We're here at Cambridge University in England where Isaac Newton formulated much of his mechanics. In the last unit we talked about motion, but without really discussing what causes an object to move. In this unit, we're going to talk about Newtonian mechanics and the role force plays in governing the motion of objects.
Let's do a quick recap of what we learned last time. Well, we talked about Aristotle and Galileo, and they each had different thoughts on how objects move when they're in free fall. Remember, Aristotle thought that the velocity of an object is proportional to its mass and that it falls with constant velocity. Galileo thought that the distance an object covers is proportional to the square time it's been falling or, in the case of the experiments, rolling. In order to distinguish between these two viewpoints, we had to do an experiment, and we found out that data is really power in physics. How did we distinguish between these two viewpoints? We did an experiment, because in the world of science evidence rules, not the ideas of brilliant thinkers. We also learned some of the equations of motion. For example, the equation which describes how far an object travels in a certain time given an initial velocity and an acceleration. So, all these together gave us the ability to predict how an object will move or to quantify its motion. The goal for this unit is to explain the cause of motion. We already know how to quantify and predict an object's motion, but we still don't know why an object is moving the way it is. To do this, we'll have to understand what it means for an object to be at rest. It seems like a straightforward concept but actually a lot to learn here. We'll also have to learn Newton's three Laws of Motion. These are the same laws of motion that you can use to send a rocket into space or put a person on the moon. We'll also be able to answer the feather problem by the end of this unit. What's the feather problem? Well, one problem with feathers is that they're very hard to draw, so I apologize, but the problem I'm referring to is that when you drop a feather, it actually does fall at more or less constant speed, which seems to align with Aristotle's viewpoint. So, we need to figure out what's going on there. Finally, we'll learn about gravity. I'll teach you how you can use your physics knowledge to lift enormous objects. Before we get started, let's go to Cambridge University in England and talk to Dr. Patricia Fara who's an expert in Isaac Newton and his book, The Principia Mathematica. [Patricia Fara] Isaac Newton symbolizes two great changes in the way that science was done, and in particular the way that physics was done. He didn't do this single-handedly. He was just one of many people who did it--particularly Galileo. There are two main reasons why his book, The Principia, The Mathematical Principles of Natural Philosophy, why it's important.
Before we get into the reasons why his work is so important-- let's take a minute to try to put ourselves in Newton's shoes and imagine what his life was like. So, here we are, leaning against an apple tree reading one of our favorite books, and all around us things are happening. Apples are falling off trees. Rocks are rolling down mountains. And as nature unfolds around it, we probably never really stop to ask what's happening. We just take it for granted. If we want to think about something profound, maybe we look at the moon or the stars or the sun. Surely, the motions of these perfect celestial bodies are somehow more perfect or different than the motions of commonplace items like apples and rocks. Now, with the benefit of hindsight, I'm going to ask you a question. What do the motions of stars and apples have in common? Do they have nothing in common at all? Are they totally distinct and separate? Or are these motions completely identical? Or are they somehow different but governed by the same set of laws? Which means that the same rules which affect how an apple falls to the ground also affect how the earth orbits the sun or the moon orbits the earth. I haven't taught you this yet, so I don't expect you to know the answer, but I do want to get you thinking. Choose what you think is the best answer.
And I said this was the best answer, but don't worry if you put something different. Let's see what Dr. Fara has to say. [Patricia Fara] There is one thing that everybody knows about Isaac Newton-- that he sat beneath an apple tree and had a flash of inspiration. We don't know, and I don't think we can ever know, whether it really happened. What he said was that he was sitting in his garden, in his orchard where he'd had to retreat from Cambridge because of the plague, so he was sitting in the orchard watching the apples fall around him and suddenly he thought to himself, why is it that that apple falls to the ground? Why doesn't it move sideways or upwards? And is it possible that there is one single force that makes the apple fall to the ground and makes the moon go 'round the earth and makes the earth go 'round the sun? And that is allegedly the great "Eurkea" moment, the flash of inspiration, and that's the significance of the apple.
Now, this is a huge claim for Newton to be making-- that there is some mysterious entity pulling apples to the ground, which he calls some force--iin fact, it calls it "gravity"--and that that same force somehow causes the moon to go in a circular orbit around the earth. Now, Newton calls this force gravity, and there's a few properties he claims it has. He calls it a mutually attractive force, which means that all objects, anything with mass, attracts every other object with mass. So, right now my hand is attracted to this pen with some gravitational force, just as he claims the earth is attracted to the moon. Next, he claims this force is proportional to 1/r². When I say "r", I mean the separation distance between the center of the two things we're talking about. We talked about this idea of proportionality last time, and we know that 1/r² means that if we were to, say, double this distance what would happen to the force? Use your knowledge of proportionality to answer this question as best you can.
Well, if we double the distance, we can use this ratio to figure out what's going on with f₂. Here I've called the radius for f₂ twice the initial radius r₁. I've made the exponent -2, because it's proportional to 1/r², so the 2 comes from it being r² and the negative comes from this being in the denominator. When we work this out we find that f₂ is only 1/4 of f₁. But I'm still confused. How could a force be mutually attractive? That means that this apple will be drawn to the center of the earth, which is exactly below it-- so the force will go that way--and the force on the moon will also be towards the center of the earth. How could a force pointing inward causing an object to move in a circle? Seems like maybe you'd want a force that pointed that way or that way. Let's do a little thought experiment with this apple, and maybe that will help us see how an attractive force could actually cause and object to perhaps move in a circular path.
