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Equilibrium with Multiple Supports

Patrick Ford
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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 gonna look into problems with multiple support points, and I'm gonna go over 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. All right, So when an object in equilibrium has multiple support, such as here got two ropes, we can think of each support as a potential access of rotation. We're gonna just right axis there so you can think of this point is being axis and in this point is being axis. One way to sort of visualize this is imagine that if you cut this rope here, the the bar is going to spin around this point where the rope touches. Same thing is, if you cut the rope here, um, 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 zero 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. They stand there. There is essentially no axis of rotation because it's not rotating, but you still treated as an axis, meaning when you write your torque equation, remember, torque equals F R sine of theta and R is the distance to access in these cases, if you're writing the equation for this, points are will be the distance to this point. We'll do this. So I wanna right here that some of all talks about a point P equals zero. And I could write this equation for this 0.1 here. Or I could write this equation 4.2 So I can say some of all talks about 0.1 equals zero, and I can say some of all talks about 00.2 equals zero. Cool. You could do either one. Um, sometimes you have to write both equations. Sometimes you're gonna get away with writing just one of them. Okay, Now, what's even more important is that we can actually write some of all talks about a point. P equals zero for any point in this bar. Any point, even if this point, um, even if the points are not the access or support points. Okay, so let me show you. Um, there's two points of interest here to additional points of interest. Um, there's this point here where little mg acts and somewhere, let's say, in the middle of the bar, there's a big M G. So let's call these points three and four. So you could also write that the sum of all talks about 40.3 equals zero in the sum of all talks about 00.4 equals zero even though they are not, um, supports. And you could go even one step further and you can pick a random point that has nothing going on there, 0.5. Nothing happens at five, and you can say the sum of all tricks that 0.5 equals zero. You can use any of these equations to solve this problem. Okay, any of them will work now, since you can choose your reference axis. I'm calling this reference access because it's not really an axis. You're just picking it and treating as an axis while you're writing the equation. So since you can do this when you're right, you're some of all talks about a point P equals zero. Okay. And you're gonna write one or more of these. If you can pick, you wanna pick the easier ones, the ones that are gonna make your life simpler. It's gonna make easier to solve this question. Um, eso how do you know which ones are the easiest? Well, you use the fact that forces acting on on an axis produce no torque if a force acts on an axis produces no torque. So if one is my axis of rotation, this tension here will not produce a torque on 10.1. Right? Which means when you write the equation, you're gonna have fewer fewer things on the equation. All right, fewer terms. So what you wanna do is you wanna pick points with the most forces on it. There's a force acting on 0.1. So that's a good point. There is a force acting on 0.0.3, so that's a good choice. There's one force acting on 10.4. That's a good choice. There's one force acting on point to there's a good choice. All of these points 123 and four have exactly one force acting on them. So they're all equally good choices. Five is a bad choice because there no forces acting on it. So you can't cancel anything. So none of these points are better than the other. Except five is for sure, the worst one. So you wanna 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 aboard 6 m in length, 12 kg in mass. So what I wanna do is I wanna move, um, this board down here and I'm gonna put this little M here. All right, So the board has a length of 6 m and it has 12 kg in mass. Mass equals 12 has uniformed mass distribution is held by two light ropes one and it's left edge. So I'm gonna call this one. There's a tension one here. Let's make it a different color. There's attention one here and The other one is 1 m away from the right edge. So this is t two. And it is a distance 1 m from the edge. Um, it says that a Nate kilogram object is placed. So this mass here is 8 kg. That's right. Little M equals eight placed 1 m from the left end. So this distance here is 1 m. 