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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

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

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

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

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

so what do these numbers mean

so let me give an example that's going

to illustrate this

let's make a table

between the applied force

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

now even though kinetic friction is 9.8

that value only applies if the box

slides

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

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

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

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

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

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

is the 15 kilogram box

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

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

so mu s is 0.35

m is 15

and g is 9.8

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

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

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

let's multiply 0.2 times 15 times 9.8

so that's 29.4

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

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

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

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

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

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

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

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

so the sum of all forces in the x

direction

is going to be this value

minus that one

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

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

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

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

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

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

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

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

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

so let's calculate the normal force

first

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

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

you