3.2.4 - Density, pressure and upthrust

Density

Density is a property of a material or object that compares its mass with the volume it occupies. A small object can have a large mass if its material is dense; a large object can have a small mass if it contains a lot of empty space.

The density and pressure relationships in this lesson are supplied in the OCR data booklet, but you still need to choose the correct equation, use consistent SI units, and explain the physical meaning of the symbols.

Here, $\rho$ is density in $\mathrm{kg} \mathrm{m}^{-3}$, $m$ is mass in $\mathrm{kg}$, and $V$ is volume in $\mathrm{m}^{3}$. The same equation can be rearranged to $m = \rho V$ or $V = m/\rho$.

The SI unit $\mathrm{kg} \mathrm{m}^{-3}$ matters. Many density values are first measured using grams and cubic centimetres, especially in school laboratory work, but OCR calculation answers normally need SI units unless the question states otherwise.

Useful conversions:

• $1\,\mathrm{g} = 1 \times 10^{-3}\,\mathrm{kg}$
• $1\,\mathrm{cm}^{3} = 1 \times 10^{-6}\,\mathrm{m}^{3}$
• $1\,\mathrm{g}\,\mathrm{cm}^{-3} = 1000\,\mathrm{kg}\,\mathrm{m}^{-3}$

Worked example: finding density from mass and volume

A metal cube has mass $0.432\,\mathrm{kg}$ and volume $54.0\,\mathrm{cm}^{3}$. Calculate its density in $\mathrm{kg} \mathrm{m}^{-3}$.

First convert the volume:

Now use the density equation:

The density is $8.00 \times 10^{3}\,\mathrm{kg}\,\mathrm{m}^{-3}$.

In practical work, the method depends on the shape.

• For a regular solid, measure dimensions with a ruler, vernier caliper or micrometer, calculate volume from geometry, and measure mass with a balance.
• For an irregular solid that does not dissolve or absorb liquid, measure mass with a balance and find volume from the rise in water level in a measuring cylinder, or from displaced water.
• For a liquid, measure mass of a known volume. Use the mass of the empty container and the mass of container plus liquid, then subtract.

Common uncertainty points are meniscus reading, parallax, zero error in measuring instruments, trapped air bubbles around an object, and judging the volume of an irregular object from a scale with limited resolution.

Pressure

Pressure is not just another word for force. It tells you how concentrated a normal force is over a surface area. A force spread over a large area gives a smaller pressure than the same force acting over a small area.

Here, $p$ is pressure in pascals, $\mathrm{Pa}$, $F$ is the force normal to the surface in newtons, $\mathrm{N}$, and $A$ is the area in square metres, $\mathrm{m}^{2}$. One pascal is one newton per square metre:

The force must be perpendicular to the area. If a book rests flat on a table, it exerts a normal contact force on the table. This equals its weight when the book is in vertical equilibrium; the weight itself acts on the book. If a gas pushes on the wall of a container, the pressure force on each small part of the wall is normal to that wall.

Worked example: pressure on a floor

A box has weight $240\,\mathrm{N}$. It rests on a rectangular base of area $0.080\,\mathrm{m}^{2}$. Calculate the pressure on the floor.

The pressure is $3.0 \times 10^{3}\,\mathrm{Pa}$.

Now imagine the same box placed on a smaller face with area $0.020\,\mathrm{m}^{2}$.

The weight has not changed, but the pressure has increased because the area is smaller.

This same idea applies to solids, liquids and gases, but the mechanism differs. A solid may press because of its weight or an applied force. A fluid exerts pressure because its particles transfer force to surfaces, and in a gravitational field the weight of fluid above a point affects the pressure at that depth.

Pressure In A Fluid

In a fluid at rest, pressure increases with depth. The deeper point has more fluid above it, so the weight of that fluid column is larger. This is why a diver feels increasing pressure under water and why a dam must be stronger near the bottom than near the top.

The pressure due to a column of fluid is:

Here, $p$ is the pressure due to the fluid in $\mathrm{Pa}$, $h$ is the vertical depth below the fluid surface in $\mathrm{m}$, $\rho$ is the fluid density in $\mathrm{kg} \mathrm{m}^{-3}$, and $g$ is gravitational field strength in $\mathrm{N} \mathrm{kg}^{-1}$ or $\mathrm{m} \mathrm{s}^{-2}$.

[DIAGRAM: asset_name: 3.2.4 - Density, pressure and upthrust - diagram 1; asset_slug: 3_02_4_density_pressure_and_upthrust__diagram_01; recommended_method: drawn_physics; caption: The weight of a fluid column produces a pressure difference with depth.; description: A fluid column of depth h and area A is used to derive pressure difference from its weight.]

You can derive the equation from $p = F/A$.

For a fluid column of height $h$ and cross-sectional area $A$:

The mass of fluid in the column is:

Its weight is:

The pressure on the base of the column is force divided by area:

The area cancels. That cancellation is important: in a static fluid, the pressure at a depth depends on $h$, $\rho$ and $g$, not on the shape or total width of the container. Two points at the same depth in the same static fluid have the same pressure due to the fluid, even if one side of the container is wide and the other is narrow.

