Equilibrium of Rigid Bodies

Applying the two conditions for equilibrium to solve problems with uniform and non-uniform rods

SPEC 9.1 • MOMENTS

Equilibrium of a Rigid Body

Definition

When a rigid body is in equilibrium, the resultant force in any direction is 0 N and the resultant moment about any point is 0 Nm.

FORCE EQUILIBRIUM

Σ F = 0

MOMENT EQUILIBRIUM

Σ M = 0

Hint

If the resultant moment is zero, then the sum of clockwise moments equals the sum of anticlockwise moments.

Key Strategy

When you take moments about a given point, you can ignore the rotational effect of any forces acting at that point (since their perpendicular distance is zero).

Uniform and Non-Uniform Rods

Uniform RodNon-Uniform Rod

Mass distributed evenly along lengthMass distributed unevenly

Weight acts at midpointWeight acts at centre of mass (given in question)

Centre of mass at

\frac{L}{2}

from either endCentre of mass position specified as distance from one end

Uniform Rod on Two Supports (showing all forces):

Problem-Solving Strategy

Method

Draw a diagram showing all forces acting on the rod

Mark distances from a chosen reference point

Take moments about a point and set ΣM = 0

Resolve vertically using ΣF = 0 if needed

Problem-Solving Tip

Take moments about the point that makes the algebra as simple as possible. Taking moments about a support results in an equation with just one unknown.

Sign Convention

Choose a positive direction for moments (e.g., anticlockwise positive). Then clockwise moments are negative. Sum all moments and set equal to zero.

Common Error

Don't forget to include the weight of the rod! The weight acts at the centre of mass (midpoint for uniform rods).

01Worked Example

A uniform rod AB of length 4 m and weight 24 N rests horizontally on supports at A and C, where AC = 3 m. Calculate the reaction at each support.

1

Weight acts at midpoint: 2 m from A.

2

Resolving vertically: R_A + R_C − 24 = 0

3

Taking moments about A (↺ +ve):

R_C(3) − 24(2) = 0

3R_C − 48 = 0

R_C = 16 N

4

Substituting into ΣF = 0:

R_A + 16 − 24 = 0

R_A = 8 N

Answer: R_A = 8 N, R_C = 16 N

02Worked Example

A uniform beam AB of mass 50 kg and length 6 m rests horizontally on supports at C and D, where AC = 1 m and DB = 1 m. A child of mass 30 kg stands at point E. The reaction at D is three times the reaction at C. Find AE.

1

Let R_C = R and R_D = 3R. Let AE = x m. Beam weight at 3 m from A.

2

Resolving vertically:

R + 3R − 50g − 30g = 0

4R = 80g → R = 20g

3

Taking moments about A (↺ +ve):

R(1) + 3R(5) − 50g(3) − 30g(x) = 0

20g + 300g − 150g − 30gx = 0

170g − 30gx = 0

4

30gx = 170g

x = 170/30 = 5.67 m (to 3 s.f.)

Answer: AE = 5.67 m

03Non-Uniform Rod

A non-uniform rod AB is 5 m long and has weight 30 N. It rests on supports at C and D, where AC = 1 m and AD = 4 m. The reaction at C is twice the reaction at D. Find the distance of the centre of mass from A.

1

Let R_D = R, so R_C = 2R. Let centre of mass be x m from A.

2

Resolving vertically:

2R + R − 30 = 0

3R = 30 → R = 10 N

3

Taking moments about A (↺ +ve):

2R(1) + R(4) − 30(x) = 0

20 + 40 − 30x = 0

60 − 30x = 0

4

30x = 60

x = 2 m

Answer: Centre of mass is 2 m from A

04Non-Uniform with Particle

A non-uniform rod AB of length 6 m and weight 40 N has its centre of mass 2 m from A. It is suspended horizontally by vertical strings at A and B. A particle of weight 20 N is attached at C, where AC = 4 m. Find the tensions.

1

Rod weight (40 N) acts at centre of mass: 2 m from A. Particle (20 N) at 4 m from A.

2

Taking moments about A (↺ +ve):

T_B(6) − 40(2) − 20(4) = 0

6T_B − 80 − 80 = 0

6T_B = 160

T_B = 26.7 N (to 3 s.f.)

3

Resolving vertically:

T_A + T_B − 40 − 20 = 0

T_A + 26.7 − 60 = 0

T_A = 33.3 N (to 3 s.f.)

Answer: T_A = 33.3 N, T_B = 26.7 N

05Tension Ratio

A uniform beam AB of length 4 m and mass 40 kg has a 20 kg load at B. It is held horizontally by vertical wires at A and C. The tension at C is five times the tension at A. Find CB.

1

Let T_A = T, so T_C = 5T. Let AC = d.

