1 00:00:00,570 --> 00:00:04,680 So I have got this block of wood here that has a mass of 5 kilograms 2 00:00:04,680 --> 00:00:08,830 and it is sitting on some dirt and we are near the surface of the earth 3 00:00:08,830 --> 00:00:16,720 and the coefficient of static friction between this type of wood and this type of dirt is 0.60 4 00:00:16,720 --> 00:00:23,030 and the coefficient of kinetic friction between this type of wood and this type of dirt is 0.55 5 00:00:23,030 --> 00:00:25,630 This was measured by someone else long ago 6 00:00:25,630 --> 00:00:28,060 or you found it in some type of a book someplace 7 00:00:28,060 --> 00:00:32,299 And let's say we push on this side of the block with a force of a 100 N 8 00:00:32,299 --> 00:00:34,440 What is going to happen? 9 00:00:34,440 --> 00:00:37,250 So the first thing you might realize is if there is no friction 10 00:00:37,250 --> 00:00:39,310 if this was a completely frictionless boundary and there is 11 00:00:39,310 --> 00:00:42,640 no air resistance, we are assuming that there is no air resistance in this example 12 00:00:42,640 --> 00:00:45,320 That in this dimension, in the horizontal dimension 13 00:00:45,320 --> 00:00:48,220 there would only be one force here, this 100 N force 14 00:00:48,220 --> 00:00:51,810 It would be completely unbalanced and that would be the net force 15 00:00:51,810 --> 00:00:56,350 and so you would have a force going in that direction of a 100 N on a mass of 5 kilograms 16 00:00:56,350 --> 00:00:59,540 Force = Mass times acceleration 17 00:00:59,540 --> 00:01:01,580 acceleration and force are vector quantities 18 00:01:01,580 --> 00:01:04,110 So you would have the force divided by the mass 19 00:01:04,110 --> 00:01:08,640 would give you 20 meters per second of acceleration in the rightward direction 20 00:01:08,640 --> 00:01:10,950 That is if there were no friction 21 00:01:10,950 --> 00:01:13,040 but there is friction in this situation 22 00:01:13,040 --> 00:01:15,160 So let's think about how we'll deal with it 23 00:01:15,160 --> 00:01:17,950 So the coefficient of friction tells us 24 00:01:17,950 --> 00:01:22,740 So this right here is the ratio between the magnitude of the force 25 00:01:22,740 --> 00:01:24,460 that I have called the budging force 26 00:01:24,460 --> 00:01:28,820 The amount of force you need to apply to get this thing to budge 27 00:01:28,820 --> 00:01:32,670 to get this thing to start moving. So we can start using the coefficient of kinetic friction 28 00:01:32,670 --> 00:01:38,200 It's the ratio between that and the magnitude of the force of contact 29 00:01:38,200 --> 00:01:41,100 between this block and the floor or ground here 30 00:01:41,100 --> 00:01:44,040 And the magnitude of that force of contact is the same thing 31 00:01:44,040 --> 00:01:48,410 as the normal force that the ground is applying on the block 32 00:01:48,410 --> 00:01:52,810 the magnitude of the normal force the ground is applying on the block 33 00:01:52,810 --> 00:01:54,150 Then once its moving 34 00:01:54,150 --> 00:01:58,650 then we can say that this is going to be--this will then be equal to 35 00:01:58,650 --> 00:02:01,710 this over here will be equal to the force of friction 36 00:02:01,710 --> 00:02:05,140 So this is the force that really overcome friction 37 00:02:05,140 --> 00:02:08,360 and this over here will be equal to the force of friction 38 00:02:08,360 --> 00:02:13,280 The magnitude of the force of friction over the force of contact 39 00:02:13,280 --> 00:02:16,720 the contact force between those two, so over the normal force 40 00:02:16,720 --> 00:02:18,140 and it makes sense 41 00:02:18,140 --> 00:02:21,030 that the larger the contact force 42 00:02:21,030 --> 00:02:23,050 the more that these are being pressed together 43 00:02:23,050 --> 00:02:26,210 the little at the atomic level, they kind of really get into each others grooves 44 00:02:26,210 --> 00:02:28,430 the more budging force you would need 45 00:02:28,430 --> 00:02:31,620 or the more friction force would go against your motion 46 00:02:31,620 --> 00:02:33,650 And in either situation 47 00:02:33,650 --> 00:02:35,920 the force of friction is going against your motion 48 00:02:35,920 --> 00:02:37,320 So even if you push it in that way 49 00:02:37,320 --> 00:02:40,420 sounds like force of friction is all of a sudden going to help you 50 00:02:40,420 --> 00:02:44,160 So let's think about what the