A lot of students have asked how I got 1.155W at 4:37. Please see the breakdown here: bit.ly/3kGNcyW 🔹How to solve the 2 equations at 6:44 : th-cam.com/users/shortsHe7lrJEB04U 🔹When to use sine or cosine: th-cam.com/users/shortsvynnKlJD_Jo Here is question 1 solved step by step (simultaneous equations solved using substitution): www.dropbox.com/s/fp3d9sepod9kbu9/Question%201%20solved.pdf?dl=0 The rest of the problems can be solved in the exact same way.
Your channel is the holy grail. You use the same book we use at my university, and the problems you work through cover each topic perfectly. It correlates to class work and exams to a tee. Not to mention your way of teaching is so easily understandable and perfect. Looking forward to watching all your videos throughout my engineering career. Much love bro
Wow, thank you! This was a really nice comment to read and I am super happy these videos are helping you. Keep up the good work and best wishes with your engineering career.
I am in tears right now!!! Since I graduated 3 years ago I’m struggling with statics of rigid bodies. Now I’m preparing for licensure exam this November. I need a refresh of basic fundamentals and you just appeared!!!! You are an angel . God bless you more. You deserve a million subscribers. Will definitely inform my colleagues to subscribed your channel. ❤😊
Wow, well done and congrats on graduating. Now you can go and ace the licensure exam too! You got this. Do as many questions as you can and I wish you the best on your exam :) Keep up the great work! ❤
Easily the best TH-camr for engineering statics, holy hell you actually carried me. My professor SHOULD HAVE NEVER BEEN HIREDD, You taught me the entire semester in what 2-4 hrs of videos? You deserve everything 🎉
You are a life saver man I was so confused that from where I should revise all my questions and you made it really easy for me man really neat and smooth explanation 💪
You're very welcome :) If you're doing statics, there is a whole playlist of videos covering pretty much everything need for your course, so I hope that will be helpful to you. Best wishes with your studies!
I love your organization methods and way of teaching, it is very helpful. I am hoping I can turn my mark around in statics so I do not have to retake it. I struggle most with identifying where the forces are and how to write out the equations of equilibrium, but this video made it much more clear thank you so much!
Statics is a course that really builds from one step to the other. I think there is 2 things everyone needs to learn. The first is breaking forces into components. The second is expressing forces in cartesian form. After that, you will be solid. If you have the time, and you need a refresh, I highly encourage you to watch the statics playlist from the beginning. I tried to make them as concise as possible while building up all the fundamental knowledge required for advanced topics covered in the later videos. I wish you the absolute best on your course, don't give up, do your best! :)
@@QuestionSolutions You’re my saviour dude!! Thanks to your clear and concise teaching I managed to achieve a 76.98% on my midterm after only getting a 37.5% on the first part of the test. Thank you so much I got 90.38% on part two because of you ❤️❤️
@@anonrelentless1311 That is AMAZING!!! Thank you so much for taking the time to write your comment, this made my day. It's really nice knowing these videos help students. Now keep up the awesome work and do your best on the final exams. You got this!
Hello Kind Sir ! I hope you are doing well. I would like to ask you a question for which I have not been able to find an answer, despite spending considerable time reflecting on it and reviewing the available comments. I thought it was appropriate to reach out directly. In your video at 8:53 seconds, you mention that cos(θ) equals x divided by H(AB). However, I am having trouble identifying what H(AB) specifically refers to, as well as H(AC). Could you kindly clarify this for me? Thank you very much for your assistance.
First, thank you using a very respectful way of asking a question. It's really nice to see! Also, thank you very much for actually looking through the comments to see if your question was answered. Very well done! To answer your question, please look at 7:41, and the diagram shown. Notice how ABC is the notation used to denote the cable given to us in the question. So HAB denotes the hypotenuse of the triangle formed at 8:53, with points AB being the hypotenuse of the smaller triangle, so left side of the pulley, and HAC being the hypotenuse of the big triangle, the one right of the pulley. Let me know if that clears it up, or if you need help with anything else. P.S. in hind sight, I now see that I should have labeled the points ABC on the diagram at the end as well, for that, I apologize.
Thank you for your clarification. I finally understood where my confusion lay. When you wrote H(AB) and H(AC), I mistakenly thought it referred to the height. It was actually the letter H that caused my misunderstanding. It turns out that HAB is simply a label you used to describe the hypotenuse. I hope am right x)
You're very welcome. Convert all the sine and cosine into decimal values. So cos(30)=0.866, cos(40)=0.766, etc.. Then solve the two equations using the substitution method. See: www.cymath.com/answer?q=-0.866a%2B0.766b%3D0%2C%200.5a%2B0.642b-98.1%3D0 I replaced F_CA = a, and F_CB = b and used decimal values for cosine and sine.
this man single-handedly carried me through my degree. thank you, good sir! may both sides of your pillow be cold tonight, and may you only have skippable youtube ads!
I am really happy to hear I helped, and thank you coming back and writing your comment. It was a pleasure to read and my pillows were nice and cold last night! Best wishes with all your future endeavors.
Hi! I love your videos, especially this one in particular. However, I need help understanding how, in the first example, you got 79.97N. Do you have a video that thoroughly breaks down how you solved it? Or could you give me a brief explanation? Other than that, fantastic video!
Hi, thanks for the nice comment! Glad to hear you love the videos :) Here is a link (pinned comment) that goes through how it was solved: www.dropbox.com/s/fp3d9sepod9kbu9/Question%201%20solved.pdf?dl=0
@@QuestionSolutions You're a literal lifesaver! I have an exam this Wednesday, and my professor is terrible at explaining things! I've been lost ever since school started, lol. I appreciate it again. Thanks! :)
5:50 In question 2, why is it when I get all the mass of each forces, FBC has the least mass. I got FBA = 44.1399kg, FAE = 88.2799kg, FBC = 26.6755kg, and FBD = 30.5810kg.
I am not really understanding your question. 😅 You get the least amount for FBC because the number in front of the W is the biggest. Assume you use 44.1399 kg (FBA weight you got). If you use that on force FBC, it exceeds 500 N, so you can't use it. The same goes for all the rest.
