Moments & Couples

The turning effect of a force is not a property of the force alone. Choose a point O, draw r from O to any point on the force's line of action, and the moment is M_O = r × F. For forces in a plane that vector points straight out of or into the page, so it reduces to one signed number, M_O = x F_y − y Fₓ, where x and y are the components of r. Take F = (30 N, −40 N) acting at (0.50 m, 0.20 m). About the origin, M = 0.50 × (−40) − 0.20 × 30 = −26 N m. About B = (0.50 m, 0), r is only (0, 0.20 m), and M = −0.20 × 30 = −6.0 N m. Same force, different point, different moment. The size is always F times the perpendicular distance from the point to the line of action, 0.52 m and 0.12 m here, and it is zero for any point on that line.

The sign of a planar moment is its sense of rotation. The usual convention takes counterclockwise as positive, because that is the +z direction when x points right and y points up, and it is exactly what x F_y − y Fₓ produces. Clockwise-positive is equally valid. What is not valid is switching partway through one sum, or writing every term as a positive F d and assigning signs afterwards by eye. Settle each term on its own. A 120 N force pushing down 0.40 m to the right of O turns the body clockwise and contributes −48 N m. A 70 N force pulling up 0.90 m to the right contributes +63 N m. The net moment is +15 N m, counterclockwise. A negative total is not a mistake to be fixed: it is the sum telling you the rotation runs the other way.

Put F at point A and −F at point B, on parallel lines a perpendicular distance d apart. The net force is zero. The net moment about any point O is r_A × F + r_B × (−F) = (r_A − r_B) × F, and O has dropped out. Test it with two 50 N forces on lines 0.40 m apart. About the midpoint each contributes 50 × 0.20 = 10 N m, so 20 N m in total. About a point on one line, that force contributes nothing and the other gives 50 × 0.40 = 20 N m. About a point 1.00 m outside the pair, the far force gives +70 N m and the near one −50 N m: 20 N m again. Because its moment is the same everywhere, a couple is a free vector, and because its net force is zero it is a pure torque. A 0.40 kg, 0.60 m rod floating free under a 0.12 N m couple has I_cm = mL²/12 = 0.012 kg m², so it spins up at α = 10 rad s⁻² while its centre stays where it was.

On a rigid body you may slide a force along its own line of action, because r × F does not change. You may not shift it sideways for free. Add F and −F at a new point B, which changes nothing. The original F at A and the −F at B now form a couple, leaving F acting at B plus C = (r_A − r_B) × F. A shelf bracket makes this concrete. A 180 N plant pot hangs from a bracket 0.35 m to the right of the wall. At the wall, the pot is equivalent to 180 N downward plus a 63 N m clockwise couple, and that pair is what the wall fixing has to resist: 180 N of shear and 63 N m of turning that tries to pull the top screw out. Leave out the couple and the fixing appears to carry nothing but shear, which is exactly how brackets get under-designed.

Any set of forces in a plane reduces to R = ΣF and M_O = Σ(x F_y − y Fₓ) about one chosen point. If R ≠ 0, a single force R reproduces both, provided it acts on the right line, the one satisfying x R_y − y Rₓ = M_O. Three loads hang from a horizontal beam: 100 N at x = 0.50 m, 300 N at 1.50 m and 200 N at 2.50 m. They give R_y = −600 N and M_O = −50 − 450 − 500 = −1000 N m, so one 600 N downward force crossing the beam at x = M_O / R_y = 1.67 m does the same job. If instead R = 0 while M_O ≠ 0, the reference-point formula makes the moment identical about every point, and no single force can match that: the system is irreducibly a couple. The only remaining case, R = 0 with M = 0, is equilibrium.

Equivalence is a claim about a rigid body's external response. The same net force and net moment produce the same acceleration, the same angular acceleration and the same support reactions. A 2.0 m beam supported at both ends and loaded by a 30 N m couple needs vertical reactions of 15 N, one pushing up and one pulling down, whether the couple sits at an end or at mid-span, because only C/L enters. Cut the beam, though, and the internal moment carried across the cut depends on which side of the cut the couple sits. Applied at the left end, that internal moment is largest, the full 30 N m, right at the loaded end and fades to zero at the far support. Applied at mid-span, it peaks at only 15 N m either side of the middle and is zero at both ends. Use equivalence for motion and reactions, and return to the real load positions before asking where a beam bends or breaks.