Trusses Method of Sections | Mechanics Statics | (Solved examples)

[@QuestionSolutions]

At 9:08, there is a typo. F_HI = 42.5 NOT 45.2.

[@remyfru]

I thought I was losing my mind when I solved it myself and watched it back, thanks for clarifying, and thanks for the great video.

[@QuestionSolutions]

@@remyfru You're very welcome :)

[@davidaum07]

@@QuestionSolutions thanks

[@yanskiejusain9611]

My professor's lecture video was 45 minutes long, but this 11-minute video was elaborated well. Thank you.

[@QuestionSolutions]

You're very welcome! I try to make them as concise as possible.

[@zach.frederick]

Wish my $2500 a semester could go to this guy instead of the professor that blabbers about how easy it was for them when they took the class 30 years ago. This guy is goated fr fr :0

[@NOTONtechsx]

@@zach.frederickMine is $3000 a semester and our professor just blabbers useless stuffs 🥲

[@katomiyamashita2870]

My prof sent a video lecture that's like 8 hrs

[@PunmasterSTP]

How'd the rest of your class go?

[@رياضالصالحين-ل5ب]

انا عربي اشكرك على جهودك هذه ولك التوفيق ان شاء الله في الدنيا والأخرة I am an Arab. I thank you for your efforts. May God grant you success in this world and the hereafter

[@QuestionSolutions]

You're very welcome! keep up the awesome work with your studies.

[@DirkdeZwijger]

Great video, clear explanation of how and why to use this method. Basically a time-saver for convenience sake is what I get from it. Understanding this method also boosts your insights and understanding of statics

[@QuestionSolutions]

Glad it was helpful! Keep up the great work and best wishes with your studies.

[@earnstein7607]

Thank you so much. You don't know how much you've helped me tonight ♥️

[@QuestionSolutions]

You're very welcome! I am really glad to hear this helped :) ♥️

[@xyz-hz9zc]

Thats what she said💀

[@musabbirahmed4591]

thank you so much for existing. I promise you one day when I become a successful civil engineer I will contact you surely and thank you for every help u did throughout my engineering program. Just remember me. I am taking a screenshot of this comment.

[@QuestionSolutions]

You’re very welcome. I’m glad to hear that you’re pursuing your dream of becoming a civil engineer. I’m sure you’ll do great things in the future. I appreciate your kind words it means a lot to me. Best wishes :)

[@garrettsanford1493]

I should be paying you my college tuition How am I paying some boring dude to tell me about physics in the most unstructured way at 8am for $1800 when you just make it so simple and quick. Hats off

[@QuestionSolutions]

Thank you very much. I hope you do amazingly on your courses and I wish you the best! Keep up the great work :)

[@muhammadsaadkhan9701]

Oh My God . This Lockdown means free time and your channel means free great quality explanation . Thank u so much 🌹. Enjoying Mechanics . ❤️🥰

[@QuestionSolutions]

Glad to hear you enjoy mechanics. ❤️ You're very welcome and I hope you learn lots!

[@Alicia-de4hj]

This was so helpful! Thank you for your easy explanations

[@QuestionSolutions]

You're very welcome. Glad to hear it was helpful

[@PunmasterSTP]

Method of sections? More like "Magnificent and knowledgeable explanations!"

[@QuestionSolutions]

Haha, thanks! :)

[@israfeelaheadmhadi102]

It is the best channel I have seen. Greetings from Iraq 🥀♥️🎉

[@QuestionSolutions]

Thank you very much! :)

[@Lostwolf16]

Thank you so much, while cutting section, can I make a slanted (diagonal) cut to avoid solving for member that I really don't need for problem?

[@QuestionSolutions]

Yes, you can cut vertically, horizontally, diagonally, etc. 👍

[@Lostwolf16]

@@QuestionSolutions Thank you

[@kon8024]

Im happy you engage with the people in the comments. It seems all my answers have been answered! Thank you.

[@QuestionSolutions]

Thank you, I try my best to make sure I answer all the comments I get. :)

[@ElCrankoPunko]

@9:12 Force in HI member would be 42.5 kN

[@QuestionSolutions]

Yup, there is a typo. Thank you for pointing it out.

[@joannabell3826]

At 6:03 you have set the 5kN force at point E to be going in the positive direction and F(GH) to be negative for finding the moment around point B. Can you elaborate on the best way to choose the direction of the forces when setting up the moment equation? I would have set it to -5kN and +F(GH) based on the direction of the forces pictured and gotten the incorrect answer.

[@QuestionSolutions]

It doesn't make any difference as to how you set the directions of your forces. You will get the same answer. The reason you are getting incorrect answers is because you're writing your moment equation incorrectly. So the direction of force doesn't matter, it's whether it creates a clockwise or a counter-clockwise moment that matters. Here, the 5kN force is pointing down, that doesn't mean it's negative. What we look for, is to see whether that force would turn the truss clockwise or counter-clockwise. Imagine you hold the truss between your fingers at B, and we apply a force exactly where the 5kN force is, downwards. How would the truss turn? It would turn clockwise, right? Now notice next to the big sigma sign, we show a clockwise positive sign. That means we assume any force that creates a clockwise moment to be positive. So since the 5kN force makes the truss go clockwise, it's going to be a positive moment. Now look at force FG. If we apply that force, how would the truss turn between your fingers? It would go counter-clockwise. So it'll be negative. When you write moment equations, the directions of the forces do not determine whether something is negative or positive, it's how this force will act upon the object itself, and what sort of moment it creates, that determines the positive or negative sign. I hope that helps, if you're still confused, please see: kzbin.info/www/bejne/h3-xn4OQa52Giqs

[@smartybayan6646]

For the 3rd problem. The Force HI is 42.5. Thank me later.

