Hello, I live in Thailand, I study in faculty of Civil engineering in Suranaree University of Technology, I wanna thank you very much that helps me cut off my confused, I following your course such as STATICS, FLUIDS etc. keep going on your teaching and share your course to global, I wise you get the good things like that you give free course to the others.
Easy to see: rho*g*h will be the pressure at the bottom, if a force diagram is drawn it will be a triangle, area of the traingle (1/2*bh) will be your total pressure as a pointload. Adding width to the area equation you would get 1/2*rho*g*h*h*w (the base of the triangle is rho*g*h) which is the same answer as the integral but just done from knowing your equations and force shapes. The height of the pointload will be 1/3H from the bottom following the centroid rule for triangles, if your wall can withstand this pressure at that height, it is most like strong enough. I'm not a mechanical engineer...rather civil diploma student, if this should not be hard to see. Just draw it out. Edit: From a structural analysis perspective I used pointload, in fluid mechanics it is your resultant force.
Since the change in pressure is a linear function of depth. The same answer (force on the wall) can be found by multiplying the total area of the wall with the average pressure (halfway down from the surface).
Thanks for the answer. Your teachings are better than my textbook! I pointed out the error in methodology to the publisher of the text book! Again thanks for the great channel. I really is a great help.
That 1/2 also confused me when our lecturer taught us. He didn't show us how to calculate '1/2' instead of saying that was the resultant force that acting on a wall. After all, I realized that force should be in the centroid of the planar and that is why '1/2' comes out for the wall and the force at the base without '1/2'.
I'm trying to reason through this. I have a garage door that gets slammed with occasional flash flooding. I want to calculate the static pressure of the water against the wall to figure out how I might reinforce it from collapsing again (I'll figure out the dynamic pressure later). If the water level reaches 2 ft high against a 16' wide door, I calculate a pressure per square foot to be 124.85 lbs. Multiply that by 32 sq ft and I get 3995 pounds and then divide it by two to average out the weight top to bottom or 1997.6 lbs of force against my garage door. Where I'm stuck is applying this to real life. If I have a swimming pool full of water 20' long so that my garage door is at one end, don't I have more pressure because I have more weight (20 ft length of water) to push against my garage door? Now imagine a 1' wide tank with the same dimension as my garage door. According to this formula (I think) I get the exact same pressure. But how can that be? what am I missing. On a side note, there is nothing better than bringing the whole concept together than using an example with real numbers. Symbols and letters just become gobbly goop for the weaker math student (me). I don't know if that makes any sense but if you can help me figure it out, I'd be very grateful!!!!
A dam will experience the same force if the lake behind is 20 meters long and when it is 20 km long as long as the depth of the water behind the dam is the same. Pressure only depends on the depth of the water and the forces against the dam only depend on the pressure and the surface area of the dam.
Mind blowing! In that case, (and thanks for responding so quickly!!) using the same garage door example, if i have 1 inch of water pressing against the whole length of the door and still 2 feet high, my pressure is still about 2000lbs BUT the total weight of the water is only about 165lbs. (4608 cubic inches of water, 231 cubic inches in a gallon, almost 20 gallons at 8.3lbs per gallon). The only way I could reconcile this is my mind was to suppose that my "width" of water would have to be equal to or greater than its height as a minimum volume, but that's just me trying to make sense of it all. As you look at that, does that seem possible to you? Greater force than the entire weight of the water? Weird!
Sir, you're better than awesome. I'm falling short of the needful word. You're no lesser than Albert Einstein for me. Sir, please reply something as it'll be like your autograph for me which I'll have all my life. Please Sir
I am just an average person who enjoys teaching and who with the help of my wife has been fortunate to share what I have learned over the years with the rest of the world.
so basicly what's the difference between this solution (submerged) and the one on your last video which is to calculate the pressure on a pool? nice video though really helpful! :)
On this video, the pressure changes with depth and thus this is how you calculate the force on a side wall. In the previous video we showed how to determine the pressure in a fluid.
Really enjoy your channel. One request on HSF though... Can you do an example of differing density liquids acting on a tank wall. For instance oil floating on water acting on the wall.
Hello sir, while calculating lateral pressure on wall due to water, why we do not consider length? We only consider the height and width of water for pressure applied on wall.
