Bro , literally you cleared me the concept in like 10 freakin minutes , the things I understood are more than what I understood in my uni lectures Keep it up
I’m thrilled to hear that you found the explanation helpful and it cleared up your concept in such a short time! If you have any more questions or need further clarification, feel free to ask. Do your best!
Thank you for the feedback! They are on my to-do list, but I am unsure when I will get to them. Currently, trying to finish off some thermodynamics videos. @@SanthoshKumar-p9f
Got it, to see the ratio of the both opposite and adjacent to hypotenuse as vectors when multiplied with applied force is fun and after all these years i finally get it. Thanks for the video!
So the y-component is given by the opposite side to the 60 degree angle, which means we need to use sine. If you use the 30 degree angle (not shown but it's the small angle inside the triangle), then you can use cosine, since from that angle, it will be the adjacent side that gives the y-component.
5:02 in the equation fbc-fbacos60-fbecos60=0 we can isolate fbc to get fbcos60+fbecos60=fbc. Since fbe=fab, we can write 2fabcos60=fbc. The 2 and cos60 multiply out to 1 thus leaving fab=fbc but fbc is half of 9.24 therefore fab and fbc should be 4.618 kN
I am sorry, I don't see the equation you're referring to at 5:02. Regardless, all values shown on the video are correct, there are no errors so somewhere in your equation, you made a mistake. Please double check your work.
@@QuestionSolutions Your FDE is shown to be positive when I think it needs to be negative since you assume it as tension. If you look closely at your Y and X-axis drawing, on the X-axis, the arrow of FDE is on the left side and its direction is going to the left, which is why I think it should be negative.
@@QuestionSolutions To clarify it to you again, I'm talking about the Joint D (Solving for Fx). As I've said, you assume FDE as tension, which is away from the PIN; if you draw that one, the FDE is going to the left side of the X-axis from the PIN, which is negative. Your answer for FDE is 4.62 kN (C), and mine is 4.62 kN (T). PS: I assumed mine as tension too; that's why I'm asking this because I don't know if mine was wrong or there was an error on yours.
@@deadlystann So initially, we assumed the force to go away from the pin. When we solve the equations, we get a negative value, which indicates the assumed direction is wrong. That means the force is actually coming towards the pin. So it's a positive 4.62 kN, coming towards the pin, which means the force is in compression. If you're confused about the equilibrium equation, when we wrote it, we assumed any force to the left to be positive. The outcome of that equation determines whether our assumption was right or not, however, you are free to choose forces going to the right as positive, you will still end up with the same answer. If you got 4.62 positive facing to the left, your answer is incorrect and there is most likely an error in your force equation with positive and negative signs. In actuality, the force is coming towards the pin. If you drew your force already coming towards the pin, you will get a positive value. Again, this is why the force is shown to be flipped so students understand what needs to be done if an assumption was wrong and I explain this at 3:30.
In the last Question the isoscales triangle in which at joint D The Fda= 3.34 C and Fdb=1.15P T because in FBD the inclined force tht we don't know we aways make tension but why you produce FDA as Comp in FBDin 9:14 kindly, reply please!
You can assume them to be in tension or compression. It's completely up to you. At the end, if you get a positive value, then your assumption was correct. If you get a negative value, it's opposite to your assumption. 👍
@@maazansari871 okay, I'll check tomorrow and let you know. It shouldn't make any difference since the y-force will turn to positive and x-force will be negative. And then in the next step, the arrow would be pointing in the opposite direction. 😅
In the intro about the pin exerting a force, It’s almost like saying the pins on both end of a rod are going to had equal and opposite force. Like for instance, if the pin on the right side exerts a positive force, the pin on the left side will exert the same amount of force As the right pin, but it will be negative, the same works the opposite way around.
Yes, so when you find the reaction at one end, the whole member has that force. The direction of the force from the pin determines whether the member is in compression or tension.
I’m glad to hear that the video explanation on trusses has helped you develop your understanding on the topic. It’s always great to receive positive feedback and know that the content has been beneficial to you. :)
Ok! I am confused! You have forces on members that act in the inward direction and your labeling them as tension. Shouldn’t that be in compression? Then you have forces acting in the opposite direction on a member and calling it compression. Shouldn’t that be called tension.Please explain that to me.
So it's because you're writing equations about point A. It's the same as when we write equations about point D, we didn't care about the forces at C. We haven't done that for any other point or force, so why would we do it for Bx? 😅I hope that makes sense, but if it doesn't, let me know and I will try to explain it differently.
Really you are the best explainer I have ever seen✨. Thanks 😍 for this amazing content, not just a content u gave me a skill as I am persuing mechancial engineering. Best wishes! ❤💫
You can use the substitution method to solve these since its 2 equations with 2 unknowns. Here are some sample problems solved that way: - th-cam.com/users/shorts86uENomd53U?feature=share - th-cam.com/users/shortsHe7lrJEB04U?feature=share - th-cam.com/users/shorts4euH1289_Kg?feature=share - th-cam.com/users/shortsrAlhrq5hWFc?feature=share
Ive a doubt on 6:39. you took sin for horizontal force and cos for vertical. I thought we take sin for vertical forces and cos for horizontal only so does that mean we can change it accordingly? If yes than what's the reason you changed it.
The answer to my question was that sin is always the opposite side of theta. In this problem the angle (theta) was at the top corner, so the bottom side will be sine and the other side will be cosine and the longest side is always the hypotenuse.
@@benshapirohamburgerhelper1239 The x-component of force F_DE is along the x-axis. That is opposite to the angle, which means it's sine. The y-component of force FDE is along the y-axis, which is adjacent to the angle, which means cosine. 👍
I think the really best way to think if it theres a compression or tension is to assume there must be a reactionary component of the force at the point. Like the first example, it was right to assume that Fdc is going top left of the point since there must be a y component reacting to the 8kN, and subsequently for Fde point right to counteract the x component of the Fdc
Once you do enough questions, you can actually see the forces in the truss with just mental math and pretty much know whether it's in tension or compression. 👍
Hi, could you explain something for me? At 4:05, If we take the the 8kN force to the same coordinate of the green arrow at D, it should be F_DC - 8*cos30 = 0, but it gives the wrong result (6.9kN). Why this does not work?
I am not entirely sure what you are asking. How are you moving the 8 kN force? It's straight down, so it will only have a y-component. There is no cos 30. Maybe I am not understanding what you are asking :( Could you reword the problem?
