Lesson Notes By Weeks and Term v4 - SHS 3
BASIC PHYSICS
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BASIC PHYSICS
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Subject: Physics
Class: SHS 3
Term: 1st Term
Week: 2
Grade code: 3.1.1.LI.4
Strand code: 1
Sub-strand code: 1
Content standard code: 3.1.1.CS.1
Indicator code: 3.1.1.LI.4
Theme: MECHANICS AND MATTER
Subtheme: BASIC PHYSICS
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This lesson explores the fundamental principles of static equilibrium, focusing on how to calculate the upward forces, or "reactions," that supports provide to keep an object balanced. This is a crucial concept in engineering and everyday life. Imagine a wooden bridge over a gutter in your community, a scaffolding on a building site in Accra, or even two people carrying a heavy load on a plank. Understanding reaction forces helps us predict if these structures are safe and will not collapse. By mastering this, you will be able to analyse and design simple stable structures.
A. What is a 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 plank or a metre rule) rests on a support (like a knife-edge or a pillar), the beam exerts a downward force (its weight plus any load on it) on the support. This is the action. In response, the support exerts an equal and opposite upward force on the beam. This upward force is called the reaction force. It is what prevents the beam from falling through the support. We typically denote reaction forces with letters like R or F. B. Conditions for Static Equilibrium For an object like a beam to be completely stationary (not moving up/down and not rotating), it must be in static equilibrium. Two conditions must be met: First Condition (Translational Equilibrium): The vector sum of all forces acting on the object must be zero. In simple terms for our vertical problems: 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 (pivot) must be zero. In simple terms: The sum of all clockwise moments must equal the sum of all anticlockwise moments about the same pivot. Mathematically: Σ(Clockwise Moments) = Σ(Anticlockwise Moments) Recall: Moment = Force × Perpendicular distance from the pivot (Unit: Newton-metre, Nm) C. Solving Problems with Two Supports This is the core skill for this indicator. When a beam is on two supports, we need to find two unknown reaction forces (e.g., R₁ and R₂). We will use both conditions of equilibrium to solve this.
The Strategy: Draw a clear diagram: Show the beam, the supports, all the loads (weights), and the reaction forces (as upward arrows). Label all distances clearly. Apply the First Condition: Write an equation: R₁ + R₂ = Total Downward Forces. This will give you one equation with two unknowns. You can't solve it yet. Apply the Second Condition (The KEY STEP): Choose one of the supports as your pivot. This is a clever trick! Why? Because the reaction force at that pivot has a distance of zero from the pivot, its moment is zero (R × 0 = 0), and it disappears from the moment equation. This leaves you with an equation with only *one* unknown reaction force, which you can easily solve. Calculate the first reaction: Solve the moment equation from Step 3 to find the value of one of the reaction forces. Calculate the second reaction: Substitute the value you just found back into the force equation from Step 2 to find the other reaction force.
Worked Examples *(Assume acceleration due to gravity, g = 10 m/s²)*
Example 1: Symmetrical Loading A uniform metre rule of mass 0.5 kg is supported by two knife-edges placed at the 10 cm and 90 cm marks. A 2 kg mass is hung at the centre (50 cm mark). Calculate the reactions at the supports.