| Start | Alright, we'll see, let's do a physics problem right.
Alright, so you have energy in physics, energy is never created or destroyed, work can change the energy on something. So when you change an object's energy, you work on it. You had an amount of energy before hand, you had an amount of energy afterwards, the change is the work done. The kind of work we're going to look at is found by doing the Force * distance. So how hard you push it and how long you push it for. |
| 0:32 | So when you, let's see, throwing a tennis ball, requires a certain force and makes contact with the ball along a certain distance.
You're changing the energy of the ball. Just ask a football player, theres a change in force for the amount of distance that you go. [?] So we need very long distances or it'll be behind us. That's always a bad thing when it hits you. |
| 1:04 | I'd recommend not being hit by the projectile. Ok, so of course this is a calculus class, so how do I calculate this stuff.
Generally, work is F*d, force times distance, however the problem is the amount of force could change over a distance. For example, you have gravitational force, the gravity here isn't the same as gravity higher up. |
| 1:31 | You change this, so your work, the force that gravity exerts on the object changes over a distance.
If you look under water, there would be pressure from the water on the surface, and then a force down deeper. So again that could change. So you could have a variable force, you could have a constant force. For the constant force you just do F*d, but for a variable, then you have to figure out what's going on as the force changes at any moment, and that's where the calculus shows up. |
| 2:02 | So work is also the integral of Force*dx.
Ok? Where the force is some equation that is always changing. For example, An object located x feet from the origin. And the force at that equation is F=x^2+4x, so the force is different depending on where you are. |
| 3:31 | Ok, at the origin, the x is 0, so theres no force, as you move away from the origin, your force gets bigger. Ok?
How much work is done moving the object from x=1, to x=3. So it's just as simple as integrating that from 1-3. |
| 4:04 | That's an easy integral you should all be able to do that.
This is, x^3/3 +2x^2, from 1-3, |
| 4:38 | Ok?
Everybody got that? You should be able to get that, ok? Of course we have to make this harder. It can't be this easy. So let's do a harder problem. Do you guys know what springs are? |
| 5:00 | So if you stretch a spring with an object attached to the spring, you stretch it, it wants to go back like a bungee cord.
Or if you push it, it wants to go back. So the spring has a natural resting point, and then you push it in or you pull it, and then it has a restoring force, a force that makes it go back to where it was. When you pull it out, it wants to go in, when you push it in it wants to go out. Thats the fun thing about springs. And the springiness of a spring is determined by a spring constant. Its how hard it is to push it, or how hard it is to pull it apart. |
| 5:33 | Ok, and the spring constant is, it's really just a measurement of like I said how hard it is to push or pull it.
And we call that k. And the force required to either stretch it a distance x or push it a distance x is k*x. It's very simple, it's linear. |
| 6:09 | So let's see, a force, of 50N is required to stretch a spring from x=10cm to x=20cm.
|
| 6:38 | So far so good?
How much work is required to stretch it from x=10, to x=40 cm. |
| 7:17 | Ok. So again, remember we do the integral from a-b of the force*x, we just need to figure out the force.
And I told you a force of 50N is required to stretch a spring from x=10 to x=20. |
| 7:35 | So, we know f=kx.
So 50N is required to stretch this thing from 10 to 20, So that's a distance of 10cm, or .1 m. This is why people get stuff wrong in [?] because they don't do their units correctly. And you get another 5 slices of your pie, and you say "but I understood it", yeah but [?]. |
| 8:10 | So you get k=500, it doesn't matter what the units are, but they're in N/m,
That is your spring constant.
So now if I want to stretch it from 10cm to 40cm, Im going to stretch it from .1 to .4 of Force, which is kx, which is 500x(dx) |
| 8:44 | Does everyone understand that? So you have to do two phases in this problem, first you have to figure out the what spring constant is so you can figure out the force.
If I give you this piece of information, 50 N is required to stretch from 10cm to 20 cm, so I use that here to figure out the spring constant. |
| 9:06 | 50N=k(.1) and I get 500, so now I know what k is, so I know what F is.
