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

Jacky LeiPost author27:10 shouldn't that actually be a value of 1 instead of 0

shaikh zoyaPost authorgreat video.

Renato PessoaPost authorNice class!

Alfredo RanucciPost authorForZa Napoli

asa felixPost authorhis the best teacher have seen so far

Mirhebib SeydeliyevPost authorperfect video… now i understand what stress and strain are. great thanks to uuuuu

A.NATARAJAN ANGAMUTHUPost authornic sir i understand this consept tq send me a vdo like this vdo sir

Naraimon JapanesePost authorWhy does the force always act on a place instead of a point?

Sitesh RoyPost authorSir,how and why is the restoring force being created?