Moments – A Level Maths Year 2

1 The Moment of a Force

In Year 1 every object was a particle: all its mass sat at a single point, so forces could only make it slide. A real door, beam or ladder can also turn. DfE content area S · Moments asks you to measure that turning effect and use it to solve problems about rigid bodies.

Moment of a force about a point

Moment $=$ force $\times$ perpendicular distance from the point to the line of action of the force
$M = Fd$ measured in newton metres (N m)

Direction matters. A moment is either clockwise or anticlockwise. Choose one sense to be positive at the start of a question, and state it. A force whose line of action passes through the point has zero moment about that point — this is the key to choosing a clever point to take moments about.

If the force is not perpendicular to the rod, either use the perpendicular component of the force, or the perpendicular distance — both give the same answer.

Force at an angle $\theta$ to a rod, applied at distance $d$ from the pivot

$M = (F\sin\theta) \times d \;=\; F \times (d\sin\theta)$

Worked Example 1 — A single force

(a) A force of $20$ N acts perpendicular to a spanner at $0.6$ m from the nut. (b) A force of $30$ N acts on a rod at $1.5$ m from a pivot, at $40^\circ$ to the rod. Find each moment.

①(a) $M = 20 \times 0.6 = 12$ N m.

②(b) The perpendicular component is $30\sin 40^\circ$.

③$M = 30\sin 40^\circ \times 1.5 = 28.9$ N m (3 s.f.).

Sanity check: a force at an angle always has a smaller moment than the same force applied perpendicularly — here $28.9 < 30 \times 1.5 = 45$.

Worked Example 2 — The resultant moment of several forces

A light rod $AB$ of length $4$ m has an upward force of $10$ N at $A$, a downward force of $15$ N at $1$ m from $A$ and a downward force of $8$ N at $B$. Find the resultant moment about $A$.

①Take clockwise as positive (with $A$ on the left).

②The $10$ N force acts through $A$, so its moment is $0$.

③$15 \times 1 + 8 \times 4 = 15 + 32 = 47$

④Resultant moment $= 47$ N m clockwise about $A$.

2 Equilibrium of a Rigid Body

Conditions for a rigid body to be in equilibrium

① the resultant force is zero (resolve in two directions)
② the resultant moment about any point is zero
i.e. total clockwise moments $=$ total anticlockwise moments

Model word	What it means for the calculation
Rigid body / rod / beam	Has length and does not bend. Forces at different points create moments.
Uniform	Mass evenly spread, so the weight acts at the midpoint.
Non-uniform	The weight acts at the centre of mass, which you usually have to find.
Light	Mass ignored — no weight to include.
Particle (a person, a load)	Its weight acts at a single point on the beam.

Choose the point wisely. Take moments about a point where an unknown force acts. That force then has no moment, and you have one fewer unknown in the equation.

3 Beams on Supports

A beam resting on supports (or hanging from vertical strings) has only vertical forces acting on it, so you get one equation from resolving vertically and one from each point you take moments about.

Worked Example 3 — Finding two reactions

A uniform plank $AB$ of length $6$ m and mass $20$ kg rests horizontally on supports at $C$ and $D$, where $AC = 1$ m and $DB = 2$ m. A child of mass $30$ kg stands on the plank $3.5$ m from $A$. Find the reaction at each support.

①Weight of plank $20g$ at the midpoint, $3$ m from $A$. $D$ is $4$ m from $A$, so $CD = 3$ m.

②Moments about $C$ (removes $R_C$): $R_D \times 3 = 20g \times 2 + 30g \times 2.5 = 115g$

③$R_D = \dfrac{115g}{3} = 375.7\ldots = 376$ N (3 s.f.)

④Resolve vertically: $R_C + R_D = 50g = 490$, so $R_C = 490 - 375.7 = 114$ N (3 s.f.).

Check by taking moments about $D$: $R_C \times 3 = 20g \times 1 + 30g \times 0.5 = 35g$, so $R_C = 35g/3 = 114$ N ✓

Worked Example 4 — Unequal reactions

A uniform rod $AB$ of length $4$ m and mass $12$ kg rests on supports at $A$ and $B$. A particle of mass $m$ kg is placed $1$ m from $A$. The reaction at $A$ is twice the reaction at $B$. Find $m$.

