Hey, guys. So up until this point, we've dealt with equilibrium questions that had a single support point, such as a bar that's held by a single rope. But now we're going to look into problems with multiple support points. And I'm going to go over a really important property of equilibrium questions, a really important technique you can use to solve these questions in an easier way. Let's check it out.

Alright. So when an object in equilibrium has multiple supports, such as here, I got 2 ropes. We can think of each support as a potential axis of rotation. I'm going to just write axis there. You can think of this point as being an axis and then this point as another axis. One way to sort of visualize this is imagine that if you cut this rope here, the bar is going to spin around this point where the rope touches. The same thing goes if you cut the rope here, the whole bar is going to spin this way around this point right there. Okay. So that's one way to sort of visualize that. Therefore, we can write that the sum of all torques equals 0 for either one of those two points, for any support point. And in doing this, you're effectively treating it as the axis, even if it's not the axis, right? So, if you don't cut it, they just stand there. There is essentially no axis of rotation because it's not rotating, but you still treat it as an axis. Meaning, when you write your torque equation, remember torque τ = F r sin θ and r is the distance to the axis. In these cases, if you're writing the equation for this point, r will be the distance to this point.

We'll do this. I want to write here that sum of all torques about a point p equals 0. And I could write this equation for this point 1 here, or I can write this equation for point 2. So I can say sum of all torques about point 1 equals 0. And I can say sum of all torques about point 2 equals 0. You can do either one. Sometimes you have to write both equations. Sometimes you're going to get away with writing just one of them.

Now what's even more important is that we can actually write the sum of all torques about a point p equals 0 for any point in this bar, any point, even if the points are not the axis or support points. There are 2 points of interest here, 2 additional points of interest. There's this point here where little mg acts. And somewhere, let's say in the middle of the bar, there's a big Mg. So let's call these points 3 and 4. You could also write that the sum of all torques about point 3 equals 0 and the sum of all torques about point 4 equals 0, even though they are not supports. And you can go even one step further and you can pick a random point that has nothing going on there, Point 5, nothing happens at 0.5. And you can say the sum of all torques at 0.5 equals 0. You can use any of these equations to solve this problem. Any of them will work.

Now, since you can choose your reference axis, I'm calling this a reference axis because it's not really an axis. You're just picking it and treating it as an axis, while you're writing the equation. Since you can do this when you write your sum of all torques about a point p equals 0, okay, and you're going to write one or more of these, if you can pick, you want to pick the easier ones, the ones that are going to make your life simpler. It's going to make it easier to solve this question.

So, how do we know which ones are the easiest? Well, you use the fact that forces acting on an axis produce no torque. If a force acts on an axis, it produces no torque. So if one is my axis of rotation, this tension here will not produce a torque on point 1, right, which means when you write the equation, you're going to have fewer, fewer things on the equation. Alright? Fewer terms. So what you want to do is you want to pick points with the most forces on them. There's a force acting on point 1. So, that's a good point. There's a force acting on point 3. So that's a good choice. There's one force acting on point 4. That's a good choice. There's one force acting on point 2. That's a good choice. All of these points, 1, 2, 3, and 4 have exactly one force acting on them. So they're all equally good choices. 5 is a bad choice because there are no forces acting on it, so you can't cancel anything. So, none of these points are better than the other, except 5 is certainly the worst one. So you want to pick points where there's a lot of stuff going on. A lot of forces are acting there so that the equations you end up with will be simpler.

Let's do an example. So we have a board and this example is just describing the board up top. A board, 6 meters in length, 12 kilograms in mass. So what I want to do is I want to move this board down here. And I'm going to put this little m here. The board has a length of 6 meters and it has 12 kilograms in mass, mass equals 12, has uniform mass distribution, is held by 2 light ropes, one on its left edge. So I'm going to call this one, there's a tension one here. Let's make it a different color._invites you to get involved-answer questions, collaborate and explore additional resources beyond the classroom setting as we dive deep into these fascinating topics . There's a tension one here. And the other one is 1 meter away from the right edge. So this is t 2, and it is a distance 1 meter from the edge. It says that an 8 kilogram object is placed. So this mass here is 8 kilograms. Let's write little m equals 8, placed 1 meter from the left end. So this distance here is 1 meter. All right. And that's it. What else do we have? We have one more force. So we have little mg. We also have big Mg that acts right in the middle right here. Big Mg that's in the middle, that's the weight of the bar. So the weight of the bar happens in the middle, that means that this is 3 meters and this here is 3 meters as well. This extra distance, 1, this has to be a 2 so that this whole thing is a 3, and this has to be a 2 as well. So, I have 1, 2, 2, 1.

