Equilibrium of rigid bodies

Further Mathematics· Unit 3 Further Mechanics, Topic 4· 25 min read

1. Conditions for Coplanar Equilibrium★★☆☆☆⏱ 8 min

📘 Definition

Equilibrium of a rigid body

A rigid body is in static equilibrium when two conditions are satisfied: 1) The vector sum of all external forces is zero, 2) The sum of moments of all external forces about any point is zero.

Example:

A stationary ladder leaning against a wall is in static equilibrium.

For coplanar forces, we can split force equilibrium into horizontal and vertical components, giving three independent scalar equations: $\sum F_{x} = 0$ , $\sum F_{y} = 0$ , $\sum M_{P} = 0$ for any point $P$ .

📐 Worked Example

A uniform rod of length 5 m has weight 10 N acting at its centre. It is supported at $x = 0$ (reaction 10 N upwards) and $x = 5$ m (reaction $R$ N upwards), with a weight $W$ hanging at $x = 2$ m. Find $W$ for equilibrium.

1
Apply force equilibrium in the vertical direction first:
2
$ext{Sum of upward forces} = ext{Sum of downward forces} \\ 10 + R = 10 + W \\ R = W$
3
Take moments about $x = 0$ to eliminate the 10 N reaction at the pivot:
4
$\ ext{Sum } M_0 = 0 \\ -W(2) - 10(2.5) + R(5) = 0$
5
Substitute $R = W$ and solve for $W$ :
6
$-2W -25 + 5W = 0 \\ 3W = 25 \\ W = \frac{25}{3} \approx 8.33 \text{ N}$

2. Equilibrium of Suspended Rigid Bodies★★★☆☆⏱ 7 min

📘 Definition

Suspended rigid body equilibrium

When a rigid body is freely suspended from a fixed pivot and is in equilibrium, its centre of mass lies directly vertically below the pivot point.

This rule lets us calculate the angle a suspended body makes with the vertical or horizontal, using trigonometry on the line connecting the pivot to the centre of mass.

📐 Worked Example

A uniform rectangular lamina with sides 4 cm (longer) and 3 cm (shorter) is suspended from one top corner. Find the angle between the longer side and the vertical.

1
The centre of mass of a uniform rectangle is at its geometric centre. From the pivot corner, this is 2 cm along the longer side, and 1.5 cm along the shorter side.
2
Let $θ$ be the angle between the longer side and the vertical. The line connecting the pivot to the centre of mass is vertical, so:
3
$\tan\theta = \frac{\text{horizontal offset from pivot}}{\text{vertical offset from pivot}} = \frac{1.5}{2} = 0.75$
4
Calculate the angle:
5
$\theta = \arctan(0.75) \approx 36.9^\circ$

3. Equilibrium of a Leaning Ladder★★★★☆⏱ 10 min

Leaning ladders are one of the most common exam problems for this topic. For any contact surface, add a normal reaction perpendicular to the surface, and friction parallel to the surface if the surface is rough.

A smooth wall has no friction, only a normal horizontal reaction
Rough ground has both a vertical normal reaction and horizontal friction to stop slipping
The weight of a uniform ladder acts at its midpoint

📐 Worked Example

A uniform ladder of length 4 m and mass 10 kg leans against a smooth vertical wall, standing on rough horizontal ground. The ladder makes 60° with the horizontal. Find the friction force at the ground.

1
Label all forces: $R$ (normal reaction at wall, horizontal right), $N$ (normal reaction at ground, vertical up), $F$ (friction at ground, horizontal left), weight $m g = 10 g$ (vertical down at midpoint).
2
Apply force equilibrium:
3
$\sum F_x = 0 \implies R - F = 0 \implies F = R \\ \sum F_y = 0 \implies N - 10g = 0 \implies N = 10g$
4
Take moments about the base of the ladder to eliminate $F$ and $N$ :
5
$\sum M_{\text{base}} = 0 \implies R(4\sin 60^\circ) - 10g(2\cos 60^\circ) = 0$
6
Substitute $sin 6 0^{\circ} = \frac{3}{2}$ and $cos 6 0^{\circ} = 0.5$ :
7
$R(2\sqrt{3}) = 10g(0.5) \\ F = R = \frac{5g}{\sqrt{3}} \approx 28.3 \text{ N}$

Exam tip:

Always label all forces before writing equations. It is very common to forget friction at a rough contact, which breaks all equilibrium calculations.

