I failed Fluids, but now that I've raised 7 kids, I feel I'm ready to finish my engineering degree. Your explanations are easy to follow. If there's something I don't understand, I can pause or rewind. I actually understood this explanation! Thank you! Excellent!
+William Gallant Wow, 7 kids! Good that you are picking up the books again. Congratulations for your life's work of raising 7 kids! (That is the most important job in the world)
It would be great if you mention the direction of that Force, since all Forces have direction. I think it is normal to the slanted wall. Also, it would be interesting to talk about the horizontal and vertical forces.
In this case there would be a vertical force on the wall due to the water's weight, and yes, a normal force exerted by the wall on the water perpendicular to the surface of the wall.
You use 1/2 for the equivalent point. Others use 1/3 for the equivalent force point. Perhaps this is because this exercise is usually used to determine the bending moment on that surface, and you are determining total force on the dam?
Is this force just the X component (to the left), or is this the NET force, taking both the horizontal and vertical parts into consideration? Because it's slanted, there is some force due to the weight above it right???
Pressure only depends on depth. Therefore you use the vertical height for pressure. But if you want to calculate the force against a slanted surface, you need to take into account the slant to find the area.
thank you for these videos and only have 1 question (as opposed to 1 thousand after my lectures), what would change if the dam was not submerged into the water and had a length l at the end exposed out of it?
Force on a submerged wall can be found by integrating over the whole wall or by simply multiplying the area of the wall by the average pressure since the pressure in linearly proportional to the depth of the water.
I failed Fluids, but now that I've raised 7 kids, I feel I'm ready to finish my engineering degree. Your explanations are easy to follow. If there's something I don't understand, I can pause or rewind. I actually understood this explanation! Thank you! Excellent!
+William Gallant
Wow, 7 kids!
Good that you are picking up the books again.
Congratulations for your life's work of raising 7 kids! (That is the most important job in the world)
How is it going?
yeaaaaaaaaa how was it
Hay here in India we have to study it in 11 th grade only
And I am amused to hear that you are taking a college degree after studying this😂😂
It would be great if you mention the direction of that Force, since all Forces have direction. I think it is normal to the slanted wall. Also, it would be interesting to talk about the horizontal and vertical forces.
In this case there would be a vertical force on the wall due to the water's weight, and yes, a normal force exerted by the wall on the water perpendicular to the surface of the wall.
@Michel van Biezen I am a fan! Everybody please support this gem of a professor!
Sir you’re great!!! And your bow ties look very nice !!!
Thank you. Glad you found our videos! 🙂
In my Fluid Mechanics class we're being instructed to use Gamma, not pgh
That is interesting.
thanks, for all beneficial lessons
you are a blessing!
How could the average pressure be the halfway point when the pressure is not constant as you submerge?
The pressure is a linear function of depth. Therefore the average pressure is half way down to the bottom.
Why isn't the centroid used to find the pressure, since it's slanted the centroid should be 1/3 from the bottom of the triangle, correct??
You use 1/2 for the equivalent point. Others use 1/3 for the equivalent force point. Perhaps this is because this exercise is usually used to determine the bending moment on that surface, and you are determining total force on the dam?
It depends on what you are trying to calculate.
@@MichelvanBiezen 1/2 for x axis force and 1/3 for moment about bottom restraint?
thanks sir ....very well explained ... thanks for all your efforts !! :)
Just wanna say Thank you!
Much appreciated
Excellent
Is this force just the X component (to the left), or is this the NET force, taking both the horizontal and vertical parts into consideration? Because it's slanted, there is some force due to the weight above it right???
Use P = F/A Therefore F = P x A
Sir in pressure you have taken the vertical height while in area you have taken the Slant height, plz can you answer.
Pressure only depends on depth. Therefore you use the vertical height for pressure. But if you want to calculate the force against a slanted surface, you need to take into account the slant to find the area.
@@MichelvanBiezen Thank you sir
thank you for these videos and only have 1 question (as opposed to 1 thousand after my lectures), what would change if the dam was not submerged into the water and had a length l at the end exposed out of it?
Thank you
I very glad too. Thk
thax for your lecture
Thank you for the videos!
why did we take the average pressure
Force on a submerged wall can be found by integrating over the whole wall or by simply multiplying the area of the wall by the average pressure since the pressure in linearly proportional to the depth of the water.