1 00:00:00,178 --> 00:00:02,053 - [Instructor] What's Newton's first law say? 2 00:00:02,053 --> 00:00:04,690 Newton's first law states that objects don't change 3 00:00:04,690 --> 00:00:08,060 their velocity unless there's an unbalanced force. 4 00:00:08,060 --> 00:00:09,843 So, if there was no force on an object, 5 00:00:09,843 --> 00:00:12,621 or the forces are balanced, then the object 6 00:00:12,621 --> 00:00:15,441 will continue moving with a constant velocity. 7 00:00:15,441 --> 00:00:16,386 Or, if it was at rest, 8 00:00:16,386 --> 00:00:18,303 it'll continue sitting at rest. 9 00:00:18,303 --> 00:00:19,457 In other words, there doesn't have 10 00:00:19,457 --> 00:00:22,079 to be a net force for something to have motion, 11 00:00:22,079 --> 00:00:23,674 there only has to be a net force 12 00:00:23,674 --> 00:00:25,641 for something to have acceleration. 13 00:00:25,641 --> 00:00:27,354 And it's really important to note 14 00:00:27,354 --> 00:00:30,579 that Newton's first law does not apply to single objects. 15 00:00:30,579 --> 00:00:33,619 It applies to systems of objects as well. 16 00:00:33,619 --> 00:00:35,879 In other words, if you consider a system of objects, 17 00:00:35,879 --> 00:00:38,546 and look at the center of mass of that system, 18 00:00:38,546 --> 00:00:40,009 the center of mass of the system 19 00:00:40,009 --> 00:00:43,191 will remain at rest or remain in constant motion 20 00:00:43,191 --> 00:00:45,461 as long as there's no external 21 00:00:45,461 --> 00:00:47,763 unbalanced forces on the system. 22 00:00:47,763 --> 00:00:50,558 So these objects may be exerting forces on each other, 23 00:00:50,558 --> 00:00:53,089 but the center of mass will remain at rest, 24 00:00:53,089 --> 00:00:55,279 or with constant velocity unless 25 00:00:55,279 --> 00:00:57,786 there's an unbalanced external force 26 00:00:57,786 --> 00:00:59,553 on this system of particles. 27 00:00:59,553 --> 00:01:01,044 So what's an example problem involving 28 00:01:01,044 --> 00:01:02,554 Newton's first law look like? 29 00:01:02,554 --> 00:01:04,260 Say you were told that a heavy elevator 30 00:01:04,260 --> 00:01:07,552 is lifted upward by a cable exerting a force Fc, 31 00:01:07,552 --> 00:01:10,010 and the elevator moves up with a constant velocity 32 00:01:10,010 --> 00:01:11,350 of five meters per second. 33 00:01:11,350 --> 00:01:13,427 We wanna know how the force from the cable 34 00:01:13,427 --> 00:01:15,796 compares to the force of gravity. 35 00:01:15,796 --> 00:01:17,751 The mistake many people make is they think 36 00:01:17,751 --> 00:01:19,940 that since the object was moving upward, 37 00:01:19,940 --> 00:01:22,078 the upward force must be larger, 38 00:01:22,078 --> 00:01:23,116 but that's not true. 39 00:01:23,116 --> 00:01:25,720 Since this is moving upward with constant velocity, 40 00:01:25,720 --> 00:01:28,179 the forces actually have to balance. 41 00:01:28,179 --> 00:01:29,682 Since Newton's first law states 42 00:01:29,682 --> 00:01:31,758 that when the net force is zero, 43 00:01:31,758 --> 00:01:34,420 the object maintains a constant velocity. 44 00:01:34,420 --> 00:01:35,823 And for the net force to be zero, 45 00:01:35,823 --> 00:01:37,506 these forces have to cancel. 46 00:01:37,506 --> 00:01:39,080 So, even though it's non-intuitive, 47 00:01:39,080 --> 00:01:42,018 this cable force has to equal the force of gravity, 48 00:01:42,018 --> 00:01:45,101 so that the elevator can move with constant velocity. 49 00:01:45,101 --> 00:01:46,931 What's Newton's second law mean? 50 00:01:46,931 --> 00:01:48,228 Newton's second law states that 51 00:01:48,228 --> 00:01:50,533 the acceleration of an object is proportional 52 00:01:50,533 --> 00:01:53,823 to the net force and inversely proportional to the mass. 53 00:01:53,823 --> 00:01:55,362 Which, written in equation form, 54 00:01:55,362 --> 00:01:57,492 states that the acceleration of an object 55 00:01:57,492 --> 00:01:59,849 is equal to the net force on that object 56 00:01:59,849 --> 00:02:02,340 divided by the mass of the object. 