Fluid Mechanics: Forces on Submerged Surfaces I (3 of 34)

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)."

[@Mapexdrummer28]

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.

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/j4nRnHVrZ91jis0

[@randallmcgrath9345]

Holy fucking shit, that's sad. Did this think their research matters more than teaching?

[@MaryAngier]

My professors at my new college do this too! Its so hard to follow and stay engaged when they just flip though slides. Also, some problems are so long they are on multiple slides and it makes me dizzy when they flip back and forth to see the given values or original schematic.

[@NK-iy6if]

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!

[@lucasfelix8976]

these lectures are pure cinema

[@cindydiaz9945]

You Cal Poly students are lucky to have a good professor! Thanks for sharing

[@CPPMechEngTutorials]

Our pleasure. We have a lot more videos in production.

[@cindydiaz9945]

Thank you much!

[@Knicks3005]

Cindy Diaz 1

[@dianafregoso-sanchez9239]

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!

[@jaxxonleighton4105]

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!

[@angeloyehuda335]

@Jaxxon Leighton instablaster =)

[@jaxxonleighton4105]

@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.

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[@angeloyehuda335]

@Jaxxon Leighton Glad I could help :)

[@nathanblizzard4103]

SOOOOOOO much better than my current professor!! May actually survive Fluid Mechanics! Thanks for sharing CPPMechEngTutorials!!

[@BaldurKhr]

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 !!

[@KhenethBraganza]

how to get Ixc =6.36?

[@hiwazz]

Thank you very much! I was confused about the calculation, now I understood

[@BaldurKhr]

My pleasure,@@hiwazz

[@antoniuslouis00]

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 :)

[@joshuarillera2269]

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!

[@needraf1]

After a week of studying, watching your video finally made the concept clear. Thank you, professor.

[@NielKlauss]

I'm teaching physics to an architecture undergrad, his professors don't bother explaining any of the formulas! the moment of area explanations were of great help to me, thank you Dr Biddle!!!

[@rtrt2889]

Biddle was my favorite professor when I was at CPP, glad you have these videos up. Would have been helpful back in the day

[@alapi6180]

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

[@fatimaezzahrahilal6699]

i wish this professor can teach every engineering class till i get my degree he s awesome

[@CPPMechEngTutorials]

Thanks!

[@mohammadtalha6328]

Absolutely right

[@francielefaria92]

I am very thankful for this Professor. Saving my life while taking Fluid Mechanics.

[@ahmednasser2131]

can you help me to do my homework, if you understand this lecture??

[@Burrahboy]

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.

[@georgekaras7162]

Shout out to Cal Poly Pomona for releasing these lectures for public consumption!

[@arsalankhan9560]

I am very thankful to Professor . These online lectures are helping me in my academic.

[@CPPMechEngTutorials]

Great!

[@CyberVenomAenima]

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.

[@dimitrisnentidis797]

4 sqrt2 is correct as the triangle has two 4 meter sides thus the hypotenus is 4sqrt2 (explanation in next video)

[@nusretalikok823]

a good 10 minutes was spent thinking about this... thanks for the clarification! :)

[@jonathansanchez2854]

sorry why is it 3sqrt2/2

[@crankfrank3274]

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

[@CPPMechEngTutorials]

We're glad you found the videos helpful. :)

[@mushy6290]

I love these videos. They are saving my grades in my fluid mechanics class!

[@abhishek0o7]

Sir, you are awesome!!! Thanks for sharing such valuable knowledge on KZbin.. Respect from India 🇮🇳

[@CPPMechEngTutorials]

Thanks!

[@ahmedal-doori2704]

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 🙏

[@andyjunior]

at 52:41 how do you get yc(distance to centroid ) please explain

[@madhavgambhir2620]

Just one word: Beautiful! Thanks to this amazing professor, sir you have left an impression in my mind forever.

[@edwintan65]

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?

[@CPPMechEngTutorials]

Thanks. We made a correction in the comments.

[@g_rr_tt]

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 KZbin and the people who record/upload these lectures.

