define and use density

Topic 4 – Forces, Density & Pressure

Learning objectives

• Define density, pressure and related quantities.
• Derive and use the formulas for weight, buoyant force and hydrostatic pressure.
• Explain turning effects of forces (moments, couples, torque) and locate the centre of gravity.
• Apply the conditions for translational and rotational equilibrium (∑F = 0 and ∑τ = 0).
• Carry out simple experiments to determine density and to verify Archimedes’ principle.
• Solve quantitative problems involving density, pressure and equilibrium.

1. Density

Density (ρ) is the mass of a material per unit volume.

• ρ – density (kg m⁻³)
• m – mass (kg)
• V – volume (m³)

Re‑arranging the definition gives two very useful forms:

1.1 Typical densities (20 °C unless otherwise stated)

Material	Density (kg m⁻³)
Water (4 °C)	1000
Aluminium	2700
Steel	7850
Air (STP)	1.29
Wood (average)	600 – 800
Concrete	2400

2. Pressure

Pressure is the normal force exerted per unit area.

• P – pressure (Pa = N m⁻²)
• F – normal force (N)
• A – area on which the force acts (m²)

2.1 Hydrostatic pressure

In a fluid at rest the pressure increases linearly with depth because the weight of the fluid above exerts a force on the fluid below.

Integrating from the free surface (where the pressure is p₀) to a depth h gives

In most Cambridge exam questions the atmospheric pressure at the surface is taken as zero, so the working form is simply p = ρgh.

2.2 Pressure on a flat inclined surface

If a flat surface of area A is inclined at an angle θ to the horizontal, the pressure at the centroid depth h_c is still p = ρg h_c. The resultant force acts normal to the surface:

For a thin rectangular plate the line of action passes through the centre of the surface (the “centre of pressure”). For more accurate work the centre of pressure is at

where t is the plate thickness. In the Cambridge syllabus it is sufficient to state that the force acts through the centre of the plate.

3. Turning effects of forces

3.1 Moment (torque)

The turning effect of a force about a point (or an axis) is called a moment or torque.

• τ – torque (N m)
• F – magnitude of the force (N)
• d – perpendicular distance from the line of action of the force to the pivot (m)
• α – angle between the force direction and the line joining the pivot to any point on the line of action (often α = 90°, so sin α = 1).

3.2 Couples

A pair of equal, opposite, parallel forces whose lines of action do not coincide forms a couple. The torque of a couple is independent of the reference point:

where d is the perpendicular separation of the two forces.

3.3 Centre of gravity (CG)

The centre of gravity of a rigid body is the point through which the weight of the body may be considered to act. For a uniform body the CG coincides with the geometric centre. In equilibrium problems the weight is treated as a single vertical force acting at the CG.

4. Equilibrium of forces

A rigid body is in static equilibrium when both the resultant force and the resultant torque about any point are zero.

4.1 Translational equilibrium

No linear acceleration – the sum of the components of all forces in each perpendicular direction must be zero.

4.2 Rotational equilibrium

No angular acceleration – the sum of the torques about any convenient axis must be zero.

4.3 Systematic method for solving equilibrium problems

1. Draw a clear free‑body diagram (FBD) showing every force, its line of action and the chosen pivot (or axis).
2. Resolve forces into perpendicular components (usually horizontal and vertical).
3. Write the ΣF = 0 equations for each set of components.
4. Write the Στ = 0 equation about the chosen pivot; pick a pivot that eliminates as many unknown forces as possible.
5. Solve the simultaneous equations for the unknown quantities.

5. Using density in force calculations

5.1 Weight

5.2 Buoyant force – Archimedes’ principle

When an object is wholly or partially immersed, the fluid exerts an upward force equal to the weight of the fluid displaced.

Derivation (concise): The pressure at depth h is p = ρgh. Integrating this pressure over the submerged surface yields a net upward force equal to the weight of the displaced volume.

5.3 Pressure forces on surfaces

• Flat vertical surface of area A at depth h: \(F = \rho g h A\).
• Flat inclined surface: use the pressure at the centroid depth, then multiply by the area; the resultant acts normal to the surface.
• Arbitrary shape: divide the surface into small elements dA, each experiencing \(dF = p\,dA\). The total force is the surface integral
  \(\displaystyle F = \int p\,\mathrm{d}A\).

6. Practical skills

6.1 Measuring the density of an irregular object (water‑displacement method)

1. Weigh the dry object on a balance → obtain mass m (kg).
2. Fill a graduated cylinder with a known volume of water and record the initial reading V₁ (m³).
3. Submerge the object completely (no air bubbles) and record the new reading V₂.
4. Displaced volume \(V_{\text{disp}} = V_{2} - V_{1}\).
5. Calculate density: \(\displaystyle \rho = \frac{m}{V_{\text{disp}}}\).

6.2 Verifying Archimedes’ principle

Set‑up: a spring balance measures the apparent weight of a solid block in air and then when fully immersed in water. The loss of weight equals the buoyant force. Plot the measured buoyant force against the displaced volume; the graph should be a straight line with gradient \(\rho_{\text{water}} g\).