Let's do a pomaceous thought experiment to find out whether the motion of an apple really is somehow similar to the motion of a moon and if an attractive force could actually somehow produce a circular orbit. Since this is such a cool experiment, our experimenter gets a suitably cool haircut. First, she's going to drop the apple. Well, we all know what's going to happen. It's going to fall to the ground. Now let's say she throws the apple with some initial velocity--V₀. Well, we learned in the last unit exactly how to quantify this motion. The apple will follow some trajectory like this. Now let's say she throws the apple with a velocity of 2V₀. Which trajectory best describes the path the apple will take?
The answer is this trajectory. Remember the apple will spend the same time in the air regardless of how fast you throw it, if you're throwing horizontally, and so doubling the horizontal velocity will just double the distance it covers, making this the best answer. It seems like we still have a problem. None of these trajectories look like they're even close to orbiting. Let's zoom out a bit.
When we zoom out we can start to see the curvature of the earth. Interesting. Now our experimenter, who is sort of gigantic in comparison to the earth now, when she drops her apple, same as before, it goes straight to the center of the earth, but now when she throws the apple, it goes all the way over here. Before when we were zoomed in and we were looking at a flat earth, the apple would have stopped right here--but it keeps going a little extra distance due to the curvature of the earth. Let's zoom out one more time. Here's our experimenter once again, and here's the center of the earth. So, now she's really quite a giant. Of course, if she drops the apple, it goes straight towards the center. If she throws the apple, well, it all depends on what speed she throws it at. If she doesn't throw it too fast, it'll follow some curved trajectory and strike the earth. If she throws it really, really fast, it'll leave the earth and be gone before the earth has a chance to pull it back. Now, somewhere in between these extremes if she throws it at just the right speed, the ball will travel a certain distance and, of course, fall a little bit towards the center of the earth. It will travel another certain distance, and of course fall again and keep up this process of traveling and falling and traveling and falling in just such a way that the path it takes is a circular orbit around the earth, and that's all due to an attractive force that wanted to pull the apple towards the center of the earth.
As long as we're doing thought experiments, let's do a quiz where our experimenter has access to control the laws of physics. Let's say she chooses to double the strength of gravity. This object is now getting pulled to the center of the earth with twice as much force as it was before. My question is if gravity were doubled, would our experimenter have to throw the apple at a different speed if she wanted it to follow this circular orbit? Remember the point of these is just to get you thinking, so don't worry if you don't know the answer just yet.
The answer is yes. She'd have to throw the object faster. When I approach questions like this in physics, I think of what I call limiting cases. For example, I ask myself what if gravity were 0? Well, if there was no gravity, the object wouldn't be attracted to the earth at all. If you threw it in some direction it would just go floating forever. As I increase that gravity, I can see that, okay, maybe the object will start to get pulled towards earth. If I make gravity huge, almost infinite, the object will always be attracted to the center of the earth. It seems like the trend is that if gravity is stronger, you have to throw it a little harder. If it's weaker, you don't have to throw it as hard to maintain the circular orbit. Now, remember when Dr. Fara said that there were 2 reasons why the Principia Mathematica was truly revolutionary? We can now go to the first one. [Dr. Fara] It establishes one single law of gravitation that extends right through the whole universe. Before that--well, a couple of hundred years before that-- everybody followed the Aristotelian idea that there was one set of laws on earth and a completely different set of laws up in the heavens, so that in the heaves everything is perfection and unchanging and eternal, and on earth everything is changing and corrupt. Galileo first but the Newton brought all those changing ideas together and said, no, there's one force that extends right through the universe.
Now, depending on how you look at this, this is either a very humbling or very awe-inspiring moment. One one hand, the laws that govern the chaos and disorder and confusion that people see on earth are the same that govern the motions of the heavens. But looking at it another way, the laws that govern the perfection of the motion of the heavens are the same that control the chaos and disorder on earth. It's either very scary or very beautiful depending on how you look at it, but either way it must have been a very exciting time in physics. Now that we've examined gravity a little bit, we can start to think more about motion. Now we're accepting gravity as a law of nature. All objects are mutually attracted to one another. And so when we think of a falling apple, we can draw an arrow which represents the force of gravity, and I've used the subscript g is indicate that this force is due to gravity. From Galileo, we know exactly how this object will move. It's going to accelerate downwards at 10 m/s². What's really fascinating, though, is that eventually this apple will hit the ground and come to rest. So, let's think about this apple, this motionless apple at rest and think about what does it mean for an object to be motionless?
