Edexcel IAL - Mechanics 3- 5.2 Equilibrium of Rigid Bodies- Study notes - New syllabus
Edexcel IAL – Mechanics 3- 5.2 Equilibrium of Rigid Bodies -Study notes- New syllabus
Edexcel IAL – Mechanics 3- 5.2 Equilibrium of Rigid Bodies -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 5.2 Equilibrium of Rigid Bodies
Simple Cases of Equilibrium of Rigid Bodies
A rigid body is said to be in equilibrium when it is both:
- In translational equilibrium
- In rotational equilibrium
This means:
Resultant force \( = 0 \)
Resultant moment about any point \( = 0 \)
Only simple, uncomplicated systems are required by the syllabus.
Conditions for Equilibrium
For a rigid body under coplanar forces:
- Sum of horizontal forces \( = 0 \)
- Sum of vertical forces \( = 0 \)
- Sum of moments about any point \( = 0 \)
(i) Suspension of a Body from a Fixed Point
When a rigid body is suspended freely from a fixed point and allowed to come to rest:
- The only forces acting are the weight and the tension at the point of suspension
- The body rotates until its centre of mass lies vertically below the point of suspension
At equilibrium:
- The line of action of the weight passes through the point of suspension
- The moment of the weight about the suspension point is zero
This principle is often used to locate the centre of mass experimentally.
(ii) A Rigid Body on a Horizontal Plane
When a rigid body rests on a horizontal plane, the forces acting typically include:
- Weight \( mg \) acting vertically downward through the centre of mass
- Normal reaction \( R \) acting vertically upward from the plane
- Friction, if the surface is rough
For equilibrium on a horizontal plane:
- The normal reaction balances the weight
- The line of action of the reaction must pass through the centre of mass to prevent rotation
(iii) A Rigid Body on an Inclined Plane
When a rigid body rests on a rough or smooth inclined plane, the forces acting include:
- Weight \( mg \), acting vertically downward
- Normal reaction \( R \), acting perpendicular to the plane
- Friction \( F \), acting along the plane (if rough)
The weight is usually resolved into components:
- Parallel to the plane: \( mg\sin\theta \)
- Perpendicular to the plane: \( mg\cos\theta \)
Moments are often taken about the point of contact with the plane to simplify calculations.
Example :
A uniform rod of length \( 2 \) m is suspended from one end. Describe the position of the rod when it is in equilibrium.
▶️ Answer/Explanation
The centre of mass of a uniform rod is at its midpoint.
When suspended, the rod rotates until the centre of mass lies vertically below the point of suspension.
Conclusion: The rod hangs vertically downward in equilibrium.
Example :
A uniform rectangular lamina rests on a horizontal table. Explain why it is in equilibrium.
▶️ Answer/Explanation
The weight of the lamina acts vertically downward through its centre of mass.
The normal reaction from the table acts vertically upward.
The forces are equal and opposite, so the resultant force is zero.
Since the line of action of the reaction passes through the centre of mass, there is no turning effect.
Conclusion: The lamina is in equilibrium.
Example :
A uniform rod of length \( 3 \) m rests in equilibrium on a rough inclined plane, with its lower end on the plane and its upper end against a smooth vertical wall. State the forces acting on the rod.
▶️ Answer/Explanation
Forces acting on the rod are:
Weight acting vertically downward through the centre of mass
Normal reaction from the inclined plane at the lower end
Friction at the lower end acting along the plane
Normal reaction from the smooth wall acting horizontally at the upper end
Conclusion: These forces balance to keep the rod in equilibrium.
Example :
A uniform rod rests in equilibrium on a rough inclined plane with one end against a smooth vertical wall. State how equilibrium is maintained.
▶️ Answer/Explanation
The forces acting are:
Weight acting through the centre of mass
Normal reaction and friction at the plane
Normal reaction from the smooth wall
These forces balance both translationally and rotationally.
Conclusion: The rod remains in equilibrium due to balanced forces and moments.