Wish my $2500 a semester could go to this guy instead of the professor that blabbers about how easy it was for them when they took the class 30 years ago. This guy is goated fr fr :0
I'm here after 3 years, i completed my degree about 1 year ago and now I'm preparing for a competitive exam, and boy did i miss this channel, i studied mechanics from you and i did great in it, recently i was having trouble with trusses then i remembered that you had made a couple of videos on it and after watching these two vids trusses feels so easy, god bless you brother
It's great to hear that my videos are still helping, even after you finished your degree, I'm glad you found it helpful for your exam prep! Keep up the awesome work and best wishes with your competitive exam :)
انا عربي اشكرك على جهودك هذه ولك التوفيق ان شاء الله في الدنيا والأخرة I am an Arab. I thank you for your efforts. May God grant you success in this world and the hereafter
thank you so much for existing. I promise you one day when I become a successful civil engineer I will contact you surely and thank you for every help u did throughout my engineering program. Just remember me. I am taking a screenshot of this comment.
You’re very welcome. I’m glad to hear that you’re pursuing your dream of becoming a civil engineer. I’m sure you’ll do great things in the future. I appreciate your kind words it means a lot to me. Best wishes :)
I should be paying you my college tuition How am I paying some boring dude to tell me about physics in the most unstructured way at 8am for $1800 when you just make it so simple and quick. Hats off
Great video, clear explanation of how and why to use this method. Basically a time-saver for convenience sake is what I get from it. Understanding this method also boosts your insights and understanding of statics
@@QuestionSolutions I watched like 10 of ur videos already and it's so damn informative. U got a new subscriber my friend...keep up the good work on the videos bro ✅❤️...I rlly hope one day I'll make it as a civil engineer and be able to donate cash to ur channel because I feel ur teachings are highly underrated and I wouldn't want you to stop 😭. Keep the videos coming bro...imma be here 4 a while❤️
@@OGTennyson Thank you very much! Don't worry, you will make it as a civil engineer, you'll do great. Keep up the great work, do as many questions as you can, and use every resource out there to help you succeed. Especially those office hours from professors. 👍
I'm glad I'm watching this video, I'm having my mechanics exam next week. But I feel you're a lil bit fast in explaining, seems quite difficult to catch up mentally. Like, before I process a part, you'd have gone 3 parts. Also, can you help write the equation one by one. You have real good content, but just to avoid confusion.
I try to keep these videos concise, so it doesn't take up too much time. It can be difficult, especially when being introduced to the subject for the first time. So I will keep what you said in mind, but try to watch the video a few times, or try to solve these questions by yourself first, and if you get stuck, then try to see what I did to get to the next part. These videos are meant as a supplemental tool, and not a substitution for a textbook/professor. Thank you for the feedback!
My approach to his every vid is to try to deeply understand the underlying concept and pause right at the very beginning of each example and try to think thoroughly on how to solve it myself. Give yourself some engagement on critical thinking rather than just "watching". Critical thinking will give you a solid comprehension of the topic that will stuck on your mind that would greatly help you on your exam and even after semester, you will still understand the topic. Works for me 100% I usually answer his examples on all statics vids 95% before he explains. Try it urself :)
That's right, but even more easier to remember is to keep in mind that any force starting where the moment is calculated, cannot create a moment at that point. The same as force BC.
Really really great content..... I also want to learn such kind of animation ....can you please tell me how you do it? It would be great help to me....love from India.
@@QuestionSolutions thanks a lot... The very first channel which I shared on my social networks.... Keep doing...... You will certainly hit the market.. 😀
@@mohammadashu8984 You're very welcome. Sometimes, Udemy has free courses on after effects as well, so keep an eye out. Thank you so much for the share, I appreciate it. 😁
In real life, no, because they are there to help with rigidity and stability. Sometimes, they will carry a force if the conditions of the bridge changes. lastly, they add to the aesthetics as well. People don't like seeing parts "missing" off of a bridge 😅
@@QuestionSolutions Hello! How come F bf is equal to 0 when the x component of Fbg is acting on BF and also 10 kN is also acting at BF? I want a solution to find Fbf using moment method.
@@rpian1999 So BF is a zero force member, that means it doesn't carry a force. The forces applied at B are carried by the other members, not BF. If you want, you can look at point B and do the equations, you will still get 0 for member BF.
@@ElCrankoPunko Unfortunately, Strength of Materials won't be done for some time. The next topic is thermodynamics. I think there are a lot of videos on UA-cam about material science, I hope they can be helpful.
I really like you videos, you explain better than my professor 😅🔥. I have a question, why did you cut the trusses like this in 10:00, why didn't you cut just vertically? It will give you the same number of unknowns
If you cut the truss straight down, you will need to work on the left side of the truss. But we figured out the reactions on the right side, so cutting it the way I did means you work on the right side of the truss. You can cut straight down though, it's totally up to you.
Great video, but I got a question at 5:11, shouldn't the forces 2kN and 9.5kN only apply moment on the components perpendicular to the truss AH, because you also did that for Fbg?
The 2kN force and the 9.5kN force are perpendicular with respect to where the moment is calculated. If you imagine your finger at the location where the 2kN force is, and push, will the truss spin about our moment point? It would right? So it will create a moment. I think you're assuming we are doing the moment about the whole member AH, which is not true. We are calculating the moment about the pink dot, just point H. When we calculated the moment about point A, only the y-component of force BG creates a moment. But notice that the 2kN and 9.5kN force only have y-components, they are vertical, so it will create a moment about point H.
I don't understand how do you know the direction of the force based of the positive/negative value? In the first example the value was positive 2 times in a row. In the first situation it was compression, in the second it was tension. How did you assumed those?
Please kindly watch this video first: ua-cam.com/video/_rK02neOF18/v-deo.html Especially the introduction, where I explain how to determine whether things are in tension or compression. If, after you have watched it, and still have trouble, please let me know and I will do my best to help you out.