To examine motionlessness, let's do an experiment in space. Since we know that the laws of physics on earth are the same as those in the heavens, any experiment we can contrive that occurs in space should teach us about something that could also occur on the earth. Now, for this thought experiment, I want you to imagine drifting out alone in the cold depths of space. Actually, for this experiment, I want us to be so deep in space that we can't even see the stars or the sun or the earth. We are truly floating in the middle of nowhere. As far as we can tell, we are completely motionless. Now, let's say you turn on your flashlight, because every good scientist floating in the depths of space remembers to bring their flashlight. You turn on your flashlight, and off in the distance you see another astronaut floating floating in the depths of space. You're not alone. In fact, this other astronaut is slowing drifting towards you. Now, I wonder what this astronaut was thinking before you turned the light on. I wonder if she thought she was motionless or if she thought she was moving. It's clear that relative to each other you're moving towards each other. But it's not clear to me who's moving. Maybe you know. So, I'm going to think about this question. Which of these two astronauts is moving? Is astronaut 1 the one whose truly moving? Is astronaut 2 the one who's actually moving? Maybe they're both moving Maybe neither of them is moving. Maybe they're both motionless. Or, is it impossible to say? I want you to really think about this question, and don't worry if you get it wrong because this is not a completely intuitive concept.
Well, the answer is that given this information, it's completely impossible to say. Even though I explicitly said that astronaut 1 was at rest and astronaut 2 was moving was moving towards astronaut 1, but that situation could equally well be described as astronaut 1 going towards astronaut 2 with astronaut 2 at rest, so what can we say about the motion of astronauts 1 and 2. Well, one thing we can definitely say that both of these space explorers will agree on is that the astronauts are definitely getting closer, but this is interesting that this argument can even show you at all. Astronaut 1 thought that he was motionless. Astronaut 2 thought she was motionless. Yet, clearly, there is some relative motion between them. This points to something funny going on between motion at a constant speed and this idea of being at rest. Why don't we go back to earth to probe this question a little more deeply?
And the answer is actually this one. If you don't believe me, you should absolutely try this out. Get in a car or on a train and while going along at constant speed, drop an object in front of yourself. You will see that it falls straight down. Now, this may cause you to ask yourself what is going on? I'm seeing the same results when I drop an apple on a train that's moving pretty fast as I see when I drop an apple on the ground, and the ground is stationary, right? Well, in fact, in physics motion at a constant speed is absolutely indistinguishable from being at rest. If you were on this train and you boarded up all the windows, and again remember it's moving perfectly smoothly but at a constant speed, there is no experiment that you could do to determine whether you're moving or at rest. That is pretty mind-blowing, I think.
We saw with our thought experiment in space that our two astronauts with no gravity or any forces acting on them couldn't distinguish between motion and rest. Now we're ready to learn Newton's First Law of Motion. Remember our space example. We saw our two astronauts with no forces acting on them, because there was no gravity in our empty universe. They couldn't distinguish between motion in a straight line at constant speed and being at rest. On the train we saw that there was absolutely no experiment that we could do which could distinguish between being at rest and traveling at a constant speed. Newton took these kind of thoughts and formulated his first law, which states that every object will always continue in its state of rest or constant velocity motion, unless it's acted on by an external force. Now, you may be saying to yourself, "What about an apple on the ground?" Surely, that's an object at rest, but it's gravity pulling down on it? Well, the answer to both of your questions is, yes, it is at rest, and, yes, gravity still is pulling down. This is a very interesting question and one that we're going to cover a little bit later in the unit. Stay tuned to find out what's going on here.
Before we move on, let's do one more thought experiment. What if instead of moving at a constant velocity, the train was moving at a constant acceleration? So, every second it's increasing by the same speed as the previous second. What would the trajectory of the apple be then? Select the most appropriate trajectory.
The answer is the apple will actually appear to move backwards. Now, this was a tricky one, but you can verify it for yourself. If you drive or if you know someone who does, get in the car, take an apple and wait at a red light. Once that light turns green, accelerate forwards and drop the apple. You'll see that it actually appears to land behind where you released the apple from. When we accelerate this way, the apple appears to move backwards. When we accelerate that way, likewise. The apple moves backwards. Of course, if you're going to do this, make sure you're safe about it and make sure your car has wheels. What's really interesting about what we found here is that acceleration seems to be somehow important. Objects at rest behave in the same way as objects moving at constant speed but acceleration is important. Things no longer behave the same when what we call our reference frame starts accelerating. We also saw that acceleration was somehow special when we looked at Galileo who found that all objects fall at a constant acceleration. Clearly, acceleration is a very important quantity, and by the end of today you will understand it very deeply.
We know that acceleration is somehow special, but how is acceleration special? Let's think of a cart, and we want to accelerate this cart. Accelerating it would make us happy. If we want to design an experiment to test what factors affect acceleration--let's think. When I push this cart what factors will control at what rate it accelerates? Does it matter how hard I push? Does it matter what material the cart is made of? Does the mass of the cart matter? Will a heavier cart maybe accelerate at a different rate? Does the color of the cart matter? Will a red paint job make the cart accelerate faster? To get you thinking about this, I want you to choose all that you think apply. When you see the answer, you maybe surprised at whether or not it matches with your intuition.
Well, how hard we push definitely matters. If I push really hard, the cart could accelerate really fast. The mass of the cart also matters. It's much harder to get a very massive cart, a heavy cart, moving than it is to get a light cart. The color, of course, doesn't matter, and the material it turns out doesn't matter either. You might have thought that a lightweight material, well, maybe that will accelerate faster. But then we're changing two variables. You'd be changing both the material and the mass of the cart. When we're thinking in terms of designing an experiment, we want to make sure that the variables we alter we're altering in isolation. If I have two carts that are the same mass, it doesn't matter what the material is made of.
