Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

This page supports the lesson note with a companion video and a short classroom-ready summary.

For class groups and homework, share this lesson page so learners also get the summary, objectives, and full lesson context.

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 of forces in equilibrium.
• Define and apply the concepts of equilibrant and net force.
• Perform vector addition and subtraction calculations.

Lesson summary

Welcome to Week 5 of our Physical Sciences journey! This week, we delve deeper into the world of Mechanics, specifically focusing on vectors in two dimensions. Building upon your understanding of scalars and vectors from Grade 10, we will now learn how to deal with forces, velocities, and displacements that act not just along a straight line, but at angles to each other. Why is this important? Imagine a taxi trying to navigate the streets of Durban, dealing with both its own engine power and the force of the wind. Or consider a soccer player kicking a ball at an angle to score a goal.

Lesson notes

What is a Vector in Two Dimensions? A vector is a quantity that has both magnitude (size) and direction. In two dimensions, we can represent a vector on a Cartesian plane (x-y plane).

Common examples include: Displacement: The change in position of an object (e.g., 5 meters at 30° North of East).

Velocity: The rate of change of displacement (e.g., 10 m/s at 45° to the horizontal).

Force: A push or pull acting on an object (e.g., 20 N at 60° downwards). Resolving Vectors into Components The key to working with vectors in two dimensions is understanding how to resolve them into their horizontal (x) and vertical (y) components. This means breaking down the vector into two perpendicular vectors that, when added together, give the original vector. Consider a vector F with magnitude F and direction θ (angle with respect to the x-axis).

The components are: Horizontal component (Fx): Fx = F cos(θ)

Vertical component (Fy): Fy = F sin(θ) Why does this work? This comes directly from trigonometry. Imagine a right-angled triangle where the original vector F is the hypotenuse, Fx is the adjacent side, and Fy is the opposite side. Then, cos(θ) = Adjacent / Hypotenuse = Fx / F, and sin(θ) = Opposite / Hypotenuse = Fy /

F. Example 1: A security guard pulls a trolley with a force of 80 N at an angle of 35° to the horizontal. Determine the horizontal and vertical components of the force. F = 80 N θ = 35° Fx = 80 N cos(35°) = 80 N 0.819 ≈ 65.5 N Fy = 80 N sin(35°) = 80 N 0.574 ≈ 45.9 N Therefore, the horizontal component of the force is approximately 65.5 N, and the vertical component is approximately 45.9 N. Vector Addition To find the resultant of two or more vectors, we need to add them together.

There are two main methods: Graphical Method (Head-to-Tail): Draw the vectors to scale, placing the tail of the second vector at the head of the first, the tail of the third at the head of the second, and so on. The resultant vector is the vector drawn from the tail of the first vector to the head of the last vector. This method is useful for visualization but less accurate for calculations.

Component Method (Algebraic Method): This is the more accurate method. Resolve each vector into its x and y components. Add all the x-components together to get the x-component of the resultant (Rx). Add all the y-components together to get the y-component of the resultant (Ry). The resultant vector R has components Rx and Ry.

To find the magnitude and direction: Magnitude: R = √(Rx² + Ry²) (Pythagorean theorem)

Direction: θ = arctan(Ry / Rx) (Inverse tangent function). You may need to adjust the angle based on the quadrant Rx and Ry are in.

Example 2: Two workers are trying to move a stalled taxi. Worker 1 applies a force of 150 N eastwards, and worker 2 applies a force of 200 N at 60° north of east. What is the resultant force on the taxi?

Worker 1: F1 = 150 N, θ1 = 0° F1x = 150 N cos(0°) = 150 N F1y = 150 N sin(0°) = 0 N Worker 2: F2 = 200 N, θ2 = 60° F2x = 200 N cos(60°) = 200 N * 0.5 = 100 N F2y = 200 N sin(60°) = 200 N * 0.866 ≈ 173.2 N Rx = F1x + F2x = 150 N + 100 N = 250 N Ry = F1y + F2y = 0 N + 173.2 N = 173.2 N R = √(Rx² + Ry²) = √(250² + 173.2²) = √(62500 + 29998.24) = √92498.24 ≈ 304.1 N θ = arctan(Ry / Rx) = arctan(173.2 / 250) = arctan(0.6928) ≈ 34.7° Therefore, the resultant force is approximately 304.1 N at an angle of 34.7° north of east. Equilibrant and Net Force Net Force (Resultant Force): The vector sum of all forces acting on an object. If the net force is zero, the object is in equilibrium.

Equilibrant: The force that, when applied to an object, will bring it into equilibrium. The equilibrant is equal in magnitude but opposite in direction to the resultant force.

Example 3: Using the result from Example 2, the equilibrant force would be 304.1 N at 34.7° South of West. This force would perfectly counteract the combined forces of the two workers, keeping the taxi stationary (in equilibrium). Forces in Equilibrium An object is in equilibrium when the net force acting on it is zero.

This means: The sum of the forces in the x-direction is zero (ΣFx = 0). The sum of the forces in the y-direction is zero (ΣFy = 0). This principle is fundamental for understanding the stability of structures like bridges and buildings. Guided Practice (With Solutions)

Question 1: A sign weighing 50 N is suspended by two ropes. One rope makes an angle of 30° with the horizontal, and the other makes an angle of 60° with the horizontal. Determine the tension in each rope.

Solution: Draw a free-body diagram: This is crucial! Draw the sign, the two ropes (T1 and T2), and the weight (W = 50 N) acting downwards. Label the angles.