Need help with fluids and pressure problem

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The discussion revolves around calculating the forces on two faces of an L-shaped tank filled with water. For face A, the correct approach involves integrating the pressure over the height of the water, considering that pressure varies with depth. The initial attempt yielded an incorrect force of 2.85x10^6 N, while the correct force on face A, after applying the integration method and excluding atmospheric pressure, is determined to be 1.25x10^6 N. The problem emphasizes the importance of understanding how pressure affects force on submerged surfaces. Proper integration is essential for accurately solving fluid mechanics problems involving varying pressure.
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[SOLVED] need help with fluids and pressure problem

Homework Statement


The L-shaped tank shown below is filled with water and is open at the top.
(a) If d = 4.0 m, what is the force on face A due to the water?
N (up)

(b) What is the force on face B due to the water?
N (right)
hrw7_14-33.gif


Homework Equations


F=PA


The Attempt at a Solution


Fa=PaAa+Patmosphere
Pa=pgHa
Ha=2d
Fa=pg2d^3+Patmosphere
d=4
I got 2.85x10^6N but that was wrong
I was given this hint when i got it wrong:Can you find the pressure at a given depth? How is force related to pressure for a given horizontal surface area? On the vertical wall, how does pressure vary with depth along the wall? Do you see that you need to first consider a strip that has the wall's width but only a differential height dy? What is the force on the strip? Do you see that you next must integrate to find the force on the full wall (all the horizontal strips)?
is this how it should be done:
Fa=Integral(pgd^2 dy)+Patmosphere
 
Last edited:
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(a) If d = 4.0 m, what is the force on face A due to the water?
N (up)
i was not suppose to add the Patmosphere
Fa=1.25E6N
 
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