You are incorrect in thinking that sine is for x or sine is for y components or cos is for x components or y components. It doesn't work that way, and you are not alone, countless students make the same mistake on exams, and points get taken off. Please watch this video, it's less than 60 seconds and it will help you not make the same mistake I see so many students making. ua-cam.com/users/shortsvynnKlJD_Jo?feature=share
Why during 4:13 your value of 2kN that is downward is considered negative but then in 4:55 your value of 5kN which is also downward is considered as positive. This confuses me a lot
So my conclusion is that downward and left force is considered as negative value and vice versa. But in 5:11, the force BC which is going right is considered positive?? Why is that? Mind explain to me
So it's not the direction of the force that matters when it comes to moments. It is the direction of the moment created by the force that matters. So let's look at the 2kN force at 4:13. We assumed clockwise to be positive. Now imagine the whole thing can rotate about point B. When the 2kN force is applied, which way will it turn? It will turn counter-clockwise. This is why it's negative. Now let's look at the 5kN force at 4:53. Now the moment is about point A, and again, we took clockwise to be positive. When the 5kN force is applied, which way will the truss rotate? It will turn clockwise, so it's positive. This is all to do with moments and you might be missing those fundamentals. If you have the time, please watch this video: ua-cam.com/video/QNNnPZ68STI/v-deo.html @@syamilariq2222
Where did you get those forces from? So when you look at a joint, you only look at that joint, you don't care about what happens at the next joint over. There are no other forces applied at joint F other than vertical forces. I think what you're doing is translating forces from other joints to this joint, but remember, when we look at a joint, it's isolated. That's why you draw a separate coordinate system about that joint. At 7:33, focus on the right side of the screen and look at the forces at that joint. :)
Thanks question solution for this video it really helped me with my quiz tomorrow just a quick question wouldn’t it be easier to take the moment at B first then take the sum of Fy to get Bg isn’t that easier?
Please kindly watch this video: ua-cam.com/video/QNNnPZ68STI/v-deo.html So initially, when writing a moment equation, I pick clockwise to be positive (this is an assumption). So you're looking to see, if we apply a force at a location, will that member turn clockwise about the point where we are calculating the moment at, or counter clockwise. If it's clockwise, its positive, if its counter clockwise, its negative (because it's opposite to our assumption). The first 2 examples in the link I provided will help out a lot with directions.
Throughout all the Truss videos you talk about assuming the way the forces are acting and just back tracking if they are negative. Is there an intuitive way to always guess the direction correctly?
The intuition comes from doing a bunch of questions, in other words, experience. By the end, you should be able to look at a truss and know which way the forces are acting and be able to locate 0-force members. Also, it's not really backtracking per say when you guess wrong, you just change the direction of the arrow and make your value positive. The magnitude of the force is always correct, it's just a matter of whether it's positive or negative.
@QuestionsSolutions gotcha its just on my exam, we have to draw a free body diagram for a grade, and its online so its kind of a pain in the ass. So I guess a better strategy would be to mess with the free body diagram last and solve the problem first so I already know what direction the forces should be before putting them all in.
That's a support reaction. So it's a pin support there, and that means there is an Ay component along with an Ax component. However, the ax component do not create a moment about point E.
6:28 in your explanation you said that only the x component of force BG can cause a moment about D but on the working you used Sin instead of Cos. May you please clarify
So my guess is, you assume cos is for x and sine is for y. This is fundamentally wrong, and you must, absolutely must, remove that idea from your head. You have to look from the perspective of the angle every single time. Here, the opposite side to the angle gives us the x-component, and sine is opposite over hypotenuse. So you need to use sine. To get the y-component, which is now adjacent to the angle, you need to use cosine. See this video, it's less than 60 seconds and you will see what I mean: ua-cam.com/users/shortsvynnKlJD_Jo?feature=share
At 6:03 you have set the 5kN force at point E to be going in the positive direction and F(GH) to be negative for finding the moment around point B. Can you elaborate on the best way to choose the direction of the forces when setting up the moment equation? I would have set it to -5kN and +F(GH) based on the direction of the forces pictured and gotten the incorrect answer.
It doesn't make any difference as to how you set the directions of your forces. You will get the same answer. The reason you are getting incorrect answers is because you're writing your moment equation incorrectly. So the direction of force doesn't matter, it's whether it creates a clockwise or a counter-clockwise moment that matters. Here, the 5kN force is pointing down, that doesn't mean it's negative. What we look for, is to see whether that force would turn the truss clockwise or counter-clockwise. Imagine you hold the truss between your fingers at B, and we apply a force exactly where the 5kN force is, downwards. How would the truss turn? It would turn clockwise, right? Now notice next to the big sigma sign, we show a clockwise positive sign. That means we assume any force that creates a clockwise moment to be positive. So since the 5kN force makes the truss go clockwise, it's going to be a positive moment. Now look at force FG. If we apply that force, how would the truss turn between your fingers? It would go counter-clockwise. So it'll be negative. When you write moment equations, the directions of the forces do not determine whether something is negative or positive, it's how this force will act upon the object itself, and what sort of moment it creates, that determines the positive or negative sign. I hope that helps, if you're still confused, please see: ua-cam.com/video/QNNnPZ68STI/v-deo.html
Great video, thanks! Just a quick question, I tried solving the 4th problem but using the other side of the cut and the answer is different. Is my answer wrong or is it possible to have different answers when using different equations?
Regardless of the cut you use, or which side you use, you should end up with the same answers. This can be verified by solving the whole truss. There is probably some numerical error. :)
Hey Again! Well explained like always. I'm wondering why at 2:44 you make all member force face away from system except one and assume that it's compression. Is it just based on overlook and common sense? Thx, again!
You can pretty much cut it anyway you want (as long as you have enough info to solve it). The cut should be made to figure out the unknowns you want, so try to cut the ones you need to find. So yes, you can cut it diagonally 👍
Only the x-component of FGH can create a moment about point B, since the y-component's line of action goes through point B. So looking from the perspective of the angle, the adjacent side gives us the x-component, so we use cosine. If it's a matter of choosing cosine or sine, please see: ua-cam.com/users/shortsvynnKlJD_Jo?feature=share
i have a question. do we always have to assume that the rotation is clock wise and positive? would changing that matter? im confused that if F turns out to be negative, my thought process is to change Tension to Compression (or vice versa) or change the rotation to counterclockwise (or clockwise)
I think you are confusing writing moment equations with forces in truss members. You can pick whatever direction you want for the moment equation to be positive, so clockwise or counter-clockwise, you will end up with the same answer. What determines if a member is in compression or tension is whether a force is coming towards the pin or leaving the pin.
hello i have a question at 2:04, why do you calculate Ay with a moment, Can't we not just work with the sum of y=0 -> Ay-2-5-5-5-2=0 -> Ay=19kN , i know this isn't the same answer but why does this not work and why do we need to use the moment
it it because the is also a single force Ey in E, so Ey + Ay=19, PS: i have been watching your whole series on statics and i these kind of videos have never been so helpfull as your channel, the fact that you answer to every question shows your passion keep up the good work!!
@@toondebeule8377 Yes, that's correct. We don't know the value of Ey, so you can't use a summation. Instead, we take a moment about E, which eliminates EY, and we can the directly solve for Ay. Also, thank you very much for your kind words. Best wishes with your studies!
and i also have another question, in my class i have to work with counter clock wise moment is positive what does this change to the solution, when i do this my forces are negative is this possible? so when i calculate Ay i get -9,5kN but in the next step it doesn't work because the 2kN and 9,5kN are in the same direction, what would be your solution?
