Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Resolve a vector into horizontal and vertical components.
• Calculate the resultant vector from two or more vectors.
• Apply vector principles to solve problems involving forces and motion.
• Define and explain equilibrium.
• Solve problems involving objects in equilibrium.

Lesson summary

This week, we delve deeper into the world of vectors, specifically focusing on vectors in two dimensions. Building upon our previous understanding of scalars and vectors, we will explore how to represent vectors in a plane, resolve them into components, and perform calculations with them. Understanding vectors is crucial for describing motion, forces, and many other physical phenomena. Think about the Springboks rugby team – the force and direction of a pass, the velocity and direction of a player running down the field – all can be accurately described and analyzed using vectors.

Lesson notes

2.1 Vector Representation in Two Dimensions: A vector in two dimensions can be represented graphically as an arrow in a plane. The length of the arrow represents the magnitude of the vector, and the direction of the arrow represents the direction of the vector relative to a chosen coordinate system (usually the Cartesian coordinate system).

Magnitude: The magnitude is the "size" or "strength" of the vector. It is always a non-negative scalar quantity.

Direction: The direction is the angle the vector makes with a reference axis (usually the positive x-axis). This angle is typically measured in degrees. Vectors can also be represented in component form. A vector A can be expressed as A = (Ax, Ay), where Ax is the x-component and Ay is the y-component of the vector. 2.2 Resolving Vectors into Components: Resolving a vector into its components involves finding the projections of the vector onto the x and y axes. We use trigonometric functions to find these components: Ax = A cos θ Ay = A sin θ Where: A is the magnitude of the vector A. θ is the angle the vector A makes with the positive x-axis.

Example 1: A farmer pulls a plough with a force of 500 N at an angle of 30° to the horizontal. Determine the horizontal and vertical components of the force.

Solution: Fx = F cos θ = 500 N cos(30°) = 500 N * 0.866 = 433 N Fy = F sin θ = 500 N sin(30°) = 500 N * 0.5 = 250 N Therefore, the horizontal component of the force is 433 N, and the vertical component is 250 N. The horizontal component is what actually contributes to pulling the plough forward. 2.3 Adding Vectors Using Components: To add vectors, we add their corresponding components. If we have two vectors, A = (Ax, Ay) and B = (Bx, By), their sum R = A + B is given by: Rx = Ax + Bx Ry = Ay + By The magnitude of the resultant vector R is then calculated using the Pythagorean theorem: R = √(Rx² + Ry²) The direction of the resultant vector R is given by: θ = arctan(Ry / Rx)

Example 2: A delivery driver drives 30 km east and then 40 km north. Determine the driver's resultant displacement (magnitude and direction) from the starting point.

Solution: Let A = (30 km, 0 km) be the eastward displacement and B = (0 km, 40 km) be the northward displacement. Rx = Ax + Bx = 30 km + 0 km = 30 km Ry = Ay + By = 0 km + 40 km = 40 km The magnitude of the resultant displacement is: R = √(30² + 40²) = √(900 + 1600) = √2500 = 50 km The direction of the resultant displacement is: θ = arctan(40 / 30) = arctan(1.333) ≈ 53.1° Therefore, the driver's resultant displacement is 50 km at an angle of approximately 53.1° north of east. 2.4 Equilibrium: An object is in equilibrium when the net force acting on it is zero. This means that the vector sum of all forces acting on the object is zero. Mathematically, this is expressed as: ΣFx = 0 and ΣFy = 0 2.5 Inclined Planes: When an object is placed on an inclined plane, the force of gravity acting on the object can be resolved into two components: one parallel to the plane (F||) and one perpendicular to the plane (F⊥). F|| = mg sin θ F⊥ = mg cos θ Where: m is the mass of the object. g is the acceleration due to gravity (approximately 9.8 m/s²). θ is the angle of the incline.

Example 3: A 5 kg block rests on an inclined plane that makes an angle of 25° with the horizontal. Determine the component of the gravitational force acting parallel to the plane.

Solution: F|| = mg sin θ = (5 kg)(9.8 m/s²) sin(25°) = (49 N)(0.423) ≈ 20.7 N Therefore, the component of the gravitational force acting parallel to the plane is approximately 20.7 N. This is the force that tends to pull the block down the incline. Guided Practice (With Solutions)

Question 1: A kite is flying at the end of a string. The tension in the string is 20 N, and the string makes an angle of 40° with the horizontal. Find the horizontal and vertical components of the tension force.

Solution: Horizontal component: Tx = T cos θ = 20 N cos(40°) = 20 N * 0.766 ≈ 15.3 N Vertical component: Ty = T sin θ = 20 N sin(40°) = 20 N * 0.643 ≈ 12.9 N

Commentary: This question directly applies the concept of resolving a vector into its components using trigonometric functions. Understanding the context (tension in a string) helps visualize the forces involved.

Question 2: A boat is being pulled by two tugboats. Tugboat 1 exerts a force of 3000 N at an angle of 20° north of east, and tugboat 2 exerts a force of 4000 N at an angle of 30° south of east. What is the magnitude and direction of the resultant force on the boat?