Density (ρ) is the mass of a material per unit volume.
\[
\boxed{\rho = \frac{m}{V}}
\]
Re‑arranging the definition gives two very useful forms:
\[
m = \rho V ,\qquad V = \frac{m}{\rho}
\]
| Material | Density (kg m⁻³) |
|---|---|
| Water (4 °C) | 1000 |
| Aluminium | 2700 |
| Steel | 7850 |
| Air (STP) | 1.29 |
| Wood (average) | 600 – 800 |
| Concrete | 2400 |
Pressure is the normal force exerted per unit area.
\[
\boxed{P = \frac{F}{A}}
\]
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.
\[
\Delta p = \rho_{\text{fluid}}\,g\,\Delta h
\]
Integrating from the free surface (where the pressure is p₀) to a depth h gives
\[
\boxed{p = p{0} + \rho{\text{fluid}}\,g\,h}
\]
In most Cambridge exam questions the atmospheric pressure at the surface is taken as zero, so the working form is simply p = ρgh.
If a flat surface of area A is inclined at an angle θ to the horizontal, the pressure at the centroid depth hc is still p = ρg hc. The resultant force acts normal to the surface:
\[
F = pA = \rho g h_c A
\]
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
\[
h{cp}=hc+\frac{t^{2}}{12h_c}
\]
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.
The turning effect of a force about a point (or an axis) is called a moment or torque.
\[
\boxed{\tau = F\,d\,\sin\alpha}
\]
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:
\[
\boxed{\tau_{\text{couple}} = F\,d}
\]
where d is the perpendicular separation of the two forces.
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.
A rigid body is in static equilibrium when both the resultant force and the resultant torque about any point are zero.
\[
\boxed{\sum\mathbf{F}=0 \qquad\text{and}\qquad \sum\boldsymbol{\tau}=0}
\]
No linear acceleration – the sum of the components of all forces in each perpendicular direction must be zero.
No angular acceleration – the sum of the torques about any convenient axis must be zero.
\[
W = mg = \rho V g
\]
When an object is wholly or partially immersed, the fluid exerts an upward force equal to the weight of the fluid displaced.
\[
\boxed{F{\text{b}} = \rho{\text{fluid}}\,V_{\text{disp}}\,g}
\]
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.
\(\displaystyle F = \int p\,\mathrm{d}A\).
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 \(\rho_{\text{water}} g\).
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
\[
\tau_W = W\,(2.00\cos\theta),\qquad
\tau_N = N\,(4.00\sin\theta)
\]
Set \(\sum\tau = 0\):
\[
W\,(2.00\cos\theta) = N\,(4.00\sin\theta)
\]
\[
N = \frac{W\cos\theta}{2\sin\theta}
= \frac{12.0\times9.81\times0.954}{2\times0.300}
\approx 1.83\times10^{2}\,\text{N}
\]
\[
\Sigma Fx = 0 \;\Rightarrow\; Ff = N = 1.83\times10^{2}\,\text{N}
\]
\[
\Sigma F_y = 0 \;\Rightarrow\; R = W = 1.18\times10^{2}\,\text{N}
\]
Thus the ground supplies a vertical reaction of 118 N upward and a horizontal frictional force of 183 N opposing the wall’s push.
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
\[
\rho{\text{wood}} V{\text{b}} = \rho{\text{water}} V{\text{sub}}
\]
\[
\frac{V{\text{sub}}}{V{\text{b}}}= \frac{\rho{\text{wood}}}{\rho{\text{water}}}
\]
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
\[
F = pA = 1.96\times10^{3}\times0.24 \approx 4.71\times10^{2}\;\text{N}
\]
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