Couple Moments | Mechanics Statics | (Learn to solve any question)
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- čas přidán 7. 08. 2024
- Learn what a couple moment is, how to solve for them using both scalar and vector analysis with solve problems. We learn about free vectors, using position vectors and more.
🔹Basics of a moment: • Moment of a Force | Me...
🔹How to find a position vector: • Force Vectors Along a ...
🔹How to solve a triple scalar product (cross product): • How to Solve a Scalar ...
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Intro(00:00)
The man tries to open the valve by applying the couple forces (01:46)
The ends of the triangular plate are subjected to three couples. (02:02)
Express the moment of the couple acting on the pipe (02:51)
Determine the resultant couple moment of the two couples (03:33)
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Book used: R. C. Hibbeler and K. B. Yap, Engineering Mechanics Statics.
Hoboken: Pearson, 2017.
Please wait, before you write a comment asking why clockwise is positive, or counter-clockwise is positive, please watch this video first: czcams.com/users/shortsP029mqnp4XY thanks!
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Thank you for the suggestion. I think I did it before, in some older videos, and then stopped. I will see, I appreciate the feedback!
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you make is seem so easy!!!
When you do a few questions, it'll become really easy for you too :)
I don't get where you get these numbers for your cross product.
So I assume you're asking about the position vectors and expressing forces in cartesian form? If yes, please watch: czcams.com/video/CCeWy1kmxMs/video.html and czcams.com/video/mz7gPpIL0Gk/video.html
Those 2 videos will cover how to get the values for the cross product. Let me know if you have further questions. Thanks!
Hello Question Solutions, hope you're doing well!
I have a question, at 3:25 you found the Force in cartesian form to be {0i+0j+125k} I'm a little confused by how you found that, I thought we usually do f*(unit vector along Rab) to find the Force in cartesian form, could you please explain to me how you were able to find the force in cartesian form by just looking at the force magnitude?
Thank you!
Sure, so if you look at the 125N force, it faces straight up. So that means it's not at an angle, which also means that it lies directly on the z-axis. So it can only have a k-component and its 125N straight up along the z axis, giving us {0i+0j+125k}. If a force lies directly along an axis, we don't need to go through the trouble of multiplying it by a unit vector. Even if you do, remember that the z-axis unit vector is {0i+0j+1k}, so it still leads us back to the same place.
really helpful
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Helpful ❤️
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You are magician
🌟 Many thanks!
Great video. But on the second example why did we use clockwise direction for 600N but not the others
You pick the direction you want. It's up to you 👍
do u have any tutorial on centriod and center of gravity of 2d and 3d problem solving?
I don't think I do, just mass moment of inertia.
Hi, quick question abt the steering wheel. if the force is at an angle, would i find the x and y component and then find the x-y component of the distance? then do M=Fd in the x component and then M=Fd in the y component
In the steering wheel as in the ships wheel? So you have the forces applied at the 9 and 3 o'clock positions? If the force is at an angle, then you break the force into x and y components. The x component won't create a moment since it would just be pulling the wheel apart. It's like grabbing the 2 sides and one side pulls to the left and the other to the right. Only the y-component will create a moment. But in general, the steps you outlined are correct.
in the last example, how did you get the x component from A to B. I thought it would only have y and z components.
Are you talking about the position vector rAB? If so, it doesn't have an x-component (4:28). Maybe I am not understanding your question, please let me know a timestamp so I can take a look at the equation you're referring to. Many thanks!
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Thanks!
For the problem at 2:51, does the vector Mc act perpendicular to the position vector, rab? also, where would you draw the couple moment?
The vector for the couple moment would be directed along the x-axis. The direction of the vector has to be found using the right hand rule. So here, a counter-clockwise moment would mean it points along the positive x-axis.
For the last question, if we were to use Rac & Rba (instead of Rca & Rba like shown in the video) are we supposed to get the same answer? was wondering since Im getting {-12.11i+10j+17.3k} should I consider the direction of the moment of each individual moment (Mc1 & Mc2)
You will get the same answer. If you go from A to C for your position vector, the only difference is that instead of ignoring the force at C, you ignore the force at A.
Nice animation.
THE rBA is A-B right?
Is the MC1 and Mc2
Add together? As your last example
Thanks!
Yes for the position coordinates, and yes for the moments. MC1 and MC2 are added together.
