Published by Patrick Mutisya · 14 days ago
Define density and use it to solve equilibrium problems involving forces.
Density (\$\rho\$) is the mass per unit volume of a material.
\$\rho = \frac{m}{V}\$
Re‑arranging the formula allows us to find mass or volume when the other quantities are known:
\$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 |
A body is in static equilibrium when both the net force and the net torque acting on it are zero:
\$\sum \mathbf{F} = \mathbf{0},\qquad \sum \boldsymbol{\tau} = \mathbf{0}\$
Density often appears when dealing with forces that depend on mass or volume, such as weight, buoyancy, and pressure.
The weight \$W\$ of an object is the force due to gravity:
\$W = mg = \rho \cdot g\$
where \$g = 9.81\ \text{m s}^{-2}\$.
When an object is submerged in a fluid, it experiences an upward buoyant force equal to the weight of the displaced fluid:
\$F{\text{b}} = \rho{\text{fluid}} V_{\text{disp}} g\$
This principle is used to determine whether an object will float, sink, or remain suspended.
Pressure \$P\$ is force per unit area. For a fluid at depth \$h\$, the pressure is:
\$P = \rho_{\text{fluid}} g h\$
The resultant force on a flat surface of area \$A\$ is \$F = PA\$.
Solution:
1. Volume of the cube:
\$V = (0.10\ \text{m})^{3} = 1.0\times10^{-3}\ \text{m}^{3}\$
Mass using \$\rho_{\text{Al}} = 2700\ \text{kg m}^{-3}\$:
\$m = \rho V = 2700 \times 1.0\times10^{-3} = 2.7\ \text{kg}\$
2. Weight of the cube:
\$W = mg = 2.7 \times 9.81 = 26.5\ \text{N}\$
Buoyant force in water (\$\rho_{\text{water}} = 1000\ \text{kg m}^{-3}\$):
\$F{\text{b}} = \rho{\text{water}} V g = 1000 \times 1.0\times10^{-3} \times 9.81 = 9.81\ \text{N}\$
Since \$W > F{\text{b}}\$, the cube sinks. For equilibrium (floating), the condition \$W = F{\text{b}}\$ would require a material density equal to that of water.