[@QuestionSolutions]

👍

[@iyakintv5144]

yeah.. this should be pinned. It can also be obtained by applying summation of moment at A.

[@makristaamada-yo2uy]

thanks i thought i got it wrong

[@toondebeule8377]

hello i have a question at 2:04, why do you calculate Ay with a moment, Can't we not just work with the sum of y=0 -> Ay-2-5-5-5-2=0 -> Ay=19kN , i know this isn't the same answer but why does this not work and why do we need to use the moment

[@toondebeule8377]

it it because the is also a single force Ey in E, so Ey + Ay=19, PS: i have been watching your whole series on statics and i these kind of videos have never been so helpfull as your channel, the fact that you answer to every question shows your passion keep up the good work!!

[@QuestionSolutions]

@@toondebeule8377 Yes, that's correct. We don't know the value of Ey, so you can't use a summation. Instead, we take a moment about E, which eliminates EY, and we can the directly solve for Ay. Also, thank you very much for your kind words. Best wishes with your studies!

[@toondebeule8377]

and i also have another question, in my class i have to work with counter clock wise moment is positive what does this change to the solution, when i do this my forces are negative is this possible? so when i calculate Ay i get -9,5kN but in the next step it doesn't work because the 2kN and 9,5kN are in the same direction, what would be your solution?

[@QuestionSolutions]

@@toondebeule8377 The direction you choose for positives or negatives does not matter in 2D problems. You will always get the same answer regardless of the direction you pick. It's completely up to you. Please see: kzbin.infoP029mqnp4XY?feature=share

[@HaymanotShitie]

Thanks very much,qm from Ethiopian 🇪🇹 Keep it up

[@QuestionSolutions]

You're very welcome!

[@jeffkwin3129]

Great video, but I got a question at 5:11, shouldn't the forces 2kN and 9.5kN only apply moment on the components perpendicular to the truss AH, because you also did that for Fbg?

[@QuestionSolutions]

The 2kN force and the 9.5kN force are perpendicular with respect to where the moment is calculated. If you imagine your finger at the location where the 2kN force is, and push, will the truss spin about our moment point? It would right? So it will create a moment. I think you're assuming we are doing the moment about the whole member AH, which is not true. We are calculating the moment about the pink dot, just point H. When we calculated the moment about point A, only the y-component of force BG creates a moment. But notice that the 2kN and 9.5kN force only have y-components, they are vertical, so it will create a moment about point H.

[@REMythic7]

For the vertical truss, does that mean member BF doesn't need to be there since the force is 0?

[@QuestionSolutions]

In real life, no, because they are there to help with rigidity and stability. Sometimes, they will carry a force if the conditions of the bridge changes. lastly, they add to the aesthetics as well. People don't like seeing parts "missing" off of a bridge 😅

[@rpian1999]

@@QuestionSolutions Hello! How come F bf is equal to 0 when the x component of Fbg is acting on BF and also 10 kN is also acting at BF? I want a solution to find Fbf using moment method.

[@QuestionSolutions]

@@rpian1999 Please give me a timestamp so I know where to look, thanks!

[@rpian1999]

@@QuestionSolutions 7:38 time in that video where Fbf is being determined.

[@QuestionSolutions]

@@rpian1999 So BF is a zero force member, that means it doesn't carry a force. The forces applied at B are carried by the other members, not BF. If you want, you can look at point B and do the equations, you will still get 0 for member BF.

[@milesbrack9188]

For the question at 5:25, why does the force in member BA not equal the reaction force from point A? I found a reaction force from joint A of 45kN when taking moments about point G, but the force in member BA is 50kN.

[@QuestionSolutions]

The cut was made above the reactions so they were not needed. No calculations were made to figure out the reactions at A, or G. Also, where did you get that BA is equal to 50kN? At 7:07, we find that BA = 45kN.

[@milesbrack9188]

@@QuestionSolutions Thanks for your quick response! I wanted to verify that the force in member AB equaled the vertical reaction force at A. I understand that it is not directly relevant to the section method. I just checked my calculations for some of forces in the Y axis, and I totally forgot to include the 5kN on point E. Now I have the correct solution for AB. Thanks again for the brilliant video!

[@QuestionSolutions]

@@milesbrack9188 Okay, glad to hear it was cleared up. Keep up the good work and best wishes with your studies :)

[@swaggyseth1454]

huge video, got a silly little quiz tomorrow, this will definitely help!

[@QuestionSolutions]

Glad to hear. Best wishes with your quiz tomorrow!

[@mastfamastfa1256]

7:33 can you explain how ? we have 3 other forces in the x direction 10 + 10 + 5 + BF = 0 BF = - 25

[@QuestionSolutions]

Where did you get those forces from? So when you look at a joint, you only look at that joint, you don't care about what happens at the next joint over. There are no other forces applied at joint F other than vertical forces. I think what you're doing is translating forces from other joints to this joint, but remember, when we look at a joint, it's isolated. That's why you draw a separate coordinate system about that joint. At 7:33, focus on the right side of the screen and look at the forces at that joint. :)

[@mastfamastfa1256]

@@QuestionSolutions ❤

[@QuestionSolutions]

@@mastfamastfa1256 ❤

[@OptimusXZero]

Really helpful. The visual is really helpful for my learning!

[@QuestionSolutions]

I am really glad to hear that!