I think you are probably overthinking it, so to simplify, convert the sin and cosine values to decimal form. Then you just have 2 unknowns with 2 equations and you can solve them using any method like substitution, elimination, graphing, etc.
So at 9:14, you need to solve for θ. To do that, you need to take the inverse of cos. So cos^-1(3.5/5)=45.57. To get 1.38, you need to solve for x at 9:48. So just convert all the tan45.57 to decimal values and then solve it like you would any equation. Tan45.57 = 1.02. So you replace all the tan45.57 with 1.02. Let me know if you need further clarification :)
Sorry, what do you mean? It's solved and the final answers are also shown. Are you referring to a specific step? Please let me know of a timestamp so I can take a look and help you out better. Thanks!
@@jaki8002I don't have a seperate video but if you have a specific equation you're having a hard time solving, you can tell me and I can try to help you out 👍
Does it apply to all problems that if it's only one cable, the tension is just the same to the starting point and the end point especially if it's a pulley.
@@QuestionSolutions thank youuuu! Just letting you know your example was my examination yesterday. Tension of the cable including the spring . You’re an angel!!!
What if the problem does not specify the x/y geometry but the lengths of the links and the unsprung length of the spring? How could I tackle such a problem?
So you're solving for x. Convert the tan45.57 into a decimal value. So tan45.57 = 1.02. This will make it easier to see. Here is a step by step solution after you convert the tan45.57 to a decimal value. www.cymath.com/answer?q=1.02x%2B0.75%3D1.02(3.5-x)
2D equilibrium problems can be done pretty quickly. Probably 5 mins or less. But all of that depends on your comfort level with these types of problems and how many practice problems you've done before. The more you do, the faster you'll get at it.
So once you establish/write your equations, completely forget about statics and just focus on algebra. Think of the variables as x,y,z and solve the equations as you would in algebra class. The key part is just writing down the proper equations, after that, don't think about opposite directions or anything else because it will be accounted for when you were writing your equations.
This one isn't hard to do a home experiment with. You just need a string and 2 little hooks. Tie the string like the same way shown on the diagram and then hang something on the string around the middle. Then using a protractor, you can measure the angle on both sides. You will see that it is the same :)
So let's assume we write an equation for x-axis forces. We have Fcosθ-Fcosθ = 0. It doesn't really do anything for us (there is no way for us to solve it). So we write an equation of equation for the y-axis forces instead.
You're simply isolating for hAC. Multiply both sides by hAC and then divide both sides by cosθ. To answer the second part of your question, what is the total length of the rope? 5m, so if we add up the two sections, it must equal 5 m.
@@FAdamTennyson So I am looking and trying to understand your question but I can't figure out what you're asking :( Are you asking why FBC is negative or that it's not negative? I just want to point out that FBC is written as negative in the equation. Notice we wrote a little arrow pointing to the right next to the sigma sign. That indicates, for this equation, any force pointing to the right will be considered positive while any force pointing to the left will be considered negative.
@@QuestionSolutions what i meant was for the final answer of Fbc should be negative but I think I got it now, you moved Fbc from left to right to make it a positive Is that right?
So you're isolating for FAB. Add W to both sides and convert sin60 to decimal form. You get 0.866FAB = W. Divide both sides by 0.866. FAB=(1/0.866)(W), convert to decimal form. You get 1.155W. Remember that W divided by sin60 is the same as 1/sin60 👍
@@jm--nf7sz Did you convert to decimals? So for example, cos30 = 0.866, cos 40 = 0.766. Once you do that, you can write your equation like this: -0.866F_CA+0.766F_CB=0. You do the same for the other equation. Then you isolate for one variable in the first equation and plug it back into the second equation and solve. Let me know if it works out 👍
@@QuestionSolutions I got it! It just kind of confused me converting everything to decimals first so I left it at the end but I ended up getting 80N and 90.4N thank you!
@@jm--nf7sz The best way to do it is actually not convert to decimals and just keep sine and cosine. The answer is more accurate. Weirdly, a lot of students get scared or confused when they see sine and cosine terms, because they tend to forget that they are actual numerical values. So to make it easier, I tell them to convert it to decimal form. I think seeing just numbers and variables is easier then seeing sine and cosine terms 😅
This one isn't hard to do a home experiment with. You just need a string and 2 little hooks. Tie the string like the same way shown on the diagram and then hang something on the string around the middle. Then using a protractor, you can measure the angle on both sides. You will see that it is the same :)
Howzit bro. Can you please make videos on the sections centroids & centre of masses,moment of inertia, and friction. If the videos are made ASAP it would be much appreciated because I am going to be writing my paper soon and I need to study these sections thoroughly. Please can you help me out
I probably won't be able to make that video ASAP, but I saw lots of videos on youtube about the subjects mentioned. Hopefully, those will help for now. I am really sorry, but in time, I will get to those as well, just not going to be the next few videos coming up.
Or videos on some of the mentioned sections atleast, will also help ...... can you leave all of the statics before those sections and make these section videos quickly... do you have a gmail or some sort of mail account ?
@@arjungovender3248 Well, I don't think I can do that because I want people to be able to go from one video to the next. So I finished off dynamics, and then it's statics. If I stop half way and then jump to a random topic, it will throw off the sequence of videos. Did you search youtube for the topics you mentioned? There are a lot of videos for those :) Also, email is contact @ questionsolutions.com.
@@QuestionSolutions I have but honestly not understanding them as well as I understand the videos made by question solutions, maybe you could send me the links to the videos you think are 100% worth it , that would be appreciated
I just started learning static and I would like to ask why is it that the highest tension should be the basis on what will be the maximum mass that it could carry Its still puzzling me
I think you're referring to the second question. Let's assume we pick the lowest of the set, which was cable AE, with a maximum tension of 0.5775W. That means if we divide 500 N by 0.5775, we get 865.8 N. Dividing that value by 9.81 gives us the mass, which is 88.26 kg. So just for cable AE, we can hang a pot with a mass of 88.26 kg. Setting that aside, let's think about the maximum tension cable, which is cable BC, with 1.911W. What would happen if we hung that 88 kg pot on that cable? So we get 1.911(865.8)=1654.5 N. Notice how this far exceeds the maximum each cable can hold, which was 500 N. This is why we base it on the cable that carries the highest tension. If we base it on the cable that carries the lowest tension, then that weight would exceed maximums of the cable that carries the highest tension. Play around with these values to get a better idea of what I mean and it will help you understand it better. 👍
@@daryldiaz7194 Once you read the chapter, try to do as many questions as you can, you will gain a lot of inside that way. The golden rule in statics is: Everything is in equilibrium, which means all the forces added together must equal zero. :)
a uniform rod of a mass 3 kg and length 2m has one end attach by smooth hinge to a ceiling at a. the other end b ais attached to a light inelastic string. the other end of the string is attached to the ceiling at c. the rod makes an angle of 30° with the horizontal and the string an angle of 60° to the horizontal qs shown. find the tension in the string. Can you make a video tutorial for this problem. Thank you
I can't really create videos on specific problems because then I'd have to create a video for every problem that students ask. I think it's better to ask these types of problems on forums that help with physics/statics. Please check them out.
