Okay in this problem were given two different scenarios where we have a weight hanging off of some chords and you're asked to calculate the tension and each one of the chords here. So let's draw our situation. We have, um, our first diagram here, and we have our weight hanging off. And we have, um, see, eh? And B, and the record is at a thirty degree angle. And ah, record here is at a forty five degree angle to the ceiling. So my mark points one and to here, and we're going to drive force body diagrams for each point. So the first body force body diagram for one is pretty simple. You just have cord. See the tension in that pointed upwards and you'd have, ah, the weight of our object pointed downwards. And our second force body diagram. We have a few different forces pointing in different directions. So we have a pointed off like so he directed upwards and now see is pointed directly below. And just so we have throwing our reference frame here. So I'm gonna break down and be here into the components of sea just s. So we stay within the um, X y reference frame that I have established. So, um, we see that sea we'Ll just stay the same. And if we break the beyond and the components on the vertical component of the attention do the court B will be be sign of forty five degrees and the horizontal component that is be co sign a forty five degrees, and we're going to do the same thing with cord A. We see the vertical component is going to be a time's the sign of thirty degrees, and the horizontal component is a times the CO Sino verity degrees. Okay, so we have a fourth body diagrams. Um, we have three unknowns and our tensions in the three different chords. And let's see. Ah, we're gonna have three equations were going to set up here. So from everybody but diagram, number one, we just have the tension to seem minus the weight sequel to zero. Um, assuming no acceleration. So we see that sea is just equal to W um, and our second part, let's do the horizontal components first. We have being times to co sign of forty five degrees minus and co. Sign of thirty degrees equals zero. And here, um, we have ah, bee sign of forty five degrees plus hey, sign of thirty degrees minus C is equal to zero. Um, so first things first I see econ substitute wn for sea here and move it over to the other side on DH. That will give me this equation here and then. Now I have a system of equation with two unknowns and two equations, Um, here and here. So I want to eliminate one of my, um, unknowns here and solve for the other. So we can use our knowledge of trig to say that co sign of forty five degrees is equal to sign and forty five degrees. So if we just subtract these two equations from each other, um, the first terms and each one will cancel each other out. So, um, that would leave us with negative, eh, co Sign of thirty degrees. Um, plus sign of thirty degrees is equal, Tio, um, negative W So those negative signs and either side cancel out. And in the end, you get that is equal to, um w over co sign of thirty degrees plus the sign, Ah, thirty degrees. And, um, when you plug in the numbers for coast on thirty and sign of thirty degrees you'd get that is equal to zero point seven three two Ah W Now let's move over here toe. Now solve for B S O. From this equation, we can get B in terms of a So it would just be a times the co sign of thirty degrees, long times the co sign of forty five degrees. And I mean probably inappropriate values for A and B. I get that This is equal Teo zero point, um eight, nine seven W So in the end, we saw further tensions and all the chords we found C um A in terms of W and, um be here. So with that, we are ready to move on to the next part. So let's look at scenario B. Here. You're given something that's hanging in the corner, the ceiling, and we have one chord coming down from the ceiling, one coming from the wall than at the intersection Point of two chords. We have one cord hanging down and again, we have some weight hanging off of this, and our angles have changed a little bit. Here we have a sixty degree angle at the bottom and a forty five degree angle on top. So just like before, I'm going to make a force that body diagram for two points we have here we have See, W again can't forget my reference frame here. Always a good thing. The label. So that's one. And then two, we're gonna have, um, another free body diagram where we have the forces pointing and all different direction. So we need to break down some of them into their components so that we can write out Newton's second law equations. So, um, when I break a down into the components, the vertical component will be a Times Co sign of sixty degrees here. And, um so go had in label. That and the horizontal component will be eight times the sign of sixty. And Virgil Cole Component be will be co sign twenty five degrees, and the vertical component will be equal to B times a co sign or sign of forty five degrees. But those are synonymous anyways, so just like last time on what I set up my noon second law equations. I've seen mice w equals zero. So, like in the previous part. We have just c is equal to w um Then we have, uh, church. Hey, Sign of forty five degrees minus a coz I know sixty minus c and I'm just gonna insert in w now, since we are, I know that equals zero. So be signed. Forty five degrees Linus, a co sign of sixty degrees is equal to W Then another equation we have be co sign of forty five degrees is ah, the positive part and their report of as a sign of sixty degrees and that would all be equal to zero. So again we see that we have two terms that equivalent since sign and co signed forty five degrees or equivalent. So if I subtract these from each other, we would get um eh and ah, I'm just going Tio, keep this coefficient positive for now And you'LL see I'll leave the negative that you get here inside or parents sies so isolating a Here we get that is equal to this expression And when plugging in the proper values for the sign and co sign of sixty degrees, I find that a is equal to two point seven three W and just like before, we can use one of our equations here The the cosign forty five degrees minus a sign of sixty degrees equals zero to put use. Ah, defined be in terms of a So we have He is a time sixty degrees co sign Ah, forty five degrees. And so this is going toe, um increase the value of a so that he is ah larger attention force and is equal to three point three. Bye w seven. We have now once again solved for the tension in the three cords kind of crossed out the answer there in this picture.