# Is the buoyant force decreasing in upper layer the deeper water?

I'm asking about a real experiment. As in this I have read in an answer in this site which is What atomic forces are acting to resist me pushing an air filled bottle underwater?. As in ben51's answer buoyant force is due to higher density of water and high repulsive force of water particles in the deep of water for the overlying water's weight. So if I push a block whose density is less than water to the lowest level of water providing block wouldn't break in that pressure. I now remove the down-directed force of my hand. So as the block is getting up would the buoyant force decrease as the block rises up and a decreasing force would act on the block as it rises from the lowest layer of water?

The bouyant force is given by:

$$F(d) = \rho_w(d) \, V(d) \, g(d)$$

where all of the three quantities in the equation are functions of the depth $d$. They are all functions of the depth because:

• the density of the water $\rho_w$ increases with depth because the pressure compresses the water. This isn't a bit effect because water isn't very compressible but it is significant. I think the water density increase is around 5% at the bottom of the deepest ocean.

• the volume displaced decreases with depth because the pressure compresses your object as well as compressing the water around it. We can write the volume of the object as $m/\rho_o(d)$, where $m$ is the (constant) mass of the object, to move all the density dependence into the density of the object.

• the acceleration due to gravity increases with depth because we are getting closer to the centre of the Earth. It increases because the rock and iron core below us has a higher density than the water above us.

So rewriting our equation as suggested above gives:

$$F(d) = \frac{\rho_w(d)}{\rho_o(d)} \, m \, g(d)$$

The trouble is that all the depth dependent quantities vary in complicated ways so there is no simple answer as to how $F$ varies with depth. If your object is incompressible then $\rho_o$ is constant, and since both $\rho_w(d)$ and $g(d)$ increase with depth that means the buoyant force increases with increasing depth. I suspect this is the answer that you had in mind.

But suppose your object is highly compressible (like an air bubble). That means $\rho_o(d)$ increases strongly with depth, and this means that the buoyant force decreases with increasing depth.

The buoyant force acting on an object is $\rho_{water}\times V_{object} \times g$, where $V$ is the volume of the object. Here, you can see that pressure has no effect on the buoyant force acting on the object. What determines whether the object floats or sinks is the $F_{net}$ acting on the object. $F_{net}=F_{buoyant}-F_{hand}-F_{weight}$. Hence, if $F_{net}>0$, then the object floats, while if $F_{net}<0$, then the object rises up. If $F_{net}=0$, the object remains stationary in the water.

• @qulckmaths I'm asking about does the buoyant force decrease as the block rises in water? – Nobody recognizeable May 19 '18 at 3:04
• please see my previous comment – Nobody recognizeable May 19 '18 at 3:13
• No as Quickmaths says it only depends on the local density of the water (and the size of the object). Although there is more pressure pushing up at greater depth, there is also a column of water above the object pushing down. – Martin Beckett May 19 '18 at 3:35
• ps. In real life the buoyancy of say a submarine does vary with depth because of changes in water density due to temperature and salinity changes – Martin Beckett May 19 '18 at 3:36
• Yes, but in theory we always assume that density is uniform throughout :) – QuIcKmAtHs May 19 '18 at 4:43