Professor, at 9:30 you said "the sum of the forces at that pin have to be zero". It makes so much sense the moment you said that so many things clicked! I have been struggling so much because I wasn't flipping the directions. I don't know if neither you or my professor never explicitly said that, or assumed it was obvious, or I never heard it before - but for me that was the missing link. Thank you very much!
my google search history: why is statics so hard? Can you finish a frame question in under 20 minutes? (my quizzes are 20 minutes) Frame problem shortcuts in conclusion, there are no shortcuts and my professor is wildin
got confused on the distances for moments & triangles mainly because of the complex measurements but Mr. Hanson's way of teaching is so effective Fab= 1750# Cx= 193.75# Cy= 2422.46#
Note, I am quite sure there are a few errors on this problem. Here are my calculations. (Take it with a grain of salt, but test for yourself). The main problem is that the angle which jeff used as 59.06 degrees on diagram should actually have been 30.95 degrees. He just mis placed the angle in the wrong position. For diagram 3 the answers would then be: Fef = 225.9 # Dx = 193.7 # Dy = 466.2 # Then the second problem was on the distances between EC and AH. In jefs textbook the distance between B and C is also 0.983 and not 0.833. The vertical distance between CD is 4.017 not 4.517. I believe Jeff just made a calculation mistake here. This gives you an angle of 58.1 degrees using arctan(4.017/2.5) in trangle CD (below member BCD), which would correspond to the angle in traingle BC on member BCD. You can then use this to find the horizontal distance AH which would then be 0.6118. Using this information on part 1 my answers are then as follows: Fab = 791.7 # (using moments). Cx = -193.7 # (assumed wrong direction) or 193.7 # if assumed correct direction. Cy = 1257.9 # Maybe, I'm wrong, but I'm quite sure this is correct.
Jeff Hanson, that was one of the finest statics problem solutions I've seen in a long time. Yeah, one minor mistake on Fab, but that was NOT an issue. In fact it added credibility! The process was just stellar! Well done. You have a great teaching style! Keep it up.
Dr. Hanson's book has points E and G on the same horizontal line. Both are 10' above the ground. The distance from the ground to point C is 5.483. Therefore, the distance from point C to E is 10 - 5.483 = 4.517. The Fex moment calculation, according to his book, should be 376.8 lb (cos59.06)*4.517.
These videos are the best but just one correction. To skin a cat refers to catfish. When cleaning catfish you have to skin them, unlike most fish. Just a fun fact! Can't wait to use these videos in dynamics!
Thank you so much for these videos, they are incredibly helpful and 100% the reason I'm doing well in statics. I have a question about how you solved for the Moment about C in FBD #1. I got the same answer when I did 1262.5 - 376.8(cos(59.06))2. But don't you have to multiply that by 2 because AB is 0.5 ft away?
It seems that you're right. If you take moments about C on the part BCD you get Fab=1750 lb which is 2x the answer given in this example. Mc = 0.5Fab + 4.17Dx - 2.5Dy = 0 ========> Fab = (2.5Dy-4.17Dx)/0.5 Plug in to that equation, Dx=193.7 lb and Dy=673.2 lb, and you get Fab = 1750.5lb = 2 x 875lb
I'm glad this helped everyone else, but I think this is way too messy of an example for introducing these kinds of frame problems, especially when so many basic elements like lengths aren't given and everything becomes convoluted with trying to figure those out.
Thank You Professor! but I think the angle for F EF should be tan-1(1.5/2.5)= 30.96 NOT tan-1( 4.17/2.5)=59.06 for EF, the height is already given, which is 1.5. but you did the math to find the angle for BCD Instead.
Why were you able to add the moment about H that you calculated in your global equilibrium to the moment about C given that they are some distance apart?
in lesson 42 there were no moments considered at E ( that was however a truss problem ) why do we consider moment here at H ? How would we know when to and when not to consider moments ?
Can we just appreciate the Professor’s sick drawings.
