#### Static Friction and Kinetic Friction Physics Problems With Free Body Diagrams - By The Organic Chemistry Tutor

00:00 | in this video , we're going to focus on static | |

00:03 | friction and kinetic friction . Now let's say if this | |

00:08 | is a carpet floor and there's a big box and | |

00:15 | imagine if you're trying to push the box , so | |

00:18 | you're trying to apply a force to move it . | |

00:22 | Now , initially , as you begin to push it | |

00:25 | , the box doesn't move . But eventually , if | |

00:29 | you continue to push it with even more force , | |

00:32 | it will begin to slide and once begin sliding it's | |

00:36 | easier to continue pushing it . But once you stop | |

00:39 | it , it's going to be hard to start it | |

00:41 | up again . The force that prevents the box from | |

00:46 | sliding at the first place or in the first place | |

00:48 | is the static frictional force . Now , once it | |

00:53 | begins to slide , the force that impedes the motion | |

00:57 | is kinetic friction , static means not moving , kinetic | |

01:03 | has to deal with motion . So the frictional force | |

01:06 | that opposes motion when the object is sliding against the | |

01:10 | carpet , that's kinetic friction , the frictional force that | |

01:13 | prevents you from moving it when you first try to | |

01:16 | push it . That static friction static friction . Let | |

01:21 | me write the equations on the right . Static friction | |

01:24 | is less than or equal to New S . Times | |

01:28 | the normal force kinetic friction . It's equal two in | |

01:32 | UK times dinner , more force . So notice that | |

01:37 | the static frictional force is represented by the inequality . | |

01:41 | Which means that it's not just a fixed number , | |

01:45 | it can be a range of numbers up to a | |

01:47 | maximum point . The kinetic frictional force , however , | |

01:51 | is not represented by an inequality . So therefore F | |

01:56 | . K . Represents a fixed number . Now let's | |

02:00 | say that this is a five kg box . What | |

02:05 | is the normal force And calculate these two values with | |

02:09 | this information ? Now this box exerts a downward weight | |

02:14 | force and the normal force has to support the way | |

02:17 | force keep in mind , the normal force is force | |

02:21 | that the surface exerts on the box and it's perpendicular | |

02:25 | to the surface . So in this example for horizontal | |

02:30 | surface , the normal force is equal to the way | |

02:32 | force , or MG . So it's gonna be five | |

02:36 | kg Times 9.8 meters per second squared . So the | |

02:42 | normal force is 49 Nunes . Now , I need | |

02:46 | to give you values from us and UK um us | |

02:51 | is typically greater than you haven't seen example when it's | |

02:55 | less . So we're going to say that um us | |

02:58 | is 0.4 and we're going to say in UK , | |

03:04 | this points you So let's calculate F cane and FS | |

03:12 | . So the kinetic frictional force is going to be | |

03:14 | point to times 49 Which is 9.8 . The static | |

03:23 | fictional force is going to be less than or equal | |

03:24 | to 0.4 times 49 Which is 19.6 . So what | |

03:35 | do these numbers mean ? So , let me give | |

03:38 | an example that's going to illustrate this . Let's make | |

03:46 | a table between the applied force , the static frictional | |

03:54 | force , the kinetic frictional force and the net force | |

04:03 | . So if the applied force is zero , The | |

04:06 | frictional forces will be zero and the net force will | |

04:08 | be zero . If the person doesn't push the box | |

