Mathematics | Statics of rigid bodies

Welcome to Statics of Rigid Bodies!

Hello future engineers and mathematicians! This chapter, Statics of Rigid Bodies, is a cornerstone of Mechanics 2. It takes everything you learned about forces and equilibrium in M1 and adds a powerful new concept: Moments (or turning effects).

Why is this important? Statics is the science of stability. When you build a bridge, design a stable structure, or even just lean a ladder against a wall, you rely on these principles to ensure nothing moves, breaks, or topples over. We will learn how to analyze structures that are held in place by balancing both linear forces and rotational forces.

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Section 1: Revisiting Equilibrium (The Foundation)

1.1 What is a Rigid Body?

In M1, you mostly dealt with particles (objects where size doesn't matter). In M2, we introduce the Rigid Body.

A rigid body is an object whose shape and size do not change when forces are applied. Think of a solid plank of wood or a metal beam – it won't bend, stretch, or compress under the forces we apply. This means forces acting at different points can cause the body to rotate, which leads us to our main topic: Moments!

1.2 Conditions for Equilibrium (Review)

If a body (particle or rigid) is in equilibrium, it is not accelerating. It is either stationary or moving at a constant velocity. Since we are dealing with statics, we focus on the body being stationary.

For any object to be in equilibrium under a system of coplanar forces (forces acting in the same 2D plane), the net forces must cancel out:

• Condition 1: The sum of forces in the horizontal direction must be zero.
  \[ \sum F_x = 0 \]
• Condition 2: The sum of forces in the vertical direction must be zero.
  \[ \sum F_y = 0 \]

Quick Review: These two equations were enough for particles. However, a rigid body could satisfy these two conditions and still spin! Imagine two opposing forces pushing a seesaw—the total force is zero, but the seesaw certainly turns.

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Section 2: Introducing the Moment of a Force

2.1 Defining the Moment

The Moment of a force (often called Torque in physics) is the measure of its turning effect about a specific point (the pivot or axis of rotation).

Analogy: Opening a Door
Imagine opening a heavy door. You instinctively push far away from the hinges. Why? If you push near the hinges (pivot), you need huge force. If you push far away (large distance), you need much less force.

The strength of the turning effect depends on two things:

1. The size of the Force (\(F\)).
2. The Perpendicular Distance (\(d\)) from the pivot to the line of action of the force.

2.2 The Moment Formula

The moment (\(M\)) is calculated as:

\[ M = F \times d \]

Where \(d\) must be the perpendicular distance.

The unit for the moment is the Newton-metre (Nm).

Important Rule: Perpendicular Distance

If the force is not perpendicular to the lever arm (the line connecting the pivot to the application point), you must use trigonometry to find the perpendicular distance, or resolve the force components.

Tip for Calculations: It is usually easier to extend the line of action of the force and drop a perpendicular line from the pivot to that extended line.

2.3 Direction of Moments (Sign Convention)

Moments cause rotation. We classify them as either:

• Clockwise (CW) moments (e.g., turning a screw tight).
• Anti-Clockwise (ACW) moments (e.g., unscrewing).

In problems, you must consistently assign one direction as positive (e.g., ACW = Positive) and the other as negative (e.g., CW = Negative).

Did You Know? If a force acts directly through the pivot point, its perpendicular distance \(d\) is zero. Therefore, the moment caused by that force about that pivot is \(M = F \times 0 = 0\). This trick is essential for solving problems!

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Section 3: Equilibrium of a Rigid Body

For a rigid body to be in complete statics (not moving and not turning), it must satisfy three conditions:

3.1 The Three Conditions for Rigid Body Equilibrium

1. Linear Equilibrium (Horizontal): \(\sum F_x = 0\)
2. Linear Equilibrium (Vertical): \(\sum F_y = 0\)
3. Rotational Equilibrium (The New Condition): The sum of all moments about any point must be zero. \[ \sum M = 0 \]

Condition 3 is known as the Principle of Moments.

3.2 The Power of Choosing the Right Pivot

You can take moments about any point (P) on the rigid body or even outside it, and if the body is in equilibrium, \(\sum M_P = 0\) will hold true.

The Smart Trick: When you have unknown forces (like a reaction force or friction), choose your pivot point (P) to lie directly on the line of action of one or more unknown forces.

Why? Because the moment created by those forces will instantly become zero (\(d=0\)), eliminating them from your moment equation. This leaves fewer unknowns to solve for, simplifying the calculation immensely!

3.3 Step-by-Step Problem Solving Strategy

Don't worry if this seems tricky at first. Follow these steps for solving typical rigid body problems (e.g., a plank supported by two supports):

1. Draw a Diagram: Sketch the body, showing all forces acting (weight, reactions, applied forces). Label distances clearly.
2. Resolve Forces: Apply Condition 1 (\(\sum F_x = 0\)) and Condition 2 (\(\sum F_y = 0\)). This gives you two equations, usually with multiple unknowns.
3. Choose a Pivot: Select a pivot point (P) strategically (usually where an unknown force acts).
4. Calculate Moments: Apply Condition 3 (\(\sum M_P = 0\)). Write down the moment equation, being careful with positive (ACW) and negative (CW) directions.
5. Solve: Use the moment equation to find one unknown force. Then substitute this value back into your force equations (from Step 2) to find the remaining unknowns.

