Transcript for:
Statics Review for FE Exam

hey well thanks for joining in today and joining us for the second fe real review session here and uh it's so glad to have you joining us for after you review session 2022 and this session is static so if you want links to the rest of the playlist you can look down below where each week this semester we'll be going through different series or different topics so today we're doing statics last week we did math we're sort of jumping over a little bit here like engineering economics and ethics with the idea that we'll come back to those later but statics is so foundational here that i want to get this in and i kind of want to go through some of the pieces of it to make sure that this is like this is there for the rest of your review right so as you get in we started with math we kind of went through the whole idea of how to use your calculator and i brought that up a bunch last time we'll deal with some calculator things today not quite as much today what we're really focusing on is is like three major things and one here is uh where'd my pen go there it is one is just going to be sum of forces in the x direction equals zero if you ever took a statics course with me i mean this is like uh sum of the four is the x direction sum of the forces in the y direction equals zero so you have to know how to sum uh forces right so you have to know you know how to do this and this isn't just like vector mechanics you might have taken a class with vector mechanics and you learned ijks and how to do all that this is just kind of the bare bones this is what i used when i was practicing as a consulting engineer for over a decade this is the basic stuff of how to make things balance so they work all right and then in addition we have to know summon moment so we're going to deal with some moments about a point like over and over again today and that has to equal zero so these this is just like kind of balancing right this is the balancing act like all the forces need to balance all the rotations need to balance and that's what we're trying to do in statics you might be saying okay that that's good mark um and honestly i i went through and and tried to explain this to my wife at one point and she's like mark that looks like greek i was like yeah you're right this is kind of greek here but that's okay you guys can still learn this uh you can still do this we use sigma uh for for what's going on here but but these basic equations right where a moment equals a force times a perpendicular distance right if you can do those those things and throw in like a little bit of like like like some basic trig here like for example the cosine of an angle uh equals the adjacent over the hypotenuse right or you can do the sine of an angle equals the opposite uh over the hypotenuse if you can get those things down you can master this static section i i 100 percent believe it and i i i know that this is going to help you master and pass and crush the fee when you get there so to do that and the way that all these review sessions work is i kind of just start with the fe spec so i just come over here and i start and say okay this is what ncs puts out there i'm just gonna take it and look at it and look at these different pieces here and say let's come up with some problems for it all right so that's what we're going to do we're going to work through each of these sections resulting in four systems i've also talked to some people that have recently taken the test and those people have given me some feedback on hey i saw this i didn't see that i saw this idea that you know i'm not taking exact questions but like i get their feedback on what was on the test to try to bring that in and you'll see that a little bit today as we go through as well so you know let's get started i mean this is fun so this is great so we're just going to jump in here this is the first you know first question and honestly this is where i start my first statics class where i where i start my first statics class is just this basic idea this is just the basic idea of forces as vectors right and when we start to look at these forces i put these down just two basic forces and the question asks okay i've got a force this force one right and that's 220 kilonewtons so this is just you know 220 kilonewtons uh we've got force two and force two here is 240 kilonewtons okay so we've got these forces and there's going to be some third force right these forces if you add them together are going to make some third force here and this third force is going to come up here i i don't have the whole you know i can't show it to to scale here just because the way the problem's set up but they're going to create some third force and that's what we're asking for we're asking for that resultant force the magnitude of the result and really one of the things that you want to do is you want to be able to sort of read these questions quickly you get what three minutes per question so you want to be able to sort of look at the question maybe circle or highlight the important things and then just parse out what they're asking for so when these forces are added together the magnitude of the resultant is most nearly right so what do we have to do we just need to use some basic trig here and the way that i like to do that is i like to show uh force components so i like to show force components like this so i you know i might call this one f one y and i have another component here that's that would be called f you know one x and then similarly on this side i'm gonna have some components here so f two y and f to x so these components when you add them together is this this works so ibrahim um moments we'll talk we'll talk moments in a little bit so we will get there i just saw that chat box you know so feel free to put things in the chat um it's just me here so i just do this just for fun kind of and for my class here uh where i'm teaching this stuff so you know so so if i don't see your chat right away just let me know just keep you know keep pinging me so um yeah so so moments we'll get there we'll get there don't worry but moment's just a force times the distance the whole idea of statics let me just take a step back for a second the whole idea with statics is we're trying to balance things we're trying to balance forces vertically we're trying to balance forces horizontally and we're trying to balance them so they don't rotate right because that's what static is it's not moving right so we just want those to balance and once they balance we're happy right in here one of the things is we're not looking to make these balance i mean we could say okay maybe we have a resultant force that's acting down but we're not even worried about that we just we're adding these two vectors together and when they add together they're going to create some third vector right so i mean if we were to look at this as well we could say like okay well we have f1 adding here and we have f2 adding here and we're going to come up with some third vector right so if we if we did like vector notation uh we could do that right but it's gonna be easier uh for you on the test i think is is to use just basic trig okay so i start with this and i start with some different angles here so i start with you know you have a 60 degree angle for one side here you have a 30 degree angle on another you have this slope triangle thing going on so let's just jump in and take a look at what that looks like so we're going to start with this and we're going to let's just start with f1 so f1 we said was 220 kilonewtons f1 x equals what well the way that i start is i just like to i just like to think that what's the cosine the cosine the adjacent over the hypotenuse right and you don't need ijs and k's to do that right but we can we can just start there adjacent over hypotenuse so so in this whole force triangle thing going on f1 we have an adjacent f1 x is adjacent to uh this angle or to uh this this f1 here f1x is adjacent by an angle 60 degrees so what i can say is i i can say that this is basically just me f1 times the cosine of 60 degrees and similarly i could write f1 y is just going to equal f1 times what well i can either say the sine of 60 or since this f1y is adjacent here with this 30 degree angle i can say that it's going to be the cosine of 30 degrees okay so we can do those out with the calculator when i did that out in the calculator the cosine of 30 is one-half so i'm just going to get 220 times one-half and that's going to be what i think 110 kilonewtons and if i make mistakes i i i do make mistakes i i did proof these a little bit more than uh maybe the last time i did the last time i kind of just threw it together but this time i i tried to put a little bit more effort into um not making as many mistakes and i think i already made one so there we go so what's what do we got we have um f1 which is 220 times the cosine of 60 is 220 cosine 60 is 110 kilonewtons and 220 times the cosine of 30 is going to be what 190 uh 0.5 okay so i mean this is this is the start of it right and then similarly oops similarly if we come over here we can do f2 we know f2 equals uh 200 and what 40 kilonewtons and again i'm not i'm not breaking this into ij's the case i'm not using unit vectors here i'm just using basic trig right but you might be looking at this and saying okay i don't have an angle anymore and how do we do it well the thing that i like to look at are these these these uh you know these triangles where we have essentially a three four and five triangles so if you take the square root of three squared plus four squared pythagorean theorem we have a three four five triangle and what we know here is basically what's let's look at this the cosine the cosine of this angle here of this angle right here right and i don't know if you can see my red mouse there but the cosine of this angle down here is going to be what the cosine is just equal to the adjacent which is the 3 over the five okay so what that means is we can say well f2x equals f2 times what times the cosine of that angle which is just three over five right so 240 times three over five is equal to 144 kilonewtons if i did it right and know how to use my my calculator so read the question to make it clear please okay so when the forces are added together so basically what we're saying is i'm coming back here for you over here but f1 plus f2 when the forces are added together so they're added we get a resultant force some result in force f3 what's the magnitude of f3 right so that's what we're looking for we're looking for the magnitude of this num of this vector so f3 the magnitude of that vector that's what we're looking for and all i'm doing to start here is trying not to move all these things around because that didn't work so well so let me let me back up here so all we're doing to start with is trying to find components okay so we're just trying to find components and i'm just using basic trig to get there and one of the reasons i want to start here is because you have to know kind of where the cosine and where these angles go and how to use them okay so that's just the starting point but f2y is going to be similar it's going to be f2 times well if we're looking at this triangle here right we're going to have essentially the sine component so sine is four fifths so that's what we're going to use four fifths here and if we take four fifths of 240 we're going to get like kilonewtons and honestly that almost is enough to finish the problem because when we look at this what you kind of see is these f1s and f1x and f2x kind of cancel out they're not exactly the same but you get this 110 and you get 144 they're pretty close i mean if we look at you know if we call this vector f3 right the question said just the magnitude it just the question just said the magnitude so this is just the magnitude and that's all that we want so if we're looking at f3 right basically that's this the result and we're going to start with f3x and what's that we're basically going to say i'm just going to manually assign a sign here i'm going to say anything to the right is positive so that means i'm going to have minus f1x right this this f1 is is going to the left i'm going to call it negative i'm going to write i'm going to say it's positive so f2x i'm going to say is is positive here f2x okay so um if it's going to the left we're going to call negative it's going to the right we're going to call it positive so that's that's where we start so what does this work out to if this is just you know negative 110 kilonewtons plus 144 kilonewtons and 110 from 144 is what 34 kilonewtons so not a not a huge amount but f f3y here you will see is bigger right in here what