04:11 | , the box will move , nothing's going to happen | |

04:14 | . But now let's say if the person applies a | |

04:17 | force of Let's say 10 newtons , what's going to | |

04:22 | happen ? Well , the box begin to slide . | |

04:25 | And what are the values of the static and kinetic | |

04:27 | frictional forces ? Now , even though kinetic friction is | |

04:34 | 9.8 , that value only applies . If the box | |

04:38 | slides . Now , what is the static frictional force | |

04:44 | ? Will the box move notice that the apply force | |

04:48 | is less , then the maximum static frictional force , | |

04:51 | so it's not going to move . Which means that | |

04:55 | there is no kinetic frictional force . The kinetic frictional | |

04:59 | force only exists if the box is sliding against the | |

05:03 | carpet , you can't have static friction and kinetic friction | |

05:09 | present at the same time . If the box is | |

05:11 | not sliding against the carpet , static friction is present | |

05:15 | . If it is sliding , kinetic friction is present | |

05:19 | . So what is static friction in this example ? | |

05:24 | What number should you put here ? Now ? If | |

05:27 | you're thinking about putting 19.6 , that will not be | |

05:30 | correct . Because imagine If the person applies a force | |

05:34 | of 10 newtons to the right to push the box | |

05:37 | And if static friction applies a force of 19.6 Nunes | |

05:41 | , that means that there's gonna be a net force | |

05:44 | of 9.6 Nunes towards the left . So imagine pushing | |

05:47 | the box only to find out the box is pushing | |

05:50 | you back to the left . That just doesn't happen | |

05:53 | . So static friction can be 19.6 . It turns | |

05:57 | out that static friction is going to match the applied | |

06:02 | force until you exceed its maximum value . So if | |

06:06 | you push it with 10 units it's going to push | |

06:08 | back on you with 10 units . And so the | |

06:11 | box doesn't move . You're you're trying to push it | |

06:13 | doesn't move . So the net force is zero . | |

06:16 | So let's say if you try to push it with | |

06:18 | 15 unions Then it's going to push back on you | |

06:21 | with 15 newtons . That means you haven't implied enough | |

06:25 | force to start moving it . So that's what happens | |

06:29 | when you try to push the box initially , you're | |

06:32 | pushing hard against it . But it's not moving because | |

06:35 | static friction is matching your applied force until it reaches | |

06:39 | a maximum value . So this is still gonna be | |

06:42 | zero . So let's say if you apply force in | |

06:44 | 19.6 newtons It's still gonna be 19.6 . No When | |

06:50 | you exceed 19.6 , that's when it begins to slide | |

06:54 | . And so you no longer have static friction but | |

06:56 | you have kinetic friction . So let's say if we | |

06:59 | go just above 19.6 , let's say if we increase | |

07:02 | it to 20 newtons now the box begins to slide | |

07:07 | and so there is no more static friction because the | |

07:10 | surfaces are sliding past each other . But there is | |

07:14 | Kinetic fiction which is always going to be 9.8 once | |

07:17 | the box begins to slide , So it's 20 -9.8 | |

07:21 | Which will give you a net force of 10.2 . | |

07:24 | Now , if you decide to increase the applied force | |

07:27 | , The started fictional force will still be zero . | |

07:30 | F . K . is going to still be 9.8 | |

07:33 | And the net force is now 30 -9.8 , Which | |

07:36 | is 20.2 . So hopefully this example help you to | |

07:41 | understand the difference between static friction and kinetic friction and | |

