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Equilibrium of Rigid Bodies 3D force Systems | Mechanics Statics | (solved examples)
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- čas přidán 7. 08. 2024
- Let's go through how to solve 3D equilibrium problems with 3 force reactions and 3 moment reactions. We go through multiple examples step by step and find reactions for different supports like journal bearings, fixed supports, and ball and socket joints.
🔹Express forces along a rope: • Force Vectors Along a ...
🔹Breaking forces into components: • Addition of Cartesian ...
🔹Finding moments: • Moment of a Force | Me...
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Intro (00:00)
The sign has a mass of 100 kg with center of mass at G. (01:18)
Determine the components of reaction at the fixed support A. (04:41)
The shaft is supported by three smooth journal bearings at A, B, and C. (06:46)
Find more at www.questionsolutions.com
Book used: R. C. Hibbeler and K. B. Yap, Engineering Mechanics Statics.
Hoboken: Pearson, 2017.
i have been struggling lately in mechanics rigid body statics ... however life has been a little bit easier lately since after finishing my learning module, i go straight to this channel and watch the corresponding lessons ... this channel is very helpful keep it up bro Godbless you
Really glad to hear these videos are helpful to you. Take it one step at a time, do as many questions as time allows. I wish you the best with your studies :)
Damn!!! There is not a single channel which teaches mechanics better than you , your animations are like every concept is getting arranged in my mind and it become impossible to forget the concepts!!
Thank you so much for the nice comment. I am really glad to hear that these videos help you out. Keep up the good work and best wishes with your studies!
I've been searching for videos that explain this well for days. Your last example finally conveyed the idea in a way I could understand. Really appreciate your work.
Glad it was helpful! Best wishes with your studies :)
I came across your channel by chance while studying for exam and wow!! Such a simple and clear explanation. Thank you so much!! I wish I could've seen this earlier. But I'm even more amazed on how you really look into and reply to each question on the comments. You deserve more recognition. Thank you thank you 🙌
Thank you very much for your nice comment! I hope you do well on your exam and I wish you the best. Keep up the great work.
And yeah, I do my best to reply to all of them. So far, I don't think I missed a comment yet :)
wow!! I was struggling with this topic in my mechanics class but you cleared up all the issues i had! Thank u:)
That's awesome, really happy to hear that this video helped you. Best wishes with your mechanics class!! :)
Incredible work, thank you so much! These videos are more helpful than all my course notes put together.
You're very welcome and I am glad these were helpful. Keep up the great work and best wishes with your studies.
thank you for this video, i’m studying for my statics midterm tomorrow and this helped me so much!!
I am really glad to hear that. I wish you the best with your midterm! Do your best.
Thanks please keep it up you are a hero to a victim of the worst education system ever.
I hope these videos help you out and I wish you the best with your studies!
Hi
I love your videos which is very helpful for me thank you very much. I look forward to watch new videos , such as Strength of meterials, Differantial Equation etc . :)
❤
You're very welcome and I am really glad you like them! :)
OH YOU BEAUTIFUL MAN THANK YOU THIS IS EXACTLY WHAT I NEEDED TO KNOW
Glad to hear :) Keep up the great work!
awesome video especially with the 3 different sockets mentioned
Thank you very much!
Thank You ❤️❤️
You're very welcome! ❤️❤️
Yeahh, I'm with my saviour again. God bless you!
Thank you! Best wishes with your studies.
@@QuestionSolutions just got out of the midterm, I destroyed it.. thx
you have saved me and my degree thank youuuuuuuuuuuuu
You're very welcomeeeeeeeeeee
Sir I'm giving u a cup when it comes to rigid bodies in 3d 💪🤘🔌. You are the best one
Thank you very much. I am glad you think these videos are good :)
Thanks for this video! May I ask if you have any other sample problems like this where you need to find the missing distance ?
You're very welcome. I don't think I have other examples for this chapter.
I just discovered this channel I wish I found it before my midterm...
I wish you the best with your studies. Keep up the good work!
For the last question I just want to make sure I understood the general rule of moment and rotation around an axis correctly, so only z-component forces can cause for a rotation to occur around the y-axis, and only x-component forces can cause for a rotation to around the z-axis, and only y-components can cause for a rotation to occur around the x-axis right?
