Moment of a Force | Mechanics Statics | (Learn to solve any question)

[@QuestionSolutions]

⭐⭐If you'd like to see more examples covered and at the same time support the channel, please see: kzbin.info/www/bejne/pIHZeGSYjquVodE ⭐⭐ 🔹Clockwise moments positive, or counter-clockwise positive? kzbin.infoP029mqnp4XY 🔹Please note there is a typo at 6:23. The "k" term is negative, not positive, so it's {-128.34i + 128.34j - 256.68k} 🔹At 6:31, there is a typo in the question, we are finding the moment about point B, NOT A. This is a typo in the question only, all steps are correct. 🔹When to use sine or cosine: kzbin.infovynnKlJD_Jo

[@moonbunnies5216]

Hi sorry for bothering. For the j term in the same timestamp you've provided, I can't seem to get positive j. 4(-42.78)-(-2)(21.39) = -128.34

[@moonbunnies5216]

btw thank you so much for these videos. Its really helping me a lot you have no idea how thankful I am to have found this channel.

[@QuestionSolutions]

@@moonbunnies5216 Since this is the second multiplication, it will have a negative in front first, so it will be like this: -[4(-42.78)-(-2)(21.39)]. You can always double check using this website: onlinemschool.com/math/assistance/vector/multiply1/ . Simply plug in the 2 vectors. And you are very welcome, glad to hear these videos are helping :)

[@moonbunnies5216]

@@QuestionSolutions oh my god I totally forgot about the formula for that, should've checked myself before commenting, thank you so so soo much for letting me know.

[@QuestionSolutions]

@@moonbunnies5216 Happens to everyone I think :) Best wishes on your studies!

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Completing my Engineering from KZbin University of Engineering and Technology

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Best wishes with your studies!

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Which engineering program?

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Thank you so much for your kind words! Really nice of you :) I will do my best to continue the channel.

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Wow, thank you very much, that was a really nice comment. I wish you the best in your studies and all your future endeavors!

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[@QuestionSolutions]

Haha, yes, perpendicular distance :) Glad to hear this helped!

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Many thanks! I wish you the best on your studies.

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[@QuestionSolutions]

I try my best to make them as concise as possible :) Best wishes with your finals!

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You are very welcome! I hope you gain a firm grasp of the materials and do amazingly on your exams!

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You're welcome! Glad it's still helpful :)

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I am really glad to hear that :) Best wishes with your studies!

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I’m really glad to hear that the video was helpful for you! It’s not uncommon to feel lost when learning something new, and it’s great that you didn’t give up. Do you best, keep going, you’re doing great! 😊

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Thank you very much! :)

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You're very welcome. You're doing exactly what I hoped students would do :)

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You Deserve a Globe's Best Teacher Award 🤩

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[@QuestionSolutions]

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[@QuestionSolutions]

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[@QuestionSolutions]

Glad it helped! Best wishes with your studies.

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[@QuestionSolutions]

I am glad to hear it helped you out. Keep up the great work, and best wishes with your studies!

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Best Teacher ever!

[@QuestionSolutions]

Thank you, you are too kind!

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The best video on KZbin, nothing better than this but could have been the best video in the galaxy if video was a bit slow. Edit: by bit slow i mean 10x slower

[@QuestionSolutions]

Thank you for your feedback. :)

[@kentpimentel9300]

I got -128.34i +128.34j -256.68k for the cross product in 6:24.

[@QuestionSolutions]

You are right, there is a typo, the k term should be negative. 👍

[@shaqur___5775]

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[@janecm6136]

Thank you. I think I get it now. I’ve been stressing and struggling summation of forces and moments. I think I finally got the principle. It’s just that the figure is overwhelming me but I think I’m almost at it now. I’m definitely subscribing ❤

[@QuestionSolutions]

Thank you very much. I am sure you will get it, just practice a few problems and follow along the examples :) I wish you the absolute best with your studies!

