Revise Equilibrium of Forces for Cambridge 9702 (4.2)

$\sum F = 0 \qquad \sum M = 0$

A moment is the turning effect of a force about a pivot. Only the perpendicular distance from the pivot to the line of action counts. On a spanner, if the force is at angle θ to the handle, use the perpendicular component: $d_\perp = d\cos\theta$.

$M = F \times d_\perp$

Reference diagram (Senpai Corner): perpendicular force on a beam gives $M = F \times d$. Angled force on a spanner gives $M = F \times d\cos\theta$. See the moments reference image in resources — the live animation above shows a balanced beam.

For a body in rotational equilibrium, the total clockwise moment about any chosen pivot equals the total anticlockwise moment about that same pivot.

A couple is two equal, parallel, opposite forces acting along different lines. They produce rotation only — no resultant force. Examples: turning a steering wheel, using a screwdriver.

$\tau = F \times d_{\text{couple}}$

When three coplanar forces act on an object in translational equilibrium, draw them tip-to-tail as vectors. If they form a closed triangle, the resultant is zero. This is the graphical test for equilibrium in one plane.

Use the live diagram (Triangle of forces tab) to see vectors a, b and c closing tip-to-tail. The Senpai Corner reference below matches this layout.

The centre of gravity is the single point where the entire weight of an object may be considered to act. For a uniform object in a uniform gravitational field, it coincides with the geometric centre — place the weight there when modelling beams and rods.