Edexcel IAL - Mechanics 2- 5.1 Moments of Forces- Study notes - New syllabus
Edexcel IAL – Mechanics 2- 5.1 Moments of Forces -Study notes- New syllabus
Edexcel IAL – Mechanics 2- 5.1 Moments of Forces -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 5.1 Moments of Forces
Moment of a Force
The moment of a force measures the turning effect produced by a force about a fixed point or axis. It plays a key role in problems involving equilibrium and rotation.
Moments are sometimes referred to as torques, but in this syllabus the term moment is used.
Definition of Moment
The moment of a force about a point is defined as
Force \( \times \) perpendicular distance from the point to the line of action of the force
If a force of magnitude \( F \) acts at a perpendicular distance \( d \) from a point \( O \), then the moment about \( O \) is
\( M = Fd \)
The SI unit of moment is the newton metre (N m).
Direction of a Moment
Moments can act in two directions:
- Clockwise moment
- Anticlockwise moment
By convention:
- Anticlockwise moments are taken as positive
- Clockwise moments are taken as negative
Principle of Moments
For a body to be in equilibrium under coplanar forces,
Sum of clockwise moments about any point \( = \) sum of anticlockwise moments about that point
Equivalently, the algebraic sum of moments about any point is zero.
Choosing a Pivot
Moments may be taken about any convenient point, often chosen to:
- Eliminate unknown forces
- Simplify calculations
Example :
A force of \( 12 \) N acts at a perpendicular distance of \( 0.5 \) m from a fixed point. Find the moment of the force about the point.
▶️ Answer/Explanation
Using the moment formula:
\( M = Fd = 12 \times 0.5 \)
\( M = 6 \)
Conclusion: The moment of the force is \( 6 \,\text{N m} \).
Example :
A uniform rod of length \( 4 \) m and weight \( 20 \) N is hinged at one end and held horizontal by a vertical force \( F \) applied at the other end. Find the value of \( F \).
▶️ Answer/Explanation
The weight of the rod acts at its centre, \( 2 \) m from the hinge.
Taking moments about the hinge:
Anticlockwise moment \( = F \times 4 \)
Clockwise moment \( = 20 \times 2 \)
Equating moments:
\( 4F = 40 \)
\( F = 10 \)
Conclusion: The force required is \( 10 \) N.
Example :
A uniform rectangular lamina rests in equilibrium about a pivot at one corner. Its weight \( W \) acts at its centre. State the condition required for rotational equilibrium.
▶️ Answer/Explanation
The moment of the weight about the pivot depends on the perpendicular distance from the pivot to the line of action of the weight.
For equilibrium, the total clockwise moments must equal the total anticlockwise moments.
Conclusion: The lamina is in equilibrium when the algebraic sum of moments about the pivot is zero.