Forces, density and pressure

4.1 Turning effects of forces

Centre of gravity

The centre of gravity of an object is the single point at which its whole weight may be taken to act. For a uniform beam it is at the middle.

The moment of a force

$\text{moment} = \text{force} \times \text{perpendicular distance from the pivot to the line of action of the force}$

The unit is N m. If the force is at an angle, use the perpendicular distance, or use the component of the force perpendicular to the beam.

Couples and torque

A couple is a pair of equal and opposite forces whose lines of action are different. It produces rotation only, with no resultant force, so there is no acceleration of the object's centre.

$\text{torque of a couple} = \text{one of the forces} \times \text{perpendicular distance between the forces}$

4.2 Equilibrium of forces

The principle of moments

Tip: take moments about the point where an unknown force acts. That force then has zero moment and disappears from the equation.

Conditions for equilibrium

An object is in equilibrium when both:

1. the resultant force on it is zero, and
2. the resultant torque about any point is zero.

Vector triangles

If three coplanar forces keep an object in equilibrium, then drawn head to tail they form a closed triangle, because their resultant is zero. Their lines of action also meet at a single point, otherwise there would be a resultant torque.

4.3 Density and pressure

Definitions

Density is mass per unit volume, $\rho = m/V$, measured in kg m⁻³. Water is about 1000 kg m⁻³.

Pressure is the normal force per unit area, $p = F/A$, measured in pascals (1 Pa = 1 N m⁻²).

Hydrostatic pressure

Consider a column of fluid of height $\Delta h$ and cross-section $A$. Its volume is $A\Delta h$, its mass is $\rho A \Delta h$, and its weight is $\rho A \Delta h\, g$. Dividing this force by the area gives:

$\Delta p = \rho g \Delta h$

The pressure in a fluid depends on the depth and density, not on the shape of the container. To find the total pressure at a depth, add the atmospheric pressure at the surface.

Upthrust

Pressure increases with depth, so the fluid pushes up harder on the bottom of a submerged object than it pushes down on the top. The difference is the upward force called upthrust.

For a cylinder of height $h$ and end area $A$: upthrust $= \rho g h A = \rho g V$, where $V$ is the volume displaced.

$F = \rho g V$