07:45 | how to calculate it based on the applied force . | |

07:50 | Let's work on this problem . A 15 kg Box | |

07:53 | rests on a horizontal surface . What is the minimum | |

07:57 | horizontal force that is required to cause the box to | |

08:00 | begin to slide . If the coefficient of static friction | |

08:04 | is .35 so you can pause the video if you | |

08:07 | want to work on this problem as well . But | |

08:09 | let's start with the picture . So this is the | |

08:12 | 15 kg box . So we wish to apply a | |

08:19 | horizontal force which we're going to call capital F . | |

08:23 | And we know that friction is going to oppose it | |

08:28 | . Now if we want the minimum horizontal force that | |

08:30 | is required to cause the box to begin to slide | |

08:34 | , we need to use static friction because until the | |

08:39 | force exceeds static friction only then can it slide ? | |

08:42 | If it doesn't exceed the static frictional force then it | |

08:46 | won't slide . So the threshold is the maximum static | |

08:49 | frictional force . So we need to set F . | |

08:53 | Equal to F . S . And that's when it | |

08:57 | begins to slide when these two have the same magnitude | |

09:01 | . Now the static frictional force , its maximum value | |

09:04 | is mu . S . Times the normal force . | |

09:07 | And in this example the normal force is going to | |

09:11 | be MG . So um us is .35 mm is | |

09:20 | 15 And G . is 9.8 . So the applied | |

09:30 | force has to be 51.45 newtons . In order for | |

09:36 | the box to begin to slide . If it's less | |

09:39 | than its value , the box will not slide , | |

09:41 | it will not move . Even if it equals this | |

09:45 | value , The net force will still be zero , | |

09:48 | it has to be just above so it's 51.46 . | |

09:51 | It will move . If it's 51.44 it's not going | |

09:55 | to move 51.45 , that's the threshold . So it | |

10:00 | really doesn't move at that point . So technically the | |

10:03 | applied force Has to be just above 51.45 . Before | |

10:09 | all practical purposes we're going to say this is the | |

10:11 | threshold value . So we'll go with that . Now | |

10:15 | , what about part B ? What is the acceleration | |

10:18 | of the system ? If a person Pushes the box | |

10:22 | with the force and 90 So 90 is greater than | |

10:25 | 51.45 . So the box will begin to slide . | |

10:29 | So therefore we no longer have static friction present because | |

10:35 | the box is sliding . Now we have kinetic friction | |

10:39 | . Whenever you want to find the acceleration , write | |

10:41 | an expression for the net force in this case , | |

10:43 | in the X . Direction . So this is going | |

10:46 | to be positive because it's directed towards the positive X | |

10:50 | . Axis and this is gonna be negative since it's | |

10:53 | directed towards the negative X axis . The net force | |

10:56 | , based on Newton's 2nd law , is mass times | |

10:59 | acceleration and F K . Is mu K . Times | |

11:04 | normal force where the normal forces MG . So now | |

11:09 | we can calculate the acceleration . So the mass is | |

11:11 | 15 . The applied force is 99 UK is 990.20 | |

11:18 | And is still 15 and G is 9.8 . Let's | |

11:29 | multiply .2 times 15 times 9.8 . So that's 29.4 | |

11:40 | 90 -29.4 is 60.6 60.6 which is F -F . | |

11:49 | K . And that's the net force by the way | |

11:51 | if you needed to find it . So the acceleration | |

11:54 | is going to be the net force divided by the | |

11:56 | mass , 60.6 divided by 15 . So that will | |

12:02 | give us an acceleration 4.04 meters per second squared . | |

12:09 | And so that's it for this problem . Now , | |

12:13 | let's look at the second example , A force of | |

12:16 | 65 newtons is needed to start an eight kg box | |

12:20 | moving across the horizontal surface , calculate the coefficient of | |

12:25 | static friction . So let's draw a picture . So | |

12:30 | here's the box , It's eight kg in mass and | |

12:36 | we need to apply a force and that force is | |

12:43 | going against static friction . So if this is the | |

12:47 | minimum force that is necessary to cause it's move , | |

12:50 | then we can say that F is equal to the | |

12:52 | maximum value of ss I mean Fs so the maximum | |

12:57 | value of F . F . S is um us | |

13:00 | times the normal force . So just like before it's | |

13:03 | going to be mu s times MG The applied force | |

13:07 | is 65 . In this example we're looking from us | |

13:11 | mm eight G is 9.8 . So let's multiply eight | |

13:19 | Times 9.8 and you should get 78.4 . So um | |

13:26 | us is going to be 65 divided by 78.4 , | |

13:32 | Which is it's pretty high .8-9 . And so that's | |

13:36 | the coefficient of static friction . Now let's move on | |

13:41 | to part B . If the box continues to move | |

13:51 | With an acceleration of 1.4 m/s squared what is the | |