@@toondebeule8377 The direction you choose for positives or negatives does not matter in 2D problems. You will always get the same answer regardless of the direction you pick. It's completely up to you. Please see: ua-cam.com/users/shortsP029mqnp4XY?feature=share
i have a question, in the first question when you were calculating moment about point H, why did you include the 9.5(2) and -2(2)? I thought the only force perpendicular to point H would be fBC and the x component of BG? Please clarify, thanks!!!
So both the 9.5kN force and the 2kN force are perpendicular to point H. Imagine you move those forces up to be aligned with point H, so straight up. Now you can see it a bit easier that they are indeed perpendicular. Another way to see it is to realize these forces are vertical, and will cause the object to rotate about point H, so it must create a moment. Force FBC and the x-component of BG is NOT perpendicular to point H. That's why they aren't in the equation. I think you are confusing perpendicular with parallel. Please draw a big diagram on a sheet of paper and really carefully look at what forces are perpendicular.
For the question at 5:25, why does the force in member BA not equal the reaction force from point A? I found a reaction force from joint A of 45kN when taking moments about point G, but the force in member BA is 50kN.
The cut was made above the reactions so they were not needed. No calculations were made to figure out the reactions at A, or G. Also, where did you get that BA is equal to 50kN? At 7:07, we find that BA = 45kN.
@@QuestionSolutions Thanks for your quick response! I wanted to verify that the force in member AB equaled the vertical reaction force at A. I understand that it is not directly relevant to the section method. I just checked my calculations for some of forces in the Y axis, and I totally forgot to include the 5kN on point E. Now I have the correct solution for AB. Thanks again for the brilliant video!
For question one I know u take thr moment at Point B but if we took the moment at point H, we would have two unknowns, Fbc and Fbg horizontal component both causing a moment. How would u solve it then?
The whole point of taking a moment about a specific point is to eliminate as many unknowns as possible. If you have 2 unknowns after writing a moment equation, then you'd need another equation, probably for vertical forces or horizontal forces. Try to take moments about points that lead to direct answers.
Thanks for the video! But I have a question for finding force BF in the second problem. If we would have isolated joint B, equilibrium along forces at x would be 10kN - 45Kcos(45) - F(bf) = 0. Hence i thought that force BF would not be zero. Im confused how this works ://
I encourage you to actually isolate for joint B and solve this problem. For your equation, you should have 4 forces. The 10kN, BG, BF, and BE for horizontal components (along x).
You can find the values using any moment location but try to use simple ones since you want to make your life as easy as possible. So go for the one that eliminates the most unknowns or leads to a direct force value.
Really nice vedio Sir I also want to learn such kind of annimation ....can you so please tell me Or provide some vedio leink .I will be very thankful of you
Their lines of action go through point B. The same as the 5 kN force, and the y-component of force HG. If you need a refresh on lines of action, please kindly take a look at this video: ua-cam.com/video/QNNnPZ68STI/v-deo.html
Glad I found these videos of yours. It was a very big help knowing I am having a hard time with our subjects right now because it was not explained properly but yours is very concised and well elaborated🤍
@@QuestionSolutions i mean, for example, in this video, after you got the force DC, you used summation of forces to get force HI, when i thought i was supposed to use moment equation and i thought i would have to do it at point F to get rid of the force HC and thus getting me the force HI
@@JohnRovicAlejandrino You definitely can use a moment equation to get force HI, but my assumption is, you're not writing a proper moment equation. There will be no difference in the answer. They all will result in the same answer. Also, once you already use a moment equation to find a force, you can do the rest with just summation of forces, there isn't a need to write a moment equation most of the time, but if you did, you can still get the same answer.
@@JohnRovicAlejandrino Please write your moment equation so I can go over it and let you know what happened. Even from the answer, we can easily see there is a mistake, you got a force of 5 kN, when all of the forces on the truss are 40s, 50s, etc. So there is an error in your moment equation and I will help you out if you write it here.
Check me if I am wrong,you said if we chose the moment at 1 point then we can eliminate the action force acting on it,but I don understand in 2nd example at 6.30 ,which moment at point D ,the 5 kN still considered?Tq
If you're referring to the top most force, applied at D, that is NOT considered when writing a moment about D. Please carefully look at the color coded boxes and lines, each corresponding to which moment is being calculated. The 5 kN used for the calculation is the one applied at point E.
If you use a section that you can solve without needing support reactions, then you don't need it. For example, at 5:35, notice we cut the supports off, and then we had enough givens to solve the top values. But notice at 8:24, even after the cut, we still needed the reaction at roller F. So in most cases, the only time you don't need to know support reactions is if you make a cut and on the side you selected, there are no supports.
8:42 i'm confused at this part. i thought sin DC should be eliminated because it will go thru pin H, and instead use cos DC? or does it depend on where the angle is placed?
So you have to always look at it from the perspective of the angle when doing trigonometry. Sine is going to give us the opposite side (opposite to the angle), which in this case would give us the x-component of force F_DC (horizontal line). That will create a moment since it's line of action doesn't go through point H. Cosine on the other hand will give us the adjacent component, (y-component, vertical line), which will go through point H, so it's eliminated.
@@QuestionSolutions ahh so we're getting the x component here but we just used sin because of the angle. I'm used seeing sin as y, and cos as x so i was confused😆. Thank u very much
You are wrong, do not fall into that pitfall. I am not sure why students tend to this think way, but they are NOT related to Fy and Fx. It's all based on the side opposite to the angle. Please take a look at this video: ua-cam.com/video/NrL5d-2CabQ/v-deo.html
Thanks for the video, you explained this so well in a way my professor never could! 😭😭 Just a quick question though. At 6:16, you say that only the x-component of force BG creates a moment about point D but in the equation it says sin45.. I thought it would cos45 since it's the x-component. Can you please explain this part? It's the only thing I'm confused about
You're very welcome. You have to make sure you look at the force from the angle. So if you were at the angle, and you looked from that perspective, the opposite side to the angle would be sine, which is the x-component, and the adjacent side is the y-component, which would be cosine. Or are you saying that sine is correlated with x-components and cosine with y-components? If so, this isn't true, please don't think that, it is completely incorrect. If you need a video on components with sine and cosine, please see this video, especially the first example: ua-cam.com/video/NrL5d-2CabQ/v-deo.html
@@QuestionSolutions Omg I thought that x-components were always associated with cosine and that y-components were always associated with sine.. I'm going to check out the video that you linked, thank you so much!! 🙏🏽
@@joonfanatic2140 Thank you for asking this question. I was stumped for a while as well and also always assumed X was Cos and Y was Sin. Game changer for me!