2

Resolving vertically:

T + 5T − 40g − 20g = 0

6T = 60g → T = 10g

3

Taking moments about A (↺ +ve):

5T(d) − 40g(2) − 20g(4) = 0

50g(d) − 80g − 80g = 0

50gd = 160g

d = 3.2 m

4

CB = 4 − 3.2 = 0.8 m

Answer: CB = 0.8 m

06Finding Reactions

A uniform rod AB of length 6 m and weight 18 N rests on supports at C and D, where AC = 1.5 m and DB = 1.5 m. Find the reactions at C and D.

1

Weight acts at midpoint: 3 m from A, which is 1.5 m from C.

2

Taking moments about C (↺ +ve):

R_D(3) − 18(1.5) = 0

3R_D − 27 = 0

R_D = 9 N

3

Resolving vertically:

R_C + R_D − 18 = 0

R_C + 9 − 18 = 0

R_C = 9 N

Answer: R_C = 9 N, R_D = 9 N

(Equal because weight is at the midpoint of CD)

Tilting Problems

Key Concept

When a rod is "on the point of tilting" about a pivot point, the reaction at the other support becomes zero.

Problem-Solving Approach

When told the system is "on the point of tilting about C":

Set the reaction at the support that would lift off (i.e. the other support) to zero

Take moments about the tilting point (pivot)

All the weight must be balanced by moments about the pivot

Important

The rod tilts about the support that is closer to the heavier side. The support on the lighter side loses contact first (reaction → 0).

07Tilting Problem

A uniform plank AB of mass M kg and length 5 m rests on supports at C and D, where AC = 1 m and DB = 1 m. A child of mass 25 kg stands at B. The plank is on the point of tilting about D. Find the mass M of the plank.

1

"On the point of tilting about D" means R_C = 0.

2

Plank weight (Mg) acts at midpoint: 2.5 m from A, which is 1.5 m to the LEFT of D.

3

Child (25g) is at B, which is 1 m to the RIGHT of D.

4

Taking moments about D (↺ +ve):

Mg(1.5) − 25g(1) = 0

1.5Mg − 25g = 0

1.5M = 25

5

M = 25 ÷ 1.5 = 16.7 kg (to 3 s.f.)

Answer: Mass of plank = 16.7 kg

Key Results

Equilibrium Conditions

Σ F = 0

and

Σ M = 0

Sum all moments about a point and set equal to zero

SituationApproach

Find one reactionTake moments about the other support, set ΣM = 0

Given relationship between reactionsUse ΣF = 0 first, then take moments

Find centre of massLet centre of mass be x from A, set ΣM = 0

On point of tilting about PReaction at other support = 0

Check Your Answer

The support closer to the centre of mass should have the larger reaction force.

Practice

Question 1: A uniform rod AB of length 8 m and weight 32 N rests horizontally on supports at C and D. Given AC = 2 m and AD = 6 m, find the reactions at C and D.

1

Weight at midpoint: 4 m from A (2 m from C).

2

Taking moments about C (↺ +ve):

R_D(4) − 32(2) = 0

4R_D = 64 → R_D = 16 N

3

ΣF = 0: R_C + 16 − 32 = 0 → R_C = 16 N

Answer: R_C = 16 N, R_D = 16 N

Practice

Question 2: A non-uniform rod AB of length 5 m and weight 10 N rests horizontally on supports at A and B. The centre of mass is 3 m from A. Find the reactions.

1

Taking moments about A (↺ +ve):

R_B(5) − 10(3) = 0

5R_B = 30 → R_B = 6 N

2

ΣF = 0: R_A + 6 − 10 = 0 → R_A = 4 N

3

Check: Centre closer to B (2 m) than A (3 m), so R_B > R_A ✓

Answer: R_A = 4 N, R_B = 6 N

Practice

Question 3: A uniform plank AB of length 4 m and mass 20 kg rests on supports at C and D where AC = 0.5 m and DB = 0.5 m. A box of mass 30 kg is placed at B. The plank is on the point of tilting. About which point does it tilt, and what is the reaction at that point?

1

Box at B makes the B-end heavier, so it tilts about D (R_C = 0).

2

Plank weight at 2 m from A = 1.5 m left of D. Box at 0.5 m right of D.

3

Taking moments about D: 20g(1.5) − 30g(0.5) = 30g − 15g = 15g ≠ 0

4

Since anticlockwise > clockwise, it doesn't tilt. R_D takes all weight.

5

R_D = 20g + 30g = 50g = 490 N

Answer: Tilts about D, R_D = 490 N

Summary

Key Concepts

ConceptKey Point

EquilibriumΣF = 0 AND ΣM = 0 (both must hold)

Uniform RodWeight acts at midpoint (L/2 from each end)

Non-Uniform RodWeight acts at centre of mass (given in question)

Tilting PointReaction at support about to lift = 0

Exam Skills

Draw clear diagrams with all forces marked

Choose your pivot point to simplify calculations

State "taking moments about [point]" and set ΣM = 0

Check: larger reaction at support closer to centre of mass