Concept: Equilibrium with Multiple Supports15m
Hey guys! 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. When an object in equilibrium has multiple supports, such as here you got two ropes, we can think of each support as a potential axis of rotation. You can think of this point as being an axis and then this point as being an 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. Same thing is if you cut the rope here, the whole bar is going to spin this way around this point. That's one way to sort of visualize that. Therefore, we can write that the sum of all torques equal zero for either one of those two points, for any support point. In doing this, you're effectively treating it as the axis even if it's not the axis. 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 = Frsin_ and r is the distance to 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 zero and I could write this equation for this point one here or I can write this equation for point two. I can say sum of all torques about point one equals zero and I can say sum of all torques about point two equals zero. 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. What's even more important is that we can actually write sum of all torques about a point p equals zero for any point in this bar, any point, even if the points are not the axis or support points. Let me show you. There's two points of interest here, two additional points of interest. There's this point here where little mg acts and somewhere let's say in the middle the bar, there's a big Mg. Let's call these points three and four. You could also write at the sum of all torques about point three equals and the sum of all torques about point four equals zero even though they are not supports. You can go even one step further and you can pick a random point that has nothing going on there. Point five, nothing happens at point five and you can say the sum of all torques at point five equals zero. You can use any of these equations to solve this problem. Any of them will work. Since you can choose your reference axis, I'm calling this 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 zero 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 and it's going to make easier to solve this question. How do we know which ones are the easiest? You use the fact that forces acting on an axis produce no torque. If a force acts on an axis, it produces no torque. If one is my axis of rotation, this tension here will not produce a torque on point one which means when you write the equation, youÕre gonna have fewer things on the equation, fewer terms. What you want to do is you want to pick points with the most forces on it. There's a force acting on point one, so that's a good point. There's a force acting on point three so that's a good choice. There's one force acting point four, that's a good choice. There's one force acting on point two, there's a good choice. All of these points Ð one, two, three and four Ð have exactly one force acting on them so they're all equally good choices. Five is a bad choice because there are no forces acting on it so you can't cancel anything. None of these points are better than the other except five is for sure the worst one. 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. We have a board and this example is just describing the board up top. A board 6 m in length, 12 kg in mass. 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 = 12), has uniform mass distribution is held by two light ropes one on its left edge so there's a tension one here and the other one is one meter away from the right edge, so this is T2 and it is a distance 1 meter from the edge. It says that an 8 kg object is placed so this mass here is 8 kilograms. Let's write little m = 8, placed 1 meter from the left end. This distance here is 1 meter 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 act right in the middle right here, big Mg that happens in the middle. That's the weight of the bar. 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. This has to be a 2 as well. I have 1, 2, 2, 1. Four forces, bunch of distances. We want to calculate what is T1 and what is T2. 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 are no forces in the x-axis. This is just going to be T1 positive + T2 positive + mg negative + big Mg negative as well. All of this equals zero. If I move the negatives to the right side, I end up with T1 + T2 = mg + 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, I have g, but I don't have T1 or T2 therefore this equation is not enough. However, check it out, once I know T1, IÕll be able to find T2 and vice versa. As long as I can get one of them, I can get the other one from this equation. Try to remember that because we're going to need that a little later. But weÕre stuck. Sum of all forces equals zero only gets as far as this equation which for now, it's useless. We're going to have to write that the sum of all torques equals zero. Previously, what we did is we just did this sum of all 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. If you want to find T1 or T2, the best thing to do is to write this at pointsÉ Let's call this point one. Let's call this point just in order here; two, three and four. The best thing to do this is to use points one or four. The reason being, if you write a equation for point one, the sum of all torques at point one equals zero, T1 is not going to show up in that equation so you're going to be able to solve for T2. If you write the sum of all torques equals zero for point T2 right here, then when you write the equation, T2 is going to be zero and you may be able to find T1. But if you write in a separate point, then if you right let's say the sum of all torques about this point here is zero, you're going to have a T1 and T2. You can still solve it, it's just more work. We're going to say the sum of all torques about point one right here is zero. Think of this as the axis of rotation. I'm actually going to redraw this. Here's the entire bar. I have T1 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. T2 is 1 meter from the edge so this distance here is 5 meters. 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. Both of them are pushing down so they're trying to spin this this way which is clockwise therefore negative. T2 is trying to spin it this way. Torque of T2 will be positive. They all have to cancel so IÕm going to write that torque of mg negative + torque of big Mg negative + torque of T2 positive = zero and then I can send both of these guys to the right side. Then I get the torque T2 = torque mg + 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. IÕm going to write that this is T2, whatever the r vector is, sin_ = mg, whatever the length of the r vector is, sine of the angle + big Mg r vector sin_. I drew all three our 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. What's the distance between the axis we picked, and the axis we picked was 1. What's the distance between that and T2? Most of the distance here which is a 5. The distance to little mg is 1 and to big Mg is 3. Getting these distances right is obviously key part of these problems. 5T2 = 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. There's a 3 here for the distance. When I plug all of this together, I have 80 + 360. This is going to be 440. Tension 2 will be 440 / 5 is 88. T2 is 88N. 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 and that's what we're going to do here. T1 + 88 = mg + Mg. Little m is 8, so 8(10) + big M is 12, so 12(10). T1 is going to be 80 + 120 Ð 88. T1 will be 112 N. HereÕs my T1, 112. T1 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. You would have to have more force because this block is closer to you. 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 keep going.
Problem: A board 8 m in length, 20 kg in mass, and of uniform mass distribution, is supported by two scales placed underneath it. The left scale is placed 2 m from the left end of the board, and the right scale is placed on the board’s right end. A small object 10 kg in mass is placed on the left end of the board. Calculate the reading on the left scale.
BONUS: Calculate the reading on the right scale.16m
A mechanical lift is used to raise a heavy load. The lift consists of a horizontal platform of length L drawn by vertical cables on either end of the platform. The platform has a mass m. The load has mass 4m and is placed a distance L/4 from the right end of the lift. The acceleration due to gravity is g. What is the tension in the right vertical cable?
1. (2/3) mg
2. (2/7) mg
3. (7/2) mg
4. (3/2) mg
5. (7/4) mg
6. (4/5) mg
8. (1/3) mg
9. (5/4) mg