Free Body Diagram: Engineering Mechanics

[@kolankprof]

A great review of FBD concepts that I had nearly 50 years ago. Thanks, Dr. Joji!

[@phinephanotube6793]

The simplest way to understand, Thanks Dr.

[@JUIEKADAM-s3c]

Thank you so much sir, finally a video which cover all the concepts of FBD

[@dejustomariel8305]

Thank you so much, this is so informative and direct to the point video. Please keep making videos like this❤

[@AdnanAhmad-Sun]

Thanks Doc, it’s a good Help

[@GravisBea]

Thank you sir nice presentation ❤

[@akankwasapeter3311]

Good presentation

[@anandshriwas2396]

Nice and very informative video sir

[@drjojithomas]

Thank you sir

[@danis620]

Big fan sir thank you for these amazing lessons ❤

[@Ajinkya-16]

Thanks for best explanation

[@kalanidias9196]

very useful pls keep it up😇

[@Pylon_Design]

Thank You Sir

[@lavanyachitturi2232]

Thank you so much sir ☺️

[@SanthiyaaMuthusamy]

What about normal force ?

[@fisolamiazendah2623]

Thank you very much🙏🏾

[@suraj_naik_7]

Easy to understand ✨✨🔥🔥

[@onthelore]

❤❤

[@hell33rro]

رحمه الله الديك ❤❤❤

[@yigitcan824]

Professor,I have a question.For an object which is in equilibrium ,it doesn't matter at where we take the moments ,right?It must be zero.Could you help me to understand

[@drjojithomas]

For equilibrium resultant force should be zero. Therefore its moment would be zero about any point.
According to Varignons theorem sum of moments of several forces about some point shall be equal to moment of there resultant about same point.

[@yigitcan824]

​@@drjojithomas Oh thanks professor.Plus,in the equilibrium questions we need to select the point which has the most unknowns right?It's gonna make easier to find other unknowns

[@rayonzray8357]

Why is T sine theta being considered an horizontal force, shouldn't T cos theta be the horizontal component of tension force

[@drjojithomas]

Angle theta is taken from vertical. Thats why vertical component would be cos and horizontal component should be sin

[@enjoy0.434]

👍🙅