Thank you so much for your videos. My professor does not give us examples in class and his hand written notes/diagrams are hard to understand. Your stuff is super straightforward with organized diagrams and I greatly appreciate your work
Amazing video, something that seems so weird and complicated made simple and intuitive. You are so good at explaining the theory and “how to attack” a problem in a easy to understand way. Please post more similar content, thank you! =)
Dear sir, at 11:55 you say that there is a uniform distribution load and therefore the Fbc acting on R/2. Why is that? Isn't the load distribution also acting in a circular way? which implies the Fbc from C is 4R/3Pi ?
The load on BC is NOT "acting in a circular way". Surface BC is a plane (flat) surface that is at a constant depth of R. The pressure vectors on this surface are uniform (gamma*R) and all point upward. This uniform pressure distribution is replaced by the total force at the center of the pressure distribution (at R/2). I hope that helps.
The main reason is so students can focus on fluid mechanics. Including in the gate weight is trivial statics, which adds unhelpful complexity when it comes to learning hydrostatic analysis. Also it will be generally true that the gate weight will be quite low compared to the hydrostatic forces. You can easily check this -- assume a 0.5 inch curved steel plate, for example.
Hello! I'm sorry I am a little confused about F_BC, what exactly is this force. Is it the reactant force due to F_A. Furthermore, why does it act at the center of BC and not at C if it is the reactant force from F_BC. I know this has to due to the depth that it is at, but I am a little confused why it is at the center and if this is the equal and opposite reaction of F_A. Thank you!
F_BC is not a reaction to F_A. Surface BC is at a depth R. So there will be a uniform pressure (upward) on this surface of our control volume. Since the pressure is uniform (=gamma*R), it will act at the center of BC. (Alternately, if you use theta 90 degrees, you get the y_cp=0 for a horizontal surface.) I hope that helps.
Thanks a lot for your explanation Sir. I am stil confused when we put the value of gate's weight and in the book they said the centroid of weight is 2R/pi. Could you help me please
2R/pi is the centroid location of a quarter ring, with all the mass (idealized as) at the same radius. The centroid of a solid quarter circle is different, of course.
@@FluidMatters I have reviewed the method to define the centroid of volume, area, and line also Sir. Your support has ledluidMatters I have reviewed the method to define the centroid of volume, area, and line also Sir. Your support has led me to much new knowledge. Thanks a lot Sir.
how is the F bc force not cancelled out by the weight force of the water? Im trying to wrap my head around how there is left over force when you take the weight away. So there is force down because of the weight, and also force down because of the fluid pressure, but isnt the fluid pressure just because of the weight?
OOHH the height x Area for the fluid pressure is different from the volume of the actual water! so there is a discrepancy there! Equal force on horizontal, so it is like a square of hydrostatic pressure. Thanks
no. F_BC is the a fluid weight. F_BC is the upward pressure force on plane surface BC. The entire surface BC is a depth R. The pressure at this depth is rho*g*R, which gets multiplied by the surface area. I hope that helps.
Surface BC is at depth R in the water. So, there will be hydrostatic pressure at this depth, equal to rho*g*R. That's how we know there is a force F_BC. Note that the weight of the water is downward and the force of the water on the gate is clearly upward, as seen from the pressure distribution (at 1:08). So, F_v cannot be equal to the weight of the water. I hope that helps.
@@HashemAljifri515 Not "due to the ground". It's the pressure in the water at that depth (at the ground-water interface). Local pressure acts in all directions. Once you define a control volume (as done in this problem) the pressure force acts inward and normal to the boundary (see video 1, Chapter 2) --- hence upward in this case.
Thank you for the explanation, Sir. I have a question, when you calculate the X bar, you take CW as (-) and CCW as (+). However, when you did the sum of moment at B you did the other way around. Why is it?
Here's the theory lecture for hydrostatic forces on curved surfaces: th-cam.com/video/LdbEpRXUpOQ/w-d-xo.htmlsi=DoySt3zv_nYTipHn You can find all the lectures as www.drdavidnaylor.net
Dear Sir, why do you count Fbc pressure force.? Usually, I saw teachers concern about the weight of the water up to free surface to find vertical force. I only find in your all of the video you count vertical pressure force, why sir?
