Moments and beams · Statics: forces, equilibrium & moments

A force can turn a body as well as push it. Its turning effect about a point is its moment:

\begin{gathered}\text{moment} = \text{force} \times \text{distance}\\ \text{(the perpendicular distance from the point)}\end{gathered}

The principle of moments

A beam balances when the clockwise moments about any point equal the anticlockwise moments about that point:

Balancing on a pivotF₁ × d₁ = F₂ × d₂

The weight of a uniform beam acts at its middle. Take moments about the pivot, so its reaction (unknown) has no moment.

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q6

A uniform beam $WX$ , of length $90\text{ cm}$ and weight $50\text{ N}$ , is suspended on a pivot $35\text{ cm}$ from $W$ . It is kept in equilibrium by means of forces $T$ and $20\text{ N}$ applied at $Y$ and $Z$ respectively. $|WY| = 10\text{ cm}$ and $|XZ| = 10\text{ cm}$ . Find the value of $T$ .

Distances from the pivot
- $T$ at $Y$ : ${35 - 10 = 25}$ cm on the $W$ side.
- The weight, at the middle: ${45 - 35 = 10}$ cm on the $X$ side.
- The 20 N at $Z$ : ${80 - 35 = 45}$ cm on the $X$ side.
Think first. The pivot is 35 cm from W. How far is each force from it?
Moments about the pivot
- ${25T = 50 \times 10 + 20 \times 45}$ .
- ${25T = 1400}$ , so ${T = 56}$ N.

Beams on two supports

With two supports there are two unknown reactions. Take moments about one support to find the other reaction; then use up = down for the first.

A beam on two supportsSet the load and where it hangs

26.25 Nreaction at A63.75 Nreaction at B

load (N) = 50load from left end (m) = 4.5

Moments about A: B's reaction × 4 = 40 × 2 + 50 × (3.5), so B's reaction is 63.75 N. Up = down: A's reaction = 40 + 50 − 63.75 = 26.25 N. Both reactions are positive, so the beam balances.

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q18 (a)

A uniform bar $PQ$ of length $60\text{ cm}$ and weight $20\text{ N}$ is supported at two points $C$ and $D$ such that $|PC| = 10\text{ cm}$ and $|QD| = 15\text{ cm}$ . Two forces $10\text{ N}$ and $15\text{ N}$ are placed at $P$ and $Q$ respectively. If the system remains in equilibrium under the action of these forces, calculate the reactions at $C$ and $D$ .

Distances from D
- From $D$ : $P$ is 45 cm, $C$ is 35 cm and the centre is 15 cm away on one side.
- $Q$ is 15 cm away on the other side.
Think first. PQ is 60 cm, with C 10 cm from P and D 15 cm from Q.
Moments about D
- ${35R_C + 15 \times 15 = 20 \times 15 + 10 \times 45}$ .
- ${35R_C = 300 + 450 - 225 = 525}$ , so ${R_C = 15}$ N.
Think first. R_C and the 15 N turn one way; the 10 N and the weight turn the other.
Up = down
- ${R_C + R_D = 10 + 20 + 15 = 45}$ , so ${R_D = 30}$ N.

Non-uniform beams

A non-uniform beam’s weight acts at its centre of gravity, which need not be the middle. Otherwise the method is the same.

Worked example · WAEC 2023

WAEC 2023 · Paper 2 · Q14 (b)

A non-uniform beam $PQ$ of length $15\text{ m}$ and weight $120\text{ N}$ rests horizontally on supports at $P$ and $Q$ . If the centre of gravity of the beam is $3.5\text{ m}$ from $Q$ , find the reaction at $Q$ .

Where the weight acts
- ${15 - 3.5 = 11.5}$ m from $P$ .
Think first. The centre of gravity is 3.5 m from Q. How far is it from P?
Moments about P
- ${15R_Q = 120 \times 11.5 = 1380}$ , so ${R_Q = 92}$ N.

More: beams

Your turn

WAEC 2018 · Paper 2 · Q15 (a)

Statics: forces, equilibrium & moments

A uniform beam $XY$ , $4\text{ m}$ long and weighing $350\text{ N}$ , rests on two pivots $P$ and $Q$ . It is kept in equilibrium by weights of $80\text{ N}$ attached at $X$ and $1000\text{ N}$ attached at a point between $P$ and $Q$ such that it is $0.6\text{ m}$ from $Q$ . $|XP| = 0.8\text{ m}$ and $|PQ| = 2.2\text{ m}$ .

(a)
Calculate the reactions at $P$ and $Q$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

From $X$ : $P$ at $0.8$ m, the centre at $2$ m, the $1000\text{ N}$ weight at $2.4$ m and $Q$ at $3$ m.
Moments about $P$ : $2.2R_Q + 80(0.8) = 350(1.2) + 1000(1.6)$ .
$2.2R_Q = 420 + 1600 - 64 = 1956$ , so $R_Q \approx 889.09\text{ N}$ .
Up = down: $R_P = 80 + 350 + 1000 - 889.09$
$\approx 540.91\text{ N}$ .

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