Fluid Mechanics: Forces on Submerged Surfaces II (4 of 34)

0:00:09 - Revisiting example problem from last lecture (submerged rectangular surface)
0:09:18 - Example: Resultant force due to layered liquids
0:19:05 - Example: Resultant force on submerged rectangular surfaces; pressure prism
0:40:18 - Resultant force on submerged curved surfaces
0:56:10 - Example: Resultant force on a submerged curved surface
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)."

[@revo1835]

Would like to add, 8 Years later, these are still helping me in my fluid mechanics class, thank you so much!

[@CPPMechEngTutorials]

The internet is amazing!

[@Moi_12344]

@@CPPMechEngTutorials here too

[@jorgecancio7554]

it's so easy when someone passionate about the subject teaches you.

[@georgegithinji935]

Facts!!

[@151kkk]

I got the concept clearly thanks to you. I couldn't do the homework assignment with just my professors notes.Thank you so much! Regards, A last-minute student

[@CPPMechEngTutorials]

Your welcome.

[@AqworldsAsh]

you're*

[@151kkk]

man, I tapped your notification and the video started in the middle of the class with full volume. you're or your, I understood what he meant and that was 4 MONTHS AGO!

[@AqworldsAsh]

Isik Okur tbh I just did it as a troll😂

[@151kkk]

If there is a consolation, professor told me to silence my phone and I got looks from my classmates.

[@penguiz23]

Error at 1:10:00 , Fv adds to 588 lb - correct value was used in calculation tho. Thanks for the lecture u a real one ong

[@nicolasbatistelribeiro7400]

From Brazil, I'm very grateful for those videos. It's gonna save me for my test on monday.

[@CPPMechEngTutorials]

Hooray!

[@hayley9546]

Dr. Biddle, you are helping me so much, thank you! Watching your videos makes everything seem so easy whereas my professor confuses me. I hope whoever runs this channel lets him know how many students he has helped!!

[@dianafregoso-sanchez9239]

Thank you for the detailed example for curved surfaces!

[@utkarshpuri3739]

Where will the net force act? I mean coordinates of application of net force

[@edutechguruengineeringlear4280]

@@utkarshpuri3739 kzbin.info/www/bejne/m6LOgaFue9irfpI

[@mahvishsyed5823]

why couldn't he be my fluids teacher :(...life is so unfair!!! BAAAHHH!!!!

[@CPPMechEngTutorials]

Well, at least you live in a time where you can watch instructors who resonate with you better.

[@ΔιονυσιοςΑσημακοπουλος-ζ2λ]

These lectures covers my missing clases. Thank you!

[@qwerty-gc8tn]

Ti spoudazeis?

[@prabakaran451]

i want to correct respected Prof that Frc should be in opposite direction due to water pressure acting against the atm presure. Tq sir for the wondwrful lecture which i appreciate so much

[@pedrorodriguez9920]

I'm studying for the PE exam and I find these videos very helpful to recap. thanks a lot.

[@charnettadenae]

Were these videos helpful during PE? I am taking PE next year

[@comicalastranomical2780]

Phenomenal professor. absolutely would apologize to you if I failed anything you taught.. You’re so clear about explanations for solutions and what is expected of the student. It is an absolute nuisance some professors almost make you feel like you’re in a guessing game of what’s needed in a project, what’s needed for clearest solutions for homework/exams… Anyways, great professor. Thank you.

[@CPPMechEngTutorials]

Thanks!

[@JessieEason-s9f]

Thank you Dr.Biddle!

[@thembalethusoka3541]

at 2:13 sir how do you came out with sqrt of 2?

[@muhammedeminural3333]

hypotenuse, 45°

[@ken-cf7tf]

sin45 = sqrt(2)/2 , 4/sin(45) = 4/(sqrt(2)/2) = 4sqrt(2)

[@linnhtet561]

45 degree triangle C^2=a^2+b^2 , since a = b because same length. So c^2=2a^2 c=a*sqrt (2) C= hypotenuse sin(45)= a/a*sqrt(2) sin(45)= sqrt(2)

[@manuelsanchez9782]

It was very helpful when the homework assignments were on the board as they were in the first video. It helped to follow along the lecture as if you were in it. Any way to get the problem sets?

[@CPPMechEngTutorials]

I will check. If it is possible, they will be posted on fluid mechanics section of ME Online (www.cpp.edu/meonline).

[@bhaskarbhattacharya4662]

He is a wonderful Teacher....Heartiest thanks to CPP for uploading such educational videos.....:)

[@CPPMechEngTutorials]

We're glad they are helping people around the world. :)

[@rexylem]

Curved surface started @43:30

[@CPPMechEngTutorials]

Thanks. We have added timestamps to all fluid mechanics videos.

[@christinejewelsilvano2245]

This is awesome, thank you.

