Static Friction and Kinetic Friction Physics Problems With Free Body Diagrams - Free Educational videos for Students in K-12 | Lumos Learning

Static Friction and Kinetic Friction Physics Problems With Free Body Diagrams - Free Educational videos for Students in k-12


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



Transcript
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 .
Summarizer

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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