The easiest way to see this is to actually get a box and place it on the table. Push with your finger to see which makes the box turn about each specific axis 😅
Can ı ask something ı couldnt really understand when we use unit vectors. I thought we use them only for different axis ( like other than x-y-z plane) but here we use them for x-y-z plane too
So a unit vector, in the most simplest sense, is just a vector with a value of 1. You can write unit vectors for any position vector you want, in other words, for any vector you draw, you can have a unit vector. What you're referring to are the axis unit vectors. So x-axis is 1,0,0, y is 0,1,0, and z is 0,0,1. That also is just a vector that extends from the origin along that the axis you're looking at. So to give a simple answer, you can have unit vectors wherever you want, whenever you want. I encourage you to take a look at this video, it might give you some insight. czcams.com/video/CCeWy1kmxMs/video.html
Hi i have a question, at 4:29 how did you get the Fbc and Fbd from the two equations when you equate them into 0, i mean how did you get the 588.6 N and -294.3 N, please that's the part where I dont know how to solve it, please help me :))
Please see: www.cymath.com/answer?q=(4%2F3)C%2B(2%2F3)D-981%3D0%2C%20(-2%2F3)C%2B(4%2F3)D%3D0
Hello, just a small clarification, is it wrong if you include the k coordinate of point G (if given, in my assignment's case) when using it as a moment arm for MA3?
timestamp: 3:58
@@70proof It's fine, you will still get the same answer since it would be a position coordinate from the origin to point G.
Hi sir at 9:05 , I do understand how you get -900(0.9) due to the clockwise direction. What about 600? I did the same but i thought it should have been -ve as well. Can you explain why its positive?
Notice how the 900 N force is to the right of the y-axis, where as the 600 N force is to the left of the y-axis. The 900 N force creates a clockwise moment, but the 600N force creates a counterclockwise moment. And that's because one force is to the left and one to the right of the y-axis.
have a question. How do we decide whether to use scalar method or cross product when finding the moment?
If the question involves cartesian coordinates, or it involves all 3 dimensions with forces not lying on x/y/z axes, then you would need to use the cross product. If it's a 2D problem, or forces don't need to be expressed in Cartesian form, you can do it using scalar methods. This is just a nutshell answer, but generally, after doing a few questions, you can determining which the method to use easily.
In the 2nd question, for calculating moments, you can also use the formula M = Fd, right, instead of doing the cross product?
Yes, in this case you can because the forces all have a single component.
At 4:29 Im still confused when keying in the values to solve moment equations to get FBC= 588.6N FBD=294.3N . I get different results . How do I solve for them
So I think you asked a similar question in a previous video too so please refer to the answer I gave to you on that video. If you're getting different results, you're probably not plugging in your values properly. Don't rush, take it step by step. Isolate for one variable, plug it into the next, and then solve. Replace F_BC as x and F_BD as y. This will allow you to see it better.
Please see: opentextbc.ca/businesstechnicalmath/chapter/solve-systems-of-equations-by-substitution
Here it is solved step by step: www.cymath.com/answer?q=4%2F3x%2B2%2F3y-981%3D0%2C%20-2%2F3x%2B4%2F3y%3D0
How would you do the last problem using the cross-product method?
The same way as example number 2.
what happens if for example at problem 2, on of the forces f1,f2,f3 are not on the x,y,z axis ?
The process is the same, and if you were given the forces in cartesian form, nothing changes. If you were given scalar values, you'd need to figure out the forces in cartesian form, but everything else remains the same.
can i know what do you mean by 4 unknown at 3:10 because i count there are 5 uknown which is Ax, Ay, Az, Fbd and Fbc
It's just a misspoken word, there are indeed 5 unknowns. 👌
Great Video! I still do not understand why the reactionary support moments were removed from the equation.
Please give a timestamp to the location you're talking about so I can take a look. Many thanks!
@@QuestionSolutions 7:13
@@karimabouelela3665 They are not applied if the object is supported at other places. We only care about them if that support is the only one holding everything together. In your textbook, it goes into detail how different types of supports work and what forces we can look for.
Hi sir I would like to ask why is timestamp 9.35 when calculating the moment about x axis the force Cy is not included.
CY won't create a moment since it's 0m above the z-axis. That's why only the 450 N force was written in the equation, with a height of 0.6 m. You have to visualize this problem as a 3D problem with 3 supports holding it in place. A good way to visualize a problem like this is to imagine it free spinning about the x-axis. So this isn't clockwise or counter-clockwise in a 2D plane, this is going into the screen or out of the screen. Focus just on CZ and forget every support. If CZ pulls up, the whole thing would rotate out of the screen about the x-axis. Now focus on the 450 N force, which is 0.6 m in the z-axis. That will also cause it to rotate out of the screen. Now imagine the CY reaction, which is flat. Pulling at the location in that direction will not cause the object to rotate about the x-axis, not into the screen or out of the screen. It can't create a moment. I hope that helps. If you need a better visual, see this: czcams.com/video/rdK0c9YZRYw/video.html at 20 seconds.
