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.
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?
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Hey, guys, let's check out this problem here. So we've got this traffic signal that's suspended by these three cables. We've got them labeled 12 and three. We want to do this. Problem is, we want to figure out the tension that's in cable number one. So this is gonna be some tension t one here. And to do that, I'm gonna have to first draw the free body diagram. So let's get to the steps. Right. So this traffic signal, which I've got 8 kg over here, I'm gonna draw this free body diagram. So I've got the m g force that acts straight down. And what happens is that this is tension. This is cable number three. I'm not gonna have any applied forces or normals or friction causes hanging, but I'm only just going to have this one tension force, which is t three. Now, this is kind of tricky, because if I'm trying to solve for t one, how am I going to solve this by using this free body diagram if t one doesn't even show up. So what happens in this problems is sometimes it's it's really good to look at this junction point between all the three cables, and even though there's no mass that's actually there, we can still kind of draw a free body diagram for this junction points. So basically what this would look like you're gonna have this dot right here and then downwards. You're not gonna have awaits because there is no mass or anything like that. But downwards there is going to be attention for us. This is T three now. What happens is if this traffic signal is just suspended by these three cables. What this means is that this is actually an equilibrium problem. So this thing is suspended in place in the X and Y axis. So it's an equilibrium. And what that means is that t three is actually equal to mg, right? These forces have to cancel outs. So what happens is it's as if you actually have this traffic signal was actually hanging right here, right? It's kind of if you just moved it up to that junction points. All right, so the other two forces remember there's gonna be no normals or frictions or anything like that are really just gonna be the two tensions that were interested in uh, so we have t to like this, and then we're gonna have t one that's going to act like this. So this free body diagram has t one. This is going to be the right one to use, so let's move on. We have to decompose are two dimensional forces. So basically, I want to figure out what are the components of this t two in the X and y. So this is gonna be my t two x and my t two y. And to do that, I'm gonna need the angle relative to the X axis the horizontal. What I'm told here is that this angle is 60 and this angle is 22. So what I can do here is I actually can draw the horizontal like this. And if you think about this, these angles here are actually gonna be opposite. They're gonna be alternating interior or whatever. Basically, what that means is that if this is 60 degrees relative to the ceiling, that means this is 60 degrees relative to the horizontal, alright? And using the exact same logic, this is going to be 22 degrees. All right, so we're gonna we can split up t one. Now, this is really gonna be t one x and then t two x is going to point up like this. Sorry. T one Y. Yeah. So we've got our forces like that, right? So basically, what I can do is I actually don't know what the magnitude or direction. Sorry. I don't know what the magnitudes of t one and t two are. In fact, that's what I'm trying to find. So what I can do here is I can just say the t one X is gonna be t one times the co sine of the angle, which is 22 t one y is gonna be t one times the sine of 22. I can do the exact same thing for T two, right? So t two X is just gonna be t two times the co sign of and then t two y is going to be t two times the sine of 60. All right, so I'm just writing the expressions for them because I actually know what the angles are. So now if we want to figure out the magnitude of t one, we're going to have to actually get into our F equals m A. So let's go ahead and do that. So we have our f equals m a and the X and the Y axis. So all the forces in the X axis all the forces in the y axis. But remember, what we just said is that this object is an equilibrium, right? Suspended in place by all these three cables. That just means that the Emma is going to be zero in both the X and the y. All right? And so if you want to find t one, let's just start off with the X axis. It's going to be in both the X and y, so I'm gonna have to pick a direction of positive, right? So it's gonna be up into the rights because I'm dealing with these two dimensional forces here. So I've got my t two X, which allies in the direction of positive minus my T one X is equal to zero. So I'm just going to replace these. I know this is t two times the co sign of 60 minus t. One times the co sign of 22 is equal to zero. So if I want to solve this. I want to solve for this t one here. Unfortunately, I can't because I don't know what this T two is. So I've got these two unknown variables and that's okay because I can just go over to the y axis and sulfur there and the y axis. I have my t one and t two y that both point in the same direction and they're both positive. So I've got t u two y plus t one y and then I've basically got my mg that t three which I'm just gonna call MG because it's simpler as downwards and that equals zero. So just like we did before, I'm just going to replace these with the values that I know. This t two y is going to be t two times the sine of 60. And this is gonna be t one times the sine of 22. And when I move this mg over to the other side, it's actually gonna be equal to mg. Alright, so here I've got two unknowns again, my t one and t two. And so I've got a situation where I have two equations into unknowns. This is equation number one. This is Equation number two. Alright, so I've got a box them like this just so we don't get confused. I remember in the past what we've done is we either use equation, addition or equations substitution to solve these systems of equations. Equation Edition is always the easier one to do, but it's actually not going to work here. I'll show you why. If you write equation number one, you're gonna start off with T two times the coastline of 60 and then blah, blah blah, and you're gonna get like T one co sign of 22. Right? When you write an equation Number two, you're gonna get t two times the sine of 60 and then blah, blah, blah t one times the sine of 22. So what happens in these problems is we usually have to add these equations straight down, right? But we can't do that because you can't just add a T to co sign 60 and a T to sign of because they're not gonna cancel out. Right. So we usually added these so that we could cancel out one of our non target variables like the T two here, but there's no way to do that by adding it here. So equation Edition is actually not going to work. So what's, uh my equation addition here is not going to work. So what I have to do is I'm gonna have to do equation substitution. So what? I remember that equation substitution. You're basically just gonna come up with an equation for t two and you're gonna substitute or plug it into the more complicated expressions so you can get rid of that other variable here. So that's what we're gonna do. So basically T two times the co sign of 60 when you move this to the other side is going to be equal to t one times the coastline of 22. So t two is equal to t one co sign of 22 divided by co sign of 60. So now what I can do here is every time I see t two inside of this equation over here, I'm just going to replace it with this expression over here. So basically what this becomes is it becomes t one co signed 22 divided by co signed 60 times the sine of 60 plus. And then I got t one times the sine of 22. This is equal to mg. Notice how now we only have one unknown right R One unknown is just t one. And the rest of these signs and coastlines just become numbers. Right? So when you do coastline 20 to sign is 60 divided by coastline is 60. Really? What this becomes is 1.6 times t one. And then when you go when you plug in sign of 22 what you're gonna get is you're gonna get 0.3. So this is equal to is that this is equal to the mass, which is eight times. Oh, I'm sorry. T one here. Right? So the massive is eight times G, which is 9.8, right? So this is M G. All right, so we simplify everything. Remember, we can just actually add these things together, and actually, I forgot when you plug in signed 22 you're gonna get 0.37 t one. All right. So we can just add these two t ones together, right? 1.6 and 0.37 You can just add those together you get 1. 71 is equal to 78.4. So when you solve this, you're gonna get T one is equal to 78.4, divided by 1.97 you're gonna get 39 point eight. All right, So it's 39.8 Newtons for your tension. So you go through your answer choices and that's going to be answer choice. Be alright, guys. That's it for this one.
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?