[@eurydice8963]

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[@QuestionSolutions]

I'm really happy you got it. Keep up the great work and I wish you the best with your studies.

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I hope it was helpful to you. Best wishes with your studies!

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Glad to hear :) Best wishes with your studies!

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I have been struggling with this chapter in Applied Mathematics, Thank you very much, now i trust myself fully in Moments, i also subscribed to your channel and sent the link to my fellow varsity mates

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I am really happy to hear that and I really appreciate your subscription and the share. Means a lot and I wish you the absolute best with your future studies! Keep up the good work 👍

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[@QuestionSolutions]

Thank you so much 💓 I really appreciate your comment and the shares, super nice of you! 💓💓💓💓

[@ZeyadAlhilali]

I really appreciate the information you provided, I found a small mistake in this video in the minute 6:25 The +256.68k should be -256.68k Thank you

[@QuestionSolutions]

You're very welcome and you are absolutely right. There is a pinned comment with that correction.

[@nataliatothemoon]

Praying Tanaka has a suggestion: may you include sample problems where we are given angles and not just lengths? It would be very helpful 🌝 🙏

[@QuestionSolutions]

I am not sure I understand what you mean? You need distances to calculate a moment, or you use the angles given to figure out the distances first. Could you elaborate on what you would like to see? :)

[@jesustheodore5126]

Thank you so much, Great Video! it was very helpful for my statics exam!

[@QuestionSolutions]

I am very glad to hear that :) I hope you did well on your exam.

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Thank you very much! :)

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Thank you :)

[@0nub1s]

first of all thanks for the beautiful explanation, but I have a question, why'd you multiply the horizontal component of F3 with 4? shouldn't the perpendicular distance from point A to Fx be 5 as well? or is every point along the horizontal part of the pipe considered as point A? (in the 1st problem)

[@QuestionSolutions]

So you can actually draw two lines at point A (perpendicular to each other like a coordinate system), extending towards the edges of your page. Then it's easier to see the perpendicular distances better. So for Fx, we are looking to see the distance towards the horizontal line, which is just 4m.

[@0nub1s]

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[@QuestionSolutions]

Congrats on graduating and well done for remembering!

[@halleighfaulkner3963]

for the fourth example, why are you using the position vector for BC if you're taking the moment about point A? would you not use AC?

[@QuestionSolutions]

There is a typo in the question, the moment was suppose to be about point B. I wrote a pinned comment about this too because I get asked it so many times. 😅

[@mustakinahmedantuplasticst3912]

I love you as a teacher

[@QuestionSolutions]

Thank you very much! :)

[@nacure360]

For example 3, if r is the position vector from the point where we take the moment about to any point along the line of action of the force, can r be the position vector from O to A? if I do that i get -128.4i +128.4j -256.08k? (the only difference is the sign of the z component to your answer).

[@QuestionSolutions]

You absolutely can. I have a typo in my answer (see pinned comment). Your answer is correct.

[@nacure360]

@@QuestionSolutions ohh thank u

[@balajivelpula2435]

Crystal clear explanation

[@QuestionSolutions]

Many thanks!

[@sujalbhatu999]

thank you so much brother

[@QuestionSolutions]

You are most welcome.

[@DivshaR-hx9im]

Thanks man this was so helpful

[@QuestionSolutions]

I'm glad to hear it!

[@kriistyy.]

in 6:13 , I used the position vector OA. Why the direction of Z axis of my moment is -256.62 and not 256.2. Is it wrong?

[@QuestionSolutions]

It should be negative. Please show me the position vector you got. Thanks!

[@viva7416]

Sir,here is the moment when I use the position vector OA. Mo={ -128i + 128.34j - 256.68k } rOA= { 3i +3j +0k } So if there's not any mistake the Moment must be the same for both position vector rOA and rOC. isn't sir Thanks a lot for ur help 😊😊

[@eiadm2047]

Thx for your great effort! In the last problem, If we wanted to use position vector OA, what force should we use when doing the cross product?

