Fluid Mechanics: Forces on Submerged Surfaces I (3 of 34)
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- Опубликовано: 13 сен 2024
- Correction: At 53:35 the answer for yR should be 3.96, not 3.54.
0:00:10 - Revisiting hydrostatic pressure distribution
0:04:06 - Resultant force on a submerged planar surface
0:14:03 - Location of center of pressure
0:35:00 - Pressure prism
0:38:03 - Example: Submerged circular surface
0:45:13 - Example: Submerged rectangular surface
0:57:21 - Tips for solving homework problems
Want to see more mechanical engineering instructional videos? Visit the Cal Poly Pomona Mechanical Engineering Department's video library, ME Online: www.cpp.edu/meo...
This lecture series was recorded live at Cal Poly Pomona during Fall 2014. The textbook is Munson et al., "Fundamentals of Fluid Mechanics (7th edition)."
My fluid mechanics professor sat behind a podium in a lecture hall, and pointed at power point slideshows with a laser pointer. The entire semester. Not one single worked through example, or ANY written info on his behalf. I'm now in gas dynamics and environmental controls in my final semester, and am so lost regarding any fluids principles. Thank you for actually TEACHING this crucial engineering topic, and making it available for free.
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Holy fucking shit, that's sad. Did this think their research matters more than teaching?
You Cal Poly students are lucky to have a good professor! Thanks for sharing
Our pleasure. We have a lot more videos in production.
Thank you much!
Cindy Diaz 1
Your explanation is amazing, thank you for explaining the points very clearly and elaborating further on them, some professors don't do this because they think the students "should already know this" but in reality, some students don't because they weren't taught the basics thoroughly. So, thank you for giving detailed explanations on the topics that you present, it definitely goes a long way! God bless you!
There are several steps he skips on this video but he explains it on the following video (4/34). Thank you Cal Poly Pomona and Dr. Biddle for your clarity and enthusiasm!
sorry to be so offtopic but does any of you know a method to log back into an Instagram account?
I was stupid lost my login password. I love any help you can give me!
@Jaxxon Leighton instablaster =)
@Angelo Yehuda I really appreciate your reply. I got to the site through google and Im trying it out now.
Takes a while so I will reply here later when my account password hopefully is recovered.
@Angelo Yehuda It worked and I finally got access to my account again. I am so happy!
Thank you so much you really help me out !
@Jaxxon Leighton Glad I could help :)
SOOOOOOO much better than my current professor!! May actually survive Fluid Mechanics! Thanks for sharing CPPMechEngTutorials!!
omg i wish i knew about this series 2 years ago...
thank you professor, and ofc the one that keeps on recording these vids, God bless u all :)
After a week of studying, watching your video finally made the concept clear. Thank you, professor.
As a recent graduate, I thank this course and professor Biddle for helping me pass my classes. This course and the heat transfer course help me get an A on my exams. Thank you Cal Poly for helping so many students with these lectures!
Biddle was my favorite professor when I was at CPP, glad you have these videos up. Would have been helpful back in the day
The students of years ago were very lucky
My current fluid mechanics prof at CPP has online only lectures that I can’t bear to sit through, and can never admit when he makes a mistake
I’m so glad these are here, Biddle makes me like the subject instead of hating it
I am very thankful for this Professor. Saving my life while taking Fluid Mechanics.
can you help me to do my homework, if you understand this lecture??
Shout out to Cal Poly Pomona for releasing these lectures for public consumption!
i wish this professor can teach every engineering class till i get my degree he s awesome
Thanks!
Absolutely right
Actually, at 53:37 the yR should be 3.959 m (calculated from the surface of the water).
To find the distance from the hinge O, to the point of yR where the Fr force acts, we must calculate the total length from O to the surface which is 5.65 m along the gate plane. yoR =(5.65 - 3.959)=1.697 m (distance from O to the point where Fr acts).
Then moments about O are as this: Fr * yoR - G * ((L * cos (theta) )/2) = 176508 - 135000 =41508 CounterClockwise about O.
Where: L - length of the gate and
(L * cos (theta) )/2 - horizontal arm from O on the direction of the Weight force.
I've worked out this problem in three different ways and I always arrived to the same results.
Thanks Professor !!
how to get Ixc =6.36?
