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Static vs Kinetic Friction · Acedra
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Static vs Kinetic Friction
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0:01
in this video we're going to focus on static friction and kinetic friction now let's say if this is a carpet floor and there's a big box and imagine if you're trying to push the box so you're trying to apply a force to move it now initially as you begin to push it the box doesn't move but eventually if you continue to push it with even more force it will begin to slide and once it begins sliding it's easier to continue pushing it but once you stop it it's going to be hard to start it up again
0:44
the force that prevents the box from sliding at the first place or in the first place is the static frictional force now once it begins to slide the force that impedes the motion is kinetic friction static means not moving kinetic has to deal with motion so the frictional force that opposes motion when the object is sliding against the carpet that's kinetic friction the frictional force that prevents you from moving it when you first try to push it that's static friction static friction
1:21
let me write the equations on the right static friction is less than or equal to mu s times the normal force kinetic friction it's equal to mu k times the normal force so notice that the static frictional force is represented by an inequality which means that it's not just a fixed number it can be a range of numbers up to a maximum point the kinetic frictional force however is not represented by an inequality so therefore fk represents a fixed number now let's say
2:01
that this is a five kilogram box what is the normal force and calculate these two values with this information now this box exerts a downward weight force and the normal force has to support the weight force keep in mind the normal force is a force that the surface exerts on the box and is perpendicular to the surface so in this example for a horizontal surface the normal force is equal to the wave force or mg so it's going to be 5 kilograms times 9.8 meters per second squared
2:42
so the normal force is 49 newtons now i need to give you values from us and uk mu s is typically greater than mu k i haven't seen an example when it's less so we're going to say that mu s is 0.4 and we're going to say mu k is 0.2 so let's calculate fk and fs so the kinetic frictional force is going to be 0.2 times 49 which is 9.8 the static frictional force is going to be less than or equal to 0.4 times 49 which is 19.6
3:35
so what do these numbers mean so let me give an example that's going to illustrate this
3:46
let's make a table between the applied force
3:53
the static frictional force the kinetic frictional force and the net force so if the applied force is zero the frictional forces will be zero and the net force will be zero if the person doesn't push the box the box won't move nothing's going to happen but now let's say if the person applies a force of let's say 10 newtons what's going to happen will the box begin to slide and what are the values of the static and kinetic frictional forces
4:32
now even though kinetic friction is 9.8 that value only applies if the box slides
4:43
now what is the static frictional force will the box move notice that the applied force is less than the maximum static frictional force so it's not going to move which means that there is no kinetic frictional force the kinetic frictional force only exists if the box is sliding against the carpet you can't have static friction and kinetic friction present at the same time if the box is not sliding against the carpet static friction is present if it is sliding kinetic friction is present
5:20
so what is static friction in this example what number should you put here now if you're thinking about putting 19.6 that will not be correct because imagine if the person applies a force of 10 newtons to the right to push the box and if static friction applies a force of 19.6 newtons that means that there's going to be a net force of 9.6 newtons towards the left so imagine pushing the box only to find that the box is pushing you back to the left that just doesn't happen
5:54
so static friction can't be 19.6 it turns out that static friction is going to match the applied force until you exceed its maximum value so if you push it with 10 newtons it's going to push back on you with 10 newtons and so the box doesn't move you're trying to push it but it doesn't move so the net force is zero so let's say if you try to push it with 15 unions then it's going to push back on you with 15 newtons that means you haven't applied enough force to start moving it
6:29
so that's what happens when you try to push the box initially you're pushing hard against it but it's not moving because static friction is matching your applied force until it reaches the maximum value so this is still going to be zero so let's say if you apply a force of 19.6 newtons it's still going to be 19.6 now when you exceed 19.6 that's when it begins to slide and so you no longer have static friction but you have kinetic friction so let's say if we go just above 19.6
7:02
