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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