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/m6LOgaFue9irfpI

[@ejminava407]

I don't understand where he keeps pulling that square root of 2 from

[@kyleoconnell1235]

If the angle of the triangle is 45 degrees you can find the length of the hypotenuse by multiplying square root of two with the length of the sides

[@jeffdvr6]

the king of zoom university

[@ahmadmosbah5520]

Can someone please explain how can we find the direction of the force in the curved surface example (the cylinder) ?

[@sreekumaran8534]

sir thank you so much it helped me a lot ...............i mean i learned lot

[@brookep6729]

time: 1:10 --> how can u tell when Fv is pointing down versus up? is it just top half of circle is down and bottom half is up? is that always true or just in this instance?

[@jonathansanchez2854]

question professor, where did the 562 come from for Frbc

[@makijulio1920]

Great vid men Thanks alot!!!!

[@CPPMechEngTutorials]

No problem.

[@rafeza1]

where the 562 at 18:29 comes from?

[@Evan-rl1rn]

(Specific weight of water)(height from surface to centroid of BC)(Area)=(62.4)(4+5)(A)

[@simong1666]

I just noticed the cylinder buoyancy problem for Fv can be solved by replacing the 'submerged' parts of the cylinder with a static water volume. Since the volume would not move, the buoyancy force is equal to the weight of water displaced, 1/2 cylinder volume on the left, and 1/4 cylinder volume on the right. Net Fv is indeed 3/4 * Vol_cylinder * gamma_water

[@aldrinadriano8036]

He helped me since the statics ti dyanmics to thermodynamics to heat transfer and now I am taking Fluid Mechanics, he was able to guide me although out, and thank you for that.

[@nicoleng4181]

I barely even learn anything from class and this video literally helped me so much .

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/m6LOgaFue9irfpI

[@kamil.lionaire1321]

i really enjoy the corse am a 3rd year mechanical engineering student from the university of nairobi am yet to take the unit this semester but with your online help am way ahead of my prof at least i will be paying attention to the workings because i have your notes

[@CPPMechEngTutorials]

It's always good to be ahead.

[@locom16deen78]

you learn this in 3rd year?? Im from the uk and we learn this in year 1...yikes

[@fatimaezzahrahilal6699]

why don't you multiply by the breath for the vertical forces?

[@paolatbk3907]

Dr Biddle you are the best. Im studying civil engineering in Greece .and you have been so helpful.thank you!!!!!!

[@CPPMechEngTutorials]

No problem!!!!!!

[@A.Hisham86]

Hi, again, I didn't get your answer on my last comment, so I'll ask again: " the resulting force you'll find using the prism method is in N/M, so you have to multiply it by the wide, to get the force Fr. 31:52

[@harmanaulakh7529]

For the layered liquids problem, does anyone else get different numbers? I am doing 1500*9.81*9*10 sqrt2

[@muhammedeminural3333]

ADOPT ME

[@joonfanatic2140]

So amazing, thanks for these videos! 🥺💗

[@Evan-rl1rn]

28:53 magic!

[@CPPMechEngTutorials]

Apparently Professor Biddle is also a magician.

[@lokahnac]

Why did professor draw Fr in the oposite direction from the previous lecture ? at 6:26

[@CPPMechEngTutorials]

It looks like he accidentally put the arrow in the wrong direction. The students should have corrected him. :)

[@lokahnac]

Ok, thanks :)

[@shadesofalpha5366]

how does square root of 2 comes in Yc

[@jessecallanan1506]

since it's a 45 degree angle the bottom and sides are equal, so the hypotenuse is sqrt( h^2 + h^2 ) = h*sqrt(2) = y_c

[@reidelleman7641]

Yc is measured along the path of the plate to where it intersects the free surface. So even though the depth is 2.5m Yc is a little linger since its at angle 45

[@Krishna-t9v9n]

From Kathmandu University(Nepal)

[@dvsalim09]

Can anyone tell where the sqrt2 comes from in 2:09? or is it missing a 1/2?

[@jmmifsud1]

Students at this university cannot tell the time !!

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/m6LOgaFue9irfpI . .

[@ieurir]

Fr. How can do many people stroll in late like that??? So distracting 🙄 and rude!

[@حسينعائدخشانالخزاعي]

What is the source name

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/m6LOgaFue9irfpI ..

[@kupferknochen]

快去休息！！😉🤭

[@majroi]

I listened this subject for various sources but none of them helped me(Vertical force). But when I watched this video I said wow I really understood this. Thanks proffesor, you are really a good teacher.

[@A.Hisham86]

But when using the pressure prism method you don't get the same result of YsubR as you found it using the equation method.

[@brookep6729]

can someone explain the difference between h and hc? h is the length of the surface (can be a diagonal). while hc is vertical difference from the free surface to the centroid of interest. can someone confirm this?

