Sections | |||
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Torque & Equilibrium | 23 mins | 0 completed | Learn |

Review: Center of Mass | 14 mins | 0 completed | Learn |

Equilibrium with Multiple Objects | 30 mins | 0 completed | Learn |

Equilibrium with Multiple Supports | 31 mins | 0 completed | Learn |

Center of Mass & Simple Balance | 31 mins | 0 completed | Learn |

Equilibrium in 2D - Ladder Problems | 40 mins | 0 completed | Learn |

Beam / Shelf Against a Wall | 53 mins | 0 completed | Learn |

More 2D Equilibrium Problems | 29 mins | 0 completed | Learn |

Example #1: Position of second kid on seesaw

**Transcript**

Hey guys! In this example we're going to solve a problem where we have two kids playing a seesaw, which is a classic problem in rotational equilibrium. In this particular one, we have a kid all the way to the left and we want to know how far we should place the kid on the right so that the system balances. It says here the seesaw is 4 meters long, so IÕm gonna write L = 4 meters. The mass of the seesaw, IÕm going to call this big M is 50. The seesaw has uniform mass distribution that means that the mg of this seesaw is in the middle and it's pivoted on a fulcrum at its middle. The fulcrum is at the middle and the mg is at the middle as well. This means you have an mg down here and you also have a normal force pushing this thing up at the fulcrum. The two kids sit at opposite side, so one on the left, the other on the right. The kid on the left has a mass of 30 and the kid on the right has a mass of 40. First of all, you might imagine that if this is a perfectly symmetric system, you don't have to draw this, but if you have the bar held at the middle and then you got the 30 here and the 40 here, this will tilt his way because they have the same distances but they have your kid has a bigger force pushing down so it tilts. The solution then is to move the heavier kid closer to the middle. That's what we want to figure out is how far from the fulcrum should the kid be. What we're doing here is we're looking for a situation where we have no rotation, so the sum of all torques will be equals zero. There are only two torques happening here. I have m1g at the left tip which will cause a Torque 1 over here and then I have an m2G on the right side somewhere in the middle that's going to cause a Torque 2. Note that the torque of mg is 0. mg does not produce a torque and that's because mg acts on the axis of rotation. Normal also doesn't produce a torque for the same reason. They both act on the axis of rotation. Forces that apply on the axis of rotation cause no torque. Since we want torques to cancel, we're just going to say that Torque 1 has to equal Torque 2. TheyÕre in opposite directions so as long as they have the same magnitude, they will cancel. I can write Torque 1 = Torque 2. Once we expand this equation, that's where we'll be able to find our target variable. Torque 1 is due to m1G so it's m1g r1 sin_1, and Torque 2 is due to m2G, so it's going to be m2g r2 sin_2. Just using the torque equation there. Notice the gravities cancel because I have gravity on all terms. What I'm looking for is r2. Let's plug in all the numbers and then get r2 out of the way. First thing however I want to show you is that once I draw the r vectors, this is r1. r vector remember it's an arrow from the axis of rotation to the point where the force happens, so that's r1 and then this is r2. Notice that the mg's will make an angle of 90 degrees with the r vector. mgs and rs are making 90 degrees which means both of these guys go away. This simplifies to just this: m1r1 = m2r2. This is what you're going to have in almost all seesaw problems. All of these seesaw problems will come down to this very basic ratio. We're looking for r2 and if I want to, I can even solve this with letters. r2 is just (m1 / m2) r1. That's an expression for the solution. Now let's plug in numbers. m1 is 30, m2 is 40, r1 is this distance here. We're in the middle the so this distance here has to be 2. The whole bar is 4 so the distance from the very middle to the very left end is half of the length so it's 2. If you multiply and divide this whole thing, you get 1.5 meters is r2. r1 is 2, and r2 is 1.5 and that should make sense. That is consistent with what we said in the beginning that's the heavier kid has to be closer to the axis of rotation. It's actually a pretty simple ratio. That's it for this one. Hope it makes sense and let's keep going.

