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In this module we are going to talk about the effects forces on objects.
Now in physics we often
treat objects is perfect rigid bodies
which means that the never deform and they never break
no matter how large or where we apply forces to them
now of course in real life we know that this simply
just isn't true if you've ever been unfortunate enough to drop your mobile
phone
you'll know that the metal case can dent
or scratch and the glass can actually
shatter and break and so different materials
glass and metal respond differently to the same
applied forces. So clearly
we're going to need to look at material properties
because different materials behave differently
and so we need to be able to quantify how a material will behave
in a particular set of circumstances. So let's start looking at something
a lot simpler than a mobile phone and will take a rubber band here
now if I apply a tension to this rubber band
you can see that it gets longer so I simply pull
on each and to apply a tension force and
the rubber band stretches so we might think
well perhaps we should use force as one of our
quantities to determine how materials behave
but supposing I take a different rubber band
and so here I got the same material, it's rubber
but now I've got a far thinner rubber band
and if I pull on it it also stretches
however I'm using far less force
to produce the same amount of stretch in this
rubber band than I did in the thicker
rubber band. So here if I take the same length of material
and I apply the same force I get far less stretch
and to get same amount of stretch I have to apply a far larger force
so force alone is not going to be
a useful quantity when we determine the deformation
of a particular material and so what we're going to do
is we're going to define a new quantity called
stress and this will take account of the
cross-sectional area of the material so in this case
I have a large cross-sectional area and so I need to provide a large force
and in this case I have a small
cross-sectional area and so I need to provide only a small force to get the
same amount of stretch
so stress is going to be a force per unit area
and what we're interested in is the forces
inside the material because it's the force inside the material
that determines how the material will deform so the question is now
how do we calculate force per unit area inside the material
and what exactly does that mean and so we will
have to switch to the computer to see so here is our rubber band
and we've applied to it two external forces
this is one the external forces and its in
equilibrium
so what we're going to do to calculate the stresses we want to know the
internal forces inside the rubber band
it's not the external forces that are important for stress it's the
internal forces so to find these out we're going to cut it
in two along plane here and then we're going to
insert here a thin plane
I'm going to put thin in inverted commas because
what we actually mean by this is an infinitely thin plane
so it's an infinitely thin plane that we're going to insert here
and we can cut it here it doesn't have to be in the mid point, we could cut it
anywhere along here
and wherever we cut it is where we will find the stress
at that point so if we cut along the middle here will find
that the the stress in the middle. In this case what we will find is that the
stresses in fact constant throughout this
rubber band I'm but in general
you cut it where you want to find the stress so we cut it in two
we insert a thin plane and
we consider the forces that act on that plane
so this is what we've done here's a picture now
the rubber band with our thin plane in blue here
and these are the forces acting on the plane
so we have one here and we have one
here so how do we get those forces why what we did
is we look at this part of the rubber band
and so here I have this half of the rubber band
and I have my external force
acting on - I'll show that with a solid arrow here
so this is my external force here - now
the rubber band is in equilibrium and so what this tells me
is there must be an equal and opposite force here
and I've shown that with an unfilled-in arrow
and this is my internal force so this is the force acting on this side to the
rubber band
and this is needed in order to be equal and opposite
the external force and keep the rubber band in
equilibrium. Well if I have here
my thin plane then
what Newton's third law says is that there must be a reaction force that
acts on this plane, I'll call that 'R'
here and that's a reaction too this force here
and Newton's third law says the reaction is
equal and opposite to the action and so that means that my reaction force here
by Newton's Third law
must be also equal to this F perpendicular
and so this is the force acting on the plane
and so it's shown here - this force here
is my reaction force, this F perpendicular, shown in blue to show
that it's acting on the plane and I can do exactly the same thing
on this side here I can resolve the forces
on this half of the rubber band and that gives me
the perpendicular force that acts on the
other side of the plane and so here we have the two forces
that act on the plane now in general
the forces on the plane will allways
cancel and in fact if you draw a diagram where they don't cancel
you got it wrong and the reason for this
is because the plane has no mass
and so therefore
cannot have any net force acting on it otherwise it would have and infinite
acceleration by Newton's second law
so it's the same argument that keeps the tension in a massless string constant
throughout the string
the same argument shows here is that we're having this theoretical
thin plane and so therefore the forces acting on either side of it
must add to zero in other words there must be no net force acting on the plane
so the forces acting on the plane
are the internal forces inside
this rubber band. So now
we can see the forces acting on the plane and we now want to look at our stress
So our stress is based on these internal forces that act on the plane
and it's just equal to the
one of the forces that act on the plane. Remember of course that the net force
will be zero so we take one of the forces acting on one side of the plane
and we divide it by the cross-sectional
area of the plane so A here is the area the plane
and this force that is acting on the plane we just take it
and we divide it by the area. Now of course in terms of units
the forces are measured in newtons in the SI system, area
is measured in square metres and so therefore stress
is measured in newtons per square metre
now the other SI unit you can use is the
Pascal, abbreviated Pa. That's a valid SI unit which has the same dimensions as
newtons per square metre. I will not use that. I don't like using it because it hides
the underlying units of the quantity you are measuring
so it's good to remember that stress is a force
per unit area and that's a lot more obvious if you write the units as
newtons per square metre so I won't use Pascal's
now you might say well
why bother going through all this rigmarole love looking at the forces
acting on part with the rubber band and then reaction forces that act on the plane?