@@AlexanderPerez-oe1gu Same here, it definitely changed my thought process and made me understand things correctly! I'm glad it helped you and hopefully you watched the other video he linked in the reply. I wish you the best in your studies 😄
@@joonfanatic2140 Thanks, yes I saw the referenced video, and totally made sense. Completely changed my perception of how to approach the x and Y components.
Regardless of where you start, you will always get the same answers. There is probably a numerical error in your solution or an error in setup of the equations. The forces in each member can't change based on where we start to solve the problem. :)
Quick question: at 2:04, how do we immediately know that there won't be any reaction on the horizontal axis? Like I can see that none of the applied forces have horizontal components, but couldn't AH (for example) cause a horizontal reaction from the support? Thanks!
So when you're solving for the reactions at a support, we don't care about internal members. They don't make a difference since we think of the object as a whole. So all the purple forces are external forces, but we have no forces applied in the horizontal direction. Let's say we had a horizontal force applied at point F, in that case, we would have an AX reaction countering the force applied at F. It's the same if we had a simple beam resting on top of supports. We just solve based on external forces, what happens inside the beam doesn't matter. Another way to think about it is to realize that while there is a force in member AH, that force goes from H to A and A to H, there by causing equilibrium. So to recap, when you are finding supports, internal forces do not matter, only external ones.
At 9:08, there is a typo. F_HI = 42.5 NOT 45.2.
I thought I was losing my mind when I solved it myself and watched it back, thanks for clarifying, and thanks for the great video.
@@remyfru You're very welcome :)
@@QuestionSolutions thanks
My professor's lecture video was 45 minutes long, but this 11-minute video was elaborated well. Thank you.
You're very welcome! I try to make them as concise as possible.
Wish my $2500 a semester could go to this guy instead of the professor that blabbers about how easy it was for them when they took the class 30 years ago. This guy is goated fr fr :0
@@zach.frederickMine is $3000 a semester and our professor just blabbers useless stuffs 🥲
My prof sent a video lecture that's like 8 hrs
How'd the rest of your class go?
I'm here after 3 years, i completed my degree about 1 year ago and now I'm preparing for a competitive exam, and boy did i miss this channel, i studied mechanics from you and i did great in it, recently i was having trouble with trusses then i remembered that you had made a couple of videos on it and after watching these two vids trusses feels so easy, god bless you brother
It's great to hear that my videos are still helping, even after you finished your degree, I'm glad you found it helpful for your exam prep! Keep up the awesome work and best wishes with your competitive exam :)
انا عربي اشكرك على جهودك هذه ولك التوفيق ان شاء الله في الدنيا والأخرة
I am an Arab. I thank you for your efforts. May God grant you success in this world and the hereafter
You're very welcome! keep up the awesome work with your studies.
thank you so much for existing. I promise you one day when I become a successful civil engineer I will contact you surely and thank you for every help u did throughout my engineering program. Just remember me. I am taking a screenshot of this comment.
You’re very welcome. I’m glad to hear that you’re pursuing your dream of becoming a civil engineer. I’m sure you’ll do great things in the future. I appreciate your kind words it means a lot to me. Best wishes :)
Thank you so much. You don't know how much you've helped me tonight ♥️
You're very welcome! I am really glad to hear this helped :) ♥️
Thats what she said💀
I should be paying you my college tuition
How am I paying some boring dude to tell me about physics in the most unstructured way at 8am for $1800 when you just make it so simple and quick.
Hats off
Thank you very much. I hope you do amazingly on your courses and I wish you the best! Keep up the great work :)
Great video, clear explanation of how and why to use this method. Basically a time-saver for convenience sake is what I get from it. Understanding this method also boosts your insights and understanding of statics
Glad it was helpful! Keep up the great work and best wishes with your studies.
Method of sections? More like "Magnificent and knowledgeable explanations!"
Haha, thanks! :)
Oh My God . This Lockdown means free time and your channel means free great quality explanation . Thank u so much 🌹. Enjoying Mechanics . ❤️🥰
Glad to hear you enjoy mechanics. ❤️ You're very welcome and I hope you learn lots!
It is the best channel I have seen. Greetings from Iraq 🥀♥️🎉
Thank you very much! :)
Xwa qazay aw sawtau basha ❤😂 This is more useful than what I got from my teacher.
Glad this is useful! Keep up the great work.👍
This was so helpful! Thank you for your easy explanations
You're very welcome. Glad to hear it was helpful
thank you so much for this, just the right time i found your channel before our quiz tomorrow!
Glad I could help! I wish you the best on your quiz tomorrow :)
Has anyone ever told u ur a life saver?🥺🥺
I think I heard it about 3 times now, from very kind people like you leaving nice comments. Thank you, it made my day! Best wishes with your studies.
@@QuestionSolutions thank you!... and ur very welcome ❤️
@@OGTennyson ❤
@@QuestionSolutions I watched like 10 of ur videos already and it's so damn informative. U got a new subscriber my friend...keep up the good work on the videos bro ✅❤️...I rlly hope one day I'll make it as a civil engineer and be able to donate cash to ur channel because I feel ur teachings are highly underrated and I wouldn't want you to stop 😭. Keep the videos coming bro...imma be here 4 a while❤️
@@OGTennyson Thank you very much! Don't worry, you will make it as a civil engineer, you'll do great. Keep up the great work, do as many questions as you can, and use every resource out there to help you succeed. Especially those office hours from professors. 👍
huge video, got a silly little quiz tomorrow, this will definitely help!
Glad to hear. Best wishes with your quiz tomorrow!
Im happy you engage with the people in the comments. It seems all my answers have been answered! Thank you.
Thank you, I try my best to make sure I answer all the comments I get. :)
Thanks very much,qm from Ethiopian 🇪🇹
Keep it up
You're very welcome!
The example my professor gave only had 1 external force, so i didn't know that I had to calculate the moments of the other downward forces
Thank you
You're very welcome. I try to pick at least 3 examples to remove any confusions in students. :)
@9:12 Force in HI member would be 42.5 kN
Yup, there is a typo. Thank you for pointing it out.
Really helpful. The visual is really helpful for my learning!
I am really glad to hear that!
Such a good video. You just earned a subscription. Keep them coming!
Thanks for the sub!
Thank you,❤❤❤❤❤❤I am excited by your explanation I was a blindness man b/c I have not seen this video.
That's great to hear! I'm glad it helped. :)
you ae the best problem solver 10Q
You're welcome.
I'm glad I'm watching this video, I'm having my mechanics exam next week. But I feel you're a lil bit fast in explaining, seems quite difficult to catch up mentally. Like, before I process a part, you'd have gone 3 parts. Also, can you help write the equation one by one. You have real good content, but just to avoid confusion.
