Sir, your videos are helping me so much, I'm in my first year of Mechanical Engineering, I'm behind on my work and lectures but I'm dedicating my time to revise as much as possible each day, your videos are helping me understand these topics so much it's unbelievable, and it's not just Fluids videos, your Applied Mechanics videos have helped me a lot too, on behalf of people who might watch your videos finding them very helpful and might not comment, Thank you so much we can't thank you enough for the help, Keep up the good work you're helping a lot of students out there
I have been struggling with this for the last few hours, I don't understand it the way my professor teaches it, and can I just say, I love that you've chosen to share this with the world and I wish you the best in life
Bonjour Monsieur, vous êtes un excellent enseignant. Je travaille en France où j'inspecte des barrages. Vos cours sont tres bien expliqués. Bravo. Patrick
@@sigourneybergraaf3381 I think the principle can be applied to any shape but the volume varies since each shapes has different formula for its volume. This is just based on my understanding. I hope this helped you :)
This series on fluids is really helpful!! Thank you :))) Oh, and I wonder if there is a scenario like this. What if the curve is the same, but the fluid is UNDER the curved wall. That means the weight is zero (because the fluid is under the curve) and the resultant force R then is equal to Fp, right? I feel like my thinking lacks something, but I don't know what it is :(
The force on the wall depends on the pressure which is a function of depth. So regardles of the curve of the dam, there will be a force pushing against the dam.
There is a whole playlist on centroids (c of g): Mechanical Engineering: Centroids & Center of Gravity (3 of 35) Centroids ruclips.net/video/tc11YEKHuFE/видео.html&index=3&list=PLX2gX-ftPVXWnfWWDNgu4x9hiCPTi0HB9
You can imagine the vertex of the parabola is at the origin. Then the parabola can be written as y = Cx^2 for some constant C. Since we know the point (A,H), we can easily find that C = H/A^2. Then the rest is just integrating. You dont have to derive it in this way of course, you can just directly use the numbers given to you.
If you work it out using integrals. y=x^2, Integral 1/3x^3. At x the height h = x^2. The total square is x^3. The integrated bottom part of the curve is 1/3 x^3. Therefore the area above the parabola is x^3-1/3x^3 = 2/3 x^3. since x^2 is h the area is therefore 2/3 h x (x). I think this is right. However, the parabola in this diagram is not y=x^2, as x=10 the height will be 100, the equation for the eg would be y=.18 x^2. The solution will yield slightly different numbers
actually using 0.18x^2 will yield the same are of 2:1 top vs bottom of parabola. Ive just worked out for all parabolas, the top area is always twice the bottom area.
Sir, your videos are helping me so much, I'm in my first year of Mechanical Engineering, I'm behind on my work and lectures but I'm dedicating my time to revise as much as possible each day, your videos are helping me understand these topics so much it's unbelievable, and it's not just Fluids videos, your Applied Mechanics videos have helped me a lot too, on behalf of people who might watch your videos finding them very helpful and might not comment, Thank you so much we can't thank you enough for the help, Keep up the good work you're helping a lot of students out there
+gemo77184
Thank you for writing. It is much appreciated. Keep working hard on your studies.
I have been struggling with this for the last few hours, I don't understand it the way my professor teaches it, and can I just say, I love that you've chosen to share this with the world and I wish you the best in life
We are glad you found us. Thank you for sharting. 🙂
This video explains everything so well. Im in Aerospace ENG and you are saving my life. Have a great day.
Thank you and keep working hard towards your degree
sir is a genius in explaining stuff ... can't thank you enough sir and God bless you!
Sir ..a bigggg thank you too... you would be glad to know that your videos are helping us to prepare for exams like JEE Advanced here in INDIA
That is great. The JEE exams are very challenging and take a lot of preparation. Keep working hard and good luck.
Bonjour Monsieur, vous êtes un excellent enseignant. Je travaille en France où j'inspecte des barrages. Vos cours sont tres bien expliqués. Bravo. Patrick
Merci beaucoup.
Absolutely nailed it. Thank you very much sir.
Sir your diagrams are so neat and beautiful
wow, now this video was EXCELLENT from beginning to end!!... nothing else to say..... :D
Hello Please i need an answer for this question. How can i measure the hydrostatic force of water under hemispherical dome on the dome ?
EXCELLENT explanation sir, thank you so much for your help!
when using a quarter circle instead of a parabolic wall, can the same principle be applied? or does the volume measurement change?
also when the situation is a half circle filled with water, can i apply the principle for a 1/4 circle times 2?
@@sigourneybergraaf3381 I think the principle can be applied to any shape but the volume varies since each shapes has different formula for its volume. This is just based on my understanding. I hope this helped you :)
These videos r lifesavers! Thank uuuu
This series on fluids is really helpful!! Thank you :))) Oh, and I wonder if there is a scenario like this. What if the curve is the same, but the fluid is UNDER the curved wall. That means the weight is zero (because the fluid is under the curve) and the resultant force R then is equal to Fp, right? I feel like my thinking lacks something, but I don't know what it is :(
The force on the wall depends on the pressure which is a function of depth. So regardles of the curve of the dam, there will be a force pushing against the dam.
Sir if the curved surface is inwards then how will we calculate force
Hi, may I know how you determine the centroid of each shape?
There is a whole playlist on centroids (c of g): Mechanical Engineering: Centroids & Center of Gravity (3 of 35) Centroids ruclips.net/video/tc11YEKHuFE/видео.html&index=3&list=PLX2gX-ftPVXWnfWWDNgu4x9hiCPTi0HB9
sir i have a question, why do u use the parabolic formula instead of a 1/4 circle? thanks :)
Because the shape of the dam is a parabola in this video.
In the Force P not to be 12 meters instead of 18 meters?
Thanks for this explains
i love you sir.. thanks
i like it so much
classic , sir is very good
Sir why the area is equal to 2/3*A*H?
The vertical area yes, the slanted area, no.
Use integration to derive the formula
You can imagine the vertex of the parabola is at the origin. Then the parabola can be written as y = Cx^2 for some constant C. Since we know the point (A,H), we can easily find that C = H/A^2. Then the rest is just integrating. You dont have to derive it in this way of course, you can just directly use the numbers given to you.
If you work it out using integrals. y=x^2, Integral 1/3x^3. At x the height h = x^2. The total square is x^3. The integrated bottom part of the curve is 1/3 x^3. Therefore the area above the parabola is x^3-1/3x^3 = 2/3 x^3. since x^2 is h the area is therefore 2/3 h x (x). I think this is right.
However, the parabola in this diagram is not y=x^2, as x=10 the height will be 100, the equation for the eg would be y=.18 x^2.
The solution will yield slightly different numbers
actually using 0.18x^2 will yield the same are of 2:1 top vs bottom of parabola. Ive just worked out for all parabolas, the top area is always twice the bottom area.
Nice sir
Good lecture... but!
You drew the resultant vector in wrong direction in very first place!
help us sir see clearly what yu wil be writing on yo board especially dimensions .....otherwise yu are a star.
Hello engineers
There are definitely engineers and engineering majors watching these videos.
Je veux un sous titrage en français pliiiiiiiiiiz