S1MechanicsStretch

Moments and Equilibrium

Every problem so far modelled objects as particles, which cannot rotate. Real beams, ladders and see-saws turn — and a $\textbf{moment}$ measures the turning effect of a force. Equilibrium now requires two conditions, not one: no resultant force $\textit{and}$ no resultant turning effect.

40 min Moments

The big picture

The strategic idea here is choosing where to take moments from. Any point works mathematically, but a force acting through your chosen point has zero moment about it — so taking moments about the line of an unknown force eliminates that unknown from the equation entirely. Choosing well can turn a pair of simultaneous equations into a single one you can solve in a line. That is a genuine problem-solving decision rather than a procedure, and it is what separates a quick solution from a laborious one.

What you'll be able to do

Calculate the moment of a force about a point, including angled forces
Apply the principle of moments to bodies in equilibrium
Choose a pivot that eliminates an unknown force
Solve problems involving rods, beams and the point of tilting

What a moment measures

The moment of a force about a point is the force multiplied by the $\textbf{perpendicular distance}$ from the point to the line of action of the force. Its units are newton-metres (N m), and it has a sense — clockwise or anticlockwise.

The word perpendicular is doing real work. A force directed straight at the pivot has zero perpendicular distance and therefore no turning effect, however large it is. Pushing a door towards its hinges does not open it.

For a force at an angle $\theta$ to the rod, the perpendicular component is $F\sin\theta$ , so the moment is $Fd\sin\theta$ where $d$ is the distance along the rod.

M = Fd

Force times perpendicular distance.

M = Fd\sin\theta

For a force at angle

\theta

to the rod.

Example 1A force of 30 N acts at the end of a 2 m rod, at

40^{\circ}

to the rod. Find its moment about the other end.

1The perpendicular component is

30\sin40^{\circ} = 19.28

2Moment

= 19.28 \times 2

Answer

38.6\ \mathrm{N\,m}

(3 s.f.)

Tip — Always identify the perpendicular distance, not the distance along the rod. Using the full length for an angled force is the standard error here.

The principle of moments

A rigid body in equilibrium satisfies two conditions. The forces balance in every direction, and the moments balance about $\textbf{any}$ point: total clockwise moment equals total anticlockwise moment.

The phrase "any point" is the useful part. The moments balance about every point, so you may choose the most convenient one — and different choices give different-looking but equally valid equations.

For a $\textbf{uniform}$ rod, the weight acts at the midpoint. For a non-uniform rod its position is usually the unknown to be found.

\Sigma M_{\text{clockwise}} = \Sigma M_{\text{anticlockwise}}

About any point, for equilibrium.

\Sigma F = 0

And forces must still balance.

Example 1A uniform 4 m plank of mass 20 kg rests on a support 1.5 m from its left end. A 50 kg child sits at the left end. How far from the support must a 30 kg child sit on the other side to balance? Take

g = 9.8

1Take moments about the support. The plank’s weight acts at its midpoint, 2 m from the left, so 0.5 m to the right of the support.

2Anticlockwise (child at left end, 1.5 m from support):

50g \times 1.5 = 735

3Clockwise: plank

20g \times 0.5 = 98

, plus second child

30g \times d

735 = 98 + 294d

, so

294d = 637

Answer

d = 2.17

m from the support (3 s.f.).

Why this works

Note that the plank’s own weight had to be included, acting at the midpoint. Forgetting the weight of the beam itself is one of the most frequent errors in moments questions.

Choosing the pivot to eliminate an unknown

A force acting through the point you take moments about has zero perpendicular distance, so it contributes nothing. That lets you delete an unknown from the equation by choosing where to take moments.

With a beam on two supports and both reactions unknown, taking moments about one support removes that reaction and leaves an equation in the other alone. Resolving vertically then gives the second reaction immediately.

That order — moments first to isolate one unknown, then resolving — is faster and less error-prone than setting up simultaneous equations.

Example 1A uniform beam of mass 40 kg and length 6 m rests on supports at each end. A 100 kg load sits 2 m from the left support. Find both reactions. Take

g = 9.8

1Take moments about the left support to eliminate

R_{L}

2Clockwise: load

100g \times 2 = 1960

; beam weight

40g \times 3 = 1176

3Anticlockwise:

R_{R} \times 6

6R_{R} = 1960 + 1176 = 3136

, so

R_{R} = 522.7

5Resolving vertically:

R_{L} + R_{R} = 140g = 1372

, so

R_{L} = 849.3

Answer

R_{L} = 849\ \mathrm{N}

;

R_{R} = 523\ \mathrm{N}

(3 s.f.).