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Section 4: Advanced Applications – Centres of Mass and Supports

4.1 Centre of Mass (Where the Weight Acts)

When dealing with a uniform rigid body (like a uniform rod), we assume the weight (\(W\)) of the body is concentrated and acts vertically downwards through a single point: the Centre of Mass (CM) or Centre of Gravity (CG).

• Uniform Rod/Beam: If the body is uniform (same density throughout and symmetrical), the CM is located exactly at the geometric centre (the midpoint).
• Composite/Non-Uniform Bodies: For objects made of different parts (or non-uniform density), the CM must be calculated using the principle of moments for masses/weights.

Calculating the Centre of Mass (For Discrete Masses)

If you have a collection of particles or sections of a rigid body (\(m_i\)) located at positions (\(x_i\)), the position of the CM (\(\bar{x}\)) is found using the formula:

\[ \bar{x} = \frac{\sum m_i x_i}{\sum m_i} \]

In essence, the total moment of mass about the origin is equal to the total mass times the distance to the CM.

4.2 Tilting and Limiting Equilibrium

Many problems involve a body resting on supports where the question asks: "What is the maximum weight that can be placed before the body tilts?"

The Tilting Condition: When a rigid body (like a plank) is supported by two supports (A and B) and is about to tilt about support B:

1. The entire plank is momentarily pivoting about B.
2. The reaction force at the other support (A) becomes zero. The plank is just about to lift off A.

To solve tilting problems: Set the reaction force at the support that is lifting off to zero, and then apply the moment equation around the remaining support (the pivot point).

4.3 Equilibrium involving Friction (The Ladder Problem)

Classic statics problems involve ladders leaning against walls. These often involve friction, which must be carefully applied.

When a body is on the point of slipping (limiting equilibrium), the friction force (\(F\)) reaches its maximum value: \[ F_{max} = \mu R \] Where \(\mu\) is the coefficient of friction and \(R\) is the normal reaction force.

Applying Forces to a Ladder:

Consider a ladder leaning against a rough floor and a smooth wall:

• Floor Reaction: The floor provides a vertical Normal Reaction (\(R_{floor}\)) and a horizontal Friction force (\(F_{floor}\)) acting towards the wall (preventing the bottom from slipping out).
• Wall Reaction: If the wall is smooth, it provides only a horizontal Normal Reaction (\(R_{wall}\)). If the wall is rough, it provides a Normal Reaction and a vertical Friction force acting upwards (preventing the ladder from slipping down).

Common Mistake to Avoid: Always ensure the friction force acts in the direction opposite to the way the object would slide. For a ladder on a rough floor, the ladder wants to slide outwards, so friction acts inwards.

Step-by-Step for a Ladder Problem:

1. Resolve Vertically: \(\sum F_y = 0\) (Often relates the vertical floor reaction to the weight). 2. Resolve Horizontally: \(\sum F_x = 0\) (Often relates the wall reaction to the floor friction). 3. Take Moments: Choose the pivot point strategically, usually the base of the ladder, as this eliminates two unknown forces (\(R_{floor}\) and \(F_{floor}\)). 4. Apply Limiting Friction: If on the point of slipping, substitute \(F = \mu R_{floor}\) or \(F = \mu R_{wall}\) into your horizontal/vertical equations.

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Section 5: Reactions at Hinges and Supports

When a rigid body is attached to a structure via a hinge, pin, or pivot, the reaction force exerted by the hinge is usually unknown in both magnitude and direction.

5.1 Dealing with Hinge Reactions

Since we don't know the direction of the hinge reaction (\(R\)), we treat it using components: \[ R = R_x \text{ (Horizontal Component)} + R_y \text{ (Vertical Component)} \]

You will therefore have two unknowns at the hinge: \(R_x\) and \(R_y\).

Solving for Hinge Reactions:

1. Apply \(\sum F_x = 0\) to find \(R_x\).
2. Apply \(\sum F_y = 0\) to find \(R_y\).
3. Use the moment equation (\(\sum M = 0\)) to find other forces acting on the body (since taking moments about the hinge eliminates \(R_x\) and \(R_y\)).
4. Once \(R_x\) and \(R_y\) are found, you can find the magnitude of the total reaction \(R\) using Pythagoras: \[ R = \sqrt{R_x^2 + R_y^2} \]

Remember: The direction (angle) of the total reaction force \(R\) can also be found using basic trigonometry (\(\tan \theta = R_y / R_x\)).

Keep practicing these three core equilibrium equations, and statics will become much clearer. Good luck!