I genuinely think that you know more than my professor does. Since you can teach this in simple words, without making it too complicated and still can apply these to much more complex situations. Thank you so much man, great lectures.
Watching these is making my statics final seem a lot more doable, really wish I would have found your channel at the start of this semester! Thank you for your videos!
Thank you for teaching us more effectively than my books or some of my professors. I also love the website!!! i don't understand why this channel is so underrated.
everytime i didnt understand what i learn from my class, i will find your channel and it very helpful. Keep making helpful video for everyone. Thank you. You are the best!!
I’m so glad to hear that the content on this channel has been helpful for you! You're very welcome and thank you for taking the time to write such a nice comment :) ❤️
thank you so much for these!!!! i got a 16% on my first statics exam 💀with the help of your vids i got a 96% on my next, thank you for creating such a great resource for this course
You can most definitely pass this course. Try to do the questions I show in the videos beforehand, and then if you get, watch the rest. Do your best, do as many questions as you can and have a positive mindset. You got this!
@@QuestionSolutions haha I have engineering friends in all 3 of the major Toronto universities: UofT, York, and Ryerson that watch your videos. Keep it up! What I personally love is that you are able to animate the problems and actually show the "dynamics" of the problem to help us visualize these problems, rather than just seeing a textbook problem that we have to infer movement, etc.
@@rehanrashid9296 That's awesome. Really glad to hear they are watching these videos. Thanks so much for taking the time to comment, I appreciate them. I will keep doing my best to help!
On the first example on the summation of the forces in the y-axis you made a mistake when adding -24 and -40. It's 64 and not 54..... By the way I love your content.
I am not sure what you mean. We didn't add -24 and -40. We solved 3 equations. The answers shown are correct. To see the steps, please see: www.cymath.com/answer?q=x-10%2Bb(4%2F5)%3D0%2C%20y-24-40%2Bb(3%2F5)%3D0%2C%2024(2)%2B40(6)-b(3%2F5)(6)%3D0 If I am still incorrect, please provide a timestamp to the location on the video so I can look and if incorrect, write a pinned comment. :) Thanks!
Hey! So all you're doing is solving for 2 unknowns with 2 equations. It might be easier to represent F_A as "x", and F_B as "y." You can use substitution, elimination, or whatever method you're comfortable with to solve them. Please see: www.cymath.com/answer?q=x(4%2F5)-y(5%2F13)%3D0%2C%20x(3%2F5)%2By(12%2F13)-100%3D0
You select wonderful problems! Question if you have time. The problem at 7:00. If there is an additional moment at the intersection of pipe A and B, what would cause it and why does it not have to be included in the computation of the X Y forces?
So with a collar, due to the constraints, it will create an opposing force in the x and y directions, along with an opposing moment. This moment isn't the same as a moment caused by a force a certain distance away from a point. You can think of it as an applied moment, in other words, a separate moment. When we do computation in the x or y forces, we do not care about moments. We only care about moments when we write a moment equation. Let me know if that clears it up 👍
@@insanehal I'm sorry to hear that. Strong backgrounds are made for any subject by doing as many questions as possible so you gain more and more experience about how questions are formulated, the method to get the answer quickest, and how you can actually go through means of solving the said problem. It really just comes down to doing as many questions as you can, with every spare minute you have. Sounds tedious but that really is the key to most subjects where memorizing things is not at the forefront. Try creating a timetable for yourself. Give more time to subjects that are hard for you, and less to those that you can easily get through. Each week, focus on doing as many textbook questions as possible during the allotted slot for your course. Use your professors/TAs office hours, go see them, ask them questions, get any confusion you have cleared up. Try your best, don't think you can't do it, because you really can. You made it into a university, probably doing engineering, which already means you have the intelligence to pull it off. Best wishes!
I am not sure what to tell you. 😅 You've made a numerical error but I don't know where. If you got TBC the same, then plugging in 80 into the FY equation directly solves for AY. See: www.cymath.com/answer?q=x-24-40%2B(80*3%2F5)%3D0 I just put AY as "x".
So you're missing the fundamentals on breaking forces into components using sine and cosine, please take a few minutes to watch this video :th-cam.com/video/NrL5d-2CabQ/w-d-xo.html If you can watch the whole thing, great, but if not, please watch from the start to the end of the first example, I go through it step by step on how to use sine and cosine.
So we need to break the force in the rope down into it's components. We are given a slope triangle and each side corresponds to the component side we need. Please see: th-cam.com/video/NrL5d-2CabQ/w-d-xo.html I cover an example with a slope triangle and show how to use them. It's the 2nd example.
May you please tell me how d owe know around which point to we calculate the moment when they dont mention in the question, such exercise 5:32 and thank you for the amazing effort.
So you want to take the moment about a point that has the most unknowns. You can take it about any point you want, but if you take it about a point that has 2 unknowns, you can eliminate those 2 forces since their lines of action go through it. See: th-cam.com/video/QNNnPZ68STI/w-d-xo.html
Lovely video. There is one thing I didn't understand though. At 6.30, the equation for calculating moments- I understand the 2nd part and the third part. I am struggling to understand why in the first part, the distance from B is not taken into consideration.
For which force was the distance from B not considered? Please let me know which, I can't figure out what you mean and I would really like to clarify for you. :)
thank you so much for the videos! I am a little confused on how you solved for theta and got 2 values in the last example problem, could you clarify? thanks again!
For a force couple moment, do we still ignore the force at one of the two points when we are calculating force moment? Just like what you did in the example?
Wonderfully made video. Could someone kindly explain the step-by-step procedure for me to arrive at the right answer? I seem to have a different values for all 3 of them once I try solving it (5:35)
@@QuestionSolutions thank you so much 🙏🏽, you probably get this a lot but you are really a big help to all of the students who stumbles upon you videos 💯. Hope you continue doing more videos, as well as engaging to the comments.