@@QuestionSolutions When we are evaluating stress or internal forces, sometimes is more useful to use other coordinate system than the regular x-y plane (y vertical, x horizontal) such as in th-cam.com/video/8JYUHt5Lqcs/w-d-xo.html . I think my doubt is more related to it. Why I cannot place the 8kN force in the same coordinate of the force F_DC to calculate it?
@@ironheart444 I think you are misunderstanding how to change coordinate systems. If you change your whole coordinate system, all the forces that no longer lie on the x-y plane must also be broken into components so that the components lie on the new coordinate system.
@@QuestionSolutions Thanks!!! That's exactly what I was doing wrong. If I changed the coordinate system, It would have a new component from the blue force at A in the equation.
@@ironheart444 Maybe you did this for practice, but in general, you want your coordinate system to include as many forces as possible without breaking them into components. That makes your life easy :) I am glad you got it though! Keep up the awesome work.
3:21 wait i understand the pattern and the logical conclusion, but isn't Fdccos60 is itself Fde? I know it will give a logical result (which is Fde=-Fdccos60, it's negative bcs of the direction), but what about the sigma Fx? Are those things (Fde and Fdccos60) two different forces, or they are just the same single force? I'm guessing they are two different forces since Fdccos60 is just Fdc pointing to the horizontal direction (it's one force broken up into two directions), while Fde is a force that always points to the horizontal direction. Is my guess correct?
In this case, yes, because there were no other forces at the pin. Also, I assume you meant cos60? Because the x-component of force FDC is equal to the magnitude of force FDE. Again, that's just in this instance. If, for example, there was another force pointing in another direction (as long as it wasn't straight up), FDE would not be equal to the x-component of force FDC.
@@jerichogaspar2559 Any variable is always a 1 (unless a value other than 1 is shown, like 2p, 0.4p, 5p etc). We just don't write the 1 in front. So if we have something like x+5=10. The x has a 1 in front, we just don't write it. Otherwise, it wouldn't exist, for example, if it was 0p, then it's just 0. I hope that makes sense :)
What do you mean by your assumpiton was right at 3:12 because of positive answer. Do you mean the correct answer for all assumption must be positive and if we got negative we must to change it o positive?
Yes, I show this at 3:30, where we assumed an incorrect direction and got a negative value. If you get a positive value, your assumption for the direction of the force was correct. If you get a negative value, that means the direction is opposite to your assumption, but the magnitude of that force is still correct.
I appreciate your time to make the video. I will grateful if you had a little time to talk about the each *term* and why it was included in the calculations. Rather you jump straight to the answer. I really love the explanation though👏
Can you give me an example timestamp where I went too fast, or "jumped" to an answer? I appreciate feedback so I can make them easier for other students. Thanks!
At 4:05. Why did you assumed Fce towards the Joint C? Why not just write Fce away from the Joint Fce? I know it doesn't matter but also why not take all the forces away from the Joint?
So that's why it's an assumption. You can assume it anyway you want. When you do a lot more questions, you will notice that you can make a very good guess as to the directions these forces will face. That makes it easier since you don't have to flip them at the end if you're wrong. Regardless, it's an assumption, pick whatever direction you want.
So notice that our force, F_CE, points up and to the right (towards the center with a slant towards the right). If we break that into components, we would have a y-component that faces straight upwards (vertical), and a force that faces to the right (horizontal). A good way to quickly see the components is to start at the end of a vector (so the opposite side to the arrow head), and mentally "walk" along the x/y axis to get to the arrow head. We have to "walk" to the right and up, or up and to the right. Either way, our x-component faces to the right, so it's positive. With the given angle, the adjacent side is the x-component, so we have cosine. Let me know if that clears it up.
Some students have asked how I solved the 2 equations at 8:02. Please see the following if you need a breakdown: bit.ly/3GGIIQM When to use sine or cosine: th-cam.com/users/shortsvynnKlJD_Jo Many thanks!
During the calculation of Fea at 5:30, would it be possible to instead calculate the force from the point A where you only have 1 unkown at that point? or is there any specific reason why you would chose point E?
@@MrJaaaboo At point A, you have 3 unknowns, a force from the pin reaction in the vertical direction, a force from the pin in the horizontal direction, and force AE/EA. Please see this video: th-cam.com/video/jQDEOwrR4UU/w-d-xo.html I explain how pins, rollers, etc. have their own reaction forces and how to account for them. 👍
@@QuestionSolutions when you're solving a pin/point, does it matter if you solve for the sum off Y or X first? i saw in your first solution you always started with Y, in the second on its always x..?
@@MrJaaaboo Completely up to you. Overtime, you will gain the ability to just look at a diagram, see the forces and know which ones will be zero. In that case, you will intuitively know if it's better to start off with x or y. But again, it makes no difference at all, just sometimes can make your life a bit easier.
At 3:05 I cannot understand how we got sin60? Isn’t should be cos60? Because Fdc is the hypotenuse so we need to use cos to find the Y component? Please if you can explain it to me because I’m already struggling with statics
@@QuestionSolutionsI’ve watched the full video and the short also but I can’t get it I still see it as cos60, and I’m talking about the triangle where Fdc is the hypotenuse and Y-component as the adjacent
So you're looking at the diagram incorrectly, which is what's leading to your confusion. I've drawn the components in dashed light green, but you need to think about how these components can be moved along their axes. So bring the y-component (vertical light green dashed line) in front of the 60 degree angle. In other words, move it to the left along the x-axis, still parallel to the y-axis. Then you can see how it's in front of the 60 degree angle. Opposite is sine, so you need to use sine to get the y-component. If you use cosine, you need to use 30 degrees, since 90 - 60 = 30. You're looking at the components from the 30 degree angle (which isn't shown). In hindsight, I think it would have been better for me to draw the components in front of the 60 degree angle, but it's almost second nature to students once you can visualize components being moved around. I think also, it was hard to show the x-component on top of the FDE force arrow 😅@@kalidalghamdi9306
How do you know the sign to use for each force member eg. At 4:30, how is Fcb negative and not positive, and how can you tell if it's negative or positive
So whenever we write an equation, we establish which sides are positive. If you look next to the sigma sign, there is a little arrow pointing to the right with a plus sign next to it. This says any force pointing to the right is positive, which means any force pointing to the left will be negative. Notice that force F_CB is pointing to the left, so it has to be negative. You can also write these equations with right being positive, in which case, F_CB will be positive and the other 2 will be negative. You will get the same answer either way.