You can use the missing "water approach". It is equivalent. I show this at the very end of the solution (about 15:00). But it can be tricking to find the line of action,. The method in this video is the most general approach. I hope that helps.
Note that F_H acts both to the left and to the right at the same horizontal location. Hence, no net moment. See the note in the box at the bottom right side of the slide. I also say it at 12:17.
Thank you so much for your videos. My professor does not give us examples in class and his hand written notes/diagrams are hard to understand. Your stuff is super straightforward with organized diagrams and I greatly appreciate your work
Glad to help. Best of luck with your studies.
i have been suffering with a single question in my book for an hour... Your video made me understand everything, thank you so so much!!
Glad to help. The video is old, but it's still a "active" in my online course. Best to luck with fluid mechanics.
I love you dont use the concept of imaginary water. such a big help sir 🎉
Amazing video, something that seems so weird and complicated made simple and intuitive. You are so good at explaining the theory and “how to attack” a problem in a easy to understand way. Please post more similar content, thank you! =)
Thanks. Glad to hear it was helpful.
Awesome and detailed solution. Thanks Dr. David
Glad to hear it was helpful. Thanks for the kind words.
Thank youuu Sirr , your way of explanation is tooo good..
Thank you so much. Glad it was helpful.
thanks dr.david for providing the solution of fluid mechanics as pdf on your website
Thanks Dr.David i understand the topic well now after watching your Video on it greetings from Germany.
Glad to help!
Thank you so much for this example, helped immensely
9:19 how is Ycp=-1.333m didn't you take your origine from the surface?
y_cp is measured downward from the centroid. See my intro video: th-cam.com/video/pIK_ywo10Qc/w-d-xo.htmlsi=_uR8Ucdri3isyM0b
Dear sir, at 11:55 you say that there is a uniform distribution load and therefore the Fbc acting on R/2. Why is that? Isn't the load distribution also acting in a circular way? which implies the Fbc from C is 4R/3Pi ?
The load on BC is NOT "acting in a circular way". Surface BC is a plane (flat) surface that is at a constant depth of R. The pressure vectors on this surface are uniform (gamma*R) and all point upward. This uniform pressure distribution is replaced by the total force at the center of the pressure distribution (at R/2). I hope that helps.
thank you for the video! really informative.
But, what about the weight of the gate, is it negligeable in the calculus? and why?
The main reason is so students can focus on fluid mechanics. Including in the gate weight is trivial statics, which adds unhelpful complexity when it comes to learning hydrostatic analysis. Also it will be generally true that the gate weight will be quite low compared to the hydrostatic forces. You can easily check this -- assume a 0.5 inch curved steel plate, for example.
Hello! I'm sorry I am a little confused about F_BC, what exactly is this force. Is it the reactant force due to F_A. Furthermore, why does it act at the center of BC and not at C if it is the reactant force from F_BC. I know this has to due to the depth that it is at, but I am a little confused why it is at the center and if this is the equal and opposite reaction of F_A. Thank you!
F_BC is not a reaction to F_A. Surface BC is at a depth R. So there will be a uniform pressure (upward) on this surface of our control volume. Since the pressure is uniform (=gamma*R), it will act at the center of BC. (Alternately, if you use theta 90 degrees, you get the y_cp=0 for a horizontal surface.) I hope that helps.
@@FluidMatters that does help, thank you so much. Shows how much you care to help people responding to a 2 year old video still :))
Thanks a lot for your explanation Sir. I am stil confused when we put the value of gate's weight and in the book they said the centroid of weight is 2R/pi. Could you help me please
2R/pi is the centroid location of a quarter ring, with all the mass (idealized as) at the same radius. The centroid of a solid quarter circle is different, of course.