[@ngsbsad5502]

At 2:00 why there's a two forces in joint A, there's already a 2kn downward load why we have to put Ay in solving moment

[@QuestionSolutions]

That's a support reaction. So it's a pin support there, and that means there is an Ay component along with an Ax component. However, the ax component do not create a moment about point E.

[@darthplague]

The example my professor gave only had 1 external force, so i didn't know that I had to calculate the moments of the other downward forces Thank you

[@QuestionSolutions]

You're very welcome. I try to pick at least 3 examples to remove any confusions in students. :)

[@egsbue]

thank you so much for this, just the right time i found your channel before our quiz tomorrow!

[@QuestionSolutions]

Glad I could help! I wish you the best on your quiz tomorrow :)

[@kingzzzmen7242]

9:07 could you have used the moment at F and still got the same Force HI?

[@QuestionSolutions]

You can find the values using any moment location but try to use simple ones since you want to make your life as easy as possible. So go for the one that eliminates the most unknowns or leads to a direct force value.

[@OGTennyson]

Has anyone ever told u ur a life saver?🥺🥺

[@QuestionSolutions]

I think I heard it about 3 times now, from very kind people like you leaving nice comments. Thank you, it made my day! Best wishes with your studies.

[@OGTennyson]

@@QuestionSolutions thank you!... and ur very welcome ❤️

[@QuestionSolutions]

@@OGTennyson ❤

[@OGTennyson]

@@QuestionSolutions I watched like 10 of ur videos already and it's so damn informative. U got a new subscriber my friend...keep up the good work on the videos bro ✅❤️...I rlly hope one day I'll make it as a civil engineer and be able to donate cash to ur channel because I feel ur teachings are highly underrated and I wouldn't want you to stop 😭. Keep the videos coming bro...imma be here 4 a while❤️

[@QuestionSolutions]

@@OGTennyson Thank you very much! Don't worry, you will make it as a civil engineer, you'll do great. Keep up the great work, do as many questions as you can, and use every resource out there to help you succeed. Especially those office hours from professors. 👍

[@XIVRedd]

QUESTION : at 8:23 why didnt we divide height over base? Like 1.5/2???

[@QuestionSolutions]

Divide it by 2 to get the angle or to find what?

[@XIVRedd]

@@QuestionSolutionsdont we usually do Tan(angle)=(height/base) ??? Our height is 1.5 and the base is 2… why in the video we did the opposite…like base over height…sorry for my bad explanation

[@QuestionSolutions]

@@XIVRedd That's incorrect and you should remove that thought from your mind. Instead, remember, tan = opposite over adjacent. So you look from the perspective of the angle, and look for the opposite side to the angle. Here, that's 2 m, and the adjacent side is 1.5 m. While we are at it, remember this too. Sine is opposite over hypotenuse and cosine is adjacent over hypotenuse. Always look from the perspective of the angle and look to see what side is opposite to it and what side is adjacent to it.

[@XIVRedd]

@@QuestionSolutions I’ll memories it right now so it doesn’t get me confused…thank you soo much sir :)

[@QuestionSolutions]

@@XIVRedd Great! If you have any other questions, let me know. Keep up the good work and best wishes with your studies :)

[@sayeedahmad8594]

Such a good video. You just earned a subscription. Keep them coming!

[@QuestionSolutions]

Thanks for the sub!

[@petrus2746]

For anyone wondering why Force BG was eliminated 3:01, it's because it's vertical component goes through B and so does its horizontal component

[@QuestionSolutions]

That's right, but even more easier to remember is to keep in mind that any force starting where the moment is calculated, cannot create a moment at that point. The same as force BC.

[@jasondinosoccer]

Another amazing video, thanks for the extra help with studying this stuff! 🙌

[@QuestionSolutions]

My pleasure! Keep up the good work with your studies.

[@ZexMutsawu-nw2ov]

6:28 in your explanation you said that only the x component of force BG can cause a moment about D but on the working you used Sin instead of Cos. May you please clarify

[@QuestionSolutions]

So my guess is, you assume cos is for x and sine is for y. This is fundamentally wrong, and you must, absolutely must, remove that idea from your head. You have to look from the perspective of the angle every single time. Here, the opposite side to the angle gives us the x-component, and sine is opposite over hypotenuse. So you need to use sine. To get the y-component, which is now adjacent to the angle, you need to use cosine. See this video, it's less than 60 seconds and you will see what I mean: kzbin.infovynnKlJD_Jo?feature=share

[@ZexMutsawu-nw2ov]

Noted thank you ​@@QuestionSolutions

[@QuestionSolutions]

@@ZexMutsawu-nw2ov Awesome. Best wishes with your studies. Let me know if you need any further clarifications.

[@stevo_cars]

6:15 is it also possible to find the moment about point C since it also helps us get rid of F-BA?

[@QuestionSolutions]

Yes, you can write a moment about any point you want. The goal is to eliminate as many unknown forces as possible.

[@quh2828]

6:31 Why the FbgSin45(8) is positive but not negative? Thank you🥺🙏🏻

[@QuestionSolutions]

So if you try to imagine the x-component of FBG, it will be pointing to the left. In other words, it will try to make this whole truss turn clockwise, and since we picked clockwise to be positive, it creates a positive moment.

