A-level · Beta

Moments and equilibrium

Move a load along a supported beam and explain how both reactions change.

R_L+R_R=F+W

Scroll the diagram sideways to see every label.

Left reaction (N): 36.66667
Right reaction (N): 23.33333
Resultant vertical force (N): 0
Sum of moments (N m): 0

Equilibrium holds about every pivot. Moving the pivot changes individual moments, but does not change the reactions.

Signed moments about the selected pivot (N m)
Force	Moment
Left reaction	0
Right reaction	140
Point load	-80
Beam weight	-60

Derive the reactions about the left support

$R_R L=Fx+W\frac L2$

$R_R=\frac{Fx}{L}+\frac W2,\quad R_L=F+W-R_R$

Moment = force × perpendicular distance. Counterclockwise moments are positive and clockwise moments are negative.

A rigid horizontal uniform beam is supported at both ends. All loads are vertical and lie on the beam; its weight acts at the midpoint. Force arrows share a scale; their displaced tails keep labels readable. Green marks the chosen pivot. Values are rounded to 5 decimal places.

Predict, test and explain

Step 1 of 3

Predict

For a supported beam in equilibrium, does changing the point about which you take moments change the support reactions?

Explore and compare

Choose the investigation above. Predict what will change before you move a control.

Predict

Try it in the simulation. Change one thing at a time.

Explain

Notes and saved values stay in this tab. They disappear when you reload.

Hints, self-check and connections

Try a self-check

Fixed values, separate from the diagram controls.

For a supported beam in equilibrium, does changing the point about which you take moments change the support reactions?

$6R_R=40(2)+20(3)$

Hint

Draw both upward reactions and both downward loads before writing equations.

Reveal answer

The supports and loads remain fixed. In equilibrium the resultant force and resultant moment are zero, so taking moments about any point gives the same reactions when all forces are included.

$R_R=\frac{70}{3}\,\mathrm N,\quad R_L=\frac{110}{3}\,\mathrm N$

Build an explanation

Hint 1

Draw both upward reactions and both downward loads before writing equations.

Hint 2

A moment is force × perpendicular distance to its line of action. Taking moments about a support removes that support reaction from the moment equation.

Hint 3

For a uniform beam, put its weight at the midpoint. Use vertical force balance as well as moment balance.

Worked example

This example uses fixed values, separate from the diagram controls.

A uniform 6 m beam weighs 20 N and is supported at both ends. A 40 N downward load is 2 m from the left end. Find the reactions.

Vertical equilibrium gives R_L + R_R = 40 + 20 = 60 N.
Take moments about the left end: 6R_R = 40 × 2 + 20 × 3 = 140 N m.
Thus R_R = 70/3 N. Subtract from 60 to obtain R_L = 110/3 N.

Answer: Left reaction = 110/3 N ≈ 36.67 N; right reaction = 70/3 N ≈ 23.33 N.

Connect this idea

Watch out: The supports and loads remain fixed. In equilibrium the resultant force and resultant moment are zero, so taking moments about any point gives the same reactions when all forces are included. The beam is rigid, uniform and stationary, with supports at its two ends. Its weight acts at the midpoint; all forces shown are vertical, so horizontal separations are perpendicular distances. Changing the point about which moments are taken does not move a support. Forces are measured in N and moments in N m.

Specification and learning route

AQA S1 · Edexcel Mechanics 9.1

A-level. This model illustrates selected content; references are not a claim of full coverage or exam-board endorsement.

Useful starting knowledge: Force diagrams; Perpendicular distance; Simultaneous equations.

AQA specification · Edexcel specification