57 00:02:02,340 --> 00:02:05,349 And this equation works for any single direction as well. 58 00:02:05,349 --> 00:02:07,448 In other words, the acceleration in the x direction 59 00:02:07,448 --> 00:02:08,898 is equal to the net force 60 00:02:08,898 --> 00:02:11,099 in the x direction divided by the mass. 61 00:02:11,099 --> 00:02:13,041 And the acceleration in the y direction 62 00:02:13,041 --> 00:02:14,247 is equal to the net force 63 00:02:14,247 --> 00:02:16,503 in the y direction divided by the mass. 64 00:02:16,503 --> 00:02:18,910 So what's an example of Newton's second law look like? 65 00:02:18,910 --> 00:02:21,112 Let's say a five kg space rock 66 00:02:21,112 --> 00:02:24,060 had the forces acting on it shown in this diagram, 67 00:02:24,060 --> 00:02:25,959 and we wanted to determine the acceleration 68 00:02:25,959 --> 00:02:27,571 in the horizontal direction. 69 00:02:27,571 --> 00:02:29,977 Since horizontal is the x direction, 70 00:02:29,977 --> 00:02:33,037 we're only gonna use forces in the x direction 71 00:02:33,037 --> 00:02:35,857 to determine the acceleration in the x direction. 72 00:02:35,857 --> 00:02:37,743 That means the 15 Newton force, 73 00:02:37,743 --> 00:02:40,354 and the five Newton force don't contribute at all 74 00:02:40,354 --> 00:02:42,509 to the acceleration in the x direction. 75 00:02:42,509 --> 00:02:43,954 The only components that contribute 76 00:02:43,954 --> 00:02:46,496 are the horizontal component of the 10 Newton force, 77 00:02:46,496 --> 00:02:48,946 which would be 10 cosine of 30, 78 00:02:48,946 --> 00:02:50,198 and the 40 Newton force. 79 00:02:50,198 --> 00:02:52,453 So the acceleration in the x direction 80 00:02:52,453 --> 00:02:55,153 would equal the net force in the x direction, 81 00:02:55,153 --> 00:02:56,949 which would be 10 cosine 30. 82 00:02:56,949 --> 00:02:58,686 That would be a positive contribution, 83 00:02:58,686 --> 00:03:00,027 since it points to the right, 84 00:03:00,027 --> 00:03:02,192 minus 40, since that's a negative 85 00:03:02,192 --> 00:03:03,953 contribution pointing to the left. 86 00:03:03,953 --> 00:03:06,536 And finally we'd divide by five kilograms, 87 00:03:06,536 --> 00:03:10,461 which gives us the correct acceleration in the x direction. 88 00:03:10,461 --> 00:03:12,077 What's Newton's third law mean? 89 00:03:12,077 --> 00:03:14,059 Newton's third law states that if an object A 90 00:03:14,059 --> 00:03:15,840 is exerting a force on object B, 91 00:03:15,840 --> 00:03:18,523 then object B must be exerting an equal 92 00:03:18,523 --> 00:03:21,198 and opposite force back on object A. 93 00:03:21,198 --> 00:03:22,720 And this is true even if the objects 94 00:03:22,720 --> 00:03:25,539 have different sizes or there's acceleration. 95 00:03:25,539 --> 00:03:27,321 In other words, let's say the Earth 96 00:03:27,321 --> 00:03:29,390 is pulling on an asteroid. 97 00:03:29,390 --> 00:03:32,375 Even though the Earth is much larger than the asteroid, 98 00:03:32,375 --> 00:03:33,946 the asteroid's gonna exert an 99 00:03:33,946 --> 00:03:36,466 equal and opposite force back on the Earth. 100 00:03:36,466 --> 00:03:38,141 And this is true whether the asteroid 101 00:03:38,141 --> 00:03:39,982 is moving with constant velocity, 102 00:03:39,982 --> 00:03:41,483 or whether it's accelerating. 103 00:03:41,483 --> 00:03:42,538 So what's an example problem 104 00:03:42,538 --> 00:03:44,506 involving Newton's third law look like? 105 00:03:44,506 --> 00:03:47,240 Let's say a metal sphere is sitting on a cardboard box, 106 00:03:47,240 --> 00:03:49,233 and we want to determine which of these choices 107 00:03:49,233 --> 00:03:52,235 constitute a Newton's third law force pair. 108 00:03:52,235 --> 00:03:53,476 The first option says that there's 109 00:03:53,476 --> 00:03:56,434 an upward force on the sphere from the box. 110 00:03:56,434 --> 00:03:58,052 So to find the third law pair, 111 00:03:58,052 --> 00:04:00,583 just reverse the order of the objects, 112 00:04:00,583 --> 00:04:02,479 which means the partner to this force 113 00:04:02,479 --> 00:04:05,076 would be the force on the box by the sphere, 114 00:04:05,076 --> 00:04:07,828 which is not what this says, so it's not option A. 115 00:04:07,828 --> 00:04:09,572 Option B refers to an upward force 116 00:04:09,572 --> 00:04:11,036 on the box from the table, 117 00:04:11,036 --> 00:04:13,062 which we know if we reverse the labels, 118 00:04:13,062 --> 00:04:15,249 should have a partner force that would be the force 119 00:04:15,249 --> 00:04:17,373 on the table by the box. 120 00:04:17,373 --> 00:04:20,024 Which is not what this says, so it's not option B. 121 00:04:20,024 --> 00:04:22,431 Option C talks about an upward force on the sphere 122 00:04:22,431 --> 00:04:24,881 from the box, which, reversing the labels 123 00:04:24,881 --> 00:04:28,449 gives us a partner force of on box by sphere. 