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 MGI that acts right in the middle, right here. Big MGI that happen in the middle. That's the mass of the, um the weight of the bar. So the way to the bar happens in the middle. That means that this is 3 m, and this here is 3 m is well, so this extra distance one, this has to be a two so that this whole thing is a three and this has to be a tube as well. So I have 12 to 1. All right. Four forces bunch of distances. We want to calculate what is t one and what is t two. First thing you can say is that the sum of all forces on the bar is zero. This is for the entire bar. So you can only write this equation once this is for the y axis because there's no forces in the X axis. Um s Oh, this is just gonna be t one, um t one positive plus t to positive plus mg negative and big mg negative as well. All of this equals zero. If I move the negatives to the right side, I end up with t one plus t two equals m g plus big M G. Um, 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. I have G, but I don't have t one or t two. Therefore, this equation is not enough. 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 could get one of them, I could get the other one from this equation. Okay, so try to remember that, because we're gonna need that a little later. Alright, But we're stuck. Some of our forces equals zero on Lee gets a Sfar is this equation which for now is useless. So we're gonna have to write that the sum of all torques equals zero. Okay, Previously, what we did is we just did this some of our talks 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 wanna find t one or t two, the best thing to do is to write this at points one. Let's call this 0.1. Let's call this point just in order here, too. Three and four. The best thing to do this is to use points one or four. The reason being if you write an equation 4.1, the sum of all topics that 0.1 equals zero t one is not going to show up in that equation. So you're gonna be able to solve for T two. And if you write the sum of all topics equals 04 point t two right here, then when you write the equation, t two is gonna be zero. You may be able to find t one. Okay, but if you write in the separate point, then if you're right, let's say the sum of all talks about this point here. Zero you're gonna have a t one and t two. You can still solve it. It's just more work. Okay, so we're going to say the sum of all talks about 0.1 right here is zero. So think of this is 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 M g here, which is at a distance. This is the our vector for M G. It's at a distance. How far is is N. G from the left. It's 1 m and then I have a distance. Two big M G, which is 3 m and then I have the distance to t two teachers here. The entire bar is three entire bar six t two is 1 m from the edge. So this distance here is a is 5 m. Okay, so this guy will produce a torque that's in this direction. That's talk of little M G. This is the torque of big MG. They both are trying to spend this. They're pushing down, right? Both of them are pushing down. So they're trying to spend this this way, which is clockwise. Therefore, negative and T two is trying to spin this this way. Let me put this somewhere else. T two is trying to spend it this way. Torque of t two will be positive. They all have to cancel. So I'm gonna write that torque of M G negative plus torque of big mg negative plus torque of t two positive equals zero, and then I can send both of these guys to the right side and then I get the torque. T two equals torque M g 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 gonna write that this is t to whatever the our vector is. A sign of data equals M g. Whatever the our vector, the length of your vector is sign of the angle. Plus big M g, our vector sign of data I drew. All three are vectors. Notice that in all of these, the are 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 one. Um, this becomes a one that's becomes a one. Okay, so what's the distance between the axis we picked and the access we picked was one right here. What's the distance between that anti too? So it's the entire, well, most of the distance here, which is a five. The distance to little M. G is one and two big M G is three. Okay, one and three. So getting these distances right is obviously key part of these problems, So five t two equals one M g. The little M is eight. We're gonna use gravity is 10 just to simplify. The Big M is 12 and gravity is 10 and there's a three here for the distance. So when I plug all of this together, I have 80 plus 3 60 right, That's 1 20 times 33 60. And this is gonna be 4 40. So attention to will be 4 40 divided by five. Um 4 40 divided by five is quote. So t two is 88 Newtons. Now that I have to to remember, I told you that as soon as you know T two, you're gonna be able to find t one. Okay? And that's what we're gonna do here. So t one plus 88 equals m g plus M g little m is eight. So eight plus 10 Big M is 12. So 12 times 10 Sorry. I said plus 10. It's times 10 obviously. Um, S O T one is going to be, um 80 plus 1 20 minus 88. So t one will be 1 12 12 Newton's. So here's my t 11 12 t two is eyes This all wanted? Yep. That's all we wanted. Notice that t one is bigger than t to. This should make sense because this mass here is closer to t one, right. Imagine instead that you are holding this here while your friend was holding this here. I'm making a mess here. Um, you would have to do You have to have more force because this block is closer to you. Okay, so it should make sense that the tension one is greater detention. 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 in 1 12. That's it for this one. Hopefully makes sense. Let me know if you have any questions and let's get going.