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When we take a spring, something that we are so familiar with now, and the spring has length l in a relaxed state, spring constant k, I can extend the spring with some force that I apply.
The spring, then, will counter it with the spring force and it will be in equilibrium there.
I call this the zero position, and let's call this now delta l instead of x, which we have done before.
If I double the force, delta l will double.
Hooke's Law says that the force is linear with delta l; in other words, delta l is proportional with F.
Nothing new, as long as Hooke's Law holds.
If I make the spring twice as long, I would get double the extension, because when I have two springs in series, each one, under the influence of this force, will get longer by this amount.
Since I have two springs in series, I will get twice delta l.
So delta l is also proportional to the length of my spring.
If I take two springs parallel, one like so and one like so, relaxed length l, both the same spring constant k, and if now I apply a force on it, then each one of these spring forces is only half this one.
Together they counter this force.
In other words, the extension delta l that I obtain from this external force is now only half as much as it would be with one spring, and if I had three springs parallel, all identical, I would only get one-third of the extension for a given force.
In other words, delta l is also inversely proportional to the number of springs that I have, assuming that they are identical springs.
Now I'm going to use a rod or wire which has cross-sectional area A and length l, and I'm going to apply a force here.
As a result of that force it will get longer by a certain amount delta l, exactly like the spring.
And clearly, when I make this force stronger, delta l will increase, and as long as Hooke's Law holds that the spring force provided by the rod balances this out--
provided that the spring force is linearly proportional with delta l--
I have again that delta l will be proportional with the force.
Double the force, I get twice delta l.
If I put two of these rods together--
so I double the length of the rod--
then clearly I will get twice delta l, because each rod will experience the force, each rod will get longer by delta l, and so two rods will get longer by two delta l.
So again, delta l is proportional with l.
Suppose, now, I have two of these rods next to each other--
notice the parallel with the two parallel springs here--
and I apply a force F.
Then the spring force on each one of them--
if I call that the spring force--
will only have to be half to counter this force.
So they both have a cross-sectional area A, and so now with double the cross-sectional area, I get only half of delta l.
And so now we have a situation that if I made a rod whereby this was 2A, just one rod, which is completely equivalent to this situation, I'm only getting half delta l for a given force.
And so now we have that delta l is proportional...
inversely proportional to the cross-sectional area of the rod.
So now we can make up the balance, and we can say, "Aha! Delta l is proportional to the force, proportional to l, "and inversely proportional to the area, the cross-sectional area." So F divided by A is then proportional to delta l over l, and that proportionality constant we give a name, and that is capital Y, and that is called Young's modulus.
So this is Young's modulus.
F over A, which has the dimension of pressure--
force per unit area--
is what we call stress.
And delta l over l, which is dimensionless, we call that strain.
If we compare two rods with different values of Young's modulus, then the one with the smaller value of Y for the same stress will give you a larger strain.
In other words, it's easier to make it longer.
If Young's modulus is very high, then the rod is extremely stiff.
Then it is very difficult to make the rod longer.
I have here for you some numbers which, of course, are on the Web so you don't have to copy them.
And you have Young's modulus there for various metals, and I also have it down there for nylon, and today we will work with that quite extensively.
We could first do a simple example just to get a feeling for what is at stake here.
I can take a rod with a radius r, which is 0.5 centimeters.
That would give me a cross-sectional area of eight times ten to the minus five square meters.
So, yay thick, the rod.
I make it very simple--
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