### Video instructions and help with filling out and completing Form 2220 Respectively

Instructions and Help about Form 2220 Respectively

Alright here we have a problem the hanging mass from some rope and two separate angles and we're told that we have a maximum tension in the rope and we want to find out what the maximum weight of the any mass would be before they would break the rope so we need to before we do this problem here we need to kind of think about what we're going to need to find in order to solve it so if we're looking we have a tension max and we need to understand what we're looking for as far as component so we know we're going to need attention one which would be right here this one would be tension one and this would be tension two and this is my weight so we'll need tension one tension to and to wait and that's pretty much it for that and we're gonna need angle 1 and angle 2 you don't want to just simply write these angles down and to start using them because you might make a mistake you want to get this in a free body diagram properly and we know the formula for this is just mass times acceleration so we know that we have the sum of the forces is equal to mass times acceleration and since the system is not moving it's staying still the acceleration is zero so that equals zero or we can say f1 plus f2 is equal to zero okay so we know that with we've we've analyzed all this we found that now we need to find out what our angles are because we don't know attention one is we don't know attention to is we got to find the weights we know that gravity is involved but for this problem we're not going to need it so I'm not gonna write it down so I need to do I need to find my components for this problem first of all so let me do that let me consider this here I have my weight going straight down and since it's going straight down I can say it's going in the negative y-direction and when that's the case we know that the x axis is perpendicular Euler to the y-direction so I can draw my x coordinate system here and I can have my positive y axis there so I'll have negative Y in my X here so let me just put an X and a y here to denote the direction and then I can recognize that spaced on trigonometry here then if I drop this down and make a right angle that this angle here is going to be 60 same as this angle and likewise over here I drop this down in this angle here is 40 same as that so now I've got two angles with respect to my x-axis and that's what I'm going for here I'm now I can extract this and make a Freebody diagram so let me do that right here just the little one I don't need nothing fancy and I can draw my vector in color I'll draw my vector here for tension one straight that way and that will be tension one and I've got a 60 degree angle here but I'm gonna wait to write that in because I'm not going to use 60 degrees and then I've got this tension here attention to and that's with the x-axis so that's my 40 degrees so I can write that so now here I can use 60 degrees but since it's in the negative direction if I use 60 degrees I got to remember to leave a negative sign I say I don't like to think about that and remembering the negative sign I like to use as little steps as possible so since we usually use the angle with respect to the x axis I'm gonna go ahead and subtract that 60 from 180 and use 120 degrees for my angle instead that way I don't have to worry about messing up my signs or anything and okay so now now we look at that and now we know that we have we have a forces going in the x-direction when we break up the components we're going to have a force going this way and a force going this way and then we have our weight going downward and then we're going to have our force in the Y direction for attention so we're gonna have our attention in the Y direction here for one and two and then we're going to have our tension in the x-direction for one and our tension in the x-direction or two right there and so these two will subtract out these two are going to add together within the equal the weight so the tension and the y-direction is going to be the same as the weights and so forth so now with that we understand all that and we've got our angles we can begin to solve the problem here so this is 120 and this is 40 and we know that tension 1 is going to be equal to well we were looking for that but we need that we need tension we need the tension in the X in the Y direction so let's look at this here and we'll remember that our sums of forces in the x-direction are going to be equal to tension 1 in the X Direction plus tension 2 in the X direction which is equal to 0 and that's going to be the same for y force in the Y Direction is equal to tension line in the Y Direction plus tension 2 in the Y direction which is equal to 0 and we know that this is nothing more than tension 1 cosine and tension 1 sine of the angle that we have here so I'll get my calculator.