There really isn't much to explain. You have 3 equations with 3 unknowns, so you can use any method you're comfortable with to solve them. I usually use substitution, but you can use elimination or even a matrix. If you don't like any of those, you can graph the three equations to get your answer :) If you forgot about substitution, I highly recommend reviewing it because you will need it for pretty much all throughout this course, dynamics, and many more. www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-systems-topic/cc-8th-systems-with-substitution/v/the-substitution-method This is for 2 equations, but it's the same process for 3 equations. You can also use wolfram alpha to check your answers.
@@QuestionSolutions But on our exams, we don't have one of those calculators. So I solved for sine theta and then used inverse sine, could that be correct?
@@StefGraafsma It's highly unlikely you will ever get an equation like this to solve. The issue with this problem is isolating for sine or cosine, because of the 600 term. It requires multiple trigonometric identity substitutions.
hi sorry i have a question for the last example at 9:59 . For Hooke's Law, to determine the stretch of the spring, is it not 1m (the stretched) minus the unstretched (3sintetha). Please explain why you only wrote 3sintetha
For hook's law, we just have F=ks, where "s" is the stretch of the spring. So imagine you have a 2 m long spring in front of you. It's not stretched, it's just 2 m. Then "s" here would be 0. Now if the spring was stretched to 3 m, then "s" would be 1m, which is just 3-2m. So here, our spring has a length of 1m. When it was stretched, it becomes 1+3sinθ, in other words, the stretch is just "3sinθ" What we found is only the stretch, not the total length. Let me know if you need further clarifications :)
@ 9:50, you used a right angle triangle but is the spring making a right angle with the board? edit: I see now that from the point extension begins to the pin forms the right angle.
At 9:30, how did you figure out the angle was the same? I mean, during a test, you can't really flip the coordinate system and verify it, so is there any other way to know that?
So I didn't "flip" the coordinate system to verify it, it was just to show students who watch this video. It should be intuitive to you that the angle is the same, but if it's not, that's okay, you just need to draw parallel lines and use alternate interior angles and co-interior angles to get a verification.
@@johnpatrickaguilar9442 Awesome :) There is a video on breaking forces into components if you need to watch, let me know, I can find it for you. It's on the statics playlist. 👍
hi there , excellent videos. quick question how did you go about finding the FA components at 6:15 ?? could you have moved the y component to the right instead of the x component up ? that is so confusing
So components can be translated along their axis and yield the same result. What I mean is, you can actually move the y-component to the right, and the x-component to the bottom. It makes no difference to your answer since either method creates a right angle triangle with a 30 degree angle.
@@QuestionSolutions Hi! I Can't thank you enough for these videos you uploaded, much appreciated. I also do have a question like this. When you brake the FA into components like the process in video, doesn't the angle to the bar make 60 degres?(between ("Fr and Fx") And in that case the Y-axis become a Sin and the X- axis a cos? I feel confused about this, in what way am I thinking wrong and do you have any tips? Thanks in advance! 🤗
You're very welcome! The way it's shown, the light green components and the dark green force, the angle is 30 degrees. If you move the components, you can draw them with a 60 degree angle. The angle doesn't matter, the components doesn't matter, none of that matters for sine and cos. In fact, whether it matters or not shouldn't be an issue to you. Forget that all and just look from the perspective of the angle. If it's in front of the angle, use sine, if it's adjacent to the angle, use cosine. Nothing else matters, don't look for patterns, or associations, or anything like that. Please see (it's less than 60 seconds): th-cam.com/users/shortsvynnKlJD_Jo @@Thecarfreak100
Hello Professor, thanks for the video. I do have a question concerning 7:22 why doesn't Point B also have a moment like Point A has? Or is the 20 Nm given is that the moment for B?
So point A has a moment because it's a collar there. Collars create a moment about it self (see 7:09). At point B, it just a smooth surface, which cannot create a moment (see 7:15). The 20 Nm is independent to both, it is simply an additional moment given/ applied to the object.
@@QuestionSolutions hii sir wanted to ask what will you do in exam taking note that you dont know the intercepts of the graph please share the solution.
So the easiest way to get the angles is to actually graph the equation. When you graph it, you will find 2 x-intercepts. Those are the 2 values you're looking for. You can solve this without graphing but you'd need to use trigonometric identities and it would take way longer than actually graphing it. There are 2 angles because at those 2 different angles, the system would be in equilibrium. In other words, at the angles 14.5 or 82.5 degrees, nothing would be moving, it's at rest. 👍
No, it creates a counter-clockwise moment. Imagine the object is free to move about point B. If I push down exactly where force B is, which way will it spin? It will spin counter-clockwise about point B.
@6:15 Im confused as to how you assumed that Fa Triangle from the provided 30 deg. I Don't understand how you constructed the other triangle and know that it is the same angle as the one provided if it's coming from another spot, I Just don't see it.
So for these types of things, if it doesn't come intuitively to you, which is perfectly fine, I encourage you to draw it out on a large piece of paper. Using a protractor can also help to understand geometric concepts as well. It's all to do with perpendicular and parallel lines (creating alternate interior angles, corresponding angles, etc.) but it can be hard to see sometimes. The best way is always to draw it out on a big piece of paper with a nice straight edge and visually seeing it. This will not only help you see things faster but give you an intuition as to how you can figure out other angles without too much trouble.
The horizontal components comes from the weight and the spring placed at an angle. So the pin counteracts those forces in both x and y directions. So the simple answer would be the x-components of the spring force and weight is countered by the x-component of the pin at A.
Can you give me a timestamp as to where you would get confused with negatives and positives? I can help if I know exactly where you're confused. In general, positives and negatives are based upon your assumptions. For example, if I pick right to be positive, and a force is facing left, then it's going to be negative. If the force faces the way you picked to be positive, then it's positive. The same with moments. If I pick clockwise to be positive, then any moment that is clockwise will be positive and any moment that is counter-clockwise will be negative.