4. Beams with Multiple Supports★★★☆☆⏱ 7 min

For a beam supported at two or more points, we use the same three equilibrium conditions to solve for unknown reaction forces. The same method applies whether the beam is uniform or non-uniform.

📐 Worked Example

A uniform beam of length 6 m and mass 20 kg rests on two supports: one at the left end, one 1 m from the right end. Find the reaction force at each support.

1
Let $R_{1}$ = reaction at left end, $R_{2}$ = reaction at the second support (5 m from left). Weight $20 g$ acts at the midpoint, 3 m from left.
2
Apply vertical force equilibrium:
3
$R_1 + R_2 = 20g$
4
Take moments about the left end to eliminate $R_{1}$ :
5
$\sum M_{\text{left}} = 0 \implies 5R_2 - 3(20g) = 0 \\ 5R_2 = 60g \\ R_2 = 12g = 117.6 \text{ N}$
6
Substitute back to find $R_{1}$ :
7
$R_1 = 20g - 12g = 8g = 78.4 \text{ N}$

5. Common Pitfalls

Wrong move:

Forgetting friction at one rough contact surface for a ladder with both rough wall and ground.

Why:

Missing any force means the equilibrium equations will not balance, leading to incorrect solutions.

Correct move:

Label every contact point: add normal reaction perpendicular to the surface, and add friction parallel to the surface for any rough contact.

Wrong move:

Using the full length along the rod as the distance for moment calculation, instead of the perpendicular distance.

Why:

Moment is defined as force multiplied by perpendicular distance from the pivot, not the distance along the body.

Correct move:

Always calculate the perpendicular component of distance, or resolve the force into components perpendicular to the rod to find the moment.

Wrong move:

Assuming the centre of mass of a non-uniform rod is at its midpoint.

Why:

Exam questions regularly use non-uniform bodies to test this, and using the midpoint gives an incorrect moment for the weight.

Correct move:

Always check if the body is stated as uniform. If it is non-uniform, the centre of mass position will be given in the question.

Wrong move:

Stopping after writing two equilibrium equations when three unknowns are present.

Why:

We have three independent equations for coplanar equilibrium, so we need all three to solve for three unknowns.

Correct move:

Write all three equations ( $\sum F_{x}$ , $\sum F_{y}$ , $\sum M$ ) before attempting to solve the system of equations.

6. Quick Reference Cheatsheet

Condition	Result/Equation
Translational Equilibrium	$\sum F_{x} = 0, \sum F_{y} = 0$
Rotational Equilibrium	$\sum M_{P} = 0$ for any point $P$
Suspended Body Equilibrium	Centre of mass lies directly below pivot
Normal Reaction	Always perpendicular to contact surface
Friction Force	Always parallel to contact surface, opposes slip

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 31
Ladder equilibrium problem
2023 · 32
Suspended lamina equilibrium
2024 · 31
Supported beam equilibrium

Going deeper

syllabusOfficial CIE 9231 SyllabusFull topic syllabus reference

What's Next

Equilibrium of rigid bodies is a core foundation for all further work in rigid body mechanics. The skills of force resolution and moment calculation you practice here are directly used in topics like rotational dynamics of rigid bodies, which is a major component of the Further Mechanics paper. This topic is consistently heavily weighted in CIE 9231 exams, so mastering all common problem types (ladders, suspended laminas, beams) is critical for a high score. Building a strong understanding of equilibrium will make more advanced dynamic topics much more approachable.