Wasn't that a lot of complication? Couldn't we just have taken the external forces
because of course this
is equal in magnitude to the external force
but it's not the external force
it the force acting on this plane, but couldn't we just take the external forces
and divided by the cross-sectional area of the rubber band?
Well in this case you could have done that and you would have been
absolutely right. However in general
you've got to be careful when doing that. Supposing
instead of having the rubber band horizontal I
hang it vertically. I'm assuming
some tension T now holding up the rubber band
and we will assume that this rubber band has some mass.
So now if I take my plane
at the top the rubber band here
then here the forces acting on this plane
are going to be quite large because if I look at this part of the rubber band here
the upward force on this part of the rubber band must support
almost all the mass
That will be almost equal to the total weight
of the rubber band because it has to support the entire weight
of the rubber band, or almost the entire weight of the rubber band
but if I take a cross-section through here
then the force acting on this plane
to support this lower half for the rubber band, or lower
little tidbit of the rubber band here, so the force acting here
will be a lot less than
'mg' and so here the stress
in this rubber band will change from the top for the rubber band where it will be
logged
because black part the rubber band to support me almost the entire way
and will become a lot less lower down here where there's very little rubber
band
so in these cases you have to do the full
you have to be I'm careful and do the fall
analysis and look at the forces acting on these planes
to make sure that you've got it correct
K now the type to stress we looked at with the rubber band
is called normal stress and that's because the forces with normal
to the internal play we look to act and normally just a fancy way of saying
perpendicular to you forces were right angles to the plane
another two types have normal stress is what we sow
with rob a bank where we have a tensile
stress so here the band is on the tension
and so we say it's a tensile stress
but we could also apply the force in the reverse motion
and so here for example if I take this book I can apply
compressive force like crime Porsche the pool
shock and you can see that the I will play
fourth normal to the plane if the plane to these pages inside the book here
the force is normal to the play but it's now a compressive forces trying to make
the book
data so are those the only two types of
stressed that the rock well let's have a look
the simple sheet of paper
now we'll just see what I was trying to compress the book
that people can stand up to a compressive stress
pretty well the book didn't squeeze down very much at all and much about for you
squeeze the area between the pages
so we can also look at tensile stress and indeed
if I along with you can see
but it stands up to tensile stress pretty well
right single sheet of paper so
if that's all there is for stress we should be able to handle any sort of
stressed pretty well
but look what happens if I do something a little different
it fell apart really quite easily
and this trying to stress well I'm applying
different forces in opposite direction so if this plane of the paper
is all stressed plane now I apply force up
once I and down at the other
and people fails almost immediately
not Piper stress his call shear stress
so let's look at my apt in a bit more detail okay
so here we have block that is
under this new type to stress called
shear stress and as you can see we have this external force
applied here which is labeled of parallel
and there is a opposite force equal
the force applied at the other end I'm in this
is putting the block im sheer stress
so once more we've divided the block in
to using theme playing and this plane has been laughed a little bit so it's
easier to illustrate the forces on it
and the question we want to know is what are the forces
marking on this plane so as we did
with normal stress we have to consider the forces
acting on this prior to the block Samaraweera just taking this problem
here
I'm gonna look at the forces acting on it while on this site here
you can see that we her their parallel
an offer letter again just like they did before because this is an external force
and then here we her wealthy in playing
this terms the second now in order for this blog to be
in equilibrium the force acting here
must be equal and opposite to the external force
and so therefore this must be bash
parallel as well so the contact force between the screenplay I'm
and the block must be better parallel as well
and so what this means is that the fourth acting
on the plane itself
right by Newton's third mall
here must be air parallel as well
so this is Newton's third law
and it has to be equal so exact parallel and opposite
certain acting upwards as opposed to the downwards
force attracting on the block itself
and so therefore this force here it's just
their parallel likewise we can do the same thing
on this block here and we'll find
this false on the other side of the plane now noticed this is faded slightly
so it's on the other side of the plane
is also their parallel and so this
is how we calculate the internal forces
acting on of them play and as you can see across these forces are
equal and opposite the equal in magnitude the perfect parallel
1x this way the other its the
in the opposite direction so they equal and opposite as we
