in this particular video we have a uniform boom uh that weighs 700 ntons and it says there is an object hanging to the right end which is 400 newtons so the weight is given to us here and it says the boom is supported by a cable and it wants us to calculate all these uh different parts A B and C all right so let's go ahead and crack out part A part A wants us to find the magnitude of the tension in the cable so I'll show you how to do this a couple different ways i prefer the first method that I'm going to show you here cuz I think it's just easier to do it's bit more work but it's easier to do just to keep track of things and then the second approach is if your geometry is on point all right so uh the first strategy requires you to just keep your x and y axis normal here uh as traditional I guess you would say so here I'm just going to go ahead and identify my tension uh the tension the cable is pointing towards the wall there it's going to point in the uh north uh west uh direction so it's going to point left and up so uh because it points left and up it has to have components because this is a diagonal that lies in between the second quadrant here where my cursor is tracing back and forth all right so it has to go up and to the left so this is my TY okay oh I do want to point out here just in case uh you're not keeping up uh this angle here is the same as this angle because it's given as 45 degrees and this is perpendicular so the other angle has to be 45 degree as well okay uh so here's ty uh according to this particular um problem because the angle is 45 you can say t y is t s of 45 or t cosine of 45 it doesn't really matter because the angle here is 45° so s of 45 and cosine of 45 is the same however if you look at this t uh properly this is the opposing uh side or the 45° so I'm just going to say that it is sign for this particular problem okay all right and then here uh the uh TX value is going to lie adjacent uh to the angle given which is 45° so TX here is just t cosine of 45° all right other than that here's the weight of the uh what is it or just the weight that's hanging that's given us 400 newtons and then I'm also going to identify the weight of the boom which is right in the center there and that is given as 700 newtons so I'll label that as 700 uh you can call it F of G1 F of G2 if you want but I'm just going to go ahead and just put the values in here just for you to see okay other than that uh once you have all these uh force and force components uh labeled out uh you just have to find the component of the distance relative to the pivot that is perpendicular to the force that you need okay all right so watch my work so this is the length of the uh boom i'm just going to call that L and then because L is also a diagonal once again cuz I'm using my X and Y axis traditionally here just pointing left or left and right up and down uh but the length L here uh is a diagonal so it has to have an L X and an L Y here l Y is just L sine of 20 because the angle here is 20 and it is sign because this is the opposite leg and likewise with this so you notice here what I'm doing is I'm just making these into um right triangles okay so lx here would just be l cosine of 20° cosine because it is the adjacent leg to this uh 20° angle so once you have all that established uh you now can look at the force here of 700 newtons uh 700 Newton is in the middle of the boom so that is a length of L over two u well but you notice here L over2 is not perpendicular to 700 m or excuse me 700 newtons so L over2 also has an LX and an L Y here I'm just going to find LX just because L Y parallels um the 700 newtons which means it generates no torque so I actually need LX um L excuse me L over 2 X uh direction so that would just be L /2* the cosine of 20° here uh so keep that in mind okay all right so now that I have all my forces and leg drawn out all you have to do is set up net torque is equal to zero meaning the torque in the counterclockwise direction equals to the torque in the counterclockwise direction at least the magnitude all right so here is my statement so hopefully keep up here with all the little information here if you look at t is a diagonal because it is a diagonal um it will have okay components here ty is pointing in the y direction but ty uh is perpendicular to lx okay so once again here's the length L but you notice here if you watch my cursor in blue here L is a diagonal and then T Y is pointing straight up those two things are not perpendicular so I don't need L Y i actually need LX and LX being L cosine of 20 so that's why in my work I have TY which is the vertical component of the tension time LX which is L cosine of 20 and that is going to pull the boom in the counterclockwise direction now if you look at the component in the horizontal direction for TX here so TX here is pointing leftward and I need the perpendicular component of L which is in this case is L Y where my cursor is tracing up and down here okay so keep in mind as you do torque make sure you focus on the word perpendicular so TX is pointing left aka in