Thank you very much! I was confused about the calculation, now I understood
My pleasure,@@hiwazz
Thanks for the great video. At 53:42 forgot to add the 3,54m to get a total Yr of 3,96m from the free surface.
I am very thankful to Professor . These online lectures are helping me in my academic.
Great!
For anyone watching this lately, the reason he uses (1.5)(weight) in the moment equation is because you can consider the hinge the IC. You can do it the normal (90Cos45)((3sqrt2)/2) and the numbers come out the same. I do the second one just to be safe. As for the 4sqrt2, it's just a simple mistake on his part.
4 sqrt2 is correct as the triangle has two 4 meter sides thus the hypotenus is 4sqrt2 (explanation in next video)
a good 10 minutes was spent thinking about this... thanks for the clarification! :)
sorry why is it 3sqrt2/2
I'm reviewing fluid mechanics on my own but got stuck on the location of the resultant force. Thank you for explaining it really well! And many thanks for the channel for posting the entire course 🙏
I love these videos. They are saving my grades in my fluid mechanics class!
Clear and informative, excellent instructor, thank you for posting
No problem. Glad it was helpful.
Sir, you are awesome!!! Thanks for sharing such valuable knowledge on RUclips.. Respect from India 🇮🇳
Thanks!
one of the great professors i've ever known :)
Thanks for this sir :)
can you help me to do my homework, if you understand this lecture??
Just one word: Beautiful!
Thanks to this amazing professor, sir you have left an impression in my mind forever.
I am very very thankful to Dr Biddle Sir.. I always had fear about this subject and I was never confident on this subject..I literally struggled a lot to understand this subject in more practical manner..but at last I found these Video lectures of Dr Biddle Sir and things changed drastically...A fear which I had about this subject over the years has been gone now...I understood the things pretty well...No one helped me to understand this subject during my struggling days....But the one way or other, Dr Biddle Sir's video lectures gave me new life.....
Dr Biddle Sir is Genius person he knows how to explain complicated things in practical and simple way....I am very thankful to you Sir....I hope we will meet someday....Love from India
Your Distant Student
We're glad you found the videos helpful. :)
my chinese professor is kind and goes on rants about random stuff, which is fun I guess, but instead of rants I'd like him to actually explain these things clearly like this Professor and not rely on office hours to show us examples in depth. Thank God for RUclips and the people who record/upload these lectures.
Thank you so much! I really appreciate these videos!! they helped me a lot for my fluids course ! Thank you !!!!!
Hooray!
This is such a life saver. THANK YOU.
You're welcome.
my fluid mech professor recommended me yours videos because you were professor of him too :)
Hello
Students are not asking questions at all, if it was my class the professor would have elaborated many more things. Happy to see they are silent unlike my class where teacher gets interrupted in middle always.
at 52:41 how do you get yc(distance to centroid ) please explain
Thank you so much, professor. You explain better than my book. LOL, God bless and please keep on saving student's grades.
:)
Quality Level A+++
The 90 KN( weight ) of the gate when we are taking moments ....shouldnt it be perpendicular to the gate ....or
Isheanesu Trevor Muchanyangi No, the weight goes straight down because weight by definition is due to the gravity of the earth, which would pull the gate towards the center of the earth/ down.
@@NicholleWillisLoves Yes the weight acts downwards. The free-body diagram is correct. However, the weight must be resolved to normal and perpendicular components relative to the gate in order to correctly solve the moment equilibrium. Hence, the weight component perpendicular to the gate will cause the gate to rotate, the normal component won't. In other words, the correct weight value is not 90 kN but only 90/sqrt(2) kN.
However, the 1.5 m value is also wrong as the distance should be along the gate. Hence, the value should be 1.5*sqrt(2).
Now, when we put the force and length together: 90/sqrt(2) x 1.5*sqrt(2) = 90*1.5 (which matches the video). Hence, two mistakes put together have, fortunately in this case, still given us the right answer...!
at 52:42 why did he multiply y_c with sqrt of 2 ?
sqrt of 2 = ( 1/sin45 )
@@sebrina2892 but sin(45) is sqrt(2)/2 and NOT just sqrt(2)
@@siten1 1/sqrt(2)/2 = sqrt(2). sine45 is in the denominator
@@ken-cf7tf why did he multiply yc by 1/sin(45) ?