Professor, at 9:30 you said "the sum of the forces at that pin have to be zero". It makes so much sense the moment you said that so many things clicked! I have been struggling so much because I wasn't flipping the directions. I don't know if neither you or my professor never explicitly said that, or assumed it was obvious, or I never heard it before - but for me that was the missing link. Thank you very much!
my google search history:
why is statics so hard?
Can you finish a frame question in under 20 minutes? (my quizzes are 20 minutes)
Frame problem shortcuts
in conclusion, there are no shortcuts and my professor is wildin
lol, definitely wildin
got confused on the distances for moments & triangles mainly because of the complex measurements but Mr. Hanson's way of teaching is so effective
Fab= 1750#
Cx= 193.75#
Cy= 2422.46#
i got the same. i decided to use the moment at c on member 2.
Thanks you
You are right I got the same
correct Ciame i got as u
Fab = 2140 Ib , check again. 😃
"Here's the base, it's all about that". Thanks for the occasional humor while teaching Dr. Hanson.
This guy is hillarious and he chill, this manz chill frfr. Thank you so much, you make a difference
"I gotch you dawg" XD
Note, I am quite sure there are a few errors on this problem. Here are my calculations. (Take it with a grain of salt, but test for yourself).
The main problem is that the angle which jeff used as 59.06 degrees on diagram should actually have been 30.95 degrees. He just mis placed the angle in the wrong position.
For diagram 3 the answers would then be:
Fef = 225.9 #
Dx = 193.7 #
Dy = 466.2 #
Then the second problem was on the distances between EC and AH. In jefs textbook the distance between B and C is also 0.983 and not 0.833. The vertical distance between CD is 4.017 not 4.517. I believe Jeff just made a calculation mistake here. This gives you an angle of 58.1 degrees using arctan(4.017/2.5) in trangle CD (below member BCD), which would correspond to the angle in traingle BC on member BCD. You can then use this to find the horizontal distance AH which would then be 0.6118.
Using this information on part 1 my answers are then as follows:
Fab = 791.7 # (using moments).
Cx = -193.7 # (assumed wrong direction) or 193.7 # if assumed correct direction.
Cy = 1257.9 #
Maybe, I'm wrong, but I'm quite sure this is correct.
At 27:31, Fab is not 875lb. It is actually 1750lb.
yup he forgot to divide by 0.5
Making use of shift solve in the calculator would eliminate a lot of errors i think
I am a first year eng student and this stuff is pretty straightforward
Which also means cy is incorrect because he implied the 875🤭
yep, very easy to make arithmetic mistakes...its brutal for that!
Jeff Hanson, that was one of the finest statics problem solutions I've seen in a long time. Yeah, one minor mistake on Fab, but that was NOT an issue. In fact it added credibility!
The process was just stellar!
Well done. You have a great teaching style! Keep it up.
Dr. Hanson's book has points E and G on the same horizontal line. Both are 10' above the ground. The distance from the ground to point C is 5.483. Therefore, the distance from point C to E is 10 - 5.483 = 4.517. The Fex moment calculation, according to his book, should be 376.8 lb (cos59.06)*4.517.
yes, I am also getting that. I was so confused because no one here seems to notice. And the angle for Fef I am getting as 30.96 degrees.
These videos are the best but just one correction. To skin a cat refers to catfish. When cleaning catfish you have to skin them, unlike most fish. Just a fun fact! Can't wait to use these videos in dynamics!
Thank you so much for these videos, they are incredibly helpful and 100% the reason I'm doing well in statics. I have a question about how you solved for the Moment about C in FBD #1. I got the same answer when I did 1262.5 - 376.8(cos(59.06))2. But don't you have to multiply that by 2 because AB is 0.5 ft away?
It seems that you're right. If you take moments about C on the part BCD you get Fab=1750 lb which is 2x the answer given in this example.