we're going to say is when we're summing vertical forces so this is summing you know horizontal forces when we're summing vertical forces or the y direction we're going to say anything is up that is positive that's just the sign convention i'm going to use so the sign invention says up is positive so f1y and f2y they're both going up they're both going to be positive so all i need to do here is say uh f 1 y plus f 2 y what's that it's just going to be 190.5 kilonewtons plus 192 kilonewtons and when we add that up i think i got like 382.5 kilonewtons okay and i i mean if if we're looking at that um rate what's the magnitude well the magnitude is just going to be the magnitude of f3 is just going to be equal to essentially the square root of of f 3 x squared plus f 3 y squared so this is essentially again just pythagorean's theorem right because if we're thinking of f3 as some some vector that comes here we're going to have some small f3x we're going to have some big f3y and this is going to be our f3 and we're just looking at essentially pythagorean's theorem to get the magnitude so nothing too crazy but when we take the square root of what 34 squared and 382.5 squared i think we get about 384 in that ballpark the range of 384 kilonewtons so let me know if i did something wrong because i make mistakes you know and it's it's nice to never make mistakes but uh we try we tried so so if that looks good i mean i don't know that somehow this got selected but but what that looks like here is the 380 kilonewtons is the appropriate magnitude but the reason i start there is just because these forces and these vectors they're sort of weird at times but if you can get this basic trig down and really this is really the basics of the cosine right the cosine or the sine i mean if you get the basics of this down you can get the basics of what's going on here and do an awesome job with it all right so that's that's that's question number one anybody has questions you know put them in the chat i'll try to check that every once in a while i gotta look off my screen here a little bit to do it but hopefully that works so so yeah i mean we could redefine this different ways maybe you learned it as we're always going to take the the measurement from you know the x-axis or something but honestly for me i just like to break it down it is simple as i can uh to make the fewest amount of mistakes that i can so let's go to question two so question two is one of those ones that i added on to this review partly because i was talking to somebody that literally just took the fe in january and he said hey i had a just a basic question give me trying to load a rectangular load and said where's the force that's the equivalent force for those you know those those uh those that set a distributed loads right so this is a this is legit uh this is a legit fe type question i mean it was it was literally something similar to this or this concept was on the fe recently so let's let's take a look at this i mean basically what you have going on here is this is that whole idea of balancing right we've got some forces and these forces are going down right so these these distributed forces are all going down and we know we have some force here that's going to go up so what we want to do is we just want to balance those so how do we balance them we have to basically use that first equation that we talked about this question reads the magnitude of force f so this force f at some unknown distance x i added in the unknown there but at some distance x as shown below that will cause vertical and rotational equilibrium again this is the vertical rotate vertical equilibrium is sum of the forces in the y direction equals zero okay i can write my sigma a little bit better but the sum of the force in the y direction equals zero that is this vertical equilibrium so we can start there rotational equilibrium is just going to be sum of the moments about some point equals zero that's it once we get that down uh we have the answer to this question okay so those are the two things that we're trying to solve for so let's jump in with that so to do that what i personal uh like to what i personally like to do when i see loads like this is i like to break them up so i'm just going to put a couple highlighters on here and i'm going to say we're going to treat this triangular load one way and we're going to treat this rectangular load another way you could probably go into your epi reference handbook and find like a centroid of a trapezoid or something like that but personally i don't remember that formula and i teach this stuff okay so i don't remember that formula i don't expect you to what i do expect you to remember is the area of a triangle formula because you learned it in probably about like third grade i mean i'm serious like my my nine year old i think knows the area of a triangle formula so if you can do an area of a triangle formula you can solve this problem okay ah not not exactly but it starts there okay so what we want to do is what we want to do is we want to find the resultant force for this triangle so i'm going to label it rt we want to find some resultant force for this rectangular load so i'm going to call it r and then we want uh some forces that's that's kind of step one and then step two is gonna be some moments so let's jump in and we'll first look at the triangular load so if we say rt what's the what's the resultant for or triangular load you already got this right so uh basically what we're going to do is one half w times l okay so that's the one half and if we look at this this is the uniform load times the length of the load right and basically here in this case all that we're going to get is one half of what 22 kilonewton meters times the length of the load and what you can see there is that's three meters so one half of 22 is 11 times three is going to be uh 11 kilonewtons okay similarly if we look at our r this is just the area of our of a rectangle so all we're doing here is we're taking the area of a rectangle and we get essentially w times l because we no longer have the one half for the triangle and we get the 22 kilonewtons per meter times its length of four meters which is a 88 uh kilonewtons and now what we can do is we can some forces right so if we sum forces in the y direction you guys are ahead of me on the chat and that that's very cool uh and honestly i take longer to talk through this i'm so glad to see you guys getting this in like the three-minute solution but basically what we're going to do is we're going to say rt is going down so we're going to subtract that our r is going down we're going to subtract that we're going to add f here and that's going to equal 0. so basically we're going to get f is just the sum of you know these values which uh if everybody else did it right where it's just gonna be 121 kilonewtons right it looks good right so that's the value of the force so right away we know that we can cross out you know these two so the next question is rotational equilibrium right so we got we you know we got this this uh vertical equilibrium the forces are balancing but we need to put it in a place like where's that point that we're gonna put it so that it can actually rotationally balance right where's kind of the fulcrum of this of the seesaw right and yeah so i i like to create points where to do that so i'm going to call this point a at the left side it's a great place to start uh because it's a it's a good starting spot you know but in general it's a great place to start because that's how x has been defined already so we defined x and uh in that that sort of should work for you so uh what we're gonna do here is yeah so if we come down here and we do kind of number two let me get the right color here we're just gonna sum moments about point a and call that equal to zero and the sign convention that i use is counterclockwise is positive clockwise negative so what we're going to do here is we're just going to say rt times its moment arm so i'm just going to call that the arm or for the triangle a for the triangle and i'm going to call that negative why am i calling it negative i'm calling it negative because if you if you rotate this thing about point a right if you look at that that direction is opposite of our positive sign convention so that's going to be negative okay so i'm again i'm not going into like like crazy formulas here i'm just using kind of basic moments of force times the distance i'll come back to what that moment arm is let's take a look at this rectangular one as well that one's going in the same direction right and it's a it's clockwise rotation so clockwise rotation is going to be negative so it's r times the a for the rectangle the the moment arm for the rectangle right and then what we're doing basically here is we're saying plus f times x y plus well it's plus because this force is coming counterclockwise which is the same direction as our sine our positive sign convention so all that needs to equal uh zero here and essentially somebody i think already put it in the the chat but but basically what we're doing here is f times x equals rt you know times the the moment arm for the triangle plus our r times the moment arm for the rectangle and what are those arms well this is where it helps to again i i don't remember the the the the centroid formula for a trapezoid off the top of my head but what i do remember is that if you have a triangle a right if you have a triangle and and maybe i'll put it over here so for for triangular loads so for you know triangular loads right basically what we have here is if we have the whole length is l this resultant is going to be at 2l over 3 from one side and l over 3 from the other side okay so what that means is our force times our x is going to equal the this rt which we said was what 11 kilonewtons times the moment arm and that moment arm we said is two-thirds of of of the distance of the the length of the triangle so this distance here is going to be two-thirds of the three meters which is going to be two meters okay and if we look at rectangular well let me let me put that in first so that's going to be the two meters okay and if we look at the rectangular load it's going to be similar so i i even have notes this time you know so this is good uh so if we have a rectangular load what do we get the rectangle is a little bit different here but let's take a look at it this is so if we look at the rectangular loads you don't want to be looking this stuff up on the on the uh on you don't this is probably in the reference handbook but you don't want to have to go look it up in the reference handbook if you can if you can avoid this because this is just kind of like basic right this is basic if you have a whole length l here uh this is going to be at l over two okay so this is just like the centroid of a of a of a symmetrical section is going to be the middle okay yeah so so what we get here is that's going to cut this four meters in half right but we have to get all the way out to the middle of it so first we have to come three meters then we have to come another two meters so our total distance here our total moment arm is going to be uh five meters so this is the plus 88 uh kilonewtons times five meters so when we get x we're going to we're going to solve all this you got to remember that this f here of 121 is the same f so when we're all said and done i think you plug all that in and i i think what you're getting is uh is uh you are totally right um so you are totally right so x is going to be uh 4.2 meters but let's go actually fix the math so for those of you that were actually doing math this is so helpful to me because i'm trying to like talk and uh and do this all at the same time my notes have 33 you probably can't see them but it says 33 in my notes so yeah you're totally right here so this this should be one half times 22 is 11. right one half times 22 is 11 times three that gets me 33 here and when i update that down here uh we should get the right answer now so does this 4.2 work you're totally totally right uh so yeah 4.18 i believe that's right so 33 times 2 plus 88 times 5 uh divided by our 121 is 4.18 yeah and honestly you can't get 121 without 33 plus 88 so thanks um i i still appreciate you guys that's this is awesome so thank you so much for that so yeah so we have a force of 121. uh we have this distance of 4.18 or 4.2 so yeah so when you look up here to get your answer we are at answer b right so that gets us to the 4.4.2 all right so again what we're doing here this is this is a legitimate fe question like no question like this is something literally somebody who just like in january said they saw something similar to this right