I was reading through a solution to the following problem:

What acceleration will a completely submerged object experience if its density is three times that of the fluid in which it is submerged?

The solution states $F_\text{down} - F_\text{buoyancy} = m_\text{object}a_\text{object}$. This is reasonable so far, but then it states that $F_\text{down} = F_\text{weight}$. This confuses me; if an object is completely submerged then doesn't it have both the force of gravity and the force of the liquid above it pushing it down? In other words, isn't the force pushing it down equal to $F_g + \rho ghA$ where $A$ is the area of the surface of the object, $h$ is the depth at which the object is submerged, and $\rho$ is the density of the liquid? What am I missing?

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    $\begingroup$ Hint: why is the buoyant force $\rho_\text{liquid} m_\text{object} g$? $\endgroup$
    – rob
    May 16, 2015 at 22:58
  • $\begingroup$ To be honest, I thought I knew, and I've been trying to write a response to that hint for a few minutes now, but after revisiting I don't actually have a thorough understanding of why the buoyant force is that in an intuitive level. $\endgroup$ May 16, 2015 at 23:04
  • $\begingroup$ What would happen if the object was replaced by water? So what does that tell you about the forces on the object in water. $\endgroup$ May 16, 2015 at 23:05
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    $\begingroup$ @rob Oh I just got it! Consider a static liquid. Any arbitrary volume of liquid within it has a weight of $\rho Vg$. Since the liquid is static, by Newton's second law of motion, there must be a force acting upwards on that volume of liquid that has equal magnitude. But since that upward force, that buoyancy, is outside the volume, when we replace the volume of liquid with an object, that buoyancy force should still be there. That makes sense! $\endgroup$ May 16, 2015 at 23:35
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    $\begingroup$ @user3002473 I wish I could tell you it was an intentional error to throw you off the scent, but actually I goofed. $\endgroup$
    – rob
    May 17, 2015 at 1:41

1 Answer 1


Expanding on @SebastianRiese's comment, the buoyant force already takes care of the downward force caused by the upper liquids. Lets consider the problem from a physical perspective. There is the downward force from gravitation, the downward force from the liquid above, and the upward force from the liquid above.

enter image description here

In the diagram, the gravitational force isn't shown but the pressures causing the upward and downward forces from the liquid are shown. The buoyant force is calculated from subtracting these two pressures and multiplying by the cross-sectional area:

$F_b = (P_B - P_T)A = (\rho g(y+h) - \rho gy)A = \rho ghA = \rho gV$

Hence, the net force is

$F_{net} = F_g - F_b = F_g - (F_T - F_B) = F_g + F_T - F_B = F_{downward} - F_{upward}$

Hence, the buoyant force already takes into account the downward force caused by the liquids above.

  • $\begingroup$ if the object was smooth and the surface of the container, If the bottom was in contact with the container, will only downwards forces exist? So will the object not rise? $\endgroup$
    – Linkin
    Nov 18, 2020 at 6:51

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