[@julialongo]

this is amazing ... I actually understand everything wow

[@CPPMechEngTutorials]

Terrific!

[@leeshanpillay1253]

if we are using the full length of 4m in height for the very first example, why is the distance that the weight makes with the hinge 1.5 and not 2?

[@Mo-ti8ym]

i think u r graduated now and u dont need the answer

[@edutechguruengineeringlear4280]

@@Mo-ti8ym kzbin.info/www/bejne/m6LOgaFue9irfpI

[@hamzaullahkhan8602]

from where 10 sqrt(2) came as height at 22:56

[@gulnarmikayilli7008]

it is not the height the prof. talked about through concept . It is one side of the plane surface we have chosen which we are working on it . actually it is length (from side view) of the figure which you see as a gate. from top view it is a rectang. and 10 sqrt 2 is the height of that rect

[@georgespir1569]

the hypotenuse length is what we are looking for, so lets name it x ... this is how you find it... sin45° = √2/2 (verical)/(hypotenuse) = sin45° = √2/2 h = 10 ft h/x = √2/2 => x = (h * 2)/√2 =(10*2 *√2 )/(√2*√2) = (20√2)/2 = 10√2

[@georgespir1569]

or you can think of it as Pythagorean, 2 equal sides of 10 ft, the hypotenuse is √(10^2+10^2)=10√2

[@sd89]

How does one get Yc ??? Thought otherwise wow already understanding this better by leaps and bounds!!!

[@MisterBinx]

The rule for vertical forces says fluids above the curved surface. The log is not a fluid though. I understand the imaginary surface but there is no actual fluid being occupied by that space. The rule doesn't say that the fluid can be imaginary only that the free surface can be.

[@kemrankarimov2399]

questions and corrections of the students are extremely silly and thoughtless 🤦🏻‍♂️

[@CPPMechEngTutorials]

In class or in the comments?

[@CPPMechEngTutorials]

@@ThaBlackBuddha :)

[@elliottseiler4828]

31:23 Casual Ebay notification

[@pritamprem1761]

Superb sir.I cleared my concept with the help of this.Thanku

[@CPPMechEngTutorials]

You're welcome.

[@noname-ic5mx]

he mentioned heavy derivation at 43:13, can someone tell me about that please? or point me to a reference pleaaaaaaaaaase, i really want to know.

[@CPPMechEngTutorials]

If you want to calculate the resultant force on a curved surface without going through the procedure shown in the lecture, you will need to utilize integrals and know how the curvature changes in space. It would be very tedious and you would not be able to obtain a general algebraic equation that can be applied to all curved surfaces. Most undergrad fluid mechanics books follow the procedure in the lecture.

[@NetSkillNavigator]

How can I use integral when getting bar h and hydrostatic force..? thnx

[@CPPMechEngTutorials]

You would need to know the orientation angle at all parts of the curved surface. This often will be difficult except for very simple geometries.

[@Moi_12344]

@@CPPMechEngTutorials Can we have a reference book pleaase siir 🙏

[@barbaroskucukpinar4125]

adamsınn lan dayi

[@saimsiddiqui1544]

where does this sqrt(2) come??

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/m6LOgaFue9irfpI ..

[@ilterkaraca3588]

allah senden razi olsun buyuk adamsin vesselam

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/m6LOgaFue9irfpI

[@ahmadayman4138]

What will be the difference if there was a different fluid in the upper half between points 1 and 2!? Would this fluid affect the vertical force between 2 and 3!?

[@utkarshpuri3739]

Can someone tell, the coordinates of point of application of the force

[@chinesemediabox3823]

How did you get 4ft on 16:16? Its top height 5 ft over there

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/m6LOgaFue9irfpI

[@rtrt2889]

Best professor!

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/m6LOgaFue9irfpI

[@danieljesusmagorodriguez6302]

MIN 18:40 WHERE DOES THE 562 COMES FROM?

[@danieljesusmagorodriguez6302]

NVM

[@zulkarnaim6697]

21:52 where did 62.4 come from

[@13julietririssukmahadi68]

I need subtitle, please :"(

[@gurleenbhatia8095]

When do we use Fr=Gamma *hc*A VS. Fr=Gamma*hc*A*sin(theta) ? Whats the difference again"?

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/m6LOgaFue9irfpI . ..

[@Destruct305]

thank you

[@Uobgraduationproject]

what does the doctor means by using 10 square root instead of using sin(45) to find Yc ?

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/m6LOgaFue9irfpI

[@Blue-gaming141]

kinda confused about that p=2(Yw) can someone explain more

[@enesyalcn3465]

What do you mean about p and Yw

[@motunbuwa9252]

My professor did the same example in class and I did not even understand anything. Thank you soo much

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/m6LOgaFue9irfpI

[@zenapsgas]

55:40 Good lecture, but are you over-complicating it with the vertical forces? You could just sum up the volume of the cylinder that IS in the water and get the same Fv result? (In other words, you sum up the volume of the water being pushed away by the cylinder. In this case, 75% of the volume of the cylinder is being pushed away.)