We’ve just seen that acceleration somehow depends on force and mass. We're going to design an experiment to figure out how. First I want to find out how does acceleration relate to force. Maybe a is proportional to the force. Maybe it's proportional to 1/F. Or maybe it's proportional to the force squared or something else. I just don't know yet. But we can figure it out with an experiment. In this experiment, we're going to take a 10 kg cart--that's its mass-- and we're going to apply differing forces on it. We're then going to time how long it takes to go 10 m, and using some of our motion equations that we learned in the last unit, we're going to figure out the acceleration. Once we get enough data, we can see how acceleration depends on force. For the first one in the experiment, we'll push with a force of 10 Newtons, and this is a new unit, but it's the unit we use to measure force, and of course named after Isaac Newton. We find that it takes a time of 4.47 sec. Now, what must the acceleration have been? You may want to use this equation, and remember that the cart is starting from rest. Enter your answer here.
Well, let's use this equation and plug in what we know. Δy, the distance covered, is 10 m. Plugging in the rest of what we know-- this is 0 because the initial velocity is 0. Acceleration is what we're solving for, and we know the time. When we solve this equation for a, we get an acceleration of 1 m/s².
Here's the rest of the data. We increase the force and the time it takes to go 10 m goes down. We keep mass fixed at 10 kg the entire time. This is a 2-part quiz. I want to know how acceleration depends on force and mass. I want you to choose the best option. The way you're going to do that is by completing this table, though you don't have to enter anything in, and determining how does acceleration scale with force. I think it's a tricky problem, but you can do it. You'll probably need this equation to calculate the acceleration, and once you've filled this out on your own, you should be able to calculate how acceleration scales with force. Is it proportional to 1/F, proportional to the force itself, or something else.
If I use this equation and calculate the accelerations, I get 2, 3, and 4. Think back on how we calculate these proportionality problems. We set them up as ratios, and we realized that sometimes this ratio needs an exponent. Maybe that exponent is 1 or 2 or -1 or -2. Let's see what exponent we need here. Let's compare this data to that data. This will be our A1 and F1--A1, F1-- and this'll be A2 and F2. Well, this just has 4/1 = 40/10, and I'm keeping in mind that I might need an exponent. These 0s cancel and now we just have 4/1 = 4/1 to something. Well, these are identical so this something is just 1. This is exactly the definition of proportionality. A is proportional to F.
We've seen how acceleration scales with force--they're proportional-- but what about mass? How does mass affect acceleration? The experiment I want to do now is slightly different. Instead of changing the force, we're going to change the mass and leave the force the same through all the trials of our experiment. Can you look at this data, keeping in mind that, of course, the force is staying the same, calculate the accelerations, and then tell me how does the acceleration scale with m or how does it depend on the mass. Which of these best describes how acceleration scales with m? And you may need this equation to help you out.
Well, using this equation I can calculate the accelerations. Notice what's happened. When I double the mass, the acceleration is cut in half. When I got from 10 to 30, tripling the mass, the acceleration is reduced by 1/3. Likewise when I quadruple it, it's 1/4. This is exactly the idea of inverse proportionality. You also could have used the same ratio method we used last time.
Now we get to discover the root cause of motion. We get to answer the question how does motion occur. Can you, using these conditions we've derived here, tell me which of these three equations describes motion?
And the answer is that F = MA. So, we have it. the equation of Newtonian physics. Force times acceleration--what a beautiful equation. It says that the cause of acceleration is force. When I apply a force to a mass, I get acceleration. This one equation when properly applied can be used to derive almost all of Newtonian physics. So, congratulations. Now that you have this equation, you're done. You might as well stop. If you can do all these things, what's left to learn. See you in the next class, I guess. But that's not entirely true, because even though this equation is underlying all of these things, we still need to know how to use it.
So, how do we use F = ma? Well, let's think of Galileo. He taught us how to describe motion mathematically. He discovered that all objects one earth fall at a constant acceleration--10 m/s². He knew that a free-falling object's distance covered is proportional to time squared. He sort of answered the how of motion. Newton, on the other hand, taught us that it's forces that cause motion, and he represented that with an equation. He sort of answered the why of motion or at least began to answer the why. Now, unfortunately, Galileo died shortly after Newton was born, but Udacity has recently uncovered a series of time traveling text messages between Isaac Newton and Galileo Galilei. We were able to recover one screen shot from this incredibly historic and terribly anachronistic text conversation. Newton starts by thanking Galileo for giving him some shoulders to stand on. "No problem," says Galieleo. Wants to know why. Newton, of course, is getting a little cocky. Galileo is impressed. Newton tells him about F = ma. Galileo, of course, immediately understands the consequences. And then Newton--oh no, we got cut off there. I guess we'll have to extrapolate and figure the rest out ourselves.
So, to see where Newton may have been going with that conversation, let's go back to Pisa. Let's quickly remind ourselves what Galileo knew about an object falling to the ground. Well, Galileo knew that objects fall at a constant acceleration. Can you tell me here, do you remember what that constant acceleration is on earth?
It's 10 m/s². We abbreviate that with the letter g. Little g means 10 m/s².
What did Newton know? He knows that the cause of any acceleration must be a what? Can you fill in the blank with the appropriate word?
