on the body."},{"Start":"03:50.975 ","End":"03:58.400","Text":"So a lot of the time we can say that our F total or net force is"},{"Start":"03:58.400 ","End":"04:05.750","Text":"going to be equal to the vector sum of all of the forces acting on the body."},{"Start":"04:05.750 ","End":"04:11.945","Text":"Now, this sign over here is the Greek letter Sigma and it represents sum,"},{"Start":"04:11.945 ","End":"04:14.930","Text":"you\u0027ll see this notation in many courses."},{"Start":"04:14.930 ","End":"04:18.125","Text":"It just means you add up everything."},{"Start":"04:18.125 ","End":"04:19.910","Text":"What does that mean?"},{"Start":"04:19.910 ","End":"04:23.765","Text":"That means that if we have 3 forces acting on the body,"},{"Start":"04:23.765 ","End":"04:32.210","Text":"so we\u0027ll have F_1 plus F_2 plus F_3."},{"Start":"04:32.210 ","End":"04:36.545","Text":"If there\u0027s more, then we\u0027ll just add on all of them."},{"Start":"04:36.545 ","End":"04:40.445","Text":"Now, when we\u0027re dealing with the vector sum of all of the forces,"},{"Start":"04:40.445 ","End":"04:45.545","Text":"that means that we have to add up every-component separately."},{"Start":"04:45.545 ","End":"04:49.640","Text":"So let\u0027s say that we\u0027re dealing with Cartesian coordinates."},{"Start":"04:49.640 ","End":"04:51.760","Text":"We\u0027re dealing with our x,"},{"Start":"04:51.760 ","End":"04:54.195","Text":"y, z coordinate system,"},{"Start":"04:54.195 ","End":"04:56.285","Text":"and just to make this a little bit quicker,"},{"Start":"04:56.285 ","End":"04:58.310","Text":"let\u0027s say that there\u0027s no z-component,"},{"Start":"04:58.310 ","End":"05:00.500","Text":"we just have x and y-components."},{"Start":"05:00.500 ","End":"05:02.930","Text":"That will mean, therefore,"},{"Start":"05:02.930 ","End":"05:10.385","Text":"that the sum of our F-components, the total,"},{"Start":"05:10.385 ","End":"05:18.004","Text":"is going to be equal to our x-component of F_1 plus"},{"Start":"05:18.004 ","End":"05:26.600","Text":"our x-component of F_2 plus our x-component of F_3,"},{"Start":"05:26.600 ","End":"05:29.540","Text":"and so on and so forth."},{"Start":"05:29.540 ","End":"05:33.680","Text":"Similarly, for the y-component,"},{"Start":"05:33.680 ","End":"05:38.735","Text":"our total force in the y-direction is going to be"},{"Start":"05:38.735 ","End":"05:45.620","Text":"our y-component of F_1 plus our y-component of F_2,"},{"Start":"05:45.620 ","End":"05:50.450","Text":"plus our y-component of F_3,"},{"Start":"05:50.450 ","End":"05:51.853","Text":"and so on and so forth."},{"Start":"05:51.853 ","End":"05:53.645","Text":"Of course, if we have a z-component,"},{"Start":"05:53.645 ","End":"05:55.990","Text":"it will look identical."},{"Start":"05:55.990 ","End":"06:00.290","Text":"In conclusion, we learned what a force is,"},{"Start":"06:00.290 ","End":"06:02.495","Text":"that it\u0027s denoted by the letter F,"},{"Start":"06:02.495 ","End":"06:04.250","Text":"usually with an arrow on top."},{"Start":"06:04.250 ","End":"06:06.020","Text":"That it\u0027s a vector quantity,"},{"Start":"06:06.020 ","End":"06:08.255","Text":"which means it has size and direction,"},{"Start":"06:08.255 ","End":"06:11.870","Text":"and that it is measured in Newtons,"},{"Start":"06:11.870 ","End":"06:14.220","Text":"okay, those are its SI units."},{"Start":"06:14.220 ","End":"06:16.760","Text":"The last important thing to remember is that"},{"Start":"06:16.760 ","End":"06:20.624","Text":"the net force or the sum of all of the forces acting on a body,"},{"Start":"06:20.624 ","End":"06:22.400","Text":"we just add them up."},{"Start":"06:22.400 ","End":"06:27.185","Text":"What\u0027s important is to break them up into the different components."},{"Start":"06:27.185 ","End":"06:30.470","Text":"Here, for instance, we have an x and y-components and add"},{"Start":"06:30.470 ","End":"06:35.060","Text":"up just the x-components and just the y-components."},{"Start":"06:35.060 ","End":"06:38.070","Text":"That\u0027s the end of this lesson."}],"ID":10601},{"Watched":false,"Name":"Common Mechanical Forces","Duration":"12m ","ChapterTopicVideoID":10272,"CourseChapterTopicPlaylistID":8960,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.935","Text":"Hello. In this lesson,"},{"Start":"00:01.935 ","End":"00:05.880","Text":"we\u0027re going to be speaking about common mechanical forces."},{"Start":"00:05.880 ","End":"00:09.855","Text":"The first force that we\u0027re going to be speaking about is"},{"Start":"00:09.855 ","End":"00:14.520","Text":"weight or due to the gravitational force."},{"Start":"00:14.520 ","End":"00:18.405","Text":"Weight is the force of an object due to gravity."},{"Start":"00:18.405 ","End":"00:24.160","Text":"Earth\u0027s gravitational force pulls the object into the Earth\u0027s center."},{"Start":"00:24.160 ","End":"00:28.530","Text":"Let\u0027s say that here we have Earth,"},{"Start":"00:28.530 ","End":"00:30.825","Text":"and here is Earth\u0027s center."},{"Start":"00:30.825 ","End":"00:33.830","Text":"So if we have a person, or an object,"},{"Start":"00:33.830 ","End":"00:36.590","Text":"or whatever standing over here on earth,"},{"Start":"00:36.590 ","End":"00:42.395","Text":"gravitational force is going to pull the person in to the center of the Earth."},{"Start":"00:42.395 ","End":"00:47.884","Text":"Similarly, if we have another person over here,"},{"Start":"00:47.884 ","End":"00:54.100","Text":"so gravity is again going to pull the person inwards to Earth\u0027s center."},{"Start":"00:54.100 ","End":"01:00.305","Text":"Now, what\u0027s important to know is that weight is a vector quantity,"},{"Start":"01:00.305 ","End":"01:05.820","Text":"so that means it has both size and magnitude."},{"Start":"01:06.080 ","End":"01:10.190","Text":"Now, the equation for weight, w,"},{"Start":"01:10.190 ","End":"01:18.290","Text":"is given by m times g. Let\u0027s talk about what each thing over here is."},{"Start":"01:18.290 ","End":"01:22.645","Text":"Our w is our weight,"},{"Start":"01:22.645 ","End":"01:26.820","Text":"and this is a vector quantity."},{"Start":"01:26.820 ","End":"01:29.900","Text":"So sometimes we can write it with an arrow on top."},{"Start":"01:29.900 ","End":"01:33.665","Text":"So let\u0027s write here, this is a vector."},{"Start":"01:33.665 ","End":"01:37.954","Text":"M is the mass of the object."},{"Start":"01:37.954 ","End":"01:42.299","Text":"If we\u0027re working in MKS,"},{"Start":"01:42.299 ","End":"01:44.510","Text":"it\u0027s given in kilograms."},{"Start":"01:44.510 ","End":"01:47.485","Text":"G over here is"},{"Start":"01:47.485 ","End":"01:51.245","Text":"the gravitational force which tells us"},{"Start":"01:51.245 ","End":"01:55.384","Text":"that our body is being pulled inwards to Earth\u0027s center,"},{"Start":"01:55.384 ","End":"01:59.190","Text":"and it\u0027s given in acceleration."},{"Start":"01:59.190 ","End":"02:04.020","Text":"So it\u0027s given in meters per second squared,"},{"Start":"02:04.020 ","End":"02:11.735","Text":"and gravitational force on Earth is usually given by 9.8 meters per second squared,"},{"Start":"02:11.735 ","End":"02:16.650","Text":"9.81, it\u0027s approximately equal to this."},{"Start":"02:17.920 ","End":"02:20.670","Text":"Because weight is a force,"},{"Start":"02:20.670 ","End":"02:23.725","Text":"it\u0027s measured in newtons."},{"Start":"02:23.725 ","End":"02:27.195","Text":"Newtons are its SI unit."},{"Start":"02:27.195 ","End":"02:35.120","Text":"Now, people in day-to-day life confuse a lot of the time their weight with their mass."},{"Start":"02:35.120 ","End":"02:38.860","Text":"A lot of the time people will weigh themselves on the scales,"},{"Start":"02:38.860 ","End":"02:42.715","Text":"and they\u0027ll say, \"I weigh 55 kilograms.\""},{"Start":"02:42.715 ","End":"02:46.674","Text":"No, kilograms is reserved for mass."},{"Start":"02:46.674 ","End":"02:48.530","Text":"When you are speaking about your weight,"},{"Start":"02:48.530 ","End":"02:55.000","Text":"you\u0027re saying that your mass times gravitational pull is equal to 55,"},{"Start":"02:55.000 ","End":"02:57.170","Text":"and then you\u0027re meant to say newtons,"},{"Start":"02:57.170 ","End":"03:01.135","Text":"but people use the term kilograms."},{"Start":"03:01.135 ","End":"03:03.450","Text":"That\u0027s more colloquial."},{"Start":"03:03.450 ","End":"03:06.650","Text":"But if we\u0027re speaking in scientific terms,"},{"Start":"03:06.650 ","End":"03:08.495","Text":"your weight is given in newtons,"},{"Start":"03:08.495 ","End":"03:11.525","Text":"and your mass is given in kilograms."},{"Start":"03:11.525 ","End":"03:14.425","Text":"Now, it\u0027s important to note that g,"},{"Start":"03:14.425 ","End":"03:16.280","Text":"our gravitational pull,"},{"Start":"03:16.280 ","End":"03:18.855","Text":"changes from place to place."},{"Start":"03:18.855 ","End":"03:23.270","Text":"If a person on Earth weighs 55 kilograms,"},{"Start":"03:23.270 ","End":"03:25.831","Text":"then on the Moon, g is less."},{"Start":"03:25.831 ","End":"03:30.230","Text":"Gravitational pull from the Moon is less than Earth\u0027s gravitational pull,"},{"Start":"03:30.230 ","End":"03:33.515","Text":"which means that their weight on the Moon will be"},{"Start":"03:33.515 ","End":"03:38.255","Text":"approximately 1/6 their weight on Earth."},{"Start":"03:38.255 ","End":"03:43.805","Text":"The next force that we\u0027re going to talk about is called the normal force."},{"Start":"03:43.805 ","End":"03:45.555","Text":"The normal force,"},{"Start":"03:45.555 ","End":"03:48.310","Text":"denoted by F_n,"},{"Start":"03:48.310 ","End":"03:50.720","Text":"or sometimes just N,"},{"Start":"03:50.720 ","End":"03:58.765","Text":"is the perpendicular component of a contact force exerted on an object by a surface."},{"Start":"03:58.765 ","End":"04:02.145","Text":"In other words, what is the normal force?"},{"Start":"04:02.145 ","End":"04:06.413","Text":"It\u0027s that if we have a body which is resting on some surface,"},{"Start":"04:06.413 ","End":"04:11.310","Text":"it\u0027s going to be a force equal and opposite to the weight."},{"Start":"04:11.660 ","End":"04:14.600","Text":"For instance, here in this diagram,"},{"Start":"04:14.600 ","End":"04:16.325","Text":"we have some box,"},{"Start":"04:16.325 ","End":"04:22.245","Text":"and it has a weight always pointing downwards towards Earth\u0027s center."},{"Start":"04:22.245 ","End":"04:26.450","Text":"So then, because our object is at rest,"},{"Start":"04:26.450 ","End":"04:30.170","Text":"that means that there must be a force equal and"},{"Start":"04:30.170 ","End":"04:34.340","Text":"opposite to w in order to make sure that our object is at rest."},{"Start":"04:34.340 ","End":"04:36.830","Text":"If there wasn\u0027t an equal and opposite force,"},{"Start":"04:36.830 ","End":"04:43.800","Text":"our object would just carry on traveling in the direction of our weight vector."},{"Start":"04:43.800 ","End":"04:49.115","Text":"The equal and opposite force is this normal force."},{"Start":"04:49.115 ","End":"04:55.040","Text":"In other words, it\u0027s a force applied by a surface on to the body."},{"Start":"04:55.040 ","End":"04:57.530","Text":"Now what\u0027s important to note is that it\u0027s always"},{"Start":"04:57.530 ","End":"05:01.895","Text":"a perpendicular component. What does that mean?"},{"Start":"05:01.895 ","End":"05:08.840","Text":"Let\u0027s say that we had a surface that was going in a slanted direction,"},{"Start":"05:08.840 ","End":"05:10.700","Text":"and then on top of that,"},{"Start":"05:10.700 ","End":"05:13.910","Text":"we had our block."},{"Start":"05:13.910 ","End":"05:18.590","Text":"Our weight would always point from the center of"},{"Start":"05:18.590 ","End":"05:23.885","Text":"the block or the center of mass of the block downwards towards Earth center,"},{"Start":"05:23.885 ","End":"05:29.550","Text":"and the normal force is the perpendicular component of the contact force."},{"Start":"05:29.550 ","End":"05:31.265","Text":"So in this case,"},{"Start":"05:31.265 ","End":"05:37.245","Text":"the normal force would be this N,"},{"Start":"05:37.245 ","End":"05:40.195","Text":"which is also equal to F_n."},{"Start":"05:40.195 ","End":"05:43.620","Text":"They can be represented in different ways."},{"Start":"05:44.080 ","End":"05:50.135","Text":"This comes from what we learned in the last lesson about Newton\u0027s first law."},{"Start":"05:50.135 ","End":"05:52.685","Text":"Because our object is at rest,"},{"Start":"05:52.685 ","End":"05:56.059","Text":"that means that it has a constant velocity,"},{"Start":"05:56.059 ","End":"05:58.055","Text":"and if its velocity is constant,"},{"Start":"05:58.055 ","End":"06:00.955","Text":"that means its acceleration is equal to 0."},{"Start":"06:00.955 ","End":"06:04.442","Text":"We saw in last lesson that if our acceleration is equal to 0,"},{"Start":"06:04.442 ","End":"06:07.310","Text":"that means the sum of the forces or"},{"Start":"06:07.310 ","End":"06:12.490","Text":"the net force on the object is also going to be equal to 0."},{"Start":"06:12.490 ","End":"06:16.760","Text":"So we can see that if we have our weight pointing downwards,"},{"Start":"06:16.760 ","End":"06:21.125","Text":"that means that our normal force is going to be pointing upwards"},{"Start":"06:21.125 ","End":"06:26.081","Text":"in the opposite direction to our weight but with the same magnitude,"},{"Start":"06:26.081 ","End":"06:29.585","Text":"which means that they will cancel each other out,"},{"Start":"06:29.585 ","End":"06:36.270","Text":"and that means that the net force on our block is going to be equal to 0."},{"Start":"06:36.850 ","End":"06:40.325","Text":"The normal force is, as we said,"},{"Start":"06:40.325 ","End":"06:43.670","Text":"dependent on what our weight is or"},{"Start":"06:43.670 ","End":"06:48.100","Text":"on the sum of all of the forces pointing in the opposite direction."},{"Start":"06:48.100 ","End":"06:51.725","Text":"In order to find what our normal force is equal to,"},{"Start":"06:51.725 ","End":"06:57.615","Text":"we have to split up our forces into their different components."},{"Start":"06:57.615 ","End":"07:01.340","Text":"For instance, in this first diagram over here,"},{"Start":"07:01.340 ","End":"07:05.240","Text":"we can see that our components are only in the y direction."},{"Start":"07:05.240 ","End":"07:08.180","Text":"We have no components going in the x direction."},{"Start":"07:08.180 ","End":"07:11.960","Text":"So then we can say that the sum of all of"},{"Start":"07:11.960 ","End":"07:16.565","Text":"our forces in the y direction is going to be equal to,"},{"Start":"07:16.565 ","End":"07:19.360","Text":"in the positive y direction we have N,"},{"Start":"07:19.360 ","End":"07:21.320","Text":"and in the negative y direction,"},{"Start":"07:21.320 ","End":"07:24.560","Text":"we have w, so negative w,"},{"Start":"07:24.560 ","End":"07:28.520","Text":"and we know that this has to be equal to 0 because we know that"},{"Start":"07:28.520 ","End":"07:33.750","Text":"our acceleration in the y direction is equal to 0 because our object is at rest,"},{"Start":"07:33.750 ","End":"07:38.882","Text":"which means that the sum of the forces in the y direction must also equal to 0,"},{"Start":"07:38.882 ","End":"07:42.500","Text":"and then we can simply rearrange this and get that"},{"Start":"07:42.500 ","End":"07:48.340","Text":"our normal force is equal to our weight force."},{"Start":"07:48.730 ","End":"07:51.305","Text":"In our second diagram,"},{"Start":"07:51.305 ","End":"07:56.150","Text":"what we\u0027d have to do is we\u0027d have to recognize that our normal force"},{"Start":"07:56.150 ","End":"08:02.555","Text":"has an x component like so and a y component like so."},{"Start":"08:02.555 ","End":"08:06.560","Text":"So we\u0027d have to split up our normal force into the x and"},{"Start":"08:06.560 ","End":"08:12.847","Text":"y components by using our cosine and sine of this angle,"},{"Start":"08:12.847 ","End":"08:18.095","Text":"and then we would just work out the same equation over here,"},{"Start":"08:18.095 ","End":"08:23.150","Text":"except then our N would be N in the y direction sine"},{"Start":"08:23.150 ","End":"08:29.505","Text":"Theta minus w. Now let\u0027s give another example."},{"Start":"08:29.505 ","End":"08:32.825","Text":"Let\u0027s say that we have our weight pulling downwards,"},{"Start":"08:32.825 ","End":"08:39.355","Text":"and someone is sitting on top or pushing our block downwards as well with some force,"},{"Start":"08:39.355 ","End":"08:44.075","Text":"F. Now let\u0027s see what our normal force is equal to."},{"Start":"08:44.075 ","End":"08:49.760","Text":"We know that the sum of all of our forces in the y direction is still going to be"},{"Start":"08:49.760 ","End":"08:55.040","Text":"equal to 0 because there is still no acceleration in either the x or the y direction,"},{"Start":"08:55.040 ","End":"09:00.295","Text":"but specifically, all of our forces only have y components."},{"Start":"09:00.295 ","End":"09:07.664","Text":"Again, the sum of all of our forces in the y direction is equal to 0."},{"Start":"09:07.664 ","End":"09:09.880","Text":"That is equal to,"},{"Start":"09:09.880 ","End":"09:12.965","Text":"so we have in the positive y direction,"},{"Start":"09:12.965 ","End":"09:15.769","Text":"we have our N, our arrows pointing upwards,"},{"Start":"09:15.769 ","End":"09:17.689","Text":"and then in the negative y direction,"},{"Start":"09:17.689 ","End":"09:21.835","Text":"we have our F and our w, so negative F,"},{"Start":"09:21.835 ","End":"09:26.810","Text":"negative w. All of this is equal to 0 because"},{"Start":"09:26.810 ","End":"09:33.125","Text":"the net force on our object is equal to 0 because our acceleration is equal to 0."},{"Start":"09:33.125 ","End":"09:35.165","Text":"So then we can rearrange,"},{"Start":"09:35.165 ","End":"09:39.650","Text":"and we\u0027ll get that our normal force is equal to F"},{"Start":"09:39.650 ","End":"09:45.925","Text":"plus w. The normal force is always a pushing force."},{"Start":"09:45.925 ","End":"09:48.430","Text":"How do we measure this normal force?"},{"Start":"09:48.430 ","End":"09:51.025","Text":"Now, usually what we do is,"},{"Start":"09:51.025 ","End":"09:54.355","Text":"if we want to measure our weight,"},{"Start":"09:54.355 ","End":"09:57.220","Text":"what we do is we stand on a scale."},{"Start":"09:57.220 ","End":"10:02.612","Text":"Now, what we\u0027re actually seeing on the scale isn\u0027t technically our weight."},{"Start":"10:02.612 ","End":"10:06.010","Text":"We\u0027re actually measuring our normal force."},{"Start":"10:06.010 ","End":"10:09.850","Text":"We\u0027re seeing what force needs to be applied once we\u0027re"},{"Start":"10:09.850 ","End":"10:13.640","Text":"standing on the scale in order to balance out our weight,"},{"Start":"10:13.640 ","End":"10:16.390","Text":"and then we\u0027re being shown the normal force,"},{"Start":"10:16.390 ","End":"10:20.740","Text":"which has an equal and opposite force to our weight."},{"Start":"10:20.740 ","End":"10:25.295","Text":"The next force that we\u0027re going to be speaking about is tension,"},{"Start":"10:25.295 ","End":"10:30.244","Text":"T. Tension is the force exerted by a string,"},{"Start":"10:30.244 ","End":"10:34.625","Text":"or a rod, or something similar to pull an object."},{"Start":"10:34.625 ","End":"10:38.435","Text":"Tension is always going to be a pulling force."},{"Start":"10:38.435 ","End":"10:46.055","Text":"Now, the direction of the tension is always going to be in the direction of the string,"},{"Start":"10:46.055 ","End":"10:47.554","Text":"in a puling direction,"},{"Start":"10:47.554 ","End":"10:51.445","Text":"so away from the object that we\u0027re trying to pull."},{"Start":"10:51.445 ","End":"10:57.700","Text":"For instance here, we can see that the direction of our string is changing,"},{"Start":"10:57.700 ","End":"11:02.090","Text":"but still, the force of tension is going to carry"},{"Start":"11:02.090 ","End":"11:07.080","Text":"on and be in the direction of the string."},{"Start":"11:07.150 ","End":"11:14.070","Text":"Our arrow is always going to be pointing away from the object which we are pulling."},{"Start":"11:14.500 ","End":"11:19.580","Text":"In some questions, they\u0027ll specify that the string"},{"Start":"11:19.580 ","End":"11:24.520","Text":"that there\u0027s tension on is an idealized string."},{"Start":"11:24.520 ","End":"11:30.860","Text":"What does that mean? That means that the string has a length, but it\u0027s massless."},{"Start":"11:30.860 ","End":"11:36.260","Text":"Now, when we\u0027re dealing with a massless string or an idealized string,"},{"Start":"11:36.260 ","End":"11:42.260","Text":"that means that the magnitude of the tension is going to be constant."},{"Start":"11:42.260 ","End":"11:46.865","Text":"The magnitude of the tension is not going to be changing."},{"Start":"11:46.865 ","End":"11:50.585","Text":"This right now might not sound like much, however,"},{"Start":"11:50.585 ","End":"11:53.885","Text":"you\u0027ll see that in future questions and exercises,"},{"Start":"11:53.885 ","End":"11:57.995","Text":"this idealized string will come in quite handy."},{"Start":"11:57.995 ","End":"12:01.110","Text":"That\u0027s the end of this lesson."}],"ID":10602},{"Watched":false,"Name":"Newton\u0027s First Law","Duration":"6m 50s","ChapterTopicVideoID":10273,"CourseChapterTopicPlaylistID":8960,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"Hello. In this lesson,"},{"Start":"00:02.145 ","End":"00:05.595","Text":"we\u0027re going to be speaking about Newton\u0027s First Law."},{"Start":"00:05.595 ","End":"00:09.390","Text":"Now, there are 2 different definitions,"},{"Start":"00:09.390 ","End":"00:11.475","Text":"which actually both mean the same thing in the end,"},{"Start":"00:11.475 ","End":"00:12.945","Text":"we\u0027ll speak about it,"},{"Start":"00:12.945 ","End":"00:18.165","Text":"but 2 different ways used to define Newton\u0027s First Law."},{"Start":"00:18.165 ","End":"00:21.060","Text":"So let\u0027s take a look at definition A."},{"Start":"00:21.060 ","End":"00:22.739","Text":"It says that,"},{"Start":"00:22.739 ","End":"00:24.510","Text":"according to Newton\u0027s First Law,"},{"Start":"00:24.510 ","End":"00:30.150","Text":"an object will continue its motion at a constant velocity and in a straight line,"},{"Start":"00:30.150 ","End":"00:35.190","Text":"as long as the net force on the object is equal to 0."