[@ashleyhallromero4336]

What i dont understand from the problem at the beginning is that both the Fr and the weight are in the same direction, therefore, the momentum would change. they would be both in the same direction, therefore, the gate would fall. Someone please help.

[@locom16deen78]

9 months old i know but to help others who may have the same problem....He has put the Fr in the wrong direction its the opposite way.

[@roonyyoouness7115]

Best Dr

[@GyeonghunPyeon]

가장 이해가 안 되는 부분이 바로 이 부분이었습니다. 일반적인 상황으로는 혹은 가장 간단한 상황으로는 직선으로 유체안에 들어가있는 물체를 상상하지만 그렇지 않은 경우에 대해서 고민을 많이 했었습니다. 압력개념으로 잘 설명해주신 것 같아 고맙습니다.

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/m6LOgaFue9irfpI

[@SeekTheTruthAndTruth]

Sir , why are you calculating area instead of volume while you solved first example on curved surfaces for vertical forces calculations ?

[@abhishek0o7]

Aqif bhat bcoz length of the cylinder is 1 ft only.

[@sahilrajeshgajbhiye4968]

ty very much sir

[@sarsharjohari8466]

God , I wish I knew in which university you are teaching sir. You are great Teacher. If I did not have your lectures, I couldn't do any of my hw problems. thank you so much.

[@CPPMechEngTutorials]

You're welcome. We are glad you enjoyed the lectures.

[@Jenipati]

California State Polytechnic University, Pomona, USA

[@BODYBUILDERS_AGAINST_FEMINISM]

hello sir, thank you for posting this, I do have a question however. Multiple times you referred to sin(45) as "square root of 2", wouldnt it be (sqrt2)/2? example 22:22, you say the length of BC is 10sqrt2, wouldnt it be 5sqrt2? Thank you again sir

[@BODYBUILDERS_AGAINST_FEMINISM]

just to be a little more clear, at 22:22, BC would be 10sin45 right? so 10sin45 is 5sqrt2 not 10sqrt2. I know im mistaken I was just hoping someone might be able to clear it up.

[@darrenmallare4057]

@@BODYBUILDERS_AGAINST_FEMINISM it would be 10/sin45 because it's the hypotenuse. That's why it's 10sqrt2

[@edutechguruengineeringlear4280]

@@BODYBUILDERS_AGAINST_FEMINISM kzbin.info/www/bejne/m6LOgaFue9irfpI

[@edutechguruengineeringlear4280]

@@darrenmallare4057 kzbin.info/www/bejne/m6LOgaFue9irfpI

[@hishammohamed984]

thnx alottttttttttttttttttttttttttttttt

[@christinejewelsilvano2245]

I tried replacing oil with water by also changing the depth that it's equivalent to, as said in the video. The final answer wasn't accurate though, where did I go wrong?

[@alexdejesus6342]

same here...curious where the 562 came from

[@angelikadetalla8787]

did you get 566A also?

[@edutechguruengineeringlear4280]

@@alexdejesus6342 kzbin.info/www/bejne/m6LOgaFue9irfpI

[@edutechguruengineeringlear4280]

@@angelikadetalla8787 kzbin.info/www/bejne/m6LOgaFue9irfpI

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/m6LOgaFue9irfpI . .

[@hussienhatem2711]

why hc is not equal to 9 + 1/2*sin45 ?

[@hussienhatem2711]

because hc is the distance from the free surface to centroid and it is 1 meter wide and inclined by angle of 45 degrees

[@hussienhatem2711]

@22:52

[@finnafitriana9148]

are there any video lectures of hydraulic classes?

[@CPPMechEngTutorials]

At Cal Poly Pomona, the Civil Engineering Dept teaches hydraulics.

[@finnafitriana9148]

CPPMechEngTutorials but does civil dept have youtube channel? i couldn't find it. sorry and thank u before.

[@CPPMechEngTutorials]

Not that we are aware of.

[@ems540]

thank you so much

[@CPPMechEngTutorials]

You're welcome.

[@13julietririssukmahadi68]

I need subtitle, please :"(

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/m6LOgaFue9irfpI

[@daniyarotynshin]

1:09:08 mistake, over 2 is correct one!

[@bilalmahmoud9183]

over 4 is correct, he calculates the upper square as r^2 and the lower quarter of the circle as pi*r^2/4

[@ericelegado1677]

I love the late comers!

[@CPPMechEngTutorials]

Immortalized on a KZbin video.

[@alulanega3828]

The bad professor I have ever seen