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Current time:0:00Total duration:4:22

2D equilibrium -- balancing games

Video transcript

I got this to balance by carefully measuring equal distances, and hanging equal weights at those distances. This balance is called equilibrium. Equilibrium is when the state of the system isn't changing. In this case, the 2D equilibrium case, the state of the system is the position of these objects, which still isn't changing, unless I throw a coin in to knock it out of equilibrium. In other types of equilibrium, the state of a system could be its temperature, or pressure, or something else entirely. But now, I'll focus on 2D equilibrium, the balanced state where the positions don't change. Some things balance in the middle, and these things tend to be symmetrical. Other things balance off to one side, always the heavier side. What if I want to move this cup a different distance away from the middle and still keep it all balanced? One way of doing this is moving the other cup to the new distance as well. Sure enough, it balances again. But there's another way of doing this. I could've left the right hand cup at the old position and just taken out some of its weight. This also balances. So we found two different ways of balancing the same thing. We can either change the weight or change the distance. In equilibrium, where we have forces F1 and F2 balancing each other at distances D1 and D2, the counterclockwise force times distance must equal the clockwise force times distance. Force times distance has a special name, torque, from the Latin word to twist. In 2D equilibrium, clockwise and counterclockwise torques are balanced. Before I calculate some torques, I need to check some of the masses. The mass of five coins is 13 grams. And the mass of the cup plus the string is also 13 grams. And since we're on Earth, this means that they both have the same weight, which I'll call W. Start it off with weights 2W one cup plus 5 coins hung on each side, distance I'll call D from the middle. And I got it balanced, and the torques 2 times W times D, are equal on both sides. And it balances. But when I moved the distance to 1/2D and kept the weight at 2W, five coins plus a cup, there were two ways that I could balance it. I could either move the other cup closer to 1/2D, or I could keep the other cup a D and just empty it of the five coins. So now its weight is only W. This leaves the counterclockwise torque, 1/2D times 2W, equal to the clockwise torque, D times W. But what if we move the left hand cup farther from the middle to 2D, and leave its weight at 2W. One way of balancing it is to move the right hand cup farther out as well. Or we can move the right hand cup back to D, and increase its weight to 4W. Let's check that this would make the torques work out. On the left hand side, we have 2D times 2W, which is 40W. And on the right hand side, we have D times 4W, which is also 40W. How many coins will we need in the cup for that? One cup is 1W, and so we need three more W, So that's 15 coins. So four W is equal to one cup plus three more W, which we get with 15 coins. I'll move the cup to a distance D from the middle and add a total of 15 coins, for a total weight of 4W. Sure enough, it balances, because in equilibrium, the counterclockwise torque has to equal the clockwise torque.