Hi, that is a very vague question. Which part did you have a hard time understanding? There aren't any hidden steps, so I really did show everything you need to solve this question. If you can be specific, I can try to help more :)
@@arinajar754 Are you not understanding the part where we solve 2 equations to get the 2 values? Instead of thinking them as FCA and FCB, think of them as x and y. You can solve 2 equations with 2 variables right? If you need a refresh, please take a look at this video: www.khanacademy.org/math/algebra-home/alg-system-of-equations/alg-equivalent-systems-of-equations/v/solving-systems-of-equations-by-elimination I highly encourage you to review solving simultaneous equations, you will need it a lot :) If this is not where you are confused, please give me a timestamp.
No, please see this video to learn about breaking forces into components using the ratio triangles: th-cam.com/video/NrL5d-2CabQ/w-d-xo.html I cover it in the 2nd example problem. Thanks!
I think you might be missing a lot of fundamentals for this course. I really encourage you to review some high school math topics like solving simultaneous equations, it will make your life very easy :) Here, you can use substitution. See: www.dropbox.com/s/fp3d9sepod9kbu9/Question%201%20solved.pdf?dl=0
You can graph the two equations or convert all the sin and cosine into decimal values. So for example, cos30 = 0.866, etc. Then it's just a simple 2 equations with 2 unknowns :)
That's not gravity, that's weight. So weight is mass times acceleration due to gravity. In this question, the mass of the cylinder is 10 kg. So we get 10 x 9.81 = 98.1 N. Remember that acceleration due to gravity has the units m/s^2 where as weight has the units N.
It's 2 simultaneous equation solving. So you have 2 equations with 2 unknowns. You can use elimination, substitution, or you can graph them. If you don't remember how to do this, please review "simultaneous equation solving."
@@saqlainraza2219 FAB = FBA, same thing, just different notation. Usually, if you look at a force from A to B, we write it as FAB. If we look at it from B to A, we write FBA. However, the sign convention is not determined by the way we write AB or BA. It's based on the direction of the force itself. You can write either and its perfectly fine. Now if a force faces left, and we got a negative value after we assumed left to be positive, then we know the force is opposite to the direction we drew it, but again, FBA or FAB notation does not matter. At the end, it's just a notation.
Add "W" to both sides. You get: FSin60° = W. Divide both sides by sin60°. You get F = 1.155W . (just wanted to add, on the right side, it's really just (1/Sin60°)(W), so you're dividing 1 by Sin60°.)
@@nbarekap You are missing fundamental algebraic knowledge. Please kindly review simplifying expressions and other forms to solving equations or you will end up using up a lot of time working on these things. I am not being mean, but it won't get any easier when the courses proceed. If we have a simple equation like this: x+2=y Here, there is 1"x" and 1"y". We just don't show the 1 in front. If we have 2x+2=y, we have 2"x" and 1"y". Again, because there are 2 x's, we put a 2 in front, but we don't put a 1 in front of the y since it's just 1. So please, review these things :)
So you're isolating for FAB. Add W to both sides and convert sin60 to decimal form. You get 0.866FAB = W. Divide both sides by 0.866. FAB=(1/0.866)(W), convert to decimal form. You get 1.155W. Remember that W divided by sin60 is the same as 1/sin60 👍 Please see: bit.ly/3kGNcyW
@@julianafairy908 If you have an equation like this: 2x = 2. You solve for it by multiplying both sides by 1/2 (or dividing by 2, which is the same thing) where did the 1/2 come from? It's the same thing. In fact, you can multiply or divide it by whatever you want as long as you do it to both sides. Don't get confused by sin/cos stuff, it's still just numerical values 👍 I hope that helps :)
If you have a string at home, tape the 2 ends to things, so there isn't a lot of tension, but it's not super lose either. Then, hang something with a paper clip or something that slides around easily, you'll see that it'll come to the center. If you measure the angle with a protractor, you'll notice they are very very close, probably not perfect since there are a few variables to consider, but it'll be close. In most cases, you'll get the same angle on both sides.
The problem with explaining how to solve simultaneous equations is that it takes away time from the majority of students who already know how to do that (and students definitely should know this stuff at this point). It extends the time of the videos by a large margin, and takes away from what I am trying to get across, which are the new material taught in a statics class. To counter this, I have created a few short videos on the channel that teach students how to solve simultaneous equations. So if you need a refresh, I encourage you to look at those to get up to speed. If you look through the pinned comment, there are already links directing students to solve the equations shown on screen as well.
Thank you so much :) I really appreciate your thought. I don't have any methods set up for support, but something you can do for me is to share any video that you found helpful. I figure maybe it'll help your classmates and friends, and then that would help the channel as well. Win win. I really appreciate your thought. Happy holidays and I wish you a very happy new year. May the new year bring you much success in all your endeavors.
So the goal of these videos isn't to teach how to solve a system of equations (this should be familiar to you from high school). At this point, you NEED to know how to do that. If you need a refresh, I created a bunch of small short videos going through how to solve them. But again, the goal of this video is to teach new concepts covered in statics, which is how to establish the equations properly to get a final answer. So I encourage you to review high school material if you don't know or don't remember how to do this. It will be of great aid to you in your courses.