7. Worked examples

Example 1 – Ladder against a smooth wall

Problem: A uniform ladder 4.00 m long, mass 12.0 kg, leans against a smooth vertical wall. The foot of the ladder is 1.20 m from the wall. Determine the horizontal and vertical components of the force exerted by the ground on the ladder (static equilibrium is assumed).

Solution

1. Free‑body diagram: weight \(W = mg\) acting at the ladder’s centre (2.00 m from each end); normal reaction \(R\) from the ground (unknown direction); frictional force \(F_f\) horizontal at the ground; normal reaction from the wall \(N\) horizontal (wall is smooth → no vertical component).
2. Geometry: Ladder makes an angle \(\theta\) with the ground where \(\sin\theta = \dfrac{1.20}{4.00}=0.30\) → \(\theta = 17.5^{\circ}\). Hence \(\cos\theta = 0.954\).
3. Choose pivot: foot of the ladder eliminates \(R\) and \(F_f\) from the torque equation.
4. Torque about the foot (counter‑clockwise positive): \[ \tau_W = W\,(2.00\cos\theta),\qquad \tau_N = N\,(4.00\sin\theta) \] Set \(\sum\tau = 0\): \[ W\,(2.00\cos\theta) = N\,(4.00\sin\theta) \] \[ N = \frac{W\cos\theta}{2\sin\theta} = \frac{12.0\times9.81\times0.954}{2\times0.300} \approx 1.83\times10^{2}\,\text{N} \]
5. Horizontal equilibrium: \[ \Sigma F_x = 0 \;\Rightarrow\; F_f = N = 1.83\times10^{2}\,\text{N} \]
6. Vertical equilibrium: \[ \Sigma F_y = 0 \;\Rightarrow\; R = W = 1.18\times10^{2}\,\text{N} \]

Thus the ground supplies a vertical reaction of 118 N upward and a horizontal frictional force of 183 N** opposing the wall’s push**.

Example 2 – Floating wooden block

Problem: A wooden block of dimensions 0.20 m × 0.10 m × 0.05 m is placed gently on water. Its density is unknown. Find the fraction of the block’s volume that is submerged at equilibrium.

Solution

• Block volume \(V_{\text{b}} = 0.20\times0.10\times0.05 = 1.0\times10^{-3}\,\text{m}^3\).
• Weight: \(W = \rho_{\text{wood}} V_{\text{b}} g\).
• Buoyant force: \(F_b = \rho_{\text{water}} V_{\text{sub}} g\) where \(V_{\text{sub}}\) is the submerged volume.
• Equilibrium → \(W = F_b\): \[ \rho_{\text{wood}} V_{\text{b}} = \rho_{\text{water}} V_{\text{sub}} \]
• Fraction submerged: \[ \frac{V_{\text{sub}}}{V_{\text{b}}}= \frac{\rho_{\text{wood}}}{\rho_{\text{water}}} \]
• Typical wood densities are 600–800 kg m⁻³, so the block is 0.60–0.80 (60–80 %) submerged.

Example 3 – Hydrostatic force on an inclined rectangular plate

Problem: A steel plate 0.30 m wide and 0.80 m long is fully submerged in water and inclined at 30° to the horizontal. The top edge lies at the water surface. Determine the magnitude and direction of the resultant hydrostatic force.

Solution

1. Depth of the plate’s centre: \(h_c = \dfrac{0.80}{2}\sin30^{\circ}=0.20\;\text{m}\).
2. Pressure at the centre: \(p = \rho g h_c = 1000\times9.81\times0.20 = 1.96\times10^{3}\;\text{Pa}\).
3. Plate area: \(A = 0.30\times0.80 = 0.24\;\text{m}^2\).
4. Resultant force magnitude: \[ F = pA = 1.96\times10^{3}\times0.24 \approx 4.71\times10^{2}\;\text{N} \]
5. Direction: the force acts normal to the plate, i.e. 30° to the vertical (or 60° to the horizontal) pointing upwards.

8. Summary checklist

• Density: \(\rho = m/V\); use to convert between mass and volume.
• Weight: \(W = mg = \rho V g\).
• Hydrostatic pressure: \(p = p_0 + \rho g h\) (often \(p = \rho g h\)).
• Buoyant force (Archimedes): \(F_b = \rho_{\text{fluid}} V_{\text{disp}} g\).
• Moment (torque): \(\tau = F d \sin\alpha\); for a couple \(\tau = Fd\).
• Equilibrium conditions: \(\sum\mathbf{F}=0\) and \(\sum\boldsymbol{\tau}=0\).
• Always start a problem with a clear free‑body diagram and list known/unknown quantities.
• Experimental tip: use water‑displacement to find the volume of irregular objects; compare measured buoyant force with \(\rho_{\text{fluid}} V g\) to test Archimedes’ principle.