ME3663 Fluid Statics 1
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- Опубликовано: 13 сен 2024
- Center of Pressure: 2:37
Vertical Surface: 5:36
Submerged Planar Surface: 11:09
Alternative Approach: 37:45
Submerged Planar Gate Example: 38:54
Submerged Curved Surface: 45:11
Curved Gate Example: 51:44
Mass and Weight Density Discussion: 1:03:59
Buoyancy & Archimedes' Principle: 1:12:15
submerged plane surfaces, center of pressure, buoyancy
My mind easily grasps your words!!! Appreciate your help.
You are just awesome. I am reviewing for the FE and your videos are making my life a lot easier. Your students are lucky to have you as a professor.
you are right
As a chemical engineering major, idk why we have to know this 🤷🏻♂️🤷🏻♂️🤷🏻♂️
Hi I am student of mechanical engineering from Korea, and I really appreciate at your clear lecture!! Thank you very much.
The best freakin video ever. I was about to give up on Fluid statics but this professor saved my ass. Love u professor
THANK YOU SO MUCH. really, my lecturer isn't all that great as he likes to just skim through the material without giving explanation for why we do the way we do it. he especially skips explanation for the advanced cases like the submerged planar surface after explaining the normal vertical scenario. i have a test today so i'm sure this will help me infinitely more. so thank you randall from 4 years ago.
thank you for clarifying concepts that can be hard to remember so clearly. Awesome video!
thank you for saving me from the past several hours of misery! impressive teaching.
True.
True.
@@lampa298 True.
Professor Randall ,daddy u re a legend! !!!..may God bless you abundantly I understood something that bothered me for a year!..billion thanks
You explained it so well!! Thank you!!!
I wish I had you as a Professor at Prairie View A&M... I've taken Fluids twice and still don't understand it at most but with your lectures so far, you've explained it very well including why we do certain process in a given example. I bet all your students exceled in your class. You are a God sent HYBRID instructor. With you and your lectures, I will succeed. Thank Remain Blessed
+Datguy obi thanks for positive feedback. Can you please edit your comment and delete name of University. I hope one day you will be a successful alumni of that University. We have all had a few difficult professors, as well as some excellent ones.
Thanks. Neither google nor hours of lecture notes and past papers could explain to me center of pressure as well as you did.
It is Very helpful. Thank you very much Professor Randall.
Very thankful, u put clarity to most concepts I had trouble with
your lecture was clear and straight to the point. thank you. I am studying for my PE in Texas.
Professor Randall, I really don't how to tell you that you're awesome. Way Too Awesome! Just Loved the lecture!
I'm glad you find the video helpful. Best wishes in all your studies.
Thank you soo much, also your video lessons are wonderful for PHD students like me.
thank you very much i have a test tomorrow in aerodynamics and this really helped
Very informative. Thank you
such a saviour! 👍. Thank you
cheers for the video, very clear explanations
Clear explaination! Thank you sir
Wow, this is good
Thanks to you I will pass the exam I have today !
your formula for the area of an equilateral triangle is wrong
He calls it equilateral, but labels it as if the "height" (measured at right angles to the vertical side) is equal to the length of the vertical side. If that's the case, then the triangle is not equilateral. His area formula is wrong for equilateral, but correct for the way he labeled the triangle.
What???? height of equilateral triangle with side b is Sqrt.3/2xb, that makes the area bsquared/4xsquare root 3. "base is equal to the height" how so??????
THANKS A LOT.
Good job sir. Very helpful!!
You said in the explanation that the triangle was EQUILATERAL, yet from the way of explanation and drawing, I think you mean Isosceles.
jffoz1 you are correct. thanks
Hi, lecturer,do you know how to solve the question？thank you
60 cubic metres of oil (RD=0.93) is poured into the tank. Calculate
(e) The depth of oil in the tank (2 marks)
60m³ of oil
RD=0.93
(f) The pressure at the oil-water interface. (2 marks)
(g) The pressure on the bottom of the tank. (2 marks)
(h) The total force in magnitude and position acting on the 6.0 m side. (10 marks)
(i) Draw and label a diagram to show the variation in pressure on the 6.0 m side of the tank after the addition of the oil. (2 marks)
There is a genius busy answering all these question that i dont know
In the first problem, why is it assumed the circular surface sinks to a depth of h1? Could it not float?
Gordon Ramsey if he taught Fluid Mechanics.
Hello Prof. Randall. At the 7:13, why you use (1/3).h from the bottom on the calculation? Why not at the centre of the red square?
There is a distributed force acting on the wall. You can represent this as a single force acting (1/3)h from the bottom.
So helpful! Thanks!
great explaine thank pro
Thanks for helping. Just a Question please. So what area dimensions do I use for calculating the Force on the vertical side? This is at about 11 minutes from start. Is it the width x 2/3h (Cp), width x 1/3h, width x h/2 or width x h?