A lot of students have asked how I got 1.155W at 4:37. Please see the breakdown here: bit.ly/3kGNcyW
🔹How to solve the 2 equations at 6:44 : th-cam.com/users/shortsHe7lrJEB04U
🔹When to use sine or cosine: th-cam.com/users/shortsvynnKlJD_Jo
Here is question 1 solved step by step (simultaneous equations solved using substitution): www.dropbox.com/s/fp3d9sepod9kbu9/Question%201%20solved.pdf?dl=0
The rest of the problems can be solved in the exact same way.
Why we multiply the W with 1/sin60
I don’t get it !?
To get rid of the sin60@@Bilal-w1x3t
thank youuuuu. I was literally fussing over that 1.155 lol
@@Hellylali You're very welcome!
@@JaynDazzy dumbass
Your channel is the holy grail. You use the same book we use at my university, and the problems you work through cover each topic perfectly. It correlates to class work and exams to a tee. Not to mention your way of teaching is so easily understandable and perfect. Looking forward to watching all your videos throughout my engineering career. Much love bro
Wow, thank you! This was a really nice comment to read and I am super happy these videos are helping you. Keep up the good work and best wishes with your engineering career.
Please can you share or name that book
@@arazaksaed4931 R.C Hibbeler Engineering Mechanics: Statics
@@arazaksaed4931 r.c hibbeler statics and dynamics 14th edition
Engineering Mechanics
Statics
14th Edition
R.C Hibbeler
I am in tears right now!!! Since I graduated 3 years ago I’m struggling with statics of rigid bodies. Now I’m preparing for licensure exam this November. I need a refresh of basic fundamentals and you just appeared!!!! You are an angel . God bless you more. You deserve a million subscribers. Will definitely inform my colleagues to subscribed your channel. ❤😊
Wow, well done and congrats on graduating. Now you can go and ace the licensure exam too! You got this. Do as many questions as you can and I wish you the best on your exam :) Keep up the great work! ❤
Newly found this page and it's underated. It is worth the watch instead of hours listening to a prof. lecture . Deserve a subscribe 👍
you probably have an exam tomorrow
@@thatsfemiin a few days
Thank you very much, I appreciate it. Best wishes your studies, and the upcoming exam as well!
I was having a hard time catching up with my instructor, your video saved me! Bless you, sir! 💗
You're very welcome! Glad you found it helpful 😊
Easily the best TH-camr for engineering statics, holy hell you actually carried me. My professor SHOULD HAVE NEVER BEEN HIREDD, You taught me the entire semester in what 2-4 hrs of videos? You deserve everything 🎉
Thank you very much for your kind comment. I wish you the absolute best with your future courses and endeavors. Keep up the good work!
You are a life saver man I was so confused that from where I should revise all my questions and you made it really easy for me man really neat and smooth explanation 💪
Thank you so much, very happy to heat these videos are helpful to you. :) Keep up the good work!
I wish you could be my in class teacher, but I'll settle for you being my new online teacher! Thank you so much! :)
You're very welcome :) If you're doing statics, there is a whole playlist of videos covering pretty much everything need for your course, so I hope that will be helpful to you. Best wishes with your studies!
Equilibrium? More like "Amazing lectures, all of 'em!" 👍
Thank you, thank you!
@@QuestionSolutions You’re most welcome!
Thank you sir! I hope your lectures help me pass my statics course.
Do your best :)
I love your organization methods and way of teaching, it is very helpful. I am hoping I can turn my mark around in statics so I do not have to retake it. I struggle most with identifying where the forces are and how to write out the equations of equilibrium, but this video made it much more clear thank you so much!
Statics is a course that really builds from one step to the other. I think there is 2 things everyone needs to learn. The first is breaking forces into components. The second is expressing forces in cartesian form. After that, you will be solid. If you have the time, and you need a refresh, I highly encourage you to watch the statics playlist from the beginning. I tried to make them as concise as possible while building up all the fundamental knowledge required for advanced topics covered in the later videos. I wish you the absolute best on your course, don't give up, do your best! :)
@@QuestionSolutions You’re the best my friend! Thank you so much for the kind words and inspiration ❤️
@@QuestionSolutions You’re my saviour dude!! Thanks to your clear and concise teaching I managed to achieve a 76.98% on my midterm after only getting a 37.5% on the first part of the test. Thank you so much I got 90.38% on part two because of you ❤️❤️
@@anonrelentless1311 That is AMAZING!!! Thank you so much for taking the time to write your comment, this made my day. It's really nice knowing these videos help students. Now keep up the awesome work and do your best on the final exams. You got this!
@@QuestionSolutions I’m so glad it made your day :) your videos are so helpful and I’m sure that so many have benefited from them.
THANK YOU for the spring problem. my prof had it worked out differently and i just couldn't comprehend it. but i like the steps you took
You're very welcome! Glad it was helpful to you :)
Fantastic crystal clear my doubts thanks a lot
You're most welcome and best wishes with your studies!
For the first question at 4:40 how did you solve for Fab?
Please see pinned comment.
Hello Kind Sir !
I hope you are doing well. I would like to ask you a question for which I have not been able to find an answer, despite spending considerable time reflecting on it and reviewing the available comments. I thought it was appropriate to reach out directly.
In your video at 8:53 seconds, you mention that cos(θ) equals x divided by H(AB). However, I am having trouble identifying what H(AB) specifically refers to, as well as H(AC). Could you kindly clarify this for me?
Thank you very much for your assistance.
First, thank you using a very respectful way of asking a question. It's really nice to see! Also, thank you very much for actually looking through the comments to see if your question was answered. Very well done!
To answer your question, please look at 7:41, and the diagram shown. Notice how ABC is the notation used to denote the cable given to us in the question. So HAB denotes the hypotenuse of the triangle formed at 8:53, with points AB being the hypotenuse of the smaller triangle, so left side of the pulley, and HAC being the hypotenuse of the big triangle, the one right of the pulley. Let me know if that clears it up, or if you need help with anything else.
P.S. in hind sight, I now see that I should have labeled the points ABC on the diagram at the end as well, for that, I apologize.
Thank you for your clarification. I finally understood where my confusion lay. When you wrote H(AB) and H(AC), I mistakenly thought it referred to the height. It was actually the letter H that caused my misunderstanding. It turns out that HAB is simply a label you used to describe the hypotenuse.