13:56 | coefficient of kinetic friction ? So let's replace this with | |

14:02 | F . K . So any time you're dealing with | |

14:05 | forces and acceleration , it's helpful to write an expression | |

14:10 | with the net force . The net force is going | |

14:12 | to be f minus F K . And then that | |

14:16 | forces Emma and F . K . We know it's | |

14:23 | in UK times the normal force , which is M | |

14:26 | . G . So our goal is to find UK | |

14:30 | . Or to find the value of the U . | |

14:31 | K . So um is eight , acceleration is 1.4 | |

14:37 | . The applied force is still 65 Because once you | |

14:41 | have a force of 65 it begins to move . | |

14:44 | and so in that force of 65 will be , | |

14:48 | will continue to apply to the a kilogram box eight | |

14:59 | times 1.4 . That's 11.2 . and we said eight | |

15:07 | times 9.8 , that's 78.4 times McCain . So now | |

15:15 | let's attract both sides by 65 . So 11.2 minus | |

15:19 | 65 That's negative 53 .8 and that's equal to negative | |

15:25 | 78.4 times McKay . So to calculate the UK , | |

15:30 | we gotta divide both sides by -78.4 . So negative | |

15:34 | 53.8 , divide by negative 78.4 Will give us um | |

15:38 | , U . K value of .686 . So as | |

15:43 | you can see us is almost always greater than UK | |

15:48 | . I haven't seen an example where UK is greater | |

15:50 | than us and so now you know how to calculate | |

15:54 | it . You can use the same formulas as what | |

15:57 | we use in the last example , number three , | |

16:01 | A force of 150 newtons pulls the 30 kg box | |

16:05 | to the right as shown below , If the coefficient | |

16:09 | of kinetic friction is .25 , what is the horizontal | |

16:13 | acceleration of the box ? So go ahead and try | |

16:18 | this problem . So now , based in this problem | |

16:23 | , we can tell that the box is moving . | |

16:24 | So there's kinetic friction . Plus the question asked us | |

16:29 | to look for the coefficient of kinetic friction . So | |

16:33 | we have to assume that the box is in motion | |

16:37 | . Now , what do we need to do in | |

16:38 | order to find the horizontal acceleration ? Well , let's | |

16:42 | write an expression for the sum of our forces in | |

16:45 | the X direction . S is not directly in the | |

16:49 | X . Direction , but a component of F which | |

16:52 | will call F of X . Is . So the | |

16:57 | sum of all forces in the extraction is going to | |

17:00 | be this value minus that one . Yeah . Mhm | |

17:07 | . Mhm . This value is going in the positive | |

17:11 | X . Direction , so it's gonna be positive ffx | |

17:14 | and this one is in a negative X direction . | |

17:16 | So negative F . K . Now we know this | |

17:21 | force based on Newton's second law , is equal to | |

17:23 | mass times acceleration F of X . Is F . | |

17:28 | Cosign data . Now what about escaping F . K | |

17:35 | . Is mu K times normal force . Now , | |

17:39 | what is the normal force in this problem ? In | |

17:43 | this example ? The normal force does not equal MG | |

17:48 | . Make sure you understand why ? Now to understand | |

17:54 | this , let's go over a few things . So | |

17:57 | let's say if we have a five kg box , | |

18:00 | The weight force of this box is going to be | |

18:03 | five times 9.8 , Which is 49 . Now , | |

18:08 | in order for the box to rest on a horizontal | |

18:10 | surface , the net force in the Y direction has | |

18:12 | to be zero , which means the normal force has | |

18:15 | to be equal to the weight force . So whenever | |

18:18 | you have a box on a horizontal surface , the | |

18:20 | normal force is equal to MG . Now , what | |

18:24 | happens if you take the same box ? And if | |

18:29 | you apply A downward force of 10 younes , what's | |

18:33 | going to happen Now ? We still have a weight | |

18:36 | force of 49 units . But what's the normal force | |

18:40 | now , before the surface ? Only needed Excuse me | |

18:46 | , to support the weight of the object , which | |

18:48 | is 49 units . But now the surface not only | |

18:52 | has to support the downward weight force of the object | |

18:55 | , but it must also support the downward force that | |

18:57 | you apply as you press down on the object . | |

19:00 | So the normal force increases any time you pressed the | |

19:03 | block against the surface . So now the normal forces | |

19:06 | , 59 unions . Now , what about if we | |

19:09 | take a rope and we pull if we try to | |

19:13 | lift up the box with a force that's less than | |

19:16 | the way for . So let's say The way forces | |

19:18 | still 49 . But the upper tension force is 20 | |

19:21 | unions . What's the normal force now ? In this | |

19:25 | case , the normal force is going to be less | |

19:27 | than 49 because it doesn't have to fully support the | |

19:31 | weight of the object on its own . The tension | |

19:33 | force supports 20 newtons out of the 49 Uh newtons | |

