14:28 - you can treat the problem as a static one. If you calculate sum of moments around point A, than: 0.4C - 0.4mg + 0.3Ff = 0 Pushing of the entire system (including AB member) by friction force Ff, happens at the wheel bearing (point B). Hence, friction force Ff acts directly on AB member at the point B. Therefore the distance for the moment around point A, caused by force Ff, should be 0.3 instead of 0.5 Such calculation yields correct result.

Thank you very much!

this helped alot!

thank you so much for the vide

great video! very helpful for my mechanics midterm!

Hi, how can you use (275 sin(30)) as the lever arm for -Pcos(60)? You are taking moment about point C, correct? The distance 275 is from line of action of P, to point B, NOT C. I noticed when you calculated -Psin(60)(275 cos(3)+70) you accounted for the extra horizontal distance needed. But you didn't account for anything in -Pcos(60)(275sin(30))

Thanks for your video. This video really helps me for my mechanic exam.

Thank you so much ❤

Wow i'm feeling as if i have passed my exam!!

exam in an hour and this JUST clicked. thank you

ilyyy ur goated

This video might've just saved my semester. Thank you a million!

You made a mistake in last question, supposed to be 40sin60(12.3) not 9.3

May I ask, from what book these problems came from?

Thank you man, Finally I understood the concept

I don't understand why the frictional force in the first example in pointing in the same direction as P.

Related Rates 2.0

Ty

You are the goat. Keep being a hero

Very good teacher. Very hard exams

Great videos.... Thanks! Although I do make Vb= 4.732i - 1j

Why he put (-) front 500 when he whan to figure out the sigma of MA???

very useful video! Found this might be the only one on KZbin So much thanks for sharing this! truly grateful! I watch this several times and played it in slow motion Really helped me a lot by seeming the real thing than just imaging it. Thanks again and hope you can keep up the good work!

i wished my professor was as clear with the course like you are and uploaded his lectures like you did :(

Finally I got someone to rely on for studying chapter 17

thank you sir

Yes frérot

Nice video

Dynamics is one of those modules that I’m glad I never have to do again 😹😹

Perfect 👌

where is solution ??

absolutely legendary. thank you for making this so easy to understand!

where 15.7b part ?

D is a thrust bearing so it will have a reaction force in the x direction as well

Hi sir, firstly thanks for you efforts. Didn't you say that when we have a vertical beam that is not oblique will have only one force this was in 6-4c truss reaction

why Fa=Fc at 10:29?

make one lecture on zero force member.

Very interesting.... Thank you!

Thank you very much! Your way of teaching is very therapeutic!

Great video

You are amazing. Thank you so much for providing the help that my professor couldn't.

The best explanation yet, Thank You

Ok

@ 11:39 (ish) is m(agx)(ybar) coming from summation of forces in X = max * radius? And in that case couldnt you use 200? And where does this equation come from?

Thanks

you are the best and the most underrated channel💖💖

hi can u make videos on Aplied maths 248 college engineering

thanks

Thank you soso much. You went into more depth than my textbook and made it easier to understand.

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