I hope am right x)
@@nabil4220 You are correct! :)
Please do more videos about statics , dynamics and thermodynamics. I hope u can cover everything soon.Thanks brader .
I will do my best :)
So good and great dr ♥️🫂
I follow you from Egypt ♥️👑
Thank you very much! ❤
Best best best wishes for best teaching style
Thank you so much! I wish you the absolute best with your studies. :)
Great explaining, thanks so much!
But how did you get the Fca and Fcb answers? I don't see how (3:40)
You're very welcome.
Convert all the sine and cosine into decimal values. So cos(30)=0.866, cos(40)=0.766, etc.. Then solve the two equations using the substitution method. See: www.cymath.com/answer?q=-0.866a%2B0.766b%3D0%2C%200.5a%2B0.642b-98.1%3D0
I replaced F_CA = a, and F_CB = b and used decimal values for cosine and sine.
I converted them to decimals then use mode 5:1 in Canon 789GSA , why i get 80 for bc and -66 ac? 😅
Thank you soooo soooo much. Appreciated
You're veryyyyyy welcome!
This is fast and easy to understand 😉 thank you so much ❤️
You're very welcome! Best wishes with your studies. ❤
this man single-handedly carried me through my degree. thank you, good sir! may both sides of your pillow be cold tonight, and may you only have skippable youtube ads!
I am really happy to hear I helped, and thank you coming back and writing your comment. It was a pleasure to read and my pillows were nice and cold last night! Best wishes with all your future endeavors.
Thanks again bro very clear and easy to understand
You're welcome! 👍 Glad it helped
Hi! I love your videos, especially this one in particular. However, I need help understanding how, in the first example, you got 79.97N. Do you have a video that thoroughly breaks down how you solved it? Or could you give me a brief explanation? Other than that, fantastic video!
Hi, thanks for the nice comment! Glad to hear you love the videos :) Here is a link (pinned comment) that goes through how it was solved: www.dropbox.com/s/fp3d9sepod9kbu9/Question%201%20solved.pdf?dl=0
@@QuestionSolutions You're a literal lifesaver! I have an exam this Wednesday, and my professor is terrible at explaining things! I've been lost ever since school started, lol. I appreciate it again. Thanks! :)
@@ThermoGuru :) thanks! Best wishes with your upcoming exam, you got this!! Keep up the great work.
5:50 In question 2, why is it when I get all the mass of each forces, FBC has the least mass. I got FBA = 44.1399kg, FAE = 88.2799kg, FBC = 26.6755kg, and FBD = 30.5810kg.
I am not really understanding your question. 😅 You get the least amount for FBC because the number in front of the W is the biggest. Assume you use 44.1399 kg (FBA weight you got). If you use that on force FBC, it exceeds 500 N, so you can't use it. The same goes for all the rest.
In the last problem: why are the angles and tension forces the same? Is this because it's a pulley?
That's correct!
Please do you have a video of cables in statics chapter 7?
Is it this one? th-cam.com/video/CCeWy1kmxMs/w-d-xo.html
Soo good! Thank you sir
You're very welcome! :)
In number 2, for Y. How did you get 1.155 W?
bit.ly/3kGNcyW
how did you solve for those 2 unknown variables Fdc and Fbd in 6:41?
I think you are probably overthinking it, so to simplify, convert the sin and cosine values to decimal form. Then you just have 2 unknowns with 2 equations and you can solve them using any method like substitution, elimination, graphing, etc.
Easy to understand.
Thanks 💙
Glad to hear :)
how did you calculate the forces for the first example?
I assume you mean how did I solve the 2 equations? If so, please see: www.dropbox.com/s/fp3d9sepod9kbu9/Question%201%20solved.pdf?dl=0
Hello! How did you get 45.57 and 1.38 on the fourth question? I hope you can answer my comment. Thank you!
So at 9:14, you need to solve for θ. To do that, you need to take the inverse of cos. So cos^-1(3.5/5)=45.57. To get 1.38, you need to solve for x at 9:48. So just convert all the tan45.57 to decimal values and then solve it like you would any equation. Tan45.57 = 1.02. So you replace all the tan45.57 with 1.02. Let me know if you need further clarification :)
@@QuestionSolutions why @9:14 the equation how to get the 45.57 by using inverse cos but based on the equation its a sin
these videos are so helpful! don't stop posting lol
Glad to hear they are helpful :)
Excelent video thank you
You're very welcome!
How did you get the Fab= 1.155W
See pinned comment.
Do you have a separate video on how to solve the last problem?
Sorry, what do you mean? It's solved and the final answers are also shown. Are you referring to a specific step? Please let me know of a timestamp so I can take a look and help you out better. Thanks!
Yeah, I mean a specific step to this 8:27 (last problem). I was a bit confused with these and those flashes of answers 😅
@@jaki8002I don't have a seperate video but if you have a specific equation you're having a hard time solving, you can tell me and I can try to help you out 👍
how did you get the angle 45.57 on the last problem , i put the exact equation on my calculator and i didn't get 45.57
Does it apply to all problems that if it's only one cable, the tension is just the same to the starting point and the end point especially if it's a pulley.
That's correct. The tension will be the same as long as we ignore stretching and assume it to be uniform (which pretty much every problem does).
@@QuestionSolutions thank youuuu! Just letting you know your example was my examination yesterday. Tension of the cable including the spring . You’re an angel!!!
@@janecm6136 Wonderful! I hope you did amazingly. Keep up the great work and thanks for letting me know :)
You can also solve for item 3 by using the mini-triangles!
👍
What if the problem does not specify the x/y geometry but the lengths of the links and the unsprung length of the spring? How could I tackle such a problem?
Is there any way you can show me a picture of the problem? Hard to say without seeing a diagram of some sorts. Thanks!
you are life saver. thank you very much
You're very welcome!
مصري؟
How would this be done with an unkown mass in the middle?
Use a variable for the mass.
Very useful
Thanks .
You're very welcome!
Why in 9:47 the answer is 1.38? Can you please explain why? Thank You
So you're solving for x. Convert the tan45.57 into a decimal value. So tan45.57 = 1.02. This will make it easier to see. Here is a step by step solution after you convert the tan45.57 to a decimal value. www.cymath.com/answer?q=1.02x%2B0.75%3D1.02(3.5-x)
Sir, at 8:55 you said that cos angle = hypotenuse over length x, but why is that shown in the vid was x/hyp?