[@gbcolby]

Clear and informative, excellent instructor, thank you for posting

[@CPPMechEngTutorials]

No problem. Glad it was helpful.

[@mmhoss]

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*

[@NicholleWillisLoves]

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

[@ken-cf7tf]

there was no mistake, besides the one addressed in the description. Length was 3sqrt(2) in his calculation. You might confuse others

[@marcelogontinas8212]

one of the great professors i've ever known :) Thanks for this sir :)

[@ahmednasser2131]

can you help me to do my homework, if you understand this lecture??

[@varshneydevansh]

Will go through all the content from this channel I mistakenly did CSE but now I know what I wanted to do.❤

[@CPPMechEngTutorials]

We hope you enjoy the content.

[@johnstark6535]

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!

[@shaheenaldhumaid2459]

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.

[@farhanbadrkiani7259]

by assuming right triangle sin45=p/h, now sin45=3\h, now h=3/sin45 final answer is 3underoot2

[@johnstark6535]

Thank you guys

[@syedmuhammadtayyeb6994]

@@farhanbadrkiani7259 answer doesn't match

[@ojoduopeyemi8082]

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

[@NetSkillNavigator]

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

[@NetSkillNavigator]

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

[@CPPMechEngTutorials]

Thanks for the explanation. Unfortunately we don't have time to answer every question about the content.

[@dianafregoso-sanchez9239]

@@NetSkillNavigator Thank you for answering!

[@ignaciolafoz6314]

At 44:12 how come yR is independent of angle? If the angle were 0 shouldnt yR be = yC?

[@lollolzi2996]

at 52:42 why did he multiply y_c with sqrt of 2 ?

[@sebrina2892]

sqrt of 2 = ( 1/sin45 )

[@siten1]

@@sebrina2892 but sin(45) is sqrt(2)/2 and NOT just sqrt(2)

[@ken-cf7tf]

@@siten1 1/sqrt(2)/2 = sqrt(2). sine45 is in the denominator

[@ken-cf7tf]

@The Adel 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.

[@ken-cf7tf]

​@The Adel i think what you were trying to do was get sum of length of the door and additional inclined length from top of door to the free surface then divide by two, which is wrong (that would be lower than yc). yc is the inclined distance of centroid to the free surface. you get yc by dividing the vertical height of door by 2 ( 3/2 = 1.5) then divide by sin45 which would be equal to 1.5sqrt2. Add this to the inclined distance at the top and youll get (1.5sqrt2 + sqrt2 = 2.5sqrt2). Alternatively, a much easier solution is to just get the vertical height of centroid to the free surface (which is hc = 1.5 + 1 = 2.5), then get the inclined distance which would be 2.5sqrt2. What you were trying before would have worked if the top of the door touches the free surface like the other sample problem. Sorry if it took me a lot of words to explain.

[@farhanbadrkiani7259]

my fluid mech professor recommended me yours videos because you were professor of him too :)

[@Sismotutors]

Hello

[@Fawad.ANizamani]

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.

[@wilkenntuazon1279]

Thank you so much, professor. You explain better than my book. LOL, God bless and please keep on saving student's grades.

[@CPPMechEngTutorials]

:)

[@rabecamohammed9518]

@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)...

[@Vito_Tuxedo]

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.

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/j4nRnHVrZ91jis0

[@Zorac777]

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

[@milosmaric6412]

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

[@lisacui1704]

Thank you so much! I really appreciate these videos!! they helped me a lot for my fluids course ! Thank you !!!!!

[@CPPMechEngTutorials]

Hooray!

[@deandordevic35]

At 54:40 why did you substract yR (3.96) from 4 square roots of 2? What is 4sqrt(2)?

[@s5960]

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

[@CPPMechEngTutorials]

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).

[@adababe27]

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?

[@radiatedracer3830]

@@adababe27 90cos45 is the perpendicular component

[@estherchua5428]

This is such a life saver. THANK YOU.

[@CPPMechEngTutorials]

You're welcome.