Mc = 0.5Fab + 4.17Dx - 2.5Dy = 0 ========> Fab = (2.5Dy-4.17Dx)/0.5
Plug in to that equation, Dx=193.7 lb and Dy=673.2 lb, and you get Fab = 1750.5lb = 2 x 875lb
This example is extraordinary, thanks for the Solid explanation, professor.👍
Thank you for creating this. It really saves my midterm exam!
DR. Hanson ,thank you for a solid introduction to Frames with 2 Force Members. This example was long and messy.
I'm glad this helped everyone else, but I think this is way too messy of an example for introducing these kinds of frame problems, especially when so many basic elements like lengths aren't given and everything becomes convoluted with trying to figure those out.
its so messy but everything is gonna be easy after this
This is my first comment on RUclips, and it is to say that Mr. Jeff Hanson is the GOAT!
Thank You Professor! but I think the angle for F EF should be tan-1(1.5/2.5)= 30.96 NOT tan-1( 4.17/2.5)=59.06
for EF, the height is already given, which is 1.5. but you did the math to find the angle for BCD Instead.
i also think so.
agree.
you should do a free body diagram drill
I'm confused how to identify a two force member and a pin connection in the diagram.
I really love your video being helping me in my school work and it’s also kinda funny 😅
I officially hate frames but loved how Mr. Hanson taught it. Imma review it again since I didn't get it the first time
for the distance DF, dont we know that from D to the bar it is 0.5 and from the bar to F it is 1.5 so we can conclude 2 right away?
Here I am again watching it clear my mind
for a moment equation doesn't the distance need to be the perpendicular distance from the point to the line of action? 24:58
In 5:58 we said that Mh=0 but then we find Mh = 1262 ft system is in equilibrium right?
where can I find the problem in the book？I want to have some practice on it
为什么要将示例问题中C点的力图绘制为CX和CY？为什么 CX 不能是箭头到 C 点，而必须是 C 点到箭头（图中）
Why were you able to add the moment about H that you calculated in your global equilibrium to the moment about C given that they are some distance apart?
as far as I can remember moments can be taken at any part of the body...so to achieve equilibrium he also had to add the moment about H
it's a burrito force. All the good stuff, including distance and guacamole, are already factored into a moment so you can just apply a moment anywhere
Am i the only one who caught that bar at 6:34 🤣 THE BASE ITS ALL ABOUT THAT🥶
in lesson 42 there were no moments considered at E ( that was however a truss problem ) why do we consider moment here at H ? How would we know when to and when not to consider moments ?
In Truss Problems, we assume the joint to be a Pin Joint. Here, the Base of the Stand is a fixed support.
Since the base is a fixed support resisting it to revolve at point H , hence the reaction at point H in form of moment of force
我是初学者，我现在最大的问题是，在绘制力图时，我无法确定箭头是从粒子延伸到延伸还是从向外的粒子延伸。
Why the point F isn't a pin, like D? It should be a pin with 2 reactions.
14:12 how do you know they are parallel? isnt that assuming the same height for each triangle which we dont know?
This was a bs problem if i ever seen a bs problem and I’ll fight anyone who says otherwise goodnight see ya tomorrow Professor Hanson
He definitely confused the measurements . The moment he said 7.33 at 24.33, a ridiculous fbd appeared.
Thank you so much!
Jeff hanson u legend
How come there is Mh, i just dont get it?
It's a fixed support. Not a pin support. Thus it prevents moments, causing a moment reaction.
To people seeing this comment, y'all can do it! (Currently preparing for my final HAHAHA)
what if the cut is between d and l?
Excellent video
well done man. thanks
The quality of the video is really low, I think it is 360p. Would you please upload 1080p?
That was because it was processing,
My hero
Wouldn't Cy be 2422lb ?
yes, ur right. I've gotten Cy 2423.
why BD is not a two force member ?
Forces in C
dad energy fr
Love from Pakistan
15 Items engineering students should have: no calipers...
completely un related but the expression more than one way to skin a cat comes from doing it while the cat is alive or dead
15:39
LOLLL you are a G
3:27