3.2 Equilibrium of a rigid body

The moment of a force about a point is force × perpendicular distance from the point to the force line: M=Fd. Choose a sign convention, usually clockwise and anticlockwise, and keep it throughout.

For several forces, take moments about a convenient point so unknown reactions through that point have zero moment. Resolve angled forces before finding their perpendicular contribution.

A 20 N force acting 0.30 m from a pivot produces a 6.0 N m moment. A 10 N force on the opposite side at 0.60 m balances it if the senses are opposite.

Distance along the rod is not automatically the moment arm; use the perpendicular distance to the force line, and do not mix force and moment units.

For particles of masses mᵢ at positions xᵢ, the centre of mass satisfies x̄=Σmᵢxᵢ/Σmᵢ, with the same weighted-average rule applied independently to y-coordinates.

Heavier masses pull the centre towards them. Set an origin and sign convention first; coordinates may be negative, and the centre can lie outside the material for a separated system.

Masses 2 kg at x=0 and 6 kg at x=4 m have x̄=(2·0+6·4)/8=3 m, much nearer the larger mass.

The centre of mass is not always the geometric centre and is not necessarily a point inside an irregular or multi-part object.

A uniform rod, rectangle, circle or sphere has its centre of mass at its geometric centre. Symmetry means equal mass is distributed at equal distances on opposite sides of the symmetry line or plane.

For a composite lamina, split the shape into standard pieces, assign each area as its mass when density and thickness are uniform, and use an area-weighted coordinate average.

A uniform L-shape can be treated as a large rectangle minus a cut-out rectangle; subtract the cut-out’s area moment rather than averaging the two visible arms.

The cut-out is negative area only in the calculation; it is not a negative physical mass, and the uniform-density assumption must be stated.

If components have masses m₁,m₂,… and centres at r₁,r₂,…, the combined centre is r̄=(Σmᵢrᵢ)/(Σmᵢ). Each component’s mass must be proportional to its area or volume only when density is common.

Choose a convenient reference point, include holes or removed pieces with negative area when appropriate, and check that the result lies between the extreme positions of positive masses.

A 3 kg block at x=1 m joined to a 1 kg block at x=5 m gives x̄=2 m, not 3 m: the lighter block should shift the centre only modestly.

Equal-sized parts do not imply equal masses when densities differ; never replace mass weights by geometry without checking the model.

A rigid body is in equilibrium when the vector sum of all forces is zero and the sum of moments about any point is zero. Translational balance alone does not prevent rotation.

Draw all external forces, resolve components, then take moments about a point that removes unknown reactions. Use the force equations afterwards to determine the remaining reactions.

A horizontal beam supported at its ends carries a central load. Equal upward reactions satisfy force balance; taking moments about either support confirms the reactions and prevents a net turn.

A pair of equal opposite forces can form a couple with zero resultant force but non-zero moment, so the body can still rotate.

A body slides when the available friction is insufficient for equilibrium; it topples when the line of action of the resultant reaction moves beyond the supporting base. Both are checked after force and moment balance.

Find the friction required and compare it with μR for sliding. For toppling, take moments about the edge that would become the pivot and set the limiting reaction there to zero.

A crate on a rough floor may remain at rest while a horizontal force increases. If the force line creates a moment that lifts one edge first, toppling occurs before the friction limit is reached.

A large friction coefficient does not prevent toppling, and “about to topple” does not mean the body has already started rotating.