Boundary note: $p = h \rho g$ gives the pressure due to the fluid column. If a question asks for total pressure in an open liquid, atmospheric pressure at the surface would also be included. In this lesson, unless a question explicitly says total or absolute pressure, treat $p = h \rho g$ as the pressure due to the fluid.

Worked example: pressure and force under water

A small horizontal hatch on a tank is $1.80\,\mathrm{m}$ below the water surface. The density of water is $1000\,\mathrm{kg}\,\mathrm{m}^{-3}$. Take $g = 9.81\,\mathrm{N}\,\mathrm{kg}^{-1}$.

Pressure due to the water:

If the hatch has area $0.0250\,\mathrm{m}^{2}$, the force due to this pressure is:

The pressure is set by depth, density and $g$; the force also depends on the area being pushed.

Upthrust And Archimedes' Principle

An object in a fluid experiences pressure on all its surfaces. The pressure is greater lower down, so the upward pressure force on the bottom of the object is larger than the downward pressure force on the top. The horizontal pressure forces on opposite sides cancel for a fully surrounded object in a static fluid. The resultant fluid force is upward: this is upthrust.

[DIAGRAM: asset_name: 3.2.4 - Density, pressure and upthrust - diagram 2; asset_slug: 3_02_4_density_pressure_and_upthrust__diagram_02; recommended_method: drawn_physics; caption: Greater pressure below an immersed body produces upthrust.; description: Pressure is greater at the bottom of an immersed body than at the top, producing upthrust; horizontal pressure forces cancel.]

Archimedes' principle gives the size of the upthrust without needing to add up all the pressure forces on every surface.

This means:

If the displaced fluid has density $\rho_{f}$ and the displaced volume is $V_{d}$, then:

This expression is not a new law separate from Archimedes' principle; it is just "weight of displaced fluid" written using $m = \rho V$ and $W = mg$.

Two details decide many answers.

1. Use the density of the fluid that has been displaced, not the density of the object.
2. Use the displaced volume. If the object is fully submerged, this is the whole object volume. If it is floating partly above the surface, this is only the submerged volume.

Worked example: upthrust on a fully submerged object

A solid object of volume $3.40 \times 10^{-4}\,\mathrm{m}^{3}$ is fully submerged in water of density $1000\,\mathrm{kg}\,\mathrm{m}^{-3}$. Calculate the upthrust.

The displaced volume is the full object volume:

Using Archimedes' principle:

The upthrust is $3.34\,\mathrm{N}$ upward.

When released from rest with no other significant forces, an object whose weight exceeds this upthrust initially accelerates downward. If upthrust exceeds weight it initially accelerates upward. Drag then develops as it moves; a rising object can settle floating partly submerged.

For a submerged cuboid of horizontal cross-section $A$, the top and bottom depths are $h_1$ and $h_2$. Subtract the downward top pressure force from the upward bottom pressure force:

Any common surface pressure cancels. This derives Archimedes' result directly for the cuboid; the general principle applies to other shapes too.

Floating, Sinking And Practical Evidence

Floating and sinking are force-equilibrium questions. For an object in a fluid, the main vertical forces are usually weight downward and upthrust upward. If the object is stationary or floating at rest, the resultant force is zero.

For a floating object:

So:

This does not mean the object has displaced its whole volume. A floating object displaces just enough fluid for the upthrust to equal its weight.

For a fully submerged object released from rest, with no other significant forces initially:

• if $W > U$, the resultant force is downward and the object sinks;
• if $U > W$, the resultant force is upward and the object rises;
• if $U = W$, the object is in vertical equilibrium.

Average density is a useful way to predict floating. If an object's average density is less than the density of the surrounding fluid, it can float. A steel ship floats because its overall volume includes air; its average density is much lower than solid steel. A helium balloon rises in air because the gasbag and helium together have lower average density than the displaced air, provided the upthrust exceeds the total weight.

Worked example: floating fraction

A wooden block has density $640\,\mathrm{kg}\,\mathrm{m}^{-3}$ and floats in water of density $1000\,\mathrm{kg}\,\mathrm{m}^{-3}$. Find the fraction of the block's volume below the water surface.

Let the block's total volume be $V$ and the submerged volume be $V_{s}$.

The block's weight is:

The upthrust is the weight of displaced water:

For floating equilibrium, $U = W$:

Cancel $g$ and rearrange:

So $64.0\%$ of the block's volume is submerged.

A practical way to test Archimedes' principle is to measure the apparent weight of an object in air and then when it is fully submerged.

1. Measure the object's weight in air with a newton meter.
2. Lower it fully into water without touching the bottom or sides of the container.
3. Record the smaller apparent weight.
4. The reduction in the newton-meter reading is the upthrust. The gravitational weight itself has not changed: the meter measures the supporting tension, and $T+U=W$.
5. Compare this with the weight of water displaced, found from the displaced water volume and $\rho V g$.

Good practical answers mention controlling trapped air bubbles, keeping the object fully submerged when required, reading the newton meter at eye level, avoiding contact with the container, repeating readings, and using a measuring cylinder or displacement can with suitable resolution.

Takeaway