Great Explanation Professor Hanson. I am also wondering the same for T(BC) components as other said. 4/5 * T(BC) for Horizontal and 3/5 * T(BC) for Vertical. Very best explanation for 2 force member sir.
Yes, you are correct, I used coordinate method, which forces you to avoid this mistake as long as you put the right coordinate points, and it comes out with Sum of FX -4/5TBC and Sum of FY for TBC as -3/5.
At 10:52 he says that it is clockwise but I got CCW. He still put positive so it should be fine. If the palm is facing point A and the fingers the direction of the force, then it would wrap upwards indicating a ccw direction.
hello hopeful for some responses. I am confused about the 3 and 4. I mean how did you end up with those numbers? Also, will 150 and 200 be ok to use and not 3 and 4 ?
The sides of the triangle are 150, 200, and 250 (you can divide each of those by 50 to get 3, 4, and 5, which is a similar triangle) and so the ration of the sides is similar to taking sin/cos as sin is Opp/Hyp and cos is Adj/Hyp so instead of taking Force * sin theta or Force * cos theta you can just do by the opp/hyp or adj/hyp (opp = opposite, adj = adjacent, hyp = hypotenuse)
Cuz, if you take the upper right corner and press it down, it's gonna rotate the body in the clockwise direction and that makes the two force member try to oppose the rotation, hence the tension in it...
@@almux4260 regardless if thats the case, you can choose whether Tbc is tension or in compression. But remember if your answer is in "-ve" the direction would be opposite to the one you obtained. So in any direction you choose it to be, keep in mind of your sign when you obtain the answer
The hypotenuse isn't actually 5, in this case it is 250mm. But that is irrelevant to finding the angle, all that matters is the ratio of the sides (and that it's a right triangle). The angles on a 3-4-5 are identical to that of a 150-200-250 because same ratios. His point in pointing that out is that the angle 36.87deg is very common in statics as they often give us 3-4-5 triangles in order to find necessary thetas. Basically, if you recognize the ratios of sides on that triangle you would know the angle is 36.87 without putting it in your calculator, similar to eyeballing two equivalent leg lengths and knowing it's a 45 degree triangle
the system is never assumed to rotate. If you say so because of the 4kN moment or "rotational force", then actually that force is just a determined force being applied to the system. It's like when you place your hand on one end of a table, the weight of your hand will generate moment, trying to fill the table up, but simply there are other forces (such as the weight of the table itself) that counter the moment of your hand and thus the system (the table) keeps its equilibrium. When a system is said to be in equilibrium it means that the system doesn't move or rotate in any direction whatsoever. It other words, the system is "static" 😉 (hence the name of the subject)
Me at 1:07 AM studying for my statics midterm in less then 12 hours. Engineering is hard but keep going guys. You got this!!!
at 01:39
Yo same
Did you pass?
thats me rn 0.0
lol me rn its 9AM and my midterm is in 4 hours 😅
As someone already mentioned, T_BC,x = (4/5)T_BC and T_BC,y = (3/5)T_BC.
Which gives,
Ax = (4/5)(15.71) = 12.57 kN
and
Ay = (3/5)(15.71)+2 = 11.426 kN.
yup you are right mate
I didn't even bother with the ratios and just did the angles instead. I got the same thing as well
13:42 I was looking for this comment. It seems I was right after all.
Great Explanation Professor Hanson. I am also wondering the same for T(BC) components as other said. 4/5 * T(BC) for Horizontal and 3/5 * T(BC) for Vertical. Very best explanation for 2 force member sir.
He just made a mistake for the horizontal and vertical.
"One hop this time", lmao. You know that Dr.Hanson gets DOWN at weddings.
Shouldn't the horizontal force be 4/5 TBC and the vertical force 3/5 TBC?
i think so too
ive spent 15min+ trying to figure out if and how i got that backwards. It definitely should be.
yea it should be as you mentioned
Yes, I kept internally screaming hoping he would notice lol
Yes, you are correct, I used coordinate method, which forces you to avoid this mistake as long as you put the right coordinate points, and it comes out with Sum of FX -4/5TBC and Sum of FY for TBC as -3/5.
Professor Hanson, thank you for a great explanation of System Equilibrium and 2D Reactions at the Supports. The Free Body Diagrams are excellent.
I got Ax=12.57kN going right and Ay=11.42kN going up for the component forces at A.
This is what I also got. Looks like we might be right, considering how others also got similar answers.
Really good explanation. Looking forward to learn more from you!
