1 00:00:00,520 --> 00:00:05,030 In the last video, we had a ten kilogram mass 2 00:00:05,070 --> 00:00:08,640 sitting on top of an inclined plane at a 30 degree angle 3 00:00:08,670 --> 00:00:11,220 And in order to figure out what would happen to this block 4 00:00:11,230 --> 00:00:16,020 we broke down the force of gravity on this block into 5 00:00:16,149 --> 00:00:19,860 the components that are parallel to the surface of the plane 6 00:00:19,900 --> 00:00:23,010 and perpendicular to the surface of the plane 7 00:00:23,020 --> 00:00:29,200 and for a perpendicular component, we got 49 times the square root of 3 N downwards 8 00:00:29,220 --> 00:00:31,950 That's 98 times this quantity over here, downwards 9 00:00:32,240 --> 00:00:37,850 But we said look! We don't see this block of ice accelerating downwards into this wedge 10 00:00:37,880 --> 00:00:39,630 because the wedge is supporting it 11 00:00:39,650 --> 00:00:44,540 So there must be a counteracting force that the wedge is exerting on the block 12 00:00:44,660 --> 00:00:49,990 And that counteracting force is the normal force of the wedge on the block of ice 13 00:00:50,010 --> 00:00:55,450 And that is exactly opposite to the force of gravity in this direction 14 00:00:55,450 --> 00:00:59,240 The normal force of gravity is the normal force of the wedge 15 00:00:59,360 --> 00:01:05,690 And these completely balance each other out in that normal, perpendicular direction 16 00:01:05,710 --> 00:01:09,620 And that is why this block is not accelerating either in that direction 17 00:01:09,880 --> 00:01:12,220 or in this direction over here 18 00:01:12,710 --> 00:01:16,740 But the one component of the force of gravity that did not seem to have any offset 19 00:01:16,740 --> 00:01:19,320 (at least the way we set up the problem in the last video) 20 00:01:19,320 --> 00:01:25,400 is the force/component that is parallel to the surface of the plane 21 00:01:25,740 --> 00:01:28,910 And we figured that out to be 49N, it was 22 00:01:29,060 --> 00:01:32,750 essentially the weight of the block times the sin of this angle 23 00:01:32,770 --> 00:01:35,020 And we said look, if there is no other forces 24 00:01:35,050 --> 00:01:37,710 then it would be accelerated in this direction 25 00:01:37,710 --> 00:01:40,870 And to figure out the rate of acceleration 26 00:01:40,990 --> 00:01:45,030 you take the force in that direction divided by the mass of the block 27 00:01:45,030 --> 00:01:47,880 and you would get 4.9 m/s^2 28 00:01:47,910 --> 00:01:50,970 Now let's say that wasn't happening. Let's say that you were to look at this system 29 00:01:50,970 --> 00:01:53,620 right over here and the block was just stationary 30 00:01:53,630 --> 00:01:56,460 And now, for the sake of argument, let's assume it is not ice on ice 31 00:01:56,460 --> 00:01:59,100 Let's assume that they are both made out of wood 32 00:01:59,130 --> 00:02:02,300 And now all of a sudden we have a situation where the block is stationary 33 00:02:02,610 --> 00:02:06,480 If it is stationary, what is necessarily the case? 34 00:02:06,860 --> 00:02:13,310 Well we already determined that if it is not accelerating in this normal direction 35 00:02:13,360 --> 00:02:15,490 there must be zero net forces on it 36 00:02:15,510 --> 00:02:17,420 But if it is stationary as a whole 37 00:02:17,450 --> 00:02:20,920 then there must be zero net forces in the parallel component too 38 00:02:20,950 --> 00:02:29,890 So there must be some force counteracting this 49 N that wants to take it down the slope 39 00:02:29,890 --> 00:02:36,110 So there must be some force counteracting 40 00:02:36,110 --> 00:02:39,920 the component of gravity that wants to accelerate it down the slope 41 00:02:39,930 --> 00:02:43,870 And the question is what is this force? We're dealing with a situation now 42 00:02:43,920 --> 00:02:49,880 where we're dealing with a stationary block, a block that is not accelerating 43 00:02:50,260 --> 00:02:52,370 So what is that force? 44 00:02:52,400 --> 00:02:56,430 I think you know from experience maybe what is the difference between 45 00:02:56,460 --> 00:03:01,000 a block of wood on top of wood and a block of ice on top of ice 46 00:03:01,100 --> 00:03:06,130 A block of ice on top of ice is much more slippery; there is no friction 47 00:03:06,240 --> 00:03:09,320 between ice and ice, but there is friction between wood and wood 48 00:03:09,320 --> 00:03:15,520 To make it a little bit more tangible, maybe we'd put some sandpaper on the surface over here 49 00:03:15,520 --> 00:03:18,070 And then it becomes a little bit clearer. The force 50 00:03:18,080 --> 00:03:25,530 that is keeping this block from sliding down in this situation is the force of friction 51 00:03:25,550 --> 00:03:29,690 and the force of friction will always act in a direction opposite 52 00:03:29,690 --> 00:03:33,090 to the motion if there was not any friction 53 00:03:33,330 --> 00:03:44,230 or the potential acceleration if there was not any action 54 00:03:44,250 --> 00:03:46,400 So what is the force of friction in this case? 55 00:03:46,500 --> 00:03:50,460 Well, this block is completely stationary. It's not accelerating down the ramp 56 00:03:50,650 --> 00:04:06,810 The force of friction over here is going to be 49 N, upwards, up the ramp 57 00:04:07,020 --> 00:04:10,820 Now I want think