So, Newton knows that F = ma, and Galileo knows that the acceleration on earth is equal to g. For an object falling in free fall, what must be the force pulling it to the ground? Which of the following could we replace the F with to have this equation agree with this equation? Select the best answer.
Well, let's see. Let's try mg. If we replace F with mg, then we get mg = ma. Well, that's easy. The m's just cancel. The mass of the object, it turns out, doesn't matter at all. Isn't that a beautiful cancellation? Disproving Aristotle with a simple mathematical cancellation. g = a, which is exactly what we wanted to find, so this force must be equal to m * g. We give this force the subscript g to remind us that the cause of this downward force is gravity, and the acceleration due to gravity is this 10 m/s². We can see that no matter what your mass is, and we can see that regardless of your mass, the mass cancels out. Acceleration on earth will always be equal to g--pretty amazing.
Now, let's think. What happens when this object hits the ground? Well, now it's no longer accelerating, so Galileo's statement doesn't make any sense. In fact, Galileo didn't spend too much time thinking about objects at rest. He was much more interested in motion. So, let's think. What would Newton be saying now? Now that the balls are at rest, the acceleration is 0. According to this equation, which is a law of physics, it must always be true. A 0 acceleration must correspond to a force of 0. So, what happened? Did gravity just turn off all of a sudden? No, of course not. The ball is still being pulled down by gravity, and I'm going to label gravity as mg from now on where m is the mass of this ball and g is the acceleration due to gravity, that 10 m/s². What's going on? According to this force diagram, the ball ought to be accelerating downwards. There's a force downwards. Well, here we're saved by the fact that force is what's called a vector. I'm representing this force with an arrow, because this force has a direction associated with it--downwards. By adding a force in the opposite direction, we can actually cancel this force. Which of the following arrows do you think best represents the force needed to cancel this downwards mg. Choose what you think is the best answer.
Well, we know the total force must be 0, because it's not accelerating, because it's not accelerating, and the acceleration is therefore 0. This means that downward forces must equal upward forces. That's the only way these could balance, and that means that this is the best answer. If this one were the answer, mg would over power it, and that would indicate and downward accelerating object. If this were the correct answer, this force is bigger than g, and it would be accelerating upward. This might be the correct answer if, let's say, we were pulling on a rope and accelerating the ball upwards.
The question now is what's causing this unknown force, which I've labeled with a question mark subscript. We know that this force is being caused by gravity and the mutual attraction of all objects, so gravity is pulling this down, but I don't know what's causing this. Is the cause of this unknown force some sort of antigravity or is it a gravitational attraction to something above the ball? Maybe Galileo's hand or something out in space. Or is the ground somehow pushing upwards on the ball Choose what you think is the best answer. And, again, don't worry if you get this one wrong. This is to get you thinking.
Before I tell you the answer, I want you to get off your chair and push down on the ground. When you do, you can actually feel the ground pushing back on your hand. It may be confusing at first to think that the ground can somehow push on something, but it can--it absolutely can, and it does, and it's this pushing of the ground that prevents this ball from falling through the earth or you from fall through your chair.
Now, one important thing to notice is that this force of the ground pushing up on the ball is always going to be perpendicular to the ground. This is some sort of perpendicular force. In fact, let's call it FN where N means normal. Normal is just another word for perpendicular. Before we move on, I want to point out something truly beautiful here. We just discovered a force. We discovered that force by having real trust in our model, F = ma. If this is true and there really is a force downwards, there has to be, must be, a corresponding force upwards. We call that the normal force. In fact, we can verify the existence of the normal force with something as simple as a scale, like the one you might find in your bathroom. Here you are, some person, your mass is m, and you're standing on a scale. I don't like this picture, because people are hard to draw. Let's simplify a bit. Here you are, standing on a scale again. This time I've drawn you as a box. No offense. I don't think you look like a box. But it's just easier and we can find that drawing things this way will actually make solving later problems much easier. We know that gravity is pulling you downwards with a strength mg. That mg, by the way, is what we colloquially call your weight. The scale, of course, is pushing up with a normal force. Now, let's think. How big is that normal force? Is it bigger than mg, equal to mg, or less than mg? Remember, the total or net force on an object is equal to its mass times the object's acceleration. Choose the best answer.
Well, in these case the normal force must be equal to mg. If the normal force were bigger, the mass would be accelerating upwards, and if it were smaller mg would win and it'd be accelerating downwards.
Now, let's do the experiment again, instead of just standing on the scale, you're standing on the scale and the ground, so your weight is distributed between the two. If I'm going to draw this force diagram and once again I'm going to represent you as a box, gravity is still pulling you down, that's your weight but now there's two points of contact. There's the normal force from the scale which I'll abbreviate FNS for scale and is the normal force from the ground, FNG. Now, I want to know how big is FNS, bigger than, equal to, or less than mg. Choose your answer.
Well, let's think acceleration is zero which means force is up, must be balanced by forces down. What are the forces up, well these two combined. Our force down is just mg and we're looking for Fns and check it out. The normal force provided by the scale which is how hard the scale is pushing up on you, is equal to the weight mg minus the normal force from the ground, and that's how hard the ground is pushing up. Since, I'm subtracting something from mg, this gets smaller. Now it turns out in this first situation, the scale would have accurately measured your weight. If you weigh 70 kg, the scale would have read 70 kg because FN was equal to mg. In this case, the scale would have read less than 70 kg. So, what the scale is actually measuring is not your weight. They're measuring how hard they have to push up, so that you don't fall through. Pretty cool.
