Submerged Cube

Classical Mechanics Level 5

The surface of a large body of water coincides with the plane $z = 0$ , and gravity is $10 \text{ m/s}^2$ in the $-z$ direction. A cube is submerged under the water, with its eight vertices $(x,y,z)$ at the locations defined below:

$\begin{aligned} \vec{P} &= (0,0,-50) \\\\ \vec{v_1} &= (3,14,18) \\ \vec{v_2} &= (6,-18,13) \\ \vec{v_3} &= (-22,-3,6) \\\\ \vec{P_1} &= \vec{P} + 0 \, \vec{v_1} + 0 \, \vec{v_2} + 0 \, \vec{v_3} \\ \vec{P_2} &= \vec{P} + 1 \, \vec{v_1} + 0 \, \vec{v_2} + 0 \, \vec{v_3} \\ \vec{P_3} &= \vec{P} + 0 \, \vec{v_1} + 1 \, \vec{v_2} + 0 \, \vec{v_3} \\ \vec{P_4} &= \vec{P} + 1 \, \vec{v_1} + 1 \, \vec{v_2} + 0 \, \vec{v_3} \\ \vec{P_5} &= \vec{P} + 0 \, \vec{v_1} + 0 \, \vec{v_2} + 1 \, \vec{v_3} \\ \vec{P_6} &= \vec{P} + 1 \, \vec{v_1} + 0 \, \vec{v_2} + 1 \, \vec{v_3} \\ \vec{P_7} &= \vec{P} + 0 \, \vec{v_1} + 1 \, \vec{v_2} + 1 \, \vec{v_3} \\ \vec{P_8} &= \vec{P} + 1 \, \vec{v_1} + 1 \, \vec{v_2} + 1 \, \vec{v_3}. \end{aligned}$

The density of water is $1000 \text{ kg/m}^3$ . Let $(F_1, F_2, F_3, F_4, F_5, F_6)$ be the magnitudes of the hydrostatic pressure forces on the six individual sides of the cube, and let $F_B$ be the net hydrostatic pressure force on the entire cube (also known as the buoyant force).

Determine the following ratio:

$\frac{F_1 + F_2 + F_3 + F_4 + F_5 + F_6 }{F_B}.$

The answer is 8.22.

1 solution

Laszlo Mihaly
Oct 3, 2018

Let us first consider a square plate of arbitrary orientation, and calculate the total hydrostatic force acting on one side of it. The center of the square is at depth $-z_c$ and the pressure there is $p_c=z_c\rho g$ , where $\rho$ is the density of water and $g$ is the gravitational acceleration. The force acting on a surface element $\Delta A$ is $\Delta F=p_c \Delta A$ . As we move away from the center in the horizontal direction the force will be the same. If we move in the vertical direction to $z_1=-z_c+\eta$ the force will be $\Delta F_1=(p_c-\eta \rho g) \Delta A$ . Using the center of the square as a symmetry center, we can always find a corresponding surface element that is at elevation $z_2=-z_c-\eta$ and the force is $\Delta F_1=(p_c+\eta \rho g) \Delta A$ . Notice that if we add up the forces so that first we pair up these surface elements, we get $\Delta F= \Delta F_1+\Delta F_2=2 p_c \Delta A$ , independent of the actual elevation of the two surface elements. Therefore the total force will be $F=p_cA$ , where $A$ is the area of the square.

Next we look at the cube, oriented so that two of the faces are horizontal and 4 of the faces are vertical. The center of the cube is at depth of $-z_c$ . Applying the ideas outlined above the total force on the 4 vertical faces is $F_1+F_2+F_3+F_4=4Ap_c$ . The top face has a force of $F_5=A (p_c -\rho g \frac{a}{2})$ and the bottom face has a force of $F_6=A (p_c +\rho g \frac{a}{2})$ The total force is $F_1+F_2+F_3+F_4+F_5+F_6= 6Ap_c$ . Notice that opposing faces each contribute $2Ap_c$ to this sum.

What happens if we rotate the cube around its center? According to the arguments outlined in the first paragraph each face will experience a force that is related to the pressure at its center. Due to the symmetry of the cube, if the rotation moves the center of one face downwards by a certain amount, the center of the opposite face will move up by the same amount. Therefore the total force on a pair of such faces remains the same, $2Ap_c$ , and the grand total is still $F_1+F_2+F_3+F_4+F_5+F_6= 6Ap_c$ . We get

$\frac{F_1+F_2+F_3+F_4+F_5+F_6}{F_B}= \frac {-6a^2 \rho g z_c}{\rho g a^3}=-\frac{6 z_c}{a}$ ,

independent of the rotations of the cube.

In our problem the center of the cube is at $\vec r_c=\vec P+\frac{1}{2}(\vec v_1+\vec v_1+\vec v_1)$ . The depth is $z_c=-31.5m$ , the edge of the cube is $23m$ and the answer is

$\frac{F_1+F_2+F_3+F_4+F_5+F_6}{F_B}= 8.217$ .

It's interesting to see your approach, which relies on physical reasoning to "compress" the problem and make it solvable without heavy computation. I actually double-integrated to find all the pressures, which is less principled, but it allows me to proceed from just a few basic primitives. I'd be interested in hearing your philosophy regarding that general dichotomy. I've noticed that it applies to almost every problem I've seen.

Steven Chase - 2 years, 8 months ago

I am very impressed by the computational skills you demonstrated by setting up the surface integrals without a mistake in the code.

In my case, I first "guessed" that the result does not depend on the orientation of the cube. (The "guess" was vaguely based on the idea that the if buoyancy force does not depend on the orientation, perhaps the sum of the forces is also independent.) I calculated the sum for a simple case and entered it as my first attempt for the solution. Since you had the same result, my guess was confirmed, and I was encouraged to justify it with symmetry arguments.

As a physicist I always look for symmetries and simplifications. I am also aware of the fact that sometimes the simplifications yield a totally wrong result. In this particular case, I think it was a kind of a collaboration between you and me. If I did not get back the confirmation from your calculation, I would have given up on finding a simple solution.

Laszlo Mihaly - 2 years, 8 months ago

That's very clever, and a good guess based on the invariance of the net force. I did happen to know that the overall x and y forces should be zero, and that the overall z force should equal the weight of the displaced water. Upon seeing the program spit out exactly that, I had some confidence that I had done it right.

Steven Chase - 2 years, 8 months ago

Submerged Cube

The answer is 8.22.

1 solution

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