15. Rotational Equilibrium

Equilibrium with Multiple Objects

1

example

## Position of second kid on seesaw

4m

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Hey, guys. So in this example, we're going to solve a problem where we have two kids playing a seesaw, which is a classic problem in rotation equilibrium. And in this particular one, we have a kid all the way to the left, and we wanna know how far we should place the kid on the right so that the system balances Alright. So it says here the seesaw is 4 m long, someone to write. L equals 4 m, the mass of the sea. So I'm gonna call this Big M is 50. Um and the seesaw has uniformed mass distribution. That means that the mg of this he saw is in the middle on bits pivoted on a fulcrum and its middle. So the focus is at the middle and the M G 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 folk. Um, okay. The two kids sit at opposite sides, so one is left on the right. The kid on the left has a massive 30. And the kid on the right has a mass of 40. Okay, So, first of all, you might imagine that if this is a, um, perfectly symmetric system, you don't have to draw this. But if you have the bar held in the middle and then you've got the 30 here in the here, this will tilt this way because they have your kid. They have the same distances, but they have. Your kid has a bigger force pushing down, so it tilts. So the solution, then, is to move the heavier kid closer to the middle. Okay, that's what I want to figure out is how far from the folk, um, should the kid be all right. So what we're doing here is we're looking for a situation where we have no rotation. So the sum of all torque who equals zero and there are only two torques happening here. I have m g m one g at the left tip, which will cause a torque one over here. And then I have an M two G on the right side, somewhere in the middle. That's going to cause a torque to note that the torque of M. G is zero. Energy does not produce a torque. And that's because M G acts on the axis of rotation. Normal also doesn't produce a torque for the same reason they both, um, act on the axis of rotation force that apply on the axis of rotation cause no torque. Since we want talks to cancel, we're just gonna say that torque one has to equal talk to their in opposite directions. So as long as they have the same magnitude, they will cancel So I can write Torque one equals torque to. And once we expand this equation, that's where we'll be able to find our target variable. Alright, so Torque one is due to M one G. So it's m one g R. One sign of theta one and talk to is due to M two G. So it's gonna be M two g are to sign of data just using the target equation there. Um, notice the gravity's cancer because I have gravity on all terms on DWhite I'm looking for is our two, all right, So let's plug in all the numbers and then get our two out of the way. First thing however I want to show you is that once I draw the are vectors, this is our one, our vector. Remember, it's the ax. A narrow from the actual rotation to the point of the force happens. So that's our one. And then this is our to notice that the MGs will make an angle of 90 degrees with the are vectors. So MGs and ours are making any degrees, which means both of these guys go away. So this simplifies to just this m one r one equals them. Two are too. And this is what you're gonna have in almost all see solve problems. So all of these see solve problems will come down to there's very basic ratio. All right, so we're looking for are too. And if I want to, I can even solve this with letters Are too is just m one over m two times are one That's an expression for the solution. Now let's plug in numbers. M one is 30. M two is 40 are one. Is this distance here? We're in the middle, the so this distance here has to be to write the whole bars four. So the distance from the very middle to the very left end is half of the length, So it's too. Um, and if you multiply, divide this whole thing you get 1.5 m is are two. So are one is 1.5 or two is one. I'm sorry R r one is to sorry about that. And our two is 1.5. And that should make sense. That is consistent with what we said in the beginning that the heavier kid has to be closer to the axis of rotation. So it's actually pretty simple ratio s So that's it for this one. Hope makes sense and let's keep going.

2

Problem

ProblemA 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?

A

2 kg

B

4 kg

C

5 kg

D

10 kg

3

Problem

ProblemTwo 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!)