If your brain is getting a little fried, stick with me for a little bit longer. We've got one more thing to talk about with motionlessness, which is actually surprisingly fascinating. I want to point out one thing from before. An object that is motionless or at rest, well, from a force point of view, it's in the exact same situation as an object moving at constant speed. From here on out, whenever I talk about an object that's at rest, we could also be talking about an object with a constant speed. It's just a little easier to visualize an object that's not moving, and the key to motionlessness is realizing that when an object is at rest, its acceleration is 0. F=ma tells us that 0 acceleration means 0 total force. All it means is that any upwards forces must be balanced by corresponding downwards forces. Likewise, leftwards forces must be balanced by rightwards forces. Let's take a look at Aayush and Brent who are each pushing on this cart. This cart has mass m and we'll say that the mass is 5 kg and right now the cart is exactly at rest. They're both pushing very hard but the cart is at rest. I should mention that most of these questions we do, we're going to ignore something called friction. Friction is the force you feel between your hands when you rub them together. It's a real force and it's something we can talk about, but to make matters simpler, we'll often ignore it. So this is a frictionless world for now--so let's go ahead and take a look at all the forces acting on this cart. I'm going to draw a force diagram. Well, the first force I'm going to draw is mg, and in this case, mg is going to equal this 5 mg*g and we know g is 10 ms² which gives me a total of 50 N which gives me a total of 50 N and Newtons are the unit of force. The ground must also be pushing upwards with a normal force. Aayush is doing his best and he's pushing to the right, we'll label that Fa for Aayush and Brent is also doing his best pushing back to the left and.we'll label that Fb. Noticed how I've drawn these arrows--you know Aayush is pushing from the left to the right--I don't draw an arrow like this. For me, it's much easier to see what's going on if the arrow is coming from the object and moving out. So we know that the cart is at rest and I've given you a couple of the forces. I told you that mg is 50 N and I've told you that Aayush is pushing with a force of 100 N. Can you use this data as well as what we learned up here to fill in these two answers.
Well let's make upwards forces have to balance downwards forces. The only downwards force we have is mg, 50 N that must be balanced by the normal force, so that has to also be 50 N. We can do the same thinking for left and right to show that this force, Brent's pushing force must equal Aayush's, so that's 100 N. Now, this problem wasn't too difficult because all of the forces where in were in what we could call the xy plane. They were all either vertical or horizontal. Let's see what happens when we make things a little more complicated.
Let's make things interesting. Let's introduce some trigonometry into the equation. So here's Robin and you can tell by her determined face that she really wants to push this cart through this brick wall and it seems like she is not having very much luck. In this example, something is different from last time. Robin isn't pushing exactly to the right or exactly to the left. She is pushing at some angle and let's call that angle here α. I'm sure you can already see the triangles forming in your head. Let's take a look at what's going on. Over here I'm going to draw a diagram for the cart, the force diagram. Of course as always we have the carts weight playing downwards and noticed that I didn't even tell you what m was, you don't even care. We have seen normal force pointing upwards. We know because the car isn't falling to the ground. We have Robin who is pushing, well looks like she is pushing down into the right. I'm actually going to put that force over here and let's label it FR for Robin, and we can't forget that the wall is pushing to the left. Of course if the wall weren't there the cart would be free to roll but since it is and it's stopping the motion, the wall itself must be providing a force. Let's call that FW for wall. Now when I look at this force diagram something jumps out to me. I've got these forces which are both vertical. The wall force which is exactly to the left and then the force of Robin pushing which is causing me some pain. It's not to the right. It's not straight down. It's some combination of right and down. Wait a second though, I know how to handle this. This is just me breaking down the vector into it's components. Let's make a right triangle with this would be the hypotenuse. Well here's a right triangle and now we see we get our angle α back again and let's label these sides. Well this is the horizontal component of Robin pushing and this is the vertical component. I'm going to label them FRx and FRy. One thing that's important to remember is that by writing FR as these two together, we can now get rid of FR. This is the exact same trick we did last unit when we had some initial velocity and we have some angle alpha. We broke that velocity into x and y components. You are doing exact same thing but with forces and just like with velocity once we have expressed it in terms of components x and y, we didn't care about this guy anymore. Likewise, we're now using the x and y component of Robin's pushing and we're not going to think about this too much anymore. But we still have to answer the question, "What are these values for the x and y components of Robin's pushing?" Let me give you some numbers so that this is less abstract. Let's say that the force Robin is pushing with is a 100 N total that's FR and this angle she's pushing at is 30°. Can you use your knowledge of trig to tell me what is FRx and what is FRy? Enter your answers here and here to one decimal place. I'll be very impress if you can get this right on your first try.
Well, let's take a look at this triangle a little more closely. Let's find FRY first. This is the opposite side of this angle and we already know the hypotenuse. It's 100 N. What trade function that opposite the hypotenuse, the sine. So if I solve this for FRY, plug in my 100 N and my 30°, I get an answer of 50 N I can use similar reasoning to find FRX. This time it's the cosine and this gives me a value of 86.6 N.