},{"Start":"00:35.190 ","End":"00:40.670","Text":"That means that as long as the sum of all of the forces on the object is equal to 0,"},{"Start":"00:40.670 ","End":"00:45.905","Text":"the object will continue its motion at a constant velocity and in a straight line."},{"Start":"00:45.905 ","End":"00:48.215","Text":"An object at rest,"},{"Start":"00:48.215 ","End":"00:50.195","Text":"which means that it has 0 velocity,"},{"Start":"00:50.195 ","End":"00:53.300","Text":"is a special case of motion with constant velocity."},{"Start":"00:53.300 ","End":"00:58.598","Text":"An object at rest will have constant 0 velocity,"},{"Start":"00:58.598 ","End":"01:01.475","Text":"so it obeys Newton\u0027s First Law."},{"Start":"01:01.475 ","End":"01:09.199","Text":"Now, the second definition is that when the net force on an object is 0,"},{"Start":"01:09.199 ","End":"01:11.600","Text":"that means the object is at equilibrium,"},{"Start":"01:11.600 ","End":"01:17.070","Text":"so then the object\u0027s acceleration is also equal to 0."},{"Start":"01:17.480 ","End":"01:24.775","Text":"An object is said to be at equilibrium when the net force on it is equal to 0."},{"Start":"01:24.775 ","End":"01:32.210","Text":"Now, how can we show that our definition B is equal to our definition A."},{"Start":"01:32.210 ","End":"01:36.170","Text":"Both of these sentences mean the same thing."},{"Start":"01:36.170 ","End":"01:41.040","Text":"Let\u0027s see. Let\u0027s take an example."},{"Start":"01:41.040 ","End":"01:46.500","Text":"We have our velocity_1 which is dependent on time,"},{"Start":"01:46.500 ","End":"01:48.420","Text":"and it\u0027s equal to,"},{"Start":"01:48.420 ","End":"01:54.030","Text":"let\u0027s say, 3t^2 in some direction."},{"Start":"01:54.030 ","End":"02:00.980","Text":"Now, we know from a previous lesson that our acceleration is also dependent on time,"},{"Start":"02:00.980 ","End":"02:07.016","Text":"and it\u0027s the time derivative of our velocity,"},{"Start":"02:07.016 ","End":"02:09.130","Text":"so dv by dt."},{"Start":"02:09.130 ","End":"02:11.235","Text":"Let\u0027s see what that is."},{"Start":"02:11.235 ","End":"02:17.410","Text":"Here, it\u0027s going to be equal to be 6t in the r direction."},{"Start":"02:18.560 ","End":"02:24.370","Text":"This isn\u0027t strictly correct in some r\u0027 direction."},{"Start":"02:24.680 ","End":"02:29.945","Text":"We can see that our acceleration isn\u0027t equal to 0,"},{"Start":"02:29.945 ","End":"02:31.295","Text":"and it isn\u0027t constant."},{"Start":"02:31.295 ","End":"02:33.755","Text":"We have a changing acceleration."},{"Start":"02:33.755 ","End":"02:38.465","Text":"Alternatively, if we take,"},{"Start":"02:38.465 ","End":"02:41.645","Text":"let\u0027s say, velocity 2,"},{"Start":"02:41.645 ","End":"02:43.460","Text":"which is also time-dependent,"},{"Start":"02:43.460 ","End":"02:46.520","Text":"but we have some constant velocity,"},{"Start":"02:46.520 ","End":"02:48.725","Text":"which is equal to, let\u0027s say,"},{"Start":"02:48.725 ","End":"02:52.080","Text":"3 in some direction."},{"Start":"02:52.080 ","End":"02:54.080","Text":"We can see that we have constant velocity."},{"Start":"02:54.080 ","End":"02:59.155","Text":"It doesn\u0027t matter if we\u0027re at t is equal to 0 or at t is equal to 1 trillion,"},{"Start":"02:59.155 ","End":"03:04.740","Text":"our velocity is always going to be in the same direction and at the same pace."},{"Start":"03:04.880 ","End":"03:09.320","Text":"Now, let\u0027s see what our acceleration is,"},{"Start":"03:09.320 ","End":"03:14.465","Text":"so dv_2 by dt."},{"Start":"03:14.465 ","End":"03:21.043","Text":"We have no T variable over here,"},{"Start":"03:21.043 ","End":"03:26.370","Text":"so the derivative of this with respect to t is going to be equal to 0."},{"Start":"03:26.390 ","End":"03:31.805","Text":"Here we can see that when we have constant velocity,"},{"Start":"03:31.805 ","End":"03:35.381","Text":"we have 0 acceleration,"},{"Start":"03:35.381 ","End":"03:37.340","Text":"or when we have 0 acceleration,"},{"Start":"03:37.340 ","End":"03:39.610","Text":"we have constant velocity."},{"Start":"03:39.610 ","End":"03:41.490","Text":"In definition 2,"},{"Start":"03:41.490 ","End":"03:45.230","Text":"we\u0027re speaking about that the net force on"},{"Start":"03:45.230 ","End":"03:49.790","Text":"an object is 0 when the object\u0027s acceleration is equal to 0,"},{"Start":"03:49.790 ","End":"03:56.885","Text":"which means that the net force on an object is equal to 0 when we have constant velocity,"},{"Start":"03:56.885 ","End":"04:00.733","Text":"which is exactly what our definition A is saying,"},{"Start":"04:00.733 ","End":"04:07.190","Text":"that as long as we have constant velocity and in a straight line,"},{"Start":"04:07.190 ","End":"04:09.455","Text":"it just means the direction isn\u0027t changing,"},{"Start":"04:09.455 ","End":"04:14.120","Text":"which is part and parcel of our velocity being constant."},{"Start":"04:14.120 ","End":"04:15.320","Text":"It\u0027s a vector quantity,"},{"Start":"04:15.320 ","End":"04:19.235","Text":"which means also its size and direction remains the same."},{"Start":"04:19.235 ","End":"04:21.635","Text":"We have constant velocity,"},{"Start":"04:21.635 ","End":"04:27.560","Text":"which means that the net force on the object is equal to 0."},{"Start":"04:28.310 ","End":"04:36.300","Text":"Now we can see that both of these definitions are equivalent to one another."},{"Start":"04:37.040 ","End":"04:43.630","Text":"Now, the next important thing to note are these 2 equations over here."},{"Start":"04:43.700 ","End":"04:51.810","Text":"This is a mathematical representation of definition B."},{"Start":"04:52.220 ","End":"04:57.083","Text":"This means this just written out mathematically,"},{"Start":"04:57.083 ","End":"04:59.190","Text":"and the sum of all of the forces,"},{"Start":"04:59.190 ","End":"05:00.415","Text":"so the net force,"},{"Start":"05:00.415 ","End":"05:02.520","Text":"is equal to 0,"},{"Start":"05:02.520 ","End":"05:08.050","Text":"if and only if the acceleration is equal to 0."},{"Start":"05:08.050 ","End":"05:12.150","Text":"If and only if means that this statement goes both ways."},{"Start":"05:12.150 ","End":"05:16.675","Text":"So we can say that if the net force on the object is equal to 0,"},{"Start":"05:16.675 ","End":"05:19.080","Text":"then the acceleration is equal to 0."},{"Start":"05:19.080 ","End":"05:21.120","Text":"We can also go the other way,"},{"Start":"05:21.120 ","End":"05:24.265","Text":"and we can say that if the acceleration is equal to 0,"},{"Start":"05:24.265 ","End":"05:27.415","Text":"then the net force is equal to 0."},{"Start":"05:27.415 ","End":"05:31.900","Text":"If and only if means that the statement can work out both ways,"},{"Start":"05:31.900 ","End":"05:36.020","Text":"and that\u0027s why it\u0027s denoted by this double-headed arrow."},{"Start":"05:36.020 ","End":"05:39.131","Text":"Now, similarly,"},{"Start":"05:39.131 ","End":"05:41.569","Text":"coming from this main statement,"},{"Start":"05:41.569 ","End":"05:45.275","Text":"is that if the sum of all of our forces in the y direction,"},{"Start":"05:45.275 ","End":"05:49.880","Text":"or the net force on the object in the y direction,"},{"Start":"05:49.880 ","End":"05:55.325","Text":"is equal to 0, then the acceleration in the y direction is equal to 0."},{"Start":"05:55.325 ","End":"05:58.835","Text":"We can say the same with the x component."},{"Start":"05:58.835 ","End":"06:01.805","Text":"If the acceleration in the x direction is equal to 0,"},{"Start":"06:01.805 ","End":"06:07.680","Text":"then the sum of all of the forces in the x direction will also equal 0."},{"Start":"06:09.140 ","End":"06:13.910","Text":"That\u0027s the end of this lesson where we spoke about Newton\u0027s First Law."},{"Start":"06:13.910 ","End":"06:16.610","Text":"It\u0027s important to remember the definitions."},{"Start":"06:16.610 ","End":"06:21.125","Text":"A quick way is to remember just this mathematical representation,"},{"Start":"06:21.125 ","End":"06:23.435","Text":"which should bring you,"},{"Start":"06:23.435 ","End":"06:25.055","Text":"if you understand it,"},{"Start":"06:25.055 ","End":"06:30.005","Text":"to understand these 2 definitions or to derive these 2 definitions."},{"Start":"06:30.005 ","End":"06:35.360","Text":"Remember that we can also split up the net forces into components in the x,"},{"Start":"06:35.360 ","End":"06:37.438","Text":"y, z, whichever direction,"},{"Start":"06:37.438 ","End":"06:41.690","Text":"and that means that if the net forces are equal to 0 then"},{"Start":"06:41.690 ","End":"06:47.240","Text":"the acceleration in the corresponding direction is also equal to 0."},{"Start":"06:47.240 ","End":"06:50.220","Text":"That\u0027s the end of this lesson."}],"ID":10603},{"Watched":false,"Name":"Exercise","Duration":"13m 47s","ChapterTopicVideoID":10274,"CourseChapterTopicPlaylistID":8960,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.370","Text":"Hello. In this question,"},{"Start":"00:02.370 ","End":"00:09.180","Text":"we\u0027re given a ball of mass m. That\u0027s this ball over here."},{"Start":"00:09.180 ","End":"00:11.490","Text":"We\u0027re given a ball of mass m,"},{"Start":"00:11.490 ","End":"00:13.590","Text":"and we\u0027re told that it\u0027s placed in a box."},{"Start":"00:13.590 ","End":"00:18.180","Text":"We\u0027re told that the box is resting at an angle of Alpha,"},{"Start":"00:18.180 ","End":"00:22.755","Text":"and we\u0027re also told that the sides of the box are perpendicular to one another."},{"Start":"00:22.755 ","End":"00:25.215","Text":"That\u0027s an angle of 90 degrees."},{"Start":"00:25.215 ","End":"00:29.610","Text":"We\u0027re being asked what is the force exerted by each face of"},{"Start":"00:29.610 ","End":"00:34.335","Text":"the box onto the ball, here and here."},{"Start":"00:34.335 ","End":"00:36.200","Text":"Let\u0027s see how we\u0027ll solve this."},{"Start":"00:36.200 ","End":"00:38.869","Text":"Let\u0027s begin by drawing a free body diagram."},{"Start":"00:38.869 ","End":"00:42.110","Text":"A free body diagram will go like this."},{"Start":"00:42.110 ","End":"00:49.880","Text":"Here we have the center of the ball with a downwards arrow of mg pointing downwards."},{"Start":"00:49.880 ","End":"00:54.560","Text":"Then we have a normal force from this phase of the box and"},{"Start":"00:54.560 ","End":"00:59.990","Text":"a normal force from this phase. Now what do I do?"},{"Start":"00:59.990 ","End":"01:05.750","Text":"I\u0027m going to treat the whole ball as if it\u0027s just 1 single point."},{"Start":"01:05.750 ","End":"01:11.550","Text":"I\u0027m going to draw all of the forces coming out of 1 point."},{"Start":"01:12.070 ","End":"01:16.780","Text":"This normal over here,"},{"Start":"01:16.780 ","End":"01:19.640","Text":"I\u0027ll draw it here."},{"Start":"01:19.640 ","End":"01:22.070","Text":"We\u0027re maintaining the same direction,"},{"Start":"01:22.070 ","End":"01:27.485","Text":"and this Normal over here in this direction."},{"Start":"01:27.485 ","End":"01:32.870","Text":"Now, let\u0027s label this Normal_1 and this Normal_2,"},{"Start":"01:32.870 ","End":"01:36.155","Text":"in which case, this is also Normal_1, and this is Number 2."},{"Start":"01:36.155 ","End":"01:41.690","Text":"Then we have, as per usual, mg going downwards."},{"Start":"01:41.690 ","End":"01:44.719","Text":"Now, important to note about vectors."},{"Start":"01:44.719 ","End":"01:50.495","Text":"Let\u0027s take Normal_2. Normal_2,"},{"Start":"01:50.495 ","End":"01:52.655","Text":"I can draw anywhere on the page."},{"Start":"01:52.655 ","End":"01:54.050","Text":"It doesn\u0027t matter where,"},{"Start":"01:54.050 ","End":"01:58.825","Text":"as long as it is the same length and pointing in the same direction,"},{"Start":"01:58.825 ","End":"02:01.580","Text":"then it represents the same vector."},{"Start":"02:01.580 ","End":"02:03.965","Text":"For instance, Normal_2,"},{"Start":"02:03.965 ","End":"02:05.585","Text":"I can draw over here,"},{"Start":"02:05.585 ","End":"02:07.235","Text":"I can draw over here,"},{"Start":"02:07.235 ","End":"02:09.800","Text":"I can draw over here, it doesn\u0027t matter."},{"Start":"02:09.800 ","End":"02:11.015","Text":"It\u0027s the same vector."},{"Start":"02:11.015 ","End":"02:14.000","Text":"However, if I was to twizzle it around, for instance,"},{"Start":"02:14.000 ","End":"02:15.650","Text":"and draw it here,"},{"Start":"02:15.650 ","End":"02:17.240","Text":"this is a different vector."},{"Start":"02:17.240 ","End":"02:19.795","Text":"This is no longer Normal_2."},{"Start":"02:19.795 ","End":"02:22.355","Text":"So that\u0027s very important to note."},{"Start":"02:22.355 ","End":"02:29.705","Text":"This rule allows us to represent the free body diagram that we have here like this."},{"Start":"02:29.705 ","End":"02:31.085","Text":"So it\u0027s very important."},{"Start":"02:31.085 ","End":"02:33.620","Text":"Now notice what is happening here."},{"Start":"02:33.620 ","End":"02:36.965","Text":"Between this Normal and the side of the page,"},{"Start":"02:36.965 ","End":"02:38.315","Text":"there\u0027s 90 degrees,"},{"Start":"02:38.315 ","End":"02:40.978","Text":"and between this Normal and the side of the page,"},{"Start":"02:40.978 ","End":"02:42.463","Text":"this is 90 degrees,"},{"Start":"02:42.463 ","End":"02:45.390","Text":"and between these 2 faces of the box,"},{"Start":"02:45.390 ","End":"02:47.460","Text":"there\u0027s also 90 degrees."},{"Start":"02:47.460 ","End":"02:51.665","Text":"Because there\u0027s 90 degrees between everything,"},{"Start":"02:51.665 ","End":"03:00.760","Text":"it means that there must also be 90 degrees between Normal_1 and Normal_2."},{"Start":"03:00.760 ","End":"03:05.660","Text":"Now, because I have to split this up into its various components,"},{"Start":"03:05.660 ","End":"03:12.800","Text":"I\u0027m going to choose axes that are easy for me to work with."},{"Start":"03:12.800 ","End":"03:15.890","Text":"The easiest thing to do right now is,"},{"Start":"03:15.890 ","End":"03:19.555","Text":"because there\u0027s 90 degrees between N_1 and N_2,"},{"Start":"03:19.555 ","End":"03:26.399","Text":"it would be easier to carry on the line with N_1 in this direction,"},{"Start":"03:26.399 ","End":"03:28.805","Text":"this being my x-axis,"},{"Start":"03:28.805 ","End":"03:31.745","Text":"and in the direction of N_2,"},{"Start":"03:31.745 ","End":"03:34.745","Text":"this being my y-axis."},{"Start":"03:34.745 ","End":"03:37.374","Text":"This is the easiest."},{"Start":"03:37.374 ","End":"03:44.140","Text":"Then all I have to do split up mg into its separate components."},{"Start":"03:44.140 ","End":"03:53.780","Text":"Now, our next task is to find where our Alpha angle is on this diagram."},{"Start":"03:53.780 ","End":"03:55.610","Text":"Now, as you recall,"},{"Start":"03:55.610 ","End":"03:59.075","Text":"as you can see here, the angle Alpha is,"},{"Start":"03:59.075 ","End":"04:01.565","Text":"in regards to this,"},{"Start":"04:01.565 ","End":"04:04.645","Text":"the horizontal line over here."},{"Start":"04:04.645 ","End":"04:10.020","Text":"So if we\u0027re to replicate this in the diagram, switch color,"},{"Start":"04:10.020 ","End":"04:16.325","Text":"this line here is the same as this line here,"},{"Start":"04:16.325 ","End":"04:20.705","Text":"which means that this angle will be Alpha."},{"Start":"04:20.705 ","End":"04:23.205","Text":"Now, it\u0027s important to concentrate here."},{"Start":"04:23.205 ","End":"04:29.230","Text":"Here, there\u0027s an angle of 90 degrees,"},{"Start":"04:29.360 ","End":"04:38.870","Text":"meaning that this angle here is 90 minus Alpha, and here,"},{"Start":"04:38.870 ","End":"04:42.650","Text":"I also have 90 degrees,"},{"Start":"04:42.650 ","End":"04:45.320","Text":"which means that this angle"},{"Start":"04:45.320 ","End":"04:54.170","Text":"here is also 90 minus 90 minus Alpha,"},{"Start":"04:54.170 ","End":"05:04.410","Text":"meaning that this angle over here is Alpha as well."},{"Start":"05:04.550 ","End":"05:07.119","Text":"To make it a bit clearer,"},{"Start":"05:07.119 ","End":"05:10.840","Text":"this angle here is Alpha as well."},{"Start":"05:10.840 ","End":"05:15.055","Text":"Now, this problem will be repeated over and over again."},{"Start":"05:15.055 ","End":"05:18.999","Text":"If you have an angle of 90 degrees,"},{"Start":"05:18.999 ","End":"05:24.475","Text":"and then another angle again of 90 degrees,"},{"Start":"05:24.475 ","End":"05:29.755","Text":"so you have 90 degrees here and 90 degrees here,"},{"Start":"05:29.755 ","End":"05:33.430","Text":"then the angle that I will have"},{"Start":"05:33.430 ","End":"05:43.010","Text":"here will be the same angle that I have here."},{"Start":"05:43.560 ","End":"05:49.510","Text":"This is because this is 90 degrees."},{"Start":"05:49.510 ","End":"05:54.055","Text":"1 sec, let\u0027s just change colors."},{"Start":"05:54.055 ","End":"06:04.340","Text":"Because this angle is 90 minus Alpha,"},{"Start":"06:04.340 ","End":"06:12.790","Text":"then this angle will be 90 minus 90 minus Alpha,"},{"Start":"06:12.790 ","End":"06:15.510","Text":"making it also Alpha,"},{"Start":"06:15.510 ","End":"06:18.025","Text":"so these 2 angles are Alpha."},{"Start":"06:18.025 ","End":"06:25.165","Text":"Now let\u0027s draw the components for mg. We have this component,"},{"Start":"06:25.165 ","End":"06:28.175","Text":"and we have this component."},{"Start":"06:28.175 ","End":"06:32.795","Text":"Now, this component, because obviously this is Alpha,"},{"Start":"06:32.795 ","End":"06:35.375","Text":"and this is the adjacent angle,"},{"Start":"06:35.375 ","End":"06:40.470","Text":"so it will be mg cosine Alpha."},{"Start":"06:40.470 ","End":"06:45.065","Text":"Just because I\u0027m drawing it up top,"},{"Start":"06:45.065 ","End":"06:47.295","Text":"I could have also drawn it over here."},{"Start":"06:47.295 ","End":"06:49.960","Text":"Because it\u0027s the opposite angle,"},{"Start":"06:49.960 ","End":"06:55.470","Text":"it\u0027s mg sine of Alpha."},{"Start":"06:55.470 ","End":"07:01.535","Text":"Now, all that\u0027s left to say is that the sum of all the forces is equal to 0,"},{"Start":"07:01.535 ","End":"07:06.395","Text":"meaning that all the forces pointing up are equal to all the forces pointing down,"},{"Start":"07:06.395 ","End":"07:11.100","Text":"and all the forces pointing right are equal to all the forces pointing left."},{"Start":"07:11.500 ","End":"07:14.585","Text":"Now, we can write some of our equations."},{"Start":"07:14.585 ","End":"07:23.705","Text":"So we can say that N_1 is equal to mg sine"},{"Start":"07:23.705 ","End":"07:33.895","Text":"of Alpha and that N_2 is equal to mg cosine of Alpha."},{"Start":"07:33.895 ","End":"07:37.970","Text":"Now, this is where we\u0027ve actually finished the question."},{"Start":"07:37.970 ","End":"07:42.950","Text":"But let\u0027s just go over something to show you how it works."},{"Start":"07:42.950 ","End":"07:46.955","Text":"If I was asked, now that I have N_1 and N_2,"},{"Start":"07:46.955 ","End":"07:52.895","Text":"what the size of these forces are, basically joint,"},{"Start":"07:52.895 ","End":"07:54.217","Text":"what their forces are,"},{"Start":"07:54.217 ","End":"08:01.430","Text":"then it would be the square root of N_1^2 plus N_2^2,"},{"Start":"08:01.430 ","End":"08:05.045","Text":"which would equal the square root"},{"Start":"08:05.045 ","End":"08:10.115","Text":"of m^2g^2sin^2Alpha"},{"Start":"08:10.115 ","End":"08:15.985","Text":"plus m^2g^2cos^2Alpha."},{"Start":"08:15.985 ","End":"08:17.925","Text":"Then as we know,"},{"Start":"08:17.925 ","End":"08:21.200","Text":"we can take the m^2g^2 outside of"},{"Start":"08:21.200 ","End":"08:25.700","Text":"brackets and put sin^2 plus cos^2 in which is equal to 1,"},{"Start":"08:25.700 ","End":"08:31.520","Text":"so then I would know that this would be the square root of m^2g^2,"},{"Start":"08:31.520 ","End":"08:34.404","Text":"which is equal to just mg."},{"Start":"08:34.404 ","End":"08:35.960","Text":"It will bring us back to mg."},{"Start":"08:35.960 ","End":"08:40.247","Text":"If there\u0027s a force of mg pointing down,"},{"Start":"08:40.247 ","End":"08:45.440","Text":"then the 2 Normals working together have to"},{"Start":"08:45.440 ","End":"08:51.680","Text":"counter this mg and have an equal and opposite upward force."},{"Start":"08:51.680 ","End":"08:56.510","Text":"Now, let\u0027s go over some intuition to make it a bit easier."},{"Start":"08:56.510 ","End":"09:02.780","Text":"Why does it make sense that we have mg sine Alpha and also mg cosine Alpha."},{"Start":"09:02.780 ","End":"09:05.815","Text":"Let\u0027s start looking at the mgs."},{"Start":"09:05.815 ","End":"09:08.795","Text":"The larger that mg is,"},{"Start":"09:08.795 ","End":"09:10.025","Text":"the heavier the mass,"},{"Start":"09:10.025 ","End":"09:12.992","Text":"the heavier the ball is,"},{"Start":"09:12.992 ","End":"09:19.130","Text":"the more force it will be applying to the sides of the box,"},{"Start":"09:19.130 ","End":"09:23.870","Text":"in which case, the Normal force will have to be larger."},{"Start":"09:23.870 ","End":"09:26.900","Text":"That\u0027s why the larger mg is,"},{"Start":"09:26.900 ","End":"09:30.755","Text":"the larger N_1 is and same with N_2."},{"Start":"09:30.755 ","End":"09:33.050","Text":"But that\u0027s pretty obvious."},{"Start":"09:33.050 ","End":"09:39.115","Text":"Now, let\u0027s look at why sine and why cosine. What is this?"},{"Start":"09:39.115 ","End":"09:41.705","Text":"We\u0027ve said if this is N_1,"},{"Start":"09:41.705 ","End":"09:44.165","Text":"so let\u0027s call the side of the box here, 1,"},{"Start":"09:44.165 ","End":"09:47.840","Text":"and this is N_2, so the side of the box, we\u0027ll call 2."},{"Start":"09:47.840 ","End":"09:53.438","Text":"Let\u0027s assume for a second that this angle Alpha is equal to 0,"},{"Start":"09:53.438 ","End":"10:00.840","Text":"so this whole diagram would actually look like this at a right angle."},{"Start":"10:00.890 ","End":"10:06.680","Text":"We\u0027ve already labeled this as side Number 2 and this as side Number 1."},{"Start":"10:06.680 ","End":"10:10.625","Text":"Now, if we place the ball into this box now,"},{"Start":"10:10.625 ","End":"10:12.350","Text":"it\u0027s very obvious that,"},{"Start":"10:12.350 ","End":"10:20.370","Text":"ball Number 2, all of its weight is being applied to side Number 2 of the box."