Fluid Dynamics - Calculate the resultant moment

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SUMMARY

The discussion focuses on calculating the resultant moment at the bottom of a vertical wall separating two liquid reservoirs: one containing oil with a density of 864 kg/m³ at a height of 6m, and the other containing water with a density of 1000 kg/m³ at a height of 5m. The resultant moment is calculated using the formula FR = PCA, leading to a final answer of 9.98 N·m after subtracting the moments due to both fluids. The calculation involves integrating the moments of horizontal slices of each fluid to arrive at the correct result.

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Mtrn_engineer
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Homework Statement



A rigid vertical wall separates two liquid reservoirs. The first reservoir contains oil of density p=864 kgm-3 at a height of 6m above the ground, while the second reservoir contains water (1000kgm-3) at a height of 5m from the reservoir bottom. Calculate the resultant moment, at the bottom of the vertical wall.

http://imgur.com/jjlf3eO

Homework Equations



FR = PCA = pgycA

The Attempt at a Solution



Area is not given so I will take that out of the equation

water = 1000*9.81*5/2 *5/3
minus
oil = 864*9.81 *6/2*6/3

Answer = 50855 - 40875 = 9.98 n.M

I think that the answer is right but I am not sure as to why it is right. Can someone please verify and explain. Thank you
 
Last edited:
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Hi Mtrn_engineer! Welcome to PF! :wink:

(I think the question is probably asking for the moment per mass, so you'll be factoring out an m = ρAL somewhere)

Call the area A, and slice each side into horizontal slices of area A thickness dh and depth h …

then find the moment for that slice, and then integrate wrt h.

Show us what you get. :smile:
 

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