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# Equilibrium with Multiple Objects

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Sections
Torque & Equilibrium
Review: Center of Mass
Beam / Shelf Against a Wall
Center of Mass & Simple Balance
Equilibrium with Multiple Objects
More 2D Equilibrium Problems
Equilibrium in 2D - Ladder Problems
Equilibrium with Multiple Supports

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