Thanks, it might be useful for me
Thankyouverymuch
Guys pressure prisms are important for the fe exam.
Whos hibler for the book ?! sorry I just want to know how to spell the name so I can find it
Hibbeler
Thanks me too I was looking for the same book
Thank you sir
can you solve this to me?
You are a team of design group, involve in designing a flood prevention system of a river. Your assignment is to propose the best position of the water gate hinge so that during heavy raining, when water level reaches 10 m from river bottom the water gate will open automatically. The water gate size is 3 m wide and 9 m high.
thank you sir for the explanation ! :D
Thanks a lot ❤️sir
Quick question. On the problem (38min-45min), Shouldn't that free body diagram have a normal force where the stop is? I know that since the only equation you used here was the moment equation, vertical forces do not contribute for this case, but then if we had to solve for Ay we would have to use the 'sum of vertical forces equal zero' equation and the numbers will be different if we put the normal force.
at 40:00 horizontal force Fs is shown. the gate is assumed to "hang" from hinge at A. So don't need vertical force acting on bottom of gate. Could the gate rest on the ground? Yes. If it did rest on the ground, would there be vertical force at bottom? Yes.
thank you
thank you so much sir
So regardless of the weight of water above the surface (for instance it could be in a container 1 square foot at bottom but narrowing from the bottom) if a waterproof weighing scale were placed on the surface it would register the computed pressure rather than actual weight of water in the container. For instance, a container with 10 feet depth of water, 1 square feet of area at the bottom, but holds only 5 pounds of water would register on the weighing scale 624 (62.4 x 10) pounds of water rather than 5 pounds. Right?
+john zimmerman pressure of water at bottom of container is not the same as "pressure" or "Force/Area" between container and scale. Force between container and scale must equal weight of container and weight of fluid in container. Sorry if this doesn't help. But that is all the time I have to try and answer your question.
+Randall Manteufel Thank you for responding. However, you misunderstand because I was not real clear. I try again. It will be simpler to reword my question using containers. Consider a container in the shape of a cylinder 10 feet depth of water, 1 square foot horizontal area. A weigh scale is placed inside the container at the bottom. One would think it would register the weight of the liquid, 624 pounds,. because that is the weight of water. But actually what the scale registers is not a function of the weight of water but the force from the water which is a function of pressure not weight, though in this example the resultant force is the same. Now take a container, 1 square foot at the bottom but that narrows and it holds 10 feet of water. Lets say it holds only 5 pounds of water. What would a scale inside the container resting on the bottom, supporting the entire liquid register 5 pounds (the weight of the water) or 624 pounds according to the formula F = 62.4lb/cu ft x 10ft x 1 sq ft? (answer is 624 pounds - right?)
2:09 pi*d/4 or pi(d/2)^2
at 11:00 "no it dont! thats why im here watching this video!!!" says my study group
why is it that when calculating the weight of cylinder, the depth used is 5ft? And area of cylinder isn't pi*r^2 but 2ft*5ft?
+SageAdam length of cylinder in third dimension (not shown) is 5 ft. Hope that helps with second question as well..
+Randall Manteufel isn't it eg(h^2/2)
legend
STRAIGHT LEGENNNNNDDDDDDDDDDDDDD
I'm trying to think of something I wouldn't do for you to be my professor. So far no luck.
Does the method used during the last part of the vidoe work for plane surfaces instead on curved surfaces?
+Dimitri Dobroskok yes, I see no reason it wouldn't work.
Thank You!
It helped
please please reply sir at 4:22 resultant force cant be applied a little bit to the left because resultant force acts through the centroid of pressure prism and it is exactly at ( b/2 ) .
discussion is confusing at this point, sorry. An equilateral triangle is drawn and incorrectly I say it is isosceles (or at least imply it is isosceles). If length of each side of equilateral is "b" then distance from middle of edge to opposite corner is "b*sin(60)" (not b as I think I say). Centroid is at (1/3) of this distance, measured from each side toward the opposite corner. To help see this, draw equilateral triangle, then 3 lines from middle of each edge to opposite corner. All 3 lines will intersect at centroid. With ruler you can verify this is (1/3) from middle of "base" to "peak"
Sir irrespective of isosceles or equilateral triangle the centre of pressure must lie on the centre of pressure prism and in this case it will be exactly at ( b /2 ) because we have a horizontal surface.
And it shouldnt be at centroid of cross section as you have mentioned..please correct me if i am wrong sir
1:06:43 somehow I got sqrt(9092^2+8736^2)=12,608.8 lb
The answer of your letter a. is wrong professor you equate the wrong value of the FRH
17:48 What was that? A yawn?
İ was confused 🙋♂️
And now?
Keşke Türkçe olsaydı.
Keske degil mi? Sanirim ders kaydi bu..
hay bin kunduz
"Not too innovative"...
My mind easily grasps your words!!! Appreciate your help.