},{"Start":"10:21.290 ","End":"10:23.985","Text":"If at angle 0,"},{"Start":"10:23.985 ","End":"10:27.835","Text":"Alpha equals 0, the maximum weight,"},{"Start":"10:27.835 ","End":"10:33.705","Text":"all of the weight of the ball is being applied onto side Number 2,"},{"Start":"10:33.705 ","End":"10:37.095","Text":"then we know that"},{"Start":"10:37.095 ","End":"10:46.240","Text":"N_2 will"},{"Start":"10:46.240 ","End":"10:47.935","Text":"have its maximum value"},{"Start":"10:47.935 ","End":"10:52.555","Text":"because we know that cosine of 0 is equal to 1,"},{"Start":"10:52.555 ","End":"10:55.795","Text":"so N_2 would just equal mg,"},{"Start":"10:55.795 ","End":"10:57.460","Text":"which is the maximum value."},{"Start":"10:57.460 ","End":"11:03.825","Text":"Now what would happen in the case that we\u0027re at 90 degrees?"},{"Start":"11:03.825 ","End":"11:07.805","Text":"We\u0027ve got a situation like this."},{"Start":"11:07.805 ","End":"11:09.860","Text":"This is now Number 2 facing up,"},{"Start":"11:09.860 ","End":"11:15.060","Text":"and Number 1 is horizontal on the ground."},{"Start":"11:15.740 ","End":"11:20.280","Text":"Once again, the ball is now here,"},{"Start":"11:20.280 ","End":"11:23.105","Text":"so now the angle Alpha,"},{"Start":"11:23.105 ","End":"11:25.100","Text":"because as we remember,"},{"Start":"11:25.100 ","End":"11:34.160","Text":"the angle Alpha is going here,"},{"Start":"11:35.810 ","End":"11:39.665","Text":"so Alpha is now 90 degrees."},{"Start":"11:39.665 ","End":"11:43.860","Text":"Now, if we plug this in to N_1,"},{"Start":"11:44.440 ","End":"11:51.427","Text":"because all the weight is leaning on the first side of the box,"},{"Start":"11:51.427 ","End":"11:58.910","Text":"then we see that sine of 90 degrees is equal to 1, which makes sense."},{"Start":"11:58.910 ","End":"12:02.030","Text":"N_1 equals mg, the maximum value,"},{"Start":"12:02.030 ","End":"12:07.925","Text":"because N_1 is holding all of the weight,"},{"Start":"12:07.925 ","End":"12:10.600","Text":"all of the mass of the ball."},{"Start":"12:10.600 ","End":"12:15.389","Text":"In other words, if we\u0027re going to look at this graphically, as we know,"},{"Start":"12:15.389 ","End":"12:18.845","Text":"cosine begins at 1,"},{"Start":"12:18.845 ","End":"12:22.936","Text":"and then as we reach Pi over 2,"},{"Start":"12:22.936 ","End":"12:25.250","Text":"cosine is reduced to 0."},{"Start":"12:25.250 ","End":"12:27.320","Text":"It goes down to 0."},{"Start":"12:27.320 ","End":"12:31.465","Text":"Conversely, the sine function begins at 0,"},{"Start":"12:31.465 ","End":"12:34.915","Text":"and as it reaches 90 degrees or Pi over 2,"},{"Start":"12:34.915 ","End":"12:38.770","Text":"it reaches the value of 1, its maximum."},{"Start":"12:38.770 ","End":"12:41.710","Text":"So if we think about that with this,"},{"Start":"12:41.710 ","End":"12:51.055","Text":"we can see that as the angle Alpha increases from 0 until 90 degrees or Pi over 2,"},{"Start":"12:51.055 ","End":"12:56.190","Text":"then the cosine is decreasing,"},{"Start":"12:56.190 ","End":"12:59.385","Text":"however, the sine is increasing."},{"Start":"12:59.385 ","End":"13:06.325","Text":"The value of N_2 is decreasing and the value of N_1 is increasing,"},{"Start":"13:06.325 ","End":"13:14.760","Text":"which means that less force is being applied on the side of the box labeled Number 2,"},{"Start":"13:14.760 ","End":"13:20.855","Text":"and more is being applied on the side of the box that we\u0027ve labeled 1."},{"Start":"13:20.855 ","End":"13:23.645","Text":"This is how you would work with intuition."},{"Start":"13:23.645 ","End":"13:25.985","Text":"Now, if you don\u0027t understand intuition,"},{"Start":"13:25.985 ","End":"13:29.000","Text":"it doesn\u0027t really matter because you can just work"},{"Start":"13:29.000 ","End":"13:33.095","Text":"everything out by just drawing it out clearly in a free body diagram."},{"Start":"13:33.095 ","End":"13:36.590","Text":"But if you do understand the intuition behind everything,"},{"Start":"13:36.590 ","End":"13:39.800","Text":"then it will make solving questions maybe a little bit easier"},{"Start":"13:39.800 ","End":"13:43.870","Text":"because you\u0027ll actually understand what is going on more clearly."},{"Start":"13:43.870 ","End":"13:47.610","Text":"Let\u0027s move on to the rest of the questions."}],"ID":10604},{"Watched":false,"Name":"Newton\u0027s Third Law","Duration":"2m 1s","ChapterTopicVideoID":10275,"CourseChapterTopicPlaylistID":8960,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.025","Text":"Hello. In this lesson,"},{"Start":"00:02.025 ","End":"00:05.070","Text":"we\u0027re going to be speaking about Newton\u0027s Third Law."},{"Start":"00:05.070 ","End":"00:07.845","Text":"First, let\u0027s start off with the Third Law."},{"Start":"00:07.845 ","End":"00:13.545","Text":"It states that for every action there\u0027s an equal and opposite reaction."},{"Start":"00:13.545 ","End":"00:17.850","Text":"What does that mean? Let\u0027s say that we have 2 bodies."},{"Start":"00:17.850 ","End":"00:19.440","Text":"Here we have body A,"},{"Start":"00:19.440 ","End":"00:21.375","Text":"and here we have body B."},{"Start":"00:21.375 ","End":"00:25.110","Text":"Let\u0027s say A is exerting some force,"},{"Start":"00:25.110 ","End":"00:31.485","Text":"so let\u0027s call it F_A, on body B."},{"Start":"00:31.485 ","End":"00:35.445","Text":"Here, specifically, let\u0027s say that it\u0027s the weight."},{"Start":"00:35.445 ","End":"00:42.070","Text":"Then that means that body B is going to be exerting some force,"},{"Start":"00:43.070 ","End":"00:48.435","Text":"F_B, onto A."},{"Start":"00:48.435 ","End":"00:52.360","Text":"So as we can see from these arrows, both our forces,"},{"Start":"00:52.360 ","End":"00:53.995","Text":"F_A and F_B,"},{"Start":"00:53.995 ","End":"00:59.085","Text":"are of equal magnitude but in the opposite directions."},{"Start":"00:59.085 ","End":"01:07.700","Text":"So then we can write that our F_A is equal to negative F_B."},{"Start":"01:07.760 ","End":"01:14.135","Text":"All of our forces are interactions between different bodies,"},{"Start":"01:14.135 ","End":"01:17.710","Text":"and therefore, there\u0027s no such thing as"},{"Start":"01:17.710 ","End":"01:22.150","Text":"a force that is not accompanied by an equal and opposite force."},{"Start":"01:22.150 ","End":"01:25.915","Text":"If there\u0027s 1 force acting on another body,"},{"Start":"01:25.915 ","End":"01:31.620","Text":"then a force is going to be acting right back on that original body."},{"Start":"01:33.740 ","End":"01:40.655","Text":"Newton\u0027s Third Law states that every action has an equal and opposite reaction."},{"Start":"01:40.655 ","End":"01:43.470","Text":"So if we have an arrow, F_A,"},{"Start":"01:43.470 ","End":"01:46.940","Text":"pointing downwards, we\u0027re going to have an arrow, F_B,"},{"Start":"01:46.940 ","End":"01:54.145","Text":"pointing upwards, and our F_B is going to be equal to negative F_A, or vice versa."},{"Start":"01:54.145 ","End":"01:58.925","Text":"It also states that the forces are going to be coming in pairs."},{"Start":"01:58.925 ","End":"02:01.860","Text":"That\u0027s the end of this lesson."}],"ID":10605}],"Thumbnail":null,"ID":8960},{"Name":"The Static Friction","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Static Friction","Duration":"16m 1s","ChapterTopicVideoID":23300,"CourseChapterTopicPlaylistID":8962,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.085","Text":"Hello. In this lesson,"},{"Start":"00:02.085 ","End":"00:05.160","Text":"we\u0027re going to be learning about the frictional force."},{"Start":"00:05.160 ","End":"00:10.080","Text":"The forces that we are already familiar with are weights,"},{"Start":"00:10.080 ","End":"00:14.519","Text":"which we know is denoted by the letter w,"},{"Start":"00:14.519 ","End":"00:20.355","Text":"the normal force, which is denoted by the letter capital N, and the tension,"},{"Start":"00:20.355 ","End":"00:22.515","Text":"which is denoted by the letter capital"},{"Start":"00:22.515 ","End":"00:29.670","Text":"T. Friction is a force just like the others that we\u0027ve already seen."},{"Start":"00:29.670 ","End":"00:32.730","Text":"But friction is a force which acts to resist"},{"Start":"00:32.730 ","End":"00:38.280","Text":"the relative motion of 2 bodies which are in contact with each other."},{"Start":"00:38.280 ","End":"00:42.755","Text":"If you take your hand and you try and slide it on a table,"},{"Start":"00:42.755 ","End":"00:45.904","Text":"you\u0027ll feel some resistance force,"},{"Start":"00:45.904 ","End":"00:50.015","Text":"which is trying to stop your hand from sliding across the table."},{"Start":"00:50.015 ","End":"00:54.810","Text":"This is friction, and this is what we\u0027re going to talk about in this lesson."},{"Start":"00:55.270 ","End":"00:59.150","Text":"There are 2 types of frictional force."},{"Start":"00:59.150 ","End":"01:07.080","Text":"There\u0027s static friction, which is denoted by f with a small s over here for static."},{"Start":"01:07.080 ","End":"01:11.225","Text":"That\u0027s the frictional force that acts between 2 bodies when there"},{"Start":"01:11.225 ","End":"01:15.865","Text":"is no relative motion between the bodies."},{"Start":"01:15.865 ","End":"01:18.035","Text":"Let\u0027s give an example of that."},{"Start":"01:18.035 ","End":"01:21.965","Text":"Let\u0027s say I have a book resting on a table,"},{"Start":"01:21.965 ","End":"01:27.630","Text":"and let\u0027s say I lift the table up a little bit such that there\u0027s an angle."},{"Start":"01:27.640 ","End":"01:32.660","Text":"But my book still doesn\u0027t start sliding down the table."},{"Start":"01:32.660 ","End":"01:34.145","Text":"Why is that?"},{"Start":"01:34.145 ","End":"01:37.620","Text":"That\u0027s because there\u0027s static friction."},{"Start":"01:38.090 ","End":"01:43.280","Text":"The static friction is stopping the book from sliding down the table."},{"Start":"01:43.280 ","End":"01:46.050","Text":"It\u0027s holding it in place."},{"Start":"01:47.300 ","End":"01:54.515","Text":"In this case we have static friction because the book remains static on the table."},{"Start":"01:54.515 ","End":"01:58.055","Text":"The second type of friction is kinetic friction,"},{"Start":"01:58.055 ","End":"02:02.570","Text":"which is denoted by an f with a small k, k for kinetic."},{"Start":"02:02.570 ","End":"02:06.935","Text":"This is the frictional force that acts between 2 bodies when there"},{"Start":"02:06.935 ","End":"02:12.914","Text":"is relative motion between the 2 bodies point of contact."},{"Start":"02:12.914 ","End":"02:16.550","Text":"Let\u0027s go back to the book on the table example."},{"Start":"02:16.550 ","End":"02:21.170","Text":"My table is at an angle and the book remains in place."},{"Start":"02:21.170 ","End":"02:24.065","Text":"It doesn\u0027t slide because there\u0027s static friction."},{"Start":"02:24.065 ","End":"02:29.299","Text":"But as I left my table to a larger and larger angle,"},{"Start":"02:29.299 ","End":"02:33.470","Text":"eventually my book will start to slide down."},{"Start":"02:33.470 ","End":"02:37.460","Text":"Now that means that if my book is sliding down the table,"},{"Start":"02:37.460 ","End":"02:42.470","Text":"that means that there is relative motion between the 2 bodies,"},{"Start":"02:42.470 ","End":"02:45.250","Text":"between the book and the table,"},{"Start":"02:45.250 ","End":"02:49.865","Text":"so that means that in this case we have kinetic friction."},{"Start":"02:49.865 ","End":"02:54.810","Text":"Kinetic because our book has kinetic energy, it\u0027s moving."},{"Start":"02:55.790 ","End":"03:00.425","Text":"Both of these types of friction are still types of friction."},{"Start":"03:00.425 ","End":"03:03.980","Text":"The static friction has to be overcome in order"},{"Start":"03:03.980 ","End":"03:07.490","Text":"for some type of body to begin its movement."},{"Start":"03:07.490 ","End":"03:12.005","Text":"The kinetic friction is going to be there,"},{"Start":"03:12.005 ","End":"03:17.810","Text":"be present during the movement of 1 or 2 of the body\u0027s relative to each other."},{"Start":"03:17.810 ","End":"03:21.940","Text":"It\u0027s something that just resists the motion."},{"Start":"03:21.940 ","End":"03:25.054","Text":"They\u0027re very similar, but there are differences,"},{"Start":"03:25.054 ","End":"03:26.750","Text":"especially in the working out."},{"Start":"03:26.750 ","End":"03:31.050","Text":"Now let\u0027s go and see what those differences are."},{"Start":"03:32.440 ","End":"03:39.680","Text":"Both types of friction always act tangentially to the direction of the surface."},{"Start":"03:39.680 ","End":"03:44.555","Text":"Now we\u0027ve already learned that the normal force always acts perpendicularly."},{"Start":"03:44.555 ","End":"03:49.850","Text":"If we have a surface over here and some mass,"},{"Start":"03:49.850 ","End":"03:55.265","Text":"so we know that the normal is always going to be perpendicular to the surface."},{"Start":"03:55.265 ","End":"04:01.220","Text":"Now, with friction, it always acts tangentially to the direction of the surface."},{"Start":"04:01.220 ","End":"04:05.480","Text":"That means that our frictional force is either going to"},{"Start":"04:05.480 ","End":"04:11.160","Text":"act in this direction or in this direction."},{"Start":"04:11.330 ","End":"04:14.600","Text":"The second point is that the direction of"},{"Start":"04:14.600 ","End":"04:20.665","Text":"kinetic friction is always opposite to the direction of travel."},{"Start":"04:20.665 ","End":"04:26.825","Text":"What does that mean? If I have some mass here on a surface and I have"},{"Start":"04:26.825 ","End":"04:34.255","Text":"some force acting to push my box in the right direction."},{"Start":"04:34.255 ","End":"04:38.075","Text":"I have a force acting on the box in the right direction."},{"Start":"04:38.075 ","End":"04:42.800","Text":"The direction of my kinetic friction is always going to be in"},{"Start":"04:42.800 ","End":"04:47.150","Text":"the opposite direction to the direction of travel."},{"Start":"04:47.150 ","End":"04:52.040","Text":"If my box is moving in this rightwards direction,"},{"Start":"04:52.040 ","End":"04:59.010","Text":"my kinetic friction is going to be acting in the leftwards direction."},{"Start":"04:59.090 ","End":"05:03.655","Text":"The third point is that the direction of static friction is"},{"Start":"05:03.655 ","End":"05:08.319","Text":"always opposite to the direction of intended motion."},{"Start":"05:08.319 ","End":"05:11.305","Text":"This is a little bit harder to understand."},{"Start":"05:11.305 ","End":"05:15.160","Text":"Let\u0027s imagine that we have again"},{"Start":"05:15.160 ","End":"05:19.450","Text":"our box and there is some force pushing against the box to move it."},{"Start":"05:19.450 ","End":"05:23.425","Text":"But this is really heavy box."},{"Start":"05:23.425 ","End":"05:27.680","Text":"This force isn\u0027t enough to actually move it."},{"Start":"05:27.680 ","End":"05:32.535","Text":"That means that our box is static but there\u0027s static friction."},{"Start":"05:32.535 ","End":"05:34.705","Text":"It\u0027s the static friction in fact,"},{"Start":"05:34.705 ","End":"05:40.255","Text":"which is stopping this force from being able to push the box."},{"Start":"05:40.255 ","End":"05:44.630","Text":"If we imagine that this force was big enough to push the box,"},{"Start":"05:44.630 ","End":"05:48.990","Text":"we can see that our box would be moving in the rightwards direction,"},{"Start":"05:48.990 ","End":"05:52.800","Text":"direction of motion would be in the right."},{"Start":"05:53.870 ","End":"05:58.310","Text":"Once again, our frictional force but"},{"Start":"05:58.310 ","End":"06:04.430","Text":"our static friction will be acting in the opposite direction to the direction of motion,"},{"Start":"06:04.430 ","End":"06:07.820","Text":"which means it will be acting in the left direction."},{"Start":"06:07.820 ","End":"06:10.294","Text":"Now of course, in static friction,"},{"Start":"06:10.294 ","End":"06:13.610","Text":"the box isn\u0027t actually moving because this force wasn\u0027t big"},{"Start":"06:13.610 ","End":"06:17.960","Text":"enough to overcome this coefficient of static friction."},{"Start":"06:17.960 ","End":"06:20.285","Text":"However, theoretically if it did,"},{"Start":"06:20.285 ","End":"06:24.570","Text":"our frictional force is acting in the opposite direction."},{"Start":"06:25.520 ","End":"06:30.140","Text":"Let\u0027s talk a little bit more about static friction."},{"Start":"06:30.140 ","End":"06:35.660","Text":"The static frictional force increases as the net force on the body increases."},{"Start":"06:35.660 ","End":"06:37.415","Text":"What does that mean?"},{"Start":"06:37.415 ","End":"06:40.055","Text":"Let\u0027s say we have this heavy box over here."},{"Start":"06:40.055 ","End":"06:45.080","Text":"Let\u0027s say it\u0027s a fridge and it\u0027s on the floor over here."},{"Start":"06:45.080 ","End":"06:51.875","Text":"Here we have someone who\u0027s either pushing or pulling the fridge in the right direction."},{"Start":"06:51.875 ","End":"06:57.215","Text":"Which means that our static friction is going to be acting"},{"Start":"06:57.215 ","End":"07:03.650","Text":"in the left direction because it\u0027s always acting in the opposite direction to the force."},{"Start":"07:03.650 ","End":"07:10.250","Text":"Now, the point here is that our fridge or our box isn\u0027t moving."},{"Start":"07:10.250 ","End":"07:13.730","Text":"It\u0027s static, which is why our static friction is acting."},{"Start":"07:13.730 ","End":"07:20.900","Text":"We can see that the net force on the fridge is equal to 0,"},{"Start":"07:20.900 ","End":"07:24.290","Text":"because the acceleration is also equal to 0."},{"Start":"07:24.290 ","End":"07:25.700","Text":"There\u0027s no motion over here,"},{"Start":"07:25.700 ","End":"07:28.955","Text":"so the sum of all of the forces is equal to 0."},{"Start":"07:28.955 ","End":"07:30.485","Text":"What does that mean?"},{"Start":"07:30.485 ","End":"07:36.675","Text":"That means that when we\u0027re dealing with static friction, our F_1,"},{"Start":"07:36.675 ","End":"07:39.210","Text":"which is our pushing or pulling force,"},{"Start":"07:39.210 ","End":"07:43.295","Text":"the force that we\u0027re trying to move this fridge with is"},{"Start":"07:43.295 ","End":"07:48.355","Text":"equal to the static frictional force."},{"Start":"07:48.355 ","End":"07:51.974","Text":"When we have an object,"},{"Start":"07:51.974 ","End":"07:54.710","Text":"which has a force being applied to it,"},{"Start":"07:54.710 ","End":"07:56.345","Text":"however, it is static."},{"Start":"07:56.345 ","End":"08:00.335","Text":"That means that the net force on that object is equal to 0,"},{"Start":"08:00.335 ","End":"08:05.644","Text":"which means that F_1 is equal to our static friction."},{"Start":"08:05.644 ","End":"08:12.025","Text":"That means that if we increase F_1 to some value,"},{"Start":"08:12.025 ","End":"08:15.545","Text":"our static friction will also increase to that value,"},{"Start":"08:15.545 ","End":"08:17.315","Text":"and if we increase F_1 more,"},{"Start":"08:17.315 ","End":"08:21.340","Text":"our static friction will increase more as well."},{"Start":"08:21.340 ","End":"08:28.640","Text":"Now of course, once our F_1 is equal to some value,"},{"Start":"08:28.640 ","End":"08:32.570","Text":"our fridge or whichever object this is over here will begin"},{"Start":"08:32.570 ","End":"08:36.984","Text":"to move and will overcome our static friction."},{"Start":"08:36.984 ","End":"08:40.865","Text":"But whilst this object is still static,"},{"Start":"08:40.865 ","End":"08:44.900","Text":"so that means that as much as we increase our F_1,"},{"Start":"08:44.900 ","End":"08:49.380","Text":"our static friction will also increase."},{"Start":"08:50.940 ","End":"08:55.525","Text":"The frictional force when we\u0027re dealing with static friction can"},{"Start":"08:55.525 ","End":"08:59.800","Text":"increase until it reaches some maximal value,"},{"Start":"08:59.800 ","End":"09:01.960","Text":"which is denoted by this f,"},{"Start":"09:01.960 ","End":"09:03.430","Text":"which is for friction,"},{"Start":"09:03.430 ","End":"09:10.510","Text":"lowercase f, s for static and max for maximum."},{"Start":"09:10.510 ","End":"09:13.540","Text":"Once we\u0027ve reached this value,"},{"Start":"09:13.540 ","End":"09:17.905","Text":"or once we\u0027ve reached a value which is greater than this value,"},{"Start":"09:17.905 ","End":"09:20.710","Text":"if we\u0027re exerting a force on a body,"},{"Start":"09:20.710 ","End":"09:24.865","Text":"it will then begin to move and then we will"},{"Start":"09:24.865 ","End":"09:31.400","Text":"be dealing with kinetic friction rather than static friction."},{"Start":"09:31.530 ","End":"09:37.750","Text":"The maximum value of static friction is equal to this f_smax."},{"Start":"09:37.750 ","End":"09:41.350","Text":"It\u0027s equal to Mu_s multiplied by"},{"Start":"09:41.350 ","End":"09:47.695","Text":"N. Where N is the normal force which the surface exerts on the body,"},{"Start":"09:47.695 ","End":"09:51.789","Text":"and Mu_s is the coefficient of static friction."},{"Start":"09:51.789 ","End":"09:56.200","Text":"Mu_s is some coefficient and it\u0027s simply"},{"Start":"09:56.200 ","End":"10:00.745","Text":"dependent on the materials that the 2 bodies are made of,"},{"Start":"10:00.745 ","End":"10:07.280","Text":"so if we\u0027re moving glass on glass or rubber on glass, etc."},{"Start":"10:08.190 ","End":"10:11.785","Text":"Now let\u0027s take a look at this equation."},{"Start":"10:11.785 ","End":"10:16.285","Text":"This is our equation for the maximal value of static friction."},{"Start":"10:16.285 ","End":"10:19.450","Text":"We can see that the equation only describes"},{"Start":"10:19.450 ","End":"10:25.300","Text":"the magnitude of this maximum value and not the direction."},{"Start":"10:25.300 ","End":"10:27.970","Text":"This isn\u0027t a vector quantity."},{"Start":"10:27.970 ","End":"10:29.980","Text":"As we know this maximum value,"},{"Start":"10:29.980 ","End":"10:31.360","Text":"the direction of it,"},{"Start":"10:31.360 ","End":"10:38.080","Text":"is always going to be opposite to the direction of intended travel."},{"Start":"10:38.080 ","End":"10:39.940","Text":"But from this equation,"},{"Start":"10:39.940 ","End":"10:42.385","Text":"this only shows us the magnitude."},{"Start":"10:42.385 ","End":"10:46.960","Text":"The second point which is very important and also very surprising,"},{"Start":"10:46.960 ","End":"10:50.440","Text":"is that the frictional force is not dependent"},{"Start":"10:50.440 ","End":"10:54.715","Text":"or it\u0027s independent of the surface area of the body."