I try to keep these videos concise, so it doesn't take up too much time. It can be difficult, especially when being introduced to the subject for the first time. So I will keep what you said in mind, but try to watch the video a few times, or try to solve these questions by yourself first, and if you get stuck, then try to see what I did to get to the next part. These videos are meant as a supplemental tool, and not a substitution for a textbook/professor. Thank you for the feedback!
My approach to his every vid is to try to deeply understand the underlying concept and pause right at the very beginning of each example and try to think thoroughly on how to solve it myself. Give yourself some engagement on critical thinking rather than just "watching". Critical thinking will give you a solid comprehension of the topic that will stuck on your mind that would greatly help you on your exam and even after semester, you will still understand the topic. Works for me 100% I usually answer his examples on all statics vids 95% before he explains. Try it urself :)
Another amazing video, thanks for the extra help with studying this stuff! 🙌
My pleasure! Keep up the good work with your studies.
For anyone wondering why Force BG was eliminated 3:01, it's because it's vertical component goes through B and so does its horizontal component
That's right, but even more easier to remember is to keep in mind that any force starting where the moment is calculated, cannot create a moment at that point. The same as force BC.
Really really great content..... I also want to learn such kind of animation ....can you please tell me how you do it? It would be great help to me....love from India.
I use after effects for animations. There are tons of really good tutorials on youtube, so please take a look :)
@@QuestionSolutions thanks a lot... The very first channel which I shared on my social networks.... Keep doing...... You will certainly hit the market.. 😀
@@mohammadashu8984 You're very welcome. Sometimes, Udemy has free courses on after effects as well, so keep an eye out. Thank you so much for the share, I appreciate it. 😁
Thank you so much, while cutting section, can I make a slanted (diagonal) cut to avoid solving for member that I really don't need for problem?
Yes, you can cut vertically, horizontally, diagonally, etc. 👍
@@QuestionSolutions Thank you
Super Helpful
Glad to hear :)
For the 3rd problem. The Force HI is 42.5.
Thank me later.
👍
yeah.. this should be pinned. It can also be obtained by applying summation of moment at A.
thanks i thought i got it wrong
For the vertical truss, does that mean member BF doesn't need to be there since the force is 0?
In real life, no, because they are there to help with rigidity and stability. Sometimes, they will carry a force if the conditions of the bridge changes. lastly, they add to the aesthetics as well. People don't like seeing parts "missing" off of a bridge 😅
@@QuestionSolutions Hello! How come F bf is equal to 0 when the x component of Fbg is acting on BF and also 10 kN is also acting at BF?
I want a solution to find Fbf using moment method.
@@rpian1999 Please give me a timestamp so I know where to look, thanks!
@@QuestionSolutions 7:38 time in that video where Fbf is being determined.
@@rpian1999 So BF is a zero force member, that means it doesn't carry a force. The forces applied at B are carried by the other members, not BF. If you want, you can look at point B and do the equations, you will still get 0 for member BF.
Thanks man, you're the best
You're very welcome!
Thank 👍you for your excellent presentation
You're very welcome!
Amazing Video, All doubts cleared
Glad to hear that :)
You are my hero! Could you also make a playlist about Strength of Materials?
Yes, that is also on my list to do :)
@@QuestionSolutions pls do it soon. love your videos btw
@@ElCrankoPunko Unfortunately, Strength of Materials won't be done for some time. The next topic is thermodynamics. I think there are a lot of videos on UA-cam about material science, I hope they can be helpful.
@@QuestionSolutions cool man. eagerly waiting for thermodynamics videos :)
@@ElCrankoPunko :)
This man is a legend
Thank you very much!
God bless you! your videos are extremely helpful! Thank you so much!
You're very welcome! :)
thank you for this, the best explanation so far.
You're very welcome! Best wishes with your studies :)
Watching this before exam is worth :)
Best wishes with your exam! :)
Happy new year everyone🥳, now back studying😆
Happy New Year! I hope this is an amazing year for you :)
I really like you videos, you explain better than my professor 😅🔥. I have a question, why did you cut the trusses like this in 10:00, why didn't you cut just vertically? It will give you the same number of unknowns
If you cut the truss straight down, you will need to work on the left side of the truss. But we figured out the reactions on the right side, so cutting it the way I did means you work on the right side of the truss. You can cut straight down though, it's totally up to you.
@@QuestionSolutions oh I understand now. Thank you for Your precious time❤
@@prog_ahmed No worries, keep up the good work!
Absolutly the best truss video woooooow great
Thank you, I hope it was helpful to you. Best wishes with your studies!
You are the best! Thank you so much brother
Happy to help and best wishes with your studies!
Thank you soo much this video really helped me !!!
Really glad to hear that! Keep up the great work :)
YOU ARE A FUCKING HERO DUDE!!!
Thank you very much!
Great video, but I got a question at 5:11, shouldn't the forces 2kN and 9.5kN only apply moment on the components perpendicular to the truss AH, because you also did that for Fbg?
The 2kN force and the 9.5kN force are perpendicular with respect to where the moment is calculated. If you imagine your finger at the location where the 2kN force is, and push, will the truss spin about our moment point? It would right? So it will create a moment. I think you're assuming we are doing the moment about the whole member AH, which is not true. We are calculating the moment about the pink dot, just point H. When we calculated the moment about point A, only the y-component of force BG creates a moment. But notice that the 2kN and 9.5kN force only have y-components, they are vertical, so it will create a moment about point H.
thank you sir for such a good explaination
You are very welcome! I hope it was helpful to you.
I don't understand how do you know the direction of the force based of the positive/negative value? In the first example the value was positive 2 times in a row. In the first situation it was compression, in the second it was tension. How did you assumed those?
Please kindly watch this video first: ua-cam.com/video/_rK02neOF18/v-deo.html Especially the introduction, where I explain how to determine whether things are in tension or compression. If, after you have watched it, and still have trouble, please let me know and I will do my best to help you out.
Thank you so much
But there was a litt'l mistake in the 3rd problem; you use sine for the X components and cosine for the Y components
You are incorrect in thinking that sine is for x or sine is for y components or cos is for x components or y components. It doesn't work that way, and you are not alone, countless students make the same mistake on exams, and points get taken off. Please watch this video, it's less than 60 seconds and it will help you not make the same mistake I see so many students making. ua-cam.com/users/shortsvynnKlJD_Jo?feature=share
inspiring solution for real
Thanks!
Thank you for ur help 🤗👏
You're very welcome!
