Submerged Cylinder

The force $\vec{F}_t$ is equal to the pressure at the center of the top times the area.

The coordinates of the center of the top is $(0, 0, -3) + \dfrac{1}{\sqrt{3}} (1, 1, 1)$

Therefore, the force on the top is $\vec{F}_t = (\pi (1)^2 ) \rho g (3 - \dfrac{1}{\sqrt{3}} ) = 10 \pi (3 - \dfrac{1}{\sqrt{3}} )) (-1, -1, -1) / \sqrt{3}$

Similarly, the force on the bottom is $\vec{F}_b = 10 \pi ( 3 + \dfrac{1}{\sqrt{3}} ) (1, 1, 1) / \sqrt{3}$

The total force on the cylinder is the weight of the displaced water (Archimedes' principle), and it is pointing upward.

$\vec{F}_{\text{Total}} = \rho g V_{\text{cylinder}} = 20 \pi (0, 0, 1)$

From this, we can find $\vec{F}_s = \vec{F}_{\text{Total}} - \vec{F}_t - \vec{F}_b = ( \dfrac{20 \pi}{3}, \dfrac{20 \pi}{3}, \dfrac{40 \pi}{3} )$

Thus $| \vec{F}_t | = 10 \pi (3 - \dfrac{1}{\sqrt{3}} ) , | \vec{F}_b | = 10 \pi ( 3 + \dfrac{1}{\sqrt{3}} )$ , and $| \vec{F}_s | = 20 \pi \sqrt{ \dfrac{2}{3} }$

And the required ratio follows from the above three figures.

$\dfrac{ | \vec{F}_t | | \vec{F}_b | | \vec{F}_s | }{ | \vec{F}_t |+ | \vec{F}_b |+ | \vec{F}_s | } = \boxed{1829.96}$

Attached below is a heavily commented simulation code. I may update this solution with analytical steps later, but the comments should provide some explanation of the steps. Any line or figment of code followed by '%' is a comment for the reader.

It was satisfying to get this problem right on the first try. This is a good one

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77

clear all
clc

R       = 1;                              % Radius of cylinder
h       = 2;                              % Height of Cylinder 
rho     = 1;                              % Fluid Density
g       = 10;                             % Gravitational acceleration (-Z)

n       = (1/sqrt(3))*[1;1;1];            % normal vector to each disk

C       = [0;0;-3];                       % Cylinder center

% Using triangle law of vector addition
% The coordinates of the disk centers can be easily found as follows:

C1      = C + 0.5*h*n;                    % Coordinates of upper disk
C2      = C - 0.5*h*n;                    % Coordinates of lower disk

% Equation of plane of disks:
% x + y + z = xci + yci + zci
% Where i is 1 or 2 depending on whether it is upper or lower disk:
% The point (0,0,zci) satusfies the equation of the plane:
% This point is used to define unit vectors on the plane
% Any other point can be chosen as well

ZC1     = [1 1 1]*C1;                     % Sum of upper disk coordinates
ZC2     = [1 1 1]*C2;                     % Sum of lower disk coordinates

PC1     = [0;0;ZC1];                      % Point lying on upper disk plane
PC2     = [0;0;ZC2];                      % Point lying on lower disk plane

% Defining unit vectors for upper disk plane:
ic1     = (PC1 - C1)/(norm((PC1 - C1)));  % Unit vector - upper disk plane
jc1     = cross(n,ic1);                   % Unit vector - upper disk plane

% Defining unit vectors for lower disk plane:
ic2     = (PC2 - C2)/(norm((PC2 - C2)));  % Unit vector - lower disk plane
jc2     = cross(n,ic2);                   % Unit vector - lower disk plane

dr      = 1e-4;                           % Numerical resoloution for r
dtheta  = pi/10000;                       % Numerical resoloution for theta


F1      = [0;0;0];                        % Bottom face force initialise
F2      = [0;0;0];                        % Top face force initialise

% Nested loops for numerical integration:
for r = 0:dr:R

    for theta = 0:dtheta:2*pi

        Pc1 = C1 + r*cos(theta)*ic1 + r*sin(theta)*jc1; % Point on upper disk
        Pc2 = C2 + r*cos(theta)*ic2 + r*sin(theta)*jc2; % Point on lower disk

        Zpc1 = [0 0 1]*Pc1; % Z coordinate of Point on upper disk
        Zpc2 = [0 0 1]*Pc2; % Z coordinate of Point on lower disk

        P1   = -rho*g*Zpc1; % Pressure at point on upper disk
        P2   = -rho*g*Zpc2; % Pressure at point on lower disk

        dA   = r*dr*dtheta; % Disk area element

        dF1  = -P1*dA*n;    % Elementary force on upper disk
        dF2  = P2*dA*n;     % Elementary force on lower disk

        F1   = F1 + dF1;    % Numerical integration for upper disk force
        F2   = F2 + dF2;    % Numerical integration for lower disk force
    end
end

Fb     = F2;
Ft     = F1;
% COmputing force on the cylinder curved surface:
Fs     = [0;0;rho*g*pi*R^2*h] - Fb - Ft; % Archimedes principle 

ANSWER = (norm(Fb)*norm(Ft)*norm(Fs))/(norm(Fb) + norm(Ft) + norm(Fs));
% ANSWER ~ 1830.3