I am not entirely sure what you mean. But if there are no other horizontal forces effecting an object, F_x wouldn't exist, since nothing is being countered. What do you mean by "does not like being zero?" In statics, all net forces must equal zero, if it doesn't the object is moving around. Maybe I am not understanding your question, sorry!
Hi! I was just wondering why at 6:01 fa in the x direction is not zero because it’s being held by a roller support meaning there would be no reaction, thanks!
That would only be true if it's along the horizontal plane. Notice that our force is applied at an angle. That means that force must be broken into components and you'd get an X and Y component. 👍
I think it's much easier to graph these problems to get a solution. That's probably what I did, I don't remember now, but usually, I try to graph them since it's faster. Otherwise, you'll have to use trigonometric identities to solve them 😅
So in a rope, the force always goes towards the support. So in this case, from B to C. Once you draw the vector, you can break it into components along the x and y components.
@@topefestus2193 Yes, that's correct, in the end, it is just an assumption. I think I cover forces along ropes and stuff on this video: th-cam.com/video/X9g4G1eBHCA/w-d-xo.html
wait! on 6:12 you broke it into x and y components. i get that however my mind is not able to understand why is y component cos and x component sin?? isnt it the other way round? i tried to understand it but i cannot :(
Watch this video, it's less than 60 seconds, and you will definitely understand :) th-cam.com/users/shortsvynnKlJD_Jo?feature=share If you still don't, let me know.
We took clockwise to be positive, so the 100N force creates a clockwise moment, which means positive. Keep in mind that this is not an equation for the summation of forces but a moment equation. So we need to keep in mind the direction of the moment created by the force.
So it's just an assumption. All we know is that there is a moment applied at the collar. The direction, we have no idea until we solve the problem. So you have to make an assumption about the direction, and if you get a negative value, then it's opposite to your assumption.
So this moment isn't created by a force that is a certain distance away. This is a counter moment created in the joint because of the collar. You can think of it as a separate force applied.
Yes, the assumed direction is totally up to you. If you get a positive answer, then your assumption is right, if you get a negative answer, then it's opposite to your assumption.
Hmm, I am not entirely sure I can explain it, but I will give it a shot. The triangle is a ratio triangle, so when the a line is perpendicular to that same force, you're just flipping it 90 degrees. At 7:31, look at the brown line at the top diagram, with the 12,13,5 triangle (right side). Now in your mind, turn that line so that it matches the green force arrow on the bottom image (right side). See how it just flips 90 degrees? I think this comes with intuition or maybe doing a lot of questions, but if you struggle with seeing this, I think an easy way to understand it would be convert the triangle into an angle. So you just use sin or cosine, use the proper sides and replace the triangle with the angle. After that, you can draw the perpendicular line and calculate the angle again. It really just boils down to knowing that a ratio triangle is just another way of giving an angle. If that doesn't work, I encourage you draw this on a blank piece of paper (draw it big), and then use a protractor to see the angle change with the perpendicular lines. That way, it will click in your mind. :)
We use both, one gives the vertical component and the other gives the horizontal component. So 3cos30 gives us the horizontal length of the left side, while the 3sin30 gives the vertical length.
Great video. Can you please explain a little better why at 7:56, when we take the sum of moments around point A, we include M_A (if its created by the forces at A) and the 20 Nm one at the curve of the body? Which forces are these moments generated from?
This is based on what type of support system it is. In this case, if you look at your textbook/ resource on what types of forces are created based on the support type, you will see that this crease a force perpendicular to the collar and a moment about it. The 20 Nm moment is already included in our problem, it's a given, so you can think of it as someone else applying a specific moment at that location. Let me know if you need more clarifications.
Using the moment equation, you can directly solve for FA since it's a single variable. Once you find FA, plug that into each of the x and y equations to solve for BY and BX. Let me know if you need further clarifications.
could you please explain how at 6:45 you concluded that the distributed force (12kN) is going to be counterclockwise and, hence negative? Is it not as if we are applying a force on the rod from the top? I feel that it would move clockwise in that sense. I am a little unsure about the signs of moments, I am having a hard time visualizing the movements, can you please explain them?
Sure, so remember we are calculating the moment about point B. That means the whole rod pivots around point B. Imagine the bar moving about point B. If we apply the 12 kN force at the location that's shown, so it's coming downwards from the top, it's impossible for it to turn clockwise, we are literally pushing something down about point B. It will 100% turn counter-clockwise. If you don't believe me, take a ruler, hold the right edge between your thumbs and push down from the top anywhere on the left side. You will see that it will try to turn counterclockwise. It's negative because we picked clockwise to be positive. So anything that goes counterclockwise becomes negative and anything clockwise becomes positive. If you picked counterclockwise to be positive, then all the clockwise moments will be negative and counter-clockwise moments will be positive. That part is up to you, but make sure to follow through. Also, I am not being mean or anything, but you are missing a lot of fundamentals when it comes to moments. Please take a few minutes to watch this video: th-cam.com/video/QNNnPZ68STI/w-d-xo.html I promise it'll help you out in the long run, even if its a tiny bit :)
@@QuestionSolutions Thank you so much! that for sure cleared a lot of my misconception. Will definitely give the video another watch! You're doing an awesome job! Thank you! (:
I assume you're talking with respect to to the FSsin30. So remember, we only care about the perpendicular distance for moments. For the sin component (which is horizontal here), the perpendicular distance, in other words, the height from the force to point B, is 3sin30, we need to break that 3 m angle slope and use the vertical distance. So we don't need the 4 m horizontal length since that's parallel to the force component.
