Did anyone notice how the 'L' in 'WdL' on the top right corner moves at 1:46, how does that happen when it was written with a board marker and there was no cut in the video? (idk why i noticed that)
I just noticed something. If you take the Cosine of 30 degrees and multiply that by the "L" , Then you get the force that you got for "h" when L = h and Theta = 0 degrees. . So, if you want to know the Total Force on a Slanted Wall.. just divide the Force you get when the Wall is Vertical by the Cosine of the angle of the slanted wall. :) In the above example... You have: (10.87 E6 x Cosine 30 = 9.41 E 6) works everytime :) ..
Thank you sir,thank you for giving us so practical lectures ,for integral，there will be generate a integration constant，in this case，the constant is 0，but we should use boundary conditions to calculate the integration constant C.
Hi Michel ... I'm revisiting this video... and I wonder, as the Angle THETA get's larger.. the Force on the wall Increases... so if Theta goes to 90 Degrees ... then the FORCE should equal INFINITY, right? and yet the COS of 90 degrees in the equation above says that the force equals zero at theta = 90 degrees. I'm confused, so I'll double check again.. lol.. I'm missing something, so it's my problem now. :) I hope all is going well during these tough times. Spring has sprung in Michigan so.. all is well.
Hi Phillip, it has been a while. When the angle approaches 90 degrees the cos(angle) does indeed go to zero, but then L will approach infinity and since l is squared it will reach infinity faster than cos(angle will reach zero). An interesting problem indeed.
Can you make an exercise where you calculate the hydrostatic force on a slanted wall that makes an angle theta of more than 90 degrees? Let's say theta=120° so the angle on the waterside is 60 degrees. Thx a lot!
Dear Sir, when calculating the vertical force on the wall, why do you ignore the atmospheric pressure acting downwards? Shouldn't P= (rho*g*y) + Atmospheric Pressure?
Interesting Problem... I would have thought that the Slanted Wall would have the same pressure as if it were a Vertical wall...... by using the Average Pressure times the total area... wow, was I wrong from an Intuitive point of view..... but I did ponder the Cos 30 degrees factor... as the angle of 30 degrees changes to 45 degrees and then eventually 90 degrees.... that Factor diminishes .... at 90 degrees the Cos 90 = 0 and hence the pressure would be Zero?????... interesting Stuff there, Michel... ofcourse an angle of 90 degrees would suggest a Dam wall that was Horizontal and laying on the bottom... to which the pressure would either be zero or Infinity... YIKES.. I have my work cut out for me.... Great video!.. as usual.. you promote thinking outside the box for me... :) ... Thanks!
I guess I was thinking of the Pressure against the wall as if it were like a ray of LIGHT... and I was assuming there was a FLUX of sort.... and with Flux you calculate the Sine to the Perpendicular but that involved the SINE of the angle.. not the Cosine.. LOL... Pressure at a given depth is always the same.. regardless of a wall that is Perpendicular or Slanted... HOWEVER.. what I learned from YOU is that F = PxA.. and P = F/A...... therefore.. if Pressure is constant at a Given Depth.. then as Area goes up, as in a Slanted wall, then the FORCE MUST GO UP TOO...... Voila!! ... thanks MICHEL!!.. I apologize for my Ramblings here...
Hey, I have three quick questions. Firstly given the picture: gyazo.com/a8643081be684d28db1ce6b548e10221 1. Is my integral actually correct? I use h instead of y, and other variables (Given from the picture) gyazo.com/31103d0edaf79031bec7eb0a827b6565 2. My question about the integral, is why u can actually: integrate the Force dF with respect to Z. Is it because there is a small area dA that has a small preasure for every little dZ and this preasure can be calculated using rho*g*h but since u can substitute h with z u can integrate with respect to z, because z and h has a realtionship (tringonometry)? 3. Real dams are often a bit arc shaped, and a bit curved/sloped? How come? Does it have anything to do with momentum or why are for instance dams thicker at the bottom? Is it because of torque or is it just because the force is bigger the deeper the water gets?
Hello, thank you for your videos. They are very instructive and clear. I have two questions though: 1. Why do not you consider a force perpendicular to the inclined surface? 2. If one is concerned about dam overturning, how would you calculate the moment arm of this force respect to the downstream toe?
There are different approaches of solving the pressure and forces on a dam with a slanted surface. I believe that I solved that with a few different techniques. Take a look at the rest of the videos for a better understanding on how to solve these types of problems.
Did anyone notice how the 'L' in 'WdL' on the top right corner moves at 1:46, how does that happen when it was written with a board marker and there was no cut in the video? (idk why i noticed that)
I noticed that too
I wish I had teacher like you when I was in college years go. I hope this comment sums it up. You are excellent
I just noticed something. If you take the Cosine of 30 degrees and multiply that by the "L" , Then you get the force that you got for "h" when L = h and Theta = 0 degrees. . So, if you want to know the Total Force on a Slanted Wall.. just divide the Force you get when the Wall is Vertical by the Cosine of the angle of the slanted wall. :) In the above example... You have: (10.87 E6 x Cosine 30 = 9.41 E 6) works everytime :) ..