let's say if we increase it to 20 newtons now the box begins to slide and so there is no more static friction because the surfaces are sliding past each other but there is kinetic friction which is always going to be 9.8 once the box begins to slide so it's 20 minus 9.8 which will give you a net force of 10.2 now if you decide to increase the applied force the static frictional force will still be zero f k is going to still be 9.8 and the net force is now 30 minus 9.8 which is 28.2
7:39
so hopefully this example help you to understand the difference between static friction and kinetic friction and how to calculate it based on the applied force let's work on this problem a 15 kilogram box rests on a horizontal surface what is the minimum horizontal force that is required to cause the box to begin to slide if the coefficient of static friction is 0.35 so you can pause the video if you want to work on this problem as well but let's start with a picture so this
8:12
is the 15 kilogram box
8:18
so we wish to apply a horizontal force which we're going to call capital f and we know that friction is going to oppose it now if we want the minimum horizontal force that is required to cause the box to begin to slide we need to use static friction because until the force exceeds static friction only then can it slide if it doesn't exceed the static frictional force then it won't slide so the threshold is the maximum static frictional force so we need to set f equal to fs
8:57
and that's when it begins to slide when these two have the same magnitude now the static frictional force its maximum value is mu s times the normal force and in this example the normal force is going to be mg
9:18
so mu s is 0.35 m is 15 and g is 9.8
9:30
so the applied force has to be 51.45 newtons in order for the box to begin to slide if it's less than its value the box will not slide it will not move even if it equals this value the net force would still be zero it has to be just above so if it's 51.46 it will move if it's 51.44 it's not going to move 51.45 that's the threshold so it really doesn't move at that point so technically the applied force has to be just above 51.45 but for all practical purposes we're
10:10
going to say this is the threshold value so we'll go with that now what about part b what is the acceleration of the system if a person pushes the box with a force of 90. so 90 is greater than 51.45 so the box will begin to slide so therefore we no longer have static friction present because the box is sliding now we have kinetic friction whenever you want to find the acceleration write an expression for the net force in this case in the x direction so this is going to be positive because
10:49
it's directed towards the positive x-axis and this is going to be negative since it's directed towards the negative x-axis the net force based on newton's second law is mass times acceleration and fk is mu k times normal force where the normal force is mg so now we can calculate the acceleration so the mass is 15. the applied force is 90 mu k is 0.20 m is still 15 and g is 9.8
11:29
let's multiply 0.2 times 15 times 9.8 so that's 29.4
11:41
90 minus 29.4 is 60.6 60.6 which is f minus fk that's the net force by the way if you needed to find it so the acceleration is going to be the net force divided by the mass 60.6 divided by 15. so that will give us an acceleration of 4.04 meters per second squared and so that's it for this problem now let's look at the second example a force of 65 newtons is needed to start an 8 kilogram box moving across a horizontal surface calculate the coefficient of static friction
12:29
so let's draw a picture so here's the box it's 8 kilograms in mass and we need to apply a force and that force is going against static friction so if this is the minimum force that is necessary to cause its move then we could say that f is equal to the maximum value of ss i mean fs so the maximum value of f s is mu s times the normal force so just like before it's going to be mu s times mg the applied force is 65. in this example we're looking from us m is eight g is nine point eight
13:17
so let's multiply eight times 9.8 and you should get 78.4 so mu s is going to be 65 divided by 78.4 which is it's pretty high 0.829 and so that's the coefficient of static friction now let's move on to part b
13:49
if the box continues to move with an acceleration of 1.4 meters per second squared what is the coefficient of kinetic friction so let's replace this with fk so anytime you're dealing with forces and acceleration it's helpful to write an expression with the net force the net force is going to be f minus fk and then that force is ma
14:23
and fk we know it's mu k times the normal force which is mg so our goal is to find mu k or to find the value of mu k so m is eight the acceleration is 1.4 the applied force is still 65 because once you have a force of 65 it begins to move and so that force of 65 will be will continue to apply to the 8 kilogram box
14:59
8 times 1.4 that's 11.2 and we said 8 times 9.8 that's 78.4 times mu k so now let's subtract both sides by 65. so 11.2 minus 65 that's negative 53.8 and that's equal to negative 78.4 times mu k so to calculate mu k we got to divide both sides by negative 78.4 so negative 53.8 divided by negative 78.4 will give us a mu k value of 0.686 so as you can see mu s is almost always greater than mu k i haven't seen an example where mu k is greater than mu s and so now you know how to calculate
15:55
it you can use the same formulas as what we used in the last example number three a force of 150 newtons pulls the 30 kilogram box to the right as shown below if the coefficient of kinetic friction is 0.25 what is the horizontal acceleration of the box