[@irekhan6247]

best teacher ever

[@CPPMechEngTutorials]

Thanks

[@josebejarano4311]

why is ysub c a product of the two different lengths? at 53:00

[@ken-cf7tf]

not really, yc = hc/sin45 = hc/(sqrt2/2) = hc sqrt2

[@GyeonghunPyeon]

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.

[@TheHonchoofAllSpearheads]

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

[@kelvinvalentine2952]

Thanks much professor receive this greetings from Tanzania

[@maniys]

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) ]?

[@radiatedracer3830]

-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

[@maniys]

@@radiatedracer3830:D great.. im in 3rd year now , following Electrical & Electronic Eng.

[@maniys]

+This video series has been a great help for me to go through Fluid dynamics exam in 1st year.. so much ❤

[@radiatedracer3830]

@@maniys 1st year? im in mechanical engineering 3rd year taking fluids, next is heat transfer

[@maniys]

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.

[@laurengalus4289]

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.

[@Vito_Tuxedo]

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.

[@alexfontaine6233]

Thank you John biddle you saved me!

[@CPPMechEngTutorials]

That's what he does.

[@guanhan898]

I think at 42:15 the lect was meant to say Ixyc is zero instead of Ixc?

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/j4nRnHVrZ91jis0

[@leeshanpillay1253]

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

[@laserdancer454]

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

[@laserdancer454]

we know its pushing up because the value is negative, and he chose the positive value to be clockwise

[@barronhoffman2000]

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).

[@joonfanatic2140]

Wish my professors were as amazing as this professor is omg 😩

[@biswaacharya8689]

he knows how engineering practically works

[@siten1]

at 51:34 , how is he getting sqrt(2) for sin(45) , shouldn't it be sqrt(2)/2 ?

[@debunkthejunk1]

he put 2.5xsqrt2 but it should be 2.5/sqrt2

[@kashkot]

perfect ,I really like the way you teach

[@leeshanpillay1253]

hi please could someone tell me where the 4(root2) comes from at 54:54?

[@laserdancer454]

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)

[@locom16deen78]

@@laserdancer454 but shouldnt opposite be just 3m as the gate doesnt extend into the 1m

[@Vito_Tuxedo]

@@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.

[@TheHonchoofAllSpearheads]

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.

[@alejandroquintero7999]

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?

[@mazphamm]

These videos are life savers

[@CPPMechEngTutorials]

Just another day of saving lives for Professor Biddle. :)

[@omarlee3095]

This was such a great lecture

[@shaahzweenyeldiablo9363]

At minute 52:09, how to find the area?

[@KiraVez]

Great video, but a mistake was made 55:48. The 1.7m represents how far down the gate the force acts, but for finding the torque you should take the 1.3m since the hinge is located at the bottom of the gate. The term for the mass of the gate as well should include a cos(theta) factor in the torque.

[@radiatedracer3830]

1.7m is how far the perpendicular force on the gate acts on the hinge

[@kunal.9587]

If only iit could give such clear lectures. Unfortunately iit only wants to eat money.

[@mezaantonio09]

At 54:42, where did 4*sqrt(2) come from? I am a little confused.

[@varunsubramanian5484]

yeah it should be 3 root 2

[@varunsubramanian5484]

@The Adel so I guess it depends on the question?

[@varunsubramanian5484]

@The Adel Oh i see ! Thanks !

[@Fillipe_Mendes]

Thanks from Brazil!

[@naishethelovebirds5145]

The 90 KN( weight ) of the gate when we are taking moments ....shouldnt it be perpendicular to the gate ....or

[@NicholleWillisLoves]

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.

[@prathikshenselvadorai2625]

@@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...!

[@teddybeyrouthy4995]

i didnt understand why he wrote at the end , the sum of moments =0 then he made the sum and turned out = -41.8. can you explain it please (55:30)

[@radiatedracer3830]

draw the components of the weight, you will find one is parallel, one is perpendicular to the gate hinge. You want the perpendicular one. Mhinge = -90kN(4.24m/2)cos45 + 104.1 kN(1.7m) = 42.05 knM. its positive and ccw is positive so gate stays up.