Tq sir
For the second problem, how did you know the direction of Ax and Ay in the beginning?
I didn't, it was just an assumption. If you get a negative value, then it's opposite to your assumption. 👍
As Salamualaikum 🤗
Thank you so much for your videos, they have been really helping me in my studies!!!
I just wanted to ask in the last question at 9:33
Why in the sum of moments in z axis we multiply 250 by 2.7☺️?
Why not 1.8?
Why would we multiply 900 by 2.7?
Would you please answer me if you could as I am really confused 💔💔💔
Thanks in Advance 💖
So we multiply 250 by 2.7 because we are looking for the perpendicular distance from the x-axis to force Cz. That's a total distance of 0.9+0.9+0.9 = 2.7m. The same applies to the 900 N force. Remember that when we are calculating the moment about an axis, we always look for the perpendicular distance from the force. Notice how both the 250N force and the 900 N force are both along the z-axis (in other words, they point straight up or down), so the perpendicular distance to those 2 forces would be the distance along the y-axis from the x-axis to where the force is being applied. I hope that helps.
@@QuestionSolutions thanks alot!
It did help
But I thought the perpendicular distance is 0.9+0.9 🥺
Would you plz clarify to me which perpendicular distance are we calculating? The one thats stuck to Cz or the one thats stuck to Az
Thanks alot again! This really helps!!!!
@@amrhelmy5795 0.9+0.9 is the perpendicular distance to the y-axis, not x-axis. So to find the perpendicular distance to the x-axis, you have to count along the y-axis. So that's three 0.9 m. Az is at the origin, which means there is no moment created about the x-axis. Please see: imgur.com/a/N6gNH3A I drew a red box around the length that is perpendicular to the Cz force.
@@QuestionSolutions Alright great!!! Thank u so much for ur support, caring, explanation and help!!! I really appreciate it and I understood everything now 🥰
Jzak Allahu khair and May Allah count it towards ur Hasanat 🙏🙏🙏
@@amrhelmy5795 Glad to hear it helped. Best wishes with your studies. 👍
at 9:35 why is the 450n force considered? And why is its perpendicular distance 0.6m?
So the 450 N force, which is the one on the top applied next to the 900 N force, creates a moment about the x-axis. The perpendicular distance is 0.6m because that's the perpendicular distance between the x-axis at the force and a moment is just force times perpendicular distance.
For the first example, is there a reason why we cannot use the Projection of F = in the direction of each cable? I got F_BC = 654N and F_BD = 327N, then the reactionary forces Ax = 0, Ay = 654N, and Az = 436N.
I have been confused when I can do projections and when I can't. Any help is massively appreciated, thanks!
You don't need to do projections when you solve 3D force equilibrium problems. You just need the forces expressed in cartesian form and then you can equate the components. Usually, you don't use projections for these types of problems, you use it mostly to figure out angles between forces or if a question specifically asks for the projection.
@@QuestionSolutions I see... So, the projections method is not the most accurate. It can give you a rough idea, but it is better to focus on it when finding angles and use the equilibrium equations to get something exact.
Thank you for responding!
Tnx
You're welcome!
quick urgent question, is this Chapter 5 EQUILIBRIUM OF A RIGID BODY
& FREE-BODY DIAGRAMS?? and do you explain in any of your videos LECTURE 15 - EQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMEBERS?? and THAANK YOU FOR THIS AMAZING VIDEOOO😍
I apologize but I am not sure of the chapters (or which book you use), though most follow a steady progress through the books I use. If you look at the statics playlist, it should have videos that cover the topics you mentioned. :)
Hii I wanna ask for the last question, taking moment at at axis, shouldn't the Cy force acting a moment to x axis? And thank you for the video =)
Please kindly provide a timestamp so I know where to look. Many thanks!
@@QuestionSolutions 9:34 thank you for your reply!
@@wendylim898 Okay, so there, Cy would not create a moment about the x axis. Notice how the 450 N force creates a moment even though it's along the y-axis, the same as Cy. We found that by multiplying 450 by the 0.6m, which is the perpendicular distance from the x-y plane to where the force is applied. Force Cy has a perpendicular distance from the x-y plane of 0 m. So no moment is created. To see this better, I encourage you to take an eraser or something, lay it flat on your table, and then apply the force at the very bottom, just like CY, and see if it tries to spin about the x-axis. You will see that it does not. I hope that helps!
Hello! I'm still confused why is the 450 force has a 0.6 distance when creating a moment at x-axis?
