Equilibrium of Rigid Bodies (2D - Coplanar Forces) | Mechanics Statics | (Solved examples)
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- čas přidán 6. 08. 2024
- Learn to solve equilibrium problems in 2D (coplanar forces x - y plane). We talk about resultant forces, summation of forces in each direction and moments.
🔹How to break forces into components: • Vector Addition of Cop...
🔹How to find moments: • Moment of a Force | Me...
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Intro (00:00)
Determine the reactions at the pin A and the tension in cord BC (03:12)
If the intensity of the distributed load acting on the beam (05:38)
Determine the reactions on the bent rod which is supported by a smooth surface (06:59)
The rod supports a cylinder of mass 50 kg and is pinned at its end A (08:52)
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Book used: R. C. Hibbeler and K. B. Yap, Engineering Mechanics Statics.
Hoboken: Pearson, 2017.
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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
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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.
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You're very welcome. I am really glad to hear these videos are helpful to you :) Best wishes with your studies!
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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!
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Glad to hear! Keep up the good work and best wishes with your studies :)
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Really glad to hear these videos help out a lot :) I wish you the best with your studies!
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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 :) ❤️
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Thank you very much. I hope all of the videos are helpful to you and your friends. Best wishes with your studies :)
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I am really happy you aced your final exam! Good job, and keep up the awesome work. Best wishes with your future studies.
Thank you for this clear explanation!
You're very welcome!
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I hope the video was helpful to you :)
@@QuestionSolutions it was helpful thank you. :)
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Glad it was helpful and thank you very much! ❤
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West hills community college student watching your vids, keep it up.
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thank you for this video! you explain better than my teacher
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!
Very much helpful..🖤
Happy to hear that! Best wishes with your studies!
Great explnation
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Thanks a lot!
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I am watching this to prepare for my FE exam , thank you!
You're very welcome! I wish you the best with your exam.
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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 :)
You're my hero sir
Thanks :)
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!
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.
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.
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 :czcams.com/video/NrL5d-2CabQ/video.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.
@ 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.
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. :)
💯 explanation
Thanks!
These videos are worth more than my rented textbook and professor combined
😅 That was a funny comment!
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.
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 :)
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!
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.
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. 👍
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.
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.
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: czcams.com/video/QNNnPZ68STI/video.html
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: czcams.com/video/NrL5d-2CabQ/video.html I cover an example with a slope triangle and show how to use them. It's the 2nd example.
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 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".
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!
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: czcams.com/video/QNNnPZ68STI/video.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! :)
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!🥰
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
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 😅
Hi, I think clockwise is a negative moment right ? why at 6:30 you said pick clockwise to become positive ?
Please see: czcams.com/users/shortsP029mqnp4XY
Do you have a tutorial video for geometry in getting the x and y components?
Yes, I do, see: czcams.com/video/NrL5d-2CabQ/video.html
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!
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.
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 :)
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: czcams.com/users/shortsP029mqnp4XY?feature=share
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.
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!
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.
thanksssss
You're welcome!
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.
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): czcams.com/users/shortsvynnKlJD_Jo @@Thecarfreak100
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.
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.
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.
@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.
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!
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!
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: czcams.com/video/NrL5d-2CabQ/video.html Especially the 2nd example, where I explain how to use a slope triangle to figure out the forces.
You're great:3 and your videos also. I can understand every step and procedure of each of the problems based on my course. The problem's from Hibbler books are easy to figure out. But my university following Analytical Mechanics by Ferries and Chambers for Mechanics Course. And the problems of that book are horrible. Too much difficult to understand. Could you please check out that book and give me some tips to solving those :))
Unfortunately, I don't know that book. But the premise of these topics are all the same. The goal isn't really to solve just problems from a single book, but rather, it is to figure out how to apply the equations learned in each section, which is why I show these examples. I also disagree that problems from Hibbler are easy to figure out, there are some very challenging ones that are very well formed by the authors of that book. I do apologize but I can't help because I don't have that book to even look at 😅
@@QuestionSolutions It's okay 😀. I do not mean the problems are easy. The figures are quite understandable and find out the required information is easy to compare the Ferris's book. The procedure is the same but I think some problems and figures are not enough informative to approach the solving method.
@@undefined.infinity3106 Sorry I couldn't be more of help. What year was your book published? I am having a really hard time locating it.
