Statics - 3D Moment about an axis example 3

[@kasra-ir5io]

5:38 mmmmkaaay mr.mackey 😅😅 Thank you

[@edgardchow]

awesome video, super simple explanation :)

[@Abdulaziz-Alsoufi]

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]

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]

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]

A small doubt does the negative sign indicate the direction is opposite to u?

[@jomanrose8382]

You are so good thanks very much

[@KilluaZoldyck-tz2sz]

thank u sir, but can you try solving, using rc? I tried it but i got a different answer

[@Unaesthetic44]

I want indian teacher here

[@fra2025]

Thank s ❤❤❤

[@laurenpadron74]

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]

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]

@@engineeringdeciphered Wow, yes! I had forgotten, thank you very much!!!

[@laurenpadron74]

I hope your day is going well!

[@jbot8108]

thank you

[@dmr5614]

Why the "j" in cross product matrix determinant got minus value?

[@yusufmoola6471]

its just a rule

[@keremgungor3745]

at R ac you didnt write 0.3 k why?

[@engineeringdeciphered]

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.