I have a quick question - for the position vectors BC and BD, i did CB and CD, meaning that my numbers are the opposite of your numbers..
this is giving me the wrong answer (i believe because W is still (0,0,-981).
so my question, SHOULD it be BC and BD (C-B) and (D-B) instead other way around?
and if yes, how do i decide which way to do it for future questions?? thanks!
Position vectors are always from the starting point to the final point. So if it's from A to B, then you subtract the points of A from B.
@@QuestionSolutions right, how do we resolve which point is the starting point and which is the ending point..? thanks again
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I just wanna ask, in problem 1 and 2, both are asking for the reaction at point A (ball point A in problem 1 and fixed support A in problem 2) but for problem 1, you didn't get the value of Max, May, and Maz unlike in problem 2. I reckon Both reaction Force and reaction Moment components are being asked for both problems so shouldn't Reaction moment be part of the final answers as well in problem 1?
Good question! So you have to remember, at a ball and socket joint, there are no moments created. That's because the object can rotate about the ball joint (which is why we use them). In a fixed support, there is no rotation, so a moment is created. Remember, when we find max, may, or ax, ay, we are finding REACTIVE forces. In simple terms, if we pull on the wooden sign, what sort of reactive forces occur at point A to keep this object from moving around? Since it can rotate about point A, there are no moments created. On the other hand, in a fixed support, it can't rotate about that point. That means a reactive moment will be created. I hope that makes sense, let me know if you need further clarifications.
I never realized that when it comes to momentum sum, we can use the sum for each axis instead than on a point. I guess it's working the same in 2d ? Because all my professors did the sum in one particular point..
Yes, absolutely. Did you see this video? czcams.com/video/rdK0c9YZRYw/video.html
Why can we assume that the Ax, Ay and Az forces are positive in the first queation?
You can assume it to be opposite, keyword being "assume." It's just an assumption. If you get a negative answer, then it's opposite to your assumption. In other words, you can assume all of these forces or some of them to be in the direction of the negative axes. It's your choice. 👍
At 8:20 why don't Ax and Bx create a moment on the y-axis?
Think of it as, if you apply force AX, would it cause the beam to rotate about the y-axis? Both Ax and Bx would cause a rotation about the z axis.
at 9:02 why is 500n force not considered for My calculation?
The 500N force can't create a moment about the y-axis. The same as how Bx, Ax, etc can't create a moment about the y-axis. An easy way to think about it is to see if a force can spin the object about the y-axis (assume it's free to spin). If we push exactly where the 500 N force is, can we cause this object to spin about the y-axis? No, right? So then it can't create a moment.
Professor is there a rule sth like.We have at least same number of equilibrium equations as unknown forces?
For example,six scalar equilibrium equations (sum of x,y and z axes=0(3 equations) ;sum of couple moments respect to x,y and z axes=0(3 equations) are enough for most 6 unknowns. Or randomly 5 equilibrium equations for 5 unknowns
I don't know if I understand your question properly, but to solve a set of equations, you need the same number of equations to the same number of unknows. So if you have 5 unknowns, you need 5 equations, and if you have 6 unknowns you need 6 equations.
@@QuestionSolutionsOh I see ,so number of equations of equilibrium must equal or greater than the number of unknowns?We have most 6 equilibrium equations so ,we can solve for most 6 unknowns ;if we have 4 unknowns,we need at least 4 equations of equilibrium
That's right, you need the same number of equations for the same number of unknowns.@@yigitcan824
at 9:39 I dont understand why is only one 450 used and why is it times 0.6? since its 0.6 meters above the z axis
Did you forget to add cy(1.8) which will make x axis rotate clockwise
No, cy won't create a moment since it's 0m above the z-axis. That's why only the 450 N force was written in the equation, with a height of 0.6 m. So I know you said it will make the x-axis rotate clockwise, but that's because you're thinking about this problem as a 2D problem. You have to visualize this problem as a 3D problem with 3 supports holding it in place. A good way to visualize a problem like this is to imagine it free spinning about the x-axis. So this isn't clockwise or counter-clockwise in a 2D plane, this is going into the screen or out of the screen. Focus just on CZ and forget every support. If CZ pulls up, the whole thing would rotate out of the screen. Now focus on the 450 N force, which is 0.6 m in the z-axis. That will also cause it to rotate out of the screen. Now imagine the CY reaction, which is flat. Pulling at the location in that direction will not cause the object to rotate about the x-axis, not into the screen or out of the screen. It can't create a moment. I hope that helps. If you need a better visual, see this: czcams.com/video/rdK0c9YZRYw/video.html at 20 seconds. @@Tyrantscandrive
at 3:32 i really couldn't understand how you calculated the MA1 (4/3i, 0j, -2/3k), please enlighten me
Please see: czcams.com/video/F8IHrg3pc7g/video.html
@1:30 , why is the coordinate of the weight (0, 0, 981), should'nt it be (0,1,981) since weight has an offset of 1 m in the y direction?