@@FluidMatters I have reviewed the method to define the centroid of volume, area, and line also Sir. Your support has ledluidMatters I have reviewed the method to define the centroid of volume, area, and line also Sir. Your support has led me to much new knowledge. Thanks a lot Sir.
how is the F bc force not cancelled out by the weight force of the water? Im trying to wrap my head around how there is left over force when you take the weight away. So there is force down because of the weight, and also force down because of the fluid pressure, but isnt the fluid pressure just because of the weight?
OOHH the height x Area for the fluid pressure is different from the volume of the actual water! so there is a discrepancy there! Equal force on horizontal, so it is like a square of hydrostatic pressure. Thanks
I think you've got it. Best of luck.
Hi, when finding F_bc, since it is a vertical force, wouldn't the formula be F_bc = pgV ?
no. F_BC is the a fluid weight. F_BC is the upward pressure force on plane surface BC. The entire surface BC is a depth R. The pressure at this depth is rho*g*R, which gets multiplied by the surface area. I hope that helps.
Should FV = the weight of the fluid?? How do you know that there is a force FBC?? Some questions never include this force I can't concentrate??!
Surface BC is at depth R in the water. So, there will be hydrostatic pressure at this depth, equal to rho*g*R. That's how we know there is a force F_BC. Note that the weight of the water is downward and the force of the water on the gate is clearly upward, as seen from the pressure distribution (at 1:08). So, F_v cannot be equal to the weight of the water. I hope that helps.
@@FluidMatters But in similar problems this is not the case ?? You mean that there is a pressure from the ground due to depth and it is upward ??!
@@HashemAljifri515 Not "due to the ground". It's the pressure in the water at that depth (at the ground-water interface). Local pressure acts in all directions. Once you define a control volume (as done in this problem) the pressure force acts inward and normal to the boundary (see video 1, Chapter 2) --- hence upward in this case.
I can not find the down load hard copy solution for this problem?
It's in the Exam Review Questions section of my website: www.drdavidnaylor.net/exam-review-questions.html
im finally able to solve my school's past quiz examples, i can go sleep now :) thankyouu
Glad to hear my video was helpful. Sleep well!
@@FluidMatters hey just wanted to let u know that i scored an A for the test! many thanks from NUS😃
@@Amanda-le8qe Congratulations! You obviously put in the work.
Pls why is normal reaction of the fluid section not taken in to account
Sorry, I don't understand your question. It helps to give the time (e.g. 2:15) in the video you are referring to.
Thank you for the explanation, Sir. I have a question, when you calculate the X bar, you take CW as (-) and CCW as (+). However, when you did the sum of moment at B you did the other way around. Why is it?
It makes no difference. The direction is an arbitrary choice. So, there no reason -- and no need to be consistent either.
@@FluidMatters I see. Thank you, Sir
Sir where can i get its theory part...
Here's the theory lecture for hydrostatic forces on curved surfaces:
th-cam.com/video/LdbEpRXUpOQ/w-d-xo.htmlsi=DoySt3zv_nYTipHn
You can find all the lectures as www.drdavidnaylor.net
Dear Sir, why do you count Fbc pressure force.?
Usually, I saw teachers concern about the weight of the water up to free surface to find vertical force.
I only find in your all of the video you count vertical pressure force, why sir?
You can use the missing "water approach". It is equivalent. I show this at the very end of the solution (about 15:00). But it can be tricking to find the line of action,. The method in this video is the most general approach. I hope that helps.
Thank you so much!
Glad it was helpful.
Notes available?
You can get a hard copy here: www.drdavidnaylor.net/exam-review-questions.html
Thank you so much!😊😊😊
Glad to hear you found it helpful.
Is the Area always (b)(h)?
It depends. Your question needs to be more specific (e.g give a time) for me to help.
Thanks a lot
Great explanation sir😂
Thanks. Glad to hear it was helpful. Good luck with your studies.
Elegance
Thanks. Glad to hear it was helpful.
Moment of F h about B. U didn't consider
Note that F_H acts both to the left and to the right at the same horizontal location. Hence, no net moment. See the note in the box at the bottom right side of the slide. I also say it at 12:17.
Thanks professor