},{"Start":"10:54.715 ","End":"10:59.380","Text":"What does that mean? Let\u0027s say we have a surface over"},{"Start":"10:59.380 ","End":"11:05.320","Text":"here and we\u0027re pushing some crates,"},{"Start":"11:05.320 ","End":"11:07.910","Text":"which looks like this."},{"Start":"11:08.430 ","End":"11:12.565","Text":"As we can see, we have all of this area"},{"Start":"11:12.565 ","End":"11:17.899","Text":"here of the box which is in contact with the ground."},{"Start":"11:17.899 ","End":"11:23.610","Text":"Let\u0027s say that someone is pushing it over here."},{"Start":"11:23.610 ","End":"11:25.785","Text":"This is F_1."},{"Start":"11:25.785 ","End":"11:28.880","Text":"This is crate number 1."},{"Start":"11:28.880 ","End":"11:31.720","Text":"Then someone says, you know what,"},{"Start":"11:31.720 ","End":"11:33.205","Text":"let\u0027s try it this way."},{"Start":"11:33.205 ","End":"11:37.705","Text":"Let\u0027s take the exact same surface and the exact same crates."},{"Start":"11:37.705 ","End":"11:39.700","Text":"This is still crate number 1."},{"Start":"11:39.700 ","End":"11:43.225","Text":"But let\u0027s put it in this formation,"},{"Start":"11:43.225 ","End":"11:46.870","Text":"such that there\u0027s only this amount of"},{"Start":"11:46.870 ","End":"11:52.700","Text":"area which is in contact between the crate and the ground."},{"Start":"11:52.710 ","End":"11:59.060","Text":"Then I begin applying a force again to push this crates."},{"Start":"11:59.310 ","End":"12:03.880","Text":"What is very surprising is that the force that I will need"},{"Start":"12:03.880 ","End":"12:08.365","Text":"to push this crate in this formation,"},{"Start":"12:08.365 ","End":"12:15.200","Text":"is going to be the exact same thing as over here, F_1."},{"Start":"12:15.210 ","End":"12:18.340","Text":"As we can see from this equation,"},{"Start":"12:18.340 ","End":"12:20.170","Text":"the only things that"},{"Start":"12:20.170 ","End":"12:27.010","Text":"our maximum static friction value is dependent on is our coefficient of static friction,"},{"Start":"12:27.010 ","End":"12:29.920","Text":"which is just dependent on what the crate is"},{"Start":"12:29.920 ","End":"12:32.965","Text":"made out of and what the ground is made out of,"},{"Start":"12:32.965 ","End":"12:35.020","Text":"and the normal force,"},{"Start":"12:35.020 ","End":"12:38.680","Text":"which as we know, is going to"},{"Start":"12:38.680 ","End":"12:42.475","Text":"be the same for the crate if it\u0027s in this position or this position,"},{"Start":"12:42.475 ","End":"12:47.900","Text":"because the total mass of the crate is still going to be the same."},{"Start":"12:47.910 ","End":"12:52.330","Text":"Whether we\u0027re pushing the crate in this formation or in this formation,"},{"Start":"12:52.330 ","End":"12:57.880","Text":"the surface area which is in contact between the 2 bodies does not"},{"Start":"12:57.880 ","End":"13:04.880","Text":"affect in any way this value for F_smax."},{"Start":"13:06.390 ","End":"13:11.140","Text":"Here we have our crate and I\u0027m pushing with this force,"},{"Start":"13:11.140 ","End":"13:20.109","Text":"F. As long as my force F doesn\u0027t increase above this value,"},{"Start":"13:20.109 ","End":"13:26.320","Text":"I\u0027m going to be still dealing with static friction and my crate will not move."},{"Start":"13:26.320 ","End":"13:32.560","Text":"If I start applying a force which is greater than this F_smax,"},{"Start":"13:32.560 ","End":"13:34.600","Text":"or rather greater or equal to,"},{"Start":"13:34.600 ","End":"13:35.860","Text":"we\u0027ll see you soon,"},{"Start":"13:35.860 ","End":"13:39.295","Text":"then my crates will begin to move,"},{"Start":"13:39.295 ","End":"13:42.025","Text":"and then will we dealing with kinetic friction,"},{"Start":"13:42.025 ","End":"13:45.415","Text":"which we will speak about during this chapter,"},{"Start":"13:45.415 ","End":"13:47.515","Text":"but in a separate lesson."},{"Start":"13:47.515 ","End":"13:51.655","Text":"Just a last word before we finish this lesson."},{"Start":"13:51.655 ","End":"13:58.510","Text":"What\u0027s important to remember is that the value for the maximum value of"},{"Start":"13:58.510 ","End":"14:01.370","Text":"our static friction is equal to"},{"Start":"14:01.370 ","End":"14:06.690","Text":"our coefficient of static friction multiplied by the normal force."},{"Start":"14:06.690 ","End":"14:10.520","Text":"If we get to any type of value which is greater"},{"Start":"14:10.520 ","End":"14:14.590","Text":"than this value for our force being applied,"},{"Start":"14:14.590 ","End":"14:18.835","Text":"then we will overcome this maximum value for static friction."},{"Start":"14:18.835 ","End":"14:25.165","Text":"Then 1 of our objects or the objects will start moving relative to 1 another,"},{"Start":"14:25.165 ","End":"14:29.935","Text":"and then we\u0027re going to be dealing with kinetic friction."},{"Start":"14:29.935 ","End":"14:33.610","Text":"However, this equation doesn\u0027t describe"},{"Start":"14:33.610 ","End":"14:37.870","Text":"our value for static friction at any point in time."},{"Start":"14:37.870 ","End":"14:41.170","Text":"It just describes our maximum value."},{"Start":"14:41.170 ","End":"14:45.775","Text":"Our actual value for static friction is always going to be"},{"Start":"14:45.775 ","End":"14:50.485","Text":"smaller or equal to our F_smax,"},{"Start":"14:50.485 ","End":"14:54.580","Text":"smaller or equal to our maximum value."},{"Start":"14:54.580 ","End":"15:03.310","Text":"As long as we are applying some force which is smaller than this value over here."},{"Start":"15:03.310 ","End":"15:08.470","Text":"We\u0027re going to have some value for static friction."},{"Start":"15:08.470 ","End":"15:13.510","Text":"Remember, because the sum of all of the forces when we\u0027re dealing with"},{"Start":"15:13.510 ","End":"15:19.120","Text":"static friction is equal to 0 just because our system is static."},{"Start":"15:19.120 ","End":"15:23.830","Text":"That means that the force that we\u0027re applying is going to be"},{"Start":"15:23.830 ","End":"15:29.020","Text":"equal to our static frictional force."},{"Start":"15:29.020 ","End":"15:34.480","Text":"I can increase this force from 5 to 10 to 15,"},{"Start":"15:34.480 ","End":"15:40.045","Text":"and then my F_s will increase subsequently from 5 to 10 to 15."},{"Start":"15:40.045 ","End":"15:44.260","Text":"But if my F_smax is equal to 20,"},{"Start":"15:44.260 ","End":"15:47.380","Text":"and then I increase my F to 21,"},{"Start":"15:47.380 ","End":"15:51.805","Text":"then I\u0027ll be larger than the value for F_smax,"},{"Start":"15:51.805 ","End":"15:54.880","Text":"in which case we enter the realm for kinetic friction,"},{"Start":"15:54.880 ","End":"15:58.060","Text":"which we will speak about in a later lesson."},{"Start":"15:58.060 ","End":"16:00.950","Text":"That\u0027s the end of this lesson."}],"ID":24211},{"Watched":false,"Name":"Exercise 1","Duration":"11m 18s","ChapterTopicVideoID":10277,"CourseChapterTopicPlaylistID":8962,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.470 ","End":"00:05.415","Text":"In this lesson, we\u0027re being asked to use the coefficient of friction,"},{"Start":"00:05.415 ","End":"00:10.620","Text":"but then a slightly different way to usual. Let\u0027s begin."},{"Start":"00:10.620 ","End":"00:13.215","Text":"Now, we have mass m,"},{"Start":"00:13.215 ","End":"00:17.820","Text":"and we also have the angle of the slope Alpha."},{"Start":"00:17.820 ","End":"00:20.055","Text":"Now, these are given to us in the question."},{"Start":"00:20.055 ","End":"00:22.050","Text":"The question is asking us,"},{"Start":"00:22.050 ","End":"00:25.020","Text":"what is the smallest value that the coefficient of"},{"Start":"00:25.020 ","End":"00:29.025","Text":"friction may be such that the system remains at rest?"},{"Start":"00:29.025 ","End":"00:32.850","Text":"In other words, let me explain it easier,"},{"Start":"00:32.850 ","End":"00:36.304","Text":"if the slope is really smooth,"},{"Start":"00:36.304 ","End":"00:40.121","Text":"then the mass will just slide down it."},{"Start":"00:40.121 ","End":"00:43.730","Text":"What we\u0027re being asked is what is the threshold,"},{"Start":"00:43.730 ","End":"00:50.150","Text":"the minimum value that the coefficient of friction is Mu,"},{"Start":"00:50.150 ","End":"00:56.120","Text":"that Mu can be so that the mass doesn\u0027t slide down."},{"Start":"00:56.120 ","End":"01:01.370","Text":"Because obviously if the coefficient of friction is higher than this value,"},{"Start":"01:01.370 ","End":"01:04.160","Text":"then the mass will still stay in place."},{"Start":"01:04.160 ","End":"01:06.590","Text":"If the surface is rougher,"},{"Start":"01:06.590 ","End":"01:11.425","Text":"the mass will stay in place and the smoother that the surface gets,"},{"Start":"01:11.425 ","End":"01:14.240","Text":"the lower the coefficient of friction,"},{"Start":"01:14.240 ","End":"01:15.845","Text":"the mass will slide down."},{"Start":"01:15.845 ","End":"01:21.515","Text":"We\u0027re trying to find the minimum threshold that the mass will stay in place."},{"Start":"01:21.515 ","End":"01:24.340","Text":"The minimum roughness, if you will."},{"Start":"01:24.340 ","End":"01:26.600","Text":"Let\u0027s see how we do this."},{"Start":"01:26.600 ","End":"01:30.845","Text":"Now, we\u0027re obviously going to start by drawing a free-body diagram."},{"Start":"01:30.845 ","End":"01:36.255","Text":"I\u0027m just reminding you if here the angle is Alpha because"},{"Start":"01:36.255 ","End":"01:38.900","Text":"the free body diagram is"},{"Start":"01:38.900 ","End":"01:43.295","Text":"the exact replica of what\u0027s happening in the actual diagram that we have,"},{"Start":"01:43.295 ","End":"01:48.145","Text":"then that means that here is also Alpha."},{"Start":"01:48.145 ","End":"01:51.615","Text":"Next, we have mg,"},{"Start":"01:51.615 ","End":"01:54.479","Text":"which as usual is pointing downwards."},{"Start":"01:54.479 ","End":"01:56.045","Text":"Then up here,"},{"Start":"01:56.045 ","End":"01:58.265","Text":"we have the normal force,"},{"Start":"01:58.265 ","End":"02:02.510","Text":"which is the force from the slope acting on the mass,"},{"Start":"02:02.510 ","End":"02:05.915","Text":"and then we also have the frictional force."},{"Start":"02:05.915 ","End":"02:07.130","Text":"Now the frictional force,"},{"Start":"02:07.130 ","End":"02:11.480","Text":"I can either draw on the purple arrow or on the red arrow."},{"Start":"02:11.480 ","End":"02:13.000","Text":"It doesn\u0027t really make a difference,"},{"Start":"02:13.000 ","End":"02:16.138","Text":"and this is because it\u0027s an unknown."},{"Start":"02:16.138 ","End":"02:19.400","Text":"Now I\u0027m going to draw it on the purple arrow just"},{"Start":"02:19.400 ","End":"02:25.850","Text":"because it\u0027s stopping the mass from rolling down."},{"Start":"02:25.850 ","End":"02:30.510","Text":"It should be in the opposite direction to the downwards motion."},{"Start":"02:30.890 ","End":"02:33.055","Text":"But having said that,"},{"Start":"02:33.055 ","End":"02:36.760","Text":"if you were to say draw the frictional force above the red arrow,"},{"Start":"02:36.760 ","End":"02:40.780","Text":"it wouldn\u0027t really matter because once we finish doing all of the calculations,"},{"Start":"02:40.780 ","End":"02:42.760","Text":"you would just get the frictional force and a minus"},{"Start":"02:42.760 ","End":"02:44.890","Text":"and you would know that it belongs on that other side."},{"Start":"02:44.890 ","End":"02:46.495","Text":"It doesn\u0027t really make a difference,"},{"Start":"02:46.495 ","End":"02:49.375","Text":"just if we\u0027re working through intuition,"},{"Start":"02:49.375 ","End":"02:53.660","Text":"it makes more sense to put it on this side for the time being."},{"Start":"02:54.240 ","End":"02:58.030","Text":"Now, because the mass is at rest,"},{"Start":"02:58.030 ","End":"03:01.840","Text":"which means that it\u0027s not accelerating or anything,"},{"Start":"03:01.840 ","End":"03:05.350","Text":"it means that the sum of all the forces is equal to 0."},{"Start":"03:05.350 ","End":"03:09.100","Text":"In order to work out what this actually means,"},{"Start":"03:09.100 ","End":"03:13.790","Text":"we have to split up the mg into its different components."},{"Start":"03:13.790 ","End":"03:17.020","Text":"How do we split this up into components?"},{"Start":"03:17.020 ","End":"03:23.725","Text":"As we remember, on the side adjacent of the triangle where Alpha is,"},{"Start":"03:23.725 ","End":"03:25.480","Text":"the side adjacent to Alpha,"},{"Start":"03:25.480 ","End":"03:31.508","Text":"we know is mg cosine Alpha,"},{"Start":"03:31.508 ","End":"03:36.300","Text":"and the side opposite which is the equivalent of this, so it\u0027s this."},{"Start":"03:36.300 ","End":"03:45.040","Text":"Because it\u0027s opposite, it\u0027s mg sine Alpha."},{"Start":"03:45.920 ","End":"03:50.379","Text":"Now we have basically all we need for the equations."},{"Start":"03:50.379 ","End":"03:59.318","Text":"We know now that N equals mg cosine Alpha,"},{"Start":"03:59.318 ","End":"04:05.880","Text":"and we know that F equals mg sine Alpha."},{"Start":"04:05.880 ","End":"04:07.415","Text":"Why is this?"},{"Start":"04:07.415 ","End":"04:12.680","Text":"Because there\u0027s no motion going up and down and there\u0027s no motion going to the sides."},{"Start":"04:12.680 ","End":"04:17.100","Text":"We know that this is why each thing equals what it does."},{"Start":"04:17.140 ","End":"04:22.700","Text":"Now, we have another equation for f static,"},{"Start":"04:22.700 ","End":"04:25.130","Text":"because this object is at rest,"},{"Start":"04:25.130 ","End":"04:26.870","Text":"we\u0027re only dealing with this in the meantime,"},{"Start":"04:26.870 ","End":"04:31.220","Text":"and that is that f is smaller or equal to"},{"Start":"04:31.220 ","End":"04:38.835","Text":"Mu N. Let\u0027s write this as also a static."},{"Start":"04:38.835 ","End":"04:43.520","Text":"Now, because we don\u0027t actually have a value for what f static is,"},{"Start":"04:44.660 ","End":"04:51.069","Text":"we can sub in this Mu N because we know what these values are."},{"Start":"04:51.069 ","End":"04:55.565","Text":"Because f static is smaller or equal to Mu N,"},{"Start":"04:55.565 ","End":"05:04.740","Text":"we can write that therefore mg sine Alpha is"},{"Start":"05:04.740 ","End":"05:09.575","Text":"smaller or equal to Mu N. But then"},{"Start":"05:09.575 ","End":"05:15.295","Text":"we know that N equals mg cosine Alpha."},{"Start":"05:15.295 ","End":"05:23.875","Text":"Then we can write that mg sine Alpha is smaller or equal to Mu,"},{"Start":"05:23.875 ","End":"05:25.460","Text":"the coefficient of friction,"},{"Start":"05:25.460 ","End":"05:32.580","Text":"multiplied by N, which is mg cosine Alpha."},{"Start":"05:32.580 ","End":"05:37.415","Text":"From here, we know that the Mu is our unknown."},{"Start":"05:37.415 ","End":"05:39.980","Text":"This is exactly what we\u0027re trying to find."},{"Start":"05:39.980 ","End":"05:43.040","Text":"We\u0027re trying to find the minimal value of"},{"Start":"05:43.040 ","End":"05:49.595","Text":"Mu such that the object remains at rest and doesn\u0027t slide down."},{"Start":"05:49.595 ","End":"05:52.355","Text":"Let\u0027s see what we do next."},{"Start":"05:52.355 ","End":"05:57.590","Text":"Now what we can do is because on each side of the equation there\u0027s an mg,"},{"Start":"05:57.590 ","End":"06:01.183","Text":"we can just cross out the mgs,"},{"Start":"06:01.183 ","End":"06:11.285","Text":"and then we\u0027re left with sine of Alpha is smaller or equal to Mu cosine of Alpha."},{"Start":"06:11.285 ","End":"06:14.705","Text":"Now, if we divide both sides by cosine,"},{"Start":"06:14.705 ","End":"06:16.730","Text":"sine over cosine, as you remember,"},{"Start":"06:16.730 ","End":"06:24.070","Text":"is tangents, tangents of Alpha is smaller or equal to Mu."},{"Start":"06:24.070 ","End":"06:26.525","Text":"Now, this is a bit of a weird answer."},{"Start":"06:26.525 ","End":"06:30.240","Text":"Let\u0027s see what this answer actually means."},{"Start":"06:30.260 ","End":"06:38.885","Text":"What this means is that as long as tangents of Alpha is smaller or equal to Mu,"},{"Start":"06:38.885 ","End":"06:40.685","Text":"the mass will stay in place."},{"Start":"06:40.685 ","End":"06:42.830","Text":"It will not slide down the slope."},{"Start":"06:42.830 ","End":"06:48.080","Text":"Now, because we\u0027re being asked what the smallest value that Mu can be,"},{"Start":"06:48.080 ","End":"06:52.915","Text":"it means that Mu can be equal to tan Alpha,"},{"Start":"06:52.915 ","End":"06:55.110","Text":"and that is the smallest value."},{"Start":"06:55.110 ","End":"06:58.075","Text":"If Mu is larger than tan Alpha,"},{"Start":"06:58.075 ","End":"07:03.020","Text":"then that\u0027s also okay, but the minimum value is when Mu is equal to tan Alpha."},{"Start":"07:03.020 ","End":"07:06.890","Text":"Now of course, if Mu is smaller than tan Alpha,"},{"Start":"07:06.890 ","End":"07:10.875","Text":"then the mass will slide down."},{"Start":"07:10.875 ","End":"07:18.725","Text":"Now, what\u0027s funny here is that the mass itself doesn\u0027t even play a part in this."},{"Start":"07:18.725 ","End":"07:22.865","Text":"Now, what\u0027s so strange about the mass not playing a part in this,"},{"Start":"07:22.865 ","End":"07:25.985","Text":"is that if you were to think intuitively,"},{"Start":"07:25.985 ","End":"07:29.000","Text":"you would have thought that the heavier the masses,"},{"Start":"07:29.000 ","End":"07:34.085","Text":"the larger the coefficient of friction would be in order to hold the mass up."},{"Start":"07:34.085 ","End":"07:37.250","Text":"But as we can see from this equation over here,"},{"Start":"07:37.250 ","End":"07:40.030","Text":"this is the final equation."},{"Start":"07:40.030 ","End":"07:41.990","Text":"The mass doesn\u0027t play a part,"},{"Start":"07:41.990 ","End":"07:46.595","Text":"It can be a mass of 1 kilogram or a mass of 1,000 kilograms."},{"Start":"07:46.595 ","End":"07:52.205","Text":"If the coefficient of friction is larger or equal to tan of Alpha,"},{"Start":"07:52.205 ","End":"07:54.665","Text":"then the mass will stay in place,"},{"Start":"07:54.665 ","End":"07:56.105","Text":"which is very strange."},{"Start":"07:56.105 ","End":"07:57.350","Text":"We wouldn\u0027t have expected it."},{"Start":"07:57.350 ","End":"08:03.410","Text":"I at least didn\u0027t. Let\u0027s explain this weird thing mathematically."},{"Start":"08:03.410 ","End":"08:11.210","Text":"Now, let\u0027s say this mass was 10 kilograms or 1,000 kilograms, doesn\u0027t matter."},{"Start":"08:11.210 ","End":"08:19.375","Text":"Then that means that the mg will be larger going in this direction, going down."},{"Start":"08:19.375 ","End":"08:27.409","Text":"However, the frictional force"},{"Start":"08:27.409 ","End":"08:29.945","Text":"would increase by the same amount."},{"Start":"08:29.945 ","End":"08:31.160","Text":"Why is that?"},{"Start":"08:31.160 ","End":"08:37.520","Text":"Because the frictional force also has mg in it."},{"Start":"08:37.520 ","End":"08:41.285","Text":"If the mg is increasing in this direction,"},{"Start":"08:41.285 ","End":"08:44.045","Text":"then it\u0027s also increasing in this direction,"},{"Start":"08:44.045 ","End":"08:46.555","Text":"which makes it balance out,"},{"Start":"08:46.555 ","End":"08:49.865","Text":"which is why it explains the fact that"},{"Start":"08:49.865 ","End":"08:55.075","Text":"the mass itself doesn\u0027t play a part in this equation."},{"Start":"08:55.075 ","End":"08:59.160","Text":"This means that when being"},{"Start":"08:59.160 ","End":"09:06.950","Text":"asked the coefficient of friction that will keep a mass in place,"},{"Start":"09:06.950 ","End":"09:10.505","Text":"the only things that are being interested in is"},{"Start":"09:10.505 ","End":"09:16.050","Text":"the coefficient of friction itself and the angle of the slope."},{"Start":"09:16.210 ","End":"09:26.360","Text":"If I have, let\u0027s say wood over here and rough stone,"},{"Start":"09:26.360 ","End":"09:32.689","Text":"then I would use the coefficient of friction between the wood and the rocks,"},{"Start":"09:32.689 ","End":"09:34.760","Text":"doesn\u0027t certain value for it."},{"Start":"09:34.760 ","End":"09:39.645","Text":"Then I would just do tangents of the angle of the slope,"},{"Start":"09:39.645 ","End":"09:41.375","Text":"and that\u0027s all I would need."},{"Start":"09:41.375 ","End":"09:44.495","Text":"I don\u0027t need to know the weight of the wood or anything."},{"Start":"09:44.495 ","End":"09:46.700","Text":"The only thing that interests me in these types of"},{"Start":"09:46.700 ","End":"09:50.570","Text":"questions is the angle and the coefficient of friction."},{"Start":"09:50.570 ","End":"09:54.065","Text":"This is now the end of the lesson."},{"Start":"09:54.065 ","End":"09:57.035","Text":"Let\u0027s just recap what we did in the lesson."},{"Start":"09:57.035 ","End":"10:01.360","Text":"We\u0027ve learned how to translate from"},{"Start":"10:01.360 ","End":"10:06.010","Text":"a diagram all the relevant information into a free body diagram."},{"Start":"10:06.010 ","End":"10:09.940","Text":"Then, because we know that the object is static,"},{"Start":"10:09.940 ","End":"10:14.815","Text":"there\u0027s no motion, which means that the sum of all the forces is equal to 0."},{"Start":"10:14.815 ","End":"10:17.890","Text":"Which means that we could say because it\u0027s not moving up and down,"},{"Start":"10:17.890 ","End":"10:21.010","Text":"we can say that N equals mg cos Alpha,"},{"Start":"10:21.010 ","End":"10:23.125","Text":"and because it\u0027s not moving left and right,"},{"Start":"10:23.125 ","End":"10:25.600","Text":"that F equals mg sine Alpha."},{"Start":"10:25.600 ","End":"10:28.405","Text":"Then we learned how to substitute"},{"Start":"10:28.405 ","End":"10:34.480","Text":"the different unknowns into the different equations in order to get our final equation,"},{"Start":"10:34.480 ","End":"10:38.620","Text":"which says that the minimum,"},{"Start":"10:38.620 ","End":"10:41.290","Text":"the smallest value that the coefficient of friction may"},{"Start":"10:41.290 ","End":"10:44.150","Text":"be such that the system remains at rest,"},{"Start":"10:44.150 ","End":"10:48.470","Text":"is when Mu is larger or equal to tan Theta."},{"Start":"10:48.470 ","End":"10:50.450","Text":"When we say minimum,"},{"Start":"10:50.450 ","End":"10:55.195","Text":"it means that Mu must equal tan Alpha."