Why during 4:13 your value of 2kN that is downward is considered negative but then in 4:55 your value of 5kN which is also downward is considered as positive. This confuses me a lot
So my conclusion is that downward and left force is considered as negative value and vice versa. But in 5:11, the force BC which is going right is considered positive?? Why is that? Mind explain to me
So it's not the direction of the force that matters when it comes to moments. It is the direction of the moment created by the force that matters.
So let's look at the 2kN force at 4:13. We assumed clockwise to be positive. Now imagine the whole thing can rotate about point B. When the 2kN force is applied, which way will it turn? It will turn counter-clockwise. This is why it's negative.
Now let's look at the 5kN force at 4:53. Now the moment is about point A, and again, we took clockwise to be positive. When the 5kN force is applied, which way will the truss rotate? It will turn clockwise, so it's positive.
This is all to do with moments and you might be missing those fundamentals. If you have the time, please watch this video: ua-cam.com/video/QNNnPZ68STI/v-deo.html @@syamilariq2222
7:33
can you explain how ?
we have
3 other forces in the x direction
10 + 10 + 5 + BF = 0
BF = - 25
Where did you get those forces from? So when you look at a joint, you only look at that joint, you don't care about what happens at the next joint over. There are no other forces applied at joint F other than vertical forces. I think what you're doing is translating forces from other joints to this joint, but remember, when we look at a joint, it's isolated. That's why you draw a separate coordinate system about that joint. At 7:33, focus on the right side of the screen and look at the forces at that joint. :)
@@QuestionSolutions ❤
@@mastfamastfa1256 ❤
You is the good teacher , god bless you inshallah 👏🏻🤍
Thank you very much, I really appreciate it.
Thanks question solution for this video it really helped me with my quiz tomorrow just a quick question wouldn’t it be easier to take the moment at B first then take the sum of Fy to get Bg isn’t that easier?
You're very welcome. Could you please give me a timestamp so I know where you're referring to, then I can take a look. Thanks!
Just wondering for moments how do you determine the direction for them?
Please kindly watch this video: ua-cam.com/video/QNNnPZ68STI/v-deo.html
So initially, when writing a moment equation, I pick clockwise to be positive (this is an assumption). So you're looking to see, if we apply a force at a location, will that member turn clockwise about the point where we are calculating the moment at, or counter clockwise. If it's clockwise, its positive, if its counter clockwise, its negative (because it's opposite to our assumption).
The first 2 examples in the link I provided will help out a lot with directions.
thank you needed this
You're very welcome! :)
how do you assume the direction of forces, like how do you determine where will the arrow point at? btw great vids man very helpful
That's completely up to you. If you watch this video, I go over how that works: ua-cam.com/video/_rK02neOF18/v-deo.html
You are the best!!!
Many thanks :)
Top much explanation sir❤
Thank you very much! ❤
Throughout all the Truss videos you talk about assuming the way the forces are acting and just back tracking if they are negative. Is there an intuitive way to always guess the direction correctly?
The intuition comes from doing a bunch of questions, in other words, experience. By the end, you should be able to look at a truss and know which way the forces are acting and be able to locate 0-force members. Also, it's not really backtracking per say when you guess wrong, you just change the direction of the arrow and make your value positive. The magnitude of the force is always correct, it's just a matter of whether it's positive or negative.
@QuestionsSolutions gotcha its just on my exam, we have to draw a free body diagram for a grade, and its online so its kind of a pain in the ass. So I guess a better strategy would be to mess with the free body diagram last and solve the problem first so I already know what direction the forces should be before putting them all in.
@@MrStepBro Ah, yes, it probably would be better to do it the way you mentioned.
At 2:00 why there's a two forces in joint A, there's already a 2kn downward load why we have to put Ay in solving moment
That's a support reaction. So it's a pin support there, and that means there is an Ay component along with an Ax component. However, the ax component do not create a moment about point E.
6:28 in your explanation you said that only the x component of force BG can cause a moment about D but on the working you used Sin instead of Cos. May you please clarify
So my guess is, you assume cos is for x and sine is for y. This is fundamentally wrong, and you must, absolutely must, remove that idea from your head. You have to look from the perspective of the angle every single time. Here, the opposite side to the angle gives us the x-component, and sine is opposite over hypotenuse. So you need to use sine. To get the y-component, which is now adjacent to the angle, you need to use cosine. See this video, it's less than 60 seconds and you will see what I mean: ua-cam.com/users/shortsvynnKlJD_Jo?feature=share
Noted thank you @@QuestionSolutions
@@ZexMutsawu-nw2ov Awesome. Best wishes with your studies. Let me know if you need any further clarifications.
At 6:03 you have set the 5kN force at point E to be going in the positive direction and F(GH) to be negative for finding the moment around point B. Can you elaborate on the best way to choose the direction of the forces when setting up the moment equation? I would have set it to -5kN and +F(GH) based on the direction of the forces pictured and gotten the incorrect answer.
It doesn't make any difference as to how you set the directions of your forces. You will get the same answer. The reason you are getting incorrect answers is because you're writing your moment equation incorrectly. So the direction of force doesn't matter, it's whether it creates a clockwise or a counter-clockwise moment that matters. Here, the 5kN force is pointing down, that doesn't mean it's negative. What we look for, is to see whether that force would turn the truss clockwise or counter-clockwise. Imagine you hold the truss between your fingers at B, and we apply a force exactly where the 5kN force is, downwards. How would the truss turn? It would turn clockwise, right? Now notice next to the big sigma sign, we show a clockwise positive sign. That means we assume any force that creates a clockwise moment to be positive. So since the 5kN force makes the truss go clockwise, it's going to be a positive moment. Now look at force FG. If we apply that force, how would the truss turn between your fingers? It would go counter-clockwise. So it'll be negative. When you write moment equations, the directions of the forces do not determine whether something is negative or positive, it's how this force will act upon the object itself, and what sort of moment it creates, that determines the positive or negative sign. I hope that helps, if you're still confused, please see: ua-cam.com/video/QNNnPZ68STI/v-deo.html
Great video, thanks! Just a quick question, I tried solving the 4th problem but using the other side of the cut and the answer is different. Is my answer wrong or is it possible to have different answers when using different equations?
Regardless of the cut you use, or which side you use, you should end up with the same answers. This can be verified by solving the whole truss. There is probably some numerical error. :)
Thank you i understod verey well
Glad to hear!
Hey Again! Well explained like always.
I'm wondering why at 2:44 you make all member force face away from system except one and assume that it's compression. Is it just based on overlook and common sense?
Thx, again!
and then you do the same in 5:44 with the second truss. I wonder :o ..