Pls at 3:27 why is f-de which is pointing towards the left not negative in the equation for fx but the 8kn which is facing downwards is negative in the fy equation
You have to pay special attention to the directions we pick to be positive when we write equations. So look next to the sigma sign (the big Greek letter E). Notice how for the x-forces, we assumed left to be positive. We show that with an arrow to the left and a small positive sign. That means any force facing left is assumed to be positive. Now look at the y-axis forces, we assumed up to be positive. So the 8 kN is facing down, which means it's negative. I hope that helps :)
First up great video, it has explained it better than my current lecturer who doesn't explain it as clear as yourself. @9:16 did you solve the two equations like at 8:02? Just making sure that I cover all of my workings to show my intent and how it was achieved.
Thank you, and yes, almost all of the simultaneous equations I've solved has been done so, using the substitution method. If there is more than 4 unknowns, I would suggest to use a matrix, but that usually comes up in dynamics, not statics.
So it's not based on negative or positive values. It's based on whether the force was coming towards the pin or going away from the pin. Force EA is coming towards pin E, so it's in compression. Force EB is going away from pin E, so it's in tension. Watch the video from 1:26 again and it should help.
Please see: th-cam.com/users/shortsvynnKlJD_Jo?si=Er6Zt0UO4NloR9jU Remember, we only care about the y-component when we write an equilibrium equation for the Y-axis.
So it has nothing, absolutely nothing, to do with whether a force is positive or not. Positives and negatives only tell you whether an assumption was right or not. What matters is the direction of the force. Is the force coming towards the pin or going away from the pin. Rewatch this part from 1:48.
Sine and cosine are NOT related x or y-axes. They are related to the angle and their opposing sides. Please see this video, it's under 60 seconds and will clear up any confusing parts. th-cam.com/users/shortsvynnKlJD_Jo?feature=share
There really isn't anything to break down. You solve directly for FCE. So plug this into your calculator (9.24sin60)/(sin60) = 9.24 You can also divide both sides by sin60, so you're left with just 9.24. Sometimes, thinking of FCE as just "x" might help you visualize it better.
I know its not correct but why can’t I use a sum of moments about c to find Bx or Ax, please see my equation, thanks! I used sum of forces in X of the whole truss and got Ax+Bx-900-600 Ax=1500-Bx Mc=600(4) +Bx(4)+Ax(4) Ay(6) Sub in Ax Mc=600(4) +Bx(4)+(1500-Bx)(4) Ay(6) This yields Bx as canceled but I don’t see why this happening, have I missed a force? If no which one. Thanks for taking to time to read this, its much appreciated!
So you divide everything by sin60 to get rid of sin60 on all the terms. That way, the only things you have left is just FBE and FBA. If your assumption is incorrect, you will get a negative answer. Then you need to flip the direction of the arrows.
So let's say you have a fraction like this: (4+5)/6. This is the same as 4/6+5/6. So in our problem, we have (1500-0.6x)/0.6. This is the same as 1500/0.6-0.6x/0.6. If you divide each term individually, you get 2500 - x.
5:34 So I solved this joint while pausing the video and noticed that I didn't even need the Y reaction since I already found the only unknown (F_ea)with the X reaction. But in what case would the Y reaction be useful? - Also lets say the roller was at Joint A, how would that affect the reactions at A and forces (F_ba) and (F_ea)? BTW, this was an amazing video. I couldn't understand a single thing in my lecture and I'm glad I can do my assignment in peace now
It's useful if it's required in your question. If it doesn't, it's just extra forces. 😅 You can't have 2 rollers as the only support. Nothing would stop the bridge from going left to right. But let's say for hypothetical reasons, you have 2 roller supports. Then the only difference is, at A, there would just be a single vertical force, and there wouldn't be an x-reaction force.
I have a question at 3:21 can we assume opposite x direction to be positive (right side)? when ı assume right to be possitive ı find -4.62 but in wrong direction. I know that F_DE should be in the right direction.
@@QuestionSolutions but i still get wrong direction my equation is -FDC*cos60 - FDE = 0 then ı find FDE= -4.62 ( ı assume right to be positive ) then in this case FDE is in the left direction
@@sevgipnar5261 No, you're getting the right answer since you got -4.62. It just means the direction we chose for the arrow F_DE is incorrect. It was facing to the right. I think you're getting 2 things mixed up. The direction of the arrows we chose can give us negative values, which means the direction we picked was wrong. When we solve problems, you can pick forces to the left or right to be positive. So there are 2 choices going on.
@@sevgipnar5261 Yes. Let's say you picked your arrow to face left, and when you solve your equations, you for a negative value. That means your arrow actually faces to the right. Please re-watch 3:26, I show how that's done.
sir why F(DE) compression ? 6:29 because you assumed is going to that direction therefore because now is positive it must be tension you have to change the direction.... can you pls explain that part to me....and thanks for the video I learn a lot from you.
So we made an assumption that force DE would be going towards pin D. We got a positive for force DE, which means our assumption was right. Any force that goes towards a pin is always in compression.
So you need to isolate for F_DC. Add 8 to both sides and then divide both sides by sin60. You have: F_DC = 8/Sin60. Plug that into your calculator and you get: F_DC = 9.24
When you use the sine to get your forces at Y, for example why don't you use angle 30 instead of 60? I have trouble making sense of that, for example at Joint C. I can also verify your answer with similar triangles, but if you can let me know why 60 degrees, thanks.
@@QuestionSolutions Would you mind showing me through an e-mail or through here? I usually struggle to choose the angle for sin from the top joints. I can follow the method, just having sense of which angle to use for sin, for cosine I know it is always the adjacent angle.
@@enrique2914 Please kindly watch this video first: th-cam.com/video/NrL5d-2CabQ/w-d-xo.html If you're tight for time, then please watch the first example. I go through the whole process, how to pick sin and cosine, break forces into components. etc. The video is only 9 mins long, and I think it can help you a lot. Let me know if that helps :)
We are trying to get the y-component of the 9.24 kN force. That's opposite to the angle, so we use sine. Please see: th-cam.com/users/shortsvynnKlJD_Jo?feature=share
Hello wanted to ask , what if i used joint A instead of join E to find F A/E. Is still goona be a correct answer or i need to use join E in order for the lecture to mark it as correct ?!!
I am not sure where you're referring to, regardless, it makes no difference as long as you end up with the same answers. Unless your professor specifically asked you to start at a certain joint, you can solve these problems however you please. Show your steps properly, and arrive at the same answer, and you should get full marks.