[@Vito_Tuxedo]

The "sum of the moments = 0" is just a convenience to make it easier to solve the problem. It's a test condition you apply in order to calculate the net force on the gate. In other words, if you assume the gate is not moving you should have a net force in excess of the weight, acting in the opposite direction. If you take the downward force of the weight as positive, then any (non-zero) amount of negative force in excess of the weight will be a net force acting to hold the gate up.

[@jonathansanchez2854]

where does the 4.24/2 come from? @@radiatedracer3830

[@dianazamudio4249]

Hello:) which area moment of inertia formula did you use? I used bh^3/36, and i keep getting Ixc=7.1111

[@syedmuhammadtayyeb6994]

Quality Level A+++

[@MohitSharma-qh4wq]

At 11:08 Sec , I didn't understand the value of Yc

[@CPPMechEngTutorials]

It comes from the definition of the centroid.

[@jonathansanchez2854]

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?

[@xhaven80]

How did he get sqrt(2) @ 51:57 in?

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/j4nRnHVrZ91jis0 .

[@13julietririssukmahadi68]

I didn't understand about moment of inertia. So, what should i do?

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/j4nRnHVrZ91jis0

[@Amirhossein-xi3vl]

thanks a lot for the all videos

[@MakandaPhilbert]

May God bless you sir

[@shechoseboone7661]

anyone know how he got yc at 52:40

[@emaadlabab1058]

same qustion i have

[@shechoseboone7661]

@@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.

[@dianafregoso-sanchez9239]

@@shechoseboone7661 Thanks a bunch. yc is the hypotenuse, the vertical height is hc which we had previously determined (2.5).

[@lilianealameddine2159]

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?

[@mahdemolan]

What is the fluid source/book name, which is used by Dr. Biddle

[@laurengalus4289]

Fundamentals of Fluid Mechanics by Bruce Munson

[@ajsh-h6m]

what that mean Fig 2.18 on 31:35

[@fun420-uv5zm]

which reference book was followed in this lecture

[@ahmadelbliwi5626]

I understood the concept Thanks prof

[@CPPMechEngTutorials]

Great!

[@RahulKumar-bx7my]

Namaste Sir ! from India I have a doubt why u didn't integrated the length y just assumed Yc ?

[@mohamedabdelkader6527]

I finished the 18 videos Can you please post the remaining up to 34 ?

[@arjungovender3248]

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

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/j4nRnHVrZ91jis0

[@nishantthaker4379]

its really nice and helpful....also not boring

[@CPPMechEngTutorials]

Glad we aren't boring.

[@pranjulsrivastava5903]

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

[@Fawad.ANizamani]

Mathematics should always be taught along with its use.

[@eloyrodriguez2617]

Isn't his y sub c incorrect. Shouldn't it be found using the equation h sub c times sin45

[@mohammedanas5455]

in the Fr equation why dont we have the gravity included ?

[@MrTik-ex2oh]

I wish you stay long live and healthy

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/j4nRnHVrZ91jis0

[@hxdistross8085]

Hardest lecture of my life , perhaps coz I’m not sober

[@funtimes6797]

I can't understand how he gets area on 51.40 as 1 × 3_/2

[@guybrushwayne4388]

Why is the hypotenuse taken to be 4sqrt(2)?? It should be 3sqrt(2) should it not? This seems like a glaring error.

[@CPPMechEngTutorials]

Are you talking about at 51:40? If so, please look up the Pythagorean theorem to find the length of the gate.

[@OmarAhmed-wv4nn]

3Sqrt2 +1 not equal 4Sqrt 2 broo!! @D28X

[@CLDJendis66]

@D28X Thank you very much, it really helped.

[@kjca7890]

If angle goes to 0 does that mean F is 0 as well?

[@SHIELDSPACES]

Thank you Professor.

[@CPPMechEngTutorials]

You're welcome!