124 00:04:28,449 --> 00:04:30,928 Which is not what this says, so it's not option C. 125 00:04:30,928 --> 00:04:34,258 And D refers to an upward force on the box from the table, 126 00:04:34,258 --> 00:04:35,837 which, if we reverse the labels, 127 00:04:35,837 --> 00:04:39,235 gives us a partner force on the table by the box, 128 00:04:39,235 --> 00:04:41,440 which is what this says, and so the forces in D 129 00:04:41,440 --> 00:04:44,492 constitute a Newton's third law force pair. 130 00:04:44,492 --> 00:04:47,353 Which means they must always be equal and opposite. 131 00:04:47,353 --> 00:04:49,632 Other pairs might be equal and opposite, 132 00:04:49,632 --> 00:04:50,918 but no matter what happens, 133 00:04:50,918 --> 00:04:53,730 these two forces have to be equal and opposite. 134 00:04:53,730 --> 00:04:54,602 How do you find the force 135 00:04:54,602 --> 00:04:56,593 of gravity on objects near the Earth? 136 00:04:56,593 --> 00:04:58,829 The force of gravity on all objects near the Earth 137 00:04:58,829 --> 00:05:01,017 is down toward the center of the Earth, 138 00:05:01,017 --> 00:05:02,362 and it's equal to the mass 139 00:05:02,362 --> 00:05:04,569 times the acceleration due to gravity. 140 00:05:04,569 --> 00:05:06,358 Another word for the force of gravity 141 00:05:06,358 --> 00:05:07,898 is the weight of an object. 142 00:05:07,898 --> 00:05:10,593 But be careful, the weight is not the mass. 143 00:05:10,593 --> 00:05:12,553 Weight is the force of gravity which means 144 00:05:12,553 --> 00:05:15,136 weight is m times g not just m. 145 00:05:16,302 --> 00:05:18,047 The force of gravity is a vector, 146 00:05:18,047 --> 00:05:19,620 and it has units of Newtons. 147 00:05:19,620 --> 00:05:20,987 So what's an example problem involving 148 00:05:20,987 --> 00:05:22,519 the force of gravity look like? 149 00:05:22,519 --> 00:05:23,870 Let's say you knew the mass and weight 150 00:05:23,870 --> 00:05:27,205 of a watermelon to be 5 kilograms and 49 Newtons 151 00:05:27,205 --> 00:05:29,206 when you measure them on the Earth. 152 00:05:29,206 --> 00:05:31,049 What might the values for mass and weight 153 00:05:31,049 --> 00:05:33,741 of that watermelon be when it's brought to the moon? 154 00:05:33,741 --> 00:05:35,990 The value of the mass isn't gonna change here 155 00:05:35,990 --> 00:05:37,528 since it's a measure of the total amount 156 00:05:37,528 --> 00:05:39,355 of substance in that object. 157 00:05:39,355 --> 00:05:41,267 But the weight of the watermelon on the moon 158 00:05:41,267 --> 00:05:43,585 is gonna be less since the gravitational pull 159 00:05:43,585 --> 00:05:45,446 is gonna be weaker on the moon. 160 00:05:45,446 --> 00:05:47,712 So the only choice consistent with those two conditions 161 00:05:47,712 --> 00:05:51,382 is A since the mass stays the same and the weight decreases. 162 00:05:51,382 --> 00:05:52,726 What's the normal force? 163 00:05:52,726 --> 00:05:54,551 The normal force is the outward force 164 00:05:54,551 --> 00:05:58,003 exerted by, and perpendicular to a surface. 165 00:05:58,003 --> 00:06:01,110 There's no formula specifically to find the normal force, 166 00:06:01,110 --> 00:06:03,562 you simply have to use Newton's second law. 167 00:06:03,562 --> 00:06:06,169 Let normal force be one of the unknowns, 168 00:06:06,169 --> 00:06:07,459 and then solve for it. 169 00:06:07,459 --> 00:06:08,705 Now, if you've just got a mass 170 00:06:08,705 --> 00:06:10,243 sitting on a horizontal surface, 171 00:06:10,243 --> 00:06:12,316 and there's no extra forces involved, 172 00:06:12,316 --> 00:06:13,834 the normal force is just gonna counter 173 00:06:13,834 --> 00:06:15,383 the force of gravity, which means 174 00:06:15,383 --> 00:06:17,417 the normal force will just be mg. 175 00:06:17,417 --> 00:06:18,996 But if there's extra forces, 176 00:06:18,996 --> 00:06:21,189 or there's acceleration in the direction 177 00:06:21,189 --> 00:06:23,283 of the normal force, then the normal force 178 00:06:23,283 --> 00:06:24,891 is not gonna equal mg, 179 00:06:24,891 --> 00:06:26,586 and you'd have to use Newton's second law 180 00:06:26,586 --> 00:06:28,331 for that direction to solve for it. 181 00:06:28,331 --> 00:06:29,965 The word normal in normal force 182 00:06:29,965 --> 00:06:31,485 refers to the fact that the force 183 00:06:31,485 --> 00:06:33,709 is always perpendicular to the surface 184 00:06:33,709 --> 00:06:35,167 exerting that force. 