4.3 - Upthrust and floating

Pressure differences give an upward resultant

How can a steel boat float when a solid piece of steel sinks? The water responds to the submerged volume and shape of the object as a whole. It does not decide whether to push upwards by checking the name of the material.

An immersed object experiences pressure forces on every surface touching the fluid. Each force acts normally into that surface. In a stationary liquid, pressure increases with depth, so a horizontal bottom face experiences a larger upward pressure force than the downward pressure force on an equal-area top face.

For a cuboid, equal sideways pressure forces on opposite vertical faces cancel. The remaining fluid-force resultant is upward. This is upthrust. It acts even on an object whose weight is too large for it to float.

Consider a fully submerged cuboid of horizontal cross-sectional area $A$ and height $h$. If its top and bottom pressures are $p_1$ and $p_2$, then

The pressure-difference relation was $p_2-p_1=\rho gh$, and $Ah=V$ is the cuboid's volume. Here $\rho$ is the fluid density. Pressure is a force per area; multiplying its difference by area gives a force.

[DIAGRAM: asset_name: Pressure forces and upthrust; asset_slug: upthrust-pressure-forces; caption: Greater pressure on the lower face gives an upward fluid-force resultant; description: A cuboid is fully submerged in water, with equal small horizontal pressure arrows at matching depths on opposite sides. A downward pressure arrow at its top is shorter than an upward pressure arrow at its bottom. A separate free-body diagram shows upward upthrust and downward weight, without counting individual pressure forces again.]

Uniform pressure added everywhere contributes equal and opposite force effects over a closed immersed object; it does not change the upthrust. The pressure difference with depth is what matters. The same principle applies to gases, although changes of gas density with height can sometimes prevent use of one constant density over a large height.

Check the mechanism rather than assuming that every upward force is drag.

The weight of displaced fluid

A fully immersed object occupies space that would otherwise contain fluid. Imagine replacing the object with the same volume of the surrounding fluid. That fluid would be at rest, with its weight balanced by the surrounding pressure forces. The surrounding fluid exerts the same pressure forces on the object's surface, so its upthrust equals the weight of the fluid displaced.

This reasoning extends the cuboid result to other shapes. It is Archimedes' principle. For a partly immersed object, use only the volume below the fluid surface, rather than the object's total volume.

At fixed fluid density, doubling displaced volume doubles upthrust. Once a rigid object is fully submerged in an incompressible liquid, moving it deeper does not increase its displaced volume. Its upthrust stays approximately constant, although the absolute pressure on every surface rises. This often feels surprising until you compare the top-to-bottom pressure difference.

An immersed object can also be held by a force meter. If it is stationary and denser than the fluid, the upward tension and upthrust balance its weight:

Here $T$, upthrust $F$, and weight $mg$ are all forces in $\mathrm{N}$. The meter reads $T=mg-F$, often called the apparent weight. The actual gravitational weight remains $mg$; the fluid shares the support.

In a liquid, a sinking object at terminal velocity may have three forces: downward weight, upward upthrust, and upward drag. The terminal balance is $mg=F+D$, not necessarily weight equals drag alone.

Floating adjusts the displaced volume

A freely floating object at rest has upthrust equal to its weight. If you place a light block on water, it sinks far enough to displace its own weight of water, then stops moving vertically once equilibrium is reached. It does not have to become fully immersed.

For a homogeneous object of total volume $V_{\mathrm{total}}$, material density $\rho_{\mathrm{object}}$, and submerged volume $V_{\mathrm{sub}}$, neglecting air upthrust and other vertical forces,

The left side is the submerged fraction; it is dimensionless, as is the density ratio on the right. This is derived from the supplied upthrust relation and weight, rather than an extra formula printed on the data sheet. If the object is denser than the fluid, the fraction required would exceed one: immersion alone cannot provide enough upthrust for floating equilibrium.

A steel hull can enclose a large volume of air while containing a relatively small volume of steel. Its external submerged shape can displace enough water to support the hull and its contents. Its average density, including the enclosed space, can be below the water density even though steel's material density is higher. A solid steel piece of the same mass has much less available displaced volume.

For example, a floating vessel with total mass $1.4\times10^{4}\,\mathrm{kg}$ must displace $14\,\mathrm{m^{3}}$ of fresh water of density $1000\,\mathrm{kg\,m^{-3}}$, because equilibrium gives $\rho_{\mathrm{water}}V=m$. Adding cargo makes it settle lower and displace more water. There is a limit: the hull must retain enough enclosed volume and a suitable waterline rather than simply filling with water.

Bring the pressure picture and the volume calculation together.

Quick Recap

• Upthrust is the upward resultant of the fluid's pressure forces, rather than a reduction in weight.
• Its magnitude is the weight of displaced fluid; use fluid density and submerged volume.
• A force meter can read weight minus upthrust while the actual gravitational weight stays the same.
• Floating equilibrium requires upthrust to equal weight, which determines the submerged fraction or displaced volume.

A steel boat floats because its hollow shape can displace its own weight of water before it is fully submerged; a solid steel piece cannot displace enough water to balance its weight.