Tip — Check the total: $849 + 523 = 1372 = 140g$ ✓. The reactions must sum to the total weight, which catches most arithmetic slips instantly.

Tilting and the point of rotation

When a beam is about to tilt over a support, it is on the point of rotating about that support — and the reaction at every $\textbf{other}$ support becomes zero, since contact there is about to be lost.

That is the extra condition tilting problems supply. Setting the other reaction to zero and taking moments about the pivot gives a single equation for the unknown.

Identify which support the beam will tilt about by asking which way the load is pushing it. A load beyond one support tips the beam about that support and lifts the far end.

Example 1A uniform 5 m plank of mass 30 kg rests on supports at 1 m and 4 m from its left end. A person walks along it from the left. How far from the left end can they get before the plank tilts? Take the person as 70 kg and

g = 9.8

1Walking left of the 1 m support would tilt the plank about that support, lifting the right side.

2At the point of tilting, the reaction at the 4 m support is zero.

3Take moments about the support at 1 m. Plank weight acts at 2.5 m, so 1.5 m to the right:

30g \times 1.5 = 441

anticlockwise.

4Person at distance

x

from the left, so

\left(1-x\right)

to the left of the support:

70g\left(1-x\right) = 441

686\left(1-x\right) = 441

, so

1-x = 0.643

Answer

x = 0.357

m from the left end (3 s.f.) — they can walk to 0.357 m before it tilts.

Why this works

The person can walk to within about 0.36 m of the left end. Beyond that, their moment about the 1 m support exceeds the plank’s and the far end lifts — which is exactly the condition "reaction at the far support becomes zero" describes.

Think like an examiner

State which point you are taking moments about, and why if it eliminates an unknown.
Include the weight of the rod or beam, acting at the midpoint if uniform.
Use perpendicular distances, resolving angled forces first.
For tilting, state that the reaction at the other support is zero.

Common misconceptions

Forgetting the weight of the beam itself.

A uniform beam has weight acting at its midpoint and it must be included.

Using the distance along the rod for an angled force.

The moment uses the perpendicular distance, giving

Fd\sin\theta

Assuming the weight of a non-uniform rod acts at the midpoint.

Only uniform rods have their centre of mass at the middle.

Keeping both reactions in a moments equation.

Take moments about one support so its reaction contributes nothing.

Moments

M = Fd

Perpendicular distance.

M = Fd\sin\theta

Angled force.

\Sigma M_{\text{cw}} = \Sigma M_{\text{acw}}

Principle of moments.

\Sigma F = 0

Forces balance too.

R = 0 \text{ at the far support}

On the point of tilting.

Stretch yourself

A non-uniform rod $AB$ of length 3 m and mass 12 kg rests horizontally on supports at $A$ and $B$ . The reaction at $A$ is 78.4 N. Find the distance of the centre of mass from $A$ . Take $g = 9.8\ \mathrm{m\,s^{-2}}$ .

Hint — Find the reaction at B by resolving, then take moments about A to locate the weight.

Reveal solution

SolutionThe rod’s weight is

12 \times 9.8 = 117.6

N. Resolving vertically,

R_{A} + R_{B} = 117.6

, so

R_{B} = 117.6 - 78.4 = 39.2

N. Now take moments about

A

, which eliminates

R_{A}

. Let the centre of mass be

x

metres from

A

. Clockwise about

A

: the weight,

117.6x

. Anticlockwise:

R_{B} \times 3 = 39.2 \times 3 = 117.6

. Setting them equal,

117.6x = 117.6

, so

x = 1

m. The centre of mass is 1 m from

A

— a third of the way along, which is consistent with

R_{A}

being exactly twice

R_{B}

: the support nearer the centre of mass carries proportionally more of the weight. That reciprocal relationship is a useful sanity check on any two-support problem.

\blacksquare

Questions students ask

Which point should I take moments about?

Usually a point where an unknown force acts, since that force then contributes no moment and drops out of the equation. With two unknown reactions, take moments about one of them.

Why must the distance be perpendicular?

Because only the component of force perpendicular to the line joining it to the pivot produces turning. A force directed at the pivot cannot rotate anything about it.

Where does the weight of a rod act?

At its centre of mass — the midpoint for a uniform rod. For a non-uniform rod the position is generally what the question asks you to find.

What happens at the point of tilting?

The body is about to rotate about one support, so contact is being lost elsewhere and the reaction at the other support is zero. That gives the extra equation.

Key takeaways