185 00:06:35,167 --> 00:06:36,183 And it's good to remember that, 186 00:06:36,183 --> 00:06:37,701 for a mass on an incline, 187 00:06:37,701 --> 00:06:40,569 that normal force is not gonna be equal to mg. 188 00:06:40,569 --> 00:06:43,430 It's gonna be mg times cosine of theta. 189 00:06:43,430 --> 00:06:45,613 The normal force is a vector, since it's a force, 190 00:06:45,613 --> 00:06:47,558 and it also has units of Newtons. 191 00:06:47,558 --> 00:06:48,641 So what's an example problem 192 00:06:48,641 --> 00:06:50,343 involving normal force look like? 193 00:06:50,343 --> 00:06:51,384 Let's say a person is pushing 194 00:06:51,384 --> 00:06:53,383 on a stationary box of mass M 195 00:06:53,383 --> 00:06:55,458 against the ceiling with a force Fp, 196 00:06:55,458 --> 00:06:57,117 and they do so at an angle theta. 197 00:06:57,117 --> 00:06:58,477 We want to know what's the magnitude 198 00:06:58,477 --> 00:07:01,426 of the normal force exerted on the box from the ceiling? 199 00:07:01,426 --> 00:07:02,909 So we'll draw a force diagram. 200 00:07:02,909 --> 00:07:04,864 There's the force Fp from the person, 201 00:07:04,864 --> 00:07:06,889 the force mg from gravity. 202 00:07:06,889 --> 00:07:08,546 If the box is stationary there'd have to be a force 203 00:07:08,546 --> 00:07:11,082 preventing it from sliding across the ceiling, 204 00:07:11,082 --> 00:07:13,123 which is most likely static friction. 205 00:07:13,123 --> 00:07:14,523 And there's also gonna be a normal force, 206 00:07:14,523 --> 00:07:17,050 but that normal force will not point upward. 207 00:07:17,050 --> 00:07:20,093 The normal force from the ceiling cannot pull up on the box. 208 00:07:20,093 --> 00:07:22,245 The normal force from the ceiling will only push 209 00:07:22,245 --> 00:07:24,809 out on the box, which will be downward. 210 00:07:24,809 --> 00:07:26,878 Since our normal force is in the vertical direction, 211 00:07:26,878 --> 00:07:29,193 we'll analyze the forces in the vertical direction. 212 00:07:29,193 --> 00:07:31,328 And we can see that the forces must be balanced 213 00:07:31,328 --> 00:07:34,313 vertically, since this box has no motion vertically. 214 00:07:34,313 --> 00:07:35,885 In other words, the normal force 215 00:07:35,885 --> 00:07:39,003 plus the gravitational force is gonna have to equal 216 00:07:39,003 --> 00:07:41,640 the vertical component of the force Fp. 217 00:07:41,640 --> 00:07:44,519 Which, since that's the opposite side from this angle, 218 00:07:44,519 --> 00:07:47,269 we can write as Fp sine of theta. 219 00:07:47,269 --> 00:07:48,897 And now we can solve for the normal force, 220 00:07:48,897 --> 00:07:52,380 which gives us Fp sine theta - mg. 221 00:07:52,380 --> 00:07:54,671 Note that we did not have to use the force of friction, 222 00:07:54,671 --> 00:07:57,169 or the horizontal component since our normal force 223 00:07:57,169 --> 00:07:58,756 was in the vertical direction. 224 00:07:58,756 --> 00:08:00,171 What's the force of tension mean? 225 00:08:00,171 --> 00:08:01,835 The force of tension is any force 226 00:08:01,835 --> 00:08:04,762 exerted by a string, rope, cable, cord 227 00:08:04,762 --> 00:08:06,704 or any other rope like object. 228 00:08:06,704 --> 00:08:07,962 And unlike the normal force 229 00:08:07,962 --> 00:08:10,904 that can only push, tension can only pull. 230 00:08:10,904 --> 00:08:13,574 In other words, ropes can't push on an object. 231 00:08:13,574 --> 00:08:14,865 But, similar to the normal force, 232 00:08:14,865 --> 00:08:16,369 there's no formula for tension. 233 00:08:16,369 --> 00:08:18,877 To find the tension, you'd insert the tension 234 00:08:18,877 --> 00:08:21,497 as an unknown variable into Newton's second law, 235 00:08:21,497 --> 00:08:22,650 and then solve for it. 236 00:08:22,650 --> 00:08:25,082 Since the force of tension is always pulling on objects, 237 00:08:25,082 --> 00:08:26,457 when you draw your force diagram, 238 00:08:26,457 --> 00:08:28,431 make sure you always draw those tension forces 239 00:08:28,431 --> 00:08:30,566 directed away from the object 240 00:08:30,566 --> 00:08:32,558 the string is exerting the tension on. 241 00:08:32,558 --> 00:08:34,234 Tension's a vector, since it's a force, 242 00:08:34,235 --> 00:08:35,701 and it has units of Newtons. 