The force, is k*x, so the force is 500*x, and the work is the integral from .1 to .4 because it's in m, 500x(dx) So this is 250x^2 from .1 to .4. |
| 9:40 | So that's equal to 40-3.5=37.5.
I think that's correct. If we're wondering what the units are, units are Joules. |
| 10:01 | Named after some french guy, Joules. Not Juuls as in...
So far so good? Can I give you one, see if you can do it?
Let's give you one of these, before I give you a hard one. It gets harder. Work problems are fun. |
| 12:04 | My arm gets tired writing all this.
Ok, a spring has a natural length of 20 cm. A 25 N force is required to stretch the spring to 30 cm. How much work is required to stretch it from 20-25 cm? |
| 12:30 | Let's see if you guys can do this.
So we know the formula, there's 2 parts to a spring problem, first we have to figure out the spring constant, unless we're nice enough to give you the spring constant. Ok and then you have to figure out how much work it takes to stretch this thing. You could also say compress it, Then you would do the integral from a bigger number to a smaller number because you're going the other way. Alright, so a spring has a natural length of 20 cm, and it takes 25 N to stretch it to 30 cm. So F=kx, So 25 n is required to stretch this thing 10 cm, or .1 m. |
| 13:11 | So you solve this and get the spring constant is 250 N/m.
Ok, now we want to do how much work is required, yes? |
| 13:30 | Ah, because it's 10 cm, 10 cm is .1 m.
Ok, alright. The units we want to use are in m. So when we have stuff in cm, we have to convert. When do you physics or chemistry, you always have to make sure your units are consistent. So when our spring constant is in N/m, you need to have m, you can't have cm. |
| 14:08 | It has a natural length of 20, and it stretches from 20 to 30, so it stretches a distance of 10 cm.
Right it starts off at 20 cm, and then you stretch it to 30 cm, so you stretched it 10 cm, ok? You pulled it out. |
| 14:33 | And then you let it go. Alright, so that's the spring constant, it's 250.
Now you want to find the work that's involved. So the work is Force*dx. So work is the integral from .2 to .25 of the force, now remember F=kx, so F=250x. |
| 15:02 | So that's 250 x* dx.
Oh .25x*dx, sorry. Thank you, so now we have to do the math. So let's see the integral of 250x is 125x^2, from .2 to .25, |
| 15:40 | So let's see,
That's 125/16 -125/25. So far so good?
|
| 16:02 | That's hard to do, so I'll just leave it like that. I can do it I guess.
I keep getting a negative number, so I'm doing something wrong. That's alright, hard to do in my head. Probably wrong, 15/16? I don't know it's probably wrong you can check in the calculator, it could be right. Did I get it? You don't know? Gosh, who are you people. |
| 16:34 | You break out your TI-84, you do the integral function on your calculator, 45/16, I was close.
Well you know, sometimes you win, sometimes you lose. Integral from .2 to .25, of .25x(dx), 45/16. Close enough. |
| 17:07 | You erase this part, and people later will discover this and think 'wow was he good'
Ok? That's the trick it's all done with mirrors. Alright, were you able to do this problem?
Some of you. Now you say, well what if you give us an obnoxious number on the test? We won't. Or we may not ask you this problem, we may ask you something harder. |
| 17:38 | Now we're going to the work required to lift a table for a change.
So imagine you have a rope, you throw it off the side of a cliff, and you're pulling up the rope, ok? It's easier at the end. It's much harder at the beginning, why? There's a lot of rope, and rope has a lot of weight, every cm of the rope has a certain amount of weight to it. |
| 18:03 | So as you pull the rope up, there's less and less hanging over, if you had to pull a person up, and they weigh a lot less when they get to the top. So it would repel it then.
You could demonstrate it for us if you want. So anyway, when they're hanging off, they weigh a lot when they're down there. Ah, that works. Anyway, so the rope itself, has weight due to it's length. I mean you could just imagine it's coiled up when you lift a coiled rope up, it weighs a lot. |
| 18:40 | If you lift a little piece, it doesn't weigh much of anything. So all that rope weighs a lot, so it's hanging off of something, rope, cable chain, whatever it is,
It's got a lot of weight to it, and a you lift it up and theres less involved, so the force is variable, which means you need calculus.