the X direction so you need a leg in the Y direction in this case it would just be L Y l Y being L sine of 20° for this particular diagram that's why in my second term here I have L excuse me I have TX times LX but LX is L sine of 20 okay so that's it those are the only things that is going to generate a counterclockwise torque and then if you look at the weight of the um uh hanging mass here uh the weight is given as 400 and that's pointing down well if it's pointing down the y direction you need a leg that is in the horizontal direction so this is the contact force here so I'm going to emphasize all this for you in just a second so here's f of g which is 400 i need l x because that is perpendicular to the 400 okay hence in my work on the right side uh the 400 newton hanging mass will generate a clockwise torque and I want to times that by l x which is l cosine of 20 and then last but not least if you look at this 700 newton the weight of the boom itself is pointing down the y direction well I don't need l over two cuz l over2 is not perpendicular i need L /2 in the X direction that I've drawn here in orange so L /2 in the X direction is simply L /2 * cossine of 20° okay so that last term in my work here that 700 pointing down the Y direction l over two cosine of 20° is pointing in the X direction which are perpendicular to each other so I can apply that and I'm adding all these based on whether they generate a clockwise torque or a counterclockwise torque okay so I'm going to emphasize the arrows here for you so hopefully you're keeping up here it uh so here is TY if that is TY I need a leg that is in the horizontal direction there's L X and then here's TX well TX is pointing left well if that's TX I need the leg that is pointing in the Y direction which is in this case is L Y and then the 400 newtons is pointing down the Y direction as you can see the arrow grows there but that is a distance of L over uh excuse me L cosine of 20° away from the pivot and then last but not least the 700 Newton there is in the center of the boom and then therefore I need the leg that is perpendicular to it which is the L over2 cosine of 20° okay uh once you have all that established the the math is pretty straightforward it's the hard part is setting this stuff up and just again just focus on perpendicular stuff okay ah uh as you notice here every term has an L in it i'm going to cancel that out and then from here it's just calculator work um as you can see on the right side I just did 400* cosine of 20 that will give me about 375.88 and then the second term 700 * 1/2 * cosine of 20 that will give me about 328.89 and then on the left side for TY well in the diagram TY is simply t * s of 45 that's why that term is there and like likewise with tx uh I just substitute out tx for t cosine of 45 so watch my cursor here earlier I made that statement tx and ty are components of t so it's just s and cosine of the hypotenuse okay uh so as you can see here the only thing really unknown is just t because s of 45 and cosine of 20 is just some number and then likewise cosine of 45 and s of 20 is just some number so I'm just going to simplify that on my calculator and add the two terms together so if you simplify this uh s of 45* cosine of 20 that'll be about 66 uh 6645 and that has a t in it because t is the variable that we're asked to find and likewise on this second term here cosine of 45 * s of 20 is simply.2418 and that t remains in the term uh here we combine like terms and then just divide it to get t so if you combine the left two terms with the house has the t you'll get about 09063t to get t by itself divide both sides by 90063 and that will give us a magnitude of about 7773 okay so that's the answer of part a now the other strategy to do part A um that I don't mind but it's up to you how good your geometry is that is to tilt your axis so here in this particular diagram uh the axis is tilted so the uh x-axis lies on the boom uh perpendicular to the boom is your y-axis um so here's f of g for the boom at 700 uh because it now is a diagonal if your incline uh knowledge is solid uh this is this uses the same kind of geometry here uh the angle given here is 20° so if you draw f of g this angle between the uh 700 and the dash component that's also 20 so therefore that would just be 700 * cosine of 20 cuz it is the adjacent leg okay and likewise with this arrow here uh this arrow points uh towards the pivot uh and that is 700 * s because it is the opposite leg of the 20° and then you notice here I'm going to go ahead and take out what I don't need okay so why you can find the component i don't need it the only thing I really need is to keep the 700 cosine of 20° cuz it is perpendicular um to the L over2 okay uh and then likewise here with the seven excuse me with the 400 uh Newton's hanging weight so here's the 400 uh but once again it lies in between the X and Y axis so the weight itself now has components that is the 400 * cosine of 20 uh again because the angle between the boom and the horizontal axis 20 uh we say the angle uh here between the y-axis