The size of the lake doesn't matter, only the depth and the width of the dam. Note that the force on the dam only depends on the pressure and the surface area of the dam. (F = P x A) The size of the lake does not affect pressure, only the depth does.
Why air pressure isn't included in here. I mean if I want to just calculate the force due to water( not the total force) shouldn't I include air pressure as well?
Good explanation! In practice though, we must know where this equiv. force is acting in order to apply this theory to a real world scenario (like designing a rectangular tank with vertical walls). We are using P(avg) in the derivation, which acts at the midpoint of the wall, but the resultant equiv. force is acting at the centroid of the triangular pressure distribution area - does it not? This would be 1/3 up from the base. Can you confirm and/or explain?
For a rectangle the centroid is at the center. For a triangle, the centroid is indeed 1/3 the distance from the base to the height of the triangle. The force acts everywhere against the surface, but we can simplify calculations by taking the total force and making it act at the centroid.
Sorry but I am still not clear. Does the equivalent point force acts at the centroid of the pressure triangle (always 1/3 up from the base), or at the centroid of the wall (which depends on the shape of the wall)?
Pressure as a function of depth is a linear function. P = rho x g x h where rho represents density of the fluid. For water the density is 1000 kg / m^3 and g = 9.8 m /sec^2 Thus the pressure is 9800 N/m^2 for every meter of depth. (98,000 N/m^2 at 10 m etc.)
The weight of the water = mg and that is directed downward, since the weight of the water is caused by the force of gravity. The weight of the water above a particular point in the water creates pressure at that point. According to Pascall's principle the pressure acts in all directions.
sir, if you had to calculate the depth of the wall for given height and length for which the wall would not slip or turn over due to the pressure of the water. how would you do that?
You would need to know information about the construction of the wall, how it was re-enforced, what type of material was used, what the shape and thickness of it was, etc.
To find the total force on a submerged gate you would change the limits. But more typically you will want to find the torque on the gate, so you'll have to integrate P * h * d where d is the distance from the position of the area strip to the hinges of the gate.
I don't believe you can calculate the torque (moment) about a hinge by using the "center of pressure". The center of pressure is at the half-way point between to top and the bottom. Try it and compare it to the result you get when you integrate over the height to get the moment and see if the results are the same. There is an example in the playlist that shows you how to do that.
@@MichelvanBiezen thank you for you reply. The hydrostatic pressure profile is a triangle with the base parallel to the wall with the fatter side at the bottom. Am I saying it correctly?
It depends on what you are looking for. If you want gauge pressure, you exclude atmospheric pressure. If you want the total pressure, you include it. Most application in science and industry work with gauge pressure.
instead of posing: dF=P*dA dF=rho*g*y*W*dy Why couldn't we instead have posed: dF=P*dA + A*dP dF=rho*g*y*W*dy+W*y*rho*g*dy dF=2*rho*g*y*W*dy ? I don't get why the 2nd set would be wrong. I guess that the way you pose the problem, P is constant over the sliver of dA while I say that it isn't. But the way I would do it leads to dF being doubled... what am I missing?
By definition, F = P x A If you now let A become an infinitesimal area dA then you'll have an infinitesimal force dF such that the pressure over that small area is constant. That is the fundamental principle of calculus. (therefore the pressure will be constant and therefor dP = 0)
@@MichelvanBiezen that makes sense, although if P is a function of y and y is varies infinitesimally then we could say that P varies infinitesimally as well?
Maybe the answer is in your next video where F =Pavg* A. F and P can be defined in every point but not A. It doesn't make sense to multiply P in one point by an area to get the force over the area therefore the definition has to be F=Pavg*A. Or Favg=Pavg*A
@@MichelvanBiezen say, it asks for the resultant force and where this acts. Would you have to take moments? Or is Force total = the greater force (the higher water level) minus the smaller force
Michel van Biezen so... the pressure will continue to increase as the point of interest goes deeper. So that means the force which is P*A will also grow.... So then if the average pressure occurs halfway to the bottom.. the average force is also halfway down? I’m sorry. I’m still confused. My fluids professor drew these distributed force triangles on the walls representing the increasing pressure?
I was trying to understand derivation of the vertical plane surface submerged in liquid and in the derivation they have first calculated the total pressure by taking elementary strip and then integrate it. And i get the concept but after that they have calculated the centre of pressure by using principle of moments. And i can't understand centre of pressure and principle of moments use. But in your lecture i have not seen anything like this.