So we'll do the problem they have in the book first. I have to do some erasing. |
| 19:07 | And I know you're saying, if I wanted to take physics, I would have signed up for that class, not this class.
Maybe this will help you when physics rolls around. I was lucky I had help at the kitchen table. A 200 pound cable, we're doing this in pounds, is 100 ft long and hangs from the top of the wall. Basically, you have a wall and the cable hangs down the wall. |
| 19:34 | And it's going to hang 100 feet, so that's the cable. There's nobody at the end of the cable, because that makes it a more difficult problem.
Ok, how much work is required to life the cable to the top of the building? So now you have to think very carefully because this is a problem with physics. As you're lifting the cable, you're lifting every little piece of the cable a certain distance, you can say let's divide this up into distances and let's call each of those distances xi. |
| 20:13 | Where i is some point on the line when you're lifting the cable up
so i is any number, 10, 11, 12 we're just going to do generic calculus with it.
|
| 20:30 | And imagine the cable is cut into little pieces each with weight (delta)x
So how much does one little piece of that weigh? well this is a 200 pound cable,
Pounds is abbreviated lb, rather than something like pd, that's a french thing.
It's 100 feet long, so therefore it weighs 2 pounds/foot, so that should make sense, you just divide them. 200 pounds of cable in 200 feet. |
| 21:16 | So every foot weighs 2 pounds. So every (delta)x, the width of the cable, is 2*(delta)x, now weight is a physics term, it has to do with the force, |
| 21:31 | So you would require a force of 2 lbs to lift this cable and the distance that you lift it, just depends on where it's located.
So it's located at location xi, and you're gonna integrate it from 0-100. Because you're gonna do all 100 feet of it. Ok, so the weight is Force* distance, That's the force, that's the distance, ok? |
| 22:02 | It's actually really the other way around, it's 2(dx)*xi but it doesn't really matter.
So again, where do we come up with that? We say that any little piece, has to be lifted up. We have a little slice here it has to be lifted. This slice here, and this slice way down there. All of those have to be lifted and each one weighs 2x, and each one has to be lifted a distance of x, but x varies so that's why we integrate from 0-100. |
| 22:39 | No? Do it again? So imagine you're lifting a cable,
The cable is divided into infinitely small pieces. You have to lift
each piece up so the pieces hanging all the way down have to be lifted the full 100 feet, the pieces at the top only get moved a little.
So the width of each piece is (delta)x. And the weight is 2*(delta)x |
| 23:04 | It's 2 pounds per foot, and the distance of the width in feet is delta x.
We're gonna use calculus to do the (delta)x part, and then you lift them each, some amount of x, so we let (delta)x get infinitely small in terms of dx, and we're lifting pieces in a range of 0-100. So the very bottom pieces have to be lifted 100 ft, and the top ones have to be lifted 0 ft. |
| 23:37 | And you add them all up, ok? And remember, calculus, the integral adds everything up.
So this is just x^2 from 0-100 so that's equal to 10,000 ft*lbs, |
| 24:00 | In the old pre-metric days, ft*lbs was a useful unit of energy.
You'll see it still when you look at engine specifications for a car. Do I have to plug x where? yes, so what happens is the work, the force, did I write that correctly, I'm sorry, I said work I should have said force, The force is 2*(delta)x, the distance, is xi. |
| 24:33 | Ok, so when you do work = f*d, its 2*(delta)x*xi, so this is the 2, this is the (delta)x, this is the xi.
So when you have to lift a rope or a chain, anything, even a person, the feet, they'll have their entire body weight at their feet, right it's easier to lift the top rather than the whole thing. I'd be very hard to pick someone up from the ankles and lift them straight up. |
| 25:05 | Ask anyone that does gymnastics or cheering.
That's why they tend to jump. Or ballet. They jump, otherwise you could never just bend down and pick them up, well maybe [?] but most of us couldn't. Got it? Can I have you try one? I'll have you try one, it's important. Wait until we do the really hard problems, these are the easy problems. |
| 25:36 | This is why, is anyone here majoring in physics?
