FE Exam Fluid Mechanics - 2.6 - Practice Problem - Fluid Statics

[@mohamedelkholy2421]

Why didn't you add the P atm as well to the water pressure in this equation? why did you neglect it?

[@annemarietowle7584]

When I approached this problem, I went to the equation in the handbook and plugged n chugged: Fr = (Patm + ρ g yc sinθ)A, with ρ = 1000kg/m^3, g = 9.81m/s^2, yc = 4m, θ = 45deg, A = 2π and Patm = 103135Pa. Why is this not the correct approach?

[@laurelshepard796]

I think the prof assumed Patm is negligible because most of the time Patm

[@ali1993sarah]

Thanks again for the videos! One thing I did not get, is how did you use hc=4m when you calculated Fp? I know you said 1 m +3 m =4m, but when you calculate Fp, hc got to be more than 4m, because you Fp is not at the center of the gate? Please explain Thanks

[@CollegeFluidMechanics]

Ali: what you are asking is the most common mistake I see in my exams. So it is a little confusing topic. Fp does NOT apply to the center of gravity. It applies to the center of pressure which is pretty much always a little higher value than the geometric center. I have my lecture videos here: kzbin.info/aero/PLCCmgp4iMmxh1S9TyRrJnxghaSjFi51TA have 2 theory part and 5 solved example questions. It may help you to clarify. Best of luck

[@sketty1011]

Hello professor, I don’t know if I’m mistaken but I’m confused on why yc is not the same as hc. I’ve been doing that but I guess I was wrong. I thought once you find hc, you use it to find the y particular eq.

[@CollegeFluidMechanics]

Hi Valerie, yc is defined from the origin placed at the intersection of inclined surface and free surface. On the other hand, hc is defined in the -z direction. So, they are related by sin(theta)=hc/yc where theta is the angle of the inclined surface makes with the free surface.