Lesson Notes By Weeks and Term v4 - SHS 3

Lesson Video

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Performance objectives

• Define reaction forces and explain their significance.
• Calculate reactions provided by a knife edge support.
• Understand and apply equilibrium conditions.
• Solve real-life problems involving supports.

Lesson summary

This lesson focuses on the fundamental principles of static equilibrium, specifically how to calculate the upward forces, known as 'reactions', that supports provide to keep an object balanced. This concept is not just for the physics classroom; it is the science behind how bridges stand, how a seesaw works at the park, and even how two people carry a heavy load like a log or a "wawa" board. Understanding these reaction forces is the first step in designing safe and stable structures, from simple shelves to massive constructions like the Adomi Bridge. By mastering this, you will be able to predict the forces at play in balanced systems.

Lesson notes

A. Fundamental Ideas Reaction Force: According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. When a beam (like a metre rule) rests on a support (like a knife edge or a pivot), the beam exerts a downward force (its weight and any load on it) on the support. In response, the support exerts an equal and opposite *upward force* on the beam. This upward force is called the reaction force. We often denote it with the letter 'R'. Equilibrium: An object is in static equilibrium when it is not moving. This means it is not translating (moving from one place to another) and not rotating. For this to happen, two conditions must be met. B. The Two Conditions for Equilibrium For any object to be in a state of stable balance (static equilibrium), the following two conditions *must* be satisfied: First Condition (Translational Equilibrium): The vector sum of all forces acting on the object must be zero. For a horizontal beam, this simplifies to: The sum of all upward forces must equal the sum of all downward forces. Mathematically: `ΣF_upward = ΣF_downward` Second Condition (Rotational Equilibrium): The sum of all moments about *any* point (or pivot) must be zero. This simplifies to: The sum of clockwise moments about any point must equal the sum of anti-clockwise moments about the same point. Mathematically: `Σ(Clockwise Moments) = Σ(Anti-clockwise Moments)` Reminder: A moment is the turning effect of a force. It is calculated as: `Moment = Force × Perpendicular distance from the pivot`. C. Worked Examples

Important Assumption: Unless stated otherwise, we will assume the metre rule or beam is uniform. This means its weight acts at its geometric centre (for a metre rule, this is the 50 cm mark). We will take the acceleration due to gravity, *g* = 10 m/s².

Example 1: Beam on a Single Support

A uniform metre rule of mass 200 g is pivoted at its centre (the 50 cm mark). A mass of 300 g is placed at the 20 cm mark, and an unknown mass *M* is placed at the 80 cm mark to balance the rule horizontally. Calculate the reaction force, R, from the pivot. Step 1: Identify all downward forces. Weight of the rule (W_rule) = 0.2 kg × 10 m/s² = 2.0 N (acting at 50 cm) Weight of the 300 g mass (W₁) = 0.3 kg × 10 m/s² = 3.0 N (acting at 20 cm) Weight of mass M (W₂). First, we need to find M using the principle of moments. Anti-clockwise moment = 3.0 N × (50 cm - 20 cm) = 3.0 N × 0.3 m = 0.9 Nm Clockwise moment = W₂ × (80 cm - 50 cm) = W₂ × 0.3 m For balance, 0.9 Nm = W₂ × 0.3 m => W₂ = 0.9 / 0.3 = 3.0 N. Step 2: Apply the First Condition for Equilibrium. The only upward force is the reaction R from the pivot. The downward forces are W_rule, W₁, and W₂. `ΣF_upward = ΣF_downward` `R = W_rule + W₁ + W₂` `R = 2.0 N + 3.0 N + 3.0 N` `R = 8.0 N` Conclusion: The reaction provided by the single support is 8.0 N.

Evaluation guide

• Calculate the reactions provided bysupports.
• Enquiry learning
• From the experiment, let learners brainstorm and come out with the reaction provided by knife
• conceptual understanding

Reference guide

• for the pillars.
• Traditional Ghanaian Activities (Community/Culture): When two people carry a heavy object like a log of wood for building, a large basin of water, or even a coffin during a funeral procession, their shoulders or heads act as the supports. The person closer to the object's centre of gravity will feel a greater reaction force (i.e., they carry more weight). This knowledge can be used to distribute the load more fairly by adjusting positions.
• Vehicle Loading (Economy/Transportation): The axles of a car or a "trotro" act as supports for the vehicle's body. When loading goods onto a cargo truck or onto the roof rack of a bus, understanding how the load is distributed affects the reaction forces on the front and rear tyres. Overloading the back of a truck can cause the front wheels to have too little reaction force, making steering unsafe and unstable.
• Differentiation, Remediation and Extension
• Remediation (For Struggling Learners)
• Hands-on First: Before calculations, give learners a metre rule, a pivot (e.g., a triangular block or an eraser), and some standard masses (or coins). Let them physically find the balance point and feel the forces.