No? We got one! Good for you. Here you go. Engineering is easier. Let's see. A very similar problem. |
| 26:01 | Ok, were we able to do this one? So this one you probably sort of memorized, you said it's just like the last one but with different numbers.
We'll that's kinda true, but it's not the key to doing well in physics however. But again, the fact the building was 120 feet high wasn't really important, what was important is that the rope hangs completely off the building. What would make it entertaining would be to take it from some spot way down the building, let's say a very thin string is holding it to a different spot on the building. |
| 26:33 | What's gonna happen is the weight of that rope at any moment
is .5*(delta)x. So the force is .5(delta)x and the distance, xi. So just like the last problem.
we have to lift this thing. So, it's a 50 ft rope, so we have to lift it 50 ft |
| 27:08 | And this is going to be .5x*dx. So .5 is also 1/2, so
so the integral of x/2 is x^2/4, from 0-50,
So that's 50^2/4 -0, which is 625.
|
| 27:35 | Right, ok. So you guys ready? Now for the fun one.
Ok, we're gonna pump water out of a tank. This is fun. Brace yourselves. |
| 28:05 | Ok, so you've got a tank, it's shaped like a cone, sure, and it's filled with water not all the way to the top.
So this is 10m high tank, And it's filled 8m up with water, which means this is 2m. |
| 28:36 | This radius of the cone is 4m, we already know the height, how much work is required to empty all the water from the tank?
Any by the way, the density of water is, does anyone know? 1 what? No what unit? |
| 29:02 | cubic cm, right. So, since we're in a m world, its about 1000 kg/m^3.
Or 1 g/cm^3, 1g/mL, good enough. You know why it's 1? why is it 1? |
| 29:34 | Because we use everything else based on water. So we called water 1, and the density of everything else is how dense is is compared to water.
Ok, so this is gonna get a little messy, but you guys can do this. So imagine that you have some spot of water, right here, which is located at some depth, xi. And this is a very thin disk of water. How thin? (delta)x. |
| 30:06 | And the radius is r.
Ok, let me do some erasing for a second. Sorta blow that up, as they say. this disk here, (delta)x width here, and this is a radius r. |
| 30:36 | And that's the water at any spot. So it's just like the chain, it's just now a disk of water rather than just a thin little strip.
So first, let's figure out what r is. We can find r by similar triangles. So if you imagine a spot right here, when we know that r, how high is this, well we're at some spot, (delta)xi, |
| 31:06 | Let me make sure I'm drawing it correctly, let me check my notes.
So the radius is to 10-x, as 4 is to 10. Ok, because if this distance is x, this distance is 10-x. And that's messy so let's do that again. So this distance is x, so the distance is 10-x, I was trying to write here the radius is to 10-x, as 4 is to 10. |
| 31:45 | Do you see the similar triangles? You have the cone, this is 4, this is 10, this is x, so this is 10-x. Right here, so r to 10-x, is 4/10. Similar triangles.
|
| 32:10 | So to solve for r, we'll call that 2/5, r=(2/5)(10-x)
Ok, so volume is (pi)*r^2(delta)x. So (pi)*((2/5)(10-x))^2*dx.
|
| 32:50 | I'm going to erase this big cone because I don't need it, so that, is 4/25*pi*(x-10)^2dx.
|
| 33:07 | So far so good? Ok, you guys remember the formula, what's the relationship between mass and volume? Density?
Sorry, so density is mass/volume, so rho*volume =m. So the density, rho, is 1000, so the mass= 1000*v, which is 1000*(4/25)*pi*(10-x)^2dx. |
| 33:55 | Which is 160*pi*(10-x)^2*dx.
|
| 34:09 | We're getting there. Ok.
