# Homework Help: Fluid Hydrostatics

1. May 31, 2015

### Steve_112

1. The problem statement, all variables and given/known data
Attached below

2. Relevant equations

3. The attempt at a solution
My first confusion is with the pressure measuring device, i just converted the units into Pa by dividing by 1.02x10-5 such that:
Pressure at the top of the tank = (0.07/1.02e-5) - (9800*0.3) = 3922.75Pa
So on the cone the average pressure is (3922.75 + 0.05*9800) = 4412.75Pa
The surface on which this force acts is the top part of the cone, where r = h*tan(30) , r = 0.0577, which means the surface is 0.031m^2
The force acts Vertically and horizontally but the horizontal components cancel each other out (right?) So the vertical force is:
4412.75*0.031*cos(60) = 68.4N, which isn't the correct result.

This is the same method i was using to solve problems related to tilted rectangular plates, however now it is a conical surface should i be solving it differently ?

Any help would be much appreciated, thank you!

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2. May 31, 2015

### Steve_112

Well i think i solved it, but i don't know why the first method didn't work!

I just considered the equilibrium of the fluid element surrounding the cone and bounded by the cylinder of radius r=0.0577.
The volume of this fluid element is
Vcyl - Vcone = 1.05e-3 - 3.49e-4 = 7e-4 m^3
Weight of fluid element = 7e-4 * 9800 = 6.83N
The upward force on this fluid element due to pressure is:
(pi*0.0577^2*(3922.75+0.1*9800) = 51.28N
The two horizontal forces cancel each other out thus the vertical force is 51.28 - 6.83N = 44.45N = S
Its the right answer but maybe its just all one big coincidence!
Could anyone confirm if this is correct, and also explain what is wrong with the first method.
Thanks!