16:18
so go ahead and try this problem so now based on this problem we can tell that the box is moving so there's kinetic friction plus the question asked us to look for the coefficient of kinetic friction so we have to assume that the box is in motion now what do we need to do in order to find the horizontal acceleration well let's write an expression for the sum of all forces in the x direction f is not directly in the x direction but a component of f which we'll call f of x is
16:57
so the sum of all forces in the x direction is going to be this value minus that one
17:09
this value is going in the positive x direction so it's going to be positive f of x and this one is in a negative x direction so negative f k now we know this force based on newton's second law is equal to mass times acceleration f of x is f cosine theta now what about f k f k is mu k times normal force now what is the normal force in this problem in this example the normal force does not equal mg make sure you understand why
17:53
now to understand this let's go over a few things so let's say if we have a five kilogram box the weight force of this box is going to be 5 times 9.8 which is 49. now in order for the box to rest on a horizontal surface the net force in the y direction has to be zero which means the normal force has to be equal to the weight force so whenever you have a box on a horizontal surface the normal force is equal to mg now what happens if you take the same box and if you apply
18:31
a downward force of 10 newtons what's gonna happen now we still have a weight force of 49 units but what's the normal force now before the surface only needed to excuse me to support the weight of the object which is 49 units but now the surface not only has to support the downward weight force of the object but it must also support the downward force that you apply as you press down on the object so the normal force increases anytime you press the block against the surface
19:05
so now the normal force is 59 newtons now what about if we take a rope and we pull if we try to lift up the box with a force that's less than the weight force so let's say the wave force is still 49 but the upward tension force is 20 newtons what's the normal force now in this case the normal force is going to be less than 49 because it doesn't have to fully support the weight of the object on its own the tension force supports 20 newtons out of the 49 newtons of weight that the object has so
19:41
the normal force has to support the other 29 so basically the sum of the upward forces must equate to the sum of the downward forces so here's what you want to take from this anytime you press down on an object you increase the normal force when you try to lift it up the normal force decreases and in this example this block is being lifted up by the y component of the force and so it is this y component that changes the normal force it decreases it so that's why the normal force doesn't
20:18
equal mg as it usually does in other examples but anytime you have a a force that partially lifts up the block it decreases the normal force so we have to come up with an expression to calculate fm we have to take this into account
20:46
now let's draw some other forces that are on this box so we have the downward weight force and we have an upward normal force plus the upward y component of the applied force so if we write an expression dealing with the sum of all forces in the y direction it's going to be f n it's upwards so it's positive plus f y and the wave force is downward now the net force in the y direction is going to be zero if this force does not exceed the weight force so we can do a quick test the weight
21:26
force is 30 times 9.8 which is 294 fy is going to be 150 times sine 30 which is 75. so fy doesn't exceed the weight if it did this object would be lifted above the ground so it remains in contact with the surface so if it's not being lifted up we could say the sum of all forces in the y direction will be equal to zero there's no acceleration in the y direction so this is zero and solving for fn we need to subtract both sides by fy and we need to add w to both sides
22:06
so therefore the normal force for this particular problem is going to be the weight force which is mg but minus this force fy as we said anytime you try to lift up the object the normal force is going to decrease and that's why we have the minus sign if we apply a downward force this will be a plus sign because the normal force would increase
22:42
so let's calculate the normal force first
22:48
so that's going to be the mass of 30 times 9.8 minus fy which is f that's 150 times sine 30. so we know 30 times 9.8 we said that's 294 and 150 times sine 30 that's 75 so 294 minus 75 that's 219. so the normal force in this example is 219 newtons now let's use that to calculate the horizontal acceleration so m is 30 f is still 150 times cosine 30 that's going to give us fx and then minus mu k which is 0.25 times the normal force of 219. let me just separate these two parts of the problem
23:45
so 150 cosine 30 that's 129.9 and .25 times 219 that's 54.75 so if we subtract those two numbers this will give you 75.15 and so that's equal to 30 times the acceleration so the acceleration in the x direction is 75.15 divided by 30 which is 2.505 meters per second squared so this is the answer
24:48
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