Why does the sidewalk perfectly support the wagon's weight? W-, we know that the wagon is perfectly supported because it's not accelerating. Its weight downward must be perfectly canceled by the upward support force that the sidewalk is exerting on the wagon. But how does a sidewalk know how to exert that amount of force on the wagon? The answer tot he question is, the wagon and sidewalk negotiate until the sidewalk perfectly supports the wagon's weight. That negotiation is real and it actually takes some time. Underlying it is the fact that there's no such thing as a truly rigid object. Everything dents slightly when you push on it and when it's pushing back on you. So when I push on the sidewalk, it dents a little as it pushes on my hands. The more I push on it the more it dents. Now the denting of a sidewalk is so minuscule it's hard to see. But it's a lot easier if I pick a softer object like this spring scale not only moves downward noticable When it's pushing, when I'm pushing on it and it's pushing on me but there's the needle that shows you how much it is dented downward. So if I push gently on it it dents a little. If I push hard on it it dents a lot. And you can see everything in between. Let's watch the negotiation when I set the wagon on the scale, as opposed to on a on the sidewalk. Here I'm going to actually drop the wagon onto the scale and watch the negotiation occur. [NOISE] That bouncing of the needle is is real, that's the negotiation. Whenever the wagon is under supported as it is before I, it even touches the scale it's accelerating downward, and therefore tends to move toward the ground, down, down. Whenever he wagon is denting the scale deeply, it's being pushed up so hard that it's accelerating upward and will soon move upward and so the, the every time tho-, the, the, dent is too deep the wagon accelerates up. Every time the dent is too small it's accelerating down. It's negotiating with the scale to find just the right amount of dent and the right amount of upward force so that the wagon can be motionless, inertial, and here at rest. And let's watch that negotiation again. Here it goes. Ready, get set, negotiate. And when it all settles down, we have a motionless, perfectly supported wagon. The same negotiation that occurs between the wagon and the spring scale also occurs between the wagon and the sidewalk. They negotiate briefly and, when the dust settles, the sidewalk is perfectly supporting the wagon's weight and the wagon is therefore inertial and motionless. Actually that same negotiation occurs whenever you set an object on a horizontal surface. For example, an egg, if I set the egg on the table, there's a brief negotiation, and then the table is perfectly supporting the egg's weight. So the net force on the egg is zero. The egg isn't accelerating. The egg is at rest and staying at rest. Voila, but that brings us to a question. Can the table ever exert and upward force on the egg that is greater an amount than the egg's weight? Although the table is prefectly supporting the egg's weight right now during the negotiation that occured when I set the egg down there were times when the egg was being pushed up harder then it's own weight. The egg can take a little bit of extra force exerted on its surface, but not a whole lot. Watch what happens if I make that negotiation extreme by dropping the egg on the table. That negotiation is an amazing thing. It ensures that the sidewalk perfectly supports the wagon. In fact it works when the wagon is motionless, or when the wagon is moving, or even when the wagon is accelerating back and forth. It's still perfectly supported because of the negotiation. Actually, if the sidewalk is rough and bumpy, the negotiation is ongoing, and the wagon is still supported on average. Well, before putting the wagon on a ramp, I have one more question to look at regarding the support of objects by surfaces. If I put a ball on the sidewalk or on the table here, the ball's pushing down on the table, the table's pushing up on the ball. Those two forces are equal in amount but opposite in direction. Do they ever cancel? Although the two forces are equal in amount in opposite directions, they act on different objects and therefore never cancel. Only one of those forces acts on the ball, namely the force that the table exerts on the ball. That affects the ball and it can easily cause the ball to accelerate. For example, [SOUND] now, the ball is accelerating. It leaps right off the table. because the table is pushing up on the ball so hard during that impact that the ball is accelerating upward. And off it goes. The table is experiencing things as well. It's attached to the whole world, so you don't notice it very much. But the bottom line is, the 2 forces in a Newton's Third Law pair never cancel because they're always on different objects. I asked that question to call attention to an important misconception people have about Newton's Third Law. While it's true forces always occur in pairs, that are equal in amount but opposite in direction, those forces never cancel because only, each, each of the forces acts on a different object. For example if I push on this wagon one of the forces acts on the wagon and causes it to accelerate. The other force in that Newton's third law pair, namely the, the, force that wagon exerts on my finger acts on me. And in principle it would cause me to accelerate. In fact, if I were on slippery ice, I would accelerate. Off I'd go in that direction, the wagon would go off to the right. Well, that's it for wagons on sidewalks. Now it's time to tip the sidewalk up to make it a ramp.