},{"Start":"10:55.195 ","End":"10:59.515","Text":"Let\u0027s say that tan Alpha is 10, just an example,"},{"Start":"10:59.515 ","End":"11:04.175","Text":"then that means that the minimum value that the coefficient of friction can be is 10."},{"Start":"11:04.175 ","End":"11:05.562","Text":"Of course, it can be 11,"},{"Start":"11:05.562 ","End":"11:07.370","Text":"12, 13, 14, and so on."},{"Start":"11:07.370 ","End":"11:13.480","Text":"But the minimum value so that the mass doesn\u0027t slide down the slope is 10."},{"Start":"11:13.480 ","End":"11:16.114","Text":"That\u0027s the end of the lesson."},{"Start":"11:16.114 ","End":"11:19.320","Text":"Let\u0027s move on to the rest of the questions."}],"ID":10607},{"Watched":false,"Name":"Exercise 2","Duration":"5m 2s","ChapterTopicVideoID":10278,"CourseChapterTopicPlaylistID":8962,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.535","Text":"In this question, we have mass that we\u0027ve called big M,"},{"Start":"00:05.535 ","End":"00:08.670","Text":"which is resting on a flat plane."},{"Start":"00:08.670 ","End":"00:13.320","Text":"Attached to that mass is a rope going up until here."},{"Start":"00:13.320 ","End":"00:15.915","Text":"Then it carries on up here,"},{"Start":"00:15.915 ","End":"00:20.790","Text":"and it\u0027s attached to a wall and goes down"},{"Start":"00:20.790 ","End":"00:25.470","Text":"here and it\u0027s attached to a mass that we\u0027ve called small m. Now,"},{"Start":"00:25.470 ","End":"00:28.200","Text":"what we\u0027re being asked in the question is what is"},{"Start":"00:28.200 ","End":"00:32.850","Text":"the minimum coefficient of friction between the plane and the mass?"},{"Start":"00:32.850 ","End":"00:35.910","Text":"Between the plane and mass,"},{"Start":"00:35.910 ","End":"00:42.900","Text":"big M, that will keep the system held in place."},{"Start":"00:42.900 ","End":"00:44.870","Text":"Now, in the question,"},{"Start":"00:44.870 ","End":"00:47.435","Text":"we\u0027re told that the system is held in place."},{"Start":"00:47.435 ","End":"00:51.665","Text":"Therefore, the sum of all the forces is equal to 0."},{"Start":"00:51.665 ","End":"01:00.425","Text":"Let\u0027s look at mass capital M. We have T_1 acting in the right direction,"},{"Start":"01:00.425 ","End":"01:05.210","Text":"which equals to F acting in the left direction."},{"Start":"01:05.210 ","End":"01:07.730","Text":"Both of these are unknown,"},{"Start":"01:07.730 ","End":"01:09.560","Text":"so we\u0027re putting a tilde on top."},{"Start":"01:09.560 ","End":"01:13.490","Text":"We have N acting upwards,"},{"Start":"01:13.490 ","End":"01:14.810","Text":"which is also unknown,"},{"Start":"01:14.810 ","End":"01:19.775","Text":"which is equal to Mg. Now M we\u0027re given,"},{"Start":"01:19.775 ","End":"01:24.080","Text":"and I\u0027ve put a tilde on top of the n because it\u0027s also an unknown."},{"Start":"01:24.080 ","End":"01:33.045","Text":"Now we can take a look at mass small m. Acting upwards,"},{"Start":"01:33.045 ","End":"01:36.540","Text":"we have T_2 because it\u0027s an unknown."},{"Start":"01:36.540 ","End":"01:41.675","Text":"It Tilde, which equals Mg pointing downwards."},{"Start":"01:41.675 ","End":"01:47.220","Text":"Now let\u0027s take a look at what is happening here in the cross-section."},{"Start":"01:47.710 ","End":"01:53.975","Text":"We have T_2 acting downwards, which again,"},{"Start":"01:53.975 ","End":"02:01.100","Text":"because the sum of the forces equals 0 equals the y component of T_3."},{"Start":"02:01.100 ","End":"02:03.825","Text":"Now because this is Alpha,"},{"Start":"02:03.825 ","End":"02:11.940","Text":"the y component of T_3 will be T_3 sine of Alpha."},{"Start":"02:11.940 ","End":"02:21.065","Text":"Then, so that\u0027s this section, T_3 sine Alpha."},{"Start":"02:21.065 ","End":"02:24.275","Text":"Then in this section,"},{"Start":"02:24.275 ","End":"02:28.610","Text":"we have that T_1 equals,"},{"Start":"02:28.610 ","End":"02:38.140","Text":"T_1 is unknown, equals T_3 cosine of Alpha because it\u0027s the x component."},{"Start":"02:38.140 ","End":"02:41.570","Text":"Now let\u0027s look at a few statements."},{"Start":"02:41.570 ","End":"02:43.725","Text":"The tension over here,"},{"Start":"02:43.725 ","End":"02:47.725","Text":"T_1 is equal to this T_1."},{"Start":"02:47.725 ","End":"02:51.040","Text":"Why? Because it\u0027s on the same string."},{"Start":"02:51.040 ","End":"02:53.370","Text":"Similarly, with T_2,"},{"Start":"02:53.370 ","End":"02:57.965","Text":"the tension here and the tension here equal because it\u0027s the same rope."},{"Start":"02:57.965 ","End":"03:01.610","Text":"But conversely, T_1 does not equal"},{"Start":"03:01.610 ","End":"03:05.785","Text":"T_2 and does not equal T_3 because they\u0027re all different ropes."},{"Start":"03:05.785 ","End":"03:08.510","Text":"That\u0027s why we\u0027ve given them different names."},{"Start":"03:08.510 ","End":"03:12.350","Text":"Now, I\u0027ve marked all of the unknowns with a tilde on top."},{"Start":"03:12.350 ","End":"03:15.565","Text":"We have that T_1, f,"},{"Start":"03:15.565 ","End":"03:19.050","Text":"n, T_2, and T_3 are all unknowns."},{"Start":"03:19.050 ","End":"03:21.270","Text":"Now we have 1, 2, 3, 4, 5,"},{"Start":"03:21.270 ","End":"03:23.880","Text":"we have 5 equations,"},{"Start":"03:23.880 ","End":"03:26.085","Text":"and we also have 5 unknowns."},{"Start":"03:26.085 ","End":"03:28.400","Text":"We could say that we finished the question."},{"Start":"03:28.400 ","End":"03:32.885","Text":"You\u0027ll just rearrange everything to get whatever you\u0027re trying to find, and that\u0027s it."},{"Start":"03:32.885 ","End":"03:36.170","Text":"But here we\u0027re being asked what is"},{"Start":"03:36.170 ","End":"03:38.450","Text":"the minimum coefficient of friction between"},{"Start":"03:38.450 ","End":"03:41.640","Text":"the plane and the mass that will keep the system held in place?"},{"Start":"03:41.640 ","End":"03:43.985","Text":"We have to find that value."},{"Start":"03:43.985 ","End":"03:50.180","Text":"Now in order to find the minimum value of the coefficient of friction,"},{"Start":"03:50.180 ","End":"03:52.775","Text":"sorry, I\u0027m going to have in one of the equations,"},{"Start":"03:52.775 ","End":"03:54.160","Text":"the coefficient of friction."},{"Start":"03:54.160 ","End":"03:56.205","Text":"Now, it doesn\u0027t appear anywhere."},{"Start":"03:56.205 ","End":"03:59.765","Text":"I can just add in the equation that we know,"},{"Start":"03:59.765 ","End":"04:03.485","Text":"which is friction equals Mu"},{"Start":"04:03.485 ","End":"04:08.710","Text":"multiplied by n. I\u0027m putting a tilde above the Mu because it\u0027s also an unknown."},{"Start":"04:08.710 ","End":"04:14.149","Text":"It\u0027s important to note because I\u0027m sure that you might have already realized that usually"},{"Start":"04:14.149 ","End":"04:20.420","Text":"we\u0027re used to the equation being written as smaller or equal to."},{"Start":"04:20.420 ","End":"04:22.460","Text":"Let me rub this out."},{"Start":"04:22.460 ","End":"04:26.270","Text":"Why have I allowed myself here to write equal to?"},{"Start":"04:26.270 ","End":"04:31.910","Text":"Because I\u0027m trying to find the minimum coefficient of friction,"},{"Start":"04:31.910 ","End":"04:37.080","Text":"that means that F equals Mu n. Because obviously,"},{"Start":"04:37.080 ","End":"04:41.490","Text":"if I choose the minimum value that Mu can be,"},{"Start":"04:41.490 ","End":"04:43.155","Text":"then the system will move."},{"Start":"04:43.155 ","End":"04:47.315","Text":"If I choose a larger value for the coefficient of friction,"},{"Start":"04:47.315 ","End":"04:49.625","Text":"then still the system won\u0027t move."},{"Start":"04:49.625 ","End":"04:52.115","Text":"That\u0027s why when we\u0027re trying to find the minimum,"},{"Start":"04:52.115 ","End":"04:54.680","Text":"we can use the equal sign."},{"Start":"04:54.680 ","End":"04:58.505","Text":"Now we can say that we finished the question,"},{"Start":"04:58.505 ","End":"05:03.270","Text":"and we can go on to the rest of the questions in this unit."}],"ID":10608},{"Watched":false,"Name":"Exercise 3","Duration":"13m 56s","ChapterTopicVideoID":10279,"CourseChapterTopicPlaylistID":8962,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.480 ","End":"00:03.580","Text":"Hello. In this question,"},{"Start":"00:03.580 ","End":"00:05.808","Text":"we\u0027re given 3 masses,"},{"Start":"00:05.808 ","End":"00:07.625","Text":"we have mass number 1,"},{"Start":"00:07.625 ","End":"00:09.230","Text":"mass number 2,"},{"Start":"00:09.230 ","End":"00:12.115","Text":"and an unknown mass number 3."},{"Start":"00:12.115 ","End":"00:17.515","Text":"We\u0027re also told that between mass number 1 and mass number 2,"},{"Start":"00:17.515 ","End":"00:20.090","Text":"there\u0027s a coefficient of friction,"},{"Start":"00:20.090 ","End":"00:24.915","Text":"which we\u0027ll label here as Mu."},{"Start":"00:24.915 ","End":"00:29.665","Text":"Now, we can see that mass number 2 is tied to a rope"},{"Start":"00:29.665 ","End":"00:34.300","Text":"here to the wall and mass number 1 is tied to"},{"Start":"00:34.300 ","End":"00:38.470","Text":"a rope that goes over some wheel as we can"},{"Start":"00:38.470 ","End":"00:42.830","Text":"see and then there\u0027s this unknown mass dangling from it."},{"Start":"00:42.830 ","End":"00:44.450","Text":"Now, the question is asking,"},{"Start":"00:44.450 ","End":"00:47.120","Text":"what is the maximum mass that this unknown can"},{"Start":"00:47.120 ","End":"00:51.905","Text":"be until it moves the system out of its static position."},{"Start":"00:51.905 ","End":"00:55.520","Text":"Now, obviously, the weight of mass number 2,"},{"Start":"00:55.520 ","End":"00:59.120","Text":"which is pushing down on the mass number 1,"},{"Start":"00:59.120 ","End":"01:02.615","Text":"it\u0027s also keeping the whole system in place."},{"Start":"01:02.615 ","End":"01:04.820","Text":"However, if this mass,"},{"Start":"01:04.820 ","End":"01:07.055","Text":"the unknown mass is great enough,"},{"Start":"01:07.055 ","End":"01:13.270","Text":"then it can pull mass number 1 from under number 2 and we\u0027ll get the system moving."},{"Start":"01:13.270 ","End":"01:16.745","Text":"Let\u0027s see how we go about solving this question."},{"Start":"01:16.745 ","End":"01:22.445","Text":"We\u0027re going to start by drawing 3 body diagrams of each of the masses."},{"Start":"01:22.445 ","End":"01:24.770","Text":"Here we have mass number 2,"},{"Start":"01:24.770 ","End":"01:28.040","Text":"which is the same as here, here,"},{"Start":"01:28.040 ","End":"01:32.460","Text":"mass number 1 and this unknown mass,"},{"Start":"01:32.460 ","End":"01:34.605","Text":"which we\u0027re going to call number 3,"},{"Start":"01:34.605 ","End":"01:38.500","Text":"put it on the side, is this 1."},{"Start":"01:38.960 ","End":"01:43.310","Text":"Let\u0027s first start by taking a look at mass number 1."},{"Start":"01:43.310 ","End":"01:50.030","Text":"Now, mass number 1 has the force m1g pointing"},{"Start":"01:50.030 ","End":"01:58.849","Text":"downwards and then also it has the normal force pointing upwards."},{"Start":"01:58.849 ","End":"02:01.763","Text":"Now we\u0027re going to call this normal force n1,"},{"Start":"02:01.763 ","End":"02:08.035","Text":"because mass number 2 is resting on mass number 1,"},{"Start":"02:08.035 ","End":"02:12.155","Text":"there\u0027s some other normal force pointing"},{"Start":"02:12.155 ","End":"02:20.115","Text":"downwards on the mass number 1 and we\u0027re going to call this n2."},{"Start":"02:20.115 ","End":"02:25.830","Text":"Now, I could easily have drawn n2,"},{"Start":"02:25.830 ","End":"02:29.790","Text":"where the m1g is as well, it doesn\u0027t matter."},{"Start":"02:29.840 ","End":"02:36.705","Text":"If I would have drawn n2 next to the m1g is just moving the vector downwards,"},{"Start":"02:36.705 ","End":"02:40.250","Text":"it remains the same, you remember this from previous lessons."},{"Start":"02:40.250 ","End":"02:43.448","Text":"Now, let\u0027s take a look at mass number 2,"},{"Start":"02:43.448 ","End":"02:46.535","Text":"mass number 2, of course,"},{"Start":"02:46.535 ","End":"02:54.843","Text":"has m2g pointing downwards and pointing upwards,"},{"Start":"02:54.843 ","End":"02:58.920","Text":"it has the normal n2."},{"Start":"02:58.920 ","End":"03:06.040","Text":"Now notice this n2 is equal to this n2 and why is this?"},{"Start":"03:06.040 ","End":"03:10.520","Text":"It\u0027s because, according to Newton\u0027s third law,"},{"Start":"03:10.520 ","End":"03:15.305","Text":"when 1 body exerts a force on a second body,"},{"Start":"03:15.305 ","End":"03:17.330","Text":"so for instance here,"},{"Start":"03:17.330 ","End":"03:19.280","Text":"mass number 2,"},{"Start":"03:19.280 ","End":"03:23.530","Text":"this body is exerting a force on body number 1."},{"Start":"03:23.530 ","End":"03:27.860","Text":"The second body simultaneously exerts a force equal"},{"Start":"03:27.860 ","End":"03:32.330","Text":"in magnitude and opposite in direction on the first body."},{"Start":"03:32.330 ","End":"03:40.135","Text":"That means when number 2 is exerting normal m2 onto mass number 1,"},{"Start":"03:40.135 ","End":"03:43.405","Text":"then according to Newton\u0027s third law,"},{"Start":"03:43.405 ","End":"03:47.225","Text":"n2 is being applied back up"},{"Start":"03:47.225 ","End":"03:53.525","Text":"to mass number 2 because the force is equal in magnitude and opposite in direction."},{"Start":"03:53.525 ","End":"03:57.535","Text":"Now, let\u0027s go back for 1 second to mass number 1."},{"Start":"03:57.535 ","End":"04:04.538","Text":"Now, you all notice that attached to mass number 1 is this rope going over this wheel,"},{"Start":"04:04.538 ","End":"04:09.245","Text":"which means, let\u0027s just change back colors,"},{"Start":"04:09.245 ","End":"04:15.780","Text":"that there\u0027s also tension acting in this direction."},{"Start":"04:17.980 ","End":"04:23.525","Text":"But seeing as there\u0027s this tension pulling it to the left,"},{"Start":"04:23.525 ","End":"04:29.855","Text":"there also is some friction pulling it to the right,"},{"Start":"04:29.855 ","End":"04:31.955","Text":"now, what is this friction?"},{"Start":"04:31.955 ","End":"04:41.305","Text":"We were told that between mass number 1 and this flat plane, there\u0027s no friction."},{"Start":"04:41.305 ","End":"04:45.776","Text":"The friction that we\u0027re talking about is where this Mu is written,"},{"Start":"04:45.776 ","End":"04:53.060","Text":"the friction is acting between the mass number 1 and mass number 2,"},{"Start":"04:53.060 ","End":"04:55.605","Text":"the friction comes from mass number 2."},{"Start":"04:55.605 ","End":"04:58.685","Text":"Now, in order to remember this,"},{"Start":"04:58.685 ","End":"05:00.455","Text":"I\u0027m going to write on the side,"},{"Start":"05:00.455 ","End":"05:02.420","Text":"I\u0027m going to write over here."},{"Start":"05:02.420 ","End":"05:07.760","Text":"F equals, remember from the equation is smaller or equal"},{"Start":"05:07.760 ","End":"05:14.720","Text":"to Mu N and because here the n is coming from mass number 2 N2."},{"Start":"05:14.720 ","End":"05:20.760","Text":"Now, this I\u0027m going to keep on the side and we\u0027re going to use it very soon."},{"Start":"05:23.600 ","End":"05:28.205","Text":"Now, just in case there\u0027s a slight bit of confusion why I wrote"},{"Start":"05:28.205 ","End":"05:32.480","Text":"f equals Mu N2 when in fact the equation goes,"},{"Start":"05:32.480 ","End":"05:35.885","Text":"F is smaller than or equal to Mu N2."},{"Start":"05:35.885 ","End":"05:37.475","Text":"Why did I write this?"},{"Start":"05:37.475 ","End":"05:39.875","Text":"In the question we\u0027re being asked,"},{"Start":"05:39.875 ","End":"05:45.665","Text":"what the maximum mass we can hang is."},{"Start":"05:45.665 ","End":"05:50.200","Text":"Which means that f, isn\u0027t going to be smaller than Mu N2."},{"Start":"05:50.200 ","End":"05:53.635","Text":"It means it\u0027s going to be equal to Mu N2 because that\u0027s the maximum,"},{"Start":"05:53.635 ","End":"05:58.355","Text":"that\u0027s why in this case it\u0027s okay to write f equals Mu N2."},{"Start":"05:58.355 ","End":"06:06.030","Text":"Let\u0027s just write that this is showing us maximum force."},{"Start":"06:06.030 ","End":"06:09.350","Text":"Now let\u0027s take a look at mass number 3."},{"Start":"06:09.350 ","End":"06:12.990","Text":"Now mass number 3, of course,"},{"Start":"06:12.990 ","End":"06:19.730","Text":"has m3 g pointing downwards and going in the upwards direction,"},{"Start":"06:19.730 ","End":"06:24.230","Text":"it has T. Now obviously this T"},{"Start":"06:24.230 ","End":"06:30.125","Text":"is equal to this T and why is that?"},{"Start":"06:30.125 ","End":"06:33.005","Text":"Because it\u0027s on the same rope,"},{"Start":"06:33.005 ","End":"06:35.725","Text":"this is the same rope."},{"Start":"06:35.725 ","End":"06:40.216","Text":"Now, let\u0027s start by writing down the equations,"},{"Start":"06:40.216 ","End":"06:47.511","Text":"we have that n2 equals m2g,"},{"Start":"06:47.511 ","End":"06:51.895","Text":"it\u0027s the forces for mass number 2."},{"Start":"06:51.895 ","End":"06:58.105","Text":"Then we have the t equals f from mass number"},{"Start":"06:58.105 ","End":"07:07.515","Text":"1 and we have that t equals m3g from mass number 3."},{"Start":"07:07.515 ","End":"07:10.945","Text":"Now, let\u0027s go back to mass number 1."},{"Start":"07:10.945 ","End":"07:14.294","Text":"Now we\u0027ve already spoken about how T equals f,"},{"Start":"07:14.294 ","End":"07:18.910","Text":"that was here,"},{"Start":"07:18.910 ","End":"07:23.115","Text":"now, let\u0027s look at the rest of the forces."},{"Start":"07:23.115 ","End":"07:28.140","Text":"We have that n1,"},{"Start":"07:28.140 ","End":"07:32.390","Text":"the only force pointing upwards is equal to"},{"Start":"07:32.390 ","End":"07:36.920","Text":"all the forces pointing downwards because it\u0027s a static system."},{"Start":"07:36.920 ","End":"07:41.520","Text":"N1 equals and pointing downwards is"},{"Start":"07:41.520 ","End":"07:49.090","Text":"n2 and also m1g, N2 plus m1g."},{"Start":"07:49.090 ","End":"07:53.540","Text":"Now, it\u0027s important to note that m2 and m1g are pointing"},{"Start":"07:53.540 ","End":"07:58.460","Text":"downwards and N1 is pointing upwards and that\u0027s why we did it like this,"},{"Start":"07:58.460 ","End":"08:02.005","Text":"the N1 equals N2 plus m1g,"},{"Start":"08:02.005 ","End":"08:05.650","Text":"m3 is, this is our unknown."},{"Start":"08:06.080 ","End":"08:09.830","Text":"Soon we\u0027re going to rearrange all of the equations in"},{"Start":"08:09.830 ","End":"08:13.570","Text":"order to find out what m3 actually is."},{"Start":"08:13.570 ","End":"08:18.214","Text":"Now, before we begin rearranging and subbing in all of the equations,"},{"Start":"08:18.214 ","End":"08:24.110","Text":"we have to check that we have an equal amount of equations as the number of unknowns."},{"Start":"08:24.110 ","End":"08:26.790","Text":"Now, the unknowns that we have,"},{"Start":"08:26.790 ","End":"08:29.794","Text":"so again change color to make this clearer,"},{"Start":"08:29.794 ","End":"08:33.910","Text":"we have N2 is our first unknown,"},{"Start":"08:33.910 ","End":"08:36.780","Text":"T is our second unknown,"},{"Start":"08:36.780 ","End":"08:42.135","Text":"f is our third unknown m3 is our fourth unknown,"},{"Start":"08:42.135 ","End":"08:48.116","Text":"and N1 is our fifth unknown and we have here 1,"},{"Start":"08:48.116 ","End":"08:54.510","Text":"2, 3, 4 and 5 equations."},{"Start":"08:54.510 ","End":"08:58.600","Text":"Perfect, now we can start."},{"Start":"08:58.900 ","End":"09:03.350","Text":"Something that it\u0027s important to note is that when we\u0027re"},{"Start":"09:03.350 ","End":"09:08.420","Text":"writing our final answer for what m3 is equal to,"},{"Start":"09:08.420 ","End":"09:12.950","Text":"we can include in that answer things that we don\u0027t know"},{"Start":"09:12.950 ","End":"09:17.915","Text":"their values such as T and 2 and 1 and so on and so forth."},{"Start":"09:17.915 ","End":"09:23.000","Text":"We can only include things like g and m1,"},{"Start":"09:23.000 ","End":"09:25.410","Text":"which we know the values for."},{"Start":"09:25.780 ","End":"09:29.350","Text":"Let\u0027s begin the process of the working out."},{"Start":"09:29.350 ","End":"09:33.470","Text":"Now, we\u0027re going to start with this equation,"},{"Start":"09:33.470 ","End":"09:37.115","Text":"because was trying to find out what m3 is."},{"Start":"09:37.115 ","End":"09:43.730","Text":"M3g equals t and then we can see that t equals"},{"Start":"09:43.730 ","End":"09:52.460","Text":"f. We can write down here f and then we put a tick here because we\u0027ve used this equation."},{"Start":"09:52.460 ","End":"09:57.275","Text":"Now, we know that f equals"},{"Start":"09:57.275 ","End":"10:06.630","Text":"mu N2 from the equation in the top right corner,"},{"Start":"10:06.630 ","End":"10:08.405","Text":"now we can put a tick here too,"},{"Start":"10:08.405 ","End":"10:10.145","Text":"because we\u0027ve used it."},{"Start":"10:10.145 ","End":"10:18.980","Text":"Then we know that N2 equals m2g equals"},{"Start":"10:18.980 ","End":"10:28.205","Text":"Mu multiplied by m2g and then we put a tick here as well because we\u0027ve just used that."},{"Start":"10:28.205 ","End":"10:32.240","Text":"Now, we can see that in this equation we have Mu,"},{"Start":"10:32.240 ","End":"10:33.770","Text":"which we\u0027re given,"},{"Start":"10:33.770 ","End":"10:35.960","Text":"we have m2 that we\u0027re given,"},{"Start":"10:35.960 ","End":"10:37.535","Text":"and g, we already know that."},{"Start":"10:37.535 ","End":"10:39.665","Text":"Here we\u0027re actually finished, now,"},{"Start":"10:39.665 ","End":"10:42.244","Text":"we can cross out the g\u0027s,"},{"Start":"10:42.244 ","End":"10:45.020","Text":"cancel it out to make it easier."},{"Start":"10:45.020 ","End":"10:55.186","Text":"Now our final answer is m3 equals Mu m2 and that\u0027s it,"},{"Start":"10:55.186 ","End":"10:56.825","Text":"that\u0027s how you work it out."},{"Start":"10:56.825 ","End":"11:00.680","Text":"Now, just a short note on what this actually means,"},{"Start":"11:00.680 ","End":"11:10.295","Text":"this means that the maximum mass that m3 can weigh is the coefficient of friction,"},{"Start":"11:10.295 ","End":"11:15.040","Text":"which is Mu, multiplied by the mass of mass number 2,"},{"Start":"11:15.040 ","End":"11:17.310","Text":"the heavier mass number 2 is,"},{"Start":"11:17.310 ","End":"11:20.190","Text":"the heavier mass number 3 can be."