Trusses Method of Joints | Mechanics Statics | Learn to Solve Questions
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- เผยแพร่เมื่อ 24 ก.ค. 2024
- Learn how to solve for forces in trusses step by step with multiple examples solved using the method of joints. We talk about determining force directions, what compression and tension is and much more.
🔹Breaking forces into components: • Vector Addition of Cop...
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Intro(00:00)
Determine the force in each member of the truss. (02:29)
Determine the force in each member of the truss and state (05:49)
The maximum allowable tensile force in the members (08:08)
Find more at www.questionsolutions.com
Book used: R. C. Hibbeler and K. B. Yap, Engineering Mechanics Statics.
Hoboken: Pearson, 2017.
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
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 :)
Literally the best video on trusses I have seen.
Thank you very much :)
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 :)
I've watched most of your lessons and you're the best
Thank you very much! I hope they were helpful to you :)
This by far is the best video on Truss!
Thank you very much :)
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. :)
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.
So good, thank you for all great videos!
You're very welcome!
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.
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. 👍
You are my hero, man! Thank you!
You're very welcome!
Please cover more topics, your videos are so far the best teaching videos on youtube.
Thank you very much! :)
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
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 :)
thank you
you express in understandable way keep it up!
You are very welcome :)
Thanks bro..i recommended your channel to my mechanical Department
Thank you for the recommendation, 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 :)
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 really appreciate this!
You're very welcome!
thank you so much brother may God reward greatly
You're very welcome and thank you so much! :)
Great explanation!
Thank you!
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. @@user-wt8po2uq6c
@@QuestionSolutions nicee ,
VERY NICELY EXPLAINED
Thank you very much!
Thank you so much! Simple and understandable.
You're very welcome! I wish you the best with your studies.
Thank you you are the best 😊.
You're very welcome! 😊
Great video!
Thank you very much!
this video saved my maths project. thanks.
Glad to hear! You're very welcome.
thanks u SIR.to clear my concept about 2nd question.
You're very welcome!
your videos are very helpful
please make more
I will do so 👍 Glad to hear they are helpful.
this video shows us how we must understand the joining of trusses
thank you for helping us
You're very welcome :)
Thank you bro you save me
Thanks a lot really helped
Glad to hear :)
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. 👍
Great work
Thanks!
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
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!
THANK YOU!!!!!!!!!!
You're very welcome!
Thank you ❤
You're welcome 😊
Straight Goat
Thanks! 👍
Bump for the algo. Love the content.
Thank you. Really appreciate it.
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
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.
thanks alot broo
You're welcome!
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.
You are amazing keep going...
Thank you very much, I appreciate it.
@@QuestionSolutions 💙
I love it
Many thanks :)
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.
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.
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.
Wow thank uuu
You're welcome! :)
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.
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.
sir, quick question. for the first question, will there not also be horizontal and vertical reaction forces at the support in A to counteract force BA and AE? and also sir, what are the different types of reaction forces provided by different support types?
Since A is a pin support, it will have horizontal and vertical reaction forces. However, the question isn't asking for those so we don't need to worry about it. We are only looking for the forces in the internal members. In this video, I talk about a few different support types: th-cam.com/video/jQDEOwrR4UU/w-d-xo.html
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
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 :)
Thx m8
You're welcome!
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
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!!!
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 :)
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 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 :)
If you have a member A to B which is upright in a truss, And there is a force on joint B going down can you move that force to joint A? (A is not a pin, member AB is somewhere in the middle of a truss)
Any chance you can send me a picture or a rough diagram of what you're talking about? It'd be easier to visualize and give a proper answer. Thanks!
Thank you , you helped me a lot🥹💙
You're very welcome! 💙
Can I ask you a specific question about kinetic energy and work?
I just sent you an email
@@kazimozel6193 Replied 👍
I didnt know what to do. Just watched this 5 minutes before quiz at my university, and I ended up acing the quiz.
I am really happy to hear you aced your quiz! Nice job. Keep up the awesome work and best wishes with your studies.
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
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.
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.
bloody brilliant amazing explanation made me stop crying
Thank you very much :) Keep up the awesome work!
if you didn't have a external force of 900N going through member CE, how would you work the force of that member hypothetically speaking. Could you use the force of member ED to help solve?