You can freely choose it. You will get the correct answer regardless. However, as you gain more experience with these problems, you should be able to mentally pick the proper directions quite easily so that you won't have to switch them at the end.
No. You need to look from the point of view of the angle. If you look straight from the angle, it's 6m straight ahead, which is considered the opposite direction, and 8 in the adjacent direction. I hope that helps. 👍
Probably the best video on Truss in YT for me. Thank you for helping us. So much respect and honor for you
You're very welcome and I am glad these help you out :)
Tum gate ke liye prepare kar rahe ho kya
@@jeetadityabiswas4819 aap kaise hain? kya aap get ke lie adhyayan kar rahe hain, ya aapane ise pahale hee le liya hai?
More students should trusst this channel to help them learn.
Literally the best video on trusses I have seen.
Thank you very much :)
My god, this is such quality content. I love it !!! Thank you so much. You helped me a ton with my statics course.
You're very welcome! Glad to hear these videos are helpful.
@@QuestionSolutions Thankyou so much as well, your youtube channel has some fantastic information on core statics material, thankyou again so much!
@@joewow1229 You're very welcome and thank you for taking the time to write your nice comment :)
i had a difficulty since my last semester in this topic but by watching it , this video clear my concepts in minutes.
I am really glad to hear that! Keep up the great work and best wishes with your studies.
Unfortunately just found this and my test on this chapter is in like 4 hours. Thanks four putting this out here I really appreciate it
Glad it helped! I hope your test went well. Keep up the good work and best wishes :)
I had the exact same experience lmao. Saved my ass
Explained it better and in less time than my prof, thanks!
I am really happy to hear this video was good for you. Keep up the great work and best wishes with your studies!
I've watched most of your lessons and you're the best
Thank you very much! I hope they were helpful to you :)
Please cover more topics, your videos are so far the best teaching videos on youtube.
Thank you very much! :)
Bro , literally you cleared me the concept in like 10 freakin minutes , the things I understood are more than what I understood in my uni lectures
Keep it up
I’m thrilled to hear that you found the explanation helpful and it cleared up your concept in such a short time! If you have any more questions or need further clarification, feel free to ask. Do your best!
@@QuestionSolutions It would be really helpful to have videos on friction , belt friction , etc
Thank you for the feedback! They are on my to-do list, but I am unsure when I will get to them. Currently, trying to finish off some thermodynamics videos. @@SanthoshKumar-p9f
@@QuestionSolutions nicee ,
Got it, to see the ratio of the both opposite and adjacent to hypotenuse as vectors when multiplied with applied force is fun and after all these years i finally get it. Thanks for the video!
You're very welcome :)
this video shows us how we must understand the joining of trusses
thank you for helping us
You're very welcome :)
This by far is the best video on Truss!
Thank you very much :)
Thanks bro..i recommended your channel to my mechanical Department
Thank you for the recommendation, really appreciate it :)
thankyou sir,
i am watching this from India sitting in my lecture class.
you are better than a PHD professor in teaching.
Thank you very much for the nice comment. Everyone has their own teaching styles, maybe mine works better for you :)
MOST WELCOME SIR...☺
Sir why at minutes 3:16 calculation of sum Fy must use sin60° instead of tan and cos?
So the y-component is given by the opposite side to the 60 degree angle, which means we need to use sine. If you use the 30 degree angle (not shown but it's the small angle inside the triangle), then you can use cosine, since from that angle, it will be the adjacent side that gives the y-component.
this video saved my maths project. thanks.
Glad to hear! You're very welcome.
Thank you! That is a great explanation in 10m that took my professor 2h
You're very welcome. I try to keep them as concise as possible :)
Excellent video, it simplified the process for me and helps me to be ready for my midterm! Thanks! 🙌
Glad to hear that! I wish you the absolute best with your midterm.
Please make more videos about trusses with examples. thanks for your work
There will be another video with the method of sections that will cover a few more examples. 👍
5:02 in the equation fbc-fbacos60-fbecos60=0 we can isolate fbc to get fbcos60+fbecos60=fbc. Since fbe=fab, we can write 2fabcos60=fbc. The 2 and cos60 multiply out to 1 thus leaving fab=fbc but fbc is half of 9.24 therefore fab and fbc should be 4.618 kN
I am sorry, I don't see the equation you're referring to at 5:02. Regardless, all values shown on the video are correct, there are no errors so somewhere in your equation, you made a mistake. Please double check your work.
Using this for exam prep for my ng exams. You are a beast.
Best wishes with your exams!
@@QuestionSolutions is there any way I can donate to the channel
@@notabot1078 Wow, thank you very much! You can donate through this link: ko-fi.com/questionsolutions
3:28 I'm confused at Joint D (Fx). Isn't the FDE supposed to be negative since the arrow of FDE from Pin in the X axis is going left?
Sorry, I am confused by your question. It is shown to be negative? Did you mean positive?
@@QuestionSolutions Your FDE is shown to be positive when I think it needs to be negative since you assume it as tension. If you look closely at your Y and X-axis drawing, on the X-axis, the arrow of FDE is on the left side and its direction is going to the left, which is why I think it should be negative.
@@QuestionSolutions To clarify it to you again, I'm talking about the Joint D (Solving for Fx). As I've said, you assume FDE as tension, which is away from the PIN; if you draw that one, the FDE is going to the left side of the X-axis from the PIN, which is negative. Your answer for FDE is 4.62 kN (C), and mine is 4.62 kN (T).
PS: I assumed mine as tension too; that's why I'm asking this because I don't know if mine was wrong or there was an error on yours.
@@deadlystann So initially, we assumed the force to go away from the pin. When we solve the equations, we get a negative value, which indicates the assumed direction is wrong. That means the force is actually coming towards the pin. So it's a positive 4.62 kN, coming towards the pin, which means the force is in compression. If you're confused about the equilibrium equation, when we wrote it, we assumed any force to the left to be positive. The outcome of that equation determines whether our assumption was right or not, however, you are free to choose forces going to the right as positive, you will still end up with the same answer. If you got 4.62 positive facing to the left, your answer is incorrect and there is most likely an error in your force equation with positive and negative signs. In actuality, the force is coming towards the pin. If you drew your force already coming towards the pin, you will get a positive value. Again, this is why the force is shown to be flipped so students understand what needs to be done if an assumption was wrong and I explain this at 3:30.