I genuinely think that you know more than my professor does. Since you can teach this in simple words, without making it too complicated and still can apply these to much more complex situations. Thank you so much man, great lectures.
Thank you very much for your kind words! I wish you the best with your studies. :)
Bro your professor also teach good but at the class you always do bla bla bla
How'd the rest of your class go?
@@lords9966 Lol
@@PunmasterSTP I passed it man, I passed it 2 years ago hahah
These are the best videos on TH-cam. The only thing getting me through statics are these videos. Thank you.
Thank you very much, I hope they are helpful to you. I wish you the best with your statics course!
Watching these is making my statics final seem a lot more doable, really wish I would have found your channel at the start of this semester! Thank you for your videos!
I am really glad to hear that. I hope they are helpful to you and wish you the best on your finals! Keep up the good work.
How'd your final end up going?
@@PunmasterSTP passed!
@@SuperBartHole That's awesome; I'm glad!
Thank you for teaching us more effectively than my books or some of my professors. I also love the website!!! i don't understand why this channel is so underrated.
You're very welcome. I am really glad to hear these videos are helpful to you :) Best wishes with your studies!
University of Pretoria student watching your videos , keep up the great work it helps a lot.
Really glad to hear these videos help out a lot :) I wish you the best with your studies!
🤣
How are/were things going in Pretoria?
everytime i didnt understand what i learn from my class, i will find your channel and it very helpful. Keep making helpful video for everyone. Thank you. You are the best!!
I’m so glad to hear that the content on this channel has been helpful for you! You're very welcome and thank you for taking the time to write such a nice comment :) ❤️
thank you so much for these!!!! i got a 16% on my first statics exam 💀with the help of your vids i got a 96% on my next, thank you for creating such a great resource for this course
Wow, that's amazing! I am really happy for you and well done! Keep up the awesome work and best wishes with your studies.
That's an impressive increase 👍
I wish we had more content like this, brief yet detailed and understandable.
I hope the video was helpful to you :)
@@QuestionSolutions it was helpful thank you. :)
I can actually see myself passing this course. Thanks a lot for the videos!
You can most definitely pass this course. Try to do the questions I show in the videos beforehand, and then if you get, watch the rest. Do your best, do as many questions as you can and have a positive mindset. You got this!
Its been 2 years since you uploaded this and its still helping students!! Thank you so much for this video, keep up the great work!
Thank you very much. I hope all of the videos are helpful to you and your friends. Best wishes with your studies :)
This just made my life 100 times easier. Thank you.💗
You're very welcome! Keep up the awesome work with your courses ❤
Your videos are the best! Keep making more sir! It helps me grasp the concepts very well!
Glad to hear! Keep up the good work and best wishes with your studies :)
thank you so much for this video. It made me ace my final exam! you are a life saver!
I am really happy you aced your final exam! Good job, and keep up the awesome work. Best wishes with your future studies.
2D coplanar? More like "Terrific lectures; there are none greater!" 👍
Thank you punmaster 👌
@@QuestionSolutions You’re most welcome.
Please do not stop. What a legend.
I will do my best!
The contents in the channel are really great. Anyone can understand easily .
Thank you!
You are a life-saver man. Thank u so much
You're very welcome! :)
I am watching this to prepare for my FE exam , thank you!
You're very welcome! I wish you the best with your exam.
West hills community college student watching your vids, keep it up.
Awesome! Same to you, keep up the awesome work :)
thank you for this video! you explain better than my teacher
You're very welcome :)
Ryerson student watching your videos, you are amazing !!!! more structural engineering topics
Thank you very much! Best wishes with your studies.
thank you for helping me understand well!! university tenaga nasional here!!
You're very welcome!
You're so smart with all these Mechanical Engineering topics. Did you study at Waterloo, UofT or MIT?
Many thanks! I hope these videos are helpful to you, and yes, UofT 👍
@@QuestionSolutions haha I have engineering friends in all 3 of the major Toronto universities: UofT, York, and Ryerson that watch your videos. Keep it up! What I personally love is that you are able to animate the problems and actually show the "dynamics" of the problem to help us visualize these problems, rather than just seeing a textbook problem that we have to infer movement, etc.
@@rehanrashid9296 That's awesome. Really glad to hear they are watching these videos. Thanks so much for taking the time to comment, I appreciate them. I will keep doing my best to help!
These videos are worth more than my rented textbook and professor combined
😅 That was a funny comment!
most beautiful and concise ways to explain. Good luck in life brother!!
Thank you very much, I really appreciate it. I wish you the same!
i really wished u were my lecturer man, i would actually enjoy it
Thank you! Hopefully, these videos will be just as helpful with your studies.
omg..thank you so much for such a good content
You're so welcome!
Thanks for your videos❤ m having a quiz and ur tutorials helped me.
I'm so glad! Keep up the great work and I hope your quiz went well.
Awesome video❤ best explanation out there😮😮
Glad it was helpful and thank you very much! ❤
@@QuestionSolutions I will get a good grade in this course just because of you😌
On the first example on the summation of the forces in the y-axis you made a mistake when adding -24 and -40. It's 64 and not 54.....
By the way I love your content.
I am not sure what you mean. We didn't add -24 and -40. We solved 3 equations. The answers shown are correct. To see the steps, please see: www.cymath.com/answer?q=x-10%2Bb(4%2F5)%3D0%2C%20y-24-40%2Bb(3%2F5)%3D0%2C%2024(2)%2B40(6)-b(3%2F5)(6)%3D0
If I am still incorrect, please provide a timestamp to the location on the video so I can look and if incorrect, write a pinned comment. :)
Thanks!
helped a lot, any books you wanna recommend to solve statics problems?
Engineering Mechanics Statics by R. C. Hibbeler and K. B. Yap
You're my hero sir
Thanks :)
Thank you for this clear explanation!