A

2.29 m

B

2.50 m

C

2.71 m

D

4.58 m

4

example

## Multiple objects hanging

11m

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Hey, guys. So in this example, we have a weird arrangement of objects that are sort of hanging together by a bunch of ropes and rods and we want to find the missing masses so we don't know the mass of A and C. We're gonna call this guy be and we're gonna see the massive B is four. So we wanna find the masses of A and C Um and we also want to know all these tensions, right? We wanna know all the tensions. So calculate the five vertical detention of the five ropes and the two missing masses. It says that the system is in linear and rotation equilibrium. This means that some of all forces equals zero. This means that some 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 is well, so we'll have to worry about those. And we're gonna use G as 10 m per second square to simplify things. Alright, So first of all, I want to name a few things. Here we have a B and C Now, this is going to be the tension of C because it holds. See, this is gonna be the tension of B because he holds B This tension here. I'm gonna call this detention of B N. C. Because he holds those two, uh, masses. This is going to be t a whips. That doesn't go there. This is gonna be t a. And this one holds a B and C So this is going to be t A B. See, Let's talk about forces first. All the forces cancel. So let's look at each object. This here has a force of M B. G. And because this objects in equilibrium, the tension up canceled the force down. I can actually calculate MBG because I have the masses. Four Gravity's 10 were using 10 so this whole thing is 40. Therefore, this has to be 40 here. I can't find I can't tell what TC is yet because I don't know m. C. So we're gonna have to use something else. Um, if I look at the stock here, you can tell that this stock is being pulled up by TBC but pulled down by these two guys here. Okay, So TBC is going to be t B plus T c? Um, if you look at a I have m a G. But I don't know what that is yet, so I can't find t a just yet. I can also see how t ABC is t A plus TBC. That's why we call it ABC, because it's all three of them. Okay, so 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. And then once we have that, we'll be able to find the whole thing. We're also missing all of this stuff up here. Okay? 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. Um, this is the lower point, the end of the problem. The reason I called the end The problem is because, um, we're going to get to this point. We're gonna need this point first, and they will contribute. And then this stuff will contribute as well. So the first thing we have to do is work our way from the bottom here, um, up right. And then once we get to this point, here will be able to figure out what a is. All right, So let's just get started. So everything's figured out for be so we're gonna go with C we can't figure out. See using using the sum of all forces equals zero because we already did that. All that tells us that t c equals M c G. But I don't know these two guys, so I'm stuck. So what you have to do is you have to use torque, and you're gonna have to say, We're going to say that the torque about this point here, let's call this 20.0.1 thes some of all talks about 0.1 is zero. Which means that all the torques acting on it will cancel out. There's two torques acting about that points. First of all, TB is doing this, So this is torque of T B and T. C is doing this torque of TC. So 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, right? And those two torques will cancel so I can write t b equals TC. Now, a torque is a force which in this case, is TV Um, the distance which gonna call RB and then sign of theta be I'm gonna write the same thing for the right side. T c r c sign of theta psi. The are vector would be this here right from the access. This is the point we're treating as the axis of rotation to this point here. So this is our B and this is our C notice that in both cases, the axes air this way, the tensions air down. So in both cases, the angles air 90. So this is going to be a one and a one, So this simplifies to t b r b equals to t c R c. And by the way, that's gonna be the case with everyone of these kinds of points. So when you go here 2.2, we're gonna be able to write the same thing there. No angles here for us to worry about. All right, so I know that t B is 40. R b is the distance. It says right there. The distance is one. Um Then I'm gonna be able to find TC because I have the distance here for r C is true. So t c equals 40. Divided by two TC is 20. Okay, TC is 20. So now that I know the TCs 20 I can plug it in here. 20 equals m c 10. That means EMC is to Okay, so not only we figured out the tension, but we also figured out that M C will be 2 kg. Okay, Now, there's a faster way you could have done this once You get good at this stuff, it's basically a little game. Um, once you realize that this is a relationship, you can sort of say, Hey, you know what this guy is? Um, this guy here is whips, See? Is twice the distance Soto have the same torque you have tow have half the mass, right? Because torque in this case of the weight is M g r. And hear the sign of data just gives us one. So if you have, if you have double the distance, you have tohave half the mass in order to have the same torque. OK, double the distance, half the mass. Cool. Now, now that we know T. C, we can actually find TBC, right. TBC is T B, which is 40 plus 2060 Newton's que Newton's. Um, Now check it out. You can think of this tension, um, this tension right here as essentially holding the four and the two, right. So you can think of this as a and six. You can combine those two before the two into a six and then there's attention here. This is one and this is four. And remember what I just told you. If you are farther, you have to be lighter. A If it's four times as far from the central point here from the axis of rotation from the support point than six. So if it's four times farther, it also has to be four times lighter. And just by using this, you can tell that the massive A has to be 6/4, which is 1.5 so you can do it like this very quickly. All right, now, I'm gonna do this the full way, which would be too right here. We wrote that all torques on 0.1 or zero. Now I'm gonna write that all torques on points to are zero and what this gives us is that the torque of a we'll cancel out with the torque of BC torque of a equals torque of BC so tension of a are a sine theta a equals tension. Bc r b c sign of theta BC We just talked about We talked about earlier how the signs will just be one because all the are vectors, air, horizontal and all the tensions Air vertical Um, the tension we're looking for here is a The distance is 4 m. So ta four t b. It's 60. The distance for TB is one. It says right here one. So t a is 60/ 15 Newton's So we know that this is 50 Newtons and these two guys have to be the same. So if ta is 15 than Emma Gravity's 10 Emma is 10 divided by 15 1.5, which is what I just mentioned here. But I wanted to show you how this stuff works out. Okay, so 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, right, you can write You could actually also right this like this M b r b equals T. C. I'm sorry, m c r c, which is essentially what I did when I shortcut of this question. The reason you could do this is because the torque equation does this mbg are b equals M c g R c. And then you're essentially canceling the arse and in the up with something simpler like this. Okay, so 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 Newtons as attention of everything. So I got the masses. The masses are 1.5 four and to the tensions are 75 15 40 in 20. Okay, and all it took was using the fact that the tensions up equals the masses of the weights down. Um, and the fact that you can use torque about points one and 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. Okay, so it's a very peculiar kind of question, but once you get the hang of this, it's really easy to solve. Ah, bunch of them. So hopefully made sense. Let me know if you guys have any questions, let's keep going.

Additional resources for Equilibrium with Multiple Objects

PRACTICE PROBLEMS AND ACTIVITIES (3)

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