I came up with a question stated below

"A vessel contains oil (density 0.8g/cc) over mercury (density 13.6g/cc). A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of the sphere in g/cc is ?"

The answer to this problem is 7.2 g/cc but this seems incorrect to me. I got 6.4 g/cc

If we consider the oil section only then, the massive force at the surface of lower part of the hemisphere is cancelling it out (Think in 3d), and small force which act on upper surface has vertical component (downwards) as horizontal component are cancelled out, there is a net downward force on sphere by oil (Buoyant force).

Sphere is homogenous, oil and mercury has constant density so dont assume out of the box

Regards enter image description here

Edit :- enter image description here

As Suggested by BIO's answer let put a small negligible rod in between two hemisphere under consideration then due to that flat surface there is a greater force on that flat surface which compensate other small force acting on curved surface to lead for an upward net force by oil, but in reality we cant assume that, as that flat surface isnt exposed to oil section (but only curved surface)

enter image description here Excuse for that drawing

Still cant figure it out help


4 Answers 4


Archimedes' principle states that the buoyant force equals the weight of the fluid displaced. This gives the solution of 7.2 g/cc. If you're not convinced, consider the following set-up

Separated hemispheres

The two hemispheres are only separated slightly but joined rigidly by a short, thin rod so that the pressures at the two flat surfaces are infinitesimally different. You should feel reassured to use Archimedes' principle now.

Regarding your original question, yes buoyant force can act downwards (as you have drawn in your diagram). The buoyant force is due to the liquid pushing the surface of the object. As pressure is higher at the bottom of the liquid due to gravity, usually the net buoyant force is pointing upwards. (A suction cup is a counterexample) However, if we just consider a part of the object's surface, the buoyant force acting on that patch is normal to the surface: $d\vec{F}_\textrm{buoy} = -p d\vec{A}$.

Your above reasoning suggests that the density of the sphere is equal to half the difference of the density of the two media. What happens when both fluids are the same?

Lastly, you can always calculate the force by direct integration, if you are familiar with vector calculus. It would be good to use spherical coordinates for this problem. Proof of Archimedes' principle uses the Gauss' theorem and can be found in this answer.

  • $\begingroup$ I got your idea but, why did you join two sphere with that small negligible rod, if the arrangement is of your type then buyoant force act upwards for oil as well as for mercury, ONLY BECAUSE OF THAT FLAT SURFACE OF YOUR HEMISPHERE as force acting on that flat surface upward with greater force , thats the crux of my whole point .That flat surface isnt exposed to oil so accordind to my diagram buoyant force must act downward in that oil section. To get upvote clarify things :)) $\endgroup$
    – 5 Dots
    Dec 27, 2021 at 4:18
  • $\begingroup$ You must not forget the force due to gravity. If we separate the spheres into two hemispheres, the buoyant force from the oil alone is not enough to support the weight of the upper hemisphere, causing it to sink downwards. On the other hand, the buoyant force from the mercury is greater than the weight of the lower hemisphere, causing the lower hemisphere to float upwards. $\endgroup$
    – Bio
    Dec 27, 2021 at 9:21
  • $\begingroup$ The result is that the two hemispheres will join back (the slice will be exactly at the same height as the oil-mercury boundary). The intention of the thin rigid rod is to allow for some oil/mercury to bathe the flat surface. This allows you to use Archimedes principle for each hemisphere separately, with hopefully less confusion as each hemisphere is now immersed only in 1 type of fluid. $\endgroup$
    – Bio
    Dec 27, 2021 at 9:26
  • $\begingroup$ Subsequently, consider the two hemispheres as 1 system. When you calculate the net force acting on the system, you should realize that the rigid rod doesn't contribute to the net force. Also, the upward buoyant force acting on the flat surface of the upper hemisphere exactly cancels the downwards buoyant force acting on the flat surface of the lower hemisphere, as the pressure is the same (remember that the rod is short). $\endgroup$
    – Bio
    Dec 27, 2021 at 9:30
  • $\begingroup$ According to your anology, everything works well as buoyant force must act upward for both the cases but my still question is that we cant assume that, as flat surface of sphere isnt exposed to oil nor mercury, so how does bouyant force act upward in oil section $\endgroup$
    – 5 Dots
    Dec 27, 2021 at 12:59

The byouyant force on an object is always opposite to the force that is applied to the fluid around object.

If you mean to get bouyant force to move inside of something instead outside of that. You need to have force on fluid in outward direction. You can do this by rotating a container containing the fluid. Now the buoyant force is toward center of motion.

Now your described question:

Let say the radius if sphere is $r$

The surface area of top hemisphere=$2\pi r^2$ =area of lower hemisphere. Convert this to a cylindrical shape so you have have $A$ area of top and $A$ area for bottom and the height equals $2r$.

Now, pressure at top(acting downward)= $\rho_o g h$

Pressure at bottom(acting upward)= $\rho_m g (h+2r)$

Force at top(acting downward)= $ \rho_o g h A$

Force at bottom(acting upward) = $\rho_m g (h+2r)A$

Difference in Force = $g A (\rho_o h - \rho_m (h +2r))$

If you are taking about gravity then there must be a net buoyant force upward in this case.


The buoyant force doesn't act downwards. There is actually a buoyant force acting upwards on the top 1/2 of the sphere. That's why the sphere has to have a higher density (7.2g/cc) than you calculate (6.4g/cc).

If the oil were replaced with air, the sphere would actually sink lower into the mercury. But the buoyant force of the displaced oil allows it to rise a bit. If the oil were denser, it would rise even more. If the density of the oil was 7.2g/cc, it would rise completely into the oil and have neutral buoyancy.

Imagine you have two separate hemispheres, one with the same density as the oil and one with the same density as the mercury. If you released them anywhere, they would migrate to the fluid with the same density. If the hemisphere that had the same density of mercury was in the oil, it would sink into the mercury, but being the same density as the mercury it would neither rise nor fall due to it's density in the mercury, it would have neutral buoyancy. Likewise if the hemisphere that had the same density as oil were in the mercury it would float up into the oil where it would have neutral buoyancy.

If we glued the two hemispheres together, the sphere would float/sink to the boundary between the oil and mercury with the 1/2 that has the density of oil on top in the oil and the 1/2 with the density of mercury on bottom in the mercury. If the sphere had a volume of 2cc, it would have a mass of 14.4g.

For a homogenous sphere to maintain the same position, it would have to have the same mass, 14.4g/2cc or 7.2g/cc. One difference from the glued hemispheres would be that it could rotate freely.


The buoyant force is the integral over the surface of the pressure times the area vector (in each infinitesimal, the area vector is the normal vector times the surface area) and the pressure. Buoyant force = $\int_S p \vec {dA}$. It's a surface integral, and Archimedes' principle can be considered an application of Gauss's theorem to that surface integral. It doesn't really make sense to talk about the buoyant force of a particular region of the surface, and doing so breaks the connection to Gauss's theorem.

Of course the integral over the top of an object will be a downward force. That's true for any fluid with positive pressure. If you imagine a pool with a hole at the bottom that is plugged with a ball (so there is no pressure at the bottom of the ball), the net force on the ball will be downward, but this won't really be a "buoyancy" force.

  • $\begingroup$ Flat surface isnt exposed to the oil section , so why does buoyant force act upward, i know that there someting wrong but cant figure it out. I want explaination on direction of buoyant force $\endgroup$
    – 5 Dots
    Dec 27, 2021 at 12:45

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