19:38 | of weight that the object has . So the number | |

19:41 | force has to support the other 29 . So basically | |

19:46 | the some of the upward forces must equate to the | |

19:50 | some of the downward forces . So here's what you | |

19:54 | want to take from this . Anytime you press down | |

19:57 | on an object , you increase the number of force | |

20:00 | . When you try to lift it up , the | |

20:02 | normal force decreases . And in this example , this | |

20:06 | block is being lifted up by the y component of | |

20:10 | the force . And so it is this white component | |

20:13 | that changes the normal force . It decreases it . | |

20:16 | So that's why the normal force doesn't equal MG as | |

20:20 | it usually dozen other examples . But anytime you have | |

20:23 | a a force that partially lifts up the block , | |

20:27 | It decreases the # four . So we have to | |

20:30 | come up with an expression to calculate . FMM We | |

20:33 | have to take this into account . Now let's draw | |

20:46 | some other forces that are on this box . So | |

20:49 | we have the downward weight force . And we have | |

20:52 | an upward no more force plus the upward Why component | |

20:58 | of the applied force ? So if we write an | |

21:01 | expression dealing with the sum of all forces in the | |

21:04 | Y direction , it's going to be fn it's upwards | |

21:09 | so it's positive plus F . Y . And the | |

21:12 | way forces downward . Now The net force in the | |

21:18 | Y Direction is going to be zero . If this | |

21:21 | force does not exceed the way force So we can | |

21:25 | do a quick test the weight forces 30 times 9.8 | |

21:29 | Which is 2 94 . F . Y . Is | |

21:31 | going to be 1 50 times sine 30 Which is | |

21:34 | 75 . So F . Y . Doesn't exceed the | |

21:37 | way . If it did , this object would be | |

21:40 | lifted above the ground so it remains in contact with | |

21:43 | the surface . So if it's not being lifted up | |

21:46 | , we can say that some of our forces in | |

21:47 | the Y direction Will be equal to zero . There's | |

21:50 | no acceleration in the Y direction , So this is | |

21:54 | zero and solvent for S . N . We need | |

21:59 | to subtract both sides by fy and we need to | |

22:03 | add W to both sides . So therefore the normal | |

22:07 | force for this particular problem is going to be the | |

22:11 | way force , which is M . G . But | |

22:14 | minus this force . F . Y . As we | |

22:20 | said , any time you try to lift up the | |

22:21 | object , the normal force is going to decrease and | |

22:25 | that's why we have the minus sign . If we | |

22:27 | apply a downward force , this would be a plus | |

22:30 | sign because the normal force would increase . So let's | |

22:42 | calculate the normal force 1st . So that's going to | |

22:48 | be the mass of 30 Times 9.8 minus F . | |

22:53 | Y . Which is F . That's 150 times signed | |

22:57 | 30 . So we know 30 times 9.8 We said | |

23:02 | that's 2 94 And 150 times signed 30 . That's | |

23:07 | 75 . So to 94 -75 , that's 2 19 | |

23:14 | . So the normal force in this example is 219 | |

23:17 | Nunes . Now let's use that to calculate the horizontal | |

23:21 | acceleration . So m . is 30 S . You | |

23:27 | still want 50 Times co sign 30 . That's going | |

23:30 | to give us FX and then minus mu k which | |

23:33 | is 0.25 Times The Normal Force of 2 19 . | |

23:38 | Let me just separate These two parts of the problem | |

23:44 | . So 150 co sign 30 , that's 1 29.9 | |

23:52 | And .25 times to 19 . That's 54 .75 . | |

23:58 | So if we subtract those two numbers , This will | |

24:04 | give you 75 0.15 . And so that's equal to | |

24:08 | 30 times acceleration . So the acceleration in the X | |

24:12 | direction is 75.15 , divided by 30 , Which is | |

24:17 | 2.505 meters per second squared . So this is the | |

24:23 | answer . |

#### DESCRIPTION:

This physics video tutorial provides a basic introduction into kinetic friction and static friction. It contains plenty of examples and physics problems that asks you to calculate the acceleration using newton's laws of motion. The static frictional force is equal to the applied force up to a maximum value. The kinetic friction force is a constant value that depends on the interaction between the horizontal surface and the object. All frictional forces is dependent on the normal of the object. You should draw a free body diagram for each practice problem if you wish to make it a lot easier.

#### OVERVIEW:

Static Friction and Kinetic Friction Physics Problems With Free Body Diagrams is a free educational video by The Organic Chemistry Tutor.

This page not only allows students and teachers view Static Friction and Kinetic Friction Physics Problems With Free Body Diagrams videos but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.

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