I misspoke. Cosine is adjacent over hypontenuse, so it's x/hypotenuse. What's shown on the screen is correct :)
Thank you so much, sir!
how long should one of these questions take to complete?
2D equilibrium problems can be done pretty quickly. Probably 5 mins or less. But all of that depends on your comfort level with these types of problems and how many practice problems you've done before. The more you do, the faster you'll get at it.
At 9:16 how did you get the value ov theta?
So 5cosθ=3.5
cosθ = 3.5/5
θ = cos^-1(3.5/5)
θ = 45.57
When you plug in Fab into Fba dont you change the sign since their facing in the opposite directions
So once you establish/write your equations, completely forget about statics and just focus on algebra. Think of the variables as x,y,z and solve the equations as you would in algebra class. The key part is just writing down the proper equations, after that, don't think about opposite directions or anything else because it will be accounted for when you were writing your equations.
@Question Solutions okay really appreciated 🙏🙏
URDU:
bhot kmal samjhtay ho yaar ap.
I don't know what that says :(
In question 4, i get that when its one cable the force is the same. but how come you can say the angle is the same.
This one isn't hard to do a home experiment with. You just need a string and 2 little hooks. Tie the string like the same way shown on the diagram and then hang something on the string around the middle. Then using a protractor, you can measure the angle on both sides. You will see that it is the same :)
@@QuestionSolutions thanks
5:17 how do you decide that its 4/5 and 3/5 for the x and y?
Please see: th-cam.com/video/NrL5d-2CabQ/w-d-xo.html
Time stamp 8:27, why does the Fx is zero again this sir and only the vertical forces are shown?
So let's assume we write an equation for x-axis forces. We have Fcosθ-Fcosθ = 0. It doesn't really do anything for us (there is no way for us to solve it). So we write an equation of equation for the y-axis forces instead.
9:02 how do you get Hac=(3.5-x)/cosa ? Even Hab+Hbc=5!
You're simply isolating for hAC. Multiply both sides by hAC and then divide both sides by cosθ. To answer the second part of your question, what is the total length of the rope? 5m, so if we add up the two sections, it must equal 5 m.
why is Fbc is not not negative? I thought moving to right hand sight would change it into negative?? for question 2
Please use timestamps, I am not sure where you're referring to. Thanks!
@@QuestionSolutions oh my bad it was at 5:34
@@FAdamTennyson So I am looking and trying to understand your question but I can't figure out what you're asking :( Are you asking why FBC is negative or that it's not negative? I just want to point out that FBC is written as negative in the equation. Notice we wrote a little arrow pointing to the right next to the sigma sign. That indicates, for this equation, any force pointing to the right will be considered positive while any force pointing to the left will be considered negative.
@@QuestionSolutions what i meant was for the final answer of Fbc should be negative but I think I got it now, you moved Fbc from left to right to make it a positive Is that right?
@@FAdamTennyson Yes, you're isolating for FBC. It's the same as solving for a variable called "x" we don't leave it negative :)
Please show the methods on how you get the answers
Which methods? Everything is shown step by step. Please be specific so I can know what you are referring to 👍
Anyone to help with how x and y was found in question 2 using the ratios
See this video, especially the 2nd example: th-cam.com/video/NrL5d-2CabQ/w-d-xo.html
How did you get 79.97N and 90.41N in question 1?
Here is the step by step solution: www.dropbox.com/s/fp3d9sepod9kbu9/Question%201%20solved.pdf?dl=0
Fcb means the force that c has as it pulls on b?
Think of it like this. The force in the cable goes from C to B.
Sir at 5:28 you write force ab equal to force ba but they are opposite to each other. Then how their symbol is same? 🙏sir please clear it.
The value of the force, in other words, the scalar value, doesn't change with notation. F_BA = F_AB.
@@QuestionSolutions thank you so much sir🙏
how did you get 1.155 for 4:37timeline
So you're isolating for FAB. Add W to both sides and convert sin60 to decimal form. You get 0.866FAB = W. Divide both sides by 0.866. FAB=(1/0.866)(W), convert to decimal form. You get 1.155W. Remember that W divided by sin60 is the same as 1/sin60 👍
@@QuestionSolutions The does the W still remain when it is divided and turned into 1.
@@lukediola5197 Yes, you aren't dividing both sides by W, only by sin60. So W remains.
Awesome content brother & A Big Thank You
I want to know that how you create physics objects like vectors ,pulleys,etc.
You mean like the diagrams? If so, I use illustrator to draw them. 👍
@@QuestionSolutions tysm bro....
How can I get 1.085 for solving B in question number 1.
Please give me a timestamp so I know where to look, thanks!
3:38 How did you get the 79.97 N & 90.41 N?
Not sure why this comment was held for review, but to answer your question, convert all the sine and cosine values into decimal values, then solve it.
@@QuestionSolutions i did this but i am still stuck. Maybe im just overthinking but can you explain a little more?
@@jm--nf7sz Did you convert to decimals? So for example, cos30 = 0.866, cos 40 = 0.766. Once you do that, you can write your equation like this: -0.866F_CA+0.766F_CB=0. You do the same for the other equation. Then you isolate for one variable in the first equation and plug it back into the second equation and solve. Let me know if it works out 👍
@@QuestionSolutions I got it! It just kind of confused me converting everything to decimals first so I left it at the end but I ended up getting 80N and 90.4N thank you!
@@jm--nf7sz The best way to do it is actually not convert to decimals and just keep sine and cosine. The answer is more accurate. Weirdly, a lot of students get scared or confused when they see sine and cosine terms, because they tend to forget that they are actual numerical values. So to make it easier, I tell them to convert it to decimal form. I think seeing just numbers and variables is easier then seeing sine and cosine terms 😅
i get that tension is same as it is the same cable, but why is angle same for last question ?
