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Moments

Rigid bodies can spin, so balance needs a second currency: the moment, force times perpendicular distance. A beam in equilibrium balances its forces and its turning effects, and the second condition is where the information hides.

Year 12-13EDEXCEL 9MA0 M9

Builds on Statics of a particle.

IN THIS TOPIC

  • Calculate moments as force times perpendicular distance, with a consistent sense of rotation.
  • Use both equilibrium conditions for beams: forces balance and moments balance about any point.
  • Model uniform beams with weight at the midpoint, and use tilting conditions where a reaction vanishes.

WHAT YOU PROBABLY THINK

A bigger force always has a bigger turning effect.

The turning effect

moment = force × perpendicular distanceIN THE FORMULAE BOOKLET
The moment of a force: 15 newtons applied 0.4 metres from the hinge turns with moment 6 newton metres15 N0.4 mmoment = 15 × 0.4 = 6 N m
FIG. 1Fifteen newtons at four tenths of a metre: a six newton metre turning effect about the hinge.

A force's moment about a point measures its turning effect there: the force times the perpendicular distance from the point to its line of action, in newton metres, labelled clockwise or anticlockwise. Forces through the point itself have no moment about it, which is the great trick of the subject: take moments about the point where the most annoying unknown acts, and it vanishes from the equation.

Beams in balance

A uniform 6 metre beam balancing on a pivot 2 metres from one end: a 20 newton child at the end balances the beam's own 40 newtons acting at the centre20 N40 N at the centre2 m1 m20 × 2 = 40 × 1
FIG. 2Twenty newtons two metres out balances forty newtons one metre out: moments, not forces, decide.

A rigid body in equilibrium satisfies two conditions at once: the forces sum to zero, and the moments about any point sum to zero. A uniform beam contributes its own weight at its midpoint.

WORKED EXAMPLE

A beam on two supports

A uniform 6 m beam of weight 120 N rests on supports at 1 m and 5 m from end A. Find the reaction at each support.

Moments about the 1 m support: 120 × 2 = R₂ × 4, so R₂ = 60 N.

Forces vertically: R₁ + 60 = 120, so R₁ = 60 N.

The symmetry check: the supports sit symmetrically about the centre, so equal reactions are exactly right.

Tilting is the boundary case: on the point of tilting about one support, the other support carries nothing, and setting that reaction to zero is the extra equation the question is fishing for.

THE EXAM BIT

  • Choose the pivot to kill an unknown: moments about a support removes its reaction from the equation.
  • Say “taking moments about A, clockwise positive” and keep every term's sense honest.
  • Uniform means weight at the centre; non-uniform beams carry their weight at an unknown point the question wants found.
  • On the point of tilting about a support: the other reaction is zero. Write that first.

CHECK YOURSELF

A uniform 4 m plank of weight 80 N rests on a support at its centre. A child of weight 400 N sits 0.5 m from one end. How far from the centre, on the other side, must a 500 N adult sit to balance it?

Show a hint

The plank's own weight acts at the pivot.

Show the answer

The child sits 1.5 m from the centre: moment 400 × 1.5 = 600 N m.

Balance: 500 × d = 600, so d = 1.2 m from the centre.

Moment = force × perpendicular distance, with a declared sense of rotation.

Beams balance twice over: forces to zero, and moments about your best pivot to zero.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

CHECK YOUR PROGRESS

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  • Calculate moments as force times perpendicular distance, with a consistent sense of rotation.
  • Use both equilibrium conditions for beams: forces balance and moments balance about any point.
  • Model uniform beams with weight at the midpoint, and use tilting conditions where a reaction vanishes.

No animated video for this topic yet; these notes stand alone.