[@quh2828]

@@QuestionSolutions​​⁠ Thank you for your reply and explanation🥺🙏🏻, but I can’t imagine the x-component of FBG pointing to the left. I thought it points to the right😢Could you please further explain why it points to the left?😭🙏🏻

[@QuestionSolutions]

@@quh2828See this diagram, I've drawn just point B so you can see the force and it's components. drive.google.com/file/d/12S2auvgrdjIPVzI8UhO1KHartq6aIhM8/view

[@Khromax]

I don't understand how do you know the direction of the force based of the positive/negative value? In the first example the value was positive 2 times in a row. In the first situation it was compression, in the second it was tension. How did you assumed those?

[@QuestionSolutions]

Please kindly watch this video first: kzbin.info/www/bejne/laOuYWWkmrR5Z5o Especially the introduction, where I explain how to determine whether things are in tension or compression. If, after you have watched it, and still have trouble, please let me know and I will do my best to help you out.

[@muhammadsyaiful67]

at 8:49 , can we know wether we need to x eithr clockwise or anti clockwise because my answer 345 for fdc

[@QuestionSolutions]

It doesn't matter which way you assume it, if you get a negative answer, it'll be opposite to your assumption. So here, I picked it going towards the pin, but assume we did the opposite. Then the only thing that changes is that instead of -F_DCsin53, you would get +F_DCsin53. Which would yield a negative answer, and that means your assumption is incorrect. If you get 345, you made a numerical error, please double check your work.

[@armanjanndajab6334]

8:42 i'm confused at this part. i thought sin DC should be eliminated because it will go thru pin H, and instead use cos DC? or does it depend on where the angle is placed?

[@QuestionSolutions]

So you have to always look at it from the perspective of the angle when doing trigonometry. Sine is going to give us the opposite side (opposite to the angle), which in this case would give us the x-component of force F_DC (horizontal line). That will create a moment since it's line of action doesn't go through point H. Cosine on the other hand will give us the adjacent component, (y-component, vertical line), which will go through point H, so it's eliminated.

[@armanjanndajab6334]

@@QuestionSolutions ahh so we're getting the x component here but we just used sin because of the angle. I'm used seeing sin as y, and cos as x so i was confused😆. Thank u very much

[@QuestionSolutions]

@@armanjanndajab6334 Yes, but don't associate sin with x and cos with y. It's all based on the location of the angle. 👍 Best wishes with your studies

[@NK-iy6if]

God bless you! your videos are extremely helpful! Thank you so much!

[@QuestionSolutions]

You're very welcome! :)

[@kjartanalmar]

Hey Again! Well explained like always. I'm wondering why at 2:44 you make all member force face away from system except one and assume that it's compression. Is it just based on overlook and common sense? Thx, again!

[@kjartanalmar]

and then you do the same in 5:44 with the second truss. I wonder :o ..

[@QuestionSolutions]

@@kjartanalmar When you do enough questions, you can pretty much "see" where the force will go :)

[@syamilariq2222]

Why during 4:13 your value of 2kN that is downward is considered negative but then in 4:55 your value of 5kN which is also downward is considered as positive. This confuses me a lot

[@syamilariq2222]

So my conclusion is that downward and left force is considered as negative value and vice versa. But in 5:11, the force BC which is going right is considered positive?? Why is that? Mind explain to me

[@QuestionSolutions]

So it's not the direction of the force that matters when it comes to moments. It is the direction of the moment created by the force that matters. So let's look at the 2kN force at 4:13. We assumed clockwise to be positive. Now imagine the whole thing can rotate about point B. When the 2kN force is applied, which way will it turn? It will turn counter-clockwise. This is why it's negative. Now let's look at the 5kN force at 4:53. Now the moment is about point A, and again, we took clockwise to be positive. When the 5kN force is applied, which way will the truss rotate? It will turn clockwise, so it's positive. This is all to do with moments and you might be missing those fundamentals. If you have the time, please watch this video: kzbin.info/www/bejne/h3-xn4OQa52Giqs @@syamilariq2222

[@EfirDop]

I confused in part 7:20 b/c the force BC=0 is it correct??what abt the external force 10kN

[@QuestionSolutions]

Sorry, I don't understand your question. We didn't calculate force BC? Did you give me the proper timestamp? Regardless, all values shown are correct, except at 9:08, where force F_HI = 42.5 NOT 45.2.

[@markshevchenko7482]

thanks for the content, why in second to last problem we only consider that in F there is Fy component why not Fx as well?

[@QuestionSolutions]

Please provide me with a timestamp at the location you're referring to. I will take a look and explain it :)

[@paranoidparadigm886]

9:28 wouldn"t that be that Fx= -Fhc - (-125)sin 53.13 = 100 (T)? Since the direction is in the opposite of the assumed direction and the Fdc is compression so its negative (-)?

[@QuestionSolutions]

If the force comes towards the pin, it's in compression, if the force leaves the pin, it's in tension. Please see: kzbin.info/www/bejne/laOuYWWkmrR5Z5o

[@abdiqanicaligalol6192]

Thank 👍you for your excellent presentation

[@QuestionSolutions]

You're very welcome!

[@Wallerrrr]

i have a question, in the first question when you were calculating moment about point H, why did you include the 9.5(2) and -2(2)? I thought the only force perpendicular to point H would be fBC and the x component of BG? Please clarify, thanks!!!

[@QuestionSolutions]

So both the 9.5kN force and the 2kN force are perpendicular to point H. Imagine you move those forces up to be aligned with point H, so straight up. Now you can see it a bit easier that they are indeed perpendicular. Another way to see it is to realize these forces are vertical, and will cause the object to rotate about point H, so it must create a moment. Force FBC and the x-component of BG is NOT perpendicular to point H. That's why they aren't in the equation. I think you are confusing perpendicular with parallel. Please draw a big diagram on a sheet of paper and really carefully look at what forces are perpendicular.