Hi Philip. That is indeed correct! 🙂
(6:24 mark) 9,410,000 divided by cos(30) = 10,870,000 also
Thank you sir,thank you for giving us so practical lectures ,for integral，there will be generate a integration constant，in this case，the constant is 0，but we should use boundary conditions to calculate the integration constant C.
Can you do a calculation fot the force acting on that strip
I believe we have a video like that.
Hi Michel ... I'm revisiting this video... and I wonder, as the Angle THETA get's larger.. the Force on the wall Increases... so if Theta goes to 90 Degrees ... then the FORCE should equal INFINITY, right? and yet the COS of 90 degrees in the equation above says that the force equals zero at theta = 90 degrees. I'm confused, so I'll double check again.. lol.. I'm missing something, so it's my problem now. :) I hope all is going well during these tough times. Spring has sprung in Michigan so.. all is well.
Hi Phillip, it has been a while. When the angle approaches 90 degrees the cos(angle) does indeed go to zero, but then L will approach infinity and since l is squared it will reach infinity faster than cos(angle will reach zero). An interesting problem indeed.
And this force is exerted at the C.G of the pressure diagram.
You can think of it that way, but in actuality the forces is acting everywhere along the surface of the dam.
@@MichelvanBiezen :) Ok
thank you so much this is exactly what i was looking for!
Can you make an exercise where you calculate the hydrostatic force on a slanted wall that makes an angle theta of more than 90 degrees? Let's say theta=120° so the angle on the waterside is 60 degrees. Thx a lot!
That would be exactly the same as a 60 degree angle. (You only need to consider angles between 0 and 90 degrees.)
Thank you. Where would this force act? Would it be a distributed load?
Yes, the force is distributed across the whole surface, and increase linearly with depth.
@@MichelvanBiezen Thank you Michel. Your videos are helping massively. I will make sure to donate at the end of my degree.
Thank you.
Dear Sir, when calculating the vertical force on the wall, why do you ignore the atmospheric pressure acting downwards? Shouldn't P= (rho*g*y) + Atmospheric Pressure?
The atmospheric pressure typically cancels out on both sides.
thank u !!!! it very helpful for our project !!
Interesting Problem... I would have thought that the Slanted Wall would have the same pressure as if it were a Vertical wall...... by using the Average Pressure times the total area... wow, was I wrong from an Intuitive point of view..... but I did ponder the Cos 30 degrees factor... as the angle of 30 degrees changes to 45 degrees and then eventually 90 degrees.... that Factor diminishes .... at 90 degrees the Cos 90 = 0 and hence the pressure would be Zero?????... interesting Stuff there, Michel... ofcourse an angle of 90 degrees would suggest a Dam wall that was Horizontal and laying on the bottom... to which the pressure would either be zero or Infinity... YIKES.. I have my work cut out for me.... Great video!.. as usual.. you promote thinking outside the box for me... :) ... Thanks!
That is a good way of looking at it (taking the angle to the limit), but you also must take L to the limit at the same time......
I guess I was thinking of the Pressure against the wall as if it were like a ray of LIGHT... and I was assuming there was a FLUX of sort.... and with Flux you calculate the Sine to the Perpendicular but that involved the SINE of the angle.. not the Cosine.. LOL... Pressure at a given depth is always the same.. regardless of a wall that is Perpendicular or Slanted... HOWEVER.. what I learned from YOU is that F = PxA.. and P = F/A...... therefore.. if Pressure is constant at a Given Depth.. then as Area goes up, as in a Slanted wall, then the FORCE MUST GO UP TOO...... Voila!! ... thanks MICHEL!!.. I apologize for my Ramblings here...
what about if there is not only water exist, but also other liquid such as oil, how do we calculate the total forces ? please make video about this ☺
Hey,
I have three quick questions. Firstly given the picture:
gyazo.com/a8643081be684d28db1ce6b548e10221
1. Is my integral actually correct? I use h instead of y, and other variables (Given from the picture)
gyazo.com/31103d0edaf79031bec7eb0a827b6565
2. My question about the integral, is why u can actually: integrate the Force dF with respect to Z. Is it because there is a small area dA that has a small preasure for every little dZ and this preasure can be calculated using rho*g*h but since u can substitute h with z u can integrate with respect to z, because z and h has a realtionship (tringonometry)?
3. Real dams are often a bit arc shaped, and a bit curved/sloped? How come? Does it have anything to do with momentum or why are for instance dams thicker at the bottom? Is it because of torque or is it just because the force is bigger the deeper the water gets?
Hello, thank you for your videos. They are very instructive and clear. I have two questions though:
1. Why do not you consider a force perpendicular to the inclined surface?
2. If one is concerned about dam overturning, how would you calculate the moment arm of this force respect to the downstream toe?
There are different approaches of solving the pressure and forces on a dam with a slanted surface. I believe that I solved that with a few different techniques. Take a look at the rest of the videos for a better understanding on how to solve these types of problems.
Thank you for your quick reply. I need to work more on that. Thanks again for your useful channel.
Thank you, Thank you, Thank you,
You are very welcome. Glad you found our videos. 🙂
Since the problem is the same in lecture 3 of 15 and 4 of 15....then why is that the total force in both the lectures different?
The force calculated in both videos is the same.
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