It’s an old rule from geometry. If you aren’t given any values for a triangle you just use the 3,4,5 rule which you can look up. Dw, that massively confused me too
since we are having 2 forces and a moment forces going down I think Ay should have been pointing up instead of pointing down. but either way the answer is the same.
the Reaction at B is already greater than the 2 applied forces so another downward force is needed for the equilibrium. also assuming the wrong direction gives a -ve value.
Random question for an old video. But I would like to build a simple crane, what would I research to find the max safe load for structural steel shapes? I'm taking a materials class atm but it doesn't look like that level of information is covered in my books.
@@Zach-yv9vz man that's a blast from the past, never did build it, though if I were to now I'd just make a model and run it threw fea, maybe hand calc verify it.
He's calculating the moment for the vertical component of Fbc....The vertical component is Fbc Sin(theta).......as sin thetha in that force is 3/5 so he has taken 3/5
why is the bc slope 4-3? should it not be 1.5 and 2. since the length of the y-axis (from point a to c) is 1.5m according to the given; and the length of the x-axis (from point a to b) is 2m.
10 months late, but since you know the base is 2m and the height is 1.5 meters, taking the pyth. theorem grants you 2.5m making it a 3/4/5 triangle (very common triangle in statics).
he means that the sides of the triangle have the ratio of 3m by 4m with 5m as the hypotenuse.With 3m height and 4m base, by pythagoras' theorem the hypotenuse is 5m. thus it is called a 3/4/5 triangle.
God bless you from Los Angeles. Seriously, you're in my prayers at night for helping me pass this damn class
this is the best lecture i have heard so far
You saved me man. From rocket science (in class) to simple math ( after this video)
If A is a fixed support shouldn't be there any moment at A
why no moment at A? It's a fixed support
It would be good if you show the forces in red - stand out
yeah im really confused here why the fixed support is being treated as a pin support
How did you get 3/5? 4/5?
It’s an old rule from geometry. If you aren’t given any values for a triangle you just use the 3,4,5 rule which you can look up. Dw, that massively confused me too
3/5 is for sine (opposite/hypotenuse) and 4/5 is cosine (adjacent/hypotenuse)
@@HiddenTalent_07 bro you replied 3 years later :0
😂😂😂😂
Atleast he reply
Thnkyou Engr Stranger
thanks alot brothor, now i can build a building from scratch
you did in five minutes what my professor said in three hours.
Thank you so much you cleared all my misconception
@ 2:50 why did you make the vertical component (3/5), shouldn't it be just 3?
because sin = 3/5
Love you man. Cheers for the help.
How did you get the 3/5 for Fbc?
that was my question as well?
Using cos right adjacent/hypotenuse ?
One year later but I can help you. The distance is horizontal so you can take the perpendicular force. Hence, 3/5 is the y component in this case.
That's equivalent to 1.5m(vertical component)/2.5m(FBC). He just scaled it two times to make them whole.
Where did the 2.5 come from?@@Blurry_Ice
since we are having 2 forces and a moment forces going down I think Ay should have been pointing up instead of pointing down. but either way the answer is the same.
the Reaction at B is already greater than the 2 applied forces so another downward force is needed for the equilibrium. also assuming the wrong direction gives a -ve value.
How do I determine the sign of the moments I didn't get the right hand rule
Thank you so much, it really helped me
How did you find the triangle for force bc
Wow god bless this guy
make g capital later example ((God)) but not (god)
why are there no reaction forces at C?
Should there be a moment at point A due to it being fixed support?
MA(0) because you started the moment at A, you can still solve it, but it wasn't asked in the problem.
Was just wondering about that too
I disagree with this approach, you can't assume it to be zero juts because you started it there. I think the problem is solved incorrectly. @@seybah
can i ask. how can we find the maximum load that can be applied on that beam?
Random question for an old video. But I would like to build a simple crane, what would I research to find the max safe load for structural steel shapes? I'm taking a materials class atm but it doesn't look like that level of information is covered in my books.
have you built your crane yet?
@@Zach-yv9vz man that's a blast from the past, never did build it, though if I were to now I'd just make a model and run it threw fea, maybe hand calc verify it.
@@Ragnar.Lothbrok.3.14 legend 😎
how you put the Ax and Ag direction ?
this is so helpful. thanks man
Why the moment for fbc is3/5 ??
He's calculating the moment for the vertical component of Fbc....The vertical component is Fbc Sin(theta).......as sin thetha in that force is 3/5 so he has taken 3/5
How is it 3/5 times 2 in momements can it also be 4/5 times 1.5
3/5 * 2 = 4/5 * 1.5....so yeah
why is the bc slope 4-3? should it not be 1.5 and 2. since the length of the y-axis (from point a to c) is 1.5m according to the given; and the length of the x-axis (from point a to b) is 2m.
Was also thinking the same cause I'm a bit confused right now
He just multiplied 1.5 and 2 by 2 to make 1.5 a whole number
And also to recognize the 3-4-5 triangle
did he even multiply 1.5*2 for perpendicular of triangle as well for base 2*2 @@oyuwa8025 if yes.. then why..?
@@isaactanko4067ratio is same
Hence you can use any number
From which book please?
Please sir the Ax is it 1950N coz I had 1950...I think there's a mistake somewhere in your calculation...no offense please...I'm humbled
you missed to multiplicate 800 x 4 ) then sum everything..
How do you get the 396.67N?
you missed to multiplicate 800 x 4 ) then sum everything..
Thanks very much bro but I will like to have Ur personal link
how do i know about the triangle ?
10 months late, but since you know the base is 2m and the height is 1.5 meters, taking the pyth. theorem grants you 2.5m making it a 3/4/5 triangle (very common triangle in statics).
@@Leo-hg6ug can you please explain it clearly
he means that the sides of the triangle have the ratio of 3m by 4m with 5m as the hypotenuse.With 3m height and 4m base, by pythagoras' theorem the hypotenuse is 5m. thus it is called a 3/4/5 triangle.
@@Ultrafats thx for this reply
@@Leo-hg6ug is this always the case? When the hypothenuse is 2.5m, this will become a common triangle?
👍
If you want to see more videos on statics, be sure to see the playlist on my channel!
I went this vedio for arbic langug...