so if this is if you know these basics about distributed loads uniform loads and you can do your math right and not and actually multiply uh in your calculator right that helps too so uh you know so so but but again to do this all we had to do was know two equations sum of forces equation sum of moments equation and that is going to be able to get us there and i say that i say that those are the only two equations yeah we had to understand how to deal with these triangular loads and how to deal with rectangular loads right these distributed loads whether they're linearly varying or they're they're you know flat all right but again understanding that the moment equation that force equation is huge all right let's keep going here so again equivalent for systems similar idea here right but but now we have forces shown can be replaced by a single equivalent moment the magnitude of an equivalent moment is most nearly so again equivalency um what we're doing here is we're saying we have some forces but if you notice if we sum forces in the y direction what's going to happen we have minus 15 kilonewtons plus i'm sorry plus 20 kilonewtons uh plus 30 kilonewtons minus 35 kilonewtons and then also equals zero and you'll notice that it does so zero equals zero so so the the forces are balanced vertically so forces are balanced vertically so what the question's saying is it's not saying like okay well give me a force it's the same kind of concept and that's what you're going to see on the fd you're going to see kind of a similar concept but it's going to be twisted a little bit maybe from the question that i put on here but it's going to use the same principle and that's where like if you can get these fundamentals down of how to do sum of forces or in this case how to do sum of moments about a point and i'm going to use point a again you're saying well where's point a i'm going to define it in this case right so we're even using sum of forces or sum of moments you can find the equivalent right so so i'm going to pick a point a here i'm just going to put this here and then i'm going to say okay well these forces i'm just going to say they probably create some moment here right they balance vertically but they're probably causing this thing to tip somehow right and let's check so if we do our sum of moments equation from point a let's see what we get so if we sum moments about point a equals zero we're doing the same thing that we did on the last one it's just a little different right in the sense that right now let's let's take a look at all the forces first so if we take a look at uh this 20 this 20 kilonewtons here all right so we've got our 20 kilonewtons uh let's let's write that in i like to write the force down right a moment is equal to a force times a perpendicular distance so 20 kilonewtons times what's the distance the distance to this force is just going to be the 1.5 meters right so nothing too crazy 1.5 meters personally i like to write those down and then kind of look back and say okay so what is uh what what direction is this going and the direction it's going here is we're saying it's it's a counter-clockwise which matches our sign convention and that counter-clockwise is going to be a positive value okay so we get a positive value here i mean that's i can show it this way as well right what's next i'm just going to do all the forces first so this 30 kilonewtons times what is its arm so i'm going to again come all the way back to this point a i'm going to go all the way to my 30 and that's if you add the 1.5 and the 4 that's going to be 5.5 meters okay i'll write that a little bit more clearly and that's going to be 30 kilometers times 5.5 meters and again that's causing kind of a counterclockwise rotation so we're going to have that positive sign and then we're going to get all the way over to this 35 kilonewtons and you'll notice that this one has a longer moment arm and that one if we add the extra 1.5 meters we're going to get a total of 7 meters so 35 times 7 meters i and you'll notice that this one's going in a clockwise direction right so if we take a look at what that looks like here right this is going in a clockwise direction and what we're going to do is because it's clockwise it's opposite of our positive side convention we're calling it negative and then i'm going to add in this moment and some of you are thinking like okay but where you madsen what what what distance is this moment at it's a moment it doesn't need a distance so this is where it's this is where you don't need a distance it's just a moment you don't need the distance and it's a moment it already has a distance in it right so you can take it anywhere you like i mean yeah you could take it at this point you could take it at this point you could take this point you could pick the middle of the beam i just personally like to start at the left hand side it doesn't matter i honestly go back and try this question a different from solving it from a different point try to solve it from this point and see if you get the same thing you should okay and this is where when you do this question out uh basically what you see is this all has to equal zero for the thing to be balanced rotationally again this is what we're trying to do is we're trying to balance this thing rotationally if you put all that in what are we going to do 20 times 1.5 plus 30 times 5.5 minus 35 times 7 and we get you know minus 50. so that's what it's we add the 50 to both sides we get 50 kilonewton meters and fortunately we have an answer uh that works here so so yeah i mean um we could do that now but honestly i'd say when you're you know when you have an extra couple minutes um take it take a minute go go solve this from a different point and see if you can get the same thing understanding these uh these these moments equations are super super critical and we'll be going over them more and more uh throughout tonight so let's keep going we'll keep going here to question four the other thing that i wanted to tell you guys if i probably mentioned this before but down below there's a link to these problems so if you want to download them and follow along print them out or use a tablet and work on them that way you can certainly do that so i make them available to you so that it's easier for you guys to be able to use them alright so question for equilibrium equilibrium again guess what we're going to do we're going to sum forces in the y direction we're going to sum forces in the x direction we're going to sum moments so these equations i mean i'm just going to keep hitting them over and over and over again because they're so important but if you can get these equations down it's uh it's it's super super helpful so here we have a rigid body the mass of a thousand kilograms and applied loads so we get the mass of this rigid body we can apply loads on it and what we're told to solve for is the magnitude of the resultant force at support a so we have a support a here and we want to maintain equilibrium of the rigid body okay that's great so this is kind of just like a beam reactions problem when i teach my statics class one of the things i tell my students is you gotta know beam reactions you just gotta know them it's gonna be like on almost i think every every test you take in my class is gonna you gotta know being reactions and statics it's just one of those things you gotta know so how do we do this this is where we're just going to jump in and we're going to we're going to you know do some some of the sum of forces equilibrium equations right so eq eq all right um so equilibrium equations we're just going to jump in so let's just do it some of the forces in the y direction and honestly i'm the fe sometimes i slack a little in my and when you know if you're taking my class and you're doing homework assignments i'm going to make you draw real free body diagrams on the fe i don't draw real free body diagrams as much what i'll do is i'll sort of pencil them in right so or or if you have your white board that you're doing you could do a quick free body diagram of this so why don't we because you honestly on the fe you're going to have a white board sort of like i have here and you're going to have to draw something right so so let's just draw on our free body diagram why not so we're going to have our our mass which is going to turn into a weight once we multiply it by gravity we're still going to have our 50 kilonewtons down we're still going to have our 20 kilonewtons to the right and then at point a we have a pin so we're going to come up with a y we're going to come up with ax and i'm just assuming directions here you can assume in many direction you want you can assume that they're going to the right or to the left or up or down and you get a negative sign it basically means you assumed wrong and that's okay i'm going to assume that you know bx is going to the right so hopefully you know what these symbols are right so these symbols a triangle typically is going to mean a pin where you have two uh two reactions uh a circle is going to mean a roller where you get just one one reaction okay so this is this should work for you and uh what we're gonna do here is we're gonna basically be able to do this so again i like to show my sign convention here anything that's up is positive so all that we're gonna have here is we're gonna have a y is up we're gonna have minus w minus 50 kilonewtons and that l has to equal zero so a y is just going to equal what 50 kilonewtons plus a thousand kilograms times 9.81 meters per second squared and here i'm going to pause and say what's wrong because i know i made a mistake here what so i didn't even make a mistake here what what else right what else do we have to do here um to make things work this is another place on the fu where you have to be careful because you have to know your units right 1 newton equals what one kilogram per meter um i'm sorry one kilogram times a meter per second squared right so if that's one kilogram yeah exactly i need to divide this by a thousand uh newtons per kilonewton right so so the units become important here where you have to you have to understand your units and what that means is a y works out to be like you know 50 was it 59. 59.81 uh kilonewtons as opposed to like five thousand you know kilo newton so yeah so so just be careful with your units so 59.81 kilonewtons and some of you're going to be tempted just to go in circle 60 but don't do it do i suggest skipping fvds i can't in good conscience suggest skipping fbds i mean fbds are just ah they're so important so i i would even for like the extra like 10 seconds it took for me to draw this i i think i think you need to probably sketch them in so yeah i think they're going to help you more than they're going to hurt you so some of you're going to be tempted just to say 60 but you're wrong if you stop there right and the reason you're wrong is because this question isn't asking for the vertical force at a it's asking for the resultant force at a and this kind of is a good segway from is from the the equilibrium or you know the other problems that we were doing here but basically what this is saying is if we look at this for a second um you're going to have some some a y force right some a y force up you're going to have some ax force in those forces are going to create some resultant force and i'll call this resultant force a okay so what this means is we kind of have to go and solve for ax okay so if we solve for ax this is this is another one of those things where now we can choose where we want to sum moments where is a convenient place to sum moments and the convenient place to sum moments on this one is going to be right at point b and the the reason i like point b for this one is because it's it cancels out the most forces right essentially a y goes right through b uh bx goes right through b so we never actually have to solve for bx if that's where we sum moments okay so let's some moments about that point i'm just going to move my screen up here a little bit move my mouse up so we still have our red dot in the right spot but let's sum moments about point b okay so same sign convention if you get used to it the most important thing is to read the questions 100 if you don't read the questions you are in trouble so definitely read the questions here and what we're going to do is uh we're going to keep going so so we're going to summon without point b and this is where i have to look for any of those forces that don't pass through point b are going to cause a moment about point b okay so for example we we know that bx doesn't or i'm sorry bx does pass through point b um we know ax does not right ax is is some distance away from