You're very welcome!
Thanks the video man,I really really appreciate ❤️❤️❤️❤️
You're very welcome! ❤
Thanks so much this really helped!!!
I am really glad to hear that! Keep up the great work and best wishes with your studies.
hey! im stuck at 7:44 where you get the answers for F-A and F-B. i dont understand how you got 39,683 and 82,54 N. Could you or anyone else explain?
Hey! So all you're doing is solving for 2 unknowns with 2 equations. It might be easier to represent F_A as "x", and F_B as "y." You can use substitution, elimination, or whatever method you're comfortable with to solve them. Please see: www.cymath.com/answer?q=x(4%2F5)-y(5%2F13)%3D0%2C%20x(3%2F5)%2By(12%2F13)-100%3D0
You select wonderful problems! Question if you have time. The problem at 7:00. If there is an additional moment at the intersection of pipe A and B, what would cause it and why does it not have to be included in the computation of the X Y forces?
So with a collar, due to the constraints, it will create an opposing force in the x and y directions, along with an opposing moment. This moment isn't the same as a moment caused by a force a certain distance away from a point. You can think of it as an applied moment, in other words, a separate moment. When we do computation in the x or y forces, we do not care about moments. We only care about moments when we write a moment equation. Let me know if that clears it up 👍
@@QuestionSolutions Thank You.
Love u man. You are the best!
Thanks 😁
I have my statics exam later. This video makes me feel less dreadful about it. Thank you so much
You're very welcome. I hope your statics exam went well!
@@QuestionSolutions i actually failed it, so I’ll be on this channel a lot more. Any tips for people who don’t have a strong background in physics?
@@insanehal I'm sorry to hear that. Strong backgrounds are made for any subject by doing as many questions as possible so you gain more and more experience about how questions are formulated, the method to get the answer quickest, and how you can actually go through means of solving the said problem. It really just comes down to doing as many questions as you can, with every spare minute you have. Sounds tedious but that really is the key to most subjects where memorizing things is not at the forefront.
Try creating a timetable for yourself. Give more time to subjects that are hard for you, and less to those that you can easily get through. Each week, focus on doing as many textbook questions as possible during the allotted slot for your course. Use your professors/TAs office hours, go see them, ask them questions, get any confusion you have cleared up. Try your best, don't think you can't do it, because you really can. You made it into a university, probably doing engineering, which already means you have the intelligence to pull it off. Best wishes!
@@QuestionSolutions thank you. this made me feel a bit better.
Thankyou so much sir❤
You're very welcome! ❤️
Thank you for summarizing my statics textbook. Godbless you. You're the best. However, do you have a lect about coeff of friction?
You're very welcome, and unfortunately, I don't have a video about coefficient of friction. Hopefully, one day :)
Very much helpful..🖤
Happy to hear that! Best wishes with your studies!
Great explnation
Thank you very much!
Hi, I got A_x and T_BC the same, but A_y = 20kN instead of 16kN for the first example.
I am not sure what to tell you. 😅 You've made a numerical error but I don't know where. If you got TBC the same, then plugging in 80 into the FY equation directly solves for AY. See: www.cymath.com/answer?q=x-24-40%2B(80*3%2F5)%3D0 I just put AY as "x".
can anyone explain what made the x component a sin and the y component cos? that was the one part I didn't understand, thank you for this video
So you're missing the fundamentals on breaking forces into components using sine and cosine, please take a few minutes to watch this video :th-cam.com/video/NrL5d-2CabQ/w-d-xo.html If you can watch the whole thing, great, but if not, please watch from the start to the end of the first example, I go through it step by step on how to use sine and cosine.
Hi can you explain to my why the tension of the rope are Tbc(4/5), Tbc(3/5)
So we need to break the force in the rope down into it's components. We are given a slope triangle and each side corresponds to the component side we need. Please see: th-cam.com/video/NrL5d-2CabQ/w-d-xo.html I cover an example with a slope triangle and show how to use them. It's the 2nd example.
May you please tell me how d owe know around which point to we calculate the moment when they dont mention in the question, such exercise 5:32 and thank you for the amazing effort.
So you want to take the moment about a point that has the most unknowns. You can take it about any point you want, but if you take it about a point that has 2 unknowns, you can eliminate those 2 forces since their lines of action go through it. See: th-cam.com/video/QNNnPZ68STI/w-d-xo.html
Lovely video. There is one thing I didn't understand though. At 6.30, the equation for calculating moments- I understand the 2nd part and the third part. I am struggling to understand why in the first part, the distance from B is not taken into consideration.
For which force was the distance from B not considered? Please let me know which, I can't figure out what you mean and I would really like to clarify for you. :)
thank you so much for the videos! I am a little confused on how you solved for theta and got 2 values in the last example problem, could you clarify? thanks again!
So the easiest way is to graph the equation and see where the x-intercepts are. Please see: www.desmos.com/calculator/zaqa0burem
@@QuestionSolutions Now I see. Thank you again for the videos and responses you are the best!
For the last question, how did you get the 82.535 deg angle, beause i can only get the 14.539 deg when using the solve for x method?
See: www.desmos.com/calculator/ph4oxvuno0
Both values satisfy the equation between 0 and 90 degrees.
i do loves the teaching but untill now i didnt know how did you find the value at ax,ay and moment
Please give me a timestamp so I can help you out better. Thanks!
GOAT
⭐
Do you have a tutorial video for geometry in getting the x and y components?
Yes, I do, see: th-cam.com/video/NrL5d-2CabQ/w-d-xo.html
What software you're using. I think it would help my students
I use illustrator to draw the diagrams and after effects to animate.
Awesome!!!
Thanks!!
how do you find the angles in q4? 10:50. i know youre supposed to graph them, but how?
www.desmos.com/calculator/7vtdo9ziaz You're looking for the x-intercepts between 0 and 90 degrees.