243 00:08:35,701 --> 00:08:37,965 So what's an example problem involving tension look like? 244 00:08:37,965 --> 00:08:40,591 Let's say two ropes are holding up a stationary box, 245 00:08:40,591 --> 00:08:42,068 and we wanna know how the magnitudes 246 00:08:42,068 --> 00:08:44,342 of the tensions in both ropes compare. 247 00:08:44,342 --> 00:08:46,216 Drawing our force diagram there'll be a downward 248 00:08:46,216 --> 00:08:48,792 force of gravity, a force of tension to the left, 249 00:08:48,792 --> 00:08:51,919 and also a diagonal force of tension up and to the right. 250 00:08:51,919 --> 00:08:53,132 Since the box is stationary, 251 00:08:53,132 --> 00:08:55,833 the forces have to be balanced in every direction. 252 00:08:55,833 --> 00:08:57,975 That means the vertical component of T2 253 00:08:57,975 --> 00:09:00,775 has to equal the magnitude of the force of gravity, 254 00:09:00,775 --> 00:09:02,841 and the horizontal component of T2 255 00:09:02,841 --> 00:09:05,846 has to equal the magnitude of the force T1. 256 00:09:05,846 --> 00:09:10,113 But if a component of T2 equals the entire T1, 257 00:09:10,113 --> 00:09:13,661 then the total tension T2 has to be bigger than T1. 258 00:09:13,661 --> 00:09:16,922 In other words, if part of T2 is equal to T1, 259 00:09:16,922 --> 00:09:19,723 then all of T2 is greater than T1. 260 00:09:19,723 --> 00:09:21,718 What's the force of kinetic friction mean? 261 00:09:21,718 --> 00:09:23,852 The force of kinetic friction is the force exerted 262 00:09:23,852 --> 00:09:27,327 between two surfaces that are sliding across each other. 263 00:09:27,327 --> 00:09:28,781 And this force always resists the 264 00:09:28,781 --> 00:09:31,653 sliding motion of those two surfaces. 265 00:09:31,653 --> 00:09:33,641 The force of kinetic friction is proportional 266 00:09:33,641 --> 00:09:36,127 to the normal force between the two surfaces, 267 00:09:36,127 --> 00:09:37,711 and it's proportional to the coefficient 268 00:09:37,711 --> 00:09:40,043 of kinetic friction between the two surfaces. 269 00:09:40,043 --> 00:09:41,905 Note that the force of kinetic friction 270 00:09:41,905 --> 00:09:44,477 does not depend on the velocity of the object. 271 00:09:44,477 --> 00:09:45,966 In other words, if the normal force 272 00:09:45,966 --> 00:09:47,819 and coefficient stay the same, 273 00:09:47,819 --> 00:09:50,803 then no matter how fast or slow the object moves, 274 00:09:50,803 --> 00:09:52,727 no matter how hard or soft you pull, 275 00:09:52,727 --> 00:09:53,851 the force of kinetic friction 276 00:09:53,851 --> 00:09:55,802 is gonna maintain the same value. 277 00:09:55,802 --> 00:09:57,168 Since kinetic friction is a force, 278 00:09:57,168 --> 00:09:59,580 it is a vector and it has units of Newtons. 279 00:09:59,580 --> 00:10:00,828 So what's an example problem 280 00:10:00,828 --> 00:10:02,819 involving kinetic friction look like? 281 00:10:02,819 --> 00:10:04,577 So you've got this question about a car 282 00:10:04,577 --> 00:10:06,451 traveling at cruising speed, 283 00:10:06,451 --> 00:10:08,778 slamming on the brakes, and skidding to a stop. 284 00:10:08,778 --> 00:10:10,104 We want to know what two changes 285 00:10:10,104 --> 00:10:12,843 could be made that would increase the distance required 286 00:10:12,843 --> 00:10:14,704 for the car to skid to a stop. 287 00:10:14,704 --> 00:10:15,875 To get some intuition about what 288 00:10:15,875 --> 00:10:17,921 would cause this car to skid farther, 289 00:10:17,921 --> 00:10:19,578 we could use a kinematic formula. 290 00:10:19,578 --> 00:10:21,211 Since the car skids to a stop, 291 00:10:21,211 --> 00:10:23,010 the final velocity would be zero. 292 00:10:23,010 --> 00:10:24,245 And if we solve for the distance, 293 00:10:24,245 --> 00:10:28,597 we get -v0 squared over two times the acceleration. 294 00:10:28,597 --> 00:10:30,810 So in order to get the car to skid further, 295 00:10:30,810 --> 00:10:33,463 we could increase the initial speed of the car, 296 00:10:33,463 --> 00:10:36,033 or reduce the deceleration. 