Yes, that is correct. The center of pressure, is the point on the submerged wall where if all of the force would be concentrated there, it would have the same effect on the wall and the same torque on the wall. We have not yet made a video about that concept.
That is the beauty if calculus. You don't have to do that, because by definition, in the limit as dy goes to zero, the pressure difference is zero. (Essentially you add up a near infinite number of strips of (almost) zero width.)
It is not the net force, but the force that would represent the total force acting on a single point, (or the equivalent force). The magnitude is the total force. The position where it acts is at the average pressure (half way down) and in the middle of the wall between left and right.
I am not sure what you mean with the force "inside" and "outside" the wall. In the video we are calculating the force on the wall caused by the fluid pressure.
![АЛАУДИНОВ у Скабеевой: эти их ВСУ нас НЕ ДОГОНЯТ 😁 [Пародия]](http://i.ytimg.com/vi/an0anqU4WGQ/mqdefault.jpg)
i read the textbook for 5 hours (still feeling confused on why its over half) and he explained in 5 mins.
Yeah, textbooks don't always help.
Having a chef show you how to make pizza is always faster than you teaching yourself through recipe books
I love how u teach from the very basics like P=F/A etc...thank you so much
Thank you for sharing.
Hello, I live in Thailand, I study in faculty of Civil engineering in Suranaree University of Technology, I wanna thank you very much that helps me cut off my confused, I following your course such as STATICS, FLUIDS etc. keep going on your teaching and share your course to global, I wise you get the good things like that you give free course to the others.
Was looking over my lecture's notes for 30mins and didn't understand a thing. Came here, understood within 2 mins. Thank you sir!
Easy to see: rho*g*h will be the pressure at the bottom, if a force diagram is drawn it will be a triangle, area of the traingle (1/2*bh) will be your total pressure as a pointload. Adding width to the area equation you would get 1/2*rho*g*h*h*w (the base of the triangle is rho*g*h) which is the same answer as the integral but just done from knowing your equations and force shapes. The height of the pointload will be 1/3H from the bottom following the centroid rule for triangles, if your wall can withstand this pressure at that height, it is most like strong enough.
I'm not a mechanical engineer...rather civil diploma student, if this should not be hard to see. Just draw it out.
Edit: From a structural analysis perspective I used pointload, in fluid mechanics it is your resultant force.
Since the change in pressure is a linear function of depth. The same answer (force on the wall) can be found by multiplying the total area of the wall with the average pressure (halfway down from the surface).
As an undergrad in mechanical engineering this video was super helpful. THANKS!
As an undergrad in mechanical engineering, I can confirm that this video is still super useful 5 years later :)
simply the best. better than all the rest. :)
Sat through two lessons at uni covering this topic and didn’t understand a word. Hopped on to this video and it clicked. You’ve got my thanks
Glad you found our videos! 🙂
I like simplicity..And this is it. Thanks
Thanks for the answer. Your teachings are better than my textbook! I pointed out the error in methodology to the publisher of the text book! Again thanks for the great channel. I really is a great help.
‘And that’s how it’s done’
DROPS MIC 🎤
That 1/2 also confused me when our lecturer taught us. He didn't show us how to calculate '1/2' instead of saying that was the resultant force that acting on a wall. After all, I realized that force should be in the centroid of the planar and that is why '1/2' comes out for the wall and the force at the base without '1/2'.
Thank you making these videos. I always come here to figure out concepts I get stuck on.
was more helpful than my own lecturers. thank you!
Glad it helped!
idk how to thank this channel I'm acing all my units in uni
being a civil engineer it helps me in analysis of dams ......Thank You From India Cheers!
You're welcome!
awesome! I never seen this simplest method ever before.
Thanks for help from student of Czech Technical University in Prague! 😁
Glad you found our videos. Welcome to the channel! 🙂
@@MichelvanBiezen Just a small question - shouldn't we add an atmospheric pressure to the total pressure too?
That miniature caterpillar truck looks good over there. Thanks for explaining this in a simple manner 😇😇
Glad it was helfful.
Simple clear direct and clean. Thanks!
Great video! I love how simple you make it.
Thank you! Glad you found our videos. 🙂
Great style comes from great understanding!
simply and Easy to understanding 👍🏼
You are very professional, thank you from my deap heart for helping me, love you prf. ❤️❤️
We are glad that the videos are helping
So very easy to understand, thank you.
