A force can make a body turn. This mini-lesson covers the moment of a force, the principle of moments, uniform and non-uniform rods, finding reactions at supports, and problems where a rod is on the point of tilting. Take g = 9.8 m s⁻².
Where this sits:Unit A2 2: Applied Mathematics — moments: the moment of a force, equilibrium of a rigid body, and problems involving rods resting on supports.
Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Press Start when you are ready.
Definition
The moment of a force
moment = force × perpendicular distance from the pivotunits: newton metres (N m) · a moment is clockwise or anticlockwise
The word perpendicular is doing all the work. If the force acts at an angle θ to the rod at distance d from the pivot, the moment is Fd sin θ.
Worked example
A force of 20 N acts perpendicular to a spanner, 3 m from the pivot.
Moment = 20 × 3 = 60 N m
Zero moment: a force whose line of action passes through the pivot has zero perpendicular distance, so it exerts no moment — which is exactly why taking moments about a support eliminates its unknown reaction.
Calculate
A single moment
1A force of 20 N acts perpendicular to a rod at a distance of 3 m from the pivot. Find its moment about the pivot.
N m
Hint: Moment = force × perpendicular distance = 20 × 3.
Equilibrium
The principle of moments
A rigid body in equilibrium satisfies two conditions:
Resultant force = 0 (it does not accelerate); and
Total clockwise moments = total anticlockwise moments about any point (it does not rotate).
Worked example — a seesaw
A 30 kg child sits 2 m from the pivot. Where must a 40 kg child sit to balance?
Moments about the pivot: 30g × 2 = 40g × d
The g cancels: 60 = 40d ⇒ d = 1.5 m
Choose your pivot cleverly: take moments about a point where an unknown force acts — that force then has zero moment and disappears from the equation.
Calculate
Balancing a seesaw
2A 30 kg child sits 2 m from the pivot of a seesaw. How far from the pivot, on the other side, must a 40 kg child sit to balance it?
m
Hint: Clockwise = anticlockwise: 30g × 2 = 40g × d. The g cancels, so d = 60 ÷ 40.
Rods
Uniform rods and reactions at supports
A uniform rod has its weight acting at the midpoint. (A non-uniform rod does not — the position of its centre of mass is often what you are asked to find.)
Worked example
A uniform rod AB of length 4 m and mass 10 kg rests on supports at A and B. A 20 kg load sits 1 m from A. Find the reactions.
Weights: rod 10g = 98 N at the midpoint (2 m from A); load 20g = 196 N at 1 m from A.
3A uniform rod AB of length 4 m and mass 10 kg rests on supports at A and B. A 20 kg load is placed 1 m from A. Taking g = 9.8, find the reaction at B.
N
Hint: Take moments about A: R(B) × 4 = (10g × 2) + (20g × 1) = 196 + 196 = 392, so R(B) = 392 ÷ 4.
Calculate
Reaction at A
4For the same rod, find the reaction at A.
N
Hint: Resolve vertically: R(A) + R(B) = total weight = 10g + 20g = 294 N. With R(B) = 98 N, R(A) = 294 − 98.
Tilting
On the point of tilting
When a rod is about to tilt about a support, the reaction at the OTHER support becomes zero. That single fact solves the whole problem.
Worked example
A uniform rod AB of length 6 m and weight 200 N rests on supports 1 m and 4 m from A. A weight W is hung at B (6 m from A). Find the greatest W before the rod tilts.
About to tilt about the support at 4 m ⇒ the reaction at the 1 m support is zero.
Take moments about the support at 4 m. The rod's weight acts at the midpoint, 3 m from A — that is 1 m to the LEFT of the support.
Anticlockwise (rod's weight): 200 × 1 = 200 · Clockwise (W at B): W × (6 − 4) = 2W
Tilting begins when 2W = 200 ⇒ W = 100 N. So the greatest W is 100 N.
Calculate
Point of tilting
5A uniform rod AB of length 6 m and weight 200 N rests on supports 1 m and 4 m from A. A weight W hangs at B. Find the greatest value of W before the rod tilts.
N
Hint: At the point of tilting about the 4 m support, the reaction at the 1 m support is zero. Moments about the 4 m support: 200 × (4 − 3) = W × (6 − 4), so 200 = 2W.
Quick check
Where is the weight?
?A uniform rod of length 6 m has weight 90 N. Where does its weight act?
Sort it
Clockwise, anticlockwise or zero?
A rod is pivoted at P. Tap a force, then tap the moment it produces about P.
↻ Clockwise
↺ Anticlockwise
0️⃣ Zero moment
Quick check
Choosing the pivot
?A rod rests on supports at A and B, and you want R(B). Where should you take moments?
Match it
Match the moments fact
Tap the term on the left, then its meaning on the right.
Statement
Answer
Quick check
Angled force
?A force of 10 N acts at 30° to a rod, 4 m from the pivot. What is its moment?
Quick check
About to tilt
?A plank rests on two supports and is about to tilt about the right-hand support. What is true?
Examiner traps
Moments pitfalls
Perpendicular distance. For an angled force use Fd sin θ, not Fd.
Uniform means weight at the midpoint — do not assume this for a non-uniform rod.
Measure distances from the pivot you chose, not from the end of the rod.
Equilibrium needs both conditions: resultant force = 0 AND resultant moment = 0.
On the point of tilting, the reaction at the other support is zero — say so explicitly.
Take moments about an unknown to eliminate it.
Recap
The big ideas to know
Moment = force × perpendicular distance (N m); Fd sin θ for an angled force
Equilibrium of a rigid body: resultant force = 0 AND clockwise moments = anticlockwise moments
Uniform rod: weight acts at the midpoint · non-uniform: it does not
Take moments about an unknown force to make it vanish from the equation
On the point of tilting: the reaction at the other support is zero
Moments complete statics; momentum and impulse handle what happens in a collision. Press Finish to see your score.
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