Equilibrium in 2D - Online Tutor, Practice Problems & Exam Prep

2D Equilibrium Problems

Hey guys. So by now, we've seen how to solve equilibrium problems, and we've seen how to solve 2D forces problems. We're really just going to put those things together in this video, and I want to show you how to solve equilibrium problems in 2 dimensions. This really just happens whenever you have forces that are acting in 2 dimensions with some angles, and you're also either told from the problem or you can figure out from the problem that this object is in equilibrium. So remember what equilibrium means. Just means that all the forces will cancel. So really, all 2-dimensional equilibrium means is that when you write out your f=ma and your x and your y axis, you know that those forces have to cancel out to 0 in the x and y axis. So you know your f=ma is just going to be 0 in the x and 0 in the y. Let me show you. We'll work this problem out together so I can show you how it works. It's very straightforward. So we've got this 5 kilogram box, and it's suspended by 2 cables. We want to calculate the tension forces of both cables. So we're just going to stick to the steps here, solve this like any other problem. We want to draw the free body diagram. So we've got our free body diagram like this, and we're going to start off with the weight force. So we know our weight force is going to pull straight down. This is our Wy=−mg, which is negative five times 9.8, and that is equal to negative 49. So that's our weight force. Now we don't have anything directly pushing or pulling this thing, so there are no applied forces. But we do have some tensions because we have cables or strings or ropes. Those are when you have tensions. Right? So we've got one that acts to the left like this. I'm going to call this T1. And we have one that is going to be at some diagonal, and I'm going to call this one T2. So those are tension forces. And we've got no surfaces in contact, so there's going to be no normal or friction force. So those are our only two forces there. So now we're going to move on to the 2nd step. We just have to decompose any 2-dimensional forces. Right? So we've got one that acts at some angle here and I want to decompose it into its x and y components. So I'm going to have to draw this little horizontal like this. This is going to be my T2x, and then this is my T2y. Now if I have the magnitude and angle, I can solve for that. So I need this angle with respect to the horizontal. Now unfortunately, with this problem, what I've got is I've got this angle which is 37 degrees which is kind of measured relative to the ceiling. What I actually need is I need this angle here. But hopefully, you guys realize that if you draw the horizontal like this, then these two angles are on opposite sides of the diagonal, and so they have to be the same. So this is also 37 degrees, and that's exactly the angle that I want. So it means that my T2x is going to be T2×cos(37), and then my T2y is going to be T2×sin(37). Alright? So I don't know what those are yet because I still don't know what T2 is. But basically, what I'm trying to find is I'm trying to find the magnitude of these tension forces. So that's what I'm trying to calculate here, T1 and T2. So what I've got to do now is I'm going to go to the 3rd step, which is writing F=ma in the x and the y axis. So here's my x axis. Here's my y axis. So what I've got to do is I'm going to look at all the forces that are acting in the x axis and in the y axis. So I'm going to look here at all the x forces. So remember, this is my T1 and then my T2x. It's everything that lies on this axis right here. And I actually know that this is equal to 0. I know that all the forces have to cancel out in both the x and y. So the sum of all forces is I'm going to start with my positive forces which is my T2x, and this is going to equal T1. Oops, I actually got plus T1, and that equals 0. So that means that my T2x is equal to negative T1 and vice versa. Just the negative sign just means that these forces are going to point in opposite directions. Right? So we know that these things point in opposite directions. But if you think about this, right, I've got my T2 one. That's my target variable. But I've got T2x here. But that's not what I'm trying to solve. I'm actually trying to solve for my T2. So what I have to do is I'm going to have to replace this T2x with T2×cos(37). So I've got T2×cos(37). This equals negative T1. So now I've got both my target variables. Unfortunately, though, what happens is I've actually got both of these unknown variables in this equation. And so whenever this happens, if you ever get stuck when you're solving for your x or y axis equations, then we just do what we've always done in these situations, which we're we're just going to go to the other axis. Whenever you get stuck, you just go to the y axis, right, or vice versa. So here, what I've got to do is look at all my y-axis forces. I'm going to look at all the forces along this line right here. And so really, that's just my T2y and then my W. So I know that my T2y, and I also know that the acceleration is equal to 0. Right? So that means my T2y plus my W are equal to 0, which just means that my T2y plus and I know this W force is actually -49. This is equal to 0 here so my T2y is equal to 49. So now what happens is just like we did before, remember, I'm not trying to solve for T2y. I'm trying to solve for T2. So I've got to basically just expand out this equation or this term here. This is going to be my T2×sin(37) is 49. So that means I can actually solve for this. I just moved the sin(37) nodes at the bottom. So what I've got here is I'm just going to move this up here. I've got my T2 is equal to 49 divided by the sin(37). If you go ahead and plug this in your calculator, what you're going to get is you're going to get 81.4. So this is 81.4 Newtons. And that is our first number. Right? T2 is 81.4. So now what we do is remember, we had one other equation that we got stuck with. We went to this axis over here, went to the y axis to solve for one of our equations. So now we just plug this back into this equation over here or rather this equation over here and then solve for T1. So what I've got here is I've got 81.4 times the cos(37) is going to equal negative T1. If I move the negative sign over to the other side, what I've got is T1 is equal to negative 65 Newtons. The negative sign just means, just like we said before, that this force is going to point to the left in our diagram. So it takes care of that sign there. So we've got T1 is equal to negative 65 Newtons. Alright? If you were asked for just the magnitudes, then you would just write those both as positive numbers. Alright. But that's it for this one, guys. Let me know if you have any questions.

A chandelier is supported from the ceiling by 2 chains. Both chains make a 30° angle with the vertical. The tension in each chain is the same because of symmetry:T₁=T₂=50 N. What is the mass of the chandelier?

• A. 2.6 kg
• B. 8.8 kg
• C. 5.1 kg
• D. 4.4 kg

A sphere hangs suspended by a light string, resting against a vertical wall. The sphere has a mass of 2 kg and the string makes an 80° angle with the horizontal. What is the force from the wall against the ball?

• A. 2.2 N
• B. 176 N
• C. 3.5 N
• D. 111 N