For a force couple moment, do we still ignore the force at one of the two points when we are calculating force moment? Just like what you did in the example?
Please provide a timestamp so I know where you are referring to. Thanks!
Wonderfully made video. Could someone kindly explain the step-by-step procedure for me to arrive at the right answer? I seem to have a different values for all 3 of them once I try solving it (5:35)
Please see: www.cymath.com/answer?q=x-10%2Bt(4%2F5)%3D0%2C%20y-24-40%2Bt(3%2F5)%3D0%2C%2048%2B40(6)-t(3%2F5)(6)%3D0
@@QuestionSolutions thank you so much 🙏🏽, you probably get this a lot but you are really a big help to all of the students who stumbles upon you videos 💯. Hope you continue doing more videos, as well as engaging to the comments.
@@babyzed7940 Always really nice to read a kind comment, thank you very much! Best wishes with your studies :)
Hi, I think clockwise is a negative moment right ? why at 6:30 you said pick clockwise to become positive ?
Please see: th-cam.com/users/shortsP029mqnp4XY
The example at 8:51, how were you able to solve the problem and get 14.359 degrees and 82.535 degrees?
You can graph them to see where it intersects. Please see: www.desmos.com/calculator/0h0abna2ty You're looking for the x-intercepts.
Hello there, could you please explain to me how you found FA at 6:55 ? I would really appreciate that.
There really isn't much to explain. You have 3 equations with 3 unknowns, so you can use any method you're comfortable with to solve them. I usually use substitution, but you can use elimination or even a matrix. If you don't like any of those, you can graph the three equations to get your answer :) If you forgot about substitution, I highly recommend reviewing it because you will need it for pretty much all throughout this course, dynamics, and many more. www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-systems-topic/cc-8th-systems-with-substitution/v/the-substitution-method This is for 2 equations, but it's the same process for 3 equations. You can also use wolfram alpha to check your answers.
Thanks a lot!
You're very welcome!
For the last question, how to solve for theta algebraically? I tried but got 17,6 degrees
It's arduous to solve this algebraically. Graphing it is easier.
@@QuestionSolutions But on our exams, we don't have one of those calculators. So I solved for sine theta and then used inverse sine, could that be correct?
@@StefGraafsma It's highly unlikely you will ever get an equation like this to solve. The issue with this problem is isolating for sine or cosine, because of the 600 term. It requires multiple trigonometric identity substitutions.
💯 explanation
Thanks!
hi sorry i have a question for the last example at 9:59 . For Hooke's Law, to determine the stretch of the spring, is it not 1m (the stretched) minus the unstretched (3sintetha). Please explain why you only wrote 3sintetha
For hook's law, we just have F=ks, where "s" is the stretch of the spring. So imagine you have a 2 m long spring in front of you. It's not stretched, it's just 2 m. Then "s" here would be 0. Now if the spring was stretched to 3 m, then "s" would be 1m, which is just 3-2m. So here, our spring has a length of 1m. When it was stretched, it becomes 1+3sinθ, in other words, the stretch is just "3sinθ" What we found is only the stretch, not the total length. Let me know if you need further clarifications :)
In 5:33, why are the values in moment + when it is clockwise. In Torque isn't + is for counterclockwise and - for clockwise?
It makes no difference to the direction you pick to be positive. You will get the same answer. See: th-cam.com/users/shortsP029mqnp4XY?feature=share
@ 9:50, you used a right angle triangle but is the spring making a right angle with the board?
edit: I see now that from the point extension begins to the pin forms the right angle.
Yes, at the initial position, when the board is resting, it will lie perfectly horizontal.
At 9:30, how did you figure out the angle was the same? I mean, during a test, you can't really flip the coordinate system and verify it, so is there any other way to know that?
So I didn't "flip" the coordinate system to verify it, it was just to show students who watch this video. It should be intuitive to you that the angle is the same, but if it's not, that's okay, you just need to draw parallel lines and use alternate interior angles and co-interior angles to get a verification.
Hi, I seem to have forgotten my past lessons. Could you tell me what I should restudy in order to solve the x and y components at 5:05? Thank you
I got it!
@@johnpatrickaguilar9442 Awesome :) There is a video on breaking forces into components if you need to watch, let me know, I can find it for you. It's on the statics playlist. 👍
@@QuestionSolutions Thanks for the reply, but I think I got it now. Thank you!
10:59 I'm confused, how'd you get the two theta angle. Can you show me how, please?
You can graph it like I did and find the 2 intersecting points. Solving it arithmetically might take quite some time to do.
hi there , excellent videos. quick question how did you go about finding the FA components at 6:15 ?? could you have moved the y component to the right instead of the x component up ? that is so confusing
So components can be translated along their axis and yield the same result. What I mean is, you can actually move the y-component to the right, and the x-component to the bottom. It makes no difference to your answer since either method creates a right angle triangle with a 30 degree angle.
@@QuestionSolutions youre absolutely right man , thanks for the reply
You're very welcome!@@eduardocarrillo930
@@QuestionSolutions Hi! I Can't thank you enough for these videos you uploaded, much appreciated. I also do have a question like this. When you brake the FA into components like the process in video, doesn't the angle to the bar make 60 degres?(between ("Fr and Fx") And in that case the Y-axis become a Sin and the X- axis a cos? I feel confused about this, in what way am I thinking wrong and do you have any tips? Thanks in advance! 🤗
You're very welcome!
