Moments
Rigid bodies can spin, so balance needs a second currency. The moment is force times perpendicular distance, and a beam in equilibrium balances its forces and its turning effects at once. The second condition is where the information hides.
Builds on Statics of a particle.
Where it earns its keep: Moments and equilibrium on InkPhysics.
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 about any point balance.
- Model uniform and non-uniform beams, and use tilting conditions where a reaction vanishes.
COMMON MISCONCEPTION
A bigger force always has a bigger turning effect.
The turning effect
A force's moment about a point measures its turning effect there. Force times the perpendicular distance from the point to the force's line of action, measured in newton metres and labelled clockwise or anticlockwise. A force acting through the point has no moment about it, and that is the great trick of the topic. Take moments about the place where the most annoying unknown acts and it disappears from the equation.
Beams in balance
A rigid body in equilibrium satisfies two conditions at the same time. 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.
Symmetry check: the supports sit symmetrically about the centre, so equal reactions are exactly right.
A non-uniform beam carries its weight at an unknown point, and that point is usually what the question wants. Take moments about one support to find it. For a 4 m rod resting on supports at both ends with reactions 30 N and 50 N, the weight is 80 N, and moments about A give 80d = 50 × 4, so the centre of mass sits 2.5 m from A, nearer the bigger reaction. It always is.
Tilting
Tilting is the boundary case. On the point of tilting about one support, the beam is about to lift off the other, so that other reaction is zero. Writing R = 0 down first is the extra equation the question needs.
WORKED EXAMPLE
How far can the child walk?
A uniform beam AB of length 5 m and weight 200 N rests on supports at C, 1 m from A, and D, 3.5 m from A. A child of weight 300 N walks from A towards B. Find how far from A the child is when the beam is about to tilt.
It tilts about D, so at that moment the reaction at C is zero and the only forces are the weight, the child and R at D.
Take moments about D. The beam's weight acts at 2.5 m from A, which is 1 m on the A side of D: 200 × 1 = 300 × (x − 3.5).
So x − 3.5 = 2/3 and x = 4.17 m from A (3 s.f.).
Sense check: the answer sits between D and B, which is the only place it could be, and a heavier child would tilt the beam sooner.
ASSESSMENT FOCUS
- Choose the pivot to eliminate an unknown. Taking moments about a support removes its reaction from the equation entirely.
- Write “taking moments about A, clockwise positive” and keep every term's sense consistent with that declaration.
- Uniform means the weight acts at the centre. Non-uniform means its position is an unknown the question wants found.
- On the point of tilting about one support, the other reaction is zero. Write that line before anything else.
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.
Or read them with their worked answers on the moments questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Calculate moments as force times perpendicular distance, with a consistent sense of rotation.
- Use both equilibrium conditions for beams: forces balance, and moments about any point balance.
- Model uniform and non-uniform beams, and use tilting conditions where a reaction vanishes.
Open the full revision checklist to see every objective in the course in one place.