and the hanging block is also the same uh due to geometry okay all right other than that here is the 400 * s of 20 but I want to go ahead and take out the 400 sign of 20 because I need stuff that's perpendicular uh this component here is parallel to the boom which generates no torque so I'm going to go ahead and take it out of the diagram just to help you kind of visualize this stuff in a bit okay i'm also going to take the 400 newtons because that's not perpendicular as well so the only thing I really need to keep is the 400 * cosine of uh 20 here because it is perpendicular to the length L okay because it lies on the y- axis l now lies on the x-axis so those two things are perpendicular and then last but not least let's go and do t uh so t here you notice again it's not in the x axis or y axis it is a diagonal but I can break that up into ty and tx so here's TY but before you do TY uh hopefully your geometry is solid uh watch my cursor uh this angle that I'm tracing back and forth here between the horizontal dash uh axis and the length of the boom is 20° so therefore this angle is the same so it's touching the boom and it's also touching the horizontal axis so if this is 20° this has to be 20° well the angle between the x axis and the y axis 90° but if you add it up 20° plus 45 degrees that already makes 65 degrees so the other angle that is left over where I'm going back and forth here is 25 degrees okay so if you need to listen to that listen to that again um therefore I can say t y in this case is the adjacent leg of this 25° that's left over so because it is the adjacent leg I'm wanting to say as t * cosine of 25 and then the other leg would just be t time s of 25 because it is now the opposite uh leg of this 25 degrees reference point all right however I want to delete stuff that's not perpendicular okay i'm only going to keep stuff that's perpendicular excuse me so here's TY and TY is perpendicular to L because L now lies on the X axis so once I once you have all this uh it's it's pretty straightforward okay once again just keep in mind take the force that's perpendicular to the uh moment of arm that you're interested in so here same statement uh torque in the counterclockwise equals torque in the clockwise direction uh so here it'll be ty * the length L because that is perpendicular that's 90° and that's my first term on the left side and that would generate a uh counterclockwise torque however if you visualize here the 400 cosine of 20° right here where I'm tracing back and forth right okay that is down in the negative y direction uh well how far is it away from the pivot it is also a length L away so on the right side uh I have 400 cosine of 20 * L because that will generate a clockwise torque and likewise with the 700 * cosine of 20 the 700 * cosine of 20 is placed at the middle uh of the boom which is a distance of L /2 but L /2 right now is already on the X axis and then 700 * cosine of 20 is on the Y axis which is perpendicular that's why the second term on the right side that generates a clockwise torque is also 700 * cosine of 20 * L over2 as you can see L is in every term so I want to cross that out i'm going to go ahead and expand TY though again because part A asks us to find T uh not the components so I'm just going to break it up uh TY is simply T cosine of 25° because that is referencing the y-axis here and the 25° earlier that I showed you how to do uh just using subtraction okay on the right side if you simplify the two terms and add them up you'll get about 704.77 and as you can see here the only term that's unknown is t you just divide both sides by cosine of 25 to get t and if you do that correctly on your calculator you should get 777.63 newtons as you can see the magnitude is the same all right so that's that uh hopefully pick a strategy that you like okay it can be a little confusing but as long as you just keep in mind perpendicular stuff your force has to be perpendicular to the moment of arm that you're interested in all right let's do part B uh part B asks us to calculate the force on the hinge on the boom and in part C in this case it's kind of related it does it just does the force uh on the hinge act along the boom so we can just justify that really quick here all right so to calculate the force on the hinge uh on the hinge on the boom uh let's go ahead and set up Newton's second law uh well nothing's moving so therefore forces in all direction must equal so net force here in the x direction has to equal to zero well the one thing you do have to identify is what are the forces in the x direction so if you look at the diagram here uh tx is the only force um in the x direction right now based on what I've drawn out well for net force equal to zero you need a minimum of at least two forces okay at least it could be 2 3 4 5 whatever however in this case you have TX pointing left which indicates that you know there is some sort of force here acting on the hinge um pointing right because you need a force left and a force right in order for things to equal to zero all right so here I'm just going