Well, we could have done that, but then we would have to find the perpendicular distances for Rcy and Rcx. We already have the perpendicular distance between Rc and point B, since Rc is normal/perpendicular to BC, so no need to break Rc into components.
You may take moments about any point you like. However, there is usually one (or maybe two) points that are most convenient for ease of solution algebra. In this case I take moments about point B because there are two unknown reactions at point B. They will have zero moment about point B which then enables me to determine the unknown reaction at C straight away (i.e. one equation with only one unknown).
@@TerryBrownMechEng If we take moment about point _A_ ,sum of the couple moments and force moments respect to _A_ also gonna equal to zero ,right?Same for point _B_ also?
Rc is perpendicular (90deg) to BC because there’s a roller support at C, and since BC is 20deg from vertical, Rc must be 20deg from horizontal. Sorry for delayed reply, only just saw your question.
![ปฏิเสธไม่ไหว (Crush on you) - LIPTA feat. No One Else [Official MV]](http://i.ytimg.com/vi/NBw1jF342vw/mqdefault.jpg)
For the moment about point b why wasn't Rc broken down into Rcy and Rcx?
Well, we could have done that, but then we would have to find the perpendicular distances for Rcy and Rcx. We already have the perpendicular distance between Rc and point B, since Rc is normal/perpendicular to BC, so no need to break Rc into components.
@@TerryBrownMechEng Thank you so much
Professor how do we know respect to which point we gonna take a moment?I'd be grateful if you could help me to understand
You may take moments about any point you like. However, there is usually one (or maybe two) points that are most convenient for ease of solution algebra. In this case I take moments about point B because there are two unknown reactions at point B. They will have zero moment about point B which then enables me to determine the unknown reaction at C straight away (i.e. one equation with only one unknown).
@@TerryBrownMechEng If we take moment about point _A_ ,sum of the couple moments and force moments respect to _A_ also gonna equal to zero ,right?Same for point _B_ also?
@@yigitcan824 yep 👍
Hello, what geometry brings about the second angle (20 degrees) which is the same as the angle between the line BC and the vertical line?
Rc is perpendicular (90deg) to BC because there’s a roller support at C, and since BC is 20deg from vertical, Rc must be 20deg from horizontal. Sorry for delayed reply, only just saw your question.
@@TerryBrownMechEng 🙏🙏🙏
why moment 500 nm become -ve ? it is clockwise
In my moment equation I assume anticlockwise is positive. Therefore clockwise moments will be negative.
so the normal force at the pin B doesn't have to be perpendicular to the segment BC? It can just be on the x and y axis?
Thien Hoang yup...me also get confused bout that.nso what the answer?..can it be like that?
Same ...
Why u dont write 500 in fx
Because it is a moment not a force
Thanks sir.
Please add an Arabic translation