necessary force will we need 51 00:02:44,160 --> 00:02:47,500 to overcome the force of friction right here in the static situation 52 00:02:47,500 --> 00:02:51,960 So the force of gravity on this block 53 00:02:51,960 --> 00:02:57,760 is going to be the gravitational field which is 9.8 m/s^2 times 5 kilograms 54 00:02:57,760 --> 00:03:05,550 9.8 m/s times 5 kilograms gives 49 kilogram meters per second or 49 newtons down 55 00:03:05,550 --> 00:03:08,390 This is the force, the magnitude of the force due to gravity 56 00:03:08,390 --> 00:03:11,540 the direction is straight down towards the center of the earth 57 00:03:11,540 --> 00:03:17,140 The normal force, and that force is there because this block is not accelerating downwards 58 00:03:17,140 --> 00:03:21,390 So there must be some force that completely balances off the force of gravity 59 00:03:21,390 --> 00:03:26,030 And in this example, it is the normal force 60 00:03:26,030 --> 00:03:30,100 So it is acting 49 newtons upward 61 00:03:30,100 --> 00:03:35,060 and so these net out. And that's why this block does not accelerate upwards or downwards 62 00:03:35,060 --> 00:03:36,970 So what we have is the budge the 63 00:03:36,970 --> 00:03:41,050 magnitude of the budging force, needs to be equal to, over the magnitude of the normal force 64 00:03:41,050 --> 00:03:43,620 well this thing right over here is going to be 49 newtons 65 00:03:43,620 --> 00:03:46,810 Is equal to 0.60 66 00:03:46,810 --> 00:03:52,370 Or we could say that the magnitude of the budging force 67 00:03:52,370 --> 00:03:56,700 is equal to 49 newtons times the coefficient of static fiction 68 00:03:56,700 --> 00:04:06,710 Or that's 49 newtons times 0.60 69 00:04:06,710 --> 00:04:09,910 And remember coefficient of friction are unitless 70 00:04:09,910 --> 00:04:13,040 So the units here are still going to be in newtons 71 00:04:13,040 --> 00:04:22,580 So this 49 times .6 gives us 29.4 newtons 72 00:04:22,580 --> 00:04:30,550 This is equal to 29.4 newtons 73 00:04:30,550 --> 00:04:34,520 So that's the force that's started to overcome static friction 74 00:04:34,520 --> 00:04:36,790 which we are applying more than enough of 75 00:04:36,790 --> 00:04:39,920 so with a 100 newtons, we would just start to budge it 76 00:04:39,920 --> 00:04:42,100 and right when we are in just in that moment 77 00:04:42,100 --> 00:04:44,190 where that thing is just starting to move 78 00:04:44,190 --> 00:04:45,580 the net force-- 79 00:04:45,580 --> 00:04:47,670 so we have a 100 newtons going in that direction 80 00:04:47,670 --> 00:04:51,090 and the force of static friction is going to go in this direction-- 81 00:04:51,090 --> 00:04:53,580 maybe I could draw it down here to show it's coming from right over here 82 00:04:53,580 --> 00:04:57,970 The force of static friction is going to be 29.4 newtons that way 83 00:04:57,970 --> 00:05:00,650 and so right when I am just starting to budge this 84 00:05:00,650 --> 00:05:03,030 just when that little movement-- 85 00:05:03,070 --> 00:05:04,930 because once I do that, then all of a sudden it's moving 86 00:05:05,030 --> 00:05:09,180 and then kinetic friction starts to matter, but just for that moment 87 00:05:09,650 --> 00:05:15,930 just for that moment I'll have a net force of 100 - 29.4 88 00:05:16,360 --> 00:05:28,660 to the right, so I have a net force of 70.6 N 89 00:05:28,660 --> 00:05:32,820 for just a moment while I budge it 90 00:05:32,820 --> 00:05:35,250 So just exactly while I'm budging it 91 00:05:35,250 --> 00:05:42,380 While we're overcoming the static friction, we have a 70.6 N net force in the right direction 92 00:05:42,690 --> 00:05:47,980 And so just for that moment, you divide it by 5 kg mass 93 00:05:48,000 --> 00:05:52,430 So just for that moment, it will be accelerating at 14.12 m/s^2 94 00:05:53,050 --> 00:06:00,280 So you'll have an acceleration of 14.1 m/s^2 to the right 95 00:06:00,300 --> 00:06:04,250 but that will just be for that absolute moment, because once I budge it 96 00:06:04,360 --> 00:06:07,420 all of a sudden the block will start to be moving 97 00:06:07,440 --> 00:06:10,880 And once it's moving, the coefficient of kinetic friction starts to matter 98 00:06:10,880 --> 00:06:12,880 We've got the things out of their little grooves 99 00:06:12,900 --> 00:06:17,160 and so they're kind of gliding past each other on the top, although there still is resistant 100 00:06:17,200 --> 