Quick question, lets say for example on exercise 3, will the result be the same if the force on the negative Z axis didnt exist? Since we dont take it into account is MA the same regarless of F = (0i,0j,-125k)?
Your position coordinate would need to be from the z-axis to the force applied. Here, the position coordinate is from the start of one force to the other force.
It must be positive when the force is in counter-clockwise direction and vice versa
No, it doesn't. It's completely up to you on which way you want it to be positive, because there aren't really any "positive or negative" moments. They have a direction. The positives and negatives help with figuring out the direction when doing problems. You can pick whichever way you want it to be positive, and if you get a negative answer, it's opposite to your assumption. In fact, I encourage you to do it opposite to what you just said so you can realize that you still get the same answer. This will bring you more insight into doing these problems, and will allow you to pick whichever direction you want to be positive, making you solve questions faster.
Hi this is Vikram from India,
Some of your videos (Absolute Dependent Motion) had been copied and re-posted in CZcams. Channel name is Hom Phaly. Please take action. I really appreciate your hard work and I am a big fan you.
Hi Vikram, thank you so much for letting me know, I really appreciate it. You are awesome! :)
@ minute 1:22 you say we have to ignore the force at A since force 1 cannot create a moment about point A. What is the force 1 in the situation? Is it the force at A cannot create a moment about point A? I just stumbled upon your channel. So far, it's really good work thank you so much!
So basically is force 1 the negative force at point A?
It could be any force, but here, I was referring to the negative force. So you just ignore the force at the location of reference, in this case, point A.@@daigosol5375
Can i ask a question about the second question, how do you get the 0.5 in the lenght of the triangle?
So the bottom length is 1 m, half of that is 0.5 m. 1/2 = 0.5. Is that what you were referring to? If not, please let me know and I will do my best to help. Thanks!
@3.32 how did you get -25j and wouldnt it be 25i +37.5j?
For the j component, it would be 0.2 x 125 = 25, and it's negative because it's the 2nd term. Please see: czcams.com/video/F8IHrg3pc7g/video.html
3.16 ( why is the moment of force a is zero and how did you know the Cartesian form of the force on point b ) ??
So when you have a couple moment applied, to calculate it, you need to forget one of the forces and write a position vector from that location. Here, I chose to forget the force at A, and then wrote a position vector from A to B. The force is along the z-axis, (so just faces vertically without any other components), which means it only has a k-component. So it's just 125k.
If i assume clockwise is negative and anticlockwise is positive, i get a different answer. Is it ok?
No, you will and should get the same exact answer. Please see: czcams.com/users/shortsP029mqnp4XY?feature=share
at 4:31 how do you get -12,11i i get that -0.346 times 35 is the -12 but why is at the i side?
Please see: czcams.com/video/F8IHrg3pc7g/video.html
Sir
At the last example
MC2 : [0 - (-10)] j
Should it be positive 10j right?
No, the j component is negative in the sequence, so you end up with (-)(-)(-) = (-)
Thank you thank you very much for answering my questions sir.
@@menglimarrero4296 You're very welcome!
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why is clockwise positive?
Use whatever direction you want to be positive, you will get the same answer. It's just a convention that counterclockwise is positive, whatever way you choose yields the same answer in the end.
moment unit is Nm or N
It's Nm, if I showed N, then it's a typo.
Why is 25j negative? 3:30 3:31
Please see this video: czcams.com/video/F8IHrg3pc7g/video.html
can u explain how couple moment can become negative or positive ?
So moments aren't really positive or negative, in 2D, you use clockwise or counter-clockwise to state the direction. In 3D, you can use the right hand rule (same with 2D), if you want to know the direction of the moment vector. But in essence, let's say you pick clockwise to be positive, and your answer is positive, then your moment is clockwise. If you get a negative answer, that means your moment is counter-clockwise because you picked clockwise to be positive but your answer was negative. The same is true for the opposite. So if you pick counter-clockwise to be positive, and your answer is positive, then your moment is counter-clockwise. If your answer is negative, then your moment is clockwise.
@@QuestionSolutions hmm. Is this the same concept for dynamics bro ? Bcs I've been searching for how to determine the sign convention for work done due to couple moment. As far as I know. It depends on the torque direction right ? So how do I know the direction of the torque ? Is it the same as the direction of the rotation ??
@@ikmalhakimharun3823 I think you might be confusing the moment vector with the direction of turning. The vector is always found using the right hand rule. Work done by moments when it comes to dynamics involves taking the mass moment of inertia. Is that something you already covered? I think you might be mixing 2 topics together. :) Is there a problem you can show me that requires what you are asking? I can take a look and let you know.
