5-11 Equilibrium of a Rigid Body (Chapter 5) Hibbeler Statics 14th Edition Engineers Academy
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- Опубликовано: 10 фев 2025
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Equilibrium of a Rigid Body (2D Equilibrium)
How to solve Equilibrium Problems | Engineers Academy
Engineering Statics by Hibbeler 14th Edition
5-11. Determine the reactions at the supports
3d equilibrium statics,
Particle equilibrium in 3d
#EngineeringMechanics #EngineeringStatics #Statics #Hibbeler #Equilibrium
Hibbeler Statics - Chapter Playlists
Chapter 2: Force Vectors - • Chapter 2: Force Vecto...
Chapter 3: Equilibrium of a Particle - • Chapter 3: Equilibrium...
Chapter 4: Force System Resultants - • Chapter 4: Force Syste...
Chapter 5: Equilibrium of a Rigid Body - • Chapter 5: Equilibrium...
Chapter 6: Structural Analysis - • Chapter 6: Structural ...
Chapter 8: Friction - • Chapter 6: Structural ...
good stuff my boi
Good. Very helpful
thank you very much sir, please do more examples from this chapter, and also please do videos on shear force and moment diagram.
Cannot agree more! Those diagrams are confusing but not extremely challenging
Hey sir, at 6:40 why 2/3? why not 1/3? Having trouble understanding this concept.
Same
See his next video usma yh barabar 1/3 use krta ha it means us ko wha galat huw ha
It is 1/3 from the highest point of triangle or 2/3 from the lowest side of traingle
Theres a lot of likes on this one so if you see this the ty0024 guy is right.
If you're still confused just think about it like you need to cut the triangle into two equal surface areas. where would you cut? you would cut higher up (exactly 2/3), and that's because the center of mass is along the line you cut, so that's where the force would exert.
@@nicholasmartinez5104have you taken electromagnetic force/electrodynamics? Would you like to be my study partner?
If it is sin theta, shouldn’t you be using 3/5 not 4/5?
Sin theta is opposite of theta over hypotenuse
yes he is very wrong
I changed my solution which is definitely incorrect but the sign of clockwise and counterclockwise even if they're opposite they still give the same answer.
good stuff indeed
I don’t know Where the 2/3 when you calculate the distance of where the force is applied come ,thks for answering me sir
The distributed force is replaced by single point load which acts at the centroid of triangle ( or any other shape which is given in the problem) the centroid of triangle is one third from one end and two third from the other end.
but doesn't the it say 400N/m? doesn't that mean 400 newton applied for every meter, why did we take the area of the triangle? if u can please explain sir
Late but for others its distributed loading therefore where divide the triangle get the area and place the force at the centroid of each
when i took the moment about point A instead of B i got a different answer ?
Then the moment arms will be different
niceee
Bruh why would you divide this triangle to two just put 1200n force in the middle of it why complicate???
Why R sin theta in the x component?
yes its wrong
Tnx
لو سمحت يا دكتور عايز ملف المسائل