@@theadel8591 yc is the hypotenous, you have the opposite side( which is hc = 2.5). Since sin= opposite/ hypotenous, you get the length of hypotenous by dividing opposite length by sin45.
One tip for this task.Put the begining of y axis at the bottom point. So the dF=pdA =& (h1+h2-y)dA it is bit more intuitive. Also u dont need to calculate rods inertia Ixc but u can calculate Ix around the orginin O at the bottom which is 1/3Ay^2
움직이면서 집단적인 행동을 보이는 여러개의 입자들을 고려할 때 그런 많은 입자들이 부딪히는 평면에 대하여 어떤 식으로 받는 힘을 계산할지에 대해 궁금했는데 이 영상을 통해 그런 궁금증을 해결할 수 있었습니다.
best teacher ever
Thanks
perfect ,I really like the way you teach
I didn't understand about moment of inertia. So, what should i do?
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Thanks much professor receive this greetings from Tanzania
thanks for this piece....sorry pls at 51:40,i do not understand how you got the area as (1)(3root2).
Also at 54:31, i do not get how u got the distance of where the weight from the hinge as 1.5, and also pls sir explain how you got 1.7 for the distance of where the force acts...THANKS IN ADV
it's the width (1m) of the gate * hypotenuse (r) or length of the gate.
To get the hypotenuse: Y=rsin(theta), then.. r=y/sin(theta)
angle: 45; y: 3m
plug in, we get r= 3/sin(45)
r=4.24 or 3root2
horizontal distance from the hinge to gate = half length of hypotenuse(r) of that gate * cos45
r: 3root2; angle: 45
hdistance = 1/2(3root2)*cos45
hdistance = 1.5
Thanks for the explanation. Unfortunately we don't have time to answer every question about the content.
@@NetSkillNavigator Thank you for answering!
he knows how engineering practically works
Not sure why it stays in place. We found that the sum of the moments does not equal zero in fact we found that it should be pushing against the gate, i think this it should "fall" (which i also dont quite understand because it is on a hinge).
This was such a great lecture
Quick question, at 53:42, is'nt the length of the gate sqrt(3^2+3^2)=3*sqrt(2), not sqrt(4^2+4^2). The triangle that formulates the hypotenuse of the gate is h2 as the height and h2 as the base; right?? sqrt(4^2+4^2) would account for the top fixture as well, which isnt part of the gate?
Thanks. We made a correction in the comments.
At 44:12 how come yR is independent of angle? If the angle were 0 shouldnt yR be = yC?
56:54 You forgot to add the second term with 3.54 m, i.e. adding the displacement between the centroid and the center of pressure, and y_c together. xDD
Random question that doesn't change the answer for the gate example, but when gravity is acting on the door, it would act straight down and wouldn't be normal to the surface like the resultant force is, so when you're calculating the moments about the hinge, when finding the moment created by the gate weight, would you multiply by Fdsin(theta), where in this case the theta would be 45 degrees?
Dr. Biddle at 51:40 in this video I had a hard time understanding how you got the area of the gate doing Pythagorean theorem when it’s a rectangular gate? Could you please just explain where you got your numbers from? Otherwise this is an outstanding lecture and the diagram you described is outstanding thank you!
Probably late comment, but he was finding the length of the gate. Remember the rectangular is tilted, and the length 3m is the projection, and not the actual length.
by assuming right triangle
sin45=p/h,
now sin45=3\h,
now h=3/sin45
final answer is 3underoot2
Thank you guys
@@farhanbadrkiani7259 answer doesn't match
Thanks from Brazil!
56:00 unaddressed mistake
length of the gate is not 4sqrt(2) it's 3sqrt(2), which visually changes the diagram quite a bit but also leads to the sum of moments at the hinge to be positive 105 KN*m as opposed to negative 41.8.. *so the gate actually falls*
Mufti Hossain I believe it’s because he’s adding the gate length plus the 1 m above the gate, which is correct, as the y(r) took that 1m above the gate into effect too, thus he has to subtract the total length, not just the gate length
there was no mistake, besides the one addressed in the description. Length was 3sqrt(2) in his calculation. You might confuse others
These videos are life savers
Just another day of saving lives for Professor Biddle. :)
why is ysub c a product of the two different lengths? at 53:00
not really, yc = hc/sin45 = hc/(sqrt2/2) = hc sqrt2
thanks a lot for the all videos
If only iit could give such clear lectures. Unfortunately iit only wants to eat money.
