What are not resolving forces as well as taking moments?

Equilibrium needs both zero resultant force and zero total moment.

EnglandMathsSyllabus dot point

How do you analyse the turning effect of forces and the conditions for a rigid body to balance?

The moment of a force about a point, the principle of moments, equilibrium of a rigid body, and problems involving rods, beams and reactions at supports.

A focused answer to the Edexcel A-Level Mathematics moments content, covering the moment of a force about a point, the principle of moments, equilibrium of a rigid body, and problems involving rods, beams and reactions at supports.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

Edexcel wants you to calculate the moment of a force about a point, apply the principle of moments, set up and solve the conditions for a rigid body in equilibrium, and handle problems with rods and beams resting on supports, including finding reaction forces and the position of supports or loads.

The answer

The moment of a force

The principle of moments

A strategy for beam problems

Most beam questions follow the same routine. First, sketch the beam and mark every force with its distance from one end: the weight of a uniform beam acts at its midpoint, point loads act where they hang, and each support exerts an upward reaction. Second, take moments about one support so that its unknown reaction disappears, leaving a single equation for the other reaction. Third, resolve vertically (total up equals total down) to find the remaining reaction. A final check is to take moments about a second point and confirm the numbers are consistent. The skill being tested is choosing the point that eliminates the most unknowns, which is almost always a support or a pivot.

Examples in context

A uniform plank

AB

of length

5

m and weight

200

N rests on supports at

A

and

B

. A child of weight

300

N stands

1

m from

A

. Find both reactions. Take

g = 9.8 \text{ m s}^{-2}

step 1 Place the forces

The plank weight $200$ N acts at the midpoint $2.5$ m from $A$ ; the child $300$ N acts $1$ m from $A$ ; the reactions $R_A$ and $R_B$ act at the ends.

step 2 Take moments about $A$ to find $R_B$

$R_B \times 5 = 200 \times 2.5 + 300 \times 1 = 500 + 300 = 800$ , so $R_B = 160$ N.

step 3 Resolve vertically for $R_A$

$R_A + R_B = 200 + 300 = 500$ , so $R_A = 500 - 160 = 340$ N.

Try this

Q1. A force of $8$ N acts $0.5$ m from a pivot. Find its moment. [2 marks]

Cue. $M = 8 \times 0.5 = 4$ N m.

Q2. A light rod $3$ m long has a $10$ N weight $1$ m from end $A$ . Find the moment about $A$ . [2 marks]

Cue. $M = 10 \times 1 = 10$ N m clockwise about $A$ .

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20197 marksA uniform rod

AB

of length

4

m and weight

80

N rests horizontally on two supports at

A

and at

C

, where

C

3

m from

A

. A particle of weight

20

N is placed at

B

. Calculate the reactions at the two supports. Take

g = 9.8

m s

^{-2}

Show worked answer →

The weight of the rod ( $80$ N) acts at the midpoint, $2$ m from $A$ ; the $20$ N particle acts at $B$ , $4$ m from $A$ (B1).

Take moments about $A$ to eliminate the reaction $R_A$ (M1): $R_C \times 3 = 80 \times 2 + 20 \times 4$ (A1).

So $3R_C = 160 + 80 = 240$ , giving $R_C = 80$ N (A1).

Resolve vertically: $R_A + R_C = 80 + 20 = 100$ (M1), so $R_A = 100 - 80 = 20$ N (A1).

Check by taking moments about $C$ (A1): consistent.

Markers reward placing the weights correctly, taking moments about a support, the second reaction by resolving, and the two reactions.

Edexcel 20225 marksA uniform beam

PQ

of length

6

m and weight

W

rests on a support at its midpoint. A load of

50

N hangs at

P

and a load of

30

N hangs at

Q

. Show that the beam cannot balance about the midpoint, and find where a single support must be placed for equilibrium with only these two loads.

Show worked answer →

Take moments about the midpoint (3 m from each end). The $50$ N load gives a moment $50 \times 3 = 150$ N m and the $30$ N load gives $30 \times 3 = 90$ N m in the opposite sense (M1).

Since $150 \ne 90$ , the moments do not balance and the beam tips toward $P$ (A1).

For balance about a support a distance $d$ from $P$ , ignoring the beam weight: $50d = 30(6 - d)$ (M1), so $50d = 180 - 30d$ and $80d = 180$ , giving $d = 2.25$ m (A1).

The support must be $2.25$ m from $P$ (A1).

Markers reward the moment comparison, the tipping conclusion, the balance equation, and the distance.

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Mathematics (9MA0) specification — Pearson Edexcel (2017)