Mini-Lesson

Moments

2 missions. Guess, learn, then check. Your place is saved.

Mission 1A 30 kg child and a 60 kg adult on a seesaw. Where does the adult sit?
Mission 2Why is a ladder more likely to slip when you climb near the top?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

A 30 kg child and a 60 kg adult on a seesaw. Where does the adult sit?

Make a guess: A 30 kg child and a 60 kg adult on a seesaw. Where does the adult sit?

Mechanics · moments

The moment of a force (A-level only)

The moment of a force about a point is its turning effect:

moment = F \times d

$d$ d is the perpendicular distance from the point to the line of action of the force. Units: N m.
State the sense: clockwise or anticlockwise. A force through the point has zero moment.
A uniform rod's weight acts at its midpoint; a light rod has no weight.

Quick check

?A light rod

A B

of length 4 m is horizontal with

A

on the left. Forces act perpendicular to it: 10 N upwards at

A

, 6 N downwards 1 m from

A

, and 8 N downwards at

B

. Find the total moment about the midpoint

M

.

Mechanics · moments

Rigid bodies in equilibrium

A rigid body is in equilibrium when the resultant force is zero and the resultant moment about any point is zero.
So: total up = total down, and clockwise moments = anticlockwise moments about any point you choose.
Take moments about the point where an unknown force acts: that force has no moment there.
A beam on the point of tilting about one support has zero reaction at the other support.

Worked example Mechanics · moments

A beam on two supports

Question

A uniform beam

A B

of length 6 m and mass 30 kg rests horizontally on supports at

C

and

D

, where

A C = 1

m and

D B = 1

m. A child of mass 40 kg stands 4 m from

A

. Find the reactions at

C

and

D

.

1
Weights: beam $30 g$ 30g at 3 m from $A$ A; child $40 g$ 40g at 4 m from $A$ A
2
Take moments about $C$ C (removes $R_{C}$ R_C): $4 R_{D} = 30 g (2) + 40 g (3) = 180 g$ 4R_D=30g(2)+40g(3)=180g
3
Solve: $R_{D} = 45 g = 441$ R_D=45g=441 N
4
Resolve vertically: $R_{C} = 70 g - 45 g = 25 g = 245$ R_C=70g−45g=25g=245 N

Answer

R_{C} = 245

N and

R_{D} = 441

N

⚠️ Watch out

distances in a moments equation are from the point you take moments about, not from the end

A

.

Calculate

?A uniform plank

A B

of length 5 m and mass 24 kg rests horizontally on supports at

C

and

D

, where

A C = 1

m and

D B = 1.5

m. A load is placed at

B

. Find the greatest mass of the load for which the plank does not tilt.

kg

Worked example Mechanics · moments

Finding the centre of mass of a non-uniform rod

Question

A non-uniform rod

A B

of length 4 m and mass 12 kg hangs horizontally from two vertical strings at

A

and

B

. The tensions are 88.2 N at

A

and 29.4 N at

B

. Find the distance of the centre of mass from

A

.

1
Check vertically: $88.2 + 29.4 = 117.6 = 12 g$ 88.2+29.4=117.6=12g
2
Let the weight act $x$ x m from $A$ A; take moments about $A$ A: $12 g \times x = 29.4 \times 4$ 12g × x=29.4 × 4
3
Solve: $117.6 x = 117.6$ 117.6x=117.6

Answer

x = 1

m from

A

⚠️ Watch out

the larger tension is at the end nearer the centre of mass, here

A

.

Calculate

?A uniform rod

A B

of length 3 m and mass 8 kg has a particle of mass 2 kg fixed at

A

and a particle of mass 6 kg fixed at

B

. The rod rests horizontally in equilibrium on a single pivot at

C

. Find the distance

A C

, as an exact decimal.

m

Sort it

?A rigid body is in equilibrium. True or false?

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?A seesaw is pivoted at its centre. A 30 kg child sits 2.4 m from the pivot. Where must a 60 kg adult sit for it to balance?

Mission 2

Why is a ladder more likely to slip when you climb near the top?

Make a guess: Why is a ladder more likely to slip when you climb near the top?

Mechanics · moments: ladders and hinges

The moment of a force at an angle (A-level only)

Moment = force × perpendicular distance to its line of action.
If a force $F$ F acts at the end of a rod of length $l$ l, at angle $θ$ θ to the rod, its moment about the other end is

Fl sin θ

Equivalent view: split $F$ F into $F sin θ$ Fsin θ perpendicular to the rod (moment $F sin θ \times l$ Fsin θ × l) and $F cos θ$ Fcos θ along the rod (no moment, as its line passes through the point).

Calculate

?A light rod

A B

has length 2 m. A force of 30 N acts at

B

at

4 0^{\circ}

to

A B

. Find the moment of this force about

A

, to 3 significant figures.

N m

Mechanics · moments: ladders and hinges

Modelling a ladder

Smooth wall: reaction $S$ S is perpendicular to the wall (horizontal). No friction there.
Rough ground: normal reaction $R$ R upwards and friction $F$ F along the ground, towards the wall (it stops the foot sliding out).
A uniform ladder's weight acts at its midpoint.
Method: resolve vertically and horizontally, then take moments about the foot, which removes $R$ R and $F$ F.
About to slip: limiting equilibrium, $F = μ R$ F=μ R. Otherwise $F \leq μ R$ F≤μ R.

