A hollow metal cube with sides 100 mm floats at the interface between a layer of water and a layer of SAE 10W oil such that 10% of the cube is exposed to the oil. What is the pressure difference between the upper and lower horizontal surfaces? What is the average density of the cube?
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
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PURPOSE OF THE EXERCISE:
This exercise is designed to test the student’s ability to apply the hydrostatic pressure formulation to a multicomponent system and use buoyancy to determine an object’s effective (average) density from its floating condition. -
ENGINEERING CONTEXT:
In many practical applications—such as designing floats, buoyant devices, or oil–water separators—one must predict the pressure loads on surfaces immersed in stratified fluids and deduce overall effective densities. Once you master this procedure, you will be able to design and analyze structures that straddle fluid interfaces. -
PHYSICAL CONTEXT:
This problem combines hydrostatic pressure differences in two immiscible fluids with Archimedes’ principle to relate the displaced volumes in each fluid to the weight (and thus effective density) of the floating body. -
ASSUMPTIONS:
Assume that (a) the oil layer is uniform with density \rho_o=850\,\mathrm{kg/m^3} and the water has \rho_w=1000\,\mathrm{kg/m^3} (consistent with previous exercises), (b) the cube is oriented so its faces are horizontal and vertical, and (c) “10% of the cube is exposed to the oil” means that 10% of its vertical dimension (i.e. 0.1\,\mathrm{m}\times0.1=0.01\,\mathrm{m}) lies in the oil and the remaining 90% is in water. These are reasonable when the cube floats with its top portion submerged in the oil and its lower portion immersed in the water. -
SOLUTION TECHNIQUE:
(a) Define a vertical coordinate with the oil free surface at z=0.
– The oil occupies 0\ge z\ge -0.01\,\mathrm{m} and the oil–water interface is at z=-0.01\,\mathrm{m}.
– The cube, with side length L=0.1\,\mathrm{m}, has its top face at z=-0.01\,\mathrm{m} and its bottom face at z=-0.1\,\mathrm{m}.
(b) Determine the pressure on the top (upper horizontal) surface.
At the oil free surface the pressure is atmospheric, p_{\infty}. At a depth h_o=0.01\,\mathrm{m} in oil, the pressure is
p_{\mathrm{upper}}= p_{\infty}+\rho_{o}\,g\,h_o = p_{\infty}+850\times9.8\times0.01\,.
Numerically,
850\times9.8\times0.01\approx83.3\,\mathrm{Pa}\,,
so
p_{\mathrm{upper}}= p_{\infty}+83.3\,\mathrm{Pa}\,.
(c) Determine the pressure on the bottom (lower horizontal) surface.
Below the oil, water begins at z=-0.01\,\mathrm{m}. The cube’s bottom is at z=-0.1\,\mathrm{m}, implying a water depth of h_w=0.09\,\mathrm{m} below the interface. The pressure at the bottom face is
p_{\mathrm{bottom}}= p_{\mathrm{upper}}+\rho_{w}\,g\,h_w\,.
Substitute the value for p_{\mathrm{upper}}:
p_{\mathrm{bottom}}= \left(p_{\infty}+83.3\,\mathrm{Pa}\right) + 1000\times9.8\times0.09\,.
Since 1000\times9.8\times0.09\approx882\,\mathrm{Pa}, then
p_{\mathrm{bottom}}= p_{\infty}+83.3\,\mathrm{Pa}+882\,\mathrm{Pa}= p_{\infty}+965.3\,\mathrm{Pa}\,.
(d) Thus, the pressure difference applied between the lower and upper faces is
\Delta p= p_{\mathrm{bottom}}- p_{\mathrm{upper}}= \left(p_{\infty}+965.3\,\mathrm{Pa}\right)-\left(p_{\infty}+83.3\,\mathrm{Pa}\right)=882\,\mathrm{Pa}\,.
(e) Determine the cube’s average density using buoyancy. The buoyant force equals the weight of displaced oil and water. With the cube’s total volume V= L^3=0.001\,\mathrm{m^3}, the displaced volumes are:
– In oil: V_o=0.10\,V=0.0001\,\mathrm{m^3}
– In water: V_w=0.90\,V=0.0009\,\mathrm{m^3}
Then the total buoyant force is
B= g\left(\rho_o\,V_o+\rho_w\,V_w\right)=9.8\Bigl(850\times0.0001+1000\times0.0009\Bigr)\,.
Calculate the displaced mass:
850\times0.0001=0.085\,\mathrm{kg}\quad\text{and}\quad1000\times0.0009=0.900\,\mathrm{kg}\,.
So the total displaced mass is 0.085+0.900=0.985\,\mathrm{kg}. Therefore,
B=9.8\times0.985\approx9.653\,\mathrm{N}\,.
At equilibrium, the cube’s weight equals the buoyant force. Thus, the cube’s mass is approximately 0.985\,\mathrm{kg} and its average density is
\rho_{\mathrm{cube}}=\frac{0.985\,\mathrm{kg}}{0.001\,\mathrm{m^3}}\approx985\,\mathrm{kg/m^3}\,.
A student would find it beneficial to sketch a diagram of the cube showing the oil layer (with thickness 0.01\,\mathrm{m}), the oil–water interface, and the water layer (with thickness 0.09\,\mathrm{m}) along with arrows indicating the pressure on the top and bottom faces.
- REFLECTION:
The analysis shows that the pressure difference between the cube’s lower and upper horizontal faces is determined solely by the water column below the oil–water interface, yielding 882\,\mathrm{Pa}. Similarly, by equating the weight of the cube to the sum of the buoyant contributions from the two fluids, the effective density of the cube is found to be approximately 985\,\mathrm{kg/m^3}. The method emphasizes the importance of relating hydrostatic pressure distributions and buoyancy through volume fraction weighting in a stratified fluid system.
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 4 Engineering Models: Hydrostatic Equation (HIGHLY RELEVANT)
- Chapter 3 Engineering Models: Hydrostatic Equation (HIGHLY RELEVANT)
- Chapter 4.1: Watercraft Part 1 (HIGHLY RELEVANT)
- Chapter 3.2: Hydrostatic Equation (HIGHLY RELEVANT)
- Chapter 4.3: Archimedes’ Law Part 1 (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
this could have been done with fewer steps by just saying that the cube weight equals the weight of the displaced fluid, and thus the pressure difference needs to be the weight difference divided by cross-sectional area. One doesn’t have to solve a fluid mechanics equation, not even the hydrostatic equation, to solve this problem.