Fluid Dynamics - Calculate the resultant moment

[Mtrn_engineer]

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

 

[tiny-tim]

Welcome to PF!

Hi Mtrn_engineer! Welcome to PF!

(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. ​