A Steel Ship Floats and a Steel Nail Sinks

The two are made of the same material, so the material cannot be the answer.

What differs is shape. The nail is solid steel throughout. The ship is mostly hollow, so the space it occupies contains far more air than steel — and a large volume with a modest mass gives a low average density.

The whole of floating and sinking follows from that one comparison, together with the upward push a liquid gives to anything placed in it. This page covers the second part of the ICSE Class 8 Physics chapter on measurement: upthrust, the principle of buoyancy, relative density, and the law of floatation.

Upthrust, or buoyant force, is the upward force a fluid exerts on a body placed in it. It exists because pressure in a liquid increases with depth.

Consider a block held under water. Its bottom face is deeper than its top face, so the liquid pressure on the bottom is greater. Pressure produces force, so the upward force on the bottom face exceeds the downward force on the top face. The sideways forces on opposite faces are at equal depths and cancel.

What is left over is a net upward force — the upthrust.

Apparent loss of weight. Because the upthrust acts against gravity, a body in water seems lighter than in air. Its true weight has not changed; only the reading on the spring balance has.

$$\text{apparent weight}=\text{true weight}-\text{upthrust}$$

so

$$\text{upthrust}=\text{weight in air}-\text{weight in water}$$

Worked example. A solid hangs from a spring balance and reads $2.5\text{N}$ in air. Lowered fully into water, the reading falls to $1.5\text{N}$:

$$\text{upthrust}=2.5-1.5=1.0\text{N}$$

The solid still weighs $2.5\text{N}$ — the water is simply carrying $1.0\text{N}$ of it.

This is what you feel in a swimming pool. Lifting a friend is far easier in water than on the poolside, and a bucket of water is easier to lift while it is still submerged in the well than after it clears the surface. In both cases the upthrust disappears the moment the object leaves the water.

Relative density is the density of a substance compared with the density of water:

$$\text{R.D.}=\frac{\text{density of the substance}}{\text{density of water}}$$

Because it is a ratio of two densities, relative density has no unit. This is the one quantity in the chapter where writing a unit is the mistake.

For the solid above:

$$\rho =\frac{250}{100}=2.5{\text{g/cm}}^3,\text{R.D.}=\frac{2.5}{1}=2.5$$

The number says the substance is $2.5$ times as heavy as an equal volume of water.

Measuring it without knowing the volume. Since the upthrust equals the weight of an equal volume of water, the ratio of weights gives the ratio of densities directly:

$$\text{R.D.}=\frac{\text{weight in air}}{\text{loss of weight in water}}$$

Using the spring balance readings:

$$\text{R.D.}=\frac{2.5}{2.5-1.5}=\frac{2.5}{1.0}=2.5$$

The same $2.5$, obtained from two balance readings alone — no measuring cylinder, no volume. This is why the method works for an irregular solid of any shape.

Reading a relative density. Compare it with $1$:

- R.D. greater than $1$ — denser than water, so it sinks
- R.D. less than $1$ — less dense than water, so it floats
- R.D. equal to $1$ — it stays wherever it is placed, fully submerged

Ice has a relative density of about $0.92$ and so floats; iron is close to $7.8$ and sinks.

A floating body displaces a weight of fluid equal to its own weight.

That is the whole law, and it follows from balance: a body floats when it is at rest, which means the upthrust exactly equals its weight. Since the upthrust is the weight of fluid displaced, the two weights must be equal.

The three cases, comparing the body's density ${\rho }_b$ with the fluid's ${\rho }_f$:

- ${\rho }_b>{\rho }_f$ — at full submersion the upthrust is still less than the weight, so the body sinks
- ${\rho }_b={\rho }_f$ — the forces balance when fully submerged, so it floats just below the surface
- ${\rho }_b<{\rho }_f$ — the forces balance before it is fully submerged, so it floats partly out of the liquid

How much sticks out. A floating body sinks only far enough to displace its own weight, so

$$\frac{\text{volume submerged}}{\text{total volume}}=\frac{{\rho }_b}{{\rho }_f}$$

Worked example. A block of wood of density $0.6{\text{g/cm}}^3$ has volume $500{\text{cm}}^3$. Its mass is

$$m=0.6\times 500=300\text{g}$$

To float it must displace $300\text{g}$ of water, which is $300{\text{cm}}^3$. So

$$\frac{300}{500}=0.6=60\%$$

sits below the surface and $40\%$ above — matching the ratio $0.6/1$ exactly.

The same calculation explains the floating ice of the previous chapter: relative density $0.92$ means about $92\%$ of an iceberg is under water and only $8\%$ shows.

And the ship. A vessel of total mass $600\text{kg}$ must displace $600\text{kg}$ of water:

$$V=\frac{m}{\rho }=\frac{600}{1000}=0.6{\text{m}}^3$$

So it settles until $0.6{\text{m}}^3$ of its hull is below the waterline. Load more cargo and it settles deeper, displacing more — which is why an overloaded boat sinks: the hull runs out of volume to submerge before the upthrust can match the weight.