This one isn't hard to do a home experiment with. You just need a string and 2 little hooks. Tie the string like the same way shown on the diagram and then hang something on the string around the middle. Then using a protractor, you can measure the angle on both sides. You will see that it is the same :)
Howzit bro. Can you please make videos on the sections centroids & centre of masses,moment of inertia, and friction. If the videos are made ASAP it would be much appreciated because I am going to be writing my paper soon and I need to study these sections thoroughly. Please can you help me out
I probably won't be able to make that video ASAP, but I saw lots of videos on youtube about the subjects mentioned. Hopefully, those will help for now. I am really sorry, but in time, I will get to those as well, just not going to be the next few videos coming up.
@@QuestionSolutions will you atleast be able to make those videos before Jan 5th 2021
Or videos on some of the mentioned sections atleast, will also help ...... can you leave all of the statics before those sections and make these section videos quickly... do you have a gmail or some sort of mail account ?
@@arjungovender3248 Well, I don't think I can do that because I want people to be able to go from one video to the next. So I finished off dynamics, and then it's statics. If I stop half way and then jump to a random topic, it will throw off the sequence of videos. Did you search youtube for the topics you mentioned? There are a lot of videos for those :) Also, email is contact @ questionsolutions.com.
@@QuestionSolutions I have but honestly not understanding them as well as I understand the videos made by question solutions, maybe you could send me the links to the videos you think are 100% worth it , that would be appreciated
I just started learning static and I would like to ask why is it that the highest tension should be the basis on what will be the maximum mass that it could carry
Its still puzzling me
I think you're referring to the second question. Let's assume we pick the lowest of the set, which was cable AE, with a maximum tension of 0.5775W. That means if we divide 500 N by 0.5775, we get 865.8 N. Dividing that value by 9.81 gives us the mass, which is 88.26 kg. So just for cable AE, we can hang a pot with a mass of 88.26 kg.
Setting that aside, let's think about the maximum tension cable, which is cable BC, with 1.911W. What would happen if we hung that 88 kg pot on that cable? So we get 1.911(865.8)=1654.5 N. Notice how this far exceeds the maximum each cable can hold, which was 500 N. This is why we base it on the cable that carries the highest tension. If we base it on the cable that carries the lowest tension, then that weight would exceed maximums of the cable that carries the highest tension.
Play around with these values to get a better idea of what I mean and it will help you understand it better. 👍
@@QuestionSolutions Ahh I see thank you very much 🥰
from the look of it I still need to study alot 😅🥲
@@daryldiaz7194 Once you read the chapter, try to do as many questions as you can, you will gain a lot of inside that way. The golden rule in statics is: Everything is in equilibrium, which means all the forces added together must equal zero. :)
a uniform rod of a mass 3 kg and length 2m has one end attach by smooth hinge to a ceiling at a. the other end b ais attached to a light inelastic string. the other end of the string is attached to the ceiling at c. the rod makes an angle of 30° with the horizontal and the string an angle of 60° to the horizontal qs shown. find the tension in the string.
Can you make a video tutorial for this problem. Thank you
I can't really create videos on specific problems because then I'd have to create a video for every problem that students ask. I think it's better to ask these types of problems on forums that help with physics/statics. Please check them out.
hello sir i hope you are doing well .. how did you find the tension in the first question plz?
i did not understand it ... thank you
Hi, that is a very vague question. Which part did you have a hard time understanding? There aren't any hidden steps, so I really did show everything you need to solve this question. If you can be specific, I can try to help more :)
@@QuestionSolutions yes sir i know
But there is something when you find out the f(FCA and FCB) i don’t how to find out them !? Thanks
@@arinajar754 Are you not understanding the part where we solve 2 equations to get the 2 values? Instead of thinking them as FCA and FCB, think of them as x and y. You can solve 2 equations with 2 variables right? If you need a refresh, please take a look at this video: www.khanacademy.org/math/algebra-home/alg-system-of-equations/alg-equivalent-systems-of-equations/v/solving-systems-of-equations-by-elimination
I highly encourage you to review solving simultaneous equations, you will need it a lot :) If this is not where you are confused, please give me a timestamp.
@@QuestionSolutions thank you so much sir for your kindness i really appreciate it 🙏🏻🙏🏻
@@arinajar754 You're welcome, I just want to know if you understood it? If not, I will do my best to help
how did you got 1.667W for FBD (question 2)
So you just solve for F_BD. In other words, we are isolating for that variable. Divide (1.155sin60) by (3/5) and you get 1.667.
@@QuestionSolutions THANK YOU!
@@jes-452 You're very welcome!
5:24 the fy isn't supposed (4/5)?
No, please see this video to learn about breaking forces into components using the ratio triangles: th-cam.com/video/NrL5d-2CabQ/w-d-xo.html I cover it in the 2nd example problem. Thanks!
i got lost at 3:41 pls help how do u solve to get Fca and Fcb
I think you might be missing a lot of fundamentals for this course. I really encourage you to review some high school math topics like solving simultaneous equations, it will make your life very easy :) Here, you can use substitution. See: www.dropbox.com/s/fp3d9sepod9kbu9/Question%201%20solved.pdf?dl=0
bro an easier way for it is to use sine law
how do you get F AB= 1.155W🤔??
See the pinned comment.
On 6:55 how did you get 282.96 and 332.96 ? You skipped over the actual solving step???
Here is a short video on how to get those 2 values you asked about: th-cam.com/users/shortsHe7lrJEB04U
thanks
You're welcome
Can do for me the steps for 3.40 ?
You can graph the two equations or convert all the sin and cosine into decimal values. So for example, cos30 = 0.866, etc. Then it's just a simple 2 equations with 2 unknowns :)
3:07 why is the gravity 98.1 not 9.81?
That's not gravity, that's weight. So weight is mass times acceleration due to gravity. In this question, the mass of the cylinder is 10 kg. So we get 10 x 9.81 = 98.1 N. Remember that acceleration due to gravity has the units m/s^2 where as weight has the units N.
got my test tomorrow and i’m fked
Well, I hope you do well on your test, and hopefully, these videos will help you out.
@@QuestionSolutions it sure did, i’ve been watching for like an hour now, thanks for making these type of videos 🙏🙏
@@devilmaycry8564 You're very welcome. Keep up the good work and try to do as many practice problems as possible.
Mine is after an hour 😭
@@mansorfathi9771 I hope your test went well!
how to you sovle the equatons at 6:50
It's 2 simultaneous equation solving. So you have 2 equations with 2 unknowns. You can use elimination, substitution, or you can graph them. If you don't remember how to do this, please review "simultaneous equation solving."
