浮体

Suppose that we have a liquid of density $\rho$ and an object of mass $M$, volume $V$, and density $D=\frac{M}{V}$.

By Archimedes's Principle, we may deduce that the object will float if $D\lt\rho$. Since we are talking about floating objects, we will restrict ourselves to this case.

The most important thing to calculate is how much of the object is immersed in the fluid.

Suppose a volume $V_2$ lies below the surface of the fluid.

By Archimedes's Principle, $$F_{buoy}=\rho g V_2$$ Calculating the weight of the object, $$F_{gr}=Mg$$ When the object is stable and floating, we have $F_{buoy}=F_{gr}$. Thus, we may solve for $V_2$: $$\rho g V_2=Mg$$ $$\rho V_2=DV$$ $$V_2=\frac{D}{\rho}V$$ This is a simple and intuitively pleasing result that states that the portion of the object that lies under the surface of the fluid is equal to the ratio of the densities of the object and fluid.

For a fun fact, this explains the common phrase "just the tip of the iceberg." Ice has a density roughly $92$ percent that of water. Thus, only about $8$ percent of an iceberg is ever above the water, which makes them appear much smaller and thus hard to detect. On a more serious note, this also provides some explanation for the Titanic disaster. About a hundred years ago, when detection technology was far weaker, spotting an iceberg when most of it was under the water would be quite difficult.