I'm trying to reason through this. I have a garage door that gets slammed with occasional flash flooding. I want to calculate the static pressure of the water against the wall to figure out how I might reinforce it from collapsing again (I'll figure out the dynamic pressure later). If the water level reaches 2 ft high against a 16' wide door, I calculate a pressure per square foot to be 124.85 lbs. Multiply that by 32 sq ft and I get 3995 pounds and then divide it by two to average out the weight top to bottom or 1997.6 lbs of force against my garage door. Where I'm stuck is applying this to real life. If I have a swimming pool full of water 20' long so that my garage door is at one end, don't I have more pressure because I have more weight (20 ft length of water) to push against my garage door? Now imagine a 1' wide tank with the same dimension as my garage door. According to this formula (I think) I get the exact same pressure. But how can that be? what am I missing. On a side note, there is nothing better than bringing the whole concept together than using an example with real numbers. Symbols and letters just become gobbly goop for the weaker math student (me). I don't know if that makes any sense but if you can help me figure it out, I'd be very grateful!!!!
A dam will experience the same force if the lake behind is 20 meters long and when it is 20 km long as long as the depth of the water behind the dam is the same. Pressure only depends on the depth of the water and the forces against the dam only depend on the pressure and the surface area of the dam.
Mind blowing! In that case, (and thanks for responding so quickly!!) using the same garage door example, if i have 1 inch of water pressing against the whole length of the door and still 2 feet high, my pressure is still about 2000lbs BUT the total weight of the water is only about 165lbs. (4608 cubic inches of water, 231 cubic inches in a gallon, almost 20 gallons at 8.3lbs per gallon). The only way I could reconcile this is my mind was to suppose that my "width" of water would have to be equal to or greater than its height as a minimum volume, but that's just me trying to make sense of it all. As you look at that, does that seem possible to you? Greater force than the entire weight of the water? Weird!
Very good method of explaining
Simple and Direct
Sir, you're better than awesome. I'm falling short of the needful word. You're no lesser than Albert Einstein for me. Sir, please reply something as it'll be like your autograph for me which I'll have all my life. Please Sir
I am just an average person who enjoys teaching and who with the help of my wife has been fortunate to share what I have learned over the years with the rest of the world.
Omg thank you very much. Finally find it!
You are welcome. Glad you found our videos! 🙂
why dont we have to consider atmospheric pressure in the pressure equation as there is normally a Po within the pressure equation?
The atmospheric pressure cancels out on both sides of the dam.
so basicly what's the difference between this solution (submerged) and the one on your last video which is to calculate the pressure on a pool? nice video though really helpful! :)
On this video, the pressure changes with depth and thus this is how you calculate the force on a side wall. In the previous video we showed how to determine the pressure in a fluid.
Really enjoy your channel. One request on HSF though... Can you do an example of differing density liquids acting on a tank wall. For instance oil floating on water acting on the wall.
There really is no difference. You find the force of each section with the different liquid as shown in this video, and then just add them.
Thanks for your smarts. Appreciate it. 💙
I am not that smart, but you are welcome.
Thank-you from India 💕
Welcome to the channel!
i like this channel very much !!!!!
WoW..!! Sir very easy explanation Thanks a lot
For non-rectangular plate (example : triangular plate), is it possible to use the formula [F = P average * A] ??
You can use it in a similar way, except the "average" pressure needs to be the pressure at the "center of mass" of the area.
this was insanely helpful! thanks!
Glad you liked it.
Brilliant video!
Hello sir, while calculating lateral pressure on wall due to water, why we do not consider length?
We only consider the height and width of water for pressure applied on wall.
The size of the lake doesn't matter, only the depth and the width of the dam. Note that the force on the dam only depends on the pressure and the surface area of the dam. (F = P x A) The size of the lake does not affect pressure, only the depth does.
@@MichelvanBiezen thank you so much sir
Thank you!
it was explained very well, thank you sir
Thanks and welcome
How to calculate the Depth where the Net Force is acting on the Gate?
That was great. Thank you.
Thanks a million.
I get the idea.
You are most welcome
Thanks for these videos
Ohh...u saved me 4rm headache
Why air pressure isn't included in here. I mean if I want to just calculate the force due to water( not the total force) shouldn't I include air pressure as well?
Air pressure is typically not included, since it exists on both sides of any barier (as in a window as well).