@@kjartanalmar When you do enough questions, you can pretty much "see" where the force will go :)
Hello, thanks a lot for this video. It was very helpful. I have a question: 5:35 can we cut the truss through BF, AG, AB? basically diagonally
You can pretty much cut it anyway you want (as long as you have enough info to solve it). The cut should be made to figure out the unknowns you want, so try to cut the ones you need to find. So yes, you can cut it diagonally 👍
@@QuestionSolutions ohh I see, thanks a lot and THANK YOU SO MUCH FOR THE INSTANT REPLY I admire your dedication💙💙💙
@@sapphireblue9209 You're very welcome. Best wishes with your studies!
@@QuestionSolutions thank you very much!
at 3:57, why the FGHcos36.87(1.5) is not Sin. Thank You so much
Only the x-component of FGH can create a moment about point B, since the y-component's line of action goes through point B. So looking from the perspective of the angle, the adjacent side gives us the x-component, so we use cosine. If it's a matter of choosing cosine or sine, please see: ua-cam.com/users/shortsvynnKlJD_Jo?feature=share
i have a question. do we always have to assume that the rotation is clock wise and positive? would changing that matter? im confused that if F turns out to be negative, my thought process is to change Tension to Compression (or vice versa) or change the rotation to counterclockwise (or clockwise)
I think you are confusing writing moment equations with forces in truss members. You can pick whatever direction you want for the moment equation to be positive, so clockwise or counter-clockwise, you will end up with the same answer. What determines if a member is in compression or tension is whether a force is coming towards the pin or leaving the pin.
@@QuestionSolutions oh i see, i get it now. i tried changing the signs at 4:10 and i still arrived at positive 12.5, thanks!
@@punpunyana Awesome :)
@10:37 shouldn't the answers be in kN not N as the original forces are in kN? Great videos and easy to understand and work through to show workings.
Yes, it's a typo. 👍
well explained
Thank you very much!
hello i have a question at 2:04, why do you calculate Ay with a moment, Can't we not just work with the sum of y=0 -> Ay-2-5-5-5-2=0 -> Ay=19kN , i know this isn't the same answer but why does this not work and why do we need to use the moment
it it because the is also a single force Ey in E, so Ey + Ay=19, PS: i have been watching your whole series on statics and i these kind of videos have never been so helpfull as your channel, the fact that you answer to every question shows your passion keep up the good work!!
@@toondebeule8377 Yes, that's correct. We don't know the value of Ey, so you can't use a summation. Instead, we take a moment about E, which eliminates EY, and we can the directly solve for Ay.
Also, thank you very much for your kind words. Best wishes with your studies!
and i also have another question, in my class i have to work with counter clock wise moment is positive what does this change to the solution, when i do this my forces are negative is this possible? so when i calculate Ay i get -9,5kN but in the next step it doesn't work because the 2kN and 9,5kN are in the same direction, what would be your solution?
@@toondebeule8377 The direction you choose for positives or negatives does not matter in 2D problems. You will always get the same answer regardless of the direction you pick. It's completely up to you. Please see: ua-cam.com/users/shortsP029mqnp4XY?feature=share
i have a question, in the first question when you were calculating moment about point H, why did you include the 9.5(2) and -2(2)? I thought the only force perpendicular to point H would be fBC and the x component of BG? Please clarify, thanks!!!
So both the 9.5kN force and the 2kN force are perpendicular to point H. Imagine you move those forces up to be aligned with point H, so straight up. Now you can see it a bit easier that they are indeed perpendicular. Another way to see it is to realize these forces are vertical, and will cause the object to rotate about point H, so it must create a moment. Force FBC and the x-component of BG is NOT perpendicular to point H. That's why they aren't in the equation. I think you are confusing perpendicular with parallel. Please draw a big diagram on a sheet of paper and really carefully look at what forces are perpendicular.
For the question at 5:25, why does the force in member BA not equal the reaction force from point A?
I found a reaction force from joint A of 45kN when taking moments about point G, but the force in member BA is 50kN.
The cut was made above the reactions so they were not needed. No calculations were made to figure out the reactions at A, or G. Also, where did you get that BA is equal to 50kN? At 7:07, we find that BA = 45kN.
@@QuestionSolutions Thanks for your quick response!
I wanted to verify that the force in member AB equaled the vertical reaction force at A. I understand that it is not directly relevant to the section method.
I just checked my calculations for some of forces in the Y axis, and I totally forgot to include the 5kN on point E. Now I have the correct solution for AB.
Thanks again for the brilliant video!
@@milesbrack9188 Okay, glad to hear it was cleared up. Keep up the good work and best wishes with your studies :)
For question one I know u take thr moment at Point B but if we took the moment at point H, we would have two unknowns, Fbc and Fbg horizontal component both causing a moment. How would u solve it then?
The whole point of taking a moment about a specific point is to eliminate as many unknowns as possible. If you have 2 unknowns after writing a moment equation, then you'd need another equation, probably for vertical forces or horizontal forces. Try to take moments about points that lead to direct answers.
Thanks for the video!
But I have a question for finding force BF in the second problem. If we would have isolated joint B, equilibrium along forces at x would be 10kN - 45Kcos(45) - F(bf) = 0. Hence i thought that force BF would not be zero. Im confused how this works ://
I encourage you to actually isolate for joint B and solve this problem. For your equation, you should have 4 forces. The 10kN, BG, BF, and BE for horizontal components (along x).
thanks for the content, why in second to last problem we only consider that in F there is Fy component why not Fx as well?
Please provide me with a timestamp at the location you're referring to. I will take a look and explain it :)
9:07 could you have used the moment at F and still got the same Force HI?
You can find the values using any moment location but try to use simple ones since you want to make your life as easy as possible. So go for the one that eliminates the most unknowns or leads to a direct force value.
Really nice vedio
Sir I also want to learn such kind of annimation ....can you so please tell me
Or provide some vedio leink .I will be very thankful of you
I use aftereffects for animations. If you google or search on youtube "after effects tutorials" you can find lots :)
@@QuestionSolutions thanks you are great 👍
@@bashirullah4512 Thank you 👍
wowowwwww thank youu so muchhh!!!!
You're very welcome!
In the first question can we use which equation we want? I mean when we use moment or Fx=0, Fy=0. We used only moment equation in this problem.
You can, but try to use ones that give you the answer the fastest.
@@QuestionSolutions okay thank you
it was helpfull thanks
You're welcome!
Can you please enlighten me on why force BC and BG would be eliminated when taking the moment at B. I am confused.
Their lines of action go through point B. The same as the 5 kN force, and the y-component of force HG. If you need a refresh on lines of action, please kindly take a look at this video: ua-cam.com/video/QNNnPZ68STI/v-deo.html
@@QuestionSolutions super helpful, thank you
Glad I found these videos of yours. It was a very big help knowing I am having a hard time with our subjects right now because it was not explained properly but yours is very concised and well elaborated🤍
Glad it was helpful! I hope you do amazingly on your exams and I wish you the best ❤
for when calculating F(BG) why didnt you do Summation(y)? why did you have to take the moment at A?
