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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Which engineering program?

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[@QuestionSolutions]

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

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

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

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

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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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[@QuestionSolutions]

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[@QuestionSolutions]

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[@QuestionSolutions]

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

[@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]

@@QuestionSolutionsoh

[@nobodywillsner1063]

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[@QuestionSolutions]

Glad it helped! Best wishes with your studies.

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Thank you :)

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[@QuestionSolutions]

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I am happy to hear these videos were helpful. I know you found them late, but it's always better late than never :)

[@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!

[@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 😊😊

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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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Thank you for your feedback. :)

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Glad to hear :) You're very welcome!

[@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.

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[@QuestionSolutions]

Thank you :)

[@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? :)

[@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.

[@clamp202]

Clear explanation, thanks

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You're very welcome!

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[@QuestionSolutions]

Thank you very much! Best wishes with your studies.

[@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]

@@QuestionSolutions thanks man appreciate it

[@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 👍

[@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. 👍

[@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.

[@avinashmbhat4256]

This is soo good

[@QuestionSolutions]

Thank you very much! I hope it was helpful.

[@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. :)

[@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

[@عمرغسان-م2خ]

Thank you so much ❤

[@QuestionSolutions]

You're welcome 😊

[@chess_speaks099]

05:39 can I also take OA instead of OC?

[@QuestionSolutions]

Yes.

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[@QuestionSolutions]

Many thanks!

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[@QuestionSolutions]

You're very welcome!

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I love this man

[@QuestionSolutions]

❤

[@dazquez]

In the curved Rod example (5:40) why did you put the position vector from O to C and not from 0 to A?

[@QuestionSolutions]

You can use either, you will get the same answer.

[@dazquez]

@@QuestionSolutions thank you!!!

[@QuestionSolutions]

@@dazquez You're very welcome!

[@chunchun3491]

Hey just wanted to ask at 6:44, why can''t I use roc instead of rbc for vector cross product??

[@QuestionSolutions]

There is a little typo in the question, we are actually finding the moment about point B and not A.

[@chunchun3491]

@QuestionSolutions I see thanks for the reply

[@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.

[@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.

[@Hussain-qt5ti]

At 5:47, why can't you just split the force into its x and y components and then solve the question like the previous two?

[@QuestionSolutions]

Excellent question! This is a 3D problem, that means the force now has x, y and z components. You cannot break it into components the same way you did with 2D problems. So you have the express that force as a cartesian vector, in other words, we must express the 80 N force in it's x, y, and z components, and then take the cross product to find the moment.

[@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.

[@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. 😅

[@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. :)

[@MubasharJameel]

Nice video keep it up

[@QuestionSolutions]

Thank you very much!

[@sewanusunday9605]

Thanks Man👍

[@QuestionSolutions]

You're very welcome! 👍👍

[@lostsky3631]

Was studying these same things for four hours and didn't understand a single thing. Can't believe I just understood everything in 8minutes😂😂I appreciate 🙏

[@QuestionSolutions]

I am really happy to hear that! Well done and keep up the awesome work :)

[@ashlynraj6939]

At 2:41 Why did you use the horizontal component of F2 for F2x instead of the vertical component like in F1?

[@QuestionSolutions]

Okay, so I think what you're actually asking is, why did I use cos for F2x, but use sin for F1x, and you are assuming cos and sin are related to horizontal and vertical components. If this is your question, please erase that thought out of your head. Sin and cos are NOT related x/y/horizontal/vertical components. They are purely based on the angle and it's corresponding side. I have no idea where students get this notion from, but please make sure you don't think like this. 😅I really encourage you to look at this video (kzbin.info/www/bejne/hKOvZpdjZ6iUmLM), it's not long, and if you're really tight on time, just watch the first example. It will give a very good foundation for working with sin/cos and how to properly use them. I break it down step by step and by the end you will never get confused again but sin and cos. If this isn't what you're asking, I am not too sure I understand your question. Maybe rephrasing it might help me, if so, I am sorry :(

[@l0velyjesse]

hi i hope you respond- at 6:55 i got +180j. please tell me why it's -180j? i thought we automatically add -in front of j? so it will be -(-)= +? i hope this makes sense. It's what my lecturer taught me :) Thank you for these awesome videos!!!

[@QuestionSolutions]

So for the second term, you'd get -[(0)(-500)-(-0.3)(600))] ==> -[0+180] ==> -180j. Please see: kzbin.info/www/bejne/fGmseaWdaNWWbck Let me know if you need any more help :) You can use this: onlinemschool.com/math/assistance/vector/multiply1/ to verify your answers.

[@l0velyjesse]

@@QuestionSolutions thank you so so much! you're getting me through my degree!!

[@Ralphvanderburg]

5:00, isn't the answer in NM? Since it is force times distance like you said before for moments?

[@QuestionSolutions]

Yup, that's a typo. Good catch!

[@humphrey6630]

Thank you so much.

[@QuestionSolutions]

You're very welcome!

[@ZMANthegreat]

great content..keep it up

[@QuestionSolutions]

Thank you very much!

[@kyordannydelvalle523]

Wow, Even after graduating. I still remember how to do it. Just needed to refresh my memory a little.

[@QuestionSolutions]

Congrats on graduating and well done for remembering!

[@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!

[@baktybek6454]

2:45 where did you get this numbers like 250cos30, 300cos60, 500*4/5 , all this numbers?

[@QuestionSolutions]

So all of those numbers are forces broken into x and y components. If you need a refresh, please see: kzbin.info/www/bejne/hKOvZpdjZ6iUmLM

[@zaynnazam2084]

hi for the question at 5:56 you found r oc how come u dont do it while finding ro a

[@zaynnazam2084]

nevermind I just seen your reply

[@QuestionSolutions]

Okay :)

[@halowiz1000]

at 6:50 wouldnt the position vector have to be "rac" if your finding moment of force about point A?

[@halowiz1000]

it seems like your taking moment of force about point b. Great video just confused if you make a error or if i just don't understand.

[@QuestionSolutions]

@@halowiz1000 Yes, so for whatever reason, I verbally say we are finding the moment about point B, but accidently wrote A. It's a typo 😅

[@DARTH-R3VAN]

Look at this legend

[@QuestionSolutions]

:) Many thanks!

[@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. 👍