①Let $R_B = R$, so $R_A = 2R$. Resolve: $3R = (12 + m)g$.

②Moments about $A$: $R \times 4 = 12g \times 2 + mg \times 1 = (24 + m)g$.

③From ① $R = \dfrac{(12+m)g}{3}$; from ② $R = \dfrac{(24+m)g}{4}$.

④$4(12 + m) = 3(24 + m) \Rightarrow 48 + 4m = 72 + 3m \Rightarrow m = 24$ kg.

Notice $g$ cancelled. Leave $g$ as a letter until the end — it often disappears.

4 On the Point of Tilting

When a beam is about to tip (or "on the point of turning") about one support, it is just losing contact with the other support. So the reaction at the other support is zero. Then take moments about the support it is tipping about.

About to tilt about $D$

$R_C = 0$ and moments about $D$ balance

Worked Example 5 — How far can the child walk?

Using the plank in Worked Example 3, the child walks towards $B$. How far from $A$ can the child stand before the plank tips?

①The plank would tip about $D$, so on the point of tipping $R_C = 0$.

②Let the child be $x$ m from $A$, with $x > 4$.

③Moments about $D$: plank weight anticlockwise, child clockwise: $20g \times (4 - 3) = 30g \times (x - 4)$

④$x - 4 = \dfrac{20}{30} = \dfrac{2}{3}$, so $x = 4.67$ m (3 s.f.).

⑤The child can stand up to $4.67$ m from $A$, i.e. $\tfrac{2}{3}$ m beyond $D$.

Do not set both reactions to zero. Only the support that the beam is lifting off has zero reaction. The other support is carrying the whole load.

5 Non-Uniform Rods

For a non-uniform rod, the weight acts at an unknown distance $x$ from one end. That distance is just another unknown in your moments equation.

Worked Example 6 — Finding the centre of mass

A non-uniform rod $AB$ of length $3$ m and mass $8$ kg hangs horizontally from two vertical strings at $A$ and $B$. The tension at $A$ is three times the tension at $B$. Find the distance of the centre of mass from $A$.

①Let $T_B = T$, $T_A = 3T$. Resolve: $4T = 8g$, so $T = 2g = 19.6$ N and $T_A = 58.8$ N.

②Let the centre of mass be $x$ m from $A$. Moments about $A$: $8g \times x = T_B \times 3 = 6g$.

③$x = \dfrac{6}{8} = 0.75$ m from $A$.

Check: the bigger tension is at $A$, so the centre of mass should be nearer $A$ — and $0.75 < 1.5$ ✓

Worked Example 7 — Two tilting conditions

A non-uniform rod $AB$ of length $4$ m and mass $M$ kg rests on supports at $C$ and $D$, where $AC = DB = 1$ m. When a $12$ kg mass is hung at $A$ the rod is about to tip about $C$. When instead an $8$ kg mass is hung at $B$ the rod is about to tip about $D$. Find $M$ and the distance $x$ of the centre of mass from $A$.

①Tipping about $C$ ($R_D = 0$), moments about $C$: $12g \times 1 = Mg(x - 1)$, so $M(x - 1) = 12$.

②Tipping about $D$ ($R_C = 0$), moments about $D$ (which is $3$ m from $A$): $8g \times 1 = Mg(3 - x)$, so $M(3 - x) = 8$.

③Add: $M(x - 1 + 3 - x) = 20$, so $2M = 20$ and $M = 10$.

④Then $x - 1 = 1.2$, so $x = 2.2$ m from $A$.

6 Rods with Forces at an Angle

When a rod is held by a hinge and an angled string, the hinge exerts a force with both a horizontal and a vertical component. Take moments about the hinge to find the tension first, then resolve to find the hinge force.

Worked Example 8 — A hinged rod held by a string

A uniform rod $AB$ of length $2$ m and weight $50$ N is freely hinged to a vertical wall at $A$. It is held horizontal by a light string attached at $B$ and fixed to the wall above $A$. The string makes $30^\circ$ with the rod. Find the tension and the force exerted by the hinge.