[@QuestionSolutions]

The forces remain the same, only the position vector changes. So when doing the cross product, you would use rOA. 👍

[@pinmycomment5239]

Dude you taught me more than my professor did in 8 minutes lmfao thanks.

[@QuestionSolutions]

I’m really glad to hear that! Keep up the great work and best wishes with your studies.

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I am glad to hear that! Keep up the good work.

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You're very welcome!

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Glad it was helpful!

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Glad to hear!

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Your videos are so amazing, thank you!

[@QuestionSolutions]

Thank you for your kind comment! I hope they are very helpful to you.

[@rjaal]

first thanks for the incredible work you've really managed to reach the ideas in such short time even for the hard learners like me but in 6:24 shouldn't the answer be (128.28i-128.28j+256.5k)? and thanks for your time 🌹

[@QuestionSolutions]

You're very welcome and thank you for the feedback! To answer your question, the correct answer is {-128.34i + 128.34j - 256.68k}. There is a typo in the video where the K term is shown as positive, when it's actually negative. So while the magnitudes of your values are correct, the signs are wrong. It's probably just a small mistake in assigning positives and negatives to your i,j,k components, so if you have some time, take a look at this video: kzbin.info/www/bejne/fGmseaWdaNWWbck

[@cemileceylan1017]

Perfect explanation!

[@QuestionSolutions]

Thank you very much!

[@tiktokshorts5450]

for the last question, could have i done it the vertical and horizontal components way to get the moments about o and solved it?

[@QuestionSolutions]

Do you mean using the scalar method? I mean you can, but it's really tedious to do. It's much easier to use a vector approach when you have forces along lines.

[@rorygame5835]

Great video! Really helped me wrap my head around this. With the curved rod question, wouldnt the k-value of the cross-product be -256.68k? ((4 x -64.17)-(0x21.39)=-256.68k) Otherwise it wouldnt be perpendicular to the plane created by the two vectors.

[@QuestionSolutions]

Glad to hear it was helpful. :) And yes, I have a pinned comment with the typo corrected. 👍

[@arjunrai8126]

1000000 times better than my college professor..

[@QuestionSolutions]

Thank you very much! Best wishes with your studies :)

[@Raniaska0306]

I'm still amazed by how you explained everything in 8 minutes which took my professor at least 2+ hours to explain XD

[@QuestionSolutions]

😅I hope it was helpful and keep up the good work!

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Great explainations 👏, hope for more videos 👍

[@QuestionSolutions]

Awesome :) Thank you so much! 👍👍👍👍

[@carolineee359]

It is soooooo helpful! Thank you!!!!

[@QuestionSolutions]

You're very welcome!

[@melyvaleriabanuelosvazquez959]

At 4:25 it was mentioned clockwise was positive, while I’m aware that counterclockwise is positive while clockwise negative, could you explain? as I’m confused now 😢

[@QuestionSolutions]

You can use whatever way you want to be positive. If you like counter-clockwise to be positive, you can solve it that way. Regardless, you will end up with the exact same answer. :)

[@johnnym500]

For the first question, when solving the moment for M3: Shouldn't it be -300n(5m)+400N(4)?

[@QuestionSolutions]

So remember, we are looking for the perpendicular distance. F_x is horizontal, so we are looking for the vertical distance from point A, that's the perpendicular distance, which is 4m. F_y is vertical, so we are looking for the horizontal distance from point A and that's 5m.

[@graceobeng4165]

I really enjoyed it, thanks

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Glad to hear :) You're very welcome!

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After 7 yrs on yt Wallahi this the only vid that deserves a like

[@DARTH-R3VAN]

Wallahi indeed

[@QuestionSolutions]

Thank you, you're very kind! :)

[@wolfchrt]

6:45 why is the reaction of BC used and not A?? Why is point B relevant?