point b so that's going to be 1.5 meters away is what the problem says so we're going to start there so ax times the 1.5 meters right and then we're gonna and then we have a couple other forces and uh what's ax ax is kind of coming this direction what else do we have i'm going to take w times what where's the w again this kind of goes back to that rectangular thing the weight is going to be again the weight's going to be half this distance because it's the rectangle so that's going to be times 2 meters and if we look at the if we look at this in terms of which way it's causing rotation it's causing rotation in this direction uh that's opposite of our positive sign convention so we're going to do a negative value here right so this is causing a rotation in a clockwise direction uh then we're gonna keep going we're gonna save the 50 kilonewtons and that's the same direction uh right so if we take a look at that that's the same direction here that's the the clockwise direction that one's gonna that one's a little bit easier that one's times four meters uh and then we have to subtract or no we have to not subtract we have to add that we have this 20 kilonewtons uh times its moment arm its moment arm again we're going from point b here so it's moment arm is going to be 1.5 meters and if we if we put that 1.5 meters in here the reason we're going to go with a positive is again this is causing rotation in a in a counterclockwise manner the counterclockwise is what we call positive so we're gonna say that's positive and this hall has to balance and equal zero so if we do all that out yikes what do we get that's that's again it's kind of long equation here but we have to remember that that we're just solving for ax so if we solve for ax let me just write this in ax times 1.5 is gonna equal this this nine point eight one uh kilonewtons times two meters you know plus fifty kilonewtons times four meters uh plus no not plus this is going to be a minus because i'm putting it to the other side minus 20 kilonewtons times 1.5 meters and i got if i did this right and you guys can you know check me on this but i see it up there i think what is it 106.4 yeah i'm liking it i'm liking that vlog you got this 12 20. uh totally like so so 106.4 uh no that's not what i got uh-oh so let's check let's check let's just put it in the calculator c 9.81 times 2 plus 50 times 4 minus 20 times 1.5 um divide that by 1.5 i got 126.41 is anybody else get 126 out there in the the checking world 126.41 uh-oh 113. i did this twice i honestly i i did this twice i even checked it and i checked it a couple times so i think i'm i think i'm right here pretty sure uh the 1 26 is is doing it right so check one of the things with this is check your signs it's so easy to screw these up on the calculator especially when you start moving things to the other side of the zero so it's really easy to screw those up but if i put in 9.81 here times 2 you know plus 50 times 4 minus 20 times 1.5 i divide all that by 1.5 i i that this i get the 126. so what i got was 190 or not not 189.62 189.62 kilonewton meters so this is essentially equal to 189.62 divided by uh 1.5 meters okay thanks nelson man you made me feel better there all right so uh so here's just you know the the check that or the intermediates that i did so if you're having issues let me know but again what we're looking for here is we're looking for the resultant of this force so i mean picture like if this thing were to be being hung by some cable right if this thing were being hung by some cable what's the resultant of that force and what we see is a is just going to equal just like we did earlier with the result in the first problem what do we have ax squared plus a y squared so a is just going to equal the square root of where we were 59.8 squared plus 126.4 squared and if i did this right which i'm pretty confident pretty confident today i get like 139.5 kilonewtons which hopefully hopefully hopefully shows up on our list and it does so i get the 140. but again what was it what did we have to do we had to know we had to know these equilibrium equations once we were able to apply those things it was good the other the other thing in here that you really have to be careful with is are these units you have to know the units you have to know the conversions i i don't want to have to go look up what a newton is okay so that's just something like if you can save time on that and not make that mistake that's gonna that's gonna help you um and again if you can know how to do these summer moments equations and what a result of force is like a pythagorean theorem type thing that will help you as well okay so let's keep going why not you know it's it's a tuesday night and it's it's still fun at least okay oh man this one this one is is interesting because it looks like i you know i appreciate doing them honestly this is one of the this is this is fun for me in some ways uh i was doing these just for my students and then i had students keep asking me like hey can we get access to those afterwards and i said i'll just do them live on youtube and then you can have access there you go but thanks for tuning in it helps me out to know that you care so with this one this is kind of like guess what we're going to use we're going to use uh summer moments and some trig on this question so remember those three things i told you some of course the summer moments intrigue again we're just going to use some moments and and some trick on this one but the question says consider the cable system below so so right away what we know is these are cables so so that's something to to consider and what's special about cables the same thing that's special about ropes you can't push them right cables and ropes they don't have they don't have compression they only have tension so that means is whatever force is in this cable whatever force is in this cable has to line up with that cable so that's a huge thing that you got to understand with forces and ropes and cables and that sort of thing you might see this question like this honestly a classic question that i give my students in first semester statics like i think every semester in some shape or form is this question and you might see this question as well if you have a you know this is something that you could totally see on the fe right so if you have a force hung by two cables and this is just uh this is just similar to what we're doing here it's just a step removed so again the fee is going to take the basic concept and maybe twist it maybe take it a little maybe give it to you straight up but you know i'll give this to my students and we'll do a homework assignment and then on the test it might look like this right and what's different basically it's rotated 180 degrees right or this question might show up where you have it like this and you have a bracket right and you have something hanging off it right and honestly the approach is so similar right and there's different techniques you can do you can use like lost signs and all sorts of weird like other methods here but i i just want to boil this down so if you can master a couple of things if you can master some moments and some trig you don't need to have special techniques right you can just use those principles and make them work okay so what we have here is is let's take a look so we know that this cable right we know this cable is at some geometry of essentially what is it we have what do we have we have one foot here and five feet here okay so the geometry of this cable we have one foot and five feet so that makes this over here the square root of five squared plus one squared i think that's the square root of 26 if i did it right okay so that's that's the geometry of the cable this is the cable geometry with cables with trusses with two bar trusses you know bigger trusses what you need to have and what you need to understand is this triangle that we just did here so this triangle that we just did here this this geometry triangle is going to be similar to the force triangle right so the force or this p value is going to share the same exact geometry so in other words we're going to have px and we're going to have a py right so so the cable geometry and the force you know the force i i like to call this the force triangle these are going to be similar right so these have a similarity to them and these are similar triangles so what that means is right this angle here and this angle here are the same angle right what it means is this angle here and this angle here are the same angle right and that's going to be super super helpful to us you might be thinking why matt's and why does that help it helps because once you think of similar triangles or if you don't like similar triangles let's think of cosine for a second okay the cosine of of let's say this green angle let's call this angle theta right this green angle right if we say the cosine of that angle what is it it's just 5 over the square root of 26 right you might be wondering madsen why are you talking about all this because it's super important for this problem and for any trust problem for any two-part cable problem you know any any stuff like that so um so what are we doing we're trying to find this force p right and if we if we keep coming back to this right with this cosine of that angle is 5 over 26 if we do the same thing on this side right so if we do the same thing over here what do we get we get the the the cosine here is going to be the cosine of this angle theta is going to equal p x over p right so so same thing same theta same cosine and if i get my head out of the way what do we get well what we see is the similar triangles tells us that you know let me write it this way 525 over the square root of 26 equals px over p which also tells us that px equals 5p over the square root of 26 and we can do the same thing with the sign right so if we do the sign uh we can do a similar ratio but one over the square root of 26 equals py over p and that's going to give us py equals uh 1p over the square root of 26 right and why is that helpful that's super helpful because now when we sum moments right what we want to do is we want to sum moments so let's let's draw that free body diagram again okay so i'm going to come back here and what we're trying to solve for again we're trying to solve for the magnitude of this force p we want to know what that force p is okay so how are we going to solve that yeah we could use the tangent and get the state angle theta but honestly i i don't like solving for angles if i don't need to so so let's look back up here and let's draw a free body diagram and i'm going to draw the free body diagram that looks like this i'm going to cable you know point b and point a okay so what i'm going to have i'm going to have a y i'm going to have ax i'm gonna have maybe this 200 pounds acting down i'm going to have what else i'm going to have px and py and in here i'm going to use the components rather than p to solve this and you can use varying young's theorem and like apply that force p somewhere else but all of a sudden when you do this problem the cool thing is let's look at this if we sum moments about a convenient point and the convenient point is always where you have a lot of unknowns that you're trying to get rid of so for example if we sum moments here at point a right what do we have all of a sudden this problem gets a little bit easier because we know this this is four feet and we know the the horizontal distance here is uh 4 feet right so if we if we know all of those what we can say is let's just sum moments right so i'm going to start with we have py times 4 feet minus 200 pounds times four feet right those are both going in the the uh counterclockwise direction and then i'm going to subtract off px times four feet equals zero that one's clockwise right so if we look at these moments the the 200 and the py are both going uh counterclockwise whereas the px is going uh clockwise around point a so do you see it yet do you see it yet uh hopefully you see that right we have we have a substitute for px here we can bring that in here we have a substitution for py here and we can bring that in here and we have one equation one unknown and we can solve this thing okay so what we can do here is i'll substitute in put them to the right spot i think so uh so we got what we got 5 p over the square root of 26 times 4 feet minus uh i'm gonna do this px was was no did i screw this up p y let me let me take these off this is messing with me um so let me just back up here so p y was p over square root of 26 times four uh px was 5p so minus 5p over the square root of 26 times 4 equals i'm just going to put this 200 over the other side