297 00:10:36,033 --> 00:10:36,942 To figure out what reduces 298 00:10:36,942 --> 00:10:39,059 the magnitude of the acceleration, 299 00:10:39,059 --> 00:10:40,826 we'll use Newton's second law. 300 00:10:40,826 --> 00:10:42,702 The force slowing the skidding car 301 00:10:42,702 --> 00:10:44,322 is the force of kinetic friction, 302 00:10:44,322 --> 00:10:46,687 and since there's no extra vertical forces, 303 00:10:46,687 --> 00:10:48,974 the normal force is just m times g. 304 00:10:48,974 --> 00:10:50,289 Since the masses cancel, 305 00:10:50,289 --> 00:10:52,696 the acceleration won't depend on the mass of the car, 306 00:10:52,696 --> 00:10:54,640 but reducing the coefficient of friction 307 00:10:54,640 --> 00:10:56,546 will reduce the deceleration, 308 00:10:56,546 --> 00:10:58,370 and reducing the deceleration 309 00:10:58,370 --> 00:11:01,561 will increase the distance the car skids to a stop. 310 00:11:01,561 --> 00:11:03,641 The force of static friction tries to prevent 311 00:11:03,641 --> 00:11:06,144 the two surfaces from slipping in the first place, 312 00:11:06,144 --> 00:11:08,021 and that force of static friction will match 313 00:11:08,021 --> 00:11:10,335 whatever force is trying to budge the object 314 00:11:10,335 --> 00:11:12,421 until that budging force matches 315 00:11:12,421 --> 00:11:14,911 the maximum possible static frictional force, 316 00:11:14,911 --> 00:11:17,262 which is proportional to the normal force 317 00:11:17,262 --> 00:11:19,389 and the coefficient of static friction. 318 00:11:19,389 --> 00:11:20,798 So if the maximum value of the 319 00:11:20,798 --> 00:11:22,941 static frictional force is 100 Newtons, 320 00:11:22,941 --> 00:11:25,829 and you try to budge the object with 80 Newtons, 321 00:11:25,829 --> 00:11:27,730 the static frictional force will just oppose you 322 00:11:27,730 --> 00:11:30,772 with 80 Newtons, preventing the object from slipping. 323 00:11:30,772 --> 00:11:33,389 If you exert 90 Newtons, the static frictional force 324 00:11:33,389 --> 00:11:35,172 will increase to 90 Newtons, 325 00:11:35,172 --> 00:11:36,715 preventing the object from slipping. 326 00:11:36,715 --> 00:11:38,883 But if you exert 110 Newtons, 327 00:11:38,883 --> 00:11:41,356 since this exceeds the maximum possible 328 00:11:41,356 --> 00:11:44,057 static frictional force, the object will budge 329 00:11:44,057 --> 00:11:46,255 and there will only be a kinetic frictional force 330 00:11:46,255 --> 00:11:47,998 now that the object is sliding. 331 00:11:47,998 --> 00:11:49,325 So what's an example problem involving 332 00:11:49,325 --> 00:11:51,169 the force of static friction look like? 333 00:11:51,169 --> 00:11:54,350 Let's say you push on a refrigerator that's 180 kilograms, 334 00:11:54,350 --> 00:11:55,883 and the coefficient of static friction 335 00:11:55,883 --> 00:11:57,881 between the floor and the fridge is .8. 336 00:11:57,881 --> 00:12:00,069 If you exert 50 Newtons on the refrigerator, 337 00:12:00,069 --> 00:12:02,852 what's the magnitude of the static friction force 338 00:12:02,852 --> 00:12:04,511 exerted on the refrigerator? 339 00:12:04,511 --> 00:12:06,658 We'll first find the maximum possible 340 00:12:06,658 --> 00:12:10,082 static frictional force using mew s times Fn. 341 00:12:10,082 --> 00:12:11,868 Since there's no extra vertical forces, 342 00:12:11,868 --> 00:12:13,994 the normal force will just be mg. 343 00:12:13,994 --> 00:12:16,485 Plugging in values, we get a maximum possible 344 00:12:16,485 --> 00:12:19,633 static frictional force of 1,411 Newtons, 345 00:12:19,633 --> 00:12:21,441 but this will not be the value 346 00:12:21,441 --> 00:12:22,804 of the static frictional force. 347 00:12:22,804 --> 00:12:24,475 This is just the maximum value 348 00:12:24,475 --> 00:12:26,089 of the static frictional force. 349 00:12:26,089 --> 00:12:28,303 So if we exert 50 Newtons to the right, 350 00:12:28,303 --> 00:12:30,103 since that does not exceed this 351 00:12:30,103 --> 00:12:32,277 maximum possible static frictional force, 352 00:12:32,277 --> 00:12:34,200 static friction will just oppose us 353 00:12:34,200 --> 00:12:36,758 with an equal 50 Newtons to the left. 354 00:12:36,758 --> 00:12:39,332 And it will continue to match whatever force we exert, 355 00:12:39,332 --> 00:12:40,919 until we exceed the maximum 356 00:12:40,919 --> 00:12:43,102 possible static frictional force. 