6. Intro to Forces (Dynamics)

Equilibrium in 2D

# 2D Equilibrium Problems

Patrick Ford

1299

15

2

Was this helpful?

Bookmarked

Hey, guys. So by now we've seen how to solve equilibrium problems, and we've seen how to solve two D forces problems. We're really just gonna put those things together in this video. And I want to show you how to solve equilibrium problems in two dimensions. This really just happens whenever you have forces that are acting in two dimensions. So 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. Uh, equilibrium just means that all the forces will cancel. So really, all two dimensional equilibrium means is that when you write out your f equals M A and your X and your y axis, you know that those forces have to cancel out to zero in the X and y axis. So you know your F equals M A is just gonna be zero in the X and zero in the UAE. Let me show you were working this problem out together so I can show you how it works. It's very straightforward. So we've got this 5 kg box and it's suspended by two cables. We want to calculate the tension forces of both cables. So we're just gonna stick to the steps here. So all this, like any other problem, we want to draw a free body diagram. So we've got our free body diagram like this. We're gonna start off with the weight force so we know our weight forces gonna pull straight down. This is our w equals negative 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's no applied forces. But we do have some tensions because we have cables or strings or ropes. Those are those are when you have tensions. Right? So we've got one that acts to the left like this, I'm gonna call this t one, and we have one that is going to be at some diagonal, and I'm gonna call this one t two. So those are attention forces, and we've got no surfaces in contact, so there's gonna be no normal or friction force. So those are only two forces there. Um, so now we're gonna move on to the second step, we just have to decompose any two 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 gonna have to draw this little horizontal like this. This is gonna be my T two X and then this is my T two y. 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 and these two angles are on opposite sides of the diagonal, 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 T two X is gonna be t two times the co sign of 37 my t two y is going to be T two times the sine of 37. All right, so I don't know what those are yet, because I still don't know what t two is, but basically what I'm trying to find as I'm trying to find the magnitude of these tension forces. So that's what I'm trying to calculate here, t one and t two. So what I've got to do now is I'm gonna go to the third step, which is writing, I think was Emma in the X and the Y axis. So here's my X axis. Here's my Y axis. So what I'm gonna do is we're gonna look at all the forces that are acting in the X axis and in the and then in the Y axis. So I'm gonna look here at all the ex forces, remember, this is my T one, and then my t two exits everything that lies on this axis right here, I actually know that this is equal to zero, and 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 gonna start with my positive forces, which is my T two X and this is going to equal t one. I actually got plus t one and that equals zero. So that means that my t two X is equal to negative t one and vice versa. Just the negative sign just means that these forces are gonna 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 t to one. That's my target variable. But I've got t two x here. But that's not what I'm trying to solve. I'm actually trying to solve for my t two. So what I have to do is I'm gonna have to replace this t two x with T two times the coastline of 37. So I got t two times the co sign of 37. This equals negative t once, and I got both my target variables. Unfortunately, though, what happens? 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 just gonna 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 gonna look at all the forces along this line right here. And so really, that's just my t t y and then my w So I know that my t two y and I also know the acceleration is equal to zero. Right? So my t two y plus my wr equal to zero, which just means that my t two y plus and I know this w forces actually negative 49 so is equal to zero here, so might teach you y is equal to 49. So now what happens is just like we did before. Remember, I'm not trying to solve for t to why I'm trying to sell for t two. So I've got to basically just expand out this equation or this term here. This is going to be my T two times the sine of 37. This is equal to 49. So that means I can actually solve for this, right? I just moved. Signed 37 hours at the bottom. So what I've got here is I'm just gonna move this up here. I've got my t two is equal to 49 divided by the sine of 37. If you go ahead and plug this in your calculator, where you're gonna get is you're gonna get 81.4. So this is 81.4 Newton's and that is our first number, right? T two is 81.4. So now we do is remember, we had one of the occasion that we got stuck with. We went to this access over here, went to the Y axis to solve for one of our equations. So now we are one of our variables so that we just plug this back into this equation over here or rather, this equation over here and then solve for t one. So what I've got here is I've got 81 4 times. The co sign of 37 is going to equal negative t one. If I move, the negative sign over to the other side would have got is t one is equal to negative Newtons. The negative sign just means just just like we said before that this voice force is going to point to the left in our diagram. So that takes care of that sign there. So we've got t two is equal to negative 65 Newtons. All right, If you were asked for just the magnitudes, then you would just write those both as positive numbers. All right, but that's it for this one. Guys, let me know if you have any questions.

Related Videos

Related Practice

Showing 1 of 3 videos