[@QuestionSolutions]

The question written is incorrect, the moment is found about point B, not A. I made a typo in the question.

[@clamp202]

Clear explanation, thanks

[@QuestionSolutions]

You're very welcome!

[@darrenwood3826]

Thanks for your help in this topic was wondering that isn't turning clockwise should be negative?

[@QuestionSolutions]

You can pick either direction, it makes no difference as long as you follow through with your positive and negative signs. So for example, assume you picked counterclockwise to be positive, then if you get a negative answer, the moment is clockwise. If you picked clockwise to be positive and you got a negative answer, then the moment is counterclockwise. If you picked counterclockwise to be positive and you got a positive answer, then the moment is counterclockwise. Lastly, if you picked clockwise to be positive and your answer is positive, then it is clockwise.

[@darrenwood3826]

@@QuestionSolutions Thank you

[@ichione732]

tyssm u ddont know how much youve helped me for this

[@QuestionSolutions]

You're very welcome! Keep up the great work and best wishes with your studies.

[@PaulOtoo-n3e]

It's good, but the speed is too much for we the beginners Learning it.😊

[@QuestionSolutions]

Thanks for the feedback :)

[@DARTH-R3VAN]

Look at this legend

[@QuestionSolutions]

:) Many thanks!

[@btw_itachi_uchiha]

amazing helpful

[@QuestionSolutions]

Thank you very much!

[@The.Wonderful.Inspiration.Of.]

Hello, I think there is a mistake at 6:23, Roc should be Roa since the starting point of the force is A no ? Or maybe I don't know the rule well ^^ Edit : Well I just heard "position vector from point O to any point of the line of force action", I learned that thank you :)

[@QuestionSolutions]

Yes, as long as the vector goes to any point on the line of action of the force, you will get the same answer 👍

[@ThecarsHQ]

please i'd like to ask if the position vector could be from anywhere like Rac instead of Roc

[@QuestionSolutions]

Please kindly provide a timestamp to the place you're referring to.

[@katavila4973]

I really liked this video and found it helpful, but it is normal to assume clockwise is positive?? In my course we have always assumed clockwise to be negative and counter clockwise to be positive. This is the only thing confusing me right now.

[@QuestionSolutions]

Yes, and it makes no difference to the answer you get. In fact, assuming clockwise to be positive when you see a lot of clockwise moments will make your flow better, simply because you're dealing with less negative signs. However, 3D problems are a completely different scenario, and you need to use the right hand rule to see where the vector points. But in 2D problems, it really makes no difference whatsoever. See: kzbin.infoP029mqnp4XY

[@footballclips5013]

Sir at 2:45,when finding the components of x and y for F1,why did you use sin for the x-component instead of cos and cos for the y-component instead of sin? Or am I mistaking please?

[@QuestionSolutions]

Please watch this: kzbin.infovynnKlJD_Jo?feature=share It'll clear up all your doubts. :)

[@divineobot1605]

I think I'll come back and watch this again 😅😅😅😅. But it's very clear🎉... maybe I dont get it yet.😅😅😅😅😅😅😅😅. Thank you very much.🎉

[@QuestionSolutions]

You're very welcome. Come back, watch it a few times, do the examples and then if you have any questions, let me know. I'll do my best to answer them. Best wishes with your studies. 👍

[@coylezxs]

@ 7:50 I think the j in MOB is - 0j since the j should be negative (not that it matters since the value is still 0)

[@QuestionSolutions]

I don't assign negatives to zeros. It's neither positive or negative, it's just zero because -0=0 also, to me, positive "looks" nicer 😅

[@harshmangukiya91]

amazing!

[@QuestionSolutions]

Thank you!

[@emanuelbutron6798]

for the pipe assembly problem, even though vector was given in cartesian form why didn't you make it a unit vector like you did previous?