You made it really simple to understand . thank u
Glad it helped!
you have a good base of the subject, thanks for your help!
Thank You
You're welcome
Good explanation! In practice though, we must know where this equiv. force is acting in order to apply this theory to a real world scenario (like designing a rectangular tank with vertical walls). We are using P(avg) in the derivation, which acts at the midpoint of the wall, but the resultant equiv. force is acting at the centroid of the triangular pressure distribution area - does it not? This would be 1/3 up from the base. Can you confirm and/or explain?
For a rectangle the centroid is at the center. For a triangle, the centroid is indeed 1/3 the distance from the base to the height of the triangle. The force acts everywhere against the surface, but we can simplify calculations by taking the total force and making it act at the centroid.
Sorry but I am still not clear. Does the equivalent point force acts at the centroid of the pressure triangle (always 1/3 up from the base), or at the centroid of the wall (which depends on the shape of the wall)?
Should not we account for the atmospheric pressure...so P = qgh + atmospheric
It depends. Are you calculating the gauge pressure or the total pressure?
@@MichelvanBiezen
Total pressure
Then you have to add the atmospheric pressure.
Isn't there a chart somewhere that shows the average pressure of water at different depths? If you have a 5 foot tall wall, what is it?
Pressure as a function of depth is a linear function. P = rho x g x h where rho represents density of the fluid. For water the density is 1000 kg / m^3 and g = 9.8 m /sec^2 Thus the pressure is 9800 N/m^2 for every meter of depth. (98,000 N/m^2 at 10 m etc.)
This was perfect ty
Wow. Thank you sir!
I am so confused direction of g is downwards , how we can calculate force on vertical wall as component of g becomes zero on vertical wall
The weight of the water = mg and that is directed downward, since the weight of the water is caused by the force of gravity. The weight of the water above a particular point in the water creates pressure at that point. According to Pascall's principle the pressure acts in all directions.
@@MichelvanBiezen thank u so much
Thank you!!
You're welcome! 🙂
thank you, thank you , thank you sooo much sir... helped me a lot.
what if the plate is irregular? does the same equation holds good?
If the plate is irregular, you have to use integration, or find the center of mass of the wall. (and then find the pressure at the center of mass)
Thank you 🙏🏼❤️
You are so welcome
Why is the small strip's height is dy? Can it be called something else because if it was different then the integration part wouldn't work, right?
Since the pressure is a function of the height, it is better to call the height of the strip dy.
you teach the way i think.
Thank you very much sir
Most welcome
You are my Legend the Savor!!!!! thanks a lot
sir, if you had to calculate the depth of the wall for given height and length for which the wall would not slip or turn over due to the pressure of the water. how would you do that?
You would need to know information about the construction of the wall, how it was re-enforced, what type of material was used, what the shape and thickness of it was, etc.
How to calculate the point where the resultant force acts on the Gate submerged inside the water ?
To find the total force on a submerged gate you would change the limits. But more typically you will want to find the torque on the gate, so you'll have to integrate P * h * d where d is the distance from the position of the area strip to the hinges of the gate.
very helpfull, thank you
amazing lecture!!!!,
Why do they say the pressure acts at 2/3 of the depth from the surface? Is that only for a hinged submerged gate?
Are they referencing the moment?
THANK YOU PROFESSOR!!!!
You are welcome!
very helpful, thank you.
Thank you sir you're the best.
Very nice
Sir please can u also derive same for torque ...plz
amazing man, thank you
You're welcome!
Do you have a video on finding torque with this similar problem
No, but that is a good question. We should add one of those examples.
at which height along the wall does the center of pressure act? I'm asking so I can calculate the moment about a hinge at the bottom of the gate/wall
I don't believe you can calculate the torque (moment) about a hinge by using the "center of pressure". The center of pressure is at the half-way point between to top and the bottom. Try it and compare it to the result you get when you integrate over the height to get the moment and see if the results are the same. There is an example in the playlist that shows you how to do that.
@@MichelvanBiezen thank you for you reply. The hydrostatic pressure profile is a triangle with the base parallel to the wall with the fatter side at the bottom. Am I saying it correctly?
