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

• State Newton's First, Second, and Third Laws of Motion.
• Apply Newton's Second Law (Fnet = ma) to solve problems.
• Draw free-body diagrams to represent forces.
• Apply Newton's Third Law to identify action-reaction pairs.
• Solve problems involving objects on inclined planes.

Lesson summary

This week, we delve deeper into Newton's Laws of Motion and their applications. Understanding these laws is crucial for explaining and predicting the motion of objects around us. From the simple act of walking to the complex movements of vehicles and machines, Newton's Laws are at play. For South African learners, this knowledge is particularly relevant as it underpins many aspects of engineering, construction, and transportation, all vital for our country's development and infrastructure. For instance, understanding Newton’s Laws is crucial for designing safe and efficient vehicles on our roads, or for constructing stable buildings that can withstand environmental forces.

Lesson notes

Newton's First Law (Law of Inertia): An object continues in a state of rest or uniform (constant velocity) motion unless acted upon by a net (resultant) force. Inertia is the resistance of an object to a change in its state of motion. The greater the mass, the greater the inertia.

Example: A soccer ball will remain at rest on the field until someone kicks it (applies a force). Once kicked, it will continue moving (ideally, at a constant velocity) until other forces like air resistance and friction from the ground slow it down.

Newton's Second Law: The net (resultant) force acting on an object is directly proportional to the rate of change of momentum of the object and is in the same direction as the change in momentum. Since momentum is mass times velocity, and assuming mass is constant, this is often written as: Fnet = ma Where: Fnet is the net force (resultant force) acting on the object (in Newtons, N). m is the mass of the object (in kilograms, kg). a is the acceleration of the object (in meters per second squared, m/s²).

Understanding Fnet: Fnet is the vector sum of all the forces acting on the object. We need to choose a direction as positive and then account for forces acting in the opposite direction.

Example: Imagine pushing a shopping cart (mass = 20 kg) with a force of 50 N. If friction opposes the motion with a force of 10 N, the net force is 50 N - 10 N = 40

N. The acceleration of the cart would be a = Fnet/m = 40 N / 20 kg = 2 m/s².

Newton's Third Law: When object A exerts a force on object B, object B simultaneously exerts a force equal in magnitude and opposite in direction on object

A. These forces act on different objects. These forces are called action-reaction pairs. Important

Note: Action-reaction pairs never act on the same object. If they did, the net force would always be zero, and there would be no acceleration, which is clearly not always the case.

Example: When you walk, your foot pushes backwards on the ground (action). The ground pushes forward on your foot (reaction), propelling you forward. The force your foot exerts on the ground and the force the ground exerts on your foot are the action-reaction pair.

Free-Body Diagrams: A free-body diagram is a diagram showing all the forces acting on an object. It is essential for solving problems using Newton's Laws.

Steps for Drawing a Free-Body Diagram: Represent the object as a point or a simple shape. Draw arrows representing each force acting on the object. The arrow's length should be proportional to the force's magnitude. Label each force clearly (e.g., Fg for weight, FT for tension, FN for normal force, Ff for friction, Fa for applied force). Indicate the direction chosen as positive.

Forces to Consider: Weight (Fg): The force of gravity acting on an object, always directed downwards. Fg = mg, where g is the acceleration due to gravity (approximately 9.8 m/s² on Earth).

Normal Force (FN): The force exerted by a surface on an object in contact with it, perpendicular to the surface.

Tension (FT): The force exerted by a string, rope, or cable, always directed along the string.

Applied Force (Fa): Any force applied to the object by an external agent.

Friction (Ff): A force that opposes motion between two surfaces in contact.

Friction: Friction is a force that opposes motion between two surfaces in contact.

There are two types of friction: Static Friction (Fs): The force that prevents an object from starting to move.

The maximum static friction is given by: Fs(max) = μs FN, where μs is the coefficient of static friction.

Kinetic Friction (Fk): The force that opposes the motion of an object that is already moving.

Kinetic friction is given by: Fk = μk FN, where μk is the coefficient of kinetic friction. Important

Note: The coefficient of static friction (μs) is generally greater than the coefficient of kinetic friction (μk) for the same two surfaces. This means it's harder to get an object moving than to keep it moving.

Inclined Planes: When an object is on an inclined plane, the weight force (Fg) needs to be resolved into components parallel (Fg||) and perpendicular (Fg⊥) to the plane. Fg|| = Fg sin(θ) = mg * sin(θ) (This component causes the object to slide down the plane). Fg⊥ = Fg cos(θ) = mg * cos(θ) (This component is balanced by the normal force). Where θ is the angle of inclination of the plane.

Worked example

Example 1: A Grade 11 learner pushes a box of mass 15 kg across a rough floor with a force of 80 N at an angle of 30° to the horizontal. The coefficient of kinetic friction between the box and the floor is 0.

2. Calculate the acceleration of the box.

Solution:

Draw a free-body diagram: Include Fg, FN, Fa (applied force), Ff (friction), and resolve Fa into horizontal (Fax) and vertical (Fay) components.

Calculate the components of the applied force:

Fax = Fa cos(30°) = 80 N * cos(30°) = 69.28 N

Fay = Fa sin(30°) = 80 N * sin(30°) = 40 N

Calculate the weight of the box:

Fg = mg = 15 kg 9.8 m/s² = 147 N

Calculate the normal force:

FN + Fay = Fg (vertical forces are balanced)

FN = Fg - Fay = 147 N - 40 N = 107 N

Calculate the kinetic friction:

Fk = μk FN = 0.2 * 