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Traffic Signal

Patrick Ford
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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.