},{"Start":"11:20.190 ","End":"11:22.970","Text":"As for a little bit of intuition with this,"},{"Start":"11:22.970 ","End":"11:25.430","Text":"if we have a really,"},{"Start":"11:25.430 ","End":"11:28.190","Text":"really heavy mass number 2,"},{"Start":"11:28.190 ","End":"11:31.520","Text":"it\u0027s going to push down more in mass number 1,"},{"Start":"11:31.520 ","End":"11:40.333","Text":"which means that in order to move this mass number 3 is going to have to be heavier,"},{"Start":"11:40.333 ","End":"11:42.454","Text":"it really makes sense."},{"Start":"11:42.454 ","End":"11:47.387","Text":"Also, we can see that Mu is in the equation,"},{"Start":"11:47.387 ","End":"11:51.050","Text":"now, Mu, which is over here,"},{"Start":"11:51.050 ","End":"11:53.975","Text":"if there\u0027s a larger coefficient of friction,"},{"Start":"11:53.975 ","End":"12:01.760","Text":"that means it will be harder to cause mass number 1 to slide from under mass number 2."},{"Start":"12:01.760 ","End":"12:04.384","Text":"Which means that again,"},{"Start":"12:04.384 ","End":"12:07.506","Text":"mass number 3 could be heavier,"},{"Start":"12:07.506 ","End":"12:10.385","Text":"it\u0027s very, very logical of this."},{"Start":"12:10.385 ","End":"12:12.320","Text":"Now, what you\u0027ll notice is"},{"Start":"12:12.320 ","End":"12:18.001","Text":"that mass number 1 doesn\u0027t even factor in into the final equation,"},{"Start":"12:18.001 ","End":"12:24.410","Text":"why is it that mass number 1 doesn\u0027t play a factor in the final equation?"},{"Start":"12:24.410 ","End":"12:27.515","Text":"Because if you remember in this question,"},{"Start":"12:27.515 ","End":"12:30.515","Text":"the plane is smooth,"},{"Start":"12:30.515 ","End":"12:35.420","Text":"meaning that there\u0027s no friction between mass number 1 and the plane,"},{"Start":"12:35.420 ","End":"12:40.505","Text":"which means that it doesn\u0027t really matter how heavy or how light mass number 1 is,"},{"Start":"12:40.505 ","End":"12:42.230","Text":"because there\u0027s no friction there,"},{"Start":"12:42.230 ","End":"12:44.223","Text":"so it doesn\u0027t make a difference,"},{"Start":"12:44.223 ","End":"12:47.180","Text":"it\u0027s just not playing in the game."},{"Start":"12:47.180 ","End":"12:53.062","Text":"The force of mass number 1 is acting downwards,"},{"Start":"12:53.062 ","End":"12:55.580","Text":"but because there\u0027s no friction over here,"},{"Start":"12:55.580 ","End":"12:57.485","Text":"it\u0027s acting on nothing."},{"Start":"12:57.485 ","End":"13:03.055","Text":"There\u0027s just nothing going on here and that\u0027s why it\u0027s just not in the equation."},{"Start":"13:03.055 ","End":"13:06.721","Text":"Now we\u0027ve gotten to the end of the lesson,"},{"Start":"13:06.721 ","End":"13:09.035","Text":"just to sum up,"},{"Start":"13:09.035 ","End":"13:16.980","Text":"it\u0027s very important that from this lesson you recognize why you could use this equation,"},{"Start":"13:16.980 ","End":"13:22.790","Text":"and that\u0027s because in the question we were being asked for, the maximum mass,"},{"Start":"13:22.790 ","End":"13:26.272","Text":"which means also the maximum friction,"},{"Start":"13:26.272 ","End":"13:28.745","Text":"after looking at the free body diagrams."},{"Start":"13:28.745 ","End":"13:32.615","Text":"If we weren\u0027t being asked what the maximum mass could be,"},{"Start":"13:32.615 ","End":"13:40.520","Text":"then this equation would have to be f is smaller than or equal to Mu N2,"},{"Start":"13:40.520 ","End":"13:43.520","Text":"in which case we couldn\u0027t have used this equation at all,"},{"Start":"13:43.520 ","End":"13:49.110","Text":"so you have to understand where you can use this equation and where you can\u0027t."},{"Start":"13:49.610 ","End":"13:55.950","Text":"Let\u0027s move on to the rest of the examples and the rest of the questions."}],"ID":10609}],"Thumbnail":null,"ID":8962},{"Name":"The Kinetic Friction","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Kinetic Friction - Lesson","Duration":"20m 42s","ChapterTopicVideoID":10423,"CourseChapterTopicPlaylistID":8961,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.505","Text":"Hello. A few lessons ago,"},{"Start":"00:02.505 ","End":"00:04.770","Text":"we learned about static friction."},{"Start":"00:04.770 ","End":"00:08.280","Text":"In this lesson, we\u0027re going to be learning about kinetic friction."},{"Start":"00:08.280 ","End":"00:09.840","Text":"Just as a reminder,"},{"Start":"00:09.840 ","End":"00:17.590","Text":"kinetic friction is when there is some type of relative motion between 2 or more bodies."},{"Start":"00:18.320 ","End":"00:23.055","Text":"Kinetic friction is a constant value,"},{"Start":"00:23.055 ","End":"00:26.160","Text":"which means that it\u0027s independent of the size of"},{"Start":"00:26.160 ","End":"00:29.385","Text":"the relative velocity between the bodies in contact."},{"Start":"00:29.385 ","End":"00:36.360","Text":"That means if you\u0027re pushing some crate along at 1 mile per hour or at 10 miles per hour,"},{"Start":"00:36.360 ","End":"00:41.140","Text":"the value for the kinetic friction will be exactly the same."},{"Start":"00:41.140 ","End":"00:43.925","Text":"Also, unlike static friction,"},{"Start":"00:43.925 ","End":"00:49.770","Text":"kinetic friction can act even if there are no other forces acting on the body."},{"Start":"00:51.140 ","End":"00:57.170","Text":"Here we can see that this is the equation for our kinetic friction f,"},{"Start":"00:57.170 ","End":"01:00.080","Text":"lowercase f denotes always frictional force,"},{"Start":"01:00.080 ","End":"01:04.235","Text":"k here denotes that we\u0027re talking about kinetic friction."},{"Start":"01:04.235 ","End":"01:07.850","Text":"Then we see that that is equal to Mu_k,"},{"Start":"01:07.850 ","End":"01:11.365","Text":"which is simply a coefficient of kinetic friction."},{"Start":"01:11.365 ","End":"01:16.040","Text":"Which is some constant value dependent on the materials that"},{"Start":"01:16.040 ","End":"01:22.690","Text":"the bodies are made out of multiplied by N which is the normal force."},{"Start":"01:22.690 ","End":"01:25.640","Text":"Here we can see that in this equation,"},{"Start":"01:25.640 ","End":"01:27.580","Text":"we have an equal sign."},{"Start":"01:27.580 ","End":"01:29.375","Text":"In the static friction,"},{"Start":"01:29.375 ","End":"01:36.185","Text":"we had that our F_s was smaller or equal to Mu_s.N."},{"Start":"01:36.185 ","End":"01:38.720","Text":"Here we can see that it\u0027s equal to."},{"Start":"01:38.720 ","End":"01:43.550","Text":"That\u0027s why kinetic friction sometimes a lot easier to calculate."},{"Start":"01:43.550 ","End":"01:45.860","Text":"All we have to do is we have to know what"},{"Start":"01:45.860 ","End":"01:49.130","Text":"the normal force is acting between the 2 bodies,"},{"Start":"01:49.130 ","End":"01:54.320","Text":"which is relatively easy to know multiplied by some coefficient of friction,"},{"Start":"01:54.320 ","End":"01:57.770","Text":"which we can find on the Internet, and in textbooks."},{"Start":"01:57.770 ","End":"02:01.790","Text":"Then we know our value for our kinetic friction."},{"Start":"02:01.790 ","End":"02:03.020","Text":"In static friction,"},{"Start":"02:03.020 ","End":"02:07.640","Text":"we always had to find out which other forces were acting on the body,"},{"Start":"02:07.640 ","End":"02:12.770","Text":"and then we can know what our value for static friction was."},{"Start":"02:12.770 ","End":"02:15.315","Text":"Here it\u0027s a lot more straightforward,"},{"Start":"02:15.315 ","End":"02:20.585","Text":"and that\u0027s an explanation of what this sentence over here means."},{"Start":"02:20.585 ","End":"02:23.810","Text":"We don\u0027t need to know what other forces are"},{"Start":"02:23.810 ","End":"02:27.230","Text":"acting on the body in order to work out this equation."},{"Start":"02:27.230 ","End":"02:29.795","Text":"Now, just like in static friction,"},{"Start":"02:29.795 ","End":"02:33.770","Text":"the direction of our kinetic friction is always going to"},{"Start":"02:33.770 ","End":"02:38.820","Text":"be in the opposite direction to the direction of motion."},{"Start":"02:40.100 ","End":"02:45.920","Text":"In most cases, the size of the coefficient of static friction,"},{"Start":"02:45.920 ","End":"02:50.510","Text":"that\u0027s our Mu_s will be larger than"},{"Start":"02:50.510 ","End":"02:56.370","Text":"the coefficient of our kinetic friction, which is Mu_k."},{"Start":"02:56.370 ","End":"02:59.270","Text":"This is in most cases, of course,"},{"Start":"02:59.270 ","End":"03:01.940","Text":"there are some special cases where"},{"Start":"03:01.940 ","End":"03:06.410","Text":"our Mu_s will be equal to our Mu_k or even smaller than."},{"Start":"03:06.410 ","End":"03:10.505","Text":"But we\u0027re not going to speak about these cases in this course."},{"Start":"03:10.505 ","End":"03:18.420","Text":"Just know that we\u0027re going to be dealing with this where Mu_s is bigger than Mu_k."},{"Start":"03:19.490 ","End":"03:22.630","Text":"Now let\u0027s look at what this means."},{"Start":"03:22.630 ","End":"03:25.720","Text":"That our coefficient of static friction"},{"Start":"03:25.720 ","End":"03:29.915","Text":"is larger than our coefficient of kinetic friction."},{"Start":"03:29.915 ","End":"03:32.425","Text":"That means that if,"},{"Start":"03:32.425 ","End":"03:35.830","Text":"let\u0027s say we have our ground over here,"},{"Start":"03:35.830 ","End":"03:40.355","Text":"and we have some crates that we\u0027re trying to push."},{"Start":"03:40.355 ","End":"03:43.585","Text":"What this means is that at the beginning,"},{"Start":"03:43.585 ","End":"03:46.600","Text":"when the crate is stationary on the ground,"},{"Start":"03:46.600 ","End":"03:50.690","Text":"we\u0027re going to have to apply some force."},{"Start":"03:50.690 ","End":"03:53.860","Text":"As we learned in the previous lesson on"},{"Start":"03:53.860 ","End":"04:00.100","Text":"static friction we\u0027re going to have to apply a force greater"},{"Start":"04:00.100 ","End":"04:03.350","Text":"than or equal to this value of"},{"Start":"04:03.350 ","End":"04:10.255","Text":"the coefficient of static friction multiplied by the normal force acting on the box."},{"Start":"04:10.255 ","End":"04:13.830","Text":"Depending on the mass of the box."},{"Start":"04:13.830 ","End":"04:24.580","Text":"We\u0027re going to have to apply a force greater than that value."},{"Start":"04:25.280 ","End":"04:30.015","Text":"Then once we apply a force greater than that value,"},{"Start":"04:30.015 ","End":"04:33.255","Text":"our crate is going to start moving."},{"Start":"04:33.255 ","End":"04:35.660","Text":"Then we\u0027ll see that if we want to"},{"Start":"04:35.660 ","End":"04:39.620","Text":"continue the motion of the crates in whichever direction,"},{"Start":"04:39.620 ","End":"04:41.570","Text":"let\u0027s say in the rightwards direction,"},{"Start":"04:41.570 ","End":"04:46.760","Text":"our force that will have to apply to the crate will have to be less"},{"Start":"04:46.760 ","End":"04:52.220","Text":"because our value for kinetic friction because now if our crate is moving,"},{"Start":"04:52.220 ","End":"04:56.105","Text":"that means that we\u0027re in the realm of kinetic friction."},{"Start":"04:56.105 ","End":"04:59.780","Text":"The force that we\u0027re going to have to apply in order to keep the crate"},{"Start":"04:59.780 ","End":"05:03.110","Text":"moving is going to be less than the force that we had"},{"Start":"05:03.110 ","End":"05:07.550","Text":"to apply in order to get the crate moving in the first place because"},{"Start":"05:07.550 ","End":"05:12.990","Text":"our value for Mu_k is smaller than our value for Mu_s."},{"Start":"05:14.410 ","End":"05:19.340","Text":"This difference in the size of our coefficients of"},{"Start":"05:19.340 ","End":"05:23.960","Text":"friction means that there\u0027s going to be some jump."},{"Start":"05:23.960 ","End":"05:26.750","Text":"We apply a force greater and greater and greater,"},{"Start":"05:26.750 ","End":"05:29.874","Text":"and yet a crate remains static."},{"Start":"05:29.874 ","End":"05:35.940","Text":"Then suddenly we reach some value of force and there\u0027s some jump, some change."},{"Start":"05:35.940 ","End":"05:42.520","Text":"Suddenly our crate starts moving and the force that we\u0027re applying can reduce."},{"Start":"05:42.520 ","End":"05:48.120","Text":"Now let\u0027s see how we can find this jump, this change."},{"Start":"05:49.180 ","End":"05:52.505","Text":"Let\u0027s see this in an example."},{"Start":"05:52.505 ","End":"05:59.030","Text":"We\u0027re being told that a body of mass M is equal to 5 kilograms is placed on a surface."},{"Start":"05:59.030 ","End":"06:05.770","Text":"We\u0027re being told that a time-dependent force F_1 is applied."},{"Start":"06:05.770 ","End":"06:09.840","Text":"Let\u0027s imagine that F_1 is some pushing for us."},{"Start":"06:09.840 ","End":"06:12.460","Text":"We\u0027re pushing this body."},{"Start":"06:12.470 ","End":"06:21.530","Text":"F_1 is given by being equal to 3 multiplied by t. That means when our time is equal to 0,"},{"Start":"06:21.530 ","End":"06:23.359","Text":"no force is being applied,"},{"Start":"06:23.359 ","End":"06:30.690","Text":"and as our time increases so our force F_1 also increases."},{"Start":"06:30.690 ","End":"06:35.205","Text":"The coefficients of friction between the body and the surface I."},{"Start":"06:35.205 ","End":"06:41.040","Text":"Our Mu_s, our coefficient of static friction is equal to 0.78,"},{"Start":"06:41.040 ","End":"06:46.640","Text":"and our Mu_k, our coefficient of kinetic friction is equal to 0.42."},{"Start":"06:46.640 ","End":"06:51.980","Text":"As per usual, we can see that our Mu_s is bigger than our Mu_k."},{"Start":"06:51.980 ","End":"06:54.680","Text":"We\u0027re being asked to draw a graph,"},{"Start":"06:54.680 ","End":"06:58.865","Text":"a frictional force as a function of time."},{"Start":"06:58.865 ","End":"07:04.540","Text":"Let\u0027s see, let\u0027s take a look at what we think is going to happen."},{"Start":"07:04.540 ","End":"07:09.750","Text":"What I would expect to see is that we\u0027ll have a body,"},{"Start":"07:09.750 ","End":"07:12.600","Text":"and when the time is small,"},{"Start":"07:12.600 ","End":"07:17.105","Text":"our F_1 will be below a certain value,"},{"Start":"07:17.105 ","End":"07:20.230","Text":"such that we\u0027ll be applying our force F_1,"},{"Start":"07:20.230 ","End":"07:22.580","Text":"and it won\u0027t move,"},{"Start":"07:22.580 ","End":"07:25.355","Text":"it will still be in the realm of static friction."},{"Start":"07:25.355 ","End":"07:27.335","Text":"Then as the time increases,"},{"Start":"07:27.335 ","End":"07:33.650","Text":"our F_1 will eventually increase above our maximum value for static friction,"},{"Start":"07:33.650 ","End":"07:37.610","Text":"and then suddenly our crate will begin"},{"Start":"07:37.610 ","End":"07:41.900","Text":"to move and then we\u0027ll be in the realm of kinetic friction,"},{"Start":"07:41.900 ","End":"07:48.475","Text":"which means that our kinetic frictional force will be much lower."},{"Start":"07:48.475 ","End":"07:52.850","Text":"We will see a decrease in our frictional force."},{"Start":"07:52.850 ","End":"07:55.820","Text":"Now let\u0027s take a look at this."},{"Start":"07:55.820 ","End":"07:59.980","Text":"Let\u0027s draw a free-body diagram."},{"Start":"07:59.980 ","End":"08:02.655","Text":"Here we have a surface,"},{"Start":"08:02.655 ","End":"08:05.025","Text":"here we have our crate,"},{"Start":"08:05.025 ","End":"08:07.530","Text":"which weighs 5 kilograms,"},{"Start":"08:07.530 ","End":"08:11.210","Text":"and then we have going in this direction,"},{"Start":"08:11.210 ","End":"08:15.800","Text":"our normal force, going in this direction, our weights,"},{"Start":"08:15.800 ","End":"08:24.330","Text":"so that\u0027s mg. Then we can say that our force F_1 is in this direction."},{"Start":"08:24.330 ","End":"08:28.955","Text":"We\u0027re pulling or pushing in the right direction which means that"},{"Start":"08:28.955 ","End":"08:31.580","Text":"our frictional force is always going to be"},{"Start":"08:31.580 ","End":"08:34.475","Text":"in the opposite direction to the direction of motion."},{"Start":"08:34.475 ","End":"08:38.350","Text":"It\u0027s going to be pointing leftwards."},{"Start":"08:39.020 ","End":"08:41.975","Text":"Here we have a frictional force."},{"Start":"08:41.975 ","End":"08:46.880","Text":"Soon we\u0027ll go into if it\u0027s static or kinetic. Let\u0027s just leave it."},{"Start":"08:46.880 ","End":"08:55.230","Text":"Now let\u0027s write out our force equations in the y-direction. Let\u0027s write this out."},{"Start":"08:55.230 ","End":"09:01.460","Text":"We have that the sum of all of our forces in the y-direction,"},{"Start":"09:01.460 ","End":"09:04.745","Text":"we can see in the positive y-direction,"},{"Start":"09:04.745 ","End":"09:06.920","Text":"we have our normal force,"},{"Start":"09:06.920 ","End":"09:08.570","Text":"and in the negative y-direction,"},{"Start":"09:08.570 ","End":"09:15.350","Text":"we have w. Negative w and this is of course equal to 0."},{"Start":"09:15.350 ","End":"09:20.225","Text":"Why is it equal to 0? We have no acceleration in the y-direction whatsoever,"},{"Start":"09:20.225 ","End":"09:21.740","Text":"we\u0027re stationary in y,"},{"Start":"09:21.740 ","End":"09:27.875","Text":"which means that the sum of all of the forces there is also equal to 0."},{"Start":"09:27.875 ","End":"09:35.570","Text":"If this is equal to 0, we can see that our normal force is equal to our weight,"},{"Start":"09:35.570 ","End":"09:37.115","Text":"which as we know,"},{"Start":"09:37.115 ","End":"09:42.480","Text":"weight is equal to mass times gravity."},{"Start":"09:42.620 ","End":"09:52.434","Text":"Now we can write in our equation for our maximum value for static friction."},{"Start":"09:52.434 ","End":"09:54.860","Text":"If we remember from a few lessons ago,"},{"Start":"09:54.860 ","End":"09:57.030","Text":"that equals to Mu_s,"},{"Start":"09:57.030 ","End":"10:03.065","Text":"the coefficient of static friction multiplied by our normal force,"},{"Start":"10:03.065 ","End":"10:13.955","Text":"which over here is mg. We have this is equal to Mu_s multiplied by mg."},{"Start":"10:13.955 ","End":"10:22.570","Text":"This is our maximum value for our static friction and any value above this value,"},{"Start":"10:22.570 ","End":"10:25.390","Text":"which we know, we can just plug in our Mu_s,"},{"Start":"10:25.390 ","End":"10:28.315","Text":"which we have over here, multiplied by our M_g."},{"Start":"10:28.315 ","End":"10:36.100","Text":"M is 5 kilograms and g is 9.81 meters per second squared."},{"Start":"10:36.100 ","End":"10:39.640","Text":"We can figure out what this value is."},{"Start":"10:39.640 ","End":"10:43.945","Text":"What we can see is that we\u0027re applying this force F_1,"},{"Start":"10:43.945 ","End":"10:46.337","Text":"which is increasing with time,"},{"Start":"10:46.337 ","End":"10:48.640","Text":"which means that our f over here,"},{"Start":"10:48.640 ","End":"10:52.315","Text":"our frictional force, is also increasing in time."},{"Start":"10:52.315 ","End":"10:57.415","Text":"It\u0027s going to be a static frictional force F_s until"},{"Start":"10:57.415 ","End":"11:05.420","Text":"our F_1 is greater than this value over here."},{"Start":"11:05.460 ","End":"11:11.005","Text":"When F_1 is greater than F_s max,"},{"Start":"11:11.005 ","End":"11:14.650","Text":"we will begin to move our crates,"},{"Start":"11:14.650 ","End":"11:17.249","Text":"our body over here will begin to move,"},{"Start":"11:17.249 ","End":"11:21.950","Text":"and then we will enter the realm of kinetic friction."},{"Start":"11:22.950 ","End":"11:30.040","Text":"Now let\u0027s see what the sum of all of the forces are on the x-axis."},{"Start":"11:30.040 ","End":"11:37.945","Text":"We can see that on the x-axis we have our F_1 and our static friction."},{"Start":"11:37.945 ","End":"11:42.310","Text":"We can say that this is the positive x-direction."},{"Start":"11:42.310 ","End":"11:48.715","Text":"That means that it equals to F_1 minus F_s."},{"Start":"11:48.715 ","End":"11:51.460","Text":"This is equal to what value?"},{"Start":"11:51.460 ","End":"11:54.803","Text":"Because right now we\u0027re using F_s,"},{"Start":"11:54.803 ","End":"11:58.254","Text":"that means that we\u0027re using static friction."},{"Start":"11:58.254 ","End":"12:00.400","Text":"Static friction, of course,"},{"Start":"12:00.400 ","End":"12:04.270","Text":"we know only exists when an object is static,"},{"Start":"12:04.270 ","End":"12:05.950","Text":"i.e. it\u0027s not moving."},{"Start":"12:05.950 ","End":"12:07.764","Text":"If it\u0027s not moving,"},{"Start":"12:07.764 ","End":"12:12.150","Text":"then that also means that obviously our acceleration is equal to 0,"},{"Start":"12:12.150 ","End":"12:19.840","Text":"which means that the sum of all of the forces in this direction is also equal to 0."},{"Start":"12:19.840 ","End":"12:27.250","Text":"Then we can say therefore that our F_1 is equal to our F_s."},{"Start":"12:27.250 ","End":"12:32.575","Text":"As we know, our F_1 given by this equation over here, so F_1,"},{"Start":"12:32.575 ","End":"12:34.045","Text":"which is equal to F_s,"},{"Start":"12:34.045 ","End":"12:39.470","Text":"is also equal to 3t given in the question."},{"Start":"12:39.770 ","End":"12:43.020","Text":"This is how we find our static friction."},{"Start":"12:43.020 ","End":"12:47.348","Text":"Notice this has nothing to do with our maximum value for our static friction,"},{"Start":"12:47.348 ","End":"12:50.135","Text":"this is just our static friction at any given time."},{"Start":"12:50.135 ","End":"12:55.150","Text":"We said that it simply is equal to the force which is doing the pushing"},{"Start":"12:55.150 ","End":"13:01.220","Text":"or the pulling on the object whilst the object is still stationary."},{"Start":"13:01.620 ","End":"13:08.785","Text":"Now the next thing that we want to do is we want to find the time."},{"Start":"13:08.785 ","End":"13:15.233","Text":"We want to find the time for which our F_1,"},{"Start":"13:15.233 ","End":"13:18.895","Text":"so that\u0027s 3t, the time,"},{"Start":"13:18.895 ","End":"13:25.670","Text":"such that F_1 is equal to F_s max."},{"Start":"13:28.020 ","End":"13:30.940","Text":"Once we find this t,"},{"Start":"13:30.940 ","End":"13:36.204","Text":"so we\u0027ll know an F_1 is equal to F_s max and then we\u0027ll know the time"},{"Start":"13:36.204 ","End":"13:42.980","Text":"when we enter the realm of F_k, kinetic friction."