Practice: A 20 kg, 5 m-long bar of uniform mass distribution is attached to the ceiling by a light string, as shown. Because the string is off-center (2 m from the right edge), the bar does not hang horizontally. To fix this, you place a small object on the right edge of the bar. What mass should this object have, to cause the bar to balance horizontally?

Practice: Two kids (m,_{LEFT} = 50 kg, m,_{RIGHT} = 40 kg) sit on the very ends of a 5 m-long, 30 kg seesaw. How far from the left end of the seesaw should the fulcrum be placed so the system is at equilibrium? (Remember the weight of the seesaw!)

Example #2: Multiple objects hanging

**Transcript**

Hey guys! In this example, we have a weird arrangement of objects that are sort of hanging together by a bunch of ropes and rods. We want to find the missing masses, so we don't know the mass of a and c. We're going to call this guy b and we're going to say the mass of b is 4. We want to find the masses of a and c. We also want to know all these tensions. We want to know all the tensions. Calculate the five vertical, the tension of the five ropes, and the two missing masses. It says that the system is in linear and rotational equilibrium. This means that sum of all forces equals zero. This means that the sum of all torques equals zero. It also tells us that all the ropes, all the vertical ropes, which are the arrows are massless and all the horizontal rods which are the horizontal lines are massless as well. We donÕt have to worry about those. We're going to use g as 10 m/s^2 to simplify things. First of all, I want to name a few things here. We have a, b and c. This is going to be the tension of c because it holds c. This is going to be the tension of b because it holds b. This tension here, IÕm going to call this the tension of b and c because it holds those two masses. This is going to be Ta and this one holds a, b and c so this is going to be Tabc. Let's talk about forces first. All the forces cancel. Let's look at each object. This here has the force of mb g. Because this object is in equilibrium, the tension up cancel with the force down. I can actually calculate mb g because I have the mass is 4, gravity is 10. We're using 10, so this whole thing is 40 therefore this has to be 40. Here I can't tell what Tc is yet because I don't know and mc, so we're going to have to use something. If I look at this dot here, you can tell that this dot is being pulled up by Tbc, but pulled down by these two guys here. Tbc is going to be Tb + Tc. If you look at a, I have ma g, but I don't know what that is yet so I can't find Ta just yet. I can also see how Tabc is Ta + Tbc, that's why we called it abc because it's all three of them. There's a bunch of stuff we don't know yet. We have to find these two masses, which will allow us to find these two tensions. And then once we have that, we're missing this that's why we can find this. Then once we have that, we'll be able to find the whole thing. We're also missing all of this stuff up here. You want to start this problem from the ends of the problem. What I mean is this is like a bunch of masses that are chained together. This is the lower point, the end of the problem. The reason I call it the end of the problem is because to get to this point, we're going to need this point first and a will contribute and then this stuff will contribute as well. The first thing we have to do is work our way from the bottom here up. Then once we get to this point here, we'll be able to figure out what a is. Let's just get started. Everything is figured out for b, so we're going to go with c. We can't figure out c using the sum of all forces equals zero, because we already did that. All that tells us is that Tc = mc g but I don't know these two guys so I'm stuck. What you have to do is you have to use torque. We're going to say that the torque about this point here, let's call this point one. The sum of all torques about point one is 0, which means that all the torques acting on it will cancel out. There's two torques acting about that point. First of all, Tb is doing this so this is torque of Tb and Tc is doing this, torque of Tc. Imagine you're the center point. There's a rope that pulls you down this way and there's a rope that pulls you down this way. Those two torques will cancel. I can write Tb = Tc. A torque is a force which in this case is Tb. The distance which weÕre going to call rb, and then sin_b. IÕm going to write the same thing for the right side, Tc rc sin_c. The r vector would be this here from the axis. This is the point weÕre treating as the axis of rotation to this point here. This is rb and this is rc. Notice that in both cases, the axes are this way, the