 ## Physics

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15. Rotational Equilibrium

# Equilibrium with Multiple Objects

1
example

## Position of second kid on seesaw 4m
Play a video:
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

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? 3
Problem

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!) 4
example

## Multiple objects hanging 11m
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