Moments, General Equilibrium, the Couple and the Centre of Gravity

• The moment (torque) of a force F about a point O is M = F × d, where d is the perpendicular distance from O to the line of action of the force. Unit: N·m.
• Only the perpendicular component contributes: if F makes an angle θ with the arm of length r, then M = F·r·sin θ.
• Moment is a turning effect; the sign convention is anticlockwise positive.
• A force has zero moment about any point on its own line of action.

Varignon's theorem

The moment of the resultant of a system of coplanar forces about any point equals the algebraic sum of the moments of its components about the same point.

Worked example

A force of 20 N acts at the end of a 0.5 m arm making 30° with the arm: M = 20 × 0.5 × sin 30° = 5 N·m. If instead the force is perpendicular to the arm, M = 20 × 0.5 = 10 N·m.

A rigid body is in equilibrium when it has no tendency to translate or to rotate. For coplanar forces this requires three independent conditions:

• ΣFx = 0 — no resultant force horizontally.
• ΣFy = 0 — no resultant force vertically.
• ΣM = 0 about any point — no resultant turning effect.

Choose the point for taking moments where the largest number of unknown forces act, because those forces then drop out of the equation.

Worked example — a uniform beam on two supports

A uniform rod of weight 60 N and length 4 m rests on supports at its ends; a 100 N load hangs 1 m from the left end.

• Moments about the left end: R₂ × 4 = 60 × 2 + 100 × 1 = 220 ⟹ R₂ = 55 N.
• ΣFy = 0: R₁ = 160 − 55 = 105 N.
• Check with moments about the right end: R₁ × 4 = 60 × 2 + 100 × 3 = 420 ⟹ R₁ = 105 N ✓.

Three-force bodies

If a body is in equilibrium under three non-parallel forces, their lines of action are concurrent (meet at one point) and coplanar — this fact is used to find the direction of an unknown reaction.

• A smooth hinge exerts two unknown components R_x, R_y (three unknowns with the direction being arbitrary); take moments about the hinge to eliminate both.
• A smooth peg or a light string over it transmits the same tension on both sides and the reaction is along the bisector of the angle between the two parts.
• A smooth wall / floor gives a reaction perpendicular to the surface only.

Worked example — hinged rod with a string

A uniform rod AB of weight W and length 2 m is hinged at A and held by a vertical string at B inclined 30° to the horizontal wall. Moments about A: T × 2 cos 30° = W × 1 ⟹ T = W/(2 cos 30°) = 0.577 W; then R_y = W − T and R_x = 0 for a vertical string.

Worked example — limiting position

A rod of length L rests with one end on a rough floor and leans against a smooth wall at angle θ. The wall reaction is horizontal, the floor gives R (vertical) and F (horizontal, F ≤ μR). Equilibrium gives F = R_wall, R = W, and moments about the floor end: R_wall·L sin θ = W·(L/2)cos θ ⟹ R_wall = (W/2)cot θ. Slipping impends when (W/2)cot θ = μW ⟹ tan θ = 1/(2μ).

• A couple is the system of two equal, opposite and parallel forces whose lines of action do not coincide.
• Its resultant force is zero, so it produces pure rotation; its moment is M = F × d, where d is the perpendicular distance between the two lines of action.
• The moment of a couple is the same about every point in the plane — this is what distinguishes it from an ordinary force system.
• A couple can only be balanced by another couple of equal moment and opposite sense; a single force can never equilibrium a couple.

Worked example

Two forces of 12 N act along parallel lines 0.4 m apart in opposite senses: M = 12 × 0.4 = 4.8 N·m. The same value is obtained about any point, including a point outside the strip between the lines.

Resultant of a system with a couple

A force F together with a couple of moment M in the same plane is equivalent to a single force F displaced perpendicular to itself by d = M/F, because moving the force off the original line creates exactly the couple M.

The centre of gravity (C.G.) is the point through which the resultant of all the gravitational forces on the body passes, whatever the body's position.

• Uniform rod → its midpoint. Uniform rectangular lamina → intersection of the diagonals. Uniform circle → the centre. Uniform triangle → the intersection of the medians (1/3 of the way up any median).
• For parallel weights W₁, W₂, … at positions x₁, x₂, …: x̄ = Σ(Wᵢxᵢ)/ΣWᵢ, and the same in y.
• Composite laminae: divide into standard shapes, tabulate area and centroid, then take weighted averages. A removed part (hole) enters with a negative area.
• Suspended-body method: when a body hangs freely from a point, its C.G. lies on the vertical line through the point of suspension; repeating from a second point fixes the C.G. at the intersection.

Worked example — a lamina with a cut-out

A circular disc of radius 6 cm has a circular hole of radius 2 cm cut from it, the hole's centre 3 cm from the disc's centre. Taking moments of area about the disc centre: x̄ = [π(36)(0) − π(4)(3)]/[π(36) − π(4)] = −12π/32π = −0.375 cm, i.e. the C.G. of the remaining part is 0.375 cm from the centre on the side opposite the hole.

Worked example — two masses on a rod

A light rod 1.2 m long carries 4 kg at one end and 8 kg at the other. The C.G. is at x = (4×0 + 8×1.2)/12 = 0.8 m from the 4 kg end.

• The angle of friction λ satisfies tan λ = μ, and the resultive reaction of a rough surface makes the angle λ with the normal (the cone of friction).
• A body on an inclined plane begins to slide when the plane's angle exceeds λ, i.e. when tan θ = μ.
• A body on the point of toppling (rather than sliding) when its C.G. passes vertically beyond the edge of its base: for a cuboid of base width 2b and C.G. height h on a rough plane, toppling impends when tan θ = b/h.
• Lowering the C.G. and widening the base increase stability — the reason a racing car is low and a lamp has a heavy base.

Worked example

A cube of side 10 cm rests on a rough plane with μ = 0.4. Slipping impends at tan θ = 0.4 ⟹ θ = 21.8°; toppling would need tan θ = 5/5 = 1 ⟹ θ = 45°. Since 21.8° < 45°, the cube slides before it topples.

1. Find the moment of a 30 N force applied at 0.6 m from a pivot, at 60° to the arm. — 30 × 0.6 × sin 60° = 15.6 N·m.
2. A uniform 5 m beam of weight 200 N carries 300 N at one end. It balances on a prop 2 m from that end. Verify by taking moments about the prop. — 300×2 = 600, weight acts 0.5 m on the other side: 200×0.5=100 ⟹ not balanced; the prop must be at x with 300x = 200(2.5 − x) ⟹ x = 4/3 m.
3. Two forces of 9 N form a couple with lines 25 cm apart. Find the moment. — 9 × 0.25 = 2.25 N·m.
4. Find the C.G. of a T-shaped lamina made of a 10×2 rectangle (top) on a 2×8 rectangle (stem) centred below it. — Areas 20 and 16; take the top edge as y = 0: ȳ = (20×1 + 16×6)/36 = 116/36 ≈ 3.22 cm below the top.
5. A 12 kg mass hangs from two strings making 30° and 60° with the ceiling. Find the tensions. — Horizontal balance T₁cos30 = T₂cos60 and vertical T₁sin30 + T₂sin60 = 12g ⟹ T₁ = 6g√3 ≈ 101.8 N, T₂ = 6g ≈ 58.8 N (with g = 9.8).
6. A body just slides down a rough 30° plane. Find μ. — μ = tan 30° = 0.577.