[@punpunyana]

i have a question. do we always have to assume that the rotation is clock wise and positive? would changing that matter? im confused that if F turns out to be negative, my thought process is to change Tension to Compression (or vice versa) or change the rotation to counterclockwise (or clockwise)

[@QuestionSolutions]

I think you are confusing writing moment equations with forces in truss members. You can pick whatever direction you want for the moment equation to be positive, so clockwise or counter-clockwise, you will end up with the same answer. What determines if a member is in compression or tension is whether a force is coming towards the pin or leaving the pin.

[@punpunyana]

@@QuestionSolutions oh i see, i get it now. i tried changing the signs at 4:10 and i still arrived at positive 12.5, thanks!

[@QuestionSolutions]

@@punpunyana Awesome :)

[@AbdeenzRock]

for when calculating F(BG) why didnt you do Summation(y)? why did you have to take the moment at A?

[@QuestionSolutions]

Please use timestamps, I don't know where you're referring to. Thanks!

[@Julianzz_19]

3:46 PLZZ tell me how do you determine if its in cos or sin??? I understand the whole thing but always messed up the cos & sin part in the equation 😭😭

[@QuestionSolutions]

Please see: kzbin.infovynnKlJD_Jo?feature=share

[@benjaminyellin5095]

Quick question: at 2:04, how do we immediately know that there won't be any reaction on the horizontal axis? Like I can see that none of the applied forces have horizontal components, but couldn't AH (for example) cause a horizontal reaction from the support? Thanks!

[@QuestionSolutions]

So when you're solving for the reactions at a support, we don't care about internal members. They don't make a difference since we think of the object as a whole. So all the purple forces are external forces, but we have no forces applied in the horizontal direction. Let's say we had a horizontal force applied at point F, in that case, we would have an AX reaction countering the force applied at F. It's the same if we had a simple beam resting on top of supports. We just solve based on external forces, what happens inside the beam doesn't matter. Another way to think about it is to realize that while there is a force in member AH, that force goes from H to A and A to H, there by causing equilibrium. So to recap, when you are finding supports, internal forces do not matter, only external ones.

[@benjaminyellin5095]

@@QuestionSolutions Wow, thanks for the detailed response! Makes a lot more sense now :)

[@QuestionSolutions]

@@benjaminyellin5095 Really glad to hear it makes more sense. Best wishes with your studies :)

[@dirtyharrysgun2020]

Thanks man, you're the best

[@QuestionSolutions]

You're very welcome!

[@chumpmark3884]

This man is a legend

[@QuestionSolutions]

Thank you very much!

[@jr1753]

Just wondering for moments how do you determine the direction for them?

[@QuestionSolutions]

Please kindly watch this video: kzbin.info/www/bejne/h3-xn4OQa52Giqs So initially, when writing a moment equation, I pick clockwise to be positive (this is an assumption). So you're looking to see, if we apply a force at a location, will that member turn clockwise about the point where we are calculating the moment at, or counter clockwise. If it's clockwise, its positive, if its counter clockwise, its negative (because it's opposite to our assumption). The first 2 examples in the link I provided will help out a lot with directions.

[@gappity]

thank you for this, the best explanation so far.

[@QuestionSolutions]

You're very welcome! Best wishes with your studies :)

[@mohammadashu8984]

Really really great content..... I also want to learn such kind of animation ....can you please tell me how you do it? It would be great help to me....love from India.

[@QuestionSolutions]

I use after effects for animations. There are tons of really good tutorials on youtube, so please take a look :)

[@mohammadashu8984]

@@QuestionSolutions thanks a lot... The very first channel which I shared on my social networks.... Keep doing...... You will certainly hit the market.. 😀

[@QuestionSolutions]

@@mohammadashu8984 You're very welcome. Sometimes, Udemy has free courses on after effects as well, so keep an eye out. Thank you so much for the share, I appreciate it. 😁

[@gims6968]

Thanks for the video! But I have a question for finding force BF in the second problem. If we would have isolated joint B, equilibrium along forces at x would be 10kN - 45Kcos(45) - F(bf) = 0. Hence i thought that force BF would not be zero. Im confused how this works ://

[@QuestionSolutions]

I encourage you to actually isolate for joint B and solve this problem. For your equation, you should have 4 forces. The 10kN, BG, BF, and BE for horizontal components (along x).

[@brianandremacaraeg6781]

You are my hero! Could you also make a playlist about Strength of Materials?

[@QuestionSolutions]

Yes, that is also on my list to do :)

[@ElCrankoPunko]

@@QuestionSolutions pls do it soon. love your videos btw

[@QuestionSolutions]

@@ElCrankoPunko Unfortunately, Strength of Materials won't be done for some time. The next topic is thermodynamics. I think there are a lot of videos on KZbin about material science, I hope they can be helpful.

[@ElCrankoPunko]

@@QuestionSolutions cool man. eagerly waiting for thermodynamics videos :)

[@QuestionSolutions]

@@ElCrankoPunko :)

[@wajdy2620]

6:23 why can't we take Moment about point C? I tried it and it doesnt work

[@QuestionSolutions]

You can take it about point C to figure out force BG, you will get the same answer. Most likely, you are making a numerical error or you're not accounting for the proper distances/forces. It's hard for me to say without seeing your steps.

[@wajdy2620]

@@QuestionSolutions it was a calculator error thanks

[@MrStepBro]

Throughout all the Truss videos you talk about assuming the way the forces are acting and just back tracking if they are negative. Is there an intuitive way to always guess the direction correctly?