so 200 pounds times four feet and the cool thing here is these four feet all drop out right so that makes this problem even a little bit easier right so i can essentially multiply both sides by square root of 26 so if i do that i get if i multiply both sides by the square root of 26 what do i get p minus 5 p equals 200 times the square root of 26 and actually i missed a negative sign didn't i something i missed some negative sign somewhere let's take a look no i didn't miss a negative sign i just messed up my sign look at that this sign should be positive did anybody see that probably not but this sign should be positive and that's in what you're doing you're screwing with me no i'm not screwing with you um what i'm i'm not trying to at least right so so what do we have we have um counterclockwise and counterclockwise these both need to be positive because they both match this plus you see that so what that means is when we subtract off we get a negative 200 over on this side you see that so we get the negative 200 negative 200 it works so so this is now we got minus 4p equals negative 200 square root of 26 and we have p equals the the negatives go away essentially 50 times the square root of 26 so p equals i think that works out to 255 pounds let me put it into my calculator somebody else put in your calculator because i i again i checked this one i'm uh i'm pretty pretty confident with this one but you know 50 square root of 26 gets us to 255 right anybody still screwed up with those uh those signs on the moments because the moments that's where that's that's where the money is i mean that's where if you can get these signs you're good but the the py and the 200 are both going in this counterclockwise direction which matches our positive sign convention so those both have to be positive px is going the opposite direction and that one has to be negative so what we get down here sammy is is we get the p minus 5p so that goes to it does go to minus 4p but i i essentially gave you a plus and a plus here uh to to make them both both positive when we're all said and done okay so so good 4p we're good i think so so i think we're good but again honestly um some sum of forces some moments and uh some good accounting with your positives negatives you know your rotational arrows and that sort of thing it's so easy to make mistakes here and you see me doing it even while we're doing this and like part of it is i'm trying to concentrate part of it is i'm not looking back at my notes as much as i is as much as i need to part of it is i'm just trying to make sure i don't screw things up and in the process you screw things up but i think we're doing all right here so so let's uh let's keep going okay frames and trusses part d is is kind of the next part here i threw on a method of joints truss here just because i i we we've been doing a lot of summon moments method of sections gets into some moments uh we'll get into methodist sections probably next week uh for sure but i i i decided to go with a method of joints question here i might be thinking like matt you need to read the question and that's okay let's let's read the question consider the trust in loading shown below ignore self weight the magnitude of the force in member ac due to the applied loads is most nearly so right away we know we're looking at this member here ac and what we want to do is we want to find that member force so there's a there's a bunch of different ways you can do this you do methodist sections you could find uh support reactions you could do all sorts of things but what i prefer with a question like this and again this is one of those questions that i'll throw on like a test in my class is just here's a here's a here's a trust go find a member force and one of the reasons i throw these on my test is because i know that something like this will very likely show up in the fe and i want you to be prepared for it but what does this look like well basically what this looks like here is with the method of joints basically all you're going to do is you're going to draw one draw draw an fbd of a joint and then two you're just gonna apply guess what this is probably getting boring at some point and you're thinking like man if i could just figure out how to do sum of forces and some of the moments i'd be an expert at statics and you're absolutely right if you could just figure out those out you would be an expert at statics with the method of joints there is no sum of moments because there is a moment arm okay so what do i mean by that well let's let's let's take a look at this and you also need to apply a healthy dosage trig okay so let's take a look at this let's draw a free body diagram here and i'm just going to draw the joint so this is joint c and what i like to do when i draw my joint is just draw the joint and then draw all the forces that i know that are on it so i know that there's 20 kilonewtons on it i know there's eight kilonewtons on it what else do i know well what else i know is that i've i in order to draw this i sort of had to cut through two members right and just like that cable trusses can only carry axial force so if you know it's this pin connected you know pin connected set of members with loads only at the joints or if in the problem statement they tell you it's a truss right you know that these forces have to line up just like our cables did the forces have to line up with the members they're either going to be in tension or compression but the force is because it's not a cable anymore but the forces have to line up i like to draw all my forces in tension some people like to do it a different way personally i like to draw my forces in tension just because that gives me a starting point so i know if i get a positive value i drew it in the right direction it's tension right it's it's pulling away from the joint and some people have an issue with that but it's like if we think about this for a second if we pull away right this is gonna be tension it doesn't matter if you're pulling to the right or the left you're gonna be in tension if you're if you're pushing towards uh it's gonna be in compression okay um so so i like to you know assume assume tension over here and that's what i that's what i'm doing but i also like to label these i'm gonna label this ac and i'm to label this bc just because those are the forces and any time i'm doing method of joints i like to also show my components and a lot of times i'll use a different color for those just to make sure that we don't get confused as to what they are but this is going to be bc y and b c x so th these are our these are our components of bc and just like with that cable problem right just like with that cable problem we know that there's some relationship between bcx and bcy okay so let's take a look here and see uh what we're doing and how we get there so i'm gonna sum the beautiful thing about the method of joints is again there's no moment arm right there's no there's no distance to any of these forces because they all are applied right at this joint so there's no moment arm so what does that look like well the first place i have to start for this one is some of the forces in the y direction so if i some force in the y direction anything up is positive so i'm going to say minus 20 kilonewtons right i'm just looking at what are the forces in the y direction so there's two forces in the y direction the minus 20 and vcy okay so if i take those here and i say minus 20 kilonewtons minus bcy equals zero i can pretty quickly get bcy equals minus 20 kilonewtons in that negative 20 right then negative means uh this means compression so the negative means i assumed incorrectly and that's okay this the direction isn't actually uh this this direction isn't actually tension and it's it's actually uh compression and that's okay so the next question is well okay let's take a look at my next equation my sum of the forces in the x direction equals zero and what do i get minus ac minus bcx plus i know i get that i have minus eight kilonewtons yeah minus eight kilonewtons so they're all going to the left right uh they're all gonna left minus eight kilonewtons equals zero okay so ac is what we're trying to solve for right that's that's what we're trying to solve for so i'm going to isolate that i'm going to put that on like the other side here and then flip everything so i'm going to ac equals you know minus bcx minus 8 kilonewtons and you might take a deep breath and say madsen that's great except what happened to bcx we don't have bcx how do we get bcx and that's again where this this uh you know this this is where we get this healthy dose of trig to come in to solve that okay so just like we did with that cable problem we're going to say we know bc right has some geometry and that geometry of bc is essentially a five and three triangle okay so we're gonna come down here and we're gonna say bc has some geometry it's three meters vertically five meters horizontally and so this is gonna be the square root of 25 plus nine what's that the square root of 25 plus 9 i don't even know what that is it's just the square root of 5 squared plus 3 squared and honestly i don't care because the fe doesn't care about all the intermediate work he just wants you to get the right answer okay so what are we going to look at we're looking at bc is this is bc we also know that these triangles are going to be similar just like on that cable problem because the force has to line up with the member so the force has to line up with the memory mike you got it man um similar triangles right so these these triangles are similar they're the same angles and what we know is if we have uh if we're trying to find bcx right bcx over b c y has to equal what five over three which means b c x equals five thirds of b c y or b c x equals five thirds of negative twenty so five over three times minus twenty is uh minus thirty three point three three kilonewtons and that interestingly enough will get us our answer so minus times a minus 33.33 kilonewtons uh minus eight kilonewtons equals actually we already have that equal to something so basically these negatives go away we get a positive when we get ac equals what's that 33.33 plus or minus eight i should say minus eight it's like 25.33 kilonewtons and you were hoping trusses went away after you got done with statics i know it's it's ah but like again the fundamental concepts here can you sum forces can you some moments can you apply that basic trig to go from one to the other and honestly what you'll notice here is i haven't used the reference handbook so far and that's okay like i mean you can use book this stuff up in the reference handbook but i would say work at the statics problems until you get to the point where you don't need the you don't need the reference handbook to know how to sum forces in some moments and use the basic trig okay there are some pieces where i honestly i'll pull it up here in a minute uh when we get to like moments of inertia and that sort of thing but so there are some pieces where it's going to help but i can't feel the bridge you know sometimes we just fall down because the bad trust is but you know we don't like to see that we try to make it work so what do we get yeah so so the answer here should be that 25 uh kilonewtons and interestingly enough if i asked you right what do we know this is a positive value so that's going to mean this one's in tension so the arrow was drawn in the correct direction okay so again kind of the basics some forces some moments and and we're pretty good all right this one is intense but uh it looks it looks harder uh than it is cb is an x component yeah it totally does so i'll just come back there for one second so cb totally has an x component or bc whatever you want to call it uh bc has that x component here that we solve for so that's where the easy answer is just to say well this eight must be taken by here and it's that's wrong i mean that's just a that's that's for people that aren't like you that are studying that are gonna get this right okay so let's go to the next one this looks intense and it is but guess what you can solve it with the sum of moments equation okay so the approach this equation questioning i mean this is just becoming a little bit of a broken record and i hate i hate it but like honestly i asked some of my students one of them that just passed the test i said what's the key to statics he's like you just got to do it over and over again you just got to do it until eventually it clicks i asked another guy who passed it last year i said so what what do you think were some of your strengths is just statics i was like really so yeah like i felt like i finally got it it's like it