@@QuestionSolutions 3rd Edition. 2016-17
hey, it's completely okay. There was a group project-based that book. I was trying to solve those problems on my own. By following the solving the procedures. Sometimes I can not end up with the same answer with the book. For that, I try to solve that again and again. And that gets frustrating to solve some problems. I think im not properly able to understand the basics and solving methods. So im trying to watch these contents to find out any solution for those. 😀
@@undefined.infinity3106 I see, are they the same exact topics that's covered in these videos? As in, the same equations and so forth?
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: czcams.com/video/X9g4G1eBHCA/video.html
@@QuestionSolutions This has been really helpful. Many 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 you tell me how to solve for teta at 11:00 since there is both sin and cos in one equation?
Thank you
Probably easiest to just graph it and find the values. Otherwise, you will have to use trigonometric identities.
@@QuestionSolutions how do you graph it? i still don't get it T_T
@@gabbylayun1756 www.desmos.com/calculator/rvdow9qi6p
We are looking at the values between 0 and 90 degrees. You will see 2 such values. :) To graph it on paper, you just plug values of theta in and get a rough idea of where the graph intersects the x-axis.
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 :) czcams.com/users/shortsvynnKlJD_Jo?feature=share
If you still don't, let me know.
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: czcams.com/video/YP2HBuBWgvA/video.html
Goated
😅
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
What software you're using. I think it would help my students
I use illustrator to draw the diagrams and after effects to animate.
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.
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.
7:29 Prof the MA is not cause by FA right?
That's correct, force FA's line of action would go through point A, so it can't create a moment about point A. The moment is created by the other forces applied to the object.
How to get the Tension BC in x and y axis
Please use timestamps so I know where you're referring to. Thanks!
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.
At 5:38 how do I solve the moment equation to get FA
It probably looks a bit confusing with all the sin and cos terms, but you can convert them to decimal form to make it easier. For example,
sin30 = 0.5 and 3sin30 = 1.5 and cos30 = 0.866 and 3cos30 = 2.598. If you use these instead of cos and sin, you'll see that it's easy to do. Let me know if you need further help. You can double check your work using cymath.com
Ok so I was doing my math wrong
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.
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.
how did you find Fa and Fb at 7:48?
Please see: czcams.com/users/shorts86uENomd53U
I made a short video for you. Thanks!
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.
can you explain how you arrived at Fa and Fb in problem 3? I cannot seem to solve it. I set Fa(4/5)=Fb(5/13) and substituted but am not getting the same answer as you. any tips?
Probably a numerical error, it's really hard to say without seeing your steps. You can always graph it or you can use wolfram alpha to double check your answers. 👍
I have the same problem!
@@loekkettering2743 I am pretty sure there is a typo on my part, sorry!
@@QuestionSolutions I have managed to get the answers, your calculations are correct. Thank you for the very informative video, helped me a lot!
@@loekkettering2743 Awesome! :)
At 5:24 , can i know how you calculate it . Where the Tbc go
You can use the moment equation to directly solve for Tbc. Think of Tbc as x, then you can easily solve for it :)
@@QuestionSolutions thankyou
@@nurulashykin9023 You're very welcome!
At 6:40, How did you get 3sin30?
So that's the perpendicular distance from point A to the line of action of B. In other words, it's the vertical distance from A to B. That can be found by using sine, since the vertical distance is opposite to the 30 degree angle, giving us 3sin30. The 3 is the 3 meter diagonal length, in this case, the hypotenuse of the triangle.
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!
5:26 best method on solving fbc ? thanks
You can directly solve for T_BC from the moment equation. After that, isolate for a single variable in eq1, then plug that value into eq2 and solve. So substitution method. If you need a refresh, I have some shorts on the channel that go through how to use the substitution method.
@@QuestionSolutions really appreciate the fast reply this is why you're the best !
@@unrelease9933 Thanks! I hope it helped you out.
@@QuestionSolutions Hello, how did you get the 80Kn value of TBC? Thanks
Professor how in the roller support rotation can occur ,could you explain me?
Please give me a timestamp so I know where you're referring to. Thanks!
@@QuestionSolutions At 2:21 for example
So one way to think about it is to imagine you placing a piece of wood or a ruler on top of a cylinder, like a can. If you push up from one end of the ruler, it would rotate about the can. In other words, the can cannot stop the ruler from rotating at all. All it can do is stop the ruler from touching your desk by creating an equal force upwards in the vertical direction. In the diagram I show at 2:20, try to imagine what would happen if you lifted up the beam from the left side end. It would rotate about the roller, right? This is why on a roller, rotation can occur. @@yigitcan824
Oh I see, thank you professor I really appreciate it
@@yigitcan824 You're very welcome. Let me know if you have any other questions