Good one
Why don't you consider atmospheric pressure
It depends on what you are looking for. If you want gauge pressure, you exclude atmospheric pressure. If you want the total pressure, you include it. Most application in science and industry work with gauge pressure.
how to find the location of it
You're awesome
Thank you👌👌👌👌
instead of posing:
dF=P*dA
dF=rho*g*y*W*dy
Why couldn't we instead have posed:
dF=P*dA + A*dP
dF=rho*g*y*W*dy+W*y*rho*g*dy
dF=2*rho*g*y*W*dy
?
I don't get why the 2nd set would be wrong.
I guess that the way you pose the problem, P is constant over the sliver of dA while I say that it isn't.
But the way I would do it leads to dF being doubled...
what am I missing?
By definition, F = P x A If you now let A become an infinitesimal area dA then you'll have an infinitesimal force dF such that the pressure over that small area is constant. That is the fundamental principle of calculus. (therefore the pressure will be constant and therefor dP = 0)
@@MichelvanBiezen that makes sense, although if P is a function of y and y is varies infinitesimally then we could say that P varies infinitesimally as well?
Maybe the answer is in your next video where F =Pavg* A.
F and P can be defined in every point but not A. It doesn't make sense to multiply P in one point by an area to get the force over the area therefore the definition has to be F=Pavg*A. Or Favg=Pavg*A
thank you my lord
What if you have water on the other side at a different height?
It depends on what the question asks.
@@MichelvanBiezen say, it asks for the resultant force and where this acts. Would you have to take moments? Or is Force total = the greater force (the higher water level) minus the smaller force
Sir i have question. Isn't force acting on wall P(total) x A
That is integral of rgh times A?
KIndly explain pls
The force is the average pressure x area, since pressure is a linear function.
I thought the resultant force occured 2/3 from the top of the wall?
There is a linear relationship between pressure and depth. So, where is the average force as a function of depth?
Michel van Biezen so... the pressure will continue to increase as the point of interest goes deeper. So that means the force which is P*A will also grow....
So then if the average pressure occurs halfway to the bottom.. the average force is also halfway down? I’m sorry. I’m still confused. My fluids professor drew these distributed force triangles on the walls representing the increasing pressure?
Take a look a the videos where the torque is calculated so you can see the difference.
In my book forces on completely submerged vertical surface they have used the concept of Moment of Inertia blah blah. Can you help me?
What is the context or what is the problem asking you to calculate?
I was trying to understand derivation of the vertical plane surface submerged in liquid and in the derivation they have first calculated the total pressure by taking elementary strip and then integrate it. And i get the concept but after that they have calculated the centre of pressure by using principle of moments. And i can't understand centre of pressure and principle of moments use.
But in your lecture i have not seen anything like this.
Yes, that is correct. The center of pressure, is the point on the submerged wall where if all of the force would be concentrated there, it would have the same effect on the wall and the same torque on the wall. We have not yet made a video about that concept.
Michel van Biezen thank you. And please make video about that topic. Btw you are the best professor in the youtube channel i have found. Thank you 🤗
What's the difference between the y & H in the diagram?
y is the variable, which represents any depth. H is a constant and represents the total depth.
Thank you so much ❤
thank u sir
SHUKRIYA
yeaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaah
how can it be solved if we decided to consider the pressure difference across dy in the equation?
That is the beauty if calculus. You don't have to do that, because by definition, in the limit as dy goes to zero, the pressure difference is zero. (Essentially you add up a near infinite number of strips of (almost) zero width.)
But where is the force acting sir?? I mean the location of the net force?
the force is acting everywhere on the wall. One can calculate the force per unit area or you can calculate the total force.
I mean to ask, the position of net force.
It is not the net force, but the force that would represent the total force acting on a single point, (or the equivalent force). The magnitude is the total force. The position where it acts is at the average pressure (half way down) and in the middle of the wall between left and right.
Thank you Soo much sir. Wish you were here in BMSCE(Banglore). May I have your WhatsApp?
in india 15 year olds derive this without help
but anyway wonderful explanation..
And in case W changes should we use double integration?
Yes, you would have to integrate over the width as well.
What if Im trying to look for the Resultant force inside and outside the wall? how would I use P_atm to calculate the outside force?
I am not sure what you mean with the force "inside" and "outside" the wall. In the video we are calculating the force on the wall caused by the fluid pressure.
the atmospheric pressures act on both side of the wall so that there should not include Patm and P=Rogy, is that u want to say?
this guy got too much cake. I couldn't focus on what he was teaching.
i love you