Wish my professors were as amazing as this professor is omg 😩
Hello:) which area moment of inertia formula did you use? I used bh^3/36, and i keep getting Ixc=7.1111
great video
Thanks!
At 54:40 why did you substract yR (3.96) from 4 square roots of 2? What is 4sqrt(2)?
Hello I love the way you teach professor but I did have a question, where did the 1.5 come from in the moment equation for the gate?
Thank you Professor.
You're welcome!
I have a question to the formula at 12:56. I understand the mathematical deviation, nontherless I am struggeling with the physical interpretation. According to this formula I can calculate the force on a given Area in dependence of the height of the centroid. Would that not suggest that it does not matter how deep the given area A is, since the hight is only relativ to hc? But that is clearly against the definition of pressure : dP/dz = -y
@54:41 Why is it 4*sqrt(2) and not 3*sqrt(2)? I thought we established it earlier that the length of the gate is 3*sqrt(2)...
You must account for the fact that the gate is still subject to the pressure at that additional depth (the additional 1 meter). In other words, y𝒸 is not measured from the centroid to the top of the gate, but rather, from the centroid all the way up to water surface. Therefore, the lineal distance from the hinge to the point at which 𝑭ʀ acts is the total distance from the hinge to the water surface (4√2) minus the lineal distance along the surface of the gate from the water surface to the point at which the resultant force acts (yʀ), or 4√2 - 3.96 = 1.70m. You would use 3√2 only if the top of the gate were at the water surface.
ruclips.net/video/YXnkB52x0Tk/видео.html
29:32 prof dodging the question of why X_R's formula is still using the y_c in the denominator of the second term like the student asked when in fact the first term is x_c (variable) 😅 by explaining that y_centerOfPressure > y_centroid 😂😂 Actually it's not from the force equation but it was derived from the Parallel Axis Theorem, sir.
which reference book was followed in this lecture
its really nice and helpful....also not boring
Glad we aren't boring.
I think at 42:15 the lect was meant to say Ixyc is zero instead of Ixc?
ruclips.net/video/YXnkB52x0Tk/видео.html
Sir as per my solution for that hinged gate problem the gate should fall,I was trying to upload my complete solution so that you can cross check and clarify my mistakes but can't find a way to do the same,please help
Isn't his y sub c incorrect. Shouldn't it be found using the equation h sub c times sin45
At 54:40 why did you use 4sqrt2 - 3.96 instead of 3sqrt2 - 3.96 ? The gate does not extend all the way up to include the 1m and the comments are unclear as to whether or not this was a mistake.
It was not a mistake. The gate does not extend all the way to the surface, true...but you must account for the fact that the gate is still subject to the *pressure* at that additional depth (the additional 1 meter). In other words, y𝒸 is not measured from the centroid to the top of the gate, but rather, from the centroid all the way up to water surface. Therefore, the lineal distance from the hinge to the point at which 𝑭ʀ acts is the total distance from the hinge to the water surface (4√2) minus the lineal distance along the surface of the gate from the water surface to the point at which the resultant force acts (yʀ), or 4√2 - 3.96 = 1.70m.
I understood the concept
Thanks prof
Great!
55:40, surely the perpendicular distance is used in moments so he should use the component of weight acting perpendicular to the gate instead of the total weight?
thanks
Since the gate is oriented 45 degrees, the gate has a horizontal projection of 3 m. The weight points downward, so the moment arm should be 1.5 m (half the projection).
I see why the moment arm is 1.5 for the weight of the plate, but why is the force of 90 used instead of the component of that force that is perpendicular? Does it not have to be the perpendicular force?
@@adababe27 90cos45 is the perpendicular component
hello
sorry for asking does anyone know how to download the solution manual of the fundamental of fluid mechanics book by munson 7th edition for free or has this book as pdf?
Namaste Sir !
from India
I have a doubt why u didn't integrated the length y just assumed Yc ?
at 51:34 , how is he getting sqrt(2) for sin(45) , shouldn't it be sqrt(2)/2 ?
he put 2.5xsqrt2 but it should be 2.5/sqrt2
anyone know how he got yc at 52:40
same qustion i have
@@emaadlabab1058 I figured it out it's 2.5/sin(45). Yc is measured from the surface. So you would add 1+3/2. That is one leg of the triangle. Then you would use the angle of 45 degrees and find the hypotenuse of the triangle.