14:46 bless you
Best teacher ever ❤️
Wonderful Lectures ! Thanks.
At 10:52 he says that it is clockwise but I got CCW. He still put positive so it should be fine. If the palm is facing point A and the fingers the direction of the force, then it would wrap upwards indicating a ccw direction.
@14:46 Bless you Doctor Handsome... I mean Hanson
😂😂
Hey Dr. Hanson, been watching all your great videos this summer! Hoped to let you know that you've got a small mistake (3/5Tbc and 4/5Tbc reversed)
bless you Professor Hanson.
I got 12560 = Ax = 4/5 BC and Ay = 11420 = 3/5 BC + 2000
What a legend
hello hopeful for some responses. I am confused about the 3 and 4. I mean how did you end up with those numbers? Also, will 150 and 200 be ok to use and not 3 and 4 ?
The sides of the triangle are 150, 200, and 250 (you can divide each of those by 50 to get 3, 4, and 5, which is a similar triangle) and so the ration of the sides is similar to taking sin/cos as sin is Opp/Hyp and cos is Adj/Hyp so instead of taking Force * sin theta or Force * cos theta you can just do by the opp/hyp or adj/hyp (opp = opposite, adj = adjacent, hyp = hypotenuse)
The only thing i noticed, why did u add the 2 in Ay? It's negative right?
He did write it as negative when the equation equals 0. If you make Ay the subject then it becomes positive.
WHY IS Tbc DOWN
At 6.38 can you explain how did you understand that it is tension and not compression?
Cuz, if you take the upper right corner and press it down, it's gonna rotate the body in the clockwise direction and that makes the two force member try to oppose the rotation, hence the tension in it...
@@beautifulday1665 What if it was pressed up? is it going to be the opposite? like how did he know it will be pressed down not up
@@almux4260 regardless if thats the case, you can choose whether Tbc is tension or in compression. But remember if your answer is in "-ve" the direction would be opposite to the one you obtained. So in any direction you choose it to be, keep in mind of your sign when you obtain the answer
Ax is not Tbc * 4 / 5?
confused on that part as well.
is thats the case ay is wrong too
we know value of Tbc so sum of X forces is Tbc(3/5) = 15.71(3/5) to balance both sides .. a lot of folks complaining. maybe I can't see issue?
i think his Ax and Ay got switched...
It should give,
Ax = (4/5)(15.71) = 12.57 kN
and
Ay = (3/5)(15.71)+2 = 11.426 kN.
Thanks doctor
Big Respects
200 mm = =0.2 m
didn't understand how the hypotenuse of the triangle is 5 didnt get that whith C=a+b^(1/2)
The hypotenuse isn't actually 5, in this case it is 250mm. But that is irrelevant to finding the angle, all that matters is the ratio of the sides (and that it's a right triangle). The angles on a 3-4-5 are identical to that of a 150-200-250 because same ratios. His point in pointing that out is that the angle 36.87deg is very common in statics as they often give us 3-4-5 triangles in order to find necessary thetas. Basically, if you recognize the ratios of sides on that triangle you would know the angle is 36.87 without putting it in your calculator, similar to eyeballing two equivalent leg lengths and knowing it's a 45 degree triangle
BCY IS COUNTER CLOCK WISE
I don't understand how you divided over 5 Tbc please
5 is the hypotenuse... so you divide the 3 and 4 by it..
why Tbc is down？
did you figure it out? i'm trying to understand this
@@chris-y9o4zbecause the force member experimentates tension and the force member is connected downwards
You're the best best best
best❤
People help please, I am confused! If the whole system is supported in two places, why do we assume that the system rotates?
the system is never assumed to rotate. If you say so because of the 4kN moment or "rotational force", then actually that force is just a determined force being applied to the system. It's like when you place your hand on one end of a table, the weight of your hand will generate moment, trying to fill the table up, but simply there are other forces (such as the weight of the table itself) that counter the moment of your hand and thus the system (the table) keeps its equilibrium. When a system is said to be in equilibrium it means that the system doesn't move or rotate in any direction whatsoever. It other words, the system is "static" 😉 (hence the name of the subject)
Nice video but the value of Ax is wrong
how?
@@nathan8472 x component of Tbc should've been 4/5(Tbc)
😂😂😂like how he tries to use the cool words
KFUPM student is passed over here .
even UQU student did.
我不明白为什么B点的力方向不是沿着CB而是BC
这个就是 two force members 的特殊点。
Heisenberg
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