[@QuestionSolutions]

The intuition comes from doing a bunch of questions, in other words, experience. By the end, you should be able to look at a truss and know which way the forces are acting and be able to locate 0-force members. Also, it's not really backtracking per say when you guess wrong, you just change the direction of the arrow and make your value positive. The magnitude of the force is always correct, it's just a matter of whether it's positive or negative.

[@MrStepBro]

@QuestionsSolutions gotcha its just on my exam, we have to draw a free body diagram for a grade, and its online so its kind of a pain in the ass. So I guess a better strategy would be to mess with the free body diagram last and solve the problem first so I already know what direction the forces should be before putting them all in.

[@QuestionSolutions]

@@MrStepBro Ah, yes, it probably would be better to do it the way you mentioned.

[@JohnRovicAlejandrino]

how do you know if you have to get the summation of moment at a certain point or simply get the summation of forces?

[@QuestionSolutions]

If you have a set of equations, and you cannot solve them using just a summation of forces, you will need a moment equation, one or more.

[@JohnRovicAlejandrino]

@@QuestionSolutions i mean, for example, in this video, after you got the force DC, you used summation of forces to get force HI, when i thought i was supposed to use moment equation and i thought i would have to do it at point F to get rid of the force HC and thus getting me the force HI

[@JohnRovicAlejandrino]

i got an answer of 5 kN and i would not have known i was wrong if you didnt show that you used summation of forces to get the force HI

[@QuestionSolutions]

@@JohnRovicAlejandrino You definitely can use a moment equation to get force HI, but my assumption is, you're not writing a proper moment equation. There will be no difference in the answer. They all will result in the same answer. Also, once you already use a moment equation to find a force, you can do the rest with just summation of forces, there isn't a need to write a moment equation most of the time, but if you did, you can still get the same answer.

[@QuestionSolutions]

@@JohnRovicAlejandrino Please write your moment equation so I can go over it and let you know what happened. Even from the answer, we can easily see there is a mistake, you got a force of 5 kN, when all of the forces on the truss are 40s, 50s, etc. So there is an error in your moment equation and I will help you out if you write it here.

[@joshuawu8687]

at 3:57, why the FGHcos36.87(1.5) is not Sin. Thank You so much

[@QuestionSolutions]

Only the x-component of FGH can create a moment about point B, since the y-component's line of action goes through point B. So looking from the perspective of the angle, the adjacent side gives us the x-component, so we use cosine. If it's a matter of choosing cosine or sine, please see: kzbin.infovynnKlJD_Jo?feature=share

[@Devil.14327]

Bro in question 3rd why have you not taken reaction at A. Please explain

[@QuestionSolutions]

Why do you want to take the reactions at A?

[@alijawad432]

For question one I know u take thr moment at Point B but if we took the moment at point H, we would have two unknowns, Fbc and Fbg horizontal component both causing a moment. How would u solve it then?

[@QuestionSolutions]

The whole point of taking a moment about a specific point is to eliminate as many unknowns as possible. If you have 2 unknowns after writing a moment equation, then you'd need another equation, probably for vertical forces or horizontal forces. Try to take moments about points that lead to direct answers.

[@Kouka2005]

Super Helpful

[@QuestionSolutions]

Glad to hear :)

[@umutsuper123]

how do we determine whether to find the support reactions or not ??

[@QuestionSolutions]

If you use a section that you can solve without needing support reactions, then you don't need it. For example, at 5:35, notice we cut the supports off, and then we had enough givens to solve the top values. But notice at 8:24, even after the cut, we still needed the reaction at roller F. So in most cases, the only time you don't need to know support reactions is if you make a cut and on the side you selected, there are no supports.

[@brienyeboah1497]

Can you please enlighten me on why force BC and BG would be eliminated when taking the moment at B. I am confused.

[@QuestionSolutions]

Their lines of action go through point B. The same as the 5 kN force, and the y-component of force HG. If you need a refresh on lines of action, please kindly take a look at this video: kzbin.info/www/bejne/h3-xn4OQa52Giqs

[@nancylu7644]

@@QuestionSolutions super helpful, thank you

[@tebarekyosef8623]

you ae the best problem solver 10Q

[@QuestionSolutions]

You're welcome.

[@steve20004]

just to be sure is it not supposed to be cosine on the second question Fbg since the Fbg sine will cancel at the pivot D.asking just to be sure

[@QuestionSolutions]

You are calculating your components incorrectly. Cosine would give you the y-component, in other words, the vertical component of force FBG. Sine would give you the x-component, the horizontal component.

[@sevgipnar5261]

In the first question can we use which equation we want? I mean when we use moment or Fx=0, Fy=0. We used only moment equation in this problem.

[@QuestionSolutions]

You can, but try to use ones that give you the answer the fastest.

[@sevgipnar5261]

@@QuestionSolutions okay thank you

[@marcoli262]

5:07 I dont get why 9.5(2)-2(2) wouldnt be 2*sin(53.13)*the length of AH and same for Ay

[@marcoli262]

since you are calculating the moment of point H

[@QuestionSolutions]

@@marcoli262 So notice how both the 9.5 kN force and the 2 kN force are vertical, they only have y-components. In simple terms, they are vertical. All we need to figure out the moment about H is the perpendicular distance from those forces. The perpendicular distance is just 2 m. The length of AH does not matter. Please watch this video first: kzbin.info/www/bejne/h3-xn4OQa52Giqs

[@marcoli262]

@@QuestionSolutions Thanks, but I still get the same final answer of Fbc by using my own method

[@QuestionSolutions]

@@marcoli262 okay 👍 My guess is, you're splitting the AH length into components, but that's costing you way more time then knowing the proper fundamentals to get your answers faster. Generally speaking, university exams are time crunches, so you should learn faster methods then using sine and cosine to break it into pieces every time, it will be more efficient for you.