①Moments about $A$ (removes both hinge components): $T\sin 30^\circ \times 2 = 50 \times 1$

②$T = \dfrac{50}{2 \times 0.5} = 50$ N

③Resolve horizontally: $H = T\cos 30^\circ = 43.3$ N (away from the wall).

④Resolve vertically: $V + T\sin 30^\circ = 50$, so $V = 25$ N.

⑤Magnitude $\sqrt{43.3^2 + 25^2} = 50$ N, at $\tan^{-1}\!\left(\tfrac{25}{43.3}\right) = 30^\circ$ above the horizontal.

The hinge force does not have to act along the rod. Always give it two unknown components and let the equations decide.

7 Quick Reference

Moment

Force $\times$ perpendicular distance, in N m.

Angled force

$M = Fd\sin\theta$, $\theta$ the angle to the rod.

Equilibrium

Resultant force $= 0$ and resultant moment $= 0$.

Uniform

Weight acts at the midpoint.

Non-uniform

Weight at an unknown centre of mass $x$ from one end.

Best point

Take moments where an unknown force acts.

About to tip about $D$

The reaction at the other support is $0$.

Hinge

Two unknown components, $H$ and $V$.

Check

Take moments about a second point.

8 Practice Questions

Use $g = 9.8$ m s$^{-2}$ and give answers to 3 significant figures where they are not exact.

Question 1

A force of $25$ N acts perpendicular to a door at $0.4$ m from the hinge. Find its moment about the hinge.

▶ Show solution

$M = 25 \times 0.4 = 10$ N m.

Question 2

A force of $40$ N is applied to the end of a spanner of length $0.25$ m, at $60^\circ$ to the spanner. Find the moment about the nut.

▶ Show solution

Perpendicular component $= 40\sin 60^\circ$.

$M = 40\sin 60^\circ \times 0.25 = 10\sin 60^\circ = 5\sqrt{3} = 8.66$ N m.

Question 3

A uniform see-saw is pivoted at its centre. A child of mass $25$ kg sits $1.6$ m from the pivot. How far from the pivot, on the other side, must a child of mass $32$ kg sit for the see-saw to balance?

▶ Show solution

The see-saw's own weight acts through the pivot, so has no moment.

$25g \times 1.6 = 32g \times d \Rightarrow d = \dfrac{40}{32} = 1.25$ m.

Question 4

A uniform beam $AB$ of length $8$ m and weight $600$ N rests on supports at $A$ and $B$. A load of $400$ N is placed $2$ m from $A$. Find the reactions at $A$ and $B$.

▶ Show solution

Moments about $A$: $R_B \times 8 = 600 \times 4 + 400 \times 2 = 3200$, so $R_B = 400$ N.

Resolve: $R_A + R_B = 1000$, so $R_A = 600$ N.

Check (moments about $B$): $R_A \times 8 = 600 \times 4 + 400 \times 6 = 4800$, so $R_A = 600$ N ✓

Question 5

A uniform plank $AB$ of length $6$ m and mass $30$ kg rests on supports $C$ and $D$ with $AC = DB = 1.5$ m. A man of mass $75$ kg walks from the middle towards $B$. How far beyond $D$ can he walk before the plank tips?

▶ Show solution

On the point of tipping about $D$, $R_C = 0$.

$D$ is $1.5$ m from the centre of the plank. Let the man be $d$ m beyond $D$.

Moments about $D$: $30g \times 1.5 = 75g \times d \Rightarrow d = \dfrac{45}{75} = 0.6$ m.

He can walk $0.6$ m beyond $D$ (i.e. to $0.9$ m from $B$).

Question 6

A non-uniform rod $AB$ of length $2.4$ m and mass $6$ kg hangs horizontally from vertical strings at $A$ and $B$. The tension at $B$ is $20$ N. Find the tension at $A$ and the distance of the centre of mass from $A$.

▶ Show solution

Resolve: $T_A + 20 = 6g = 58.8$, so $T_A = 38.8$ N.

Moments about $A$: $6g \times x = 20 \times 2.4 = 48$.

$x = \dfrac{48}{58.8} = 0.816$ m from $A$.