},{"Start":"13:44.070 ","End":"13:52.419","Text":"Now I\u0027m going to look for this value of time when we enter the realm of kinetic friction."},{"Start":"13:52.419 ","End":"13:55.855","Text":"I\u0027m going to carry this on over here."},{"Start":"13:55.855 ","End":"14:01.540","Text":"What I will say is that my F_s max has to equal to F_1."},{"Start":"14:01.540 ","End":"14:07.150","Text":"I know that my F_s max from over here is equal to the coefficient for"},{"Start":"14:07.150 ","End":"14:15.173","Text":"static friction multiplied by mg and I want to find when this is equal to F_1,"},{"Start":"14:15.173 ","End":"14:18.860","Text":"when this is equal to 3t."},{"Start":"14:19.500 ","End":"14:23.110","Text":"Now, because I\u0027m trying to find what t is,"},{"Start":"14:23.110 ","End":"14:29.305","Text":"so I\u0027m going to isolate out my t so that I can say that t is equal to,"},{"Start":"14:29.305 ","End":"14:34.810","Text":"so I\u0027ll divide both sides by 3 in order to isolate out my t. I get that t is equal"},{"Start":"14:34.810 ","End":"14:41.695","Text":"to Mu_s multiplied by M_g divided by 3."},{"Start":"14:41.695 ","End":"14:44.619","Text":"Now we can plug in all of our values."},{"Start":"14:44.619 ","End":"14:52.935","Text":"We know that our M_u s is equal to 0.78 multiplied by M_g."},{"Start":"14:52.935 ","End":"14:55.545","Text":"Our M is 5 kilograms,"},{"Start":"14:55.545 ","End":"14:57.264","Text":"multiplied by 5,"},{"Start":"14:57.264 ","End":"15:03.415","Text":"and then multiply it by g. We know that our g is equal to,"},{"Start":"15:03.415 ","End":"15:06.190","Text":"some people say that it\u0027s equal to 9.8,"},{"Start":"15:06.190 ","End":"15:09.760","Text":"some 9.81 depending on your level of accuracy."},{"Start":"15:09.760 ","End":"15:16.285","Text":"Let\u0027s make it 9.8 meters per second squared just for the sake of simplifying this."},{"Start":"15:16.285 ","End":"15:19.015","Text":"Then we divide all of this by 3."},{"Start":"15:19.015 ","End":"15:22.059","Text":"Then if we plug this into a calculator,"},{"Start":"15:22.059 ","End":"15:31.150","Text":"we\u0027ll get that our value for t is equal to 12.74 seconds."},{"Start":"15:32.360 ","End":"15:39.895","Text":"What does that mean? That means that the moment that our value for t is"},{"Start":"15:39.895 ","End":"15:47.207","Text":"greater than 12.74 seconds,"},{"Start":"15:47.207 ","End":"15:53.260","Text":"we will be in the realm of kinetic friction and then we will"},{"Start":"15:53.260 ","End":"16:00.040","Text":"have to use this coefficient of friction because we\u0027re using kinetic friction,"},{"Start":"16:00.040 ","End":"16:03.230","Text":"which is equal to 0.42."},{"Start":"16:04.170 ","End":"16:08.395","Text":"Now for our equations for kinetic friction,"},{"Start":"16:08.395 ","End":"16:09.610","Text":"I\u0027m going to scroll a bit down,"},{"Start":"16:09.610 ","End":"16:12.685","Text":"so it\u0027s going to cover up a bit of the question."},{"Start":"16:12.685 ","End":"16:19.045","Text":"I know that my equation for kinetic friction is always equal to"},{"Start":"16:19.045 ","End":"16:25.555","Text":"the coefficient of kinetic friction multiplied by my normal force."},{"Start":"16:25.555 ","End":"16:30.670","Text":"It\u0027s not smaller or equal to like what we saw in our static friction,"},{"Start":"16:30.670 ","End":"16:35.155","Text":"it\u0027s always equal to and it\u0027s a constant value, it\u0027s never changing."},{"Start":"16:35.155 ","End":"16:39.730","Text":"Now we already know that our n is equal to"},{"Start":"16:39.730 ","End":"16:44.316","Text":"M_g because the sum of the forces in the y-direction,"},{"Start":"16:44.316 ","End":"16:46.135","Text":"the equation doesn\u0027t change."},{"Start":"16:46.135 ","End":"16:48.115","Text":"We\u0027re not moving up or down."},{"Start":"16:48.115 ","End":"16:54.680","Text":"That means that our normal force is still going to be equal to M_g."},{"Start":"16:55.020 ","End":"17:02.740","Text":"This is equal to our coefficient of kinetic friction multiplied by M_g."},{"Start":"17:02.740 ","End":"17:04.570","Text":"Now let\u0027s plug in our values."},{"Start":"17:04.570 ","End":"17:09.370","Text":"Our Mu_k is equal to 0.42,"},{"Start":"17:09.370 ","End":"17:15.626","Text":"and then multiplied by our mass which is covered right now but it was 5 kilograms,"},{"Start":"17:15.626 ","End":"17:19.165","Text":"and then multiplied by our gravitational force,"},{"Start":"17:19.165 ","End":"17:23.680","Text":"which we\u0027ll again write as 9.8 meters per second squared."},{"Start":"17:23.680 ","End":"17:26.845","Text":"Once we plug this into a calculator,"},{"Start":"17:26.845 ","End":"17:31.250","Text":"we\u0027ll get that the answer is equal to 23.52."},{"Start":"17:31.920 ","End":"17:38.290","Text":"Notice that right now we\u0027re finding the units for our kinetic friction and as we know,"},{"Start":"17:38.290 ","End":"17:39.610","Text":"friction is a force,"},{"Start":"17:39.610 ","End":"17:44.350","Text":"so the units are going to be N newtons."},{"Start":"17:44.350 ","End":"17:46.450","Text":"Not N normal force,"},{"Start":"17:46.450 ","End":"17:50.090","Text":"this is Newtons, so I\u0027ll actually write Newtons."},{"Start":"17:53.160 ","End":"17:59.047","Text":"What\u0027s important to note about our frictional force in the kinetic region,"},{"Start":"17:59.047 ","End":"18:04.120","Text":"our kinetic frictional force is that this value for kinetic frictional is"},{"Start":"18:04.120 ","End":"18:10.780","Text":"only brought to life"},{"Start":"18:10.780 ","End":"18:17.500","Text":"at a value where t is greater than 12.74 seconds."},{"Start":"18:17.500 ","End":"18:20.665","Text":"When t is equal to 0,"},{"Start":"18:20.665 ","End":"18:24.535","Text":"we\u0027re not using this value."},{"Start":"18:24.535 ","End":"18:27.040","Text":"That\u0027s what\u0027s important to note,"},{"Start":"18:27.040 ","End":"18:30.385","Text":"and now what we\u0027re going to do is we\u0027re going to draw"},{"Start":"18:30.385 ","End":"18:34.788","Text":"the graph for the frictional force,"},{"Start":"18:34.788 ","End":"18:37.105","Text":"our F_s and our F_k,"},{"Start":"18:37.105 ","End":"18:40.070","Text":"as a function of time."},{"Start":"18:40.800 ","End":"18:50.320","Text":"Here we can see our graph of our frictional force as a function of time."},{"Start":"18:50.320 ","End":"18:52.675","Text":"As we saw at the beginning,"},{"Start":"18:52.675 ","End":"18:57.235","Text":"our friction is going to be static frictional force,"},{"Start":"18:57.235 ","End":"18:59.785","Text":"which means that it was equal to F_1."},{"Start":"18:59.785 ","End":"19:04.929","Text":"You can go to the previous screen to remind yourself of the equations."},{"Start":"19:04.929 ","End":"19:07.705","Text":"F_1 we said was equal to 3t."},{"Start":"19:07.705 ","End":"19:09.340","Text":"As we can see,"},{"Start":"19:09.340 ","End":"19:14.290","Text":"up until we reach the value of a maximum value for static friction,"},{"Start":"19:14.290 ","End":"19:20.999","Text":"where afterwards we move into kinetic friction because our body begins its motion,"},{"Start":"19:20.999 ","End":"19:24.340","Text":"we can see that our frictional force"},{"Start":"19:24.340 ","End":"19:29.350","Text":"increases linearly with time according to the equation for F_1."},{"Start":"19:29.350 ","End":"19:34.610","Text":"Now we worked out at which time our F_s,"},{"Start":"19:34.610 ","End":"19:37.965","Text":"our frictional force would turn from static friction to"},{"Start":"19:37.965 ","End":"19:42.645","Text":"kinetic friction and we said that that was at this time over here."},{"Start":"19:42.645 ","End":"19:48.305","Text":"We can see we reached this time over here and it reaches our F_s max."},{"Start":"19:48.305 ","End":"19:52.255","Text":"We know that after this value,"},{"Start":"19:52.255 ","End":"19:56.335","Text":"we enter the realm of kinetic friction."},{"Start":"19:56.335 ","End":"20:02.278","Text":"Then we worked out that kinetic friction is going to be always equal to,"},{"Start":"20:02.278 ","End":"20:03.868","Text":"from this moment,"},{"Start":"20:03.868 ","End":"20:07.655","Text":"from the moment when our body begins its motion,"},{"Start":"20:07.655 ","End":"20:10.045","Text":"that\u0027s from this time,"},{"Start":"20:10.045 ","End":"20:16.585","Text":"our kinetic friction is just going to be a constant."},{"Start":"20:16.585 ","End":"20:23.005","Text":"We reach this value of F_s max and then our frictional force"},{"Start":"20:23.005 ","End":"20:29.320","Text":"drops down to this value for F_k and continuous for eternity."},{"Start":"20:29.320 ","End":"20:34.025","Text":"Now we can really see this jump over here."},{"Start":"20:34.025 ","End":"20:37.880","Text":"This is the jump that we were talking about when we move"},{"Start":"20:37.880 ","End":"20:43.140","Text":"from static frictional force to kinetic frictional force."}],"ID":10778}],"Thumbnail":null,"ID":8961},{"Name":"Newton\u0027s Second Law","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Newton\u0027s Second Law","Duration":"1m 56s","ChapterTopicVideoID":10267,"CourseChapterTopicPlaylistID":8963,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.810","Text":"Now let\u0027s speak about Newton\u0027s second law."},{"Start":"00:03.810 ","End":"00:07.964","Text":"Now, Newton\u0027s second law says that the vector sum of the forces"},{"Start":"00:07.964 ","End":"00:12.660","Text":"on an object is equal to mass times acceleration."},{"Start":"00:12.660 ","End":"00:14.970","Text":"They\u0027re saying as a little side note,"},{"Start":"00:14.970 ","End":"00:18.930","Text":"acceleration is produced when a force acts on a mass."},{"Start":"00:18.930 ","End":"00:21.385","Text":"Let\u0027s see what this means mathematically."},{"Start":"00:21.385 ","End":"00:26.175","Text":"The vector sum of forces on an object is"},{"Start":"00:26.175 ","End":"00:32.500","Text":"equal to the mass of the object multiplied by acceleration."},{"Start":"00:33.020 ","End":"00:37.695","Text":"Here we can see that if we have some net force,"},{"Start":"00:37.695 ","End":"00:42.135","Text":"so some summer voices which is not equal to 0."},{"Start":"00:42.135 ","End":"00:46.845","Text":"Then over here, we\u0027re going to get some acceleration."},{"Start":"00:46.845 ","End":"00:49.295","Text":"Because we\u0027re assuming that our mass is never"},{"Start":"00:49.295 ","End":"00:52.940","Text":"0 and also when we\u0027re dealing with relativity,"},{"Start":"00:52.940 ","End":"00:55.310","Text":"then we\u0027ll be dealing with change in mass."},{"Start":"00:55.310 ","End":"00:58.770","Text":"Over here, our mass is a constant."},{"Start":"01:00.590 ","End":"01:05.150","Text":"When we have some net force acting on some mass,"},{"Start":"01:05.150 ","End":"01:11.150","Text":"that means that in order for this equation to work out or to make any sense,"},{"Start":"01:11.150 ","End":"01:15.575","Text":"that means that we\u0027re going to have some value for acceleration."},{"Start":"01:15.575 ","End":"01:20.750","Text":"Now, because we\u0027re dealing with the vector sum of force is a vector and"},{"Start":"01:20.750 ","End":"01:23.690","Text":"acceleration is a vector that means that we can"},{"Start":"01:23.690 ","End":"01:26.930","Text":"split up everything into the different components."},{"Start":"01:26.930 ","End":"01:32.010","Text":"Let\u0027s say we\u0027re working in Cartesian coordinates and we have x and y."},{"Start":"01:32.010 ","End":"01:36.620","Text":"We can say that the sum of the forces in the x-direction is"},{"Start":"01:36.620 ","End":"01:41.330","Text":"equal to mass multiplied by the acceleration in the x-direction."},{"Start":"01:41.330 ","End":"01:45.740","Text":"We can say that the sum of the forces in the y-direction is equal to"},{"Start":"01:45.740 ","End":"01:50.021","Text":"the mass multiplied by the acceleration in the y-direction."},{"Start":"01:50.021 ","End":"01:51.980","Text":"Of course, if we have a z-direction,"},{"Start":"01:51.980 ","End":"01:53.675","Text":"it\u0027s the exact same thing."},{"Start":"01:53.675 ","End":"01:56.490","Text":"That\u0027s the end of this lesson."}],"ID":10597},{"Watched":false,"Name":"Defining Mass","Duration":"6m 43s","ChapterTopicVideoID":10268,"CourseChapterTopicPlaylistID":8963,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.070","Text":"Hello. In this lesson,"},{"Start":"00:02.070 ","End":"00:06.195","Text":"we\u0027re going to be defining what exactly mass is."},{"Start":"00:06.195 ","End":"00:13.424","Text":"Mass is how much an object is resistant to acceleration when the net force is applied."},{"Start":"00:13.424 ","End":"00:16.320","Text":"Mass is also what dictates the strength of"},{"Start":"00:16.320 ","End":"00:22.150","Text":"its mutual gravitational attraction to other objects."},{"Start":"00:22.430 ","End":"00:28.080","Text":"The definition of mass actually comes from Newton\u0027s second law."},{"Start":"00:28.080 ","End":"00:30.765","Text":"Just to remind you what Newton\u0027s second law is,"},{"Start":"00:30.765 ","End":"00:35.835","Text":"it\u0027s the vector sum of all of the forces on a body,"},{"Start":"00:35.835 ","End":"00:41.190","Text":"which is equal to the mass of the body multiplied by its acceleration."},{"Start":"00:41.190 ","End":"00:47.030","Text":"As we can see, whereas we know from our everyday day-to-day life,"},{"Start":"00:47.030 ","End":"00:49.490","Text":"the larger the mass of an object,"},{"Start":"00:49.490 ","End":"00:51.305","Text":"so the heavier it is,"},{"Start":"00:51.305 ","End":"00:56.389","Text":"the harder it is to cause some change in its motion,"},{"Start":"00:56.389 ","End":"01:00.985","Text":"and the change in its motion is dictated by the acceleration."},{"Start":"01:00.985 ","End":"01:04.535","Text":"For a body to speed up,"},{"Start":"01:04.535 ","End":"01:07.790","Text":"if we let say kick a ball from rest"},{"Start":"01:07.790 ","End":"01:12.740","Text":"until it\u0027s moving with some velocity, constant or changing."},{"Start":"01:12.740 ","End":"01:16.025","Text":"We know that the heavier the ball is,"},{"Start":"01:16.025 ","End":"01:21.385","Text":"the more force we have to apply in order to get it to move."},{"Start":"01:21.385 ","End":"01:23.805","Text":"That\u0027s what the mass is."},{"Start":"01:23.805 ","End":"01:30.335","Text":"How hard it is to accelerate the object with this mass."},{"Start":"01:30.335 ","End":"01:32.750","Text":"Now, it also dictates the strength of"},{"Start":"01:32.750 ","End":"01:35.750","Text":"its mutual gravitational attraction to other bodies."},{"Start":"01:35.750 ","End":"01:40.705","Text":"As we know, there\u0027s gravitational attraction between earth."},{"Start":"01:40.705 ","End":"01:43.735","Text":"Here all of the countries,"},{"Start":"01:43.735 ","End":"01:48.450","Text":"between Earth and let\u0027s say the Moon."},{"Start":"01:49.330 ","End":"01:56.150","Text":"However, the moon is also exerting some force on the Earth,"},{"Start":"01:56.150 ","End":"01:59.005","Text":"attracting the Earth to the Moon as well,"},{"Start":"01:59.005 ","End":"02:03.140","Text":"and that comes from Newton\u0027s third law,"},{"Start":"02:03.140 ","End":"02:06.300","Text":"that every force has an equal and opposite reactions."},{"Start":"02:06.300 ","End":"02:08.495","Text":"The Earth is attracted to the Moon,"},{"Start":"02:08.495 ","End":"02:11.940","Text":"and the Moon is attracted to the Earth."},{"Start":"02:13.240 ","End":"02:19.250","Text":"Because Earth\u0027s mass is significantly larger than the Moon\u0027s mass."},{"Start":"02:19.250 ","End":"02:23.210","Text":"It\u0027s determines the gravitational attraction to other bodies."},{"Start":"02:23.210 ","End":"02:29.080","Text":"So the moon will move more towards the Earth,"},{"Start":"02:29.080 ","End":"02:31.580","Text":"and the Earth will move towards the moon just"},{"Start":"02:31.580 ","End":"02:34.925","Text":"because the Earth\u0027s mass is heavier and therefore,"},{"Start":"02:34.925 ","End":"02:39.425","Text":"a larger force has to be applied to Earth in order to move it."},{"Start":"02:39.425 ","End":"02:41.810","Text":"Whereas, a smaller force has to be applied to"},{"Start":"02:41.810 ","End":"02:46.410","Text":"the moon in order to move it slightly towards the Earth."},{"Start":"02:46.970 ","End":"02:50.405","Text":"That\u0027s how in this example, specifically,"},{"Start":"02:50.405 ","End":"02:56.700","Text":"Newton\u0027s second and third laws come into play and we can see that very clearly."},{"Start":"02:57.370 ","End":"03:01.355","Text":"In other words, to make this clearer,"},{"Start":"03:01.355 ","End":"03:08.610","Text":"everybody that has mass will cause a gravitational pull towards it."},{"Start":"03:08.800 ","End":"03:13.780","Text":"Then we know that from Newton\u0027s third law."},{"Start":"03:13.780 ","End":"03:18.670","Text":"If you\u0027re attracted to some kind of object,"},{"Start":"03:18.670 ","End":"03:25.035","Text":"that object is also going to be equally and oppositely attracted to you."},{"Start":"03:25.035 ","End":"03:29.435","Text":"Then what determines why if you jump in the air,"},{"Start":"03:29.435 ","End":"03:34.655","Text":"you\u0027re pulled down towards Earth rather than Earth being pulled towards you,"},{"Start":"03:34.655 ","End":"03:37.145","Text":"comes from Newton\u0027s second law,"},{"Start":"03:37.145 ","End":"03:40.685","Text":"which states that the larger the mass,"},{"Start":"03:40.685 ","End":"03:44.600","Text":"the harder it is or the more force you have to apply in"},{"Start":"03:44.600 ","End":"03:50.455","Text":"order to cause some change in its motion or change in its position,"},{"Start":"03:50.455 ","End":"03:53.315","Text":"and that is why when you jump up,"},{"Start":"03:53.315 ","End":"03:57.170","Text":"you are pulled down to Earth because Earth has a larger mass."},{"Start":"03:57.170 ","End":"03:59.390","Text":"It\u0027s harder to move it towards you,"},{"Start":"03:59.390 ","End":"04:03.050","Text":"whereas it\u0027s easier to just move you towards Earth."},{"Start":"04:03.050 ","End":"04:09.590","Text":"Now, an important thing to note is that mass is a constant size."},{"Start":"04:09.590 ","End":"04:13.250","Text":"It\u0027s not changing unless you\u0027re dealing with relativity."},{"Start":"04:13.250 ","End":"04:14.945","Text":"But in our day-to-day life,"},{"Start":"04:14.945 ","End":"04:17.640","Text":"mass is unchanging,"},{"Start":"04:17.930 ","End":"04:24.335","Text":"and that also means that your mass on earth is going to be your mass on the moon,"},{"Start":"04:24.335 ","End":"04:28.970","Text":"even though your weight on Earth is different to your weight on the moon."},{"Start":"04:28.970 ","End":"04:31.790","Text":"Now let\u0027s talk about the units of force."},{"Start":"04:31.790 ","End":"04:34.895","Text":"So force, as we know,"},{"Start":"04:34.895 ","End":"04:39.035","Text":"has the units of Newton\u0027s."},{"Start":"04:39.035 ","End":"04:41.390","Text":"Because force, as we know,"},{"Start":"04:41.390 ","End":"04:43.120","Text":"we\u0027ve been speaking about Newton\u0027s laws."},{"Start":"04:43.120 ","End":"04:49.445","Text":"The units of force naturally are going to Newton denoted by N. Now,"},{"Start":"04:49.445 ","End":"04:53.180","Text":"from our equation for Newton\u0027s second law of motion,"},{"Start":"04:53.180 ","End":"05:01.340","Text":"we can see what SI units the Newton is equal to."},{"Start":"05:01.340 ","End":"05:07.940","Text":"We know that force is going to be equal"},{"Start":"05:07.940 ","End":"05:14.570","Text":"to mass multiplied by acceleration."},{"Start":"05:14.570 ","End":"05:16.894","Text":"Now in SI units,"},{"Start":"05:16.894 ","End":"05:20.750","Text":"we know that mass is given by kilogram,"},{"Start":"05:20.750 ","End":"05:27.350","Text":"and then acceleration is given by meters per second squared,"},{"Start":"05:27.350 ","End":"05:31.605","Text":"and then that is equal to 1 newton."},{"Start":"05:31.605 ","End":"05:37.970","Text":"A newton is simply equal to kilogram meters per second squared."},{"Start":"05:37.970 ","End":"05:41.650","Text":"Now, if we\u0027re working with CGS,"},{"Start":"05:41.650 ","End":"05:45.695","Text":"so that means Centimeters Grams in Seconds,"},{"Start":"05:45.695 ","End":"05:48.634","Text":"instead of meters kilograms seconds,"},{"Start":"05:48.634 ","End":"05:56.370","Text":"the force is therefore measured in dyne."},{"Start":"05:56.620 ","End":"06:01.410","Text":"Then we can say that 1 dyne,"},{"Start":"06:01.610 ","End":"06:04.460","Text":"which is similar to 1 Newton,"},{"Start":"06:04.460 ","End":"06:06.260","Text":"but instead of kilogram meters per second,"},{"Start":"06:06.260 ","End":"06:09.125","Text":"we\u0027re working in centimeters gram per second."},{"Start":"06:09.125 ","End":"06:16.670","Text":"That\u0027s going to be equal to our mass in grams multiplied by acceleration,"},{"Start":"06:16.670 ","End":"06:19.985","Text":"which instead of meters, centimeters,"},{"Start":"06:19.985 ","End":"06:22.445","Text":"and then per second squared,"},{"Start":"06:22.445 ","End":"06:25.720","Text":"and that\u0027s what 1 dyne is equal to."},{"Start":"06:25.720 ","End":"06:32.675","Text":"What\u0027s important to remember here are the units for Newton\u0027s in SI units."},{"Start":"06:32.675 ","End":"06:36.740","Text":"Here we\u0027re dealing with MKS units,"},{"Start":"06:36.740 ","End":"06:40.930","Text":"and here we\u0027re dealing with CGS units."},{"Start":"06:40.930 ","End":"06:43.870","Text":"That\u0027s the end of this lesson."}],"ID":10598},{"Watched":false,"Name":"Exercise 1","Duration":"6m 58s","ChapterTopicVideoID":10269,"CourseChapterTopicPlaylistID":8963,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.820","Text":"In this question there are 3 masses that are resting on a smooth surface,"},{"Start":"00:05.820 ","End":"00:08.565","Text":"meaning that there\u0027s no coefficient of friction."},{"Start":"00:08.565 ","End":"00:11.760","Text":"They are being pushed with an acceleration of a."},{"Start":"00:11.760 ","End":"00:19.095","Text":"Now, we\u0027re being asked to calculate the force going in this direction."},{"Start":"00:19.095 ","End":"00:23.175","Text":"We\u0027re going to use our favorite equation and say"},{"Start":"00:23.175 ","End":"00:27.405","Text":"that the sum of all the forces equals ma."},{"Start":"00:27.405 ","End":"00:30.690","Text":"Now, we\u0027re going to look at mass number 1, 2,"},{"Start":"00:30.690 ","End":"00:33.325","Text":"and 3 as 1 body;"},{"Start":"00:33.325 ","End":"00:35.730","Text":"just 1 big mass."