[@QuestionSolutions]

@@marcoli262 I think I figured out what you're doing. So you're doing "2*sin(53.13)*the length of AH" where the length of AH is 2.5 m. So you're literally multiplying by 2m, but going through a long journey to get to 2 m, because sin(53.13)*2.5 = 2m. Again, ONLY the perpendicular distance matters, so you don't need to calculate the length of AH, and find the 53.13 angle, it's not needed at all. You get the same answer because you're doing what's shown on the screen but taking way longer to get there. Does that make sense?

[@sumitrohan1839]

Watching this before exam is worth :)

[@QuestionSolutions]

Best wishes with your exam! :)

[@abidemiahmed9309]

I'm glad I'm watching this video, I'm having my mechanics exam next week. But I feel you're a lil bit fast in explaining, seems quite difficult to catch up mentally. Like, before I process a part, you'd have gone 3 parts. Also, can you help write the equation one by one. You have real good content, but just to avoid confusion.

[@QuestionSolutions]

I try to keep these videos concise, so it doesn't take up too much time. It can be difficult, especially when being introduced to the subject for the first time. So I will keep what you said in mind, but try to watch the video a few times, or try to solve these questions by yourself first, and if you get stuck, then try to see what I did to get to the next part. These videos are meant as a supplemental tool, and not a substitution for a textbook/professor. Thank you for the feedback!

[@shankylezapanta1464]

My approach to his every vid is to try to deeply understand the underlying concept and pause right at the very beginning of each example and try to think thoroughly on how to solve it myself. Give yourself some engagement on critical thinking rather than just "watching". Critical thinking will give you a solid comprehension of the topic that will stuck on your mind that would greatly help you on your exam and even after semester, you will still understand the topic. Works for me 100% I usually answer his examples on all statics vids 95% before he explains. Try it urself :)

[@steve20004]

can you help me with the second question.is Fbg not supposed to be cosine for the moment because the Fbg sine is going towards the pivot which will cancel.asking just to be sure

[@QuestionSolutions]

No, it's sine because that gives you the x-component. So you need to look at the sides from the angle. The opposite side to the angle gives you sine, which in this case, is the x-component of FBG. Cosine would give you the y-component, which wouldn't create a moment.

[@imansalim3874]

Check me if I am wrong,you said if we chose the moment at 1 point then we can eliminate the action force acting on it,but I don understand in 2nd example at 6.30 ,which moment at point D ,the 5 kN still considered?Tq

[@QuestionSolutions]

If you're referring to the top most force, applied at D, that is NOT considered when writing a moment about D. Please carefully look at the color coded boxes and lines, each corresponding to which moment is being calculated. The 5 kN used for the calculation is the one applied at point E.

[@imansalim3874]

@@QuestionSolutions ohh I see it now,my mistake.Thankyou

[@QuestionSolutions]

@@imansalim3874 Glad you got it :) Best wishes with your studies!

[@tunes_nirvana]

Amazing Video, All doubts cleared

[@QuestionSolutions]

Glad to hear that :)

[@RobertoNicolasAmadorSomarr-f4j]

why in BG is eliminated?? 2:59

[@QuestionSolutions]

Any force originating from a point where a moment is calculated about, cannot create a moment. Another way of thinking about it is, if the line of action of a force goes through the point at which a moment is calculated about, it cannot create a moment. See: kzbin.info/www/bejne/h3-xn4OQa52Giqs

[@joeyni2861]

Thanks question solution for this video it really helped me with my quiz tomorrow just a quick question wouldn’t it be easier to take the moment at B first then take the sum of Fy to get Bg isn’t that easier?

[@QuestionSolutions]

You're very welcome. Could you please give me a timestamp so I know where you're referring to, then I can take a look. Thanks!

[@squareroot1698]

Can you kindly explain why you used "sin" operation for x components and "cos" operations for y components....? i always thought we should use sin for y and cos for x...

[@QuestionSolutions]

Please see this video, it's less than 60 seconds: kzbin.infovynnKlJD_Jo?feature=share

[@kahoh1052]

Does zero force members not apply when solving Method of Sections

[@QuestionSolutions]

It's not a matter of "applying" per say. You can find them when solving the whole truss or you can visually look for them too. It just depends on the truss given.

[@XIVRedd]

Thank you soo much this video really helped me !!!

[@QuestionSolutions]

Really glad to hear that! Keep up the great work :)

[@mariorodriguezpinzon3542]

I believe there is a mistake at 9:07 . Solving for HI should be 42.5, not 45.2. lmk if I am wrong but I do not believe so.

[@QuestionSolutions]

Yeah, it's a typo 👍

[@annerivera176]

Great video, thanks! Just a quick question, I tried solving the 4th problem but using the other side of the cut and the answer is different. Is my answer wrong or is it possible to have different answers when using different equations?

