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 vector of two or more vectors.
• Apply the principle of equilibrium to solve problems.
• Solve quantitative problems involving vectors.

Lesson summary

This week, we delve into the fascinating world of vectors in two dimensions, a crucial topic in mechanics. Understanding vectors is fundamental to describing motion and forces accurately. Instead of just dealing with objects moving in a straight line, we'll explore how to represent and analyze motion and forces acting at angles. This has applications in everything from understanding how a soccer ball curves in the air to how engineers design bridges and buildings that can withstand various forces. Think about trying to push a stuck car – you instinctively try to push at an angle that maximizes your effect. That’s vectors in action!

Lesson notes

2.1 Scalars vs.

Vectors: A scalar quantity has only magnitude (size).

Examples: mass (e.g., 50 kg), temperature (e.g., 25°C), time (e.g., 10 s), distance (e.g., 10 m). A vector quantity has both magnitude and direction.

Examples: 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).

Important: Direction is CRUCIAL for vectors. 10 m north is VERY different from 10 m south. 2.2 Representing Vectors: Vectors are typically represented graphically by arrows. The length of the arrow represents the magnitude, and the arrowhead points in the direction of the vector. We use symbols with arrows above them (e.g., $\vec{F}$ for force, $\vec{v}$ for velocity, $\vec{d}$ for displacement) or boldface (e.g., F, v, d) to denote vector quantities. 2.3 Components of a Vector: A vector in two dimensions can be resolved (broken down) into two perpendicular components: a horizontal component (x-component) and a vertical component (y-component). This is the cornerstone of working with vectors in 2D. Consider a vector $\vec{F}$ with magnitude F making an angle θ with the horizontal (x-axis).

The components are: Horizontal component (Fx): $F_x = F \cos θ$ Vertical component (Fy): $F_y = F \sin θ$ Why does this work? It's simple trigonometry! Think of the vector as the hypotenuse of a right-angled triangle. The adjacent side is Fx and the opposite side is Fy.

Example 1: Determining Vector Components A soccer player kicks a ball with a force of 80 N at an angle of 30° to the horizontal. Calculate the horizontal and vertical components of the force.

Solution: $F = 80 N$ $θ = 30°$ $F_x = F \cos θ = 80 N \cos 30° = 80 N 0.866 \approx 69.3 N$ $F_y = F \sin θ = 80 N \sin 30° = 80 N 0.5 = 40 N$ Therefore, the horizontal component of the force is approximately 69.3 N, and the vertical component is 40 N. This tells us that the kick propels the ball both forward and upwards. 2.4 Vector Addition (Component Method): To add vectors, we cannot simply add their magnitudes unless they are in the same direction. The component method is a precise and reliable way to add vectors: Resolve each vector into its x and y components. Add all the x-components together to get the x-component of the resultant vector (Rx). Add all the y-components together to get the y-component of the resultant vector (Ry). The resultant vector, $\vec{R}$, has components Rx and Ry. Find the magnitude of the resultant vector using the Pythagorean theorem: $R = \sqrt{R_x^2 + R_y^2}$ Find the direction (angle) of the resultant vector using the arctangent function: $θ = \arctan(\frac{R_y}{R_x})$ (Be careful about the quadrant!)

Example 2: Adding Vectors Two students are pulling a heavy box across a floor. Student A pulls with a force of 100 N at an angle of 20° to the horizontal. Student B pulls with a force of 80 N at an angle of 30° to the horizontal. Calculate the magnitude and direction of the resultant force on the box.

Solution: Student A: $F_{A} = 100 N$ $θ_A = 20°$ $F_{Ax} = 100 N * \cos 20° \approx 94.0 N$ $F_{Ay} = 100 N * \sin 20° \approx 34.2 N$ Student B: $F_{B} = 80 N$ $θ_B = 30°$ $F_{Bx} = 80 N * \cos 30° \approx 69.3 N$ $F_{By} = 80 N * \sin 30° = 40.0 N$ Resultant Vector: $R_x = F_{Ax} + F_{Bx} = 94.0 N + 69.3 N \approx 163.3 N$ $R_y = F_{Ay} + F_{By} = 34.2 N + 40.0 N = 74.2 N$ $R = \sqrt{R_x^2 + R_y^2} = \sqrt{(163.3 N)^2 + (74.2 N)^2} \approx \sqrt{26666.89 + 5505.64} \approx 180 N$ $θ = \arctan(\frac{R_y}{R_x}) = \arctan(\frac{74.2 N}{163.3 N}) \approx \arctan(0.454) \approx 24.4°$ Therefore, the resultant force on the box is approximately 180 N at an angle of 24.4° to the horizontal. 2.5 Equilibrium: An object is in equilibrium when the net force acting on it is zero. This means both the sum of the forces in the x-direction and the sum of the forces in the y-direction must be zero.

Translational Equilibrium: ΣFx = 0 and ΣFy =

0. The object is either at rest or moving with constant velocity in a straight line.

Example 3: Equilibrium A traffic light is suspended by two cables. Each cable makes an angle of 15° with the horizontal. If the traffic light weighs 120 N, what is the tension in each cable? (Assume the weight acts vertically downward.)

Solution: Let T be the tension in each cable. Because the system is in equilibrium, the sum of the vertical forces must equal the weight, and the sum of the horizontal forces must be zero.

Vertical components: 2 T sin(15°) = 120 N T = 120 N / (2 * sin(15°)) T = 120 N / (2 * 0.259) T ≈ 231.66 N Therefore, the tension in each cable is approximately 231.66 N. The horizontal components of the tension forces will be equal and opposite, canceling each other out. Guided Practice (With Solutions)

Question 1: A child pulls a toy car with a force of 20 N at an angle of 40° above the horizontal. Determine the horizontal and vertical components of the force.