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Statics - 3D Moment about an axis example 3 

Engineering Deciphered
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27 окт 2024

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Комментарии : 19   
@kasra-ir5io
@kasra-ir5io Год назад
5:38 mmmmkaaay mr.mackey 😅😅 Thank you
@edgardchow
@edgardchow Месяц назад
awesome video, super simple explanation :)
@Abdulaziz-Alsoufi
@Abdulaziz-Alsoufi 4 года назад
Why didn't you break F to its component as you did in the previous problem? Also, when do you know when to break F to its component or to just use what's given in the diagram like this problem when you used F=-300 N? Thank yoooooooooou!!!
@engineeringdeciphered
@engineeringdeciphered 4 года назад
If F is only in one direction (like this problem) then there is no component in the other directions. So F = -300k IS breaking it into its components. Don’t over complicate it if it’s pointed in one direction like this.
@shailkumarjain
@shailkumarjain 3 года назад
Check the videos number 6-21 of static series for vector components. You will understand the answer. If you see the videos in a flow then you can understand it better. It has been beautifully explained with each problem covering a new aspect. All the best.
@rameesmonk8986
@rameesmonk8986 Месяц назад
A small doubt does the negative sign indicate the direction is opposite to u?
@jomanrose8382
@jomanrose8382 Год назад
You are so good thanks very much
@KilluaZoldyck-tz2sz
@KilluaZoldyck-tz2sz 4 месяца назад
thank u sir, but can you try solving, using rc? I tried it but i got a different answer
@fra2025
@fra2025 28 дней назад
Thank s ❤❤❤
@Unaesthetic44
@Unaesthetic44 2 года назад
I want indian teacher here
@laurenpadron74
@laurenpadron74 3 года назад
I used AC for the position vector where I divided {0.6, 0, 0.3} by the magnitude (sq.root of 0.6^(2)+0.3^(2)). My triple product matix looks identical except I used this other position vector {0.8944, 0.4472, 0}. I've looked it over a considerable amount of times, but am still not getting 80.5 for the momentum's magnitude. Has anyone else run into this? The magnitude I'm getting is 119.9927.
@engineeringdeciphered
@engineeringdeciphered 3 года назад
Hmmm... You say you used AC as the position vector. That’s fine, but don’t divide that by the magnitude. The equation calls for an “r” there so don’t divide by magnitude. Does that help?
@laurenpadron74
@laurenpadron74 3 года назад
@@engineeringdeciphered Wow, yes! I had forgotten, thank you very much!!!
@laurenpadron74
@laurenpadron74 3 года назад
I hope your day is going well!
@dmr5614
@dmr5614 3 года назад
Why the "j" in cross product matrix determinant got minus value?
@yusufmoola6471
@yusufmoola6471 3 года назад
its just a rule
@keremgungor3745
@keremgungor3745 3 года назад
at R ac you didnt write 0.3 k why?
@engineeringdeciphered
@engineeringdeciphered 3 года назад
I didn't do Rac. I did an R from A to the bottom of the line of action of the force. For the R in the equation, you can go from anywhere on the axis to anywhere on the (line of action of the) force. You can definitely use Rac = .6i + .3k if you want. See what happens - you'll get the same answer.
@jbot8108
@jbot8108 8 месяцев назад
thank you
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