finally made sense to me like i was able to get it and like uh honestly like i that was that surprised me from his standpoint that like you know that was his that's what he would come back with as a strength but i think it's one of those things that practice practice practice with with your statics so here you know we have a rectangular loaded triangular load okay so um we have a rectangular load r we have a triangular load rt and we have to sum a moment and this gets a little bit more complicated because it's a frame but not much more honestly it's not much more because remember what a moment is a moment is a force times a perpendicular distance right so if we're looking at this we we all of a sudden let's actually let's read the question first uh before i just jump in i'm just excited today because i love statics right what do we get consider the rigid frame and applied loads below so again frames trusses ignore the self weight the magnitude of the vertical reaction at d so we're going to have some vertical reaction here due to the applied loads again so applied loads means no self weight due to the applied loads so we have the vertical reaction d to the applied loads is most nearly so do you see it yet basically we're going to have some you know support reactions at 8y we have some support reactions at ax but ultimately all that we're doing here is we're just doing a sum of moments equation so this question looks really terrible and if i wanted to you know if i wanted to um make it even worse abraham just to point out like also if you look down on the link below there's a link to the question so if you want to see him while i'm going you can totally download them as well if i want to make this question more complicated my my students hate me for when i do this but you could have to find these components and that gets annoying too we're not going to do that tonight but if you want to go back and try that okay so oh man that's a that's another fun question but let's take a look at this we're just trying to find that vertical reaction at d okay and what do we get we're just gonna sum moments about point a why point a because we don't need ax and a y we don't know a x and a y and honestly at this point we don't care about x and a y because all we want to do is solve this problem pass the fee get the six thousand dollar raise and move on towards our p e right so so if we sum it's about point a let's take a look at this what we're going to have is our r times its arm what's it the moment arm for r this is where you have to look that perpendicular distance and do you see it i mean we could use the components but here why would you this is it's so much easier if you could if you can look at this and come up with the perpendicular distance that load so if you can find that distance that's great but you're thinking matson we don't have that distance and you're right we don't have that distance but what do we have what we know is we know that this distance is six this distance is eight and if you remember your fancy fun triangles right we have a six eight 6 8 10 triangle which is similar to a 3 4 5 triangle and if we go halfway to the six and halfway to the eight that puts us exactly at essentially a three four five triangle right so what that means is this distance halfway through the beam is going to be five feet okay so again that kind of the basic geometry the fundamental geometry if you can get that stuff down you're doing you're doing all right so we've rr times five feet which way is that going right so this is going uh what's that that's that's clockwise right so that's a clockwise moment that's gonna be negative okay we're going to do the same thing with rtrt times its moment arm what's its moment arm again we're going to come from point a all the way over to the center of this thing where's the center well we come six feet plus three feet so that's a total of nine feet and again if we look at this thing this is kind of go which direction is that that's a clockwise direction so that's going to be negative same thing with the seven kips i'll just stay here seven kips is going clockwise so we're gonna subtract the seven if i get the right pen here uh we're gonna subtract the seven kips times the total distance here so this total distance six plus nine plus three plus three um you can put that in your calculator you can do in your head six and six and six is twelve plus nine is that twenty one i think i think that's 21 21 feet and then the last one that we we need here is this you know we have this distance to uh the distance here to point d and that's gonna be six plus nine plus three so six plus nine is fifteen plus three is eighteen right so you know if this one's coming clock uh clockwise what we see is that the d y is again it's going which way if this this d y is going counterclockwise which is the same as our positive sign convention and that's going to make that one positive so d y times 18 feet equals zero and this just becomes a calculator practice the the question looks super intimidating because it is uh but but when you break it down it's one equation in one unknown and if you notice if you know how to sum moments hopefully you can make that work so so let's go through this and add our resultant forces here our are rrt okay and we will make that work so rr is again whenever you see a load that looks like a rectangle you're just integrating the area of it essentially to get your your your your total load so rr is going to be minus what this is 2 kips per foot times 10 feet okay and then we're going to multiply that by the five feet still the rt this is a triangle so we're going to say one half two kips per foot times a total of nine feet right so that one goes this one goes a total of nine feet here okay and that one we can't forget to multiply it by the nine feet as well that's the moment arm although it's so easy to make those mistakes uh and then minus seven times you know kips times 21 feet and then plus d y times 18 b equals zero so good grief that's a nice long equation but uh against again we get something here uh oh 16.25 i don't think i got 16.25 uh what did i get so if i put in 2 times 10 times 5 plus 0.5 times 2 times 9 plus 7 times 21 i get like 256 kip feet is going to equal d y times 18 feet so if we divide that by 18 i get like d y equal to like uh 14.2 uh-oh what did i do wrong now now i did do something wrong i know i did uh so let's look 2 times 10 is 20 times 5 that's good one-half times 2 times 9 is 9 times 9 7 what do i do anybody else get a different we're getting all sorts of numbers out here including me so i did something i messed something up um let's find it this is where like you don't get an answer that works let's let's go back and check our calculator so 2 times 10 times 5. plus one half times two times nine times nine okay so the second time i put that in my calculator i think i got the right answer of 328 anybody else get 328 and then if i divide that by 18 i get a dy of 18.2 so the second time i put this in my calculator i get the 328 i don't have a calculator rule number one go back and watch math buy the either ti-36 or casio 115 you've got to get used to using the calculator because it's just man it's it's oh so i what else can i say uh because obviously like even i know how to use a calculator the first time through i did it i broke down there i got 256 which is just wrong okay so so go get a calculator uh yeah i i don't know what to say anybody else can verify that the 18.2 i mean i i verified it somewhere else here but uh honestly you know one of the things that i did honestly before before even this session here was i opened up a structural analysis program called visual analysis i use that often for checking stuff like this and i'm just pulling it up here but that's that's what i use oftentimes to check you don't get the the benefit of something like that on the fe so uh normally i don't show you how to do that but you know if we want to take a look at that here uh that's where did it go i don't even know where it is uh but yeah so that's like that's similar to this problem but you can look at the result and we get an 18.2 so uh you know i'm happy with it 5 a.m oh man this is i love this this is awesome normally like nine o'clock is good because the east coast west coast works but this is awesome where we can even go across cultural and uh across the atlantic so this is this is awesome so glad to have him have you joined us but yeah so what we get here is we have uh we should be at 18 kips okay we're just going here so centroids centroids oh you this is where you might want to understand where this comes up in the manual this one honestly do you need the manual for it no you don't but you might not remember the centroid formula if you don't remember the centroid formula that's okay but you should probably learn where it is in it you know it's it's good to understand how to use the manual so if we do a control find centroid we get this this equation here right so so this is where we get an equation and this equation i i mean we get a centroid of an area this uh what is it there's this sigma that greek term again of y times a over a and let's go let's go play that out for a second but but before we go there remember how i said you have to kind of know that sum of moments equation this is so like the sum of moments equation because it basically is the same thing right remember when we had that question earlier where we were we were trying to balance this thing what did we do we essentially took the sum of moments and divided it by the sum of forces remember that this was like a force times the distance the sum of the force times the distance divided by the sum of forces and that's really really really similar to what we're going to do here because what we're going to do here is we're going to say the centroid equals the sum of a y divided by the sum of a you see it's it's really kind of a similar thing and when i do questions like this oh man that's awesome that's so that's so cool so last week we had people from ethiopia and canada and now we get even to go to africa oh no we have to guess ethiopia too but um this is awesome so thank you uh for joining us and what do we get here so what we're doing is we got let's the way that i like to do this uh is is is is is like this right where we got one and and two right so we got two different areas i like to just break it up like this and make a table and you know basically what i'm gonna do here is i'm gonna write the part part one part two i'm gonna write a i'm gonna write y i'm going to write a y and we're just going to solve this right so what do we get we get the area of of part one here right this is the area of part one is just going to be the 4 times 8 which is going to be 32 millimeters squared and interestingly enough if we look at it down here we get the same thing we get you know that i didn't try to make this too overly complicated with the geometry although i have to admit the first time i did this question i made a mistake and it just drove me nuts but um what do we get here we got 32 millimeters squared okay area not not too crazy we sum that up we get 64 millimeters squared this is our sum of our area all right and then what do we do next well what we do next is we want to go and basically look at these these y values what's y y is the distance from a reference axis so so here i'm going to take my reference axises uh is down to the bottom of of where this y bar is that we're calculating which is the x axis up to the centroid of the part so again the centroid of a rectangle is the middle of it so i'm going to take this distance kind of like this is going to be like y2 and this distance here is kind of going to be like y1 so if we know those distances that's going to be that's going to be what we're finding uh over here so for so y one we essentially have to add four plus eight over two which is going to get us to uh eight millimeters and then y two is just going to be half of this 4 so 4 over 2 which is going to be 2 millimeters so we multiply those out at 32 times 8 and we get 32 times 8 what do we get i think 256 uh millimeters cubed and 2 times 32 is 64 millimeters cubed we add those up this is our sum of a y and i think that's 320 millimeters cubed okay so i'll move that up a little bit but again that now all we're doing is we're kind of doing this this this mass balance essentially and we're getting y bar equals 320 uh millimeters cubed divided by what was it 64 millimeters squared and 320 over 64 is actually works out to be a nice number of 5.00 millimeters so you know it's centroid not a not a super crazy concept but if you think of it it's sort of it mirrors that sum of moments