@@QuestionSolutions hmm. Probably bro. I think so too. I don't know much about couple moment. So I just Google them and found out the sign convention are related with the rotation of torque. Btw bro. Where can I give u the problem ?
@@ikmalhakimharun3823 Email me at "contact @ questionsolutions .com" without any spaces. Make sure to check your junk/spam box for my reply, for some reason, sometimes, it goes into those.
I like you brother !!!
Many thanks!
dear the and for 2:50 is 617.15 i guess
Sorry, I am unsure of what you mean? The answer of 829.7 N is correct. How did you get 617.15?
3.25 I didn't understand the R(AB) vector
See this video first: czcams.com/video/CCeWy1kmxMs/video.html
1:52
But I thought clockwise meant negative and counterclockwise meant positive..
Try it either way, you get the same answer. When you do more questions, you'll realize that your life will become easier if you pick the side that gives you the most positive signs, or all positive signs. Either way works, do what you like/ easiest for you 👍
I thought clockwise=negative? Pls help
See pinned comment.
My bad, I didn't check the comments first. Thank you very much.
That's alright. Best wishes with your studies!@@akosibrothercoolpzzle123
2:46
What about 2:46? Did you forget to write a question? 😅
clockwise is negative.
See: czcams.com/users/shortsP029mqnp4XY
So moments aren't really positive or negative, they just have directions. When we do calculations, we choose a direction to be positive. This allows us to solve the problem. Feel free to use clockwise as negative, it will give you the same answer as choosing clockwise as positive. It makes no difference.
I'm confused at the side of the triangle 2:28
I tried solving it with 1sin(40°) and my calculator gives me 0.6427
Ohh can you explain me why is it 0.5(cos40)? Im pretty bad at this trigonometry thing
And can you also explain to me where did you also get that 400? Around 2:28 also
@@archiedahili5391 Okay, so first, look at 2:20. See how I drew that pink right angle triangle? That triangle is an exact half of the big blue triangle. That means the bottom length is 0.5 m, since that's 1/2 = 0.5 m. Now that we know the bottom length, we can use the angle given to us (the 40 degree angle), to figure out the hypotenuse of the pink triangle. This hypotenuse is also the side length of the big blue triangle. So we know the adjacent length, (bottom length), which was 0.5 m. Since we have adjacent length, we need to use cosine because cosine = adjacent/hypotenuse. We can't use sine unless we know the opposite length, in this cause, that would be the height of the triangle.
So we can write: cos(40) = 0.5/length.
Isolate it for length.
length = 0.5/cos(40) ==> 0.653 m.
I hope that helps.
@@QuestionSolutions oohhh ok ok thank youuu
@@archiedahili5391 You're very welcome!
1vi9 1
Not sure what that means?
4:30 you forgot to add - to J vector when doing cross multiplication. it should be positive 12.11
I assume you mean the i component, not j, since the j component is 0, and the normal convention is to not put negative signs in front of 0s, but you can if you want, it's your choice. I mean it has no value, so it doesn't really matter. 😅In most cases, you wouldn't even write out the j component, but I did because students tend to get confused or ask what happened to it. Anyways, if you are really referring to the i component, what's shown on screen is perfectly correct. The i component is -12.11, since it's -0.346 x 35 = -12.11i. 👍
@@QuestionSolutions Ah yes silly of me... I should go for a sleep now... Thank you sincerely.
@@theturkish1373 Have a wonderful rest! :)
Why are you going through this so fast, i mean can you just slow down and show us each step in your solutions.
What steps did you want to see? Can you give me a timestamp where you thought I should have shown more steps? Then I can think about it for future videos. Thanks!
I always mess up because I mess up the first step, getting the coordinates, then from there everything is wrong🫠
Well, on the bright side, you know where you usually make a mistake. So if you spend a few more minutes at the starting step, then the rest will be good :) Don't be too hard on yourself, do as many questions as you can. You got this!
Hello, I am a little confused about the differences between the terms moments, couples, and torques. Is the total torque of a system equal to the sum of all the moments and couples acting on the system?
They are all the same, just interchangeable words when it comes to solving problems. The total summation of moments about a system, is the same as the total torque of a system, etc. Please see this video first: czcams.com/video/QNNnPZ68STI/video.html