
#1
Apr1909, 01:34 PM

P: 8

1. The problem statement, all variables and given/known data
The Lshaped tank shown below is filled with water and is open at the top. (a) If d = 7.0 m, what is the force on face A due to the water? (b) What is the force on face B due to the water? 2. Relevant equations F=PA P=pgh 3. The attempt at a solution I've got the (a) part right: pgd*A = 1000*9.8*14*49 = 6722800. However, I don't have a clue how I can get pressure on face B! I tried setting distance as 3d = 21 and 4d = 28, and area as d^2 = 49, but nothing seems to work. The answer is supposed to be 8.40e+06. I need to know this for exam on Tuesday. Please, HELP!!!!!!! picture's at http://www.webassign.net/hrw/hrw7_1433.gif 



#2
Apr1909, 03:56 PM

HW Helper
P: 5,346

No image, so I have no idea either.




#3
Apr1909, 04:15 PM

P: 8





#4
Apr1909, 04:37 PM

HW Helper
P: 5,346

HELP!! Lshaped tank problem!
Face B is a vertical face. Hence the pressure changes with depth.
So consider then the average force across the B face between 2d and 3d times the area. 



#5
Apr1909, 04:55 PM

P: 8

So, do I do 2.5d..? And 2.5d seems to do the trick since I got the correct answer with it: F = PA = pgdA = 1000*9.8*(2.5*7)(7*7) = 8403500 N Thanks a bunch :) !!! 



#6
Apr1909, 06:58 PM

HW Helper
P: 5,346

Yep. You got it.




#7
Aug2811, 04:05 PM

P: 1

well, it is actually not 2.5D. the more precise working is integrating.
let pgh be the function where p and g is constant. Hence, you can integrate h where the upper boundary is 3D and the lower boundary is 2D. The answer would be more or less the same if you use 2.5D but better luck next time, you might need this :) 


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