Please use timestamps, I don't know where you're referring to. Thanks!
Thank you bro
You're welcome!
Bro, thank you
You're very welcome!
how do you know if you have to get the summation of moment at a certain point or simply get the summation of forces?
If you have a set of equations, and you cannot solve them using just a summation of forces, you will need a moment equation, one or more.
@@QuestionSolutions i mean, for example, in this video, after you got the force DC, you used summation of forces to get force HI, when i thought i was supposed to use moment equation and i thought i would have to do it at point F to get rid of the force HC and thus getting me the force HI
i got an answer of 5 kN and i would not have known i was wrong if you didnt show that you used summation of forces to get the force HI
@@JohnRovicAlejandrino You definitely can use a moment equation to get force HI, but my assumption is, you're not writing a proper moment equation. There will be no difference in the answer. They all will result in the same answer. Also, once you already use a moment equation to find a force, you can do the rest with just summation of forces, there isn't a need to write a moment equation most of the time, but if you did, you can still get the same answer.
@@JohnRovicAlejandrino Please write your moment equation so I can go over it and let you know what happened. Even from the answer, we can easily see there is a mistake, you got a force of 5 kN, when all of the forces on the truss are 40s, 50s, etc. So there is an error in your moment equation and I will help you out if you write it here.
Check me if I am wrong,you said if we chose the moment at 1 point then we can eliminate the action force acting on it,but I don understand in 2nd example at 6.30 ,which moment at point D ,the 5 kN still considered?Tq
If you're referring to the top most force, applied at D, that is NOT considered when writing a moment about D. Please carefully look at the color coded boxes and lines, each corresponding to which moment is being calculated. The 5 kN used for the calculation is the one applied at point E.
@@QuestionSolutions ohh I see it now,my mistake.Thankyou
@@imansalim3874 Glad you got it :) Best wishes with your studies!
VERY unimportant as it has no affect on the method but on 9:13 I think the answer is 42.5 not 45.2 but correct me if I am wrong
Yes, you are correct. There is a typo there. Good catch :)
sir can you please recheck the video at 3:39 minutes because I think the answer at this point is 0.5 rather then 1,5,
The value shown on the video is correct. 2 x tan(36.87) = 1.5. I am unsure how you arrived at 0.5 but if you need help, please show your work. Thanks!
how do we determine whether to find the support reactions or not ??
If you use a section that you can solve without needing support reactions, then you don't need it. For example, at 5:35, notice we cut the supports off, and then we had enough givens to solve the top values. But notice at 8:24, even after the cut, we still needed the reaction at roller F. So in most cases, the only time you don't need to know support reactions is if you make a cut and on the side you selected, there are no supports.
8:42 i'm confused at this part. i thought sin DC should be eliminated because it will go thru pin H, and instead use cos DC? or does it depend on where the angle is placed?
So you have to always look at it from the perspective of the angle when doing trigonometry. Sine is going to give us the opposite side (opposite to the angle), which in this case would give us the x-component of force F_DC (horizontal line). That will create a moment since it's line of action doesn't go through point H. Cosine on the other hand will give us the adjacent component, (y-component, vertical line), which will go through point H, so it's eliminated.
@@QuestionSolutions ahh so we're getting the x component here but we just used sin because of the angle. I'm used seeing sin as y, and cos as x so i was confused😆. Thank u very much
@@armanjanndajab6334 Yes, but don't associate sin with x and cos with y. It's all based on the location of the angle. 👍 Best wishes with your studies
I think sine is for summation of Fy and cosine is for Fx? Correct me if im wrong.
You are wrong, do not fall into that pitfall. I am not sure why students tend to this think way, but they are NOT related to Fy and Fx. It's all based on the side opposite to the angle. Please take a look at this video: ua-cam.com/video/NrL5d-2CabQ/v-deo.html
Can you do a video about space trusses?
I will add it to my list, but it probably won't get done for some time. I am still working on thermodynamics videos.
Thanks for the video, you explained this so well in a way my professor never could! 😭😭
Just a quick question though. At 6:16, you say that only the x-component of force BG creates a moment about point D but in the equation it says sin45.. I thought it would cos45 since it's the x-component. Can you please explain this part? It's the only thing I'm confused about
You're very welcome.
You have to make sure you look at the force from the angle. So if you were at the angle, and you looked from that perspective, the opposite side to the angle would be sine, which is the x-component, and the adjacent side is the y-component, which would be cosine. Or are you saying that sine is correlated with x-components and cosine with y-components? If so, this isn't true, please don't think that, it is completely incorrect. If you need a video on components with sine and cosine, please see this video, especially the first example: ua-cam.com/video/NrL5d-2CabQ/v-deo.html
@@QuestionSolutions Omg I thought that x-components were always associated with cosine and that y-components were always associated with sine.. I'm going to check out the video that you linked, thank you so much!! 🙏🏽
@@joonfanatic2140 Thank you for asking this question. I was stumped for a while as well and also always assumed X was Cos and Y was Sin. Game changer for me!
@@AlexanderPerez-oe1gu Same here, it definitely changed my thought process and made me understand things correctly! I'm glad it helped you and hopefully you watched the other video he linked in the reply. I wish you the best in your studies 😄
@@joonfanatic2140 Thanks, yes I saw the referenced video, and totally made sense. Completely changed my perception of how to approach the x and Y components.
In the problem 2, is it also same answer if i use the bottom part because i tried it and i got different answer or did i get wrong?
Regardless of where you start, you will always get the same answers. There is probably a numerical error in your solution or an error in setup of the equations. The forces in each member can't change based on where we start to solve the problem. :)
Quick question: at 2:04, how do we immediately know that there won't be any reaction on the horizontal axis?
Like I can see that none of the applied forces have horizontal components, but couldn't AH (for example) cause a horizontal reaction from the support?
Thanks!
So when you're solving for the reactions at a support, we don't care about internal members. They don't make a difference since we think of the object as a whole. So all the purple forces are external forces, but we have no forces applied in the horizontal direction. Let's say we had a horizontal force applied at point F, in that case, we would have an AX reaction countering the force applied at F. It's the same if we had a simple beam resting on top of supports. We just solve based on external forces, what happens inside the beam doesn't matter.
Another way to think about it is to realize that while there is a force in member AH, that force goes from H to A and A to H, there by causing equilibrium. So to recap, when you are finding supports, internal forces do not matter, only external ones.
@@QuestionSolutions Wow, thanks for the detailed response! Makes a lot more sense now :)
@@benjaminyellin5095 Really glad to hear it makes more sense. Best wishes with your studies :)