Mechanics
Moments and Equilibrium of Rigid Bodies
Year 12 · Year 13
- ✓Define the moment of a force about a point.
- ✓Calculate the moment of a force, including cases where the force is not perpendicular to the distance from the pivot.
- ✓State and apply the Principle of Moments for a body in rotational equilibrium.
- ✓Apply the conditions for a rigid body to be in complete equilibrium (translational and rotational).
- ✓Solve problems involving forces and moments acting on rigid bodies in equilibrium.
Key concepts
The moment of a force about a point (often called the pivot) is a measure of its turning effect. It depends on both the magnitude of the force and its perpendicular distance from the pivot to the line of action of the force. Moments are typically described as either clockwise or anticlockwise.
For a rigid body to be in rotational equilibrium (i.e., not rotating or rotating at a constant angular velocity), the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.
A rigid body is in complete equilibrium if it is not accelerating translationally (moving in a straight line) and not accelerating rotationally (spinning). This requires two conditions to be met simultaneously: 1. Translational Equilibrium: The vector sum of all external forces acting on the body must be zero (net force in any direction is zero). 2. Rotational Equilibrium: The sum of the moments of all external forces about any point must be zero (Principle of Moments).
The weight of an object is a force acting vertically downwards. For calculations involving moments, this weight can be considered to act through a single point called the centre of mass (or centre of gravity). For a uniform rod, the centre of mass is at its geometric centre.
Key facts to remember
- 1The moment of a force is its turning effect about a pivot, calculated as Force × Perpendicular distance (M = Fd).
- 2The unit of moment is the Newton-metre (Nm).
- 3For a body to be in rotational equilibrium, the sum of clockwise moments about any point must equal the sum of anticlockwise moments about the same point (Principle of Moments).
- 4For a body to be in complete equilibrium, two conditions must be met: the net force in any direction is zero (translational equilibrium), AND the net moment about any point is zero (rotational equilibrium).
- 5The weight of a uniform object acts through its geometric centre (centre of mass).
- 6When choosing a pivot for moment calculations, selecting a point where an unknown force acts can simplify the problem by eliminating that force from the moment equation.
Worked examples
Example 1
A uniform rod AB of length 4 m and mass 5 kg is supported horizontally by two vertical strings, one at A and one at B. A particle of mass 10 kg is placed on the rod at a distance of 1 m from A. Find the tensions in the strings. (Use g = 9.8 m/s²)
Answer
The tension in the string at A is 98 N, and the tension in the string at B is 49 N.
Choosing a pivot at one of the unknown forces simplifies the calculation by removing that force from the moment equation.
Example 2
A uniform rod AB of length 3 m and mass 8 kg rests with end A on rough horizontal ground and end B against a smooth vertical wall. The rod is inclined at an angle of 60° to the horizontal. Find the magnitude of the normal reaction force from the wall at B and the friction force at A. (Use g = 9.8 m/s²)
Answer
The normal reaction force from the wall at B is approximately 22.6 N. The friction force at A is approximately 22.6 N.
Remember to use the perpendicular distance from the pivot to the line of action of the force. This often involves trigonometry.
Example 3
A uniform plank AB of length 6 m and mass 20 kg rests horizontally on two supports C and D. Support C is at A, and support D is 1 m from B. A man of mass 80 kg stands at a point P on the plank. Find the range of possible positions for P such that the plank remains in equilibrium and does not tip. (Let x be the distance of P from A. Use g = 9.8 m/s²)
Answer
The man can stand at any point P such that 0 ≤ x ≤ 5.5 m from A.
Always consider both tipping points. Sometimes one of the conditions will show that tipping is not possible in that direction within the physical constraints of the problem.
Common mistakes
- ✗Using the distance along the rod/lever arm instead of the perpendicular distance from the pivot to the line of action of the force.
- ✗Incorrectly identifying forces as causing clockwise or anticlockwise moments.
- ✗Forgetting to include the weight of the object itself in moment calculations, especially for uniform rods/beams.
- ✗Only applying the Principle of Moments and forgetting to consider translational equilibrium (sum of forces = 0), or vice-versa, for complete equilibrium problems.
- ✗Incorrectly resolving forces or distances when the force is not perpendicular to the lever arm.
Exam tips
- ★Always draw a clear, large diagram, labelling all forces (including weight and reactions) and relevant distances and angles.
- ★Choose your pivot strategically. Often, choosing a pivot at an unknown reaction force will eliminate it from the moment equation, simplifying calculations.
- ★Systematically list all clockwise moments and all anticlockwise moments about your chosen pivot to avoid errors.
- ★For complete equilibrium, remember to apply both conditions: ΣF = 0 (horizontally and vertically) and ΣM = 0.
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