},{"Start":"00:35.730 ","End":"00:39.995","Text":"Then we can just write this as F with"},{"Start":"00:39.995 ","End":"00:44.000","Text":"a tilde on top to remind us that it is unknown and because we\u0027ve"},{"Start":"00:44.000 ","End":"00:47.930","Text":"said that all the masses are just 1 big mass,"},{"Start":"00:47.930 ","End":"00:57.000","Text":"we\u0027ll do (m_1+m_2+m_3)a."},{"Start":"00:57.000 ","End":"01:00.465","Text":"That\u0027s it. We finished the 1st question."},{"Start":"01:00.465 ","End":"01:03.950","Text":"This is the force we have to apply in order to"},{"Start":"01:03.950 ","End":"01:07.805","Text":"move all of these masses at an acceleration of a."},{"Start":"01:07.805 ","End":"01:10.265","Text":"The next question is asking us,"},{"Start":"01:10.265 ","End":"01:14.405","Text":"which forces are acting between each of the masses?"},{"Start":"01:14.405 ","End":"01:17.180","Text":"This is meant to be which forces."},{"Start":"01:17.180 ","End":"01:20.570","Text":"Which forces are acting between each of the masses?"},{"Start":"01:20.570 ","End":"01:23.120","Text":"For whoever doesn\u0027t know how to do this,"},{"Start":"01:23.120 ","End":"01:27.305","Text":"I recommend going over this section of the video a few times."},{"Start":"01:27.305 ","End":"01:29.404","Text":"In order to do this section,"},{"Start":"01:29.404 ","End":"01:34.220","Text":"I\u0027m not going to think of all the masses as 1 big mass,"},{"Start":"01:34.220 ","End":"01:36.715","Text":"but I\u0027m going to think of them separately."},{"Start":"01:36.715 ","End":"01:41.160","Text":"Let\u0027s draw this picture again making it slightly more clear."},{"Start":"01:41.160 ","End":"01:43.415","Text":"Here we have all of the masses"},{"Start":"01:43.415 ","End":"01:47.270","Text":"drawn slightly further apart to make it easier to understand."},{"Start":"01:47.270 ","End":"01:50.930","Text":"Now, we know the masses and we know that each of"},{"Start":"01:50.930 ","End":"01:54.835","Text":"their acceleration is equal to the acceleration over here."},{"Start":"01:54.835 ","End":"01:57.920","Text":"This is because if we know that all 3 of"},{"Start":"01:57.920 ","End":"02:01.670","Text":"the masses are accelerating at an acceleration of a,"},{"Start":"02:01.670 ","End":"02:05.695","Text":"then each individual mass is also accelerating at an acceleration of a."},{"Start":"02:05.695 ","End":"02:10.285","Text":"Let\u0027s start drawing all of the forces acting on each of the masses."},{"Start":"02:10.285 ","End":"02:14.645","Text":"On mass number 1, we have this force."},{"Start":"02:14.645 ","End":"02:19.910","Text":"Now, this force pointing rightwards is a force acting on mass number"},{"Start":"02:19.910 ","End":"02:25.535","Text":"1 and this is the F that we worked out on top."},{"Start":"02:25.535 ","End":"02:28.700","Text":"The force acting in the leftwards direction,"},{"Start":"02:28.700 ","End":"02:35.900","Text":"this force is some force acting in the opposite direction to F,"},{"Start":"02:35.900 ","End":"02:40.775","Text":"and we\u0027ll call it N_1 and we\u0027re putting a tilde on top because it\u0027s an unknown."},{"Start":"02:40.775 ","End":"02:46.565","Text":"Now, who is applying this N_1 force onto mass number 1?"},{"Start":"02:46.565 ","End":"02:48.605","Text":"Now, that\u0027s of course mass number 2."},{"Start":"02:48.605 ","End":"02:53.330","Text":"It\u0027s applying an equal and opposite force to the F,"},{"Start":"02:53.330 ","End":"03:00.185","Text":"which means that if N_1 is going in the left direction on m_1,"},{"Start":"03:00.185 ","End":"03:06.190","Text":"then N_1 is going in the right direction on m_2."},{"Start":"03:06.410 ","End":"03:13.112","Text":"Then we can see that on mass number 2 we have a normal force as well."},{"Start":"03:13.112 ","End":"03:14.495","Text":"It\u0027s called N_2,"},{"Start":"03:14.495 ","End":"03:16.100","Text":"and it\u0027s also unknown."},{"Start":"03:16.100 ","End":"03:19.950","Text":"I\u0027m just going to rub out this tilde here so that it\u0027s not confusing."},{"Start":"03:19.950 ","End":"03:26.345","Text":"N_2 is unknown and who is applying this force onto mass number 2,"},{"Start":"03:26.345 ","End":"03:28.955","Text":"mass number 3 but in an opposite direction."},{"Start":"03:28.955 ","End":"03:30.935","Text":"This is also N_2."},{"Start":"03:30.935 ","End":"03:36.260","Text":"Now, let\u0027s work out the sum of all the forces on each 1 of the masses."},{"Start":"03:36.260 ","End":"03:37.910","Text":"Let\u0027s start with m_1."},{"Start":"03:37.910 ","End":"03:41.375","Text":"We have F pointing in the right direction."},{"Start":"03:41.375 ","End":"03:44.975","Text":"Now, we worked out above with what F was so we have"},{"Start":"03:44.975 ","End":"03:52.745","Text":"(m_1+m_2+m_3)a"},{"Start":"03:52.745 ","End":"03:55.345","Text":"and in the opposite direction"},{"Start":"03:55.345 ","End":"03:57.885","Text":"on mass number 1 we have N_1."},{"Start":"03:57.885 ","End":"04:04.890","Text":"We\u0027ll do minus N_1 equals this new force,"},{"Start":"04:04.890 ","End":"04:10.230","Text":"which is mass times acceleration so it\u0027s m_1 times a."},{"Start":"04:10.230 ","End":"04:14.940","Text":"Now, let\u0027s take a look at mass number 2."},{"Start":"04:14.940 ","End":"04:20.690","Text":"We have N_1 in the right direction minus"},{"Start":"04:20.690 ","End":"04:27.520","Text":"N_2 in the left direction equals m_2 times a."},{"Start":"04:27.520 ","End":"04:31.230","Text":"Then we have on mass number 3."},{"Start":"04:31.230 ","End":"04:38.145","Text":"We have only N_2 acting on it which equals m_2 times a,"},{"Start":"04:38.145 ","End":"04:44.740","Text":"the mass number 3 multiplied by the acceleration."},{"Start":"04:45.470 ","End":"04:49.055","Text":"All I did here was workout for each mass,"},{"Start":"04:49.055 ","End":"04:50.990","Text":"the sum of all the forces acting on it,"},{"Start":"04:50.990 ","End":"04:54.830","Text":"and equal to the equation for force which"},{"Start":"04:54.830 ","End":"04:58.580","Text":"is the mass of the mass multiplied by the acceleration,"},{"Start":"04:58.580 ","End":"05:02.300","Text":"which is the same acceleration for each 1 of the masses."},{"Start":"05:02.300 ","End":"05:11.130","Text":"Now, what I have to do is I have to find what N_1 is and what N_2 is."},{"Start":"05:11.130 ","End":"05:12.450","Text":"Now, N_2 is easy."},{"Start":"05:12.450 ","End":"05:14.565","Text":"We\u0027ve already found it."},{"Start":"05:14.565 ","End":"05:23.375","Text":"N_2=m_3 multiplied by a and now I just have to find what N_1 equals."},{"Start":"05:23.375 ","End":"05:26.150","Text":"Now, let\u0027s work out what N_1 is."},{"Start":"05:26.150 ","End":"05:36.190","Text":"From our equation for mass number 2 we have the N_1=m_2 multiplied by a plus,"},{"Start":"05:36.190 ","End":"05:40.960","Text":"I\u0027m just moving the minus N_2 to the other side of the equal side."},{"Start":"05:40.960 ","End":"05:44.075","Text":"It makes it plus N_2."},{"Start":"05:44.075 ","End":"05:46.620","Text":"We already know what N_2 is."},{"Start":"05:46.620 ","End":"05:52.350","Text":"We have m_2 times a plus m_3 times a."},{"Start":"05:52.350 ","End":"05:57.250","Text":"This is N_1."},{"Start":"05:57.380 ","End":"06:02.210","Text":"Let\u0027s have a quick little recap of what we did in this lesson."},{"Start":"06:02.210 ","End":"06:04.040","Text":"In the first section,"},{"Start":"06:04.040 ","End":"06:05.945","Text":"the first question that we had,"},{"Start":"06:05.945 ","End":"06:07.670","Text":"we looked at m_1,"},{"Start":"06:07.670 ","End":"06:10.310","Text":"m_2 and m_3 as 1 mass."},{"Start":"06:10.310 ","End":"06:11.615","Text":"Then we worked out,"},{"Start":"06:11.615 ","End":"06:17.240","Text":"that the force acting on it is m_1 plus"},{"Start":"06:17.240 ","End":"06:23.525","Text":"m_2 plus m_3 multiplied by a because they\u0027re all being accelerated at the same rate."},{"Start":"06:23.525 ","End":"06:25.415","Text":"In the 2nd question,"},{"Start":"06:25.415 ","End":"06:29.240","Text":"we were asked to look at each mass separately"},{"Start":"06:29.240 ","End":"06:33.395","Text":"and to work out the forces acting between all the masses. That\u0027s what we did."},{"Start":"06:33.395 ","End":"06:35.059","Text":"We said that with F,"},{"Start":"06:35.059 ","End":"06:38.515","Text":"we already found out from the previous question."},{"Start":"06:38.515 ","End":"06:48.365","Text":"Then using what we know the sum of all the forces"},{"Start":"06:48.365 ","End":"06:53.120","Text":"equals mass times acceleration by using that equation we came to"},{"Start":"06:53.120 ","End":"06:59.040","Text":"find out what N_1 and N_2 equals. That\u0027s it."}],"ID":10599},{"Watched":false,"Name":"Exercise 2","Duration":"5m 24s","ChapterTopicVideoID":10270,"CourseChapterTopicPlaylistID":8963,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.710","Text":"In the diagram, we\u0027re given a drawing where we have m_1,"},{"Start":"00:04.710 ","End":"00:08.100","Text":"m_2, and m_3 on a frictionless slope."},{"Start":"00:08.100 ","End":"00:12.345","Text":"We\u0027re told that the angle between the horizontal,"},{"Start":"00:12.345 ","End":"00:15.015","Text":"the slope is Alpha."},{"Start":"00:15.015 ","End":"00:18.405","Text":"We\u0027re told that the masses are accelerating."},{"Start":"00:18.405 ","End":"00:21.915","Text":"We\u0027re being asked to find what the acceleration is."},{"Start":"00:21.915 ","End":"00:26.850","Text":"Now, we\u0027re going to answer this question in a slightly different way than we usually do."},{"Start":"00:26.850 ","End":"00:30.460","Text":"Now, what do we usually have done is,"},{"Start":"00:31.820 ","End":"00:40.240","Text":"let me draw it here, we\u0027ve taken this section as 1 area."},{"Start":"00:41.450 ","End":"00:46.475","Text":"We\u0027ve worked out all the forces and everything acting on an acceleration."},{"Start":"00:46.475 ","End":"00:51.935","Text":"Then we\u0027ve looked at this as a separate section and worked out the forces."},{"Start":"00:51.935 ","End":"00:55.070","Text":"Now what we\u0027re going to do today is slightly different and"},{"Start":"00:55.070 ","End":"00:58.550","Text":"a slightly more comfortable and easy way to solve this."},{"Start":"00:58.550 ","End":"01:01.385","Text":"Instead of looking at the 2 systems,"},{"Start":"01:01.385 ","End":"01:06.145","Text":"we\u0027re going to just look at the whole thing like this."},{"Start":"01:06.145 ","End":"01:09.200","Text":"This is the positive direction."},{"Start":"01:09.200 ","End":"01:13.820","Text":"We can do this only in certain types of questions,"},{"Start":"01:13.820 ","End":"01:15.205","Text":"and this is 1 of them."},{"Start":"01:15.205 ","End":"01:16.575","Text":"Why is this?"},{"Start":"01:16.575 ","End":"01:18.780","Text":"Because m_1, m_2,"},{"Start":"01:18.780 ","End":"01:22.130","Text":"and m_3 are all connected by the same rope,"},{"Start":"01:22.130 ","End":"01:24.010","Text":"they all have the acceleration,"},{"Start":"01:24.010 ","End":"01:32.025","Text":"and therefore, we can look at it as in the same system without having to separate it up."},{"Start":"01:32.025 ","End":"01:33.935","Text":"It makes it a lot easier."},{"Start":"01:33.935 ","End":"01:36.275","Text":"Because if m_1,"},{"Start":"01:36.275 ","End":"01:39.124","Text":"for instance, moves down 1 meter,"},{"Start":"01:39.124 ","End":"01:45.380","Text":"then we\u0027d expect m_2 and m_3 to move up here 1 meter."},{"Start":"01:45.380 ","End":"01:47.390","Text":"Because it\u0027s the same rope,"},{"Start":"01:47.390 ","End":"01:50.179","Text":"we can use this system."},{"Start":"01:50.179 ","End":"01:54.050","Text":"Now let\u0027s draw the force diagrams."},{"Start":"01:54.050 ","End":"01:57.575","Text":"We have m_1,"},{"Start":"01:57.575 ","End":"02:01.585","Text":"and we have m_1g pointing downwards,"},{"Start":"02:01.585 ","End":"02:05.515","Text":"and we have T_2 pointing upwards."},{"Start":"02:05.515 ","End":"02:08.980","Text":"Now, it\u0027s important to note the row between m_1 and"},{"Start":"02:08.980 ","End":"02:13.190","Text":"m_2 is the same row and between m_2 and m_3 is a different rope."},{"Start":"02:13.190 ","End":"02:19.680","Text":"Then we have here m_2,"},{"Start":"02:19.680 ","End":"02:25.367","Text":"which has here T_2,"},{"Start":"02:25.367 ","End":"02:28.828","Text":"and here it has T_1,"},{"Start":"02:28.828 ","End":"02:34.040","Text":"which is the tension in the rope adjoined to mass number 3."},{"Start":"02:34.040 ","End":"02:38.875","Text":"Here it has m_2g."},{"Start":"02:38.875 ","End":"02:46.105","Text":"Mass number 3 has as well T_1,"},{"Start":"02:46.105 ","End":"02:50.610","Text":"which is equal to the T_1 that\u0027s coming out of m_2,"},{"Start":"02:50.980 ","End":"02:55.164","Text":"and it has m_3g."},{"Start":"02:55.164 ","End":"03:01.805","Text":"Now we\u0027re going to separate out the components of m_2g and m_3g,"},{"Start":"03:01.805 ","End":"03:05.306","Text":"making that in the x-direction."},{"Start":"03:05.306 ","End":"03:06.515","Text":"Like in T_1,"},{"Start":"03:06.515 ","End":"03:12.430","Text":"we have m_2g sine Alpha,"},{"Start":"03:12.430 ","End":"03:18.155","Text":"and in the opposite direction to T_1 and m_3,"},{"Start":"03:18.155 ","End":"03:23.745","Text":"we have m_3g sine Alpha."},{"Start":"03:23.745 ","End":"03:26.735","Text":"With regards to separating components,"},{"Start":"03:26.735 ","End":"03:29.060","Text":"if someone doesn\u0027t understand how I did it,"},{"Start":"03:29.060 ","End":"03:32.760","Text":"please go back to the videos where I explain how to do it."},{"Start":"03:34.310 ","End":"03:38.800","Text":"Now let\u0027s write down all of the equations."},{"Start":"03:39.260 ","End":"03:43.310","Text":"Because we\u0027re already told that the masses are accelerating,"},{"Start":"03:43.310 ","End":"03:45.950","Text":"we know that the sum of all the forces equals"},{"Start":"03:45.950 ","End":"03:49.948","Text":"mass of this specific mass times acceleration,"},{"Start":"03:49.948 ","End":"03:52.460","Text":"and it\u0027s the same acceleration for each of the masses,"},{"Start":"03:52.460 ","End":"03:56.480","Text":"because we\u0027ve already said that they\u0027re accelerating at the same rate."},{"Start":"03:56.480 ","End":"03:59.700","Text":"Also because of our system here,"},{"Start":"04:01.130 ","End":"04:08.450","Text":"then we know that the acceleration is uniform throughout."},{"Start":"04:08.450 ","End":"04:13.290","Text":"We have m_1g in the positive direction."},{"Start":"04:13.290 ","End":"04:15.645","Text":"Because it\u0027s going in this direction,"},{"Start":"04:15.645 ","End":"04:21.575","Text":"minus T_2, because the T_2 is going in the opposite direction."},{"Start":"04:21.575 ","End":"04:31.025","Text":"Just going to rub it out, equals mass 1 times same acceleration."},{"Start":"04:31.025 ","End":"04:33.470","Text":"Then on mass number 2,"},{"Start":"04:33.470 ","End":"04:40.730","Text":"we have that T_2 going in the positive direction because it\u0027s going up,"},{"Start":"04:40.730 ","End":"04:48.780","Text":"minus T_1 minus m_2g sine Alpha"},{"Start":"04:48.780 ","End":"04:53.025","Text":"equals mass 2 times acceleration."},{"Start":"04:53.025 ","End":"04:55.350","Text":"Then for mass number 3,"},{"Start":"04:55.350 ","End":"05:03.380","Text":"we have the T_1 minus m_3g sine Alpha"},{"Start":"05:03.380 ","End":"05:07.760","Text":"equals mass 3 times acceleration."},{"Start":"05:08.450 ","End":"05:10.920","Text":"Now as we can see,"},{"Start":"05:10.920 ","End":"05:13.145","Text":"T_2 is an unknown,"},{"Start":"05:13.145 ","End":"05:14.735","Text":"T_1 is an unknown,"},{"Start":"05:14.735 ","End":"05:15.845","Text":"and a is an unknown."},{"Start":"05:15.845 ","End":"05:18.760","Text":"We have 3 unknowns and 3 equations."},{"Start":"05:18.760 ","End":"05:22.850","Text":"We can just rearrange to find what a is, and that\u0027s it."},{"Start":"05:22.850 ","End":"05:25.590","Text":"Here we finished this question."}],"ID":10600}],"Thumbnail":null,"ID":8963},{"Name":"Additional Exercises","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise - Chain Resting On A Slope","Duration":"4m 51s","ChapterTopicVideoID":8966,"CourseChapterTopicPlaylistID":8964,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.385","Text":"In this question, we have a diagram,"},{"Start":"00:02.385 ","End":"00:04.815","Text":"we\u0027re speaking of this diagram right now."},{"Start":"00:04.815 ","End":"00:09.420","Text":"In the diagram, we have a smooth slope and resting on the slope,"},{"Start":"00:09.420 ","End":"00:14.940","Text":"we have a chain which is said to be of length L and of mass m. We\u0027re"},{"Start":"00:14.940 ","End":"00:21.180","Text":"being asked to find what the length of the part of the chain that is labeled X is."},{"Start":"00:21.180 ","End":"00:25.020","Text":"What is this length?"},{"Start":"00:25.020 ","End":"00:29.550","Text":"Now, of course, this length over here is"},{"Start":"00:29.550 ","End":"00:34.140","Text":"L minus x because the length of the chain in total is L. We know"},{"Start":"00:34.140 ","End":"00:45.215","Text":"that here it is x and so we know that this section here is L minus x."},{"Start":"00:45.215 ","End":"00:48.770","Text":"Now, what we\u0027re going to do is we\u0027re going to draw in"},{"Start":"00:48.770 ","End":"00:52.550","Text":"this diagram in a slightly easier way,"},{"Start":"00:52.550 ","End":"00:57.710","Text":"where it\u0027s easier to understand what\u0027s happening and we\u0027re going to look at it here."},{"Start":"00:57.710 ","End":"01:04.955","Text":"We\u0027re going to look at this rope as being like the chain, but without mass,"},{"Start":"01:04.955 ","End":"01:11.910","Text":"and m_2 is the mass of the chain which is on this side,"},{"Start":"01:11.910 ","End":"01:14.240","Text":"so the mass of the chain here,"},{"Start":"01:14.240 ","End":"01:18.870","Text":"and mass number 1 is the mass of the chain over here."},{"Start":"01:19.580 ","End":"01:25.770","Text":"We\u0027re going to look at this wheel."},{"Start":"01:25.770 ","End":"01:32.490","Text":"That is, if the chain is passing over it as if it\u0027s this corner over here."},{"Start":"01:32.540 ","End":"01:34.715","Text":"Now, that we\u0027ve drawn this,"},{"Start":"01:34.715 ","End":"01:40.250","Text":"we can know that the system on the right is identical to the system on the left."},{"Start":"01:40.250 ","End":"01:44.220","Text":"It\u0027s identical to the system that we\u0027re being asked about."},{"Start":"01:44.690 ","End":"01:49.685","Text":"The only difference is that on this system,"},{"Start":"01:49.685 ","End":"01:51.830","Text":"it\u0027s much easier for us to understand what\u0027s"},{"Start":"01:51.830 ","End":"01:56.050","Text":"happening and to draw all the forces in a more clear way."},{"Start":"01:56.050 ","End":"01:59.495","Text":"Let\u0027s write down all of the equations."},{"Start":"01:59.495 ","End":"02:06.620","Text":"We have the T-m_1g=0."},{"Start":"02:06.620 ","End":"02:12.933","Text":"Why zero? Because we are told that the chain is at rest,"},{"Start":"02:12.933 ","End":"02:15.610","Text":"which means that the sum of all the forces is 0."},{"Start":"02:15.610 ","End":"02:20.060","Text":"We can also just write the T=m_1g."},{"Start":"02:20.060 ","End":"02:24.155","Text":"Then we can also write the same on there,"},{"Start":"02:24.155 ","End":"02:28.080","Text":"m number 2, this was m_1."},{"Start":"02:28.080 ","End":"02:31.070","Text":"We can write the same on number 2, because again,"},{"Start":"02:31.070 ","End":"02:40.800","Text":"the sum of all the forces is equal to 0 so we can rewrite the T=m_2 g sine Alpha."},{"Start":"02:42.670 ","End":"02:48.965","Text":"Now, this leaves us with 2 equations with 3 unknowns,"},{"Start":"02:48.965 ","End":"02:52.650","Text":"T, m_1, and m_2."},{"Start":"02:52.940 ","End":"02:57.095","Text":"In this case, let\u0027s see what I do from here."},{"Start":"02:57.095 ","End":"03:03.680","Text":"Now, I have to find the link between the mass,"},{"Start":"03:03.680 ","End":"03:10.645","Text":"for instance, of m_1 with regards to the chain."},{"Start":"03:10.645 ","End":"03:13.415","Text":"What\u0027s another equation that I can add?"},{"Start":"03:13.415 ","End":"03:17.810","Text":"I can add that the mass of a chain,"},{"Start":"03:17.810 ","End":"03:20.495","Text":"because that\u0027s what mass 1 is meant to represent,"},{"Start":"03:20.495 ","End":"03:22.490","Text":"this section of the chain,"},{"Start":"03:22.490 ","End":"03:30.285","Text":"so m_1 equals density times length."},{"Start":"03:30.285 ","End":"03:32.655","Text":"The length is x,"},{"Start":"03:32.655 ","End":"03:35.338","Text":"which is unknown so I\u0027ll put a tilde on top,"},{"Start":"03:35.338 ","End":"03:38.735","Text":"multiplied by the density,"},{"Start":"03:38.735 ","End":"03:42.210","Text":"which is also unknown with a tilde on top."},{"Start":"03:43.790 ","End":"03:46.830","Text":"Of course, with mass number 2,"},{"Start":"03:46.830 ","End":"03:48.450","Text":"it\u0027s exactly the same thing."},{"Start":"03:48.450 ","End":"03:51.330","Text":"Mass number 2 is length,"},{"Start":"03:51.330 ","End":"04:01.350","Text":"which is L minus x multiplied by the density."},{"Start":"04:02.380 ","End":"04:06.485","Text":"Now, in order to find what the density is,"},{"Start":"04:06.485 ","End":"04:11.420","Text":"we know that density is the mass of the chain,"},{"Start":"04:11.420 ","End":"04:13.760","Text":"which is given in the question,"},{"Start":"04:13.760 ","End":"04:15.890","Text":"divided by the length of the chain,"},{"Start":"04:15.890 ","End":"04:18.050","Text":"which is also given in the question."},{"Start":"04:18.050 ","End":"04:23.340","Text":"We have density is mass divided by length."},{"Start":"04:23.690 ","End":"04:27.065","Text":"Now, what it didn\u0027t say in the question is"},{"Start":"04:27.065 ","End":"04:31.670","Text":"that the density is uniform throughout the chain,"},{"Start":"04:31.670 ","End":"04:39.775","Text":"which means that the density on this side is equal to the density on that side."},{"Start":"04:39.775 ","End":"04:44.370","Text":"Now, we have 5 equations, 5 unknowns."},{"Start":"04:44.370 ","End":"04:49.060","Text":"Here, it\u0027s just a matter of substitution and simple algebra."},{"Start":"04:49.060 ","End":"04:51.679","Text":"And so here we finished the question."}],"ID":9251}],"Thumbnail":null,"ID":8964}]

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