required before so
now the question is what's the stress
well we know the stress is force per unit area
the only difference between this shear stress
so
and normal stress is that the forces
on the plane mara parallel to the plane the light
in the plane itself the not perpendicular to the plane
and so therefore the shear stress is just the
force which is this parallel force
and that is divided by the cross-sectional area
of the object at the plate right surfaces the plane
that fills the object the object of cross-sectional area
a here then we just take the parallel force
acting in the plane and divide it by
the cross-sectional area the plane and so again
or Unix are going to be Newton's per square meter
or if you like Pascal's but as I said before
right don't like you Pascal's I like to use newman's
per square meter so this
is just the definition of shear stress the only difference is
is it now we have a force the lorries in the plane
rather than perpendicular to the plane
no this does raise a question
which is how do I pick the plane supposing I have
some less obvious object like a cue
which doesn't have necessarily particularly long axis
I'm for which I could take the
playing parallel a perpendicular to the long axis
or in fact I might have forces
being added in multiple directions so
I might have a force half acting like this
but I might also have a force p acting
up and down white back of this force might not exactly man's
I may have a larger force here at one corner
and a small force here this corner at the Q
so in this case how on earth can I possibly pick
the right play well of course in general
for problems like these the isn't right playing
in the problems will be leaving this course there will always be an obvious
plane or you'll be told
which plan to pick but for general problems like this were to be on the
level of discourse
their is no correct answer and so what you do
is you take three planes because you've got a three-dimensional object
you take one plane like this you take
another plane perpendicular to it
I'm you take a third plane thats mutually perpendicular
to you the other two planes because you got three dimensions you could pick
3 perpendicular planes and this is easier to see perhaps if I droid
as the XY I'm sad axes
surfaces axe this is why re:
and this is a sad I can take the
ex white playing I can take the
backs said plane and I can take
why that plane and
52 these planes I'm in a multiplier have for you to these planes I'm gonna have a
shear stress
and I'm gonna have a normal stress
and
so what I've got is this is going to give me 6
quantities
and I write these down in
the matrix and so I'm gonna end up
with three by three matrix
and it's a metric so I have
the normal stresses go along this diagonal here
and then these things are these passive calm inside the
I'm shear stresses so this is the same shear stress like we're here
this year stress component is the same report here
the shear stress components is the same and I put here and you end
up with what's called a tenser
quantity and so does
dress is in fact mine either a scaler
nor vector it's something we call the tensor
however the cancer treatment stress
is well beyond the level of discourse or you need to know
is that it is attends the quantities night the scaler nor vector
we're not going to deal with it is a tensor we're gonna do with it in simple
situations where you can calculate
shear stress and the normal stress in a single plane
and that plane will try to beat given will be
extremely obvious but in general stress is a lot more complicated but we are
gonna do with a complication
here now we've already defined a quantity calls
rest which is a measure of if you like the fourth
acting on a rigid object and
now what we need to do is we need to look at
how the optics 24 so we need to go to quantify
the defamation vortex so when I apply my stress
to this rob a bank you can see the material structures
and deforms answer the question is now how do we quantify this defamation
well the simple way to do this that might come to you
first glance is well let's just measure the extension
all the material so if I started with this size
and I structure to hear I can measure the change in length material I may be
out by itself is a useful quantity
but let's zoom in and have a look a bit more closely at this program
to see maybe why that's not such a good idea okay
and what you can see here is that I've marked one centimeter lengths
have the rubber band and you can see that when there's no tension applied
the remand these are all one centimeter apart
map what what happens when I apply
stress each of these marks
gets further apart and so
this point here I've got the more about
one and a half centimeters roughly apart
so each one centimeter length
the rubber band has stretched by half a centimeter
so if I now measure the distance between the first and the last one
instead of three centimeters it's now
four and a half centimeters or if I measure the extension of the first two
centimeters a rubber band it's now been extended by one centimeter
see you can see that the absolute extension
is not for use for quantity instead
what we want to do is to find a quantity called strain
which will be the extension her unit length
and to see how to do that will look at the diagram on the computer
okay so the diagram here shows
block that starts off with the Lancs l0
getting stretched bite this additional and
due to a force the taxing on it so we have this
