Suppose we have a ball floating in liquid A. Then we pour a liquid B on top of the ball (Assuming that the liquid is poured such that the ball doesn't oscillate) . The ball rises up. Why? According to me, the liquid on top (liquid B) will exert a force downward on the ball and it should go lower. The particles in the liquid B will exert an normal force which is acting downwards on the ball. How can adding liquid B make it rise up, that is, provide an upward force?

Taking a specific case, density of B > density of ball > density of A what would happen?

  • $\begingroup$ The result probably means that the ball rises up after pouring the liquid and once it's settled. The exact motion is too complicated to determine upon pouring, and also specification of the density of the fluids and the material of the ball is required. $\endgroup$ – GodotMisogi Mar 20 at 14:10
  • $\begingroup$ @GodotMisogi Even after the liquid has settled, the normal contact force should be downwards right? $\endgroup$ – Tejaswi Hegde Mar 20 at 14:19
  • 1
    $\begingroup$ If you have the ball standing at the bottom of a tank and you then pour liquid A over it, then it will also rise. The same thing happens when liquid B is poured - liquid A is the "bottom" this time. $\endgroup$ – Steeven Mar 20 at 14:20
  • $\begingroup$ @Steeven I think that depends on the density of the ball and whether it's larger or smaller than the density of liquid B. $\endgroup$ – GodotMisogi Mar 20 at 14:31
  • 1
    $\begingroup$ BTW, was there a typo in this question? I think it should read density A > density of ball > density B, or else the ball wouldn't have been floating in the first place. $\endgroup$ – JMac May 2 at 19:25

There's a key factor you're missing when you're considering the added weight on the ball.

The fluid B wont just be pushing down on the top of the object floating in liquid A, it also pushes down on liquid A itself. Due to the hydrostatic principle, this therefore increases the pressure of liquid A. This added pressure increases the force that liquid A is exerting on the object in the upwards direction. This is already enough to cancel out the effects of the added weight on top of the object.

As far as the additional height goes, this is because fluid B is more dense than air. Using Archimedes' principle, we can see that this should necessarily increase the buoyant force on the object because $$F_{Buoyant} = \rho_{fluid} V g$$ (where V is volume displaced by the submerged object, $\rho_{fluid}$ is the density of the fluid and $g$ is acceleration due to gravity.

We can see that when the density of the surrounding fluid increases (and it should be safe to assume fluid B is more dense than the air formerly above fluid A), the buoyant force increases, which will result in more lift on the submerged object. As it rises, the effective density of the fluid will decrease, as more and more of the object gets surrounded by the less dense fluid B, instead of fluid A. (Buoyancy then becomes $F_{buoyant} = ( \rho_{A} V_{A} + \rho_{B} V_{B} ) g$ where $V_A$ and $V_B$ are the volumes that each fluid acts on respectively) If fluid B is more dense than the object, it will continue to rise until it floats on fluid B; if fluid B is less dense than the object, it will rise until $F_{buoyant} = F_{gravity}$ and it reaches an equilibrium position between the two fluids.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.