[@QuestionSolutions]

Regardless of the cut you use, or which side you use, you should end up with the same answers. This can be verified by solving the whole truss. There is probably some numerical error. :)

[@sapphireblue9209]

Hello, thanks a lot for this video. It was very helpful. I have a question: 5:35 can we cut the truss through BF, AG, AB? basically diagonally

[@QuestionSolutions]

You can pretty much cut it anyway you want (as long as you have enough info to solve it). The cut should be made to figure out the unknowns you want, so try to cut the ones you need to find. So yes, you can cut it diagonally 👍

[@sapphireblue9209]

@@QuestionSolutions ohh I see, thanks a lot and THANK YOU SO MUCH FOR THE INSTANT REPLY I admire your dedication💙💙💙

[@QuestionSolutions]

@@sapphireblue9209 You're very welcome. Best wishes with your studies!

[@sapphireblue9209]

@@QuestionSolutions thank you very much!

[@cristinamartinez5680]

why is it 25(4) in 6:32 doesnt it have a horizontal distance of 8 :(((

[@QuestionSolutions]

Okay, let's look really carefully. Horizontal distance is across your screen (not up). So if we look at point D, and point F, what is the horizontal distance? Maybe you're looking at the vertical distance? Not sure, but it's 4m. I show the distance with a pink colored line, so look for the corresponding colors to help you out. Let me know if you don't see it, I will try to help you out another way.

[@dupitaslizaruthr.4329]

Glad I found these videos of yours. It was a very big help knowing I am having a hard time with our subjects right now because it was not explained properly but yours is very concised and well elaborated🤍

[@QuestionSolutions]

Glad it was helpful! I hope you do amazingly on your exams and I wish you the best ❤

[@PasantAssem]

inspiring solution for real

[@QuestionSolutions]

Thanks!

[@drvirusg9046]

sir can you please recheck the video at 3:39 minutes because I think the answer at this point is 0.5 rather then 1,5,

[@QuestionSolutions]

The value shown on the video is correct. 2 x tan(36.87) = 1.5. I am unsure how you arrived at 0.5 but if you need help, please show your work. Thanks!

[@prog_ahmed]

I really like you videos, you explain better than my professor 😅🔥. I have a question, why did you cut the trusses like this in 10:00, why didn't you cut just vertically? It will give you the same number of unknowns

[@QuestionSolutions]

If you cut the truss straight down, you will need to work on the left side of the truss. But we figured out the reactions on the right side, so cutting it the way I did means you work on the right side of the truss. You can cut straight down though, it's totally up to you.

[@prog_ahmed]

​@@QuestionSolutions oh I understand now. Thank you for Your precious time❤

[@QuestionSolutions]

@@prog_ahmed No worries, keep up the good work!

[@bii_gii]

You are the best! Thank you so much brother

[@QuestionSolutions]

Happy to help and best wishes with your studies!

[@bolingonraffyt.6154]

I think sine is for summation of Fy and cosine is for Fx? Correct me if im wrong.

[@QuestionSolutions]

You are wrong, do not fall into that pitfall. I am not sure why students tend to this think way, but they are NOT related to Fy and Fx. It's all based on the side opposite to the angle. Please take a look at this video: kzbin.info/www/bejne/hKOvZpdjZ6iUmLM

[@invisiblenavigateur315]

Absolutly the best truss video woooooow great

[@QuestionSolutions]

Thank you, I hope it was helpful to you. Best wishes with your studies!

[@shyazlinmushidi9214]

Hello, I would like to say that I am quite confused. can you tell me how do you determine if your direction of force is positive or negative?

[@QuestionSolutions]

The initial direction is just an assumption. If you get a positive value, then your assumption was correct. If you get a negative value, then it's opposite to your assumption.

[@wajdy2620]

How do you know when to take moment or Sum of forces in x,y to solve for unknown forces?

[@QuestionSolutions]

You usually have to do both methods to find all the unknowns. If you can find an unknown with a moment equation, then you should do that because it saves time. If you can't use a single moment equation to figure out a force, then you will have to sum forces in the x/y direction. This comes with intuition, so doing a lot of questions will help with this. I don't think there is a straight answer to your question. Try to do little as possible to get the answer to save time. That usually means starting off with a moment equation since you can eliminate a lot of the forces that go through the point where the moment is calculated.

[@cokonkiyot]

@@QuestionSolutions how to determine which point to take to create moment eq ?

[@QuestionSolutions]

@@cokonkiyotPick a point that eliminates the most unknowns.

[@cokonkiyot]

@@QuestionSolutions ahh i see . how to know which forces affect that moment ? i confuse lil bit even tho i watched ur video about it

[@la_silenciosa_maria]

so every slope needs to represent vertical and horizontal components?

[@QuestionSolutions]

Sorry, I don't understand your question. What slopes?

[@navarrocarljericcabigon7787]

In the problem 2, is it also same answer if i use the bottom part because i tried it and i got different answer or did i get wrong?

[@QuestionSolutions]

Regardless of where you start, you will always get the same answers. There is probably a numerical error in your solution or an error in setup of the equations. The forces in each member can't change based on where we start to solve the problem. :)

[@79elcapitan]

@10:37 shouldn't the answers be in kN not N as the original forces are in kN? Great videos and easy to understand and work through to show workings.

[@QuestionSolutions]

Yes, it's a typo. 👍

[@RathodAbhishek-p7b]

thank you sir for such a good explaination

[@QuestionSolutions]

You are very welcome! I hope it was helpful to you.

[@autobahnwaybab7608]

how did you get FDE or FJI times by 3? why times by 3?

[@QuestionSolutions]

I don't know where you're referring to. Please use timestamps. Thanks!

[@MrJaaaboo]

Tried to find a video about moment equations like the one at 2:00, do you have any walkthrough videos on theese? :)

[@QuestionSolutions]

Please see: kzbin.info/www/bejne/h3-xn4OQa52Giqs I go through the basics of moments.