Flotation in a Liquid

If we hold an object completely below the surface of the liquid (e.g. underwater), we have seen that the buoyant force on it depends on its volume V_object and the density of the fluid: F_B = ρ_fluid · V_object · g

The question is whether the object will sink or rise up to the surface when released. That depends on whether the object’s weight, m_object · g is greater than or less than the buoyant force.

If this weight is greater than the buoyant force, it will sink below the surface.

The condition for an object to sink is: m · g > F_B, i.e., m_object · g > ρ_fluid · V_object · g .

Rearranging and canceling the “g”, we find the condition m_object /V_object > ρ_fluid. On the left side, the mass of the object divided by its volume is just its average density, ρ_object.

So we can write for an object that sinks, ρ_object> ρ_fluid, i.e., if the object’s density is greater than the liquid’s, it will sink to the bottom.

On the other hand, if the weight of our completely immersed object is less than the buoyant force on it, the object will experience a net force upwards, causing it to rise upwards to the surface. Once at the surface, the object will float with just a fraction of its volume submerged, such that the buoyant force and its weight are in equilibrium.

The condition for an object that floats: m · g < F_B, i.e. m_object · g < ρ_fluid · V_object · g. As before, rearranging and canceling the “g” gives us the condition for flotation:

m_object /V_object = ρ_object < ρ_fluid

This means that if the object’s average density is less than the liquid’s it will rise up and float on the surface.

Figure 1: Two balls underwater, with equal volumes but different weights. On the left, the ball weighs more than the buoyant force, so it sinks. On the right, the ball weighs less than the buoyant force, so it rises to the surface.