@@QuestionSolutions Thank you so much for clarifying it! Very much appreciated, you got my subscribe!
In the last Question the isoscales triangle in which at joint D The Fda= 3.34 C and Fdb=1.15P T because in FBD the inclined force tht we don't know we aways make tension but why you produce FDA as Comp in FBDin 9:14 kindly, reply please!
You can assume them to be in tension or compression. It's completely up to you. At the end, if you get a positive value, then your assumption was correct. If you get a negative value, it's opposite to your assumption. 👍
@@QuestionSolutionsbut our answers of fda and fdb are not same if I assume the fda as tension in FBD?
@@maazansari871 You changed the signs for both the y-components and x-components?
Yes of course I do.. you also try and check please!
@@maazansari871 okay, I'll check tomorrow and let you know. It shouldn't make any difference since the y-force will turn to positive and x-force will be negative. And then in the next step, the arrow would be pointing in the opposite direction. 😅
In the intro about the pin exerting a force, It’s almost like saying the pins on both end of a rod are going to had equal and opposite force. Like for instance, if the pin on the right side exerts a positive force, the pin on the left side will exert the same amount of force As the right pin, but it will be negative, the same works the opposite way around.
Yes, so when you find the reaction at one end, the whole member has that force. The direction of the force from the pin determines whether the member is in compression or tension.
Thank you for the video explanation on trusses. It has really developed my understanding on trusses a bit more.
I’m glad to hear that the video explanation on trusses has helped you develop your understanding on the topic. It’s always great to receive positive feedback and know that the content has been beneficial to you. :)
thank you so much for this bud! Your so good at explaining.
You're very welcome! Thank you for the compliment :)
I was just curious but i still got it thank you for the good job.
Thank you very much :)
Ok! I am confused! You have forces on members that act in the inward direction and your labeling them as tension. Shouldn’t that be in compression? Then you have forces acting in the opposite direction on a member and calling it compression. Shouldn’t that be called tension.Please explain that to me.
Watch from 1:27, and if you still don't understand it, reply back and I will try to explain it in a different way.
Got it! Thanks
Wow very powerful presentation about truss..like it❤
Glad you liked it!
thank you
you express in understandable way keep it up!
You are very welcome :)
very good video. Greetings from Bangladesh
Thank you very much and greetings to you as well!
Hi, at 9:34, why isnt the supports and reactions at B considered? wouldn't it be Fab+Bx ?
So it's because you're writing equations about point A. It's the same as when we write equations about point D, we didn't care about the forces at C. We haven't done that for any other point or force, so why would we do it for Bx? 😅I hope that makes sense, but if it doesn't, let me know and I will try to explain it differently.
@QuestionSolutions Oh right, i assumed the hinge at B would act on pin A, but it doesnt, thank you very much!
@@wine7481 Awesome :)
Really you are the best explainer I have ever seen✨. Thanks 😍 for this amazing content, not just a content u gave me a skill as I am persuing mechancial engineering. Best wishes! ❤💫
You're very welcome! I'm so glad that my videos have been helpful to you in your mechanical engineering journey. 🤩
Straight Goat
Thanks! 👍
Bump for the algo. Love the content.
Thank you. Really appreciate it.
thank you so much....the explanation is so clear and easy to understand！all the best for my test tmr😂😂
You're very welcome. I wish you the best on your test tomorrow :)
8:01 Can you explain how you solved this or how it is solved in the calculator?
You can use the substitution method to solve these since its 2 equations with 2 unknowns. Here are some sample problems solved that way:
- th-cam.com/users/shorts86uENomd53U?feature=share
- th-cam.com/users/shortsHe7lrJEB04U?feature=share
- th-cam.com/users/shorts4euH1289_Kg?feature=share
- th-cam.com/users/shortsrAlhrq5hWFc?feature=share
@@QuestionSolutions Thank youu so much, I get it now
You're very welcome :)@@bntiscz8473
Ive a doubt on 6:39. you took sin for horizontal force and cos for vertical. I thought we take sin for vertical forces and cos for horizontal only so does that mean we can change it accordingly? If yes than what's the reason you changed it.
Please see this video: th-cam.com/users/shortsvynnKlJD_Jo?feature=share
You are my hero, man! Thank you!
You're very welcome!
hello. at 3:05 why is FDC sin(60) and not sin(30), becasue the force vector is 30degrees from the y axis but it gives me the wrong answer..
You are using sine and cosine incorrectly. Please see this short video: th-cam.com/users/shortsvynnKlJD_Jo?si=JoKIX808RPq7bMYu
6:36
why do you use sin for x forces and cos for y forces in this problem?
The answer to my question was that sin is always the opposite side of theta. In this problem the angle (theta) was at the top corner, so the bottom side will be sine and the other side will be cosine and the longest side is always the hypotenuse.
@@benshapirohamburgerhelper1239 The x-component of force F_DE is along the x-axis. That is opposite to the angle, which means it's sine. The y-component of force FDE is along the y-axis, which is adjacent to the angle, which means cosine. 👍
@@QuestionSolutions thanks sir
I think the really best way to think if it theres a compression or tension is to assume there must be a reactionary component of the force at the point. Like the first example, it was right to assume that Fdc is going top left of the point since there must be a y component reacting to the 8kN, and subsequently for Fde point right to counteract the x component of the Fdc
Once you do enough questions, you can actually see the forces in the truss with just mental math and pretty much know whether it's in tension or compression. 👍
Hi, could you explain something for me? At 4:05, If we take the the 8kN force to the same coordinate of the green arrow at D, it should be F_DC - 8*cos30 = 0, but it gives the wrong result (6.9kN). Why this does not work?
I am not entirely sure what you are asking. How are you moving the 8 kN force? It's straight down, so it will only have a y-component. There is no cos 30. Maybe I am not understanding what you are asking :( Could you reword the problem?
@@QuestionSolutions When we are evaluating stress or internal forces, sometimes is more useful to use other coordinate system than the regular x-y plane (y vertical, x horizontal) such as in th-cam.com/video/8JYUHt5Lqcs/w-d-xo.html . I think my doubt is more related to it. Why I cannot place the 8kN force in the same coordinate of the force F_DC to calculate it?
@@ironheart444 I think you are misunderstanding how to change coordinate systems. If you change your whole coordinate system, all the forces that no longer lie on the x-y plane must also be broken into components so that the components lie on the new coordinate system.