@@shechoseboone7661 Thanks a bunch. yc is the hypotenuse, the vertical height is hc which we had previously determined (2.5).
What is the fluid source/book name, which is used by Dr. Biddle
Fundamentals of Fluid Mechanics by Bruce Munson
in the Fr equation why dont we have the gravity included ?
Thank you so much ❤
From Kerala?
I finished the 18 videos Can you please post the remaining up to 34
?
If angle goes to 0 does that mean F is 0 as well?
At 11:08 Sec , I didn't understand the value of Yc
It comes from the definition of the centroid.
Why is the hypotenuse taken to be 4sqrt(2)?? It should be 3sqrt(2) should it not? This seems like a glaring error.
Are you talking about at 51:40? If so, please look up the Pythagorean theorem to find the length of the gate.
3Sqrt2 +1 not equal 4Sqrt 2 broo!! @D28X
@D28X Thank you very much, it really helped.
hi please could someone tell me where the 4(root2) comes from at 54:54?
sin (45) = opposite (1+3=4) / hypotenuse. so the hypotenuse is equal to 4(root2). he needs the distance from the bottom (point O) to Fr so he subtracts Yr (3.96) from the total length of the gate (4sqrt2)
@@laserdancer454 but shouldnt opposite be just 3m as the gate doesnt extend into the 1m
@@locom16deen78 The gate doesn't extend into the 1m, but the surface of the gate is still subject to the pressure from the additional 1m of water depth. You must account that additional depth. In other words, y𝒸 is not measured from the centroid to the top of the gate, but rather, from the centroid all the way up to water surface. Therefore, the lineal distance from the hinge to the point at which 𝑭ʀ acts is the total distance from the hinge to the water surface (4√2) minus the lineal distance along the surface of the gate from the water surface to the point at which the resultant force acts (yʀ), or 4√2 - 3.96 = 1.70m. You would use 3√2 only if the top of the gate were at the water surface.
At minute 52:09, how to find the area?
I wish you stay long live and healthy
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Will the 30 degree angle be considered as 30 degrees with the horizontal, and does that mean the horizontal is the y-axis in this case
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efsanesin moruk
Thanks, professor
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why we don't zdd Patm ???
I can't understand how he gets area on 51.40 as 1 × 3_/2
55:26 sec.. weight isnt normal to the gate surface. how can that distance from hinge to the centroid be 1.5 for the torque calculation ?. isnt it should be [0 = (1.5 cos 45 * 90) - (1.7*104) ]?
-90kN(4.24m/2)cos45 degrees. you want the perpendicular component of weight on the gate to the hinge. 2 years late but someone else might have same question
@@radiatedracer3830:D great.. im in 3rd year now , following Electrical & Electronic Eng.
+This video series has been a great help for me to go through Fluid dynamics exam in 1st year.. so much ❤
@@maniys 1st year? im in mechanical engineering 3rd year taking fluids, next is heat transfer
We had fluid and thermo introduction course in 1st year before the field specialization happened in the year end. Thermo and fluid are very interesting subjects. Which country u are from.? I'm from 🇱🇰 Sri Lanka.
what page in the text is on centroids?
Why is I(xc) 6.36? I(xx)=bL^3/12 right?
he explains this in next video. the L is 3sqrt2 not just sqrt2.
@@mikerosell9081 thank you!
ty very much sir
is there any notes of this lecture
ruclips.net/video/YXnkB52x0Tk/видео.html
How did he get sqrt(2) @ 51:57 in?
ruclips.net/video/YXnkB52x0Tk/видео.html
.
and also when we took moments about point O at 55:01, what moment value are we solving for? and how do we know the gate stays in place from the value we obtained? Please any help is appreciated urgently
im not sure about the summation of moments being equal to 0. i think he finds the summation of moments to be - 41.8 it is pushing the gate up. im not sure though
we know its pushing up because the value is negative, and he chose the positive value to be clockwise
Mathematics should always be taught along with its use.
REALLY NICE THANKS A LOT
No problem.