Sorry, can you give me a timestamp so I know where you're referring to. It'll allow me to help you out better. Thanks!
@@QuestionSolutions 7:00
@@Umarranis Solve it without the 900. You'd write your equations the same, just without the 900. So you can visually see in that case, force CE would be a zero force member. If you can't, that's okay too, but writing an equilibrium equation for the x-axis forces would be simply give F_CE = 0.
Hi @ 8:52 how did you calculate Fcb=1.155P (C)
So this isn't the same example, but I've solved one pretty similar here: bit.ly/3kGNcyW
Great video. One question though. I'm failing to understand how you've come up with the force for DC, I've plugged in -8 sin(60) with my calculator in degrees. It doesn't come up with the 9.24 kN you've found. Have you any idea as to what I'm doing wrong?
Are you referring to 3:04? If yes, you've made a slight mistake. So our equation looks like this (I will use "x" for F_DC):
-8 + xsin60 = 0.
You now need to isolate this equation for x. So add 8 to both sides, and then divide by sin 60 on both sides.
x = 8/(sin60)
Now, you can plug this into your calculator.
x = 9.24 kN
I hope that helps.
@@QuestionSolutions Yees!! Finally got it. Absolutely brilliant. Thank you very much!
@@TomDavid-bw6in Awesome!!! Really glad to hear you got it. Keep up the great work and best wishes with your studies.
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 🤙
Great video they help me a lot so thank you for that! I just have a small doubt , after we assume the direction of a force and get a negative answer is it okay to change the direction then to the correct orientation ? Does that effect the results in any way?
You can do that, but it's so much easier to stick to what you get and continue with the same positive and negative signs until the very end. It's just too easy to mess up the signs when you're trying to switch them. But, yes, as long as you account for all the changes and you are very careful, the answer you get will be the same. Personally, I don't recommend it, I find a lot of students already get positive and negative signs mixed up 😅
@@QuestionSolutions Alright I'll stick to your advice, thanks a lot!
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
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
Hello sir, at 5:42,may I know why Fea is in compression although it gives positive value?
If the force comes towards the pin, it's in compression. We drew the force pointing towards the pin, we got a positive value, which means our assumption was correct. So remember, if the force comes towards the pin, the member is in compression.
Understood. thank you sir 👍love your lectures btw🥰it helps me a lot 🥰😄
@@jezzyanalachaw8325 That's awesome to hear! Best wishes with your studies. 👍👍
In 6:40 why is it sin in Fx and cos in Fy? Shoudn't it be the other way around?
No, the opposite side to the angle is sine and adjacenet side is cosine. So here, F_X is sine because that's the opposite side. Please take a few minutes to watch this video: th-cam.com/video/NrL5d-2CabQ/w-d-xo.html
Sine and cosine are NOT related to x and y sides, they are related based on their angle and it's corresponding side.
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. 😂
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
Hello, why did you use sin in finding the value of Fde in 2nd figure when it says summation of Fx?
So I am thinking what you're asking is, why is sin used for x-components? Because sine and cosine are NOT related to x or y components. They are related to the side with respect to an angle. Please don't think that sin is just for y-components or cosine is just for x-components. You have to look at the angle, and see whether the component you're trying to find is opposite to the angle or adjacent to it. Here, we want the x-component of force F_DE, which is opposite to the angle, which means we need to use sine. If you need a refresh, watch this video, especially the first example: th-cam.com/video/NrL5d-2CabQ/w-d-xo.html
@@QuestionSolutions Thank you, now I get it. Cos for CAH and Sine for SOH
@@heavenskyclouds4480 That's right! Best wishes with your studies :)
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.
Just a quick question, how would I go about finding Ey in the first problem if I decide to use reference angles. I ask this because when trying to find Ey in example 1, my equation is Ey + F_BEsin(120) + F_CEsin(60) = 0. When I do this I get -16 for Ey but in the video it's equal to 16. I'm not sure what I'm doing wrong.
Why did you use sin 120 for the y-component of force F_BE? You can use either 60 or 30. If you use 30, you need to flip the components, since now you're looking at the sides of the right angle from the opposite side.
I still don't understand how you assume which forces are in tension and compression. Is it just if I get a negative value I flip the sign? Also, how do you check when you're solving systems of equations?