If force F(ba) is 1. 155W then shouldn't F(ab) be - 1.155W?
At 5:18
@@saqlainraza2219 FAB = FBA, same thing, just different notation. Usually, if you look at a force from A to B, we write it as FAB. If we look at it from B to A, we write FBA. However, the sign convention is not determined by the way we write AB or BA. It's based on the direction of the force itself. You can write either and its perfectly fine. Now if a force faces left, and we got a negative value after we assumed left to be positive, then we know the force is opposite to the direction we drew it, but again, FBA or FAB notation does not matter. At the end, it's just a notation.
@@QuestionSolutions oh yes I got it explained it well thanks a lot buddy :)
@@saqlainraza2219 You're welcome :)
How to get that 1.38m?
Please give me a timestamp so I know where to look. Many thanks!
x tan45.57 + 0.75 = (3.5-x) tan 45.57
1.02x +0.75 = (3.5 tan 45.57 - xtan45.57)
1.02x + 0.75 = 3.57 - 1.02x
1.02x + 1.02x = 3.57-0.75
2.04x = 2.82
x=1.38
At 4.40 , how do u grt Fab = -1.155W
Add "W" to both sides. You get: FSin60° = W. Divide both sides by sin60°. You get F = 1.155W . (just wanted to add, on the right side, it's really just (1/Sin60°)(W), so you're dividing 1 by Sin60°.)
But w=500 over sin60 is 577.35
Wait so we assume w is 1 ???
@@nbarekap No Aidan. 2/X= (1/X) x 2. Thus, W/sin60 = (1/sin60) x W.
@@nbarekap You are missing fundamental algebraic knowledge. Please kindly review simplifying expressions and other forms to solving equations or you will end up using up a lot of time working on these things. I am not being mean, but it won't get any easier when the courses proceed.
If we have a simple equation like this: x+2=y
Here, there is 1"x" and 1"y". We just don't show the 1 in front. If we have 2x+2=y, we have 2"x" and 1"y". Again, because there are 2 x's, we put a 2 in front, but we don't put a 1 in front of the y since it's just 1. So please, review these things :)
How I get to know that moment is not zero
You only need to calculate the moments if you have more than 2 unknowns. With moments, you can solve for 3 unknowns.
How did you get Fab=1.155W in the second question, that's where I became lost
bro pleas read comment
So you're isolating for FAB. Add W to both sides and convert sin60 to decimal form. You get 0.866FAB = W. Divide both sides by 0.866. FAB=(1/0.866)(W), convert to decimal form. You get 1.155W. Remember that W divided by sin60 is the same as 1/sin60 👍 Please see: bit.ly/3kGNcyW
@@QuestionSolutions ohh I can see it now thanks
@@QuestionSolutions @Question Solutions 4:35 why did you multiply both sides by 1/sin60? where did you get it?
@@julianafairy908 If you have an equation like this: 2x = 2. You solve for it by multiplying both sides by 1/2 (or dividing by 2, which is the same thing) where did the 1/2 come from? It's the same thing. In fact, you can multiply or divide it by whatever you want as long as you do it to both sides. Don't get confused by sin/cos stuff, it's still just numerical values 👍 I hope that helps :)
Voice 🔥
😅
@@QuestionSolutions ❤️
Trying to solve the first problem for hours just to find out my calculator was in radian mode😅🤣
Happens to the best of us!
9:52
2F sin 45.57 - 981 = 0
2F 0.7 = 981
F =981/1.14
F =860.5N
YOU WRITE 686.9N HOW?
You made a simple mistake, 2 x 0.7 = 1.4, NOT 1.14.
my bad..by the way, your videos helped me a lot in my mechanics paper.@@QuestionSolutions
thanks😀@@QuestionSolutions
@@3drendering183 you're very welcome. Glad to hear these videos helped :)
Hello
Can you please tell me why the angles in the last question are same
If you have a string at home, tape the 2 ends to things, so there isn't a lot of tension, but it's not super lose either. Then, hang something with a paper clip or something that slides around easily, you'll see that it'll come to the center. If you measure the angle with a protractor, you'll notice they are very very close, probably not perfect since there are a few variables to consider, but it'll be close. In most cases, you'll get the same angle on both sides.
@@QuestionSolutions Thank you very much
@@madihavahaj599 You're very welcome!
❤️❤️❤️
❤❤❤
@@QuestionSolutions which country u are ?
@@EveryEngMechanical Canada.
@@QuestionSolutions oh
th-cam.com/users/MechLearner
This is my channel
@@EveryEngMechanical I saw, I looked through it before :) Good tutorials on the solidworks modeling! I hope you grow really big.
i think you could have done a better job explaining at how you got at the end answers
The problem with explaining how to solve simultaneous equations is that it takes away time from the majority of students who already know how to do that (and students definitely should know this stuff at this point). It extends the time of the videos by a large margin, and takes away from what I am trying to get across, which are the new material taught in a statics class. To counter this, I have created a few short videos on the channel that teach students how to solve simultaneous equations. So if you need a refresh, I encourage you to look at those to get up to speed. If you look through the pinned comment, there are already links directing students to solve the equations shown on screen as well.
❤️
❤❤
I want to send you a fruit basket. Please let me know how I can support your chancel
Thank you so much :) I really appreciate your thought. I don't have any methods set up for support, but something you can do for me is to share any video that you found helpful. I figure maybe it'll help your classmates and friends, and then that would help the channel as well. Win win. I really appreciate your thought. Happy holidays and I wish you a very happy new year. May the new year bring you much success in all your endeavors.
pen deni a?
I don't know what that says? :(
sb
?
What sort of rubbish explanation was that….. wdym by you can solve them any way you like
So the goal of these videos isn't to teach how to solve a system of equations (this should be familiar to you from high school). At this point, you NEED to know how to do that. If you need a refresh, I created a bunch of small short videos going through how to solve them. But again, the goal of this video is to teach new concepts covered in statics, which is how to establish the equations properly to get a final answer. So I encourage you to review high school material if you don't know or don't remember how to do this. It will be of great aid to you in your courses.
do you have a seperate video on how to draw free body diagrams
No, I don't. :(