That's not the coordinate. That's the force, (weight), expressed in cartesian form. So it has just one component, which is straight down, along the z-axis. So the only component it has is the k component.
@@QuestionSolutions Does this mean that answers will be the same, even if the the weight is located with an offset 100 m in the y direction?
@@samuelsudario6762 Well not for 100 m, since the whole board is only 2 m in length. But if you put the weight at 1.5m or 0.5 m, it makes no difference to the answer.
the bz(1.8) how do we come by it
Please give me a timestamp so I know where you're referring to.
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4:29 i have struggle to solve for fbc and fbd
The method I use often is substitution. Isolate for one variable and then plug that into the 2nd equation. You can also graph the two equations to get an answer, or use wolfram alpha.
do you have any statics book recommendations?
Engineering Mechanics Statics by R. C. Hibbeler and K. B. Yap
@@QuestionSolutions Will check that out. Thanks for responding!! Love your tutorials, we appreciate you.
Why is it that the J component is negative 450j under Moment of Problem 2?
So when you calculate MA3, at 6:21, you will notice that it has -450j. It's the result of the cross product.
@@QuestionSolutions Yes its a result of cross product, but when you multiply negative 0.75 and negative 600 will result in a positive 450j.. but then its still negative 450j on the video?
@@wisdombuilds5044 The second term is always negative, so it goes (+)(-)(+) which means we have -(-0.75)(-600)=-450. Please take a look at this video where I go over how to do this step by step: czcams.com/video/F8IHrg3pc7g/video.html
@@QuestionSolutions Thank you very much! Godbless
@@wisdombuilds5044 You're very welcome!
why didn't we include the 500 N force while calculating my in 9:02
The 500N force can't create a moment about the y-axis. The same as how Bx, Ax, etc can't create a moment about the y-axis. An easy way to think about it is to see if a force can spin the object about the y-axis (assume it's free to spin). If we push exactly where the 500 N force is, can we cause this object to spin about the y-axis? No, right? So then it can't create a moment.
3:07 Aren't there 5 unknowns here? Fbd, Fbc, Az, Ax, and Ay?
Yup, I misspoke.
@@QuestionSolutions Okay, just wanted to make sure I wasn't missing something. Thank you for clarifying! :)
@@Breyerlover4ever23 You're very welcome!
In question 1, there are 5 unknowns, not 4. Why does it work to use 4 equations in this case? I thought you needed the same number of equations as unknowns.
Please give me a timestamp so I know where you're referring to.
why there is moment about axis and not about the points
Which part are you referring to?
@pattis Usually, we do this to remove as many unknowns as possible. You can take the moment about a point, an axis, or even an imaginary line.
Why didn't you solve the last question in cartesian form since it is 3D? Wouldn't have that been easier?
How do you know what method to use when - Cartesian or scalar?
It's really up to you. You can solve it anyway you like. 2D problems are solved using scalar most of the time. 3D problems can be solved using the Cartesian method or sometimes you can use the scalar method. If forces are along the axes, then you can opt to use scalar. Whatever is easier for you.
Where is Cy in 3rd question when you solve moments in x direction
CY won't create a moment since it's 0m above the z-axis. That's why only the 450 N force was written in the equation, with a height of 0.6 m. You have to visualize this problem as a 3D problem with 3 supports holding it in place. A good way to visualize a problem like this is to imagine it free spinning about the x-axis. So this isn't clockwise or counter-clockwise in a 2D plane, this is going into the screen or out of the screen. Focus just on CZ and forget every support. If CZ pulls up, the whole thing would rotate out of the screen about the x-axis. Now focus on the 450 N force, which is 0.6 m in the z-axis. That will also cause it to rotate out of the screen. Now imagine the CY reaction, which is flat. Pulling at the location in that direction will not cause the object to rotate about the x-axis, not into the screen or out of the screen. It can't create a moment. I hope that helps. If you need a better visual, see this: czcams.com/video/rdK0c9YZRYw/video.html at 20 seconds.
at 2:12 you are subtracting D-B. Its a positive 2 in the x direction not negative 2.
i actually see where i messed up
@@anthonypicciano7313 Okay, awesome! :)
İn problem 1
R'bd'= _2i _2j + k
Why j = _2 ??
Same reasoning for r_BC. From point B, it's 2 m in the negative y-axis (j-direction). In other words, you have to go backwards 2 m from point B to get to point D.