So an assumption means it's just a guess. You are guessing the direction first. Let's say you guess the force is going away from the pin, and you solve your equations and end up with -40N, then you know it's opposite to your assumption. In other words, the force is coming towards the pin and it's a positive 40N. So if you get a negative value, then you flip the assumption. I actually show this on the first example at 3:30. I encourage you to re-watch the video and do the problems along side what's shown on the screen. It will clear things up for you.
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
Sir on 6:32 why was sin used in looking for the x component not cos?
Please see: th-cam.com/users/shortsvynnKlJD_Jo
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
hi sir, when do we consider the reactions at the supports? great video👌
You can consider them if the question asks you to find them, or you have no place to start and you need them as givens.
@@QuestionSolutions okay thank you🙏 also with the method of joints, do we not consider the zero force members?
@@PulengManchidi There isn't much to consider about zero force members. What I mean is, doing a few examples will allow you to instantly realize which members are zero force members, so you can easily write them off. Other than that, when you solve your problems using the method of joints, you will get zero force members pointed out in the solution.
hello , I hope you can answer since the video is old
i just wanna ask you , Fdc on the first example the one you found = 9.24Kn isn't that Fdcy component ?? and not Fdc ? why did you plug that value in Fdcx
No, if you want the y-component, you will need to use sin60 and get the component. Otherwise, what you are getting is the resultant force FDC, which is why sin60 was already used in the equation. Again, just to clarify, what you find as FDC is FDC, NOT any component.
@@QuestionSolutions what you find as FDC is FDC, NOT any component. That explains 4 questions i have :') thanks a lot mate, i have test tomorrow lookin back at some work my professor doesn't explain sh*t I only find ready answers (results) . again thanks
@@TH-cam_vods130 You're very welcome. Your question is common to a lot of students, so I am really glad it cleared it up. :) Keep up the awesome work and best wishes with your test.
I'm just watching this now
At 6min why did you use sine to solve for the X axis forces(Fx) and cos for the Y axis forces(Fy) at point D
I think you might be under the impression that sine is for y-axis forces and cosine for x-axis forces, or vice versa. This is NOT true, and it is incredibly important that you remove this idea. Please take a few minutes to look at this video (especially the first example): th-cam.com/video/NrL5d-2CabQ/w-d-xo.html
Sine and Cosine are related to the angle and their corresponding sides, not set for x-axis forces or y-axis forces.
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. 👍
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.
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 :)
In the 1st question why did you assume that fx is going to the left in FDC? I tried assuming it would go right then I got tension
For the future, please use timestamps, thank you :)
So you're free to assume any direction you want. Let's say you assumed it would go towards pin D(which is what I assume you did), you should end up with -9.24kN. The magnitude of your force is correct, however the negative sign indicates it's opposite to your assumption. Which means it's actually going away from pin D.
@@QuestionSolutions oh so you assumed left is positive since in the fbd you made it go towards left?
The initial assumption is just a guess. So you can pick the direction of the arrows anyway you like. When you write your equations, you can pick right or left to be positive and write your equation according to your choice. You will always get the same exact answer but if you get a negative value, then your assumption for the direction of the arrow is incorrect. It's opposite to your assumption.@@akira...7819
@@QuestionSolutions thank youuu
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
How can you tell if it's sin or cos when summing forces in the X or y direction
(Ik sohcahtoa, ik sin is rise and cos is run, I'm in uni) what triangle are u visualing, cause what u declare the opposite and adjacent to be will dictate which of the two you'll decide on
See: th-cam.com/users/shortsvynnKlJD_Jo?si=VXlf4Xev7merwG8w
You can visualize the triangle whichever way you want as long as you know how to use sine and cosine properly, you'll get the same answer.
@@QuestionSolutions thank you, I really appreciate your care I didn't expect a reply let alone that quick, thank you 🙏
You're very welcome! Let me know if you need further clarifications. Best wishes with your studies.
why you use sin in summation of Fx forces instead of cos?
Please see the whole video, it's less than 60 seconds, it will help you out a lot: th-cam.com/users/shortsvynnKlJD_Jo
Because my teacher had taught some strategy i didn't remember to select the direction of force
I hope the video helped :)
Can you articulate more on convection sign
Which sign conventions are you referring to?