about, this is something that can be determined experimentally 58 00:04:11,110 --> 00:04:17,860 as long as you have some way of measuring force, you can do this experimentally 59 00:04:18,190 --> 00:04:22,090 But the interesting question here is 60 00:04:22,370 --> 00:04:27,320 how much do I have to push on this block 61 00:04:27,320 --> 00:04:32,140 until it starts to move down the ramp? How much do I have to push on it? 62 00:04:32,400 --> 00:04:35,330 Let's say you are able to experimentally determine 63 00:04:35,350 --> 00:04:55,230 that if you can apply another 1 N on this 64 00:04:55,250 --> 00:05:00,930 then all of a sudden, I can at least get the box to start accelerating down 65 00:05:01,170 --> 00:05:05,080 Not the rate which it would do naturally, but I can just start to nudge it down 66 00:05:05,080 --> 00:05:09,550 if I give it another push of one N in the parallel direction 67 00:05:10,000 --> 00:05:13,590 So what is the total force--so exactly one N 68 00:05:13,900 --> 00:05:18,740 So the total force at this point that's acting on it in order to just start to budge it 69 00:05:19,190 --> 00:05:21,780 I'll call this the budging force 70 00:05:22,510 --> 00:05:24,600 Remember here isn't a traditional class 71 00:05:24,600 --> 00:05:26,940 F sub B for the budging force 72 00:05:27,200 --> 00:05:29,790 The budging force is in the parallel direction 73 00:05:30,040 --> 00:05:35,820 If I'm applying 1 N in this direction and it already has 49 N due to 74 00:05:35,820 --> 00:05:40,920 the component of gravity in this direction, then my budging force is 50 N 75 00:05:40,950 --> 00:05:43,170 And so an interesting thing that you can determine 76 00:05:43,320 --> 00:05:46,890 based on the materials that are coming in contact with each other is 77 00:05:46,890 --> 00:05:51,080 just how much force you need to just start to overcome friction 78 00:05:51,080 --> 00:05:54,620 In this case, it's the budging force. That's a term that I made up 79 00:05:54,950 --> 00:05:59,860 And the interesting ratio which tends to hold for given materials pretty well 80 00:05:59,990 --> 00:06:02,770 is the ratio between the amount of force just to budge it 81 00:06:03,080 --> 00:06:06,810 and the amount of force between the two objects 82 00:06:06,810 --> 00:06:10,130 between how much force they are exerting on each other 83 00:06:10,280 --> 00:06:17,620 And in this case, the amount of force that is being exerted by this, by the wedge on the block 84 00:06:17,620 --> 00:06:21,090 is the normal force, is 49 sqrt of 3 85 00:06:21,100 --> 00:06:25,080 Maybe I should say the magnitude of the budging force 86 00:06:25,090 --> 00:06:29,100 over the magnitude of the force that is putting these two things in contact 87 00:06:29,480 --> 00:06:33,970 In this case, it is 49 square root of 3 N 88 00:06:34,750 --> 00:06:43,270 Over the magnitude of the normal force, and that is 49 sqrt of 3 89 00:06:43,300 --> 00:06:48,410 We call this the coefficient of static friction 90 00:06:48,430 --> 00:06:51,690 We're gonna use this a little bit more deeply in other problems 91 00:06:51,720 --> 00:06:55,010 but it tends to hold true for different materials so that 92 00:06:55,320 --> 00:06:59,560 in the future if you have a different mass, you have a different incline 93 00:06:59,600 --> 00:07:01,920 but you have the same materials 94 00:07:01,940 --> 00:07:05,270 given the normal force, you can figure out the budging force 95 00:07:05,270 --> 00:07:07,580 You can figure out exactly how much force you need to put 96 00:07:07,580 --> 00:07:10,870 if you know this, which you usually figure out experimentally 97 00:07:10,880 --> 00:07:16,240 So what would be the value in this case? You have 50 N over 49 square root of 3 N 98 00:07:16,410 --> 00:07:20,240 Let's get the calculator out 99 00:07:21,050 --> 00:07:31,840 So I have 50 divided by 40 times the square root of 3 100 00:07:32,520 --> 00:07:38,430 Gives me .72--I'll just round to two significant digits--0.72 101 00:07:38,440 --> 00:07:41,570 This is 0.72 102 00:07:41,580 --> 00:07:52,210 And then you can use this information. This is the coefficient of static friction 103 00:07:52,220 --> 00:07:55,330 We call it the coefficient of static friction because 104 00:07:55,350 --> 00:08:02,470 this deals with the ratio of the force of friction relative to the normal force 105 00:08:02,470 --> 00:08:07,130 I guess the force just to overcome the force of friction, just kind of get right over that 106 00:08:07,150 --> 00:08:12,330 the most friction that can be applied by kind of the abrasiveness of the 2 things 107 00:08:12,370 --> 00:08:14,510 when the object is stationary 108 00:08:14,520 --> 00:08:17,890 I'll do a whole video on how this is different 109 00:08:17,890 --> 00:08:20,830 when an object is stationary to when it's moving 110 00:08:20,850 --> 00:08:24,430 A lot of times, they're very very very close, but for certain materials 111 00:08:24,440 --> 00:08:28,160 you have a very--at least, a noticeably different coefficient 112 00:08:28,160 --> 00:08:33,280 of friction when the object is stationery as opposed to when it is moving 113 00:08:33,570 --> 00:08:36,730 So I'll leave you there. In the next few videos, we'll use coefficient of friction 114 00:08:36,730 --> 00:00:00,000 or calculate coefficient of friction to do some more problems