Hello, wonderful channel you have got here, sir! I was wonderig, why didn't we put the weight of the gate as well as the reaction force at point A into account? Thank you very much
The weight of the gate should be included, but it was not provided in the problem statement. The reaction force at A was not included as we are computing the force that is required to hold the gate shut. If the force was larger than required, then there would be a reaction force on the other side of the gate at A, but we are computing the force that is "just required" to hold the gate shut, and as a result the force applied is enough to balance out the hydrostatic force but nothing more than that. Glad to hear that you find my lectures helpful.
I think the Fr should come out to be 61.04 because you calculated 58 on basis of total force acting along centerline but it should act along center of pressure that is on h* where h*=h+((I*sin230)/h) Plzzz clarify my doubt sir...
+bajanalpha4 It is because you have Patm on either side of the plate. On the water side the pressure is Patm plus the hydrostatic head pressure (rho g h) while on the other side it is only Patm. The hydrostatic pressure varies with vertical position on the plate, whereas the Patm component does not vary - given that Patm can be considered a constant for the very small vertical displacement that we are considering in this problem. If you were to do a static force balance on either side of the plate (pressure times area), the force on the air side would be a constant (Patm * Area). On the water side you would have the component due to atmospheric pressure (Patm * Area) plus a component that would vary with position along the plate - the (rho g h) component. Hope that helps ... and happy to hear that you find the videos useful.
+Ron Hugo Thanks so much! and yes! they help tremendously! I've been using you to help understand things since Thermo and will continue to Heat Transfer after I pass Fluids hopefully! Thank for all that you do!
![Hunxho - Ball Forever [Official Video]](http://i.ytimg.com/vi/nki9ZTF9Uls/mqdefault.jpg)
Great explanation, I really appreciate how you expand on each step. Thanks, man.
ruclips.net/video/eqkPn8FsxH0/видео.html
.
.
Your video is incredibly helpful, God bless you! Thank you so much!
Very helpful. Thank you for taking the time to make this video.
ruclips.net/video/eqkPn8FsxH0/видео.html
...
Hello,
wonderful channel you have got here, sir!
I was wonderig, why didn't we put the weight of the gate as well as the reaction force at point A into account?
Thank you very much
The weight of the gate should be included, but it was not provided in the problem statement. The reaction force at A was not included as we are computing the force that is required to hold the gate shut. If the force was larger than required, then there would be a reaction force on the other side of the gate at A, but we are computing the force that is "just required" to hold the gate shut, and as a result the force applied is enough to balance out the hydrostatic force but nothing more than that. Glad to hear that you find my lectures helpful.
Yes, understood.
Thank you for the very quick reply.
I think the Fr should come out to be 61.04 because you calculated 58 on basis of total force acting along centerline but it should act along center of pressure that is on h* where h*=h+((I*sin230)/h)
Plzzz clarify my doubt sir...
Thanks for this great tutorial :)
Sir can you please describe a problem with a cylindrical surface it will be very useful thank you sir
Hey, thanks for all of your videos. I'm confused about why Patm drops out of the equation. Can you explain that a little more please?
+bajanalpha4 It is because you have Patm on either side of the plate. On the water side the pressure is Patm plus the hydrostatic head pressure (rho g h) while on the other side it is only Patm. The hydrostatic pressure varies with vertical position on the plate, whereas the Patm component does not vary - given that Patm can be considered a constant for the very small vertical displacement that we are considering in this problem. If you were to do a static force balance on either side of the plate (pressure times area), the force on the air side would be a constant (Patm * Area). On the water side you would have the component due to atmospheric pressure (Patm * Area) plus a component that would vary with position along the plate - the (rho g h) component. Hope that helps ... and happy to hear that you find the videos useful.
+Ron Hugo Thanks so much! and yes! they help tremendously! I've been using you to help understand things since Thermo and will continue to Heat Transfer after I pass Fluids hopefully! Thank for all that you do!
Iam not understanding this problem.. but nice way to solving problem
ruclips.net/video/eqkPn8FsxH0/видео.html
.
..
Thanks sir
Hello sir, Why did you choose Ixy=0 @04:12
Thank you sir
great video, helped a lot 6 year later hah
ruclips.net/video/eqkPn8FsxH0/видео.html
..
you should have used the cos to find the not the sin right? its not to correct you its just to know the correct solution thanks
at 8:29 where does that - sign come from?
So you basically found force of p atm and then dropped it and made everyone confused..?
But Fr = Density * g * Ycp * A * Sin 30º as per FE handbook V. 10.01 - Page 180
ruclips.net/video/eqkPn8FsxH0/видео.html
..
@@edutechguruengineeringlear4280 good but late .. I have passed my exam .. thx
@@markmatta1936 👍👍
Too many mistakes. Centroid of gate is 0.866 not 0.5 2cos30 not 2sin30