00:06:20,550 So once we budge it, we'll have that acceleration for just a moment 101 00:06:20,560 --> 00:06:23,550 Now all of a sudden, the coefficient of kinetic friction comes to play 102 00:06:23,930 --> 00:06:27,080 And the force of friction, assuming we're moving 103 00:06:27,410 --> 00:06:31,120 the magnitude of the force of friction will always go against our movement 104 00:06:31,550 --> 00:06:35,460 is going to be--remember, our normal force is 49 N 105 00:06:35,700 --> 00:06:38,850 So we can multiply both sides of this times 49 106 00:06:39,000 --> 00:06:57,780 We get 49 N times 0.55 which is equal to 26.95 N 107 00:06:57,800 --> 00:07:00,400 This is the force of friction; this is the magnitude 108 00:07:00,520 --> 00:07:02,400 and it's going to go against our motions 109 00:07:02,410 --> 00:07:05,690 So as soon as we start to move in that direction, the force of friction 110 00:07:05,890 --> 00:07:09,480 is going to be going in that direction 111 00:07:09,740 --> 00:07:15,050 So once we start moving, assuming that I'm continuing to apply this 100 newtons of force 112 00:07:15,130 --> 00:07:18,280 what is the net force? So I have 100 N going that way 113 00:07:18,380 --> 00:07:23,180 and I have 26.95 going that way 114 00:07:23,200 --> 00:07:25,080 Remember, with vectors, I don't have to draw them here 115 00:07:25,080 --> 00:07:28,890 I can draw all of their tails start at the center of mass of the 116 00:07:28,890 --> 00:07:32,810 object. I can draw them whatever, but remember this is acting on the object 117 00:07:32,810 --> 00:07:37,140 If we want to be precise, we can show it on the center of mass because 118 00:07:37,140 --> 00:07:39,940 we can view all of these atoms as one collective object 119 00:07:39,950 --> 00:07:42,150 But anyway, what is the net force now? 120 00:07:42,160 --> 00:07:46,420 We have 100 N to the right; we have 26.95 to the left 121 00:07:47,080 --> 00:07:51,750 100 minus 26.95 122 00:07:51,750 --> 00:07:53,600 100 N that I'm applying to the right 123 00:07:53,600 --> 00:07:59,260 - 26.95 N which is the force of friction to the left always acting against us 124 00:07:59,410 --> 00:08:03,410 means that there's a net force to the right of 73.05 125 00:08:03,430 --> 00:08:10,840 So once we're moving, we have a net force to the right of 73.05 N 126 00:08:10,840 --> 00:08:18,070 This is the net force and it's acting to the right 127 00:08:18,260 --> 00:08:22,850 Right after we budge it, how quickly will this accelerate? 128 00:08:22,850 --> 00:08:32,700 Well, 73.05 divided by the mass, divided by 5 kg, gives us 14.61 129 00:08:33,220 --> 00:08:40,430 So the acceleration once we're moving is going to be 14.61 m/s squared 130 00:08:40,630 --> 00:08:44,450 to the right 131 00:08:44,490 --> 00:08:47,190 So I really want to make sure you understand what's happening here 132 00:08:47,220 --> 00:08:51,270 We always have enough force to start budging it 133 00:08:51,560 --> 00:08:53,510 but right when we budged it 134 00:08:53,510 --> 00:08:56,430 we overcome the static friction for just a moment 135 00:08:56,430 --> 00:08:59,810 our acceleration was slower 136 00:08:59,820 --> 00:09:01,900 because we're overcoming that static friction 137 00:09:01,900 --> 00:09:04,900 but once we budged it and once it's moving 138 00:09:04,920 --> 00:09:08,600 and assuming that we're continuing to apply a constant force over here 139 00:09:08,790 --> 00:09:10,480 then all of a sudden, the force of friction since 140 00:09:10,480 --> 00:09:14,400 we're kind of bump it along the top now and not stuck in their grooves 141 00:09:14,920 --> 00:09:17,570 we're now using the coefficient of kinetic friction 142 00:09:17,590 --> 00:09:22,780 And so once it's moving, the net force becomes greater in the rightward direction because 143 00:09:22,930 --> 00:09:27,590 you can kind of view that force of friction will become less once it starts moving 144 00:09:27,790 --> 00:09:31,820 And so now the force of friction went down a little bit to 26.95 N 145 00:09:31,870 --> 00:09:38,630 And so now we're accelerating to right at a slightly faster rate 14.61 m/s^2 146 00:09:38,640 --> 00:09:41,860 So right when you budge it, it accelerates at 14.1 m/s^2 147 00:09:41,860 --> 00:09:44,870 but just for a moment, almost unnoticeable moment once it starts moving 148 00:09:44,920 --> 00:00:00,000 Then you're going to be going to the right with this constant acceleration