/b pls help me be a little less autistic so i can get a job and get out of the Sup Forums life.
The question states (pic related) replace the loading on the frame by a single resultant force. Specify where its line of action intersects a horizontal line along member CB, measured from the end C.
I know for a fact I got the correct resultant force and angle which is Fr = 356.1 N with its components being Frx= 220 N Fry= -280 N and its angle being -51* from horizontal, but how in the world do I specify its line of action according to horizontal member CB, measured from C? That question makes 0 sense to me and it makes me wana kms. No answers online for this and I asked my teacher and she told me to start by taking momentum along point which would be Ma= -730 so now what?!?!
You seriously came to b for answers to a statics problem. You're gonna go far child.
Michael Torres
NP. Good luck.
Oliver Cox
It's antiautism like this that causes bridges to collapse
Jaxon Walker
you'll be surprised, some people from /b are extremely smart and have a good life. I've gotten help from physics to matrix theory here within seconds mate, not everyone is a trap loving college drop out fag
Colton Adams
i'm a physicist and this doesn't make any sense at all or it's some troll shit. how the fuck does air push up on the beam with 600N of force?
Ayden James
It's a gantry for a zeppelin
Asher Baker
Momentum isn't a factor here. I'm assuming you mean moment arm. Do a force balance different points on the beam and continue
Nathaniel Wilson
the diagonal piece of wood exerts a counter force upwards against gravity. keeping the board in place
Nathan Bell
nothing is moving so you don't need momentum here
remember the resultant force is somewhere on your fbd take the moment at A and set it equal to moments due to the resultant force
basically what they're doing in this moment summation
Kevin Young
i guess everyone have a diff method, teachers method confused me. Thanks user's i think I got it from here
Owen Rodriguez
It's statics, so you have two things to remember: all the forces cancel and all the moments cancel. You only have horizontal and vertical to worry about, so no Z-component to trip you up. You got the sum of all the forces in the x and y components, but not the moment arms. Solving for those tells you where the equivalent moment arm/line of action is.
The question you are asking is where along CB is the resultant force vector located to where its moment equals the moment of all the smaller forces in the picture combined. Pretty sure.
t. Bachelors in Mech Eng.
Anthony Watson
>ut how in the world do I specify its line of action according to horizontal member CB
think of the resultant point. when taking the moment at that point. it basically creates a large cross (two perpendicular lines) extending across your drawing. These two lines will intersect at CB somewhere and will also intersect at AB somewhere the question is asking you to find where on cb its intersecting
OP HERE. TO CLARIFY FOR ANY FUTURE "Sup Forums CACHE" QUESTIONS AND ANSWERS , HERE IS THE SOLUTION I CAME TO
Find sum of Moment along point A which will equal sum of moment of resultant on A. This yeilds Moment along A COUNTER CLOCKWISE POSITIVE =400(1.5)-600(.5)-900(3/5)(2.5)+400(4/5)(1) simplified = -730 Moment of FR A COUNTER CLOCKWISE POSITIVE (pretend you have FR on middle of CB) = -220b+280a
730 = 220b - 280a
distance b is always 3 bc its on BC so we get a= -0.25 a+d= 1.5 so d = 1.75m CORRECT ANSWER according to teacher
Im a horrible explainer but thats it folks, now go back to trap posting