357 00:12:43,102 --> 00:12:45,201 How do you deal with inclines? 358 00:12:45,201 --> 00:12:47,321 Inclines are just angled surfaces that objects 359 00:12:47,321 --> 00:12:49,701 can slide up or down, and since the object 360 00:12:49,701 --> 00:12:53,442 can't move into the incline, or off of the incline, 361 00:12:53,442 --> 00:12:55,235 the motion will only be taking place 362 00:12:55,235 --> 00:12:57,535 parallel to the surface of the incline. 363 00:12:57,535 --> 00:12:59,986 There will be no acceleration perpendicular 364 00:12:59,986 --> 00:13:01,360 to the surface of the incline. 365 00:13:01,360 --> 00:13:03,679 So instead of breaking our forces into x and y, 366 00:13:03,679 --> 00:13:05,886 we break them into forces perpendicular 367 00:13:05,886 --> 00:13:08,219 to the surface and parallel to the surface. 368 00:13:08,219 --> 00:13:10,061 The component of gravity that's parallel 369 00:13:10,061 --> 00:13:13,346 to the surface is gonna equal mg sine theta, 370 00:13:13,346 --> 00:13:16,411 where theta is the angle between the horizontal floor, 371 00:13:16,411 --> 00:13:18,063 and the inclined surface. 372 00:13:18,063 --> 00:13:20,022 And the component of gravity perpendicular 373 00:13:20,022 --> 00:13:23,086 to the surface is gonna be mg cosine theta, 374 00:13:23,086 --> 00:13:25,402 where again, theta is the angle measured between 375 00:13:25,402 --> 00:13:27,931 the horizontal floor and the angled surface. 376 00:13:27,931 --> 00:13:29,969 Since there's no acceleration perpendicular 377 00:13:29,969 --> 00:13:32,087 to the surface, the net force in the 378 00:13:32,087 --> 00:13:34,547 perpendicular direction has to be zero. 379 00:13:34,547 --> 00:13:36,325 And that means this perpendicular component 380 00:13:36,325 --> 00:13:38,442 of gravity has to be exactly canceled 381 00:13:38,442 --> 00:13:40,434 by the normal force, which is why the value 382 00:13:40,434 --> 00:13:42,086 for the normal force is the same 383 00:13:42,086 --> 00:13:44,204 as the perpendicular component of gravity. 384 00:13:44,204 --> 00:13:46,777 And since those perpendicular components cancel, 385 00:13:46,777 --> 00:13:49,802 the total net force on an object on an incline 386 00:13:49,802 --> 00:13:52,393 is just gonna equal the component of the net force 387 00:13:52,393 --> 00:13:54,718 that's parallel to the surface of the incline. 388 00:13:54,718 --> 00:13:55,961 Which, if there's no friction, 389 00:13:55,961 --> 00:13:58,039 would simply be mg sine theta, 390 00:13:58,039 --> 00:13:59,154 and if there was friction, 391 00:13:59,154 --> 00:14:02,233 it would be mg sine theta minus the force of friction. 392 00:14:02,233 --> 00:14:03,353 But be careful when you're finding 393 00:14:03,353 --> 00:14:05,310 the force of friction on an incline, 394 00:14:05,310 --> 00:14:08,263 the normal force will not be m times g, 395 00:14:08,263 --> 00:14:11,470 the normal force is gonna be mg cosine theta. 396 00:14:11,470 --> 00:14:14,127 So what's an example problem involving inclines look like? 397 00:14:14,127 --> 00:14:16,208 Let's say a box started with a huge speed 398 00:14:16,208 --> 00:14:17,986 at the bottom of a frictionless ramp, 399 00:14:17,986 --> 00:14:19,400 and then it slides up the ramp 400 00:14:19,400 --> 00:14:22,251 and through these points shown w, x, y and z. 401 00:14:22,251 --> 00:14:23,784 We want to rank the magnitudes 402 00:14:23,784 --> 00:14:25,734 of the net force on the box 403 00:14:25,734 --> 00:14:27,559 for these points that are indicated. 404 00:14:27,559 --> 00:14:29,269 Well, when the box is flying through the air, 405 00:14:29,269 --> 00:14:31,338 we know the net force is simply the force 406 00:14:31,338 --> 00:14:34,288 of gravity straight down which is m times g. 407 00:14:34,288 --> 00:14:36,284 So the net force at y and z are equal. 408 00:14:36,284 --> 00:14:37,784 And, on an incline, the net force 409 00:14:37,784 --> 00:14:39,660 is the forced component that's parallel 410 00:14:39,660 --> 00:14:41,085 to the surface of the incline, 411 00:14:41,085 --> 00:14:43,586 which is gonna be mg sine theta. 412 00:14:43,586 --> 00:14:46,213 Note that the net force on the incline points down 413 00:14:46,213 --> 00:14:49,514 the incline, even though the mass moves up the incline. 