All right so let's clean this up a bit. We've broken FR down into x and y component. That means we don't need FR anymore, instead, we're going to replace it with FRx, which points to the right and FRy, which points down. Let's say, I'll tell you the mass is 10 kg. Can you fill out the rest of these five forces? We already calculated FRy and x, what's mg? What's the normal force? What's the force the wall is pushing on the cart with? Remember the acceleration is equal to zero and one thing to be very careful of is the normal force. It may not be exactly equal to mg like it was last time. This is a tricky problem but if you get it right, very impressive. Good luck!
Well let's see, to calculate mg, we multiply m, which is 10 kg by g which is 10 m/s². If the right word force is 86.6 N and the object isn't accelerating, well the left word force must be bouncing it exactly. These two forces are the only horizontal forces that must be exactly equal. So this is 86.6 N. We also know that for opposite that rests, forces up must bounce forces down and for us that means that the normal force that goes up must equal not just mg, which goes down, but also mg + FRy. Those are the two downward forces and that gives me a value of 150 N. This was a very tricky problem, if you follow along, you're doing great and if not go back and watch again
If Isaac Newton is here today, he would say "Congratulations!" This is not easy stuff and you're doing great. Now, we're about to move into motion where things are actually accelerating. If your mind is feeling a little exhausted, now is the great time to take a break, maybe go and try to push a shopping cart into a brick wall. That being said, you should still come back soon because what we were about to cover are some of the problems that I think are the most fun to solve in all of Physics.
In the last problem, we broke down Robin's pushing force. We broke it into an x and y component, we had to do some trig but that was okay. Speaking of breaking things down, what if Robin succeeds? What if she breaks this wall apart? The force provided by the wall, well, of course, it has to go away. And what do we have now? Now, we have unbalanced forces. And when I say unbalance, what I mean is that this force doesn't have a companion force to the left. We are now entering the world of acceleration. Now, how do we solve the problem of what is this cart's acceleration going to be? Well, we already know that these forces balance this force. There's not going to be any vertical acceleration, ay = 0, ax on the other hand will not equal zero. It will equal something much more interesting and let's calculate. Well, the simplest thing, F = ma, but now, the only force we're using is the unbalanced force and that's equal to 86.6 N. I know the mass is 10, and when I solve, I get a = 8.66 m/s². Sorry, I did that one on my own. You have a chance to this next one.
For this problem, we have a sad Galileo. You can tell because well he's dropped his object and now he wants to get it back. Unfortunately, the leaning tower is very high and he doesn't want to walk all the way down. Luckily, Galileo has a friend--this makes him slightly happier but he still wants this thing back up here. He's going to repeat his experiment--that's how you get good data. Good thing that Galileo also has a rope, and let's say his friend ties this rope to the object. Now, we're talking. Now, Galileo's feeling pretty weak today. He can only pull upwards with the force of 25N. Can you tell me what would be the acceleration of the apple? I highly recommend you draw a force diagram, and when you do, once you get an answer--enter here.
Well, let's try a diagram. As always, I represent the object as just a box. The first thing I'm going to try is M₁g downwards. Note that M is 2 here and g is always 10, so this is going to equal 20N. We say that he can pull with the force of 25N--that means the net upwards force would be 25 minus 20. Let's call that FTotal. 25-20=5N, and now we're making progress. We know that F=ma and the key and this is where you may have made a mistake is that this the force we used--the total force after all the balancing has worked itself out 5N. And when I worked this out, I find the acceleration is 2.5 m/s². All right! Let's do a more challenging one.
Now, this is a really tricky question and that's why we saved it for last. I always save the good stuff for last. This problem involves two masses connect by a string which is draped over this pulley, so it's free to rotate. What we're going to do is solve this question algebraically to get an equation that tells us "If we let the system go, what will happen? Will it accelerate counterclockwise or just stay in this sort of equilibrium where nothing moves?" In regardless of the motion, I want to know some numbers. I want to be able to quantify this. What exactly will the acceleration be? Let's see if we can solve this. When we look at a complicated problem like this, we don't want to get scared. The first thing we want to do is make some sense of it in our heads. once we do that, we might have some bright idea about how we approach it quantitatively. When I'm doing this sort of thought experiment in my head, one question I'm going to ask is--what happens when the masses are equal? Will this whole system be motionless? Will it accelerate clockwise? That would mean this guy goes up and this guy goes down. Or will it accelerate counterclockwise? That would mean this mass will go up, this one would go down, and the pulley would spin that way. I'm also going want to think about "What will happen when m₁ is the bigger mass? This guy is heavier than this guy. Or when M₁ is the smaller mass?" Can you, for each of these three cases, tell me what you expect the motion to look like?
Well, if the masses are equal, there's no way to favor one direction out of the other--it'll remain motionless. If M₁ is bigger, it's going to go downwards--so that would be counterclockwise. And if M₁ is smaller, then M₂ will tend to go down--that's clockwise and there we go.