Four forces, bunch of distances. We want to calculate what is t 1 and what is t 2. First thing you can say is that the sum of all forces on the bar is 0. This is for the entire bar. So you can only write this equation once. This is for the y-axis because there are no forces in the x-axis. So this is just going to be t 1, t 1 positive plus t 2 positive plus mg negative and big ("Mg negative as well. All of this equals 0. If I move the negatives to the right side, I end up with t 1 plus t 2 equals mg plus big Mg. This should make sense because all this is saying right here is that all the forces going up equal all the forces going down. I have the masses. However, check it out. Once I know t one, I'll be able to find t two and vice versa. So as long as I can get one of them, I can get the other one from this equation.

Alright. Overstock, sum of all forces equals 0, only gets us as far as this equation, which for now is useless. So we're going to have to write that the sum of all torques equals 0. Okay? Previously, what we did is just did this, some of our torques about the support point but now we have multiple support points and now we know that we can go beyond support points and really just pick anything, okay?

So if you want to find T1 or T2, the best thing to do is to write this at points 1. Let's call this point 1. Let's call this points just in order here 2, 3, and 4. Okay? The best thing to do this is to use points 1 or 4. The reason being, if you write an equation for point 1, the sum of all torques at point 1 equals 0, T1 is not going to show up in that equation. So you're going to be able to solve for T2. And if you write the sum of all torques equals 0 for point T2 right here, then when you write the equation, T2 is going to be 0, you may be able to find T1.

But if you write in a separate point, then if you write, let's say, the sum of all torques about this point here is 0, you're going to have a T one and a T two. You can still solve it. It's just more work. Okay? So we're going to say the sum of all torques about point 1 right here is 0. So think of this as the axis of rotation. I'm actually going to redraw this. Here's the entire bar. I have T one here, but it's not going to give us a torque. I have mg here, which is at a distance. This is the r vector for mg. It's at a distance. How far is mg from the left? It's 1 meter. And then I have, a distance to big Mg which is 3 meters, and then I have the distance to T2. T2 is here. The entire bar is 6 meters. T2 is 1 meter from the edge. So this distance here is 5 meters. Okay. So this guy will produce a torque that's in this direction. That's torque of little mg. This is the torque of big Mg. They both are trying to spin this. They're pushing down. Right? Both of them are pushing down, so they're trying to spin this this way, which is clockwise, therefore negative. And T2 is trying to spin this this way. Let me put this somewhere else. T2 is trying to spin it away. Torque of T2 will be positive. They all have to cancel. So I'm going to write that torque of mg negative plus torque of big Mg, negative, plus torque of T2, positive, equals 0. And then I can send both of these guys to the right side, and then I get the torque T2 equals torque mg plus torque big Mg, which should make sense. Again, all I'm saying is that all the torques going this way cancel with all the torques going this way.

The next step is to expand this equation. So I'm going to write that this is T2, whatever the r vector is, sine of θ equals mg, whatever the r vector, the length of the r vector is, sine of the angle, plus big Mg r vector sine of θ. I drew all 3 r vectors. Notice that in all of these, the r vectors are horizontal, and the forces are vertical. So that means that all the angles will be 90 degrees, which is nice. This becomes a 1, this becomes a 1, this becomes a 1. Okay. So what's the distance between the axis we picked and the axis we picked was 1 right here? What's the distance between that and T2? So it's the entire well, most of the distance here which is 5, the distance to little mg is 1 and to big Mg is 3. Okay. 1 and 3. So getting these distances right is obviously a key part of these problems. So 5 T2 equals 1mg. The little m is 8. We're going to use gravity as 10. Just to simplify, the big M is 12 and gravity is 10, and there's a 3 here for the distance. So when I plug all of these together, I have 80 + 3.60. Right? That's 120 times 3 360 and this is going to be 440. So tension 2 will be 440 divided by 5. 440 divided by 5 is 88. Cool. So T2 is 88 newtons.

Now that I have T2, remember I told you that as soon as you know T2, you're going to be able to find T1. Okay? And that's what we're going to do here. So T1 + 88 equals mg plus Mg. Little m is 8, so 8 times 10. Big M is 12, so 12 times 10. Sorry, I said plus 10. It's times 10, obviously. So T1 is going to be 80 + 120 minus 88. So T1 will be 112 112 newtons. So here's my T1, 112. T2 is 88. Is this all we wanted? Yep. That's all we wanted. Notice that T1 is bigger than T2. This should make sense because this mass here is closer to T1. Imagine instead that you are holding this here while your friend was holding this here or making a mess here. You would have to have more force because this block is closer to you. Okay. So it should make sense, that the tension one is greater. The tension one should be greater. So you can use that to sort of reason whether your answers are likely to be correct. These are the correct answers. I got 88 and 112. That's it for this one. Hopefully, it made sense. Let me know if you have any questions and let's get going.