divided by sum of forces equilibrium type of thing as well and if you remember this equation great if not i'll go look it up in the manual okay so so we'll keep going okay so centroids i i think they'll be on there honestly i do uh moment of inertia definitely gonna be on there again i talked to somebody you just took in january said definitely saw a moment of inertia and that's something you know it's not super crazy but it's something that you need to know and if we come back and take a look at moment of inertia what we're trying to find for this question is the moment of inertia about the x-axis axis so we only have a couple questions left but let's let's get into them here i'm ownership about the x x axis uh again if we come back here uh to our our equations this is where sometimes it just gets overwhelming because you look at this and you think like oh man i thought madsen said all we need to know is summon moments and some uh you know some trig and now now we got some calculus thrown in here and you can solve these with calculus you can't solve moment of inertia with calculus but but don't okay the reason i say that is because uh there's an easier way to do this right so if we come down here one of the things that you have to see in the manual is there are some standard shapes and for example the standard shape of a rectangle is one of those one of those ones that you're gonna uh possibly use so we see here that i xc the the moment of inertia about the centroid of that is bh cubed over 12. okay so this this rectangle the moment of inertia about the x-axis but its centroid is bh cubed over 12. so let's come back here and we see a channel okay and when i see this channel i see it differently than maybe you do when i i mean maybe you learn the moment of inertia formula equals this this parallel axis there where you get sum of i naught plus 80 squared and you have to go through and honestly if we come back to the manual that's probably in here right so and we'll use that later on let's come back we'll come back to friction uh let me zoom out here a little bit but yeah i mean like here's the parallax's theorem it doesn't really say the sum of it doesn't say it the same way so honestly this is one of those equations where it'd be it'd be nice to know uh oh no is that that's that's different let's look let's come up here oh here's the parallaxis there i'm sorry i was looking through a spot so basically what's this is it's the the moment of inertia about the centroid plus the distance from the centroid squared times the area so when you see something like this you might think parallel axis theorem but when i see something like this i try to simplify it and the reason i throw these on the reviews because there's like this is one of those tips and tricks that kind of help uh we just looked up the formula for rectangle which is b h and i xc about the centroid as bh cubed over 12. so we just look that up when i look at this channel what i see is i uh a big box minus i of a little box and you're thinking like what are your boxes matson and that's okay you can think that but what i think of is i think of a big box right so that's this big box minus the eye of this little box right and and what we can do is because these boxes share the same horizontal axis of symmetry we can just add and subtract them it only gets the point where you you have these terms when they don't share this same axis of symmetry so here this question actually boils down a little bit easier what we're saying is the eye of the big box is going to be what it's going to be 4 times 10 cubed over 12. right so that's this one here is the kind of the big box and then what we're going to do is we're going to subtract off what do we have we have this is like 3 so 3 times 6 cubed over 12. okay and this is kind of like if i can keep the page set sorry about that i was trying to flip my marker but this is kind of like the little box so that gets us our whole moment of inertia and it's it's kind of a shortcut don't worry if you really really wanted to see the parallel axis theorem in its full glory we will uh we'll take a look at it and you could you can go back and solve this problem uh using the parallel axis theorem uh but you know if we if we take a look at this four times ten cubed over twelve uh that's going to be like 333 minus three times six times six times six divided by twelve that one is uh 54 and the i total is going to be 333 minus the 54. it's gonna be like about 279.3 so the 280 looks good and we will circle that one make a little bit of sense so this is going to be centimeters to the fourth let's keep going because i do want to see the the parallax of serum and it's full glory here but again the same type of thing we have the moment of inertia about the x x axis is most nearly so same type of thing uh this is you know i equals the sum of i not or or i think the way the manual puts the ixc about the centroid plus uh the a d squared term and they they call this the d y term you know and that's okay so so we can use that and with this one what i see is is some shapes again i like to break this into smaller shapes so i see a couple circles i see a rectangle okay and you know the the shortcut might be to think like okay well well can i just come up with an equivalent rectangle and that's really dicey it's really dicey because this this ad squared term influences it quite a bit so let's just let's work that out so to work this out i'm just going to make a table just like we did for the centroid i'm gonna make a table for the moment of inertia so i'm gonna label some parts here i'm gonna do part one part two and part three and then i'm gonna come down here and i'm gonna say part you know one two and three okay and what do we get we get uh the the base moment of inertia ixc and let's let's do that i'll start with part two because we already know that a rectangle is bh cubed over 12. so we just looked that one up let's start with that with that one's that one's we know that's 1 times 6 cubed over 12. uh i think what do we get for that we get i get on the right page here i think that works out to 18 uh 18 centimeters to the fourth okay and now we have to come back here to ixt for a circle and this one you might not remember but if you know that you can look this up it's a lot easier than integrating a circle formula uh you can come down here and you can uh let me zoom out a little bit so that we can get down the right page kind of quickly let's look for a circle there's our circle and what do we get we get the ixc is pi a to the fourth over four where a is what the radius okay so let's come back here and we'll write that in we get a circle and we get pi a to the fourth over four and what do we get we have pi times a in this case if we have a two centimeter circle uh the radius is going to be one centimeter radius okay so what's that one uh one to the fourth over four and you'll have to apologize i'm leaving the centimeter u and it's kinda off there just just for the sake of space but basically what this works out is pi over four centimeters to the fourth okay and it's going to be matched down here because those those dumbbells on either side are the same okay so once we do it for the top it's the same for the bottom so we get this ixc term we can add all that up so pi over four plus pi over four is going to be about 19.57 centimeters to the fourth and you can see that our values in the answers are far far off of that right so what that means is this ad squared term has quite an impact okay so let's look at our area our distance and our ad squared so i like to set this up in a table it makes a lot of sense to me uh circles uh hopefully you know a circle the formula for circles pi r squared or pi a squared in this case for the formula i mean these are the ixc formulas uh circle is just pi r squared so pi times one squared okay so that's just gonna equal pi centimeters squared uh the area for the rectangle is gonna be one times six centimeters so uh it's gonna be six centimeters squared and then this one down here the other circle is just gonna be pi centimeters squared okay distance what's the distance the distance is the distance from the neutral axis or the centroid of this thing and because this thing is para or has horizontal symmetry right it's symmetric it has a horizontal axis of symmetry i should say basically what we're saying here is is we know that this distance is the distance from the centroid this is the distance from the total centroid to the centroid of the part right so if the centroid of this circle is right there and the centroid of the total is right there this distance is going to be what's going to be half of the 6 and half of this 2 so that's going to be 4 centimeters okay so that's going to be 4 centimeters and it's going to be actually the same down here it's going to be 4 centimeters and i don't care about the plus and the minus because ultimately that term gets squared anyway so i don't really care about the plus or minus the distance between right this this this middle though the distance between the the middle of part two and the center of the whole thing because it has that horizontal axis of symmetry in the middle this is just going to be this is going to be 0 here so this term just goes away for the ad squared term but what we can see here is we get essentially what 4 times 4 is 16 16 times pi so this is gonna be like what 16 pi and this is going to be like 16 pi so it's gonna be like what 32 pi okay and if we add that this is centimeters to the fourth if we add those up the total moment of inertia is just going to be 19.57 plus six there are 32 pi and that's going to equal what 19.57 if i can put it in my calculator right that helps too 0.757 plus uh 32 pi i'd also recommend getting the calculator just because sometimes like as you get used to using it you realize as you're punching things you didn't press the button quite right and it helps you know you know when to pause and when to stop and we see that this total moment inertia is 120 centimeters to the fourth all right but again that that this table is going to be useful to setting it up to making it work and and hopefully it helps you as well all right we will keep going here we got two questions left uh friction is the last question that kind of gets thrown in at statics really the biggest thing with friction is is the friction force equals mu times the normal force so we can go look that up in the for in the in the manual but if you know that friction is just the the coefficient friction times the normal force you're good so this this question is set up that you got two blocks uh this the first the top block here is restrained against the wall it's got a you know something holding against the wall so it won't move okay but what we're trying to do is just pull out this bottom block of three kips we're just trying to pull that one out and we're saying okay how how's that gonna work so we're given some information we're told that the coefficient of friction between the blocks is 0.55 between the block and the horizontal surface is 0.4 and we want to find this force p so that the bottom block can move okay so what's that going to look like well we want to find the the force p that's going to look like let's draw a free body diagram so so we have this you know three kit block here and what we have is we have essentially seven kips acting down we got some force of friction from the top we have some force of friction uh for the bottom and then we're gonna have some normal force you know down here as well last last force here is this this force p that we're trying to solve for so here right this is where we just have one equation now some of the forces in the x direction uh equals zero and what we know is essentially of minus the force of friction at the top minus the force of friction at the bottom and plus p has to equal zero and if we only knew what the forces of friction where we'd be good all right so the other thing that we can look at here is we can we can actually write our sum of forces in the y-direction equation we can say some of the forces in the y-direction equation equals zero and we can get this normal force equals what is essentially 7 kips plus 3 kips equals 10 kips so the total normal force down here is going to be that 10 kips but what we what we're looking for here is we're looking for minus the force of friction top so what's the force of normal at the top times you know the coefficient of friction for the top you know minus the the force normal at the bottom times the you know mu for the bottom the coefficient friction for the bottom and then this is going to be plus p equals 0. so i need a little bit more room here and i get the