At the end of the day what we want is an equation of motion describing how these masses will accelerate. Before we jumped into solving the problem, there's still a little more benefit we can get from doing more thought experiments. Let's see what will happen when M₂ is much bigger than M₁, for example may be M₂ is a grand piano and M₁ is a slice of apple pie. This is a sort of silly example but I'm trying to make a somewhat serious point. We can push nature to its limits in our head but we never have to set up this experiment. We just have to think what's going to happen. Well, we already know when M₂ is greater than M₁, the system accelerates clockwise, of course. The piano pulls more than the pie, of course. In fact, when M₂ is so much bigger than M₁, we can even estimate its acceleration. So in this situation now, I want you to tell me--what do you think is the best estimate for how quickly this piano works all the way downwards and therefore, how quickly the pie works all the way upwards. Is it 10, 5, 2 or 1 m/s². Now, I want you to remember that a free falling object accelerates downwards at 10 m/s².
Well, this is essentially a free-falling object. The pie is going to have almost no effect on slowing down the piano. This is the best guess. Now, let's go ahead and make an equation that describes this motion and we've to check to make sure that it fits all of these, let's call them, test cases.
How do we solve this problem? Well, there's two objects we really care about--the two masses. The pulley we're largely ignoring--the only purpose of the pulley is to redirect where this rope is going. Well, two objects means we have to do two diagrams. Let's think of M₁ first. Well, of course, there's weight going downwards. Every object has its M₁g. But then the rope is also providing a force and that force is called tension. Now remember we're dealing with the case where M₂ >> M₁. So I want to know what's the relationship between this weight force going downwards and this tension force. Are these M₁g = FT? Is the M₁g > FT? Or is the M₁g < FT? So what do you think is the correct answer.
Well, we said earlier that the system is accelerating clockwise--that means M₁ must be going upwards. If it's accelerating upwards, this force has to dominate over this force-- the FT must be bigger than M₁g.
Now, we can solve for the acceleration of block 1. Well, the acceleration is going to be caused by how much bigger this force is in this force or FT - m₁g and that's going to equal m₁ * a₁. Now, the problem is we can't just go and solve this for a₁, not yet anyways. This equation has two unknowns, the a₁ and the tension. One equation will never be able to solve for two unknowns. Maybe we can get more information by analyzing the second block. Well, let's do the diagram. And before we go any further, I'm going to move this equation over here to emphasize that the blank is at block 1. Now, previously we showed the FT is bigger than m₁g since it's accelerating that way. Well, m₂ we know must be accelerating this way, so m₂g must be bigger than FT I've crossover an important point here--is this FT here the same as this FT here? and the answer is yes, it is. A rope always maintains the same tension throughout. Maybe you prove it to yourself by trying to think about what would happen if somehow the rope's tension change. What would happen at the boundary point where the tension change? That's something you should talk about in the forums. Well, now the force doing the acceleration is how much bigger m₂g is over FT and we can write a very similar equation m₂g - FT = m₂a₂, net force = mass * acceleration, and once again we have two unknowns, a₂ and FT, well that consist a total of three unknowns, FT, a₁, and a₂. Two equations can never solve three unknowns, but hey, I realized something, these blocks are connected by a string. Any motion downwards from this guy has to correspond to motion upwards from this guy. That means this acceleration has to be identical to this one. We've done a little strange here what we called acceleration on this side, positive when it's up, and on this side, positive when it's down, but that's okay, we can do that. So, now instead of the a₁ and a₂, I can just write a. We're almost there--we're almost at something very cool because now we only have two unknowns and two equations can solve two unknowns.
I'm going to go through this algebra--so if you are an ace in solving this sort of systems of equation you can skip this part, but for those who aren't, I'm going to walk you through it. So, here we have our two equations. I'd like to write them one above the other to make thing a little simpler for me. Now my goal is to eventually know what the acceleration of the system is. I'm going to do that by, first, solving for the tension in one of these equations and then plugging in and let me show you what I mean by that, so let's take this equation. If I move this term to this side, I find that the tension is equal m₁a and a is still unknown plus m₁g. I'm going to plug this in, this value of the tension, in to here. The idea of being that, now I'll get this equation in terms of only one unknown, a, and FT won't show up anymore. So rewrite m₂g minus FT, but now I'm using this, now equals m₂a. Let's just distribute this minus sign and remove the parenthesis. Well now I'm solving for a, so let's group all the a terms together. I'm going to move this a over here, okay? So, now that we've done that, I'm going to take out the common factors of each of these terms and each of these terms. For example, each of these terms has g in it and each of these terms has an a in it, and now we're really close. Let's divide both sides by this term here, and here is our answer and this is truly a thing of beauty and let me explain why.
This is such a beautiful equation. There is so much intuition goes in here. And there is a few reasons why I love this equation. First, it says that acceleration is just your normal acceleration due to gravity times some multiplier and then we can think about, what is this multiplier. Well, this multiplier depends on the difference of the two masses. If this masses are identical, well subtracting them will give you zero and you'll have no motion. Acceleration will be zero exactly like we expected. If m₂ is bigger than m₁, well this will be positive in the denominator that means we'll have positive acceleration which we defined as clockwise Exactly what we expected when m₂ > m₁. Likewise, if m₁ > m₂, it will be negative which becomes counterclockwise and finally, in the case where let's say m₂ is >>> m₁. Let's say m₂ was 1000 kg and m₁ was 1 Likewise,1000+1, who cares about the 1, ignore it. We have the a = g * m₂/m₂, that cancels. Our acceleration is equal to g 10 m/s². I love that problem. I hope you like it as much as I do. There is going to be a lot more learning to do on the problem set. You are not done here, I promise, but I think what you've done today is really amazing. Good work.