The way it's shown, the light green components and the dark green force, the angle is 30 degrees. If you move the components, you can draw them with a 60 degree angle. The angle doesn't matter, the components doesn't matter, none of that matters for sine and cos. In fact, whether it matters or not shouldn't be an issue to you. Forget that all and just look from the perspective of the angle. If it's in front of the angle, use sine, if it's adjacent to the angle, use cosine. Nothing else matters, don't look for patterns, or associations, or anything like that. Please see (it's less than 60 seconds): th-cam.com/users/shortsvynnKlJD_Jo @@Thecarfreak100
Hello Professor, thanks for the video. I do have a question concerning 7:22 why doesn't Point B also have a moment like Point A has? Or is the 20 Nm given is that the moment for B?
So point A has a moment because it's a collar there. Collars create a moment about it self (see 7:09). At point B, it just a smooth surface, which cannot create a moment (see 7:15). The 20 Nm is independent to both, it is simply an additional moment given/ applied to the object.
I am still confused.In the last problem,how was the angle found in one equation with two unknowns?
You can graph the equation, there is only one unknown, which is theta. There aren't 2 variables.
Sir at the last example
Finding Angle between 0°-90°
Angle14.539
90-14.539
Should it be=75.461° ?
Why 82. 535?
How?
Please see: www.desmos.com/calculator/rvdow9qi6p
You are looking for x-intercepts between 0 and 90 degrees.
@@QuestionSolutions hii sir wanted to ask what will you do in exam taking note that you dont know the intercepts of the graph please share the solution.
at 11:04, can u please explain how u get the angle 82.535 and why they are two different angles? thankyou
So the easiest way to get the angles is to actually graph the equation. When you graph it, you will find 2 x-intercepts. Those are the 2 values you're looking for. You can solve this without graphing but you'd need to use trigonometric identities and it would take way longer than actually graphing it. There are 2 angles because at those 2 different angles, the system would be in equilibrium. In other words, at the angles 14.5 or 82.5 degrees, nothing would be moving, it's at rest. 👍
Why is the 12 kN force in 6:51 negative? Doesn't it create a clockwise motion?
No, it creates a counter-clockwise moment. Imagine the object is free to move about point B. If I push down exactly where force B is, which way will it spin? It will spin counter-clockwise about point B.
@6:15 Im confused as to how you assumed that Fa Triangle from the provided 30 deg. I Don't understand how you constructed the other triangle and know that it is the same angle as the one provided if it's coming from another spot, I Just don't see it.
So for these types of things, if it doesn't come intuitively to you, which is perfectly fine, I encourage you to draw it out on a large piece of paper. Using a protractor can also help to understand geometric concepts as well. It's all to do with perpendicular and parallel lines (creating alternate interior angles, corresponding angles, etc.) but it can be hard to see sometimes. The best way is always to draw it out on a big piece of paper with a nice straight edge and visually seeing it. This will not only help you see things faster but give you an intuition as to how you can figure out other angles without too much trouble.
Could you explain how the force is balanced here at 10:13? I can't find a force to balance the horizontal force F_Ax. Thank you very much!!
The horizontal components comes from the weight and the spring placed at an angle. So the pin counteracts those forces in both x and y directions. So the simple answer would be the x-components of the spring force and weight is countered by the x-component of the pin at A.
@@QuestionSolutions Thanks a lot!🥰
For the 1st problem, why is there no reaction at C? Shouldn't there be a horizontal and vertical reaction that prevent translation?
There is, but we don't care about point C since we're focusing on the beam only.
Where can i find sample problems like this?
Most statics textbooks have plenty of questions. If you need a reference, please check the description, I always list the books used.
Hello sir im having a hard time if its negative or positive what can you suggest for me to watch
Can you give me a timestamp as to where you would get confused with negatives and positives? I can help if I know exactly where you're confused. In general, positives and negatives are based upon your assumptions. For example, if I pick right to be positive, and a force is facing left, then it's going to be negative. If the force faces the way you picked to be positive, then it's positive. The same with moments. If I pick clockwise to be positive, then any moment that is clockwise will be positive and any moment that is counter-clockwise will be negative.
At 2:20, the horizontal net force does not like being zero, unless there is one more force on the opposite direction. I really get confused on this.
I am not entirely sure what you mean. But if there are no other horizontal forces effecting an object, F_x wouldn't exist, since nothing is being countered. What do you mean by "does not like being zero?" In statics, all net forces must equal zero, if it doesn't the object is moving around. Maybe I am not understanding your question, sorry!
Hi! I was just wondering why at 6:01 fa in the x direction is not zero because it’s being held by a roller support meaning there would be no reaction, thanks!
That would only be true if it's along the horizontal plane. Notice that our force is applied at an angle. That means that force must be broken into components and you'd get an X and Y component. 👍
@@QuestionSolutionsthanks!
I have problems in visualising the angles, do you know where or how can i improve? This makes it hard for me to solve questions
Could you give me an example? Like on these problems, where did you have a hard time visualizing angles? Then I can give you some tips.
do you mind showing me how to solve for theta at the last question? i really dont know how to the value of theha
I think it's much easier to graph these problems to get a solution. That's probably what I did, I don't remember now, but usually, I try to graph them since it's faster. Otherwise, you'll have to use trigonometric identities to solve them 😅
How can you solve the value of Fa and Fb?
In example number
Please use timestamps, I am not sure where you're referring to. Thanks!
3:30,we never got taught this at school i think, how do you know that you need to divide 5 by 13 Times 26 to get Fx?
Please see: th-cam.com/video/NrL5d-2CabQ/w-d-xo.html Especially the 2nd example, where I explain how to use a slope triangle to figure out the forces.
Hello, for the first question, how do determine the x and y direction of the Tension in the rope (Point C)??
So in a rope, the force always goes towards the support. So in this case, from B to C. Once you draw the vector, you can break it into components along the x and y components.
@@QuestionSolutions ok thanks for replying. One more question please. Do I assume the direction of the arrows for point C?