@@QuestionSolutions Thanks!!! That's exactly what I was doing wrong. If I changed the coordinate system, It would have a new component from the blue force at A in the equation.
@@ironheart444 Maybe you did this for practice, but in general, you want your coordinate system to include as many forces as possible without breaking them into components. That makes your life easy :) I am glad you got it though! Keep up the awesome work.
VERY NICELY EXPLAINED
Thank you very much!
Great video!
Thank you very much!
thank you so much brother may God reward greatly
You're very welcome and thank you so much! :)
3:21 wait i understand the pattern and the logical conclusion, but isn't Fdccos60 is itself Fde? I know it will give a logical result (which is Fde=-Fdccos60, it's negative bcs of the direction), but what about the sigma Fx? Are those things (Fde and Fdccos60) two different forces, or they are just the same single force? I'm guessing they are two different forces since Fdccos60 is just Fdc pointing to the horizontal direction (it's one force broken up into two directions), while Fde is a force that always points to the horizontal direction. Is my guess correct?
In this case, yes, because there were no other forces at the pin. Also, I assume you meant cos60? Because the x-component of force FDC is equal to the magnitude of force FDE. Again, that's just in this instance. If, for example, there was another force pointing in another direction (as long as it wasn't straight up), FDE would not be equal to the x-component of force FDC.
@@QuestionSolutions Yes, i meant cos60. Okay, thanks for the confirmation, i appreciate it
if u have time could anyone explain at 9:15 why he took fx (fDa COS 30 instead of sin30 )
The x-component is opposite to the 30 degree angle, and opposite is sine. See: th-cam.com/users/shortsvynnKlJD_Jo?feature=share
Hi at 9:13 I was just wondering how you simplified the equations to find the forces in terms of P, is it just simultaneous equations?
Yes, that's correct. Remember you can convert terms like cos30 to decimal values if it makes it easier to see.
Someone else asked the same question, so I solved it step by step here: drive.google.com/file/d/1_CmBjvBnuvTXX-dmYY4VpGSVfu7Jwd0p/view?usp=sharing
@@QuestionSolutions thanks man you're the goat
Please @6:33 when calculating for Efx=0 why is it DEsin36.87 and not DEcos36.87
Please see: th-cam.com/users/shortsvynnKlJD_Jo
@@QuestionSolutions thank you so much 💓
Thank you really appreciate this!
You're very welcome!
How did you gett 1.155P
(8:52)
Can we just assume the P as 1?
@@jerichogaspar2559 Any variable is always a 1 (unless a value other than 1 is shown, like 2p, 0.4p, 5p etc). We just don't write the 1 in front. So if we have something like x+5=10. The x has a 1 in front, we just don't write it. Otherwise, it wouldn't exist, for example, if it was 0p, then it's just 0. I hope that makes sense :)
What do you mean by your assumpiton was right at 3:12 because of positive answer. Do you mean the correct answer for all assumption must be positive and if we got negative we must to change it o positive?
Yes, I show this at 3:30, where we assumed an incorrect direction and got a negative value. If you get a positive value, your assumption for the direction of the force was correct. If you get a negative value, that means the direction is opposite to your assumption, but the magnitude of that force is still correct.
I appreciate your time to make the video. I will grateful if you had a little time to talk about the each *term* and why it was included in the calculations. Rather you jump straight to the answer. I really love the explanation though👏
Can you give me an example timestamp where I went too fast, or "jumped" to an answer? I appreciate feedback so I can make them easier for other students. Thanks!
on 8:02 how did you get the answer for fea=1750N, C and for feb=750N, T? Thank you so muuch
Please see: www.cymath.com/answer?q=0.6x%2B0.6y-1500%3D0%2C%200.8x-0.8y-800%3D0
Thank you you are the best 😊.
You're very welcome! 😊
At 4:05. Why did you assumed Fce towards the Joint C? Why not just write Fce away from the Joint Fce? I know it doesn't matter but also why not take all the forces away from the Joint?
So that's why it's an assumption. You can assume it anyway you want. When you do a lot more questions, you will notice that you can make a very good guess as to the directions these forces will face. That makes it easier since you don't have to flip them at the end if you're wrong. Regardless, it's an assumption, pick whatever direction you want.
So good, thank you for all great videos!
You're very welcome!
On min 04:27 when solving for the x axis, kindly explain to me why the component Fcecos60 is positive if we are taking the right side to be positive
So notice that our force, F_CE, points up and to the right (towards the center with a slant towards the right). If we break that into components, we would have a y-component that faces straight upwards (vertical), and a force that faces to the right (horizontal). A good way to quickly see the components is to start at the end of a vector (so the opposite side to the arrow head), and mentally "walk" along the x/y axis to get to the arrow head. We have to "walk" to the right and up, or up and to the right. Either way, our x-component faces to the right, so it's positive. With the given angle, the adjacent side is the x-component, so we have cosine. Let me know if that clears it up.
Some students have asked how I solved the 2 equations at 8:02. Please see the following if you need a breakdown: bit.ly/3GGIIQM
When to use sine or cosine: th-cam.com/users/shortsvynnKlJD_Jo
Many thanks!
During the calculation of Fea at 5:30, would it be possible to instead calculate the force from the point A where you only have 1 unkown at that point? or is there any specific reason why you would chose point E?
@@MrJaaaboo At point A, you have 3 unknowns, a force from the pin reaction in the vertical direction, a force from the pin in the horizontal direction, and force AE/EA. Please see this video: th-cam.com/video/jQDEOwrR4UU/w-d-xo.html I explain how pins, rollers, etc. have their own reaction forces and how to account for them. 👍
@@QuestionSolutions when you're solving a pin/point, does it matter if you solve for the sum off Y or X first? i saw in your first solution you always started with Y, in the second on its always x..?
@@MrJaaaboo Completely up to you. Overtime, you will gain the ability to just look at a diagram, see the forces and know which ones will be zero. In that case, you will intuitively know if it's better to start off with x or y. But again, it makes no difference at all, just sometimes can make your life a bit easier.
At 3:05 I cannot understand how we got sin60? Isn’t should be cos60? Because Fdc is the hypotenuse so we need to use cos to find the Y component? Please if you can explain it to me because I’m already struggling with statics
Please see: th-cam.com/users/shortsvynnKlJD_Jo?si=cZM_sz2GrFu4FTWY
This is very important to understand so watch the whole thing.