Worked example Mechanics · moments: ladders and hinges

The least coefficient of friction

Question

A uniform ladder of length 4 m and mass 15 kg rests against a smooth vertical wall with its foot on rough horizontal ground. The ladder makes

6 0^{\circ}

with the ground. Find the least possible value of

μ

.

1
Resolve vertically: $R = 15 g = 147$ R=15g=147 N
2
Moments about the foot; $S$ S is $4 sin 6 0^{\circ}$ 4sin 60° above it, the weight $2 cos 6 0^{\circ}$ 2cos 60° across: $S \times 4 sin 6 0^{\circ} = 15 g \times 2 cos 6 0^{\circ}$ S × 4sin 60°=15g × 2cos 60°
3
Solve: $S = 42.4$ S=42.4 N; resolve horizontally: $F = S = 42.4$ F=S=42.4 N
4
Equilibrium needs $F \leq μ R$ F≤μ R: $μ \geq \frac{S}{15 g} = \frac{1}{2 tan 6 0 ^{\circ}}$ μ≥S/15g=1/(2tan 60°)

Answer

least

μ = 0.289

(3 s.f.)

⚠️ Watch out

the perpendicular distance for

S

is the height of the top of the ladder,

4 sin 6 0^{\circ}

, not the length 4 m.

Quick check

?For the ladder in the example (length 4 m, at

6 0^{\circ}

to the ground), which expression is the moment of the wall reaction

S

about the foot?

Calculate

?A uniform ladder of length 5 m and mass 20 kg rests against a smooth vertical wall, with its foot on rough horizontal ground, at angle

α

to the ground, where

tan α = 2

. The coefficient of friction is 0.3. A person of mass 80 kg climbs the ladder. Find how far up the ladder they can climb before it slips, as an exact decimal.

m

Mechanics · moments: ladders and hinges

Hinged rods

A hinge can push or pull in any direction. Model its force by two unknown components, horizontal $X$ X and vertical $Y$ Y.
Take moments about the hinge first: $X$ X and $Y$ Y drop out, leaving one unknown (often a tension).
Then resolve horizontally and vertically to find $X$ X and $Y$ Y.
Magnitude of the hinge force: $X^{2} + Y^{2}$ √(X²+Y²); direction from $tan^{- 1} \frac{Y}{X}$ tan⁻¹Y/X.

Worked example Mechanics · moments: ladders and hinges

A rod held by a string

Question

A uniform rod

A B

of length 2 m and mass 5 kg is freely hinged at

A

to a vertical wall. It is held horizontal by a light string from

B

to a point on the wall above

A

, making

3 0^{\circ}

with the rod. Find the tension and the force at the hinge.

1
Moments about $A$ A: $T sin 3 0^{\circ} \times 2 = 5 g \times 1$ Tsin 30° × 2=5g × 1, so $T = 5 g = 49$ T=5g=49 N
2
Resolve horizontally: $X = T cos 3 0^{\circ} = 42.4$ X=Tcos 30°=42.4 N (away from the wall)
3
Resolve vertically: $Y + T sin 3 0^{\circ} = 5 g$ Y+Tsin 30°=5g, so $Y = 24.5$ Y=24.5 N (upwards)
4
Combine: $42. 4^{2} + 24. 5^{2} = 49.0$ √(42.4²+24.5²)=49.0 and $tan^{- 1} \frac{24.5}{42.4} = 3 0^{\circ}$ tan⁻¹24.5/42.4=30°

Answer

T = 49

N; hinge force 49.0 N at

3 0^{\circ}

above the horizontal

⚠️ Watch out

do not assume the hinge force acts along the rod; find both components.

Calculate

?A uniform rod

A B

of length 3 m and mass 4 kg is freely hinged at

A

to a vertical wall. A particle of mass 6 kg is attached at

B

. The rod is held horizontal by a light string from

B

to a point on the wall vertically above

A

, making

4 0^{\circ}

with the rod. Find the vertical component of the force at the hinge, to 3 significant figures.

N

Mission check

?A person climbs a ladder resting on rough ground against a smooth wall. Why is it more likely to slip when they are near the top?

Recap

Big ideas

Moment = force × perpendicular distance, with a sense (clockwise or anticlockwise), in N m.
Equilibrium of a rigid body: resultant force zero and resultant moment zero about any point; take moments about an unknown force's point.
On the point of tilting, the reaction at the other support is zero.
The moment of a force uses the perpendicular distance to its line of action: $Fl sin θ$ Flsin θ for a force at angle $θ$ θ to a rod.
Ladder: smooth wall gives a horizontal reaction; rough ground gives $R$ R and friction; take moments about the foot, then resolve.
Hinge: unknown components $X$ X and $Y$ Y; take moments about the hinge first.

🏆

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Mission 1A 30 kg child and a 60 kg adult on a seesaw. Where does the adult sit?
Mission 2Why is a ladder more likely to slip when you climb near the top?

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Specification points and sources for this lesson are listed at the bottom of the page.