tensions are down. In both cases the angles are 90, so this is going to be a 1 and a 1. This simplifies to Tb rb = Tc rc. By the way, that's going to be the case with every one of these kinds of points. When you go here to point two, we're going to be able to write the same thing. There are no angles here for us to worry about. I know that Tb is 40, rb is the distance it says right. There the distance is 1. Then IÕm going to be able to find Tc because I have the distance here for rc is 2. Tc = 40 / 2. Tc is 20. Now that I know that Tc is 20, I can plug it in here 20 = mc (10). That means mc is 2. Not only we figured out the tension but we also figured out that mc will be 2 kilograms. There's a faster way you could have done this once you get good at this stuff. It's basically a little game. Once you realize that this is the relationship, you can sort of say c is twice the distance, so to have the same torque you have to have half the mass because torque in this case of the weight is mg r and here the sin_ just gives us 1. If you have double the distance, you to have half the mass in order to have the same torque. Double the distance, half the mass. Now that we know Tc, we can actually find Tbc. Tbc is Tb which is 40+20 = 60 N. Check it out. You can think of this tension right here as essentially holding the 4 and the 2. You can think of this as a and 6. You can combine those two Ð the 4 and the 2 Ð into a 6 and then there's a tension here. This is 1 and this is 4. Remember what I just told you. If you are farther, you have to be lighter. a is four times as far from the central point here from the axis of rotation from the support point than 6. If it's four times farther, it also has to be four times lighter. Just by using this, you can tell that the mass of a has to be 6/4 which is 1.5. You can do it like this very quickly. I'm going to do this the full way which would be to writeÉ Here we wrote that all torques on point one are zero. Now IÕm going to write that all torques on point two are zero. What this gives us is that the torque of a will cancel out with the torque of bc. Torque of a = torque of bc, so tension of a ra sin_a = tension bc rbc sin_bc. We talked about earlier how the sines will just be one because all the revectors are horizontal and all the tensions are vertical. The tension we're looking for here is a. The distance is 4 meters, so Ta(4), Tb is 60, the distance for Tb is 1 it says right here, 1. Ta is 60/4 = 15 N. We know that this is 15 N and these two guys have to be the same. If Ta is 15, then ma, gravity is 10, ma is 10 / 15, 1.5 which is what I just mentioned here but I wanted to show you how this stuff works out. You can actually skip some steps if you just realize that you have this relationship here at any one of these points that connect stuff. You can actually also write this like this, mb rb = mc rc, which is essentially what I did when I shortcut to this question. The reason you can do this is because the torque equation does this, mb g rb = mc g rc, and then you're essentially cancelling the rÕs and then the output something simpler like this. That's a quick shortcut you can use. I almost have everything. The last thing I have to do is I have to combine the 60 here with the 15 here and I get a 75 N as the tension of everything. I got the masses. The masses are 1.5, 4, and 2. The tensions are 75, 15, 60, 40 and 20. All it took was using the fact that the tensions up equals the weights down and the fact that you can use torque about points one & two, the fact that the torques cancel and you end up with stuff like this and stuff like this that allows you to find the other ones. It's a very peculiar kind of question but once you get the hang of this, it's really easy to solve a bunch of them. Hopefully it made sense. Let me know if you guys have any questions and letÕs keep going.

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Example #1: Position of second kid on seesaw

Practice #1: Balancing a bar with a mass

Practice #2: Position of fulcrum on seesaw

Example #2: Multiple objects hanging

Two objects of masses 28 kg and 21 kg are hung from the ends of a stick that is 70 cm long and has marks every 10 cm, as shown.
If the mass of the stick is negligible, at which of the points indicated should a cord be attached if the stick is to remain horizontal when suspended from the cord?
1. B
2. D
3. G
4. F
5. C

A meter stick is supported by a knife-edge at the 50-cm mark. Doug hangs masses of 0.40 and 0.60 kg from the 20-cm and 80-cm marks, respectively. Where should Doug hang a third mass of 0.30 kg to keep the stick balanced?a. 20 cmb. 70 cmc. 30 cmd. 25 cm

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