@@QuestionSolutionsI’ve watched the full video and the short also but I can’t get it I still see it as cos60, and I’m talking about the triangle where Fdc is the hypotenuse and Y-component as the adjacent
So you're looking at the diagram incorrectly, which is what's leading to your confusion. I've drawn the components in dashed light green, but you need to think about how these components can be moved along their axes. So bring the y-component (vertical light green dashed line) in front of the 60 degree angle. In other words, move it to the left along the x-axis, still parallel to the y-axis. Then you can see how it's in front of the 60 degree angle. Opposite is sine, so you need to use sine to get the y-component. If you use cosine, you need to use 30 degrees, since 90 - 60 = 30. You're looking at the components from the 30 degree angle (which isn't shown).
In hindsight, I think it would have been better for me to draw the components in front of the 60 degree angle, but it's almost second nature to students once you can visualize components being moved around. I think also, it was hard to show the x-component on top of the FDE force arrow 😅@@kalidalghamdi9306
@@QuestionSolutions thank you so much for your explanation
Sir on 6:32 why was sin used in looking for the x component not cos?
Please see: th-cam.com/users/shortsvynnKlJD_Jo
How do you know the sign to use for each force member eg. At 4:30, how is Fcb negative and not positive, and how can you tell if it's negative or positive
So whenever we write an equation, we establish which sides are positive. If you look next to the sigma sign, there is a little arrow pointing to the right with a plus sign next to it. This says any force pointing to the right is positive, which means any force pointing to the left will be negative. Notice that force F_CB is pointing to the left, so it has to be negative. You can also write these equations with right being positive, in which case, F_CB will be positive and the other 2 will be negative. You will get the same answer either way.
@@QuestionSolutions thanks 🤙
Pls at 3:27 why is f-de which is pointing towards the left not negative in the equation for fx but the 8kn which is facing downwards is negative in the fy equation
You have to pay special attention to the directions we pick to be positive when we write equations. So look next to the sigma sign (the big Greek letter E). Notice how for the x-forces, we assumed left to be positive. We show that with an arrow to the left and a small positive sign. That means any force facing left is assumed to be positive. Now look at the y-axis forces, we assumed up to be positive. So the 8 kN is facing down, which means it's negative. I hope that helps :)
Great work
Thanks!
I still dont understand how u simplified the equation in 9:17, (Fdb = 2P ), but where did the Fda go then ?
I've solved it for you here: drive.google.com/file/d/1_CmBjvBnuvTXX-dmYY4VpGSVfu7Jwd0p/view?usp=sharing
It's just using the normal substitution.
First up great video, it has explained it better than my current lecturer who doesn't explain it as clear as yourself. @9:16 did you solve the two equations like at 8:02? Just making sure that I cover all of my workings to show my intent and how it was achieved.
Thank you, and yes, almost all of the simultaneous equations I've solved has been done so, using the substitution method. If there is more than 4 unknowns, I would suggest to use a matrix, but that usually comes up in dynamics, not statics.
@@QuestionSolutions do you have an example? confused on what to do with the p
In 8:03 , how come that FEB become tension? Since, both FEB and FEA has a Negative value
So it's not based on negative or positive values. It's based on whether the force was coming towards the pin or going away from the pin. Force EA is coming towards pin E, so it's in compression. Force EB is going away from pin E, so it's in tension. Watch the video from 1:26 again and it should help.
3:02 why is Force DC sin and no cosine in the y direction?
Please see: th-cam.com/users/shortsvynnKlJD_Jo?si=Er6Zt0UO4NloR9jU
Remember, we only care about the y-component when we write an equilibrium equation for the Y-axis.
Thank you very much
9:29 Hello at fdb How'd you get T? Do you have any video about it
Sorry, I am not seeing T at the timestamp you provided. Are you referring to the tension associated with force DB?
4:25 im confuse why left value in Compression and right value is tension but both in positive
So it has nothing, absolutely nothing, to do with whether a force is positive or not. Positives and negatives only tell you whether an assumption was right or not. What matters is the direction of the force. Is the force coming towards the pin or going away from the pin. Rewatch this part from 1:48.
Why have you used sin to calculate for the x axis instead of cos on your second example please clarify me so i can continues
Sine and cosine are NOT related x or y-axes. They are related to the angle and their opposing sides. Please see this video, it's under 60 seconds and will clear up any confusing parts. th-cam.com/users/shortsvynnKlJD_Jo?feature=share
hi, at 4:20, i am confused how Fce became 9.24kn, can you break it down?
thankss
There really isn't anything to break down. You solve directly for FCE. So plug this into your calculator (9.24sin60)/(sin60) = 9.24
You can also divide both sides by sin60, so you're left with just 9.24. Sometimes, thinking of FCE as just "x" might help you visualize it better.
@@QuestionSolutions Thank you!!!
I know its not correct but why can’t I use a sum of moments about c to find Bx or Ax, please see my equation, thanks!
I used sum of forces in X of the whole truss and got Ax+Bx-900-600
Ax=1500-Bx
Mc=600(4) +Bx(4)+Ax(4) Ay(6)
Sub in Ax
Mc=600(4) +Bx(4)+(1500-Bx)(4) Ay(6)
This yields Bx as canceled but I don’t see why this happening, have I missed a force? If no which one. Thanks for taking to time to read this, its much appreciated!
Please kindly provide a timestamp so I know where to look. Then I can look through your equations :)
Thank you so much! Simple and understandable.
You're very welcome! I wish you the best with your studies.
I don't get the part in 4:54 😭 wdy divide both with sine60 ,and in Fx what if the assumption was wrong how would you know what's wring between the two
So you divide everything by sin60 to get rid of sin60 on all the terms. That way, the only things you have left is just FBE and FBA. If your assumption is incorrect, you will get a negative answer. Then you need to flip the direction of the arrows.
At 8:02 sir you had your solution in a pdf and I’m a bit confused how did the y = 1500-0.6x/0.6 became y = 2500 - x Thank you
So let's say you have a fraction like this: (4+5)/6. This is the same as 4/6+5/6. So in our problem, we have (1500-0.6x)/0.6. This is the same as 1500/0.6-0.6x/0.6. If you divide each term individually, you get 2500 - x.
Thank you so much you made the topic so easy i am a little confused where i put the angle i found do i put it x axis or y axis ❤❤❤
You're very welcome. You can put the angle "against" any axes, as long as you use the correct sine or cosine for the components.
5:34 So I solved this joint while pausing the video and noticed that I didn't even need the Y reaction since I already found the only unknown (F_ea)with the X reaction. But in what case would the Y reaction be useful?