You're one step ahead. So again, so first we just said I have to lift this thing, So what is this thing, its a thing disk of water. The water is (delta)x thin, it's got pi*r^2 in area, so the volume is pi*r^2(delta)x. I find r from similar triangles and I plug it in, |
| 34:33 | Now the lifting of a force is mass*gravity, Since you all love physics, how much it takes to lift something, mg, ok? So we find m multiplying the density times the volume, and now to get the force, it's mg, so it's 160*pi*g*(10-x)^2dx, which happens to be 9.8, |
| 35:08 | So far so good? How far are we lifting this? Well, this piece of water gets lifted 2m, this piece of water gets lifted 10m, so we integrate from 2-10.
We integrate from 2-10, this, times one more x, because its force*distance. So it's 160*Pi*9.8*x*(10-x)^2*dx. |
| 35:53 | Why is it 2-10, ok, yes. The top of the water,
still has to get pumped out of the tank. The top of the water is 2m from the top of the tank.
|
| 36:02 | So this water gets pumped 2 m, this water at the bottom gets pumped 10 m.
The whole thing has to get lifted. So if you imagine that you just pumped the very very thin surface, it still gets lifted 2m, so everything gets lifted 2 m. All the way up to 10 at the very bottom of the water. And then its F*d, so the force is 160*pi*9.8*(10-x)^2, distance is x, dx. And when you work that out, you get, |
| 36:40 | About 3.4*10^6 Joules.
Or 3.4 MJ. (megajoules) I think these problems are very hard. You're not imagining it if you think they're hard. They're hard to set up, not hard to integrate. |
| 37:02 | Ok, it's very hard to figure out what to do because you have to have a lot of physics knowledge to imagine that you're lifting a thin disk.
And then the calculus part is that since theres an infinite number of very thin bits, that's why you have an integral rather than just multiplying fx. Ok, the force is different, the force of the here has the force of all the water on top of it. The force of the water here has much less force on it. |
| 37:31 | So at every moment theres a different weight from the water depending on where you are.
That's why its the variable in force and all that. How'd we do on this one? Eh? Should I go through it again? Ill do it one more time, Ill make sure you really get this. So you have water sitting in a tank, and imagine you're pumping the water out of the tank. So at any instant you pump some slice of water. That water has a thinness, (delta)x. |
| 38:05 | And you have to lift all of those slices up so imagine it was sliced up, not water, but you know.
Pizzas, sure All pizzas stacked up here, all slightly different radii. Ok, so you have a cone of pizza. You could have a lot of fun with that, heres a project for Friday night, I'd do it with oreos first, work your way up. So each slice of pizza has a width of (delta)x and radius of r. So the volume of the water that you're lifting is pi*r^2*(delta)x. |
| 38:41 | What you want to do is find the force required to lift it, so to find the force, its the weight. The weight is mg,
for those that haven't taken physics, g is . constant, 9.8.
It has to do with the acceleration of gravity at the surface of the earth. If you were lifting this really high, g would start to change if you were lifting it high enough. |
| 39:07 | To find the mass we need to know the volume, and multiply it by 1000 because that's the density of water.
The volume is pi*r^2*(delta)x To find r, you use similar triangles, you that the radius of this length, which is 10-x, because this is x meters from the top since its 10 m total, is 10-x to wherever you lifted it. |
| 39:37 | So you do a little similar triangle to get the radius at any moment will be (2/5)(10-x)
So the volume is pi*r^2*(delta)x which simplifies to this,
and the mass you just multiply by 1000 because the density of the water is 1000 kg/m^3.
And then you multiply it by 9.8 because that turns it into a force because you take the mass * gravity. |
| 40:06 | The work, the energy, is that times one more x.
Well you get to take in a card, right, so I don't have to give you the formula because you can fill up both sides of the card. I don't know if we're going to have anything like this on the exam. Ok, personally, I think this particular problem is too hard, I always had to do it with something like a spring. |
| 40:32 | Ok, but it's not completely up to me.
Alright, to do this integration, the x*(10-x)^2 isn't that hard, you just multiply it put to be a polynomial, but the numbers are annoying, ok? So I might just ask you to set it up. Remember, the work is the force*x, so the force was everything other than the x, so you have to multiply by x. |
| 41:02 | Now we move to the next one which is even harder. And that is Wednesday. I don't have the energy to do another one, so on that note, we'll see you Wednesday.
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