414 00:14:49,514 --> 00:14:51,638 That just means the mass is slowing down, 415 00:14:51,638 --> 00:14:55,164 but since mg is greater than mg sine theta, 416 00:14:55,164 --> 00:14:59,355 z and y are greater than the net force at x and w. 417 00:14:59,355 --> 00:15:02,429 What does treating systems as as single object mean? 418 00:15:02,429 --> 00:15:03,672 This is a trick you can use when 419 00:15:03,672 --> 00:15:06,382 two or more objects are required to move 420 00:15:06,382 --> 00:15:08,503 with the same speed and acceleration, 421 00:15:08,503 --> 00:15:09,971 which will allow us to avoid 422 00:15:09,971 --> 00:15:13,138 having to use multiple equations to find the acceleration, 423 00:15:13,138 --> 00:15:16,381 and instead use one equation to get the acceleration. 424 00:15:16,381 --> 00:15:17,790 When you treat a system of objects 425 00:15:17,790 --> 00:15:21,482 as a single object, you get to ignore internal forces, 426 00:15:21,482 --> 00:15:24,272 since the internal forces will always cancel. 427 00:15:24,272 --> 00:15:26,556 This means you can find the acceleration of the system 428 00:15:26,556 --> 00:15:30,233 by looking at only the external forces on that system, 429 00:15:30,233 --> 00:15:33,083 and then dividing by the total mass of that system. 430 00:15:33,083 --> 00:15:34,459 So what's an example of treating 431 00:15:34,459 --> 00:15:36,525 systems as a single object look like? 432 00:15:36,525 --> 00:15:39,518 Let's say a mass m1 is pulled across a rough 433 00:15:39,518 --> 00:15:43,099 horizontal table by a rope connected to mass m2. 434 00:15:43,099 --> 00:15:44,550 If the coefficient of kinetic friction 435 00:15:44,550 --> 00:15:47,216 between m1 and the table is mew k, 436 00:15:47,216 --> 00:15:48,300 then what's an expression for the 437 00:15:48,300 --> 00:15:51,071 magnitude of the acceleration of the masses? 438 00:15:51,071 --> 00:15:52,846 So instead of analyzing the forces 439 00:15:52,846 --> 00:15:54,665 on each mass individually, 440 00:15:54,665 --> 00:15:56,363 which would give us multiple equations 441 00:15:56,363 --> 00:15:58,964 and multiple unknowns, we'll use one equation 442 00:15:58,964 --> 00:16:00,979 of Newton's second law but we'll treat this system 443 00:16:00,979 --> 00:16:02,847 as if it were a single object. 444 00:16:02,847 --> 00:16:04,546 Which means we're basically just gonna ask 445 00:16:04,546 --> 00:16:07,559 what external forces are gonna make this system go? 446 00:16:07,559 --> 00:16:10,532 And what external forces are gonna make this system stop? 447 00:16:10,532 --> 00:16:12,647 The external force makes this system go 448 00:16:12,647 --> 00:16:14,856 is the force of gravity on m2. 449 00:16:14,856 --> 00:16:17,624 It's an external force, since it's exerted by the Earth, 450 00:16:17,624 --> 00:16:18,949 which is not part of our system, 451 00:16:18,949 --> 00:16:20,356 and it's making the system go, 452 00:16:20,356 --> 00:16:22,215 so we'll call that force positive. 453 00:16:22,215 --> 00:16:23,333 And we'll call forces that try 454 00:16:23,333 --> 00:16:25,163 to make our system stop negative, 455 00:16:25,163 --> 00:16:27,607 like this force of kinetic friction on m1, 456 00:16:27,607 --> 00:16:29,799 which is also external because the table 457 00:16:29,799 --> 00:16:31,736 is not part of the mass of our system. 458 00:16:31,736 --> 00:16:34,016 But we will not include the force of tension, 459 00:16:34,016 --> 00:16:35,798 since this is an internal force, 460 00:16:35,798 --> 00:16:37,499 and these forces will cancel. 461 00:16:37,499 --> 00:16:39,763 Now, since we're treating this system as a single mass, 462 00:16:39,763 --> 00:16:42,108 we'll divide by the total mass of our system. 463 00:16:42,108 --> 00:16:44,058 And then if we write the force of kinetic friction 464 00:16:44,058 --> 00:16:45,751 in terms of the coefficient, 465 00:16:45,751 --> 00:16:48,022 we get mew k times the normal force, 466 00:16:48,022 --> 00:16:52,084 and the normal force on m1 is gonna be m1 times g. 467 00:16:52,084 --> 00:16:54,451 Which, with a single equation, gives us an expression 468 00:16:54,451 --> 00:16:56,109 for the acceleration of our system, 469 00:16:56,109 --> 00:16:58,298 without having to solve multiple equations 470 00:16:58,298 --> 00:00:00,000 with multiple unknowns.