-
Also lets say the roller was at Joint A, how would that affect the reactions at A and forces (F_ba) and (F_ea)?
BTW, this was an amazing video. I couldn't understand a single thing in my lecture and I'm glad I can do my assignment in peace now
It's useful if it's required in your question. If it doesn't, it's just extra forces. 😅
You can't have 2 rollers as the only support. Nothing would stop the bridge from going left to right. But let's say for hypothetical reasons, you have 2 roller supports. Then the only difference is, at A, there would just be a single vertical force, and there wouldn't be an x-reaction force.
@@QuestionSolutions I see, we'd have a moving bridge then. 😂
On your 2nd example, why did you use sin on the summation of forces on the x axis while cosine on the y axis?
Sine and cosine are not related to x or y axes. Please see: th-cam.com/users/shortsvynnKlJD_Jo
I have a question at 3:21 can we assume opposite x direction to be positive (right side)? when ı assume right to be possitive ı find -4.62 but in wrong direction. I know that F_DE should be in the right direction.
You can assume any direction to be positive. If you end up with a negative answer, then you know it's opposite to your assumption.
@@QuestionSolutions but i still get wrong direction my equation is -FDC*cos60 - FDE = 0 then ı find FDE= -4.62 ( ı assume right to be positive ) then in this case FDE is in the left direction
@@sevgipnar5261 No, you're getting the right answer since you got -4.62. It just means the direction we chose for the arrow F_DE is incorrect. It was facing to the right. I think you're getting 2 things mixed up. The direction of the arrows we chose can give us negative values, which means the direction we picked was wrong. When we solve problems, you can pick forces to the left or right to be positive. So there are 2 choices going on.
@@QuestionSolutions so that means everytime i get negative values my assumption is incorrrect and i should change the direction?
@@sevgipnar5261 Yes. Let's say you picked your arrow to face left, and when you solve your equations, you for a negative value. That means your arrow actually faces to the right. Please re-watch 3:26, I show how that's done.
At 3:03 how’d you solve the equation to get 9.24Kn?
Isolate for F_DC
F_DC = 8/sin60
F_DC = 9.24
Why did you switch the cos and sin for the equilibrium equations at the third example?
It's based on how the angle is given. Please see: th-cam.com/users/shortsvynnKlJD_Jo
Thank you , you helped me a lot🥹💙
You're very welcome! 💙
sir why F(DE) compression ? 6:29 because you assumed is going to that direction therefore because now is positive it must be tension you have to change the direction.... can you pls explain that part to me....and thanks for the video I learn a lot from you.
So we made an assumption that force DE would be going towards pin D. We got a positive for force DE, which means our assumption was right. Any force that goes towards a pin is always in compression.
@@QuestionSolutions so if it was negative ?? Then we can change the direction??
@@totmanthescorpion Yes, I cover that scenario at 3:28, we get a negative value, which means our assumption was wrong, so we need to flip the arrow.
hey man why you used sin in the equation of x axis at 6:34(time)
Please see this video: th-cam.com/users/shortsvynnKlJD_Jo?feature=share Watch the whole thing, it's less than 60 seconds.
Thanks for your response Dear
@@QuestionSolutions
bloody brilliant amazing explanation made me stop crying
Thank you very much :) Keep up the awesome work!
2:58 how do you calculate -8 + FdcSin60 = 0
So you need to isolate for F_DC. Add 8 to both sides and then divide both sides by sin60. You have: F_DC = 8/Sin60. Plug that into your calculator and you get: F_DC = 9.24
thank you so much@@QuestionSolutions
You're very welcome!@@gary1585
hello. why is the vertical reaction in point E not included when the summation of vertical forces were taken? thank you
I don't know where you're referring to. Please use timestamps.
When you use the sine to get your forces at Y, for example why don't you use angle 30 instead of 60? I have trouble making sense of that, for example at Joint C. I can also verify your answer with similar triangles, but if you can let me know why 60 degrees, thanks.
You can use 30 degrees if you want, as long as you use the proper cosine and sine functions, you will get the same answer. 👍
@@QuestionSolutions Would you mind showing me through an e-mail or through here? I usually struggle to choose the angle for sin from the top joints. I can follow the method, just having sense of which angle to use for sin, for cosine I know it is always the adjacent angle.
@@enrique2914 Please kindly watch this video first: th-cam.com/video/NrL5d-2CabQ/w-d-xo.html
If you're tight for time, then please watch the first example. I go through the whole process, how to pick sin and cosine, break forces into components. etc. The video is only 9 mins long, and I think it can help you a lot. Let me know if that helps :)
I'm a freshmen studying engineering early, why did you plug in 9.24 in sin(60) in 4:13 timestamp.
We are trying to get the y-component of the 9.24 kN force. That's opposite to the angle, so we use sine. Please see: th-cam.com/users/shortsvynnKlJD_Jo?feature=share
Hello wanted to ask , what if i used joint A instead of join E to find F A/E. Is still goona be a correct answer or i need to use join E in order for the lecture to mark it as correct ?!!
I am not sure where you're referring to, regardless, it makes no difference as long as you end up with the same answers. Unless your professor specifically asked you to start at a certain joint, you can solve these problems however you please. Show your steps properly, and arrive at the same answer, and you should get full marks.
Hi can you do a videon rigid body systems in equilibrium but with Pulley examples? thank you
There might already be one, please check this playlist: th-cam.com/play/PLXePpKFSUW2ZxGn6VdAPY9d7l7HowWVYL.html&si=pVk8aZ9b7FjUqmN9
Great explanation!
Thank you!
your videos are very helpful
please make more
I will do so 👍 Glad to hear they are helpful.
Hi, at 5:06. how to find which force is in tension or compression?
Please re-watch from 1:27 where I explain how this is done.
Thank you bro you save me
Happy to help :)
Thank you ❤
You're welcome 😊
I have question that if you want to select direction of force you can select it freely or by analysing the whole problem
You can freely choose it. You will get the correct answer regardless. However, as you gain more experience with these problems, you should be able to mentally pick the proper directions quite easily so that you won't have to switch them at the end.
Thx for reply, now I am clear about that
at 6:14 shouldn't have been 8/6 since 8 is the opposite and 6 the adjacent ?
No. You need to look from the point of view of the angle. If you look straight from the angle, it's 6m straight ahead, which is considered the opposite direction, and 8 in the adjacent direction. I hope that helps. 👍