@@topefestus2193 Yes, that's correct, in the end, it is just an assumption. I think I cover forces along ropes and stuff on this video: th-cam.com/video/X9g4G1eBHCA/w-d-xo.html
@@QuestionSolutions This has been really helpful. Many thanks
wait!
on 6:12 you broke it into x and y components. i get that however my mind is not able to understand why is y component cos and x component sin?? isnt it the other way round?
i tried to understand it but i cannot :(
Watch this video, it's less than 60 seconds, and you will definitely understand :) th-cam.com/users/shortsvynnKlJD_Jo?feature=share
If you still don't, let me know.
There is an error at 8:17 the 100N force going downwards, shouldent it be negative ? -100(0.3)
We took clockwise to be positive, so the 100N force creates a clockwise moment, which means positive. Keep in mind that this is not an equation for the summation of forces but a moment equation. So we need to keep in mind the direction of the moment created by the force.
@@QuestionSolutions ohhh yaa, thanks for answering me
Hello sir. Question at 7:12 . Why on the black pic the moment is ccw, but at Fa, moment is cw?
So it's just an assumption. All we know is that there is a moment applied at the collar. The direction, we have no idea until we solve the problem. So you have to make an assumption about the direction, and if you get a negative value, then it's opposite to your assumption.
How is moment A there at 7:56. Isn't the distance =0 from the axis of rotation to the Normal force of the rod?
So this moment isn't created by a force that is a certain distance away. This is a counter moment created in the joint because of the collar. You can think of it as a separate force applied.
How to get the Tension BC in x and y axis
Please use timestamps so I know where you're referring to. Thanks!
At 5:03, do we get to choose if cw, or ccw is positive? Thanks.
Yes, the assumed direction is totally up to you. If you get a positive answer, then your assumption is right, if you get a negative answer, then it's opposite to your assumption.
Ohhhh ok. so regardless of the the direction (cw or ccw) they’ll both have the same answer?
Thank you for your help, I understand now ☺️ and thanks for the videos, they help a ton!
can u please explain the flipping of triangles in Q3?
Thanks! :)
Hmm, I am not entirely sure I can explain it, but I will give it a shot. The triangle is a ratio triangle, so when the a line is perpendicular to that same force, you're just flipping it 90 degrees. At 7:31, look at the brown line at the top diagram, with the 12,13,5 triangle (right side). Now in your mind, turn that line so that it matches the green force arrow on the bottom image (right side). See how it just flips 90 degrees? I think this comes with intuition or maybe doing a lot of questions, but if you struggle with seeing this, I think an easy way to understand it would be convert the triangle into an angle. So you just use sin or cosine, use the proper sides and replace the triangle with the angle. After that, you can draw the perpendicular line and calculate the angle again. It really just boils down to knowing that a ratio triangle is just another way of giving an angle. If that doesn't work, I encourage you draw this on a blank piece of paper (draw it big), and then use a protractor to see the angle change with the perpendicular lines. That way, it will click in your mind. :)
You can find your answer in this video: th-cam.com/video/YP2HBuBWgvA/w-d-xo.html
6:42 sir plz tell me why you can't take 3sin30+4 as a distance
We use both, one gives the vertical component and the other gives the horizontal component. So 3cos30 gives us the horizontal length of the left side, while the 3sin30 gives the vertical length.
Great video. Can you please explain a little better why at 7:56, when we take the sum of moments around point A, we include M_A (if its created by the forces at A) and the 20 Nm one at the curve of the body? Which forces are these moments generated from?
This is based on what type of support system it is. In this case, if you look at your textbook/ resource on what types of forces are created based on the support type, you will see that this crease a force perpendicular to the collar and a moment about it. The 20 Nm moment is already included in our problem, it's a given, so you can think of it as someone else applying a specific moment at that location. Let me know if you need more clarifications.
Sir at 6:56 how did you solve the equations?
Using the moment equation, you can directly solve for FA since it's a single variable. Once you find FA, plug that into each of the x and y equations to solve for BY and BX. Let me know if you need further clarifications.
@@QuestionSolutions Okay sir, thankyou!
could you please explain how at 6:45 you concluded that the distributed force (12kN) is going to be counterclockwise and, hence negative?
Is it not as if we are applying a force on the rod from the top? I feel that it would move clockwise in that sense.
I am a little unsure about the signs of moments, I am having a hard time visualizing the movements, can you please explain them?
Sure, so remember we are calculating the moment about point B. That means the whole rod pivots around point B. Imagine the bar moving about point B. If we apply the 12 kN force at the location that's shown, so it's coming downwards from the top, it's impossible for it to turn clockwise, we are literally pushing something down about point B. It will 100% turn counter-clockwise. If you don't believe me, take a ruler, hold the right edge between your thumbs and push down from the top anywhere on the left side. You will see that it will try to turn counterclockwise. It's negative because we picked clockwise to be positive. So anything that goes counterclockwise becomes negative and anything clockwise becomes positive. If you picked counterclockwise to be positive, then all the clockwise moments will be negative and counter-clockwise moments will be positive. That part is up to you, but make sure to follow through.
Also, I am not being mean or anything, but you are missing a lot of fundamentals when it comes to moments. Please take a few minutes to watch this video: th-cam.com/video/QNNnPZ68STI/w-d-xo.html
I promise it'll help you out in the long run, even if its a tiny bit :)
@@QuestionSolutions Thank you so much! that for sure cleared a lot of my misconception. Will definitely give the video another watch! You're doing an awesome job! Thank you! (:
@@wt7146 Awesome! :)
at 6:48 why didn’t you add the other 4m like you did at force FAcos30
I assume you're talking with respect to to the FSsin30. So remember, we only care about the perpendicular distance for moments. For the sin component (which is horizontal here), the perpendicular distance, in other words, the height from the force to point B, is 3sin30, we need to break that 3 m angle slope and use the vertical distance. So we don't need the 4 m horizontal length since that's parallel to the force component.