Question 7

A uniform beam $AB$ of length $4$ m and mass $20$ kg rests on supports at $A$ and $B$. A particle of mass $m$ kg is placed $0.5$ m from $B$. The reaction at $B$ is $150$ N. Find $m$.

▶ Show solution

The particle is $3.5$ m from $A$. Moments about $A$:

$150 \times 4 = 20g \times 2 + mg \times 3.5$

$600 = 392 + 34.3m \Rightarrow m = \dfrac{208}{34.3} = 6.06$ kg.

Question 8

A uniform rod $AB$ of length $1.2$ m and mass $5$ kg is hinged to a vertical wall at $A$ and held horizontal by a light string attached at $B$, making $40^\circ$ with the rod. Find the tension and the horizontal and vertical components of the hinge force.

▶ Show solution

Moments about $A$: $T\sin 40^\circ \times 1.2 = 5g \times 0.6 = 29.4$

$T = \dfrac{29.4}{1.2\sin 40^\circ} = 38.1$ N.

Horizontal: $H = T\cos 40^\circ = 29.2$ N (away from the wall).

Vertical: $V = 5g - T\sin 40^\circ = 49 - 24.5 = 24.5$ N upwards.

(Neat check: $T\sin 40^\circ = 24.5 = \tfrac12 \times 49$, because the weight acts at the midpoint, so the string and hinge share the vertical load equally.)

Question 9

A non-uniform plank $AB$ of length $5$ m and mass $20$ kg has its centre of mass $2$ m from $A$. It rests on supports at $C$ and $D$ where $AC = 0.5$ m and $DB = 1$ m. Find the largest mass that can be placed (a) at $B$, (b) at $A$, without the plank tipping.

▶ Show solution

$D$ is $4$ m from $A$.

(a) Tipping about $D$, $R_C = 0$. Moments about $D$: $m g \times 1 = 20g \times (4 - 2)$, so $m = 40$ kg.

(b) Tipping about $C$, $R_D = 0$. Moments about $C$: $m g \times 0.5 = 20g \times (2 - 0.5)$, so $m = 60$ kg.

Question 10

A uniform beam $AB$ of length $6$ m and mass $50$ kg rests horizontally on supports at $C$ and $D$, where $AC = 1$ m and $DB = 1.5$ m. A man of mass $80$ kg stands on the beam at $E$, where $AE = x$ m.

(a) Show that the reaction at $C$ is $R_C = \dfrac{(435 - 80x)g}{3.5}$. (b) Given that the reaction at $D$ is twice the reaction at $C$, find $x$. (c) Find the range of positions on the beam where the man can stand without the beam tipping. (d) State two modelling assumptions you have made.

▶ Show solution

(a) $D$ is $4.5$ m from $A$; the beam's weight acts $3$ m from $A$, i.e. $1.5$ m from $D$; $CD = 3.5$ m.

Moments about $D$: $R_C \times 3.5 = 50g \times 1.5 + 80g(4.5 - x) = 75g + 360g - 80gx$

$R_C = \dfrac{(435 - 80x)g}{3.5}$ as required.

(b) Resolve: $R_C + R_D = 130g$. With $R_D = 2R_C$: $3R_C = 130g$, so $R_C = \dfrac{130g}{3}$.

$\dfrac{435 - 80x}{3.5} = \dfrac{130}{3} \Rightarrow 435 - 80x = 151.67 \Rightarrow x = 3.54$ m.

(c) Need $R_C \geqslant 0$: $435 - 80x \geqslant 0 \Rightarrow x \leqslant 5.4375$.

Need $R_D \geqslant 0$. Moments about $C$: $R_D \times 3.5 = 50g \times 2 + 80g(x - 1) = (20 + 80x)g$, which is positive for every $x \geqslant 0$.

So the man can stand anywhere with $0 \leqslant x \leqslant 5.44$ m (3 s.f.) — he cannot reach the last $0.56$ m by $B$.

(d) The man is modelled as a particle (his weight acts at a single point); the beam is a uniform rigid rod (it does not bend). Either is acceptable, as is "the supports are points".

Moments · A Level Maths Year 2 · Created with MathJax