benefit of being able to do this digitally so i'm just going to move this down but you know what we're doing here is is essentially this force of normal at the top i'm just gonna because these are both negatives here i'm gonna put them to the other side i'm gonna say p equals what the force normal at the top is just the seven kips times the mu of the top if you remember that one was what point five five and then we're going to add in the force of normal at the bottom which is 10 kips times 0.4 and that's going to equal i think seven point what was it seven point something seven point i think when you do this out all right so we're doing all right there just warming up with friction don't worry it'll get more complicated you know and that's the thing it's like we take these kind of these concepts and it's like can you apply this concept right that's that's kind of the way it goes can you apply this concept last question of the night and then we will call it a night here but the last question of the night is is this right uh what do we have we have a block with a mass of 200 kilograms on an incline plane and we're told that this angle is let me get the right pen here that this angle is 15 degrees so we have a static coefficient of friction uh we have a mass right so we've got 200 kilograms we've got 0.35 we've got an angle of 15 and this force p acts horizontally through the center of mass of the body the magnitude of the force p that will cause the motion to begin up the plane is most nearly you'll love the the fe language is most nearly right so so what do we have this is where it helps to draw a good free body diagram and definitely definitely definitely helps to know some trig here so so the trig and geometry are really the the hardest part of this this question um first before we do anything else i'm just going to say force equals mass times acceleration so this is going to be 200 kilograms times 9.81 meters per second squared and that's going to be this force of this this mass is going to be 200 times 9.81 200 times 9.81 is what 1962 newtons okay so we know that this you know mass is like 19 having an effect of 1962 newton is acting down we also know that there's gonna be some normal force acting up so this is going to be the normal force we also know that we have some force p that's acting horizontally let me draw that a little bit more horizontal and the last force that we have on our free body diagram is going to be this friction force so the force of friction that acts along the surface so what i like to do when i'm solving this this type of a problem is i like to align my my coordinate system so that x follows my force of friction that's just the way that i like to do it so i have my x and my y but i like to rotate my coordinate system sometimes that helps for these problems sometimes it doesn't but for this one i like to do that so what does that mean so that means we're pretty good with like our force of friction and our normal force because those line up with our our coordinate system okay but what we see here is the forces that don't line up with the coordinate system we have to apply some trig so we know that this 1962 and this p they do not line up with our coordinate system so we need to uh deal with those okay so let's do that and what does that look like well what that looks like is i'm going to draw in some components here and i like to do those in a different color if possible but this is i'm just going to call this f y and this one f x right we call this this this force f over here so i'm going to call this f y and f x and then similarly we're going to have some component p x and some component p y okay looks complicated and honestly it is complicated are you gonna get a question this complicated on the fe i don't know you might uh is it good to know how to do i think it is i've seen this type of question this is kind of a classic friction question a block on a plane what forces whether it's going off the plane or down the plane or staying at rest of the plane or maybe there's a pulley acting down and this thing's moving and acceleration on the plane i mean there's all sorts of crazy ways you can manipulate this problem so let's just let's let's send it home and be done with this right so let's find our components right so that's what we want to do here is we want to find our components uh so what we want to find is the components of fx equals fy equals and px equals and py equals and to do that what we want to see is or what we want to figure out is what angle shares this 15 degrees and hopefully if you draw a decent free body diagram you can see that right this angle here shows that 15 in this angle here shares it shares that 15. so once you know cosine you remember cosine is is we started this is where we started right cosines adjacent over hypotenuse right so the cosine's always adjacent to the angle right so so that's that's the way i can remember it but f x if if we know that the f y f y here is adjacent to that 15 degrees i'm going to start with fy is f times the cosine of 15 so that makes that makes f x f times the sine of 15. okay and similarly if we look at this p x value down here p x is adjacent to that 15 degree angle so p x is going to be p times the cosine of 15 and p y is going to be p times the sine of 15. so the good news is the good news is we are we have f so we can we can do these out so 1962 times uh the sine of 15 is going to be like 5 0 i'm just going to round up to 508 newtons and i could go for a fig newton right now and the i you guys are probably like eating and like drinking and like hopefully nothing like alcoholic because that makes it harder to do the fe i'm just saying but maybe you'll go celebrate after i don't know like that's not my thing but maybe it's your thing so what are we do in 1895 or maybe you just make a cake when you're all done oh man i'm getting hungry um so so so we got these four we got these values but we don't know what p is yet okay but we can solve for it so let's just go down and this also if you look at it this also looks a little bit like a method of joints if you remember method of joints what did we have this is this we essentially don't have any distances here we just we have two equations we have some of the forces in the x direction equals zero and we have some of the forces in the y direction equals zero we have two two equations here sorry my we have two equations that we need to use to solve this thing all right so let's start with some of the force in the y direction just because that one's a little bit simpler so what we have is the normal force minus py uh minus fy equals zero so those are the vertical forces okay makes some sense so the normal force minus py minus uh you know the normal force is going up all right this is going up py's going down fy is going down and uh we get one equation so what else do we get here so so if we look at in the x direction what do we get i'm going to sort of try this in the same direction we're going to start with minus ff plus px minus fx equals zero and when i'm doing these problems i try to keep them in the same order here like n and p and f all in the same order just because it actually helps when you start to do either elimination or substitution because what ultimately you're going to end up with is a formula with two equations and two unknowns right the unknowns that we have are essentially uh the normal force in in p okay so so they you know when we get this p component that adds to the normal force and we don't know what p is yet we end up with an equation uh with the system of equations that we need to solve but we can start substituting in here because we also know that the friction force equals mu times the normal force so what i can do here is i can say well this is minus mu times the normal force plus and instead of px i'm going to use p cosine 15 p cosine 15 and i'm going to say this equals f x all right i'm going to put the f x on the other side i'm not trying to do smoke and mirrors here but then what i'm going to do here is i'm going to say the normal force minus p y and instead of p y i'm going to use p sine 15. and that's going to equal fy okay so what you see all of a sudden is this is a this is a system of two equations uh two unknowns and you should be able to actually just go and solve this so i'm going to put in some i'm going to start substituting in our mu what was it was 0.35 up here so minus minus 0.35 and plus what's cosine 15 i don't know cosine 15 is is 0.9659 so 0.9659 p equals fx and fx was 508. okay what else do we get we have the normal force here which is n minus uh p times the sine of 15. so what do we get there so if we do that out if this is like 0.2588 uh p equals fy what's fy 1895 newtons okay and you know i don't have my calculator set up anymore but what i want you to see here is you can if you like on the ti 36x pro it's it's not really working wonderfully here but uh if you on the ti 36x pro there is a system of equation solver so if you do like a second tangent here and if you have this calculator you can use this to solve but a second tangent you can put in a second tangent and you can do i don't know if you can see it but you can do that's backwards a two by two linear system uh linear equation so this is going to be super helpful in when we get something like this and if i put that in here and again it this is this is kind of not working i i changed my cameras around last time because it didn't work very well for me but basically you can put that in here uh and you can solve a linear system in your calculator which is a lot easier than doing uh elimination or substitution on the test so so if you want to do elimination substitution you can otherwise yeah you can't read it but um what we get here is we essentially get that the the p force equals uh i believe it's 1337 and the end force equals 2241 newtons so if you solve you know i'm just going to do dot dot dot this is the system solver on ti 36x pro or you know you use you know elimination substitution right and you'll be able to to go ahead and solve these to get those values okay but the real crux of this is is what it's it's just like i said in the beginning right it's knowing the sum of the forces equation sum of the forces equation some basic trig to meet your geometry you put all those together and you can do statics so 100 i know if you've made it this far you care enough to actually make it and i know you'll be able to so so i i think statics is so important to start with because uh well math and statics those two i mean if you can do these basic pieces this will apply towards everything else you know and what that does honestly i love seeing people from across the world it's it's so awesome here i worked with a group uh that that at one point said you know poverty isn't lacking money poverty's lack of opportunity and one thing i love about this platform is it is it gives the opportunity for my current students for my you know new students the opportunity to learn and i totally love uh the fact that that this opportunity is available for you and you know what we'll just end it right there so so what we'll say is the the beautiful thing about this is this course is all about statics right where everything is set still and this is amazing because even though it's static you're still full of potential that wasn't really funny but i tried hey um quick question my pdf 36 isn't giving me the decimal forms so decimal forms yeah so totally um decimal forms so so decimal forms there's there's the bottom button here i don't know if you see it it's right above uh right above the enter sign so right above the enter sign this button will change from uh this button will change from uh what is it called um radical form to decimal form so if or you can go and set a cl the math print you can go to classic print in your in your options so if you either go to classic print in your options or this button literally right above the enter button will change from radical form to decimal form so i don't know if that helps you sami um but yeah so what gets the ball rolling this this is your potential right and we'll talk about dynamics at some point but what gets the ball rolling i'm come on we're doing an fe review course this is fe pass right this is what you guys are gonna get and you're gonna you're gonna go there so um you're you guys are full potential i'm so excited for you hey and um this time next week we should be doing structural analysis we'll we'll keep going and i'll post some questions here probably tomorrow the next day uh probably by thursday i'll have them up but you know keep looking and uh thanks so much for tuning in until next time keep working hard moving on upward