to call this F hh hingch in the x direction and that's it uh I'm just going to say that TX minus fing equal to zero or literally you can just say TX is equal to F hing X if you want um so or you can switch this even you can say f hint x minus tx equal to zero it doesn't really matter cuz however you make the statement since there are only two forces here the magnitudes will be the same okay so here I'm just going to solve for tx real quick and say tx is equal to f hingx cuz this is negative uh f hh hinch i'm just going to move to the right side and as you can see uh tx here is just t cosine of 45 but we already found what t is from part a so t was 777.73 newtons times cosine of 45 and that's going to give us the uh force on the hinge in the x direction okay but we're not done yet okay well again because the system's at rest the net force in the y direction also has to equal to zero so uh let's go and crack this out so I'm going to make the same general statement net force in the y direction equal to zero however it's important that you identify all the forces okay so here um if you think about it uh the weight of the boom itself is 700 uh the weight of the hanging uh block here is 400 so together they already add up to 1,100 newtons but ty here uh I already know is only 549.94 newtons cuz the angle is 45 so tx is the same as ty the magnitude anyways right so without calculating it I know that TY is TX because the angle is 45 well you can't have 1,100 Ntons pulling down and only have 54994 Newtons pointing upward that that doesn't equal to zero because that doesn't equal to zero it suggests that there is a Y component uh acting on the hinge as well okay so I'm just going to call this F hinge Y it's pointing up because you need arrows up to equals arrow down you have two vectors down here that's waiting uh together as 1100 newtons but you only have 549 uh 94 pointing up therefore you need another force pointing up okay so here I'm just going to go ahead and just solve for it again if you listen me to explain you know that TY is equal to T TX tx in this one because the angle is 45 um so here I'm just going to make this general statement uh TY is pointing up i want to add that to the Fhinch Y because that's also pointing up i want to subtract it from everything that's pointing down i'm going to set that equal to zero and here I'm just going to go ahead and move the -400 and 700 to the right side just to sum it up which is essentially what I said verbally uh forces pointing up must equal the forces pointing down uh so here if we just isolate for fing in the y direction uh 400 plus 700 is 1100 and then this is positive 549.94 so move to the side you end up just subtracting so here we find that fh hinge in the y direction is about 550.06 okay well now that we have the horizontal and the vertical component we can just really just do Pythagorean theorem here so I'm just going to say Fhinch oops I forgot an E there in the hinge um but just ignore that because that's just subscripts um so Finch X already found which is 549.94 when I square that because we're doing Pythagoran theorem I just found Finch in the Y direction that's 550.06 square that and then uh to find Fhinch square root the sum of these values on the right and if you do that correctly you should get a value of about 777.82 newtons but you're not done yet because force is a vector and I didn't use the word magnitude you have to find the direction so here if you uh watch the uh left information here so this is what I just found not drawn a scale right not drawn a scale at all um I just found the diagonal the hypotenuse of these two forces by doing pagan theorem well force is a vector so I need to give it a direction uh this the direction again unless otherwise stated I'm just going to reference the horizontal axis here and that theta is super easy to find all you have to do is just do tangent okay or inverse tan excuse me so to find this uh angle that's missing I can just do net force in the y direction over net force in the x direction or in this case it'll just be fing in the y direction over fing in the x direction cuz tangent is rise over run or y overx and to solve for theta I just need to do inverse of those two values that I found fhinch y here and then fringe in the x uh other than that if you plug that in uh correctly on your calculator uh so if you do tan inverse of these two quantities you'll get 45° so the angle here is 45° so the proper way to say the answer for part B would be the magnitude is 777.82 Newtons 45° above the positive x-axis or 45° uh northeast uh in terms of direction okay so that's that so I'm just going to highlight the magnitude here all right other than that um uh let's do part C part C said or asked us does the force on the hinge act along the boom well the answer is no the boom makes an angle of 20 degrees relative to the x- axis as given in the diagram and we just found that the force acting on the hinge is 45 degrees above the horizontal axis so the answer for part C is uh definitely no all right I hope that helps stay nerdy don't be whack