[@QuestionSolutions]

You only need the unit vector to express a scalar force in cartesian form. If it's already given in cartesian form, you don't need to do anything to it. Use it as is.

[@DraftingCADemy2023]

6:23 shouldn't 256.68 be a negative (-)? Since we alternate our operations in doing matrices + - + last term should be + [(4)(-64.17) - (0)(21.39)] which results to - 256.68. Please correct me if I'm wrong.

[@QuestionSolutions]

Please see pinned comment. You are correct, it's a typo. 👍

[@DraftingCADemy2023]

@@QuestionSolutions Thank you!

[@jordanchongo3752]

absolutely fantastic 👏👏

[@QuestionSolutions]

Thank you :)

[@Hussain-e9z]

In Q2, the force was applied at point B. How did u transfer that to point A? Also which way would the pipe rotate?

[@QuestionSolutions]

Please see pinned comment. There is a typo in the question.

[@Hussain-e9z]

@@QuestionSolutions Thanks for replying. In the pinned comment you talk about Q4 having a typo. But I have a doubt in Q2 starting from 4:30 . You don't address that. Can you please help me out? Thanks!

[@QuestionSolutions]

​@@Hussain-e9zokay, I see. For future questions, please use timestamps so it's easier for me to find. So in this question, it's not really a transfer of forces. When a force is applied anywhere along the pipe, a moment is created about point B. You can visually see this if you break a force into components. For example, look at force FX. We can slide that force along the horizontal length of the pipe and it will create the same moment regardless. Why is that? That's because the moment created is only dependant on the magnitude of the force and the perpendicular distance. In this case, it's 0.9m. Now look at FY, the perpendicular distance is 0.7m. For the second part of your question, the pipe will rotate clockwise. Imagine the pipe is free to rotate about point A and you push at point B towards the left. Then we can see it would start spinning clockwise.

[@Hussain-e9z]

@@QuestionSolutions Thank you so much! You're really good at explaining. Please never stop uploading videos.

[@MubasharJameel]

Nice video keep it up

[@QuestionSolutions]

Thank you very much!

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This is soo good

[@QuestionSolutions]

Thank you very much! I hope it was helpful.

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I love this man

[@QuestionSolutions]

❤

[@yuzarsif3586]

thank you so much, u r a holly person!

[@QuestionSolutions]

You're very welcome!

[@BahexRefrigeration]

thanks so much for the tutorials but i still have challenge in position vectors coordinates

[@QuestionSolutions]

This video might help: kzbin.info/www/bejne/eXTIiKxnoNKrg9U

[@santiagoperez633]

You are my hero. Thank you

[@QuestionSolutions]

Many thanks!

[@ThunderGowPlayer]

Hi, I have a question, why in 5:45 we get the position vector from OC and in 6:41 its from BC? That wouldn't mean that our position vector in 5:45 should be from O to A? I have tried that and it will change the values of the momentum vector

[@QuestionSolutions]

You can take it from O to A, and you will get the same answer. There will be no difference. You most likely wrote down an incorrect position vector. The position vector OA would be 3i+3j+0k. Keep in mind that there is already typo in the solution for that question as mentioned in the pinned comment. The answer for that problem is -128.34i+128.34j-256.68k. Whether you use a position vector from O to C or O to A, you will get the same answer.

[@ThunderGowPlayer]

@@QuestionSolutions Oooh, I just checked, and you are right! I was messing up with the position vector, but yeah, when you do it with 3i+3j+0k, you get the same answer. Sorry for the confusion. Thank you for your help and answering my comment, even though this video is 3 years old! New subscriber

[@QuestionSolutions]

@@ThunderGowPlayer You're very welcome. I am glad you got it :) Keep up the great work and best wishes with your studies.

[@shaqur___5775]

6:24 when you get the cross product for the k part shouldn’t it be positive 256?

[@QuestionSolutions]

Please see pinned comment.