Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Resolve a vector into its horizontal and vertical components.
• Determine the resultant of two or more vectors.
• Apply vector principles to solve problems.
• Represent vector addition and subtraction.
• Distinguish between scalar and vector quantities.

Lesson summary

This week, we delve into the fascinating world of vectors in two dimensions within the realm of mechanics. Understanding vectors is absolutely crucial for describing motion, forces, and various other physical quantities that have both magnitude (size) and direction. Unlike scalars which are fully described by a magnitude alone (e.g., temperature, mass), vectors require both magnitude and direction for a complete description. This topic builds upon your understanding of scalars and vector basics from Grade

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0. Why does this matter to you, as a South African learner? Imagine planning the route for a taxi traveling between townships.

Lesson notes

2.1 Scalars vs. Vectors A scalar quantity is completely described by its magnitude (size) and unit.

Examples include: Mass (e.g., 5 kg) Temperature (e.g., 25 °C) Time (e.g., 10 seconds) Distance (e.g., 10 m) Speed (e.g., 5 m/s) A vector quantity is described by both magnitude, unit, and direction.

Examples include: Displacement (e.g., 10 m East) Velocity (e.g., 5 m/s North) Force (e.g., 20 N downwards) Acceleration (e.g., 2 m/s² West) Weight (e.g., 50 N downwards) Direction is crucial for a vector quantity. Without it, the description is incomplete. 2.2 Representing Vectors Vectors can be represented graphically using arrows. The length of the arrow represents the magnitude, and the arrowhead indicates the direction. We typically use a coordinate system (x-y plane) to define direction. In Physics, bold letters (e.g., F) are used to represent vectors but as this is markdown we will use an arrow above the letter, for example: $\vec{F}$. 2.3 Components of a Vector Any vector in two dimensions can be resolved into its horizontal (x) and vertical (y) components. This is a crucial skill for adding vectors and solving problems. We use trigonometry (SOH CAH TOA) to find these components. If we have a vector $\vec{A}$ with magnitude A and angle θ with respect to the positive x-axis, then: The x-component, A x , is given by: A x = A cos θ The y-component, A y , is given by: A y = A sin θ Example 1: A force of 50 N is applied to a wheelbarrow at an angle of 30° above the horizontal. Determine the horizontal and vertical components of the force.

Solution: Draw a diagram: (Imagine a wheelbarrow being pushed, with an arrow representing the force at 30 degrees to the ground).

Identify the given values: A (magnitude of the force) = 50 N, θ (angle) = 30°. Calculate the x-component (horizontal component): F x = F cos θ F x = 50 N cos 30° F x ≈ 43.3 N Calculate the y-component (vertical component): F y = F sin θ F y = 50 N sin 30° F y = 25 N Therefore, the horizontal component of the force is approximately 43.3 N, and the vertical component is 25 N. 2.4 Adding Vectors using Components To add vectors, we follow these steps: Resolve each vector into its x and y components. Add all the x-components together to get the x-component of the resultant vector (R x ). Add all the y-components together to get the y-component of the resultant vector (R y ). Calculate the magnitude of the resultant vector using the Pythagorean theorem: R = √(R x ² + R y ²) Calculate the direction of the resultant vector using the tangent function: θ = tan -1 (R y / R x ). Consider the quadrant when finding the direction.

Example 2: A delivery vehicle travels 10 km East and then 5 km North. Calculate the resultant displacement.

Solution: Vector 1 (A): 10 km East. A x = 10 km, A y = 0 km Vector 2 (B): 5 km North. B x = 0 km, B y = 5 km Add x-components: R x = A x + B x = 10 km + 0 km = 10 km Add y-components: R y = A y + B y = 0 km + 5 km = 5 km Calculate the magnitude of the resultant displacement: R = √(R x ² + R y ²)* R = √(10² + 5²)* R = √(125)* R ≈ 11.2 km Calculate the direction of the resultant displacement: θ = tan -1 (R y / R x )* θ = tan -1 (5 / 10)* θ ≈ 26.6°* Therefore, the resultant displacement is approximately 11.2 km at an angle of 26.6° North of East.

Example 3: Two tugboats are towing a ship. Tugboat 1 applies a force of 2000 N at an angle of 20° North of East. Tugboat 2 applies a force of 3000 N at an angle of 30° South of East. Determine the magnitude and direction of the resultant force.

Solution: Tugboat 1 (F1): 2000 N at 20° North of East F1 x = 2000 N cos(20°) ≈ 1879.39 N F1 y = 2000 N sin(20°) ≈ 684.04 N Tugboat 2 (F2): 3000 N at 30° South of East (Important: because it's south of east, the y component is negative) F2 x = 3000 N cos(30°) ≈ 2598.08 N F2 y = 3000 N sin(30°) = -1500 N Add x-components: R x = F1 x + F2 x ≈ 1879.39 N + 2598.08 N ≈ 4477.47 N Add y-components: R y = F1 y + F2 y ≈ 684.04 N - 1500 N ≈ -815.96 N Calculate the magnitude of the resultant force: R = √(R x ² + R y ²)* R = √((4477.47 N)² + (-815.96 N)²)* R ≈ 4551.56 N Calculate the direction of the resultant force: θ = tan -1 (R y / R x )* θ = tan -1 (-815.96 N / 4477.47 N)* θ ≈ -10.34°* (Note the negative sign. This indicates that the resultant is South of East).

Therefore, the resultant force is approximately 4551.56 N at an angle of 10.34° South of East. 2.5 Vector Subtraction Subtracting vectors is similar to adding them, but you need to reverse the direction of the vector being subtracted. Mathematically, $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$. To reverse the direction of a vector, change the signs of its x and y components. Then, proceed with the addition as explained above. Guided Practice (With Solutions)

Question 1: A farmer pulls a plough with a force of 80 N at an angle of 40° to the horizontal. Calculate the horizontal and vertical components of the force.