perpendicular force here scratches the
object out and causes it to get longer
so in here
the definition of Oscar rain is equal to the
change in length which of course is justice del Toro
quantity here and that's divided
by the original length l0 which is this
quantity here and so thats our
west rain to this particular case
now if you notice that the units
for the change in length of course are language
and so this isn't
meters history in SI units on the original angst in SI units is also gonna
be measured
in meters and so what this means is that strain
is I mention that she has no units
now of course when you taking this ratio you do have to be careful
that you do not have your centimeters divided by meatless
right that would be a bad thing to have happens you gotta make sure
that is sent to meet is divided by centimetres or meters divided by
meters in order to get the time mention Rt %uh the dimensionless
I'm quantity correct otherwise you have also left over 5 200 or
100 so whatever so for example if this would be
well we did this rubber band and we I am
started with and initial links I o0
problem and was three centimeters cuz reporters
3 one centimeter lengths marked out I'll final links
I'll was 4.5
centimeters on so here we want to write
all strain may be slightly differently so strained here
is going to be equal to a change in length which of course is now offline
land
minus our initial and divided by our initial and
so in this particular example it's going to be 1.5
some two meters divided by 3
centimeters another course is just 0.2
song so meet sample we looked at just before this
the strain was that rubber band was 0.5
now this is what we call tensile
strain because we're putting I'm
the object apart but you can also harm
what's called a compressive strain on that's when you push
the Enzian she applied compressive stressed
and then you get it compressive strange
now of course a compressive strain you end up with instead of an extension
to the link here you end up with a reduction
in the links so sheer through tensile
strain you end up with a I'm
Delta I'll is greater m0
on for a compressive strain here you will end up with Delta I'll
being less than syrup
okay so that's for tensile compressive
and strains a loser in response to compressive and tensile stresses
which are both types of the normal stressed with the force is perpendicular
to the cross-section but we also dealt which
Shia stress so next question is
what is the response to a shear stress
so to look at the response to shear stress we're going to use this textbook
so never let it be said that you can't learn physics from a textbook
I'm perhaps sometimes in more ways than you realize
so here we have the textbook
and what I'm going to do is I'm going to apply a shear strain
to the book I'm gonna move the top
the book that way I'm you apply four from the bottom in the book
this way or I'm gonna hold this still I'm
try move the top for the book in that direction
and you'll see what the response is so here we go
so you can see the top have the book
slides in the direction
the applied force now obviously the thicker
this book is the easier it would be for me to move it
in that direction and so the displacement will get larger
has the thickness of the book increases
but you can clearly see that there's a displacement
due to the a strain the stress
and this displacement is known as the Shia
strain but the difference is
from the strains we've seen before is that now
the thickness of the book is in this direction and the displacement
is in fact direction and so the displacement we're going to be
interested in
is how far this top and moves relative
to its original position so you can see that here
it's a displacement the perpendicular to the thickness
of the book and will switch back to the computer screen to have a look at that
in a bit more detail
should now we've got here his we've got a block and this block
is subject to a shear stress so we have these two
forces that are now and opposite but now acting in the same
plane and so this defamation is a little bit different
we get our original link here and I'll 0
but we get our dish placement are extensional change defamation whatever
you wanna call it
our extension here is perpendicular to this
original links here which is you know
0 so in this case
we have what's called he Shia strain
and this is defined in a very similar manner
to on normal strain and it's now just the
extension divided by the original link so in many ways it's the same
because this is now perpendicular to the
original likes right so there's a um this is it ninety degrees
to the original links if the object that we're concerned about
now course again this is measured in meters
and this is measured in meters so again
this is a time mention less
value and so it has no units
and this is what we call the shear strength so for example
if we take that book we said that the original sickness of the book
this is l0 was 5
centimeters on our extension
that we got white squeezing it in on moving sideways
walls from 0.5
centimeters than of shear strain
to be equal to 0.5 divided by
five centimeters and so that is just 0.1
I'm strain
and so that's our definitions for Shia string
sewing Nancy how to measure can quantify
both the internal stresses inside an object
and the defamation so those stresses calls the strain
so the next question we have to answer is
how is the stress related to the strain
and to do that we have to introduce a new topic called elastic moduli
and that will cover in the next morning