Introduction: The World of Rigid Equilibrium
Welcome to the chapter on the Equilibrium of a Rigid Body! This is where mechanics gets serious—and incredibly practical. You’re moving beyond particles (tiny objects where size doesn't matter) and stepping into the world of rigid bodies (real objects where size, shape, and how forces are applied *do* matter).
Why is this important? Everything stable around you—from the chair you’re sitting on to tall buildings and bridges—depends on these principles. You will learn the mathematical rules that determine if an object stays put or starts to move or spin. Don't worry if this seems tricky at first; we will break down the concepts of forces and turning effects (moments) step-by-step!
1. Defining Equilibrium and Rigid Bodies
1.1 What is a Rigid Body?
In mechanics, a rigid body is an object whose shape and size do not change when forces are applied to it. In this chapter, we focus on situations where the forces acting on the body are coplanar (all lie in the same plane).
1.2 What does "Equilibrium" mean?
A rigid body is in equilibrium if it meets two fundamental criteria simultaneously:
- It is not moving (no translation).
- It is not rotating (no angular acceleration).
To satisfy these two criteria, we need to balance both the forces and the turning effects.
Quick Review Box: The Two Rules of Equilibrium
1. All forces must balance (prevents sliding/moving).
2. All turning effects (moments) must balance (prevents toppling/spinning).
2. The Turning Effect: Moments of a Force
When you push a particle, it moves. When you push a rigid body, it might move, or it might rotate. The tendency of a force to cause rotation is called the moment (or torque) of the force.
2.1 Calculating the Moment
The moment of a force is calculated relative to a specific point, known as the pivot or the reference point.
The formula for the magnitude of the moment \(M\) of a force \(F\) about a point \(P\) is:
$$M = F \times d$$
Where:
- \(F\) is the magnitude of the force.
- \(d\) is the perpendicular distance from the pivot point \(P\) to the line of action of the force.
Analogy: Opening a Door
If you push a door close to the hinge (the pivot), you need a huge force to open it (small \(d\), large \(F\)). If you push far from the hinge (where the handle is), you need a small force (large \(d\), small \(F\)). The moment (the turning effect) required is the same!
2.2 Direction of Moments
Moments are classified by the direction of rotation they cause:
- Clockwise (CW) moments.
- Counter-Clockwise (CCW) moments (also called anti-clockwise, ACW).
When summing moments for equilibrium, we treat one direction as positive and the other as negative (e.g., CW = Positive, CCW = Negative).
Important Tip: Finding Perpendicular Distance
Sometimes the force \(F\) is not perpendicular to the body itself. You have two ways to calculate the moment:
- Find the actual perpendicular distance \(d\) from the pivot to the line of action of \(F\).
- Resolve the force \(F\) into components: one parallel to the body and one perpendicular to the line connecting the pivot to the point of application. Only the perpendicular component causes a moment.
If the force \(F\) acts at a distance \(r\) from the pivot, and the angle between \(F\) and the line connecting the pivot to \(F\) is \(\theta\), then the perpendicular distance \(d = r \sin \theta\).
$$M = F (r \sin \theta)$$
Key Takeaway for Moments
Moments measure the rotational effect. The calculation is always Force \(\times\) Perpendicular Distance. Remember to calculate moments about a chosen pivot and assign positive/negative directions (like CW/CCW).
3. Centre of Mass (CoM) and Gravity
When dealing with a rigid body, its weight (the gravitational force) acts through a single theoretical point called the Centre of Mass (CoM).
3.1 Weight and the CoM
For any rigid body, the effect of gravity is equivalent to a single resultant force (the total weight, \(W\)) acting vertically downwards through the CoM.
3.2 Finding the Centre of Mass (CoM) of Uniform Bodies
If a body is uniform (density is constant), its CoM can often be found by symmetry.
- Uniform Rod: CoM is exactly at the midpoint.
- Uniform Rectangular Lamina: CoM is at the intersection of the diagonals (the exact centre).
- Uniform Circular Disc: CoM is at the geometric centre.
The syllabus requires knowledge of the CoM locations for specific basic shapes, which are provided in your MF19 formula sheet:
- Triangular Lamina: \(\frac{1}{3}\) along the median from the vertex.
- Solid Cone/Pyramid (height \(h\)): \(\frac{1}{4} h\) from the base.
- Solid Hemisphere (radius \(r\)): \(\frac{3}{8} r\) from the centre.
- Hemispherical Shell (radius \(r\)): \(\frac{1}{2} r\) from the centre.
3.3 CoM of a Composite Body
A composite body is an object made up of several simpler shapes joined together (e.g., an L-shaped lamina, or a cylinder joined to a hemisphere).
To find the CoM of a composite body, we treat it as an equivalent system of particles, where the mass of each simple part is concentrated at its own CoM.
Step-by-Step: CoM Calculation (Composite Bodies)
- Divide the Body: Separate the composite body into simple uniform shapes (A, B, C...).
- Find Mass and CoM for Each Part:
- Determine the mass (\(m_i\)) of each part. For uniform laminas, mass is proportional to area. For uniform solids, mass is proportional to volume.
- Determine the coordinates \((x_i, y_i)\) of the CoM for each part (using symmetry or MF19 formulae).
- Choose an Origin: Set up a coordinate system \((x, y)\). This is crucial for consistency.
- Calculate Total Moment of Mass: Use the equivalent system of particles method:
The coordinates \((\bar{x}, \bar{y})\) of the overall CoM are given by:
$$\bar{x} = \frac{\sum m_i x_i}{\sum m_i}$$ $$\bar{y} = \frac{\sum m_i y_i}{\sum m_i}$$
Where \(\sum m_i\) is the total mass of the body.
Did you know? Finding the CoM is often the first step in solving a complex equilibrium problem, as you must know exactly where the weight acts.
Key Takeaway for CoM
The weight of a uniform body acts through its CoM. For composite bodies, treat the individual parts as particles located at their respective CoMs and use the weighted average formula \(\frac{\sum m_i x_i}{\sum m_i}\).
4. The Conditions for Full Equilibrium
A rigid body subject to coplanar forces is in equilibrium if and only if:
4.1 Condition 1: Translational Equilibrium (No Linear Motion)
The vector sum of the forces is zero. This means the forces must be balanced horizontally and vertically.
- Horizontal forces balance: \(\sum F_x = 0\)
- Vertical forces balance: \(\sum F_y = 0\)
4.2 Condition 2: Rotational Equilibrium (No Rotation)
The sum of the moments of the forces about any point is zero.
This is extremely powerful. We can choose *any* point as our pivot, but choosing a point where many unknown forces act will simplify the calculation significantly (because the moment generated by a force passing through the pivot is zero).
- Sum of Clockwise Moments = Sum of Counter-Clockwise Moments (\(\sum M = 0\))
Solving Problems Involving Equilibrium
These problems are typically solved by applying the three equations derived from the two conditions:
- Resolve forces horizontally: \(\sum F_x = 0\)
- Resolve forces vertically: \(\sum F_y = 0\)
- Take moments about a strategic point \(P\): \(\sum M_P = 0\)
By solving these three simultaneous equations, you can find up to three unknown forces or distances.
5. Limiting Equilibrium: Toppling and Sliding
Often, problems ask about the conditions under which a body is just about to lose equilibrium. This state is called limiting equilibrium, and it usually involves friction, sliding, or toppling.
5.1 Sliding (Translational Failure)
Sliding occurs when the horizontal forces are strong enough to overcome the maximum static friction force.
A body resting on a rough surface is in limiting equilibrium with respect to sliding if:
$$F = \mu R$$
Where:
- \(F\) is the friction force acting parallel to the surface.
- \(R\) is the normal contact force (perpendicular to the surface).
- \(\mu\) is the coefficient of friction.
If \(F < \mu R\), the body remains stationary (static equilibrium).
If \(F = \mu R\), the body is on the point of sliding (limiting equilibrium).
5.2 Toppling (Rotational Failure)
Toppling occurs when the total moment trying to rotate the body exceeds the restoring moment provided by gravity.
Consider a block resting on a surface and being pushed sideways. It will tend to rotate about the edge closest to the push (let's call this edge \(A\)).
Condition for Toppling (Critical Equilibrium)
When the body is on the point of toppling about an edge \(A\), the normal reaction force (\(R\)) shifts to act precisely through that edge \(A\).
Why? The normal force \(R\) is generated by the surface supporting the object. As the object tilts, the weight is carried entirely by the pivot edge. If \(R\) moves outside the base of the object, the object will definitely topple.
To solve a toppling problem:
- Identify the pivot edge \(A\) (the edge it will rotate about).
- Set up the diagram assuming the body is on the point of toppling. Crucially, the normal reaction \(R\) (and friction \(F\), if calculated about \(A\)) must act at \(A\).
- Take moments about the pivot edge \(A\). At the point of critical equilibrium, the moments causing CW rotation must balance the moments causing CCW rotation, OR, if only one external force causes rotation, the moment of the external force must equal the moment of the weight (restoring force).
5.3 The Choice: Sliding or Toppling?
In problems where a body is subject to increasing force, it will fail by whichever mechanism requires the least force.
- Calculate the external force \(P_{slide}\) needed to make \(F = \mu R\).
- Calculate the external force \(P_{topple}\) needed to make the normal reaction act on the edge.
- The body fails by the mechanism corresponding to the smaller force (\(\min(P_{slide}, P_{topple})\)).
Key Takeaway for Limiting Equilibrium
Sliding happens when friction fails (\(F = \mu R\)). Toppling happens when the normal reaction force effectively moves outside the base of the object, meaning when the body is on the point of toppling, the normal reaction acts through the rotation edge.
Chapter Summary: Three Equations to Rule Them All
To ensure a single rigid body under coplanar forces is in full equilibrium, you must satisfy three conditions, giving you three equations to solve:
- Sum of Horizontal Forces = 0: \(\sum F_x = 0\)
- Sum of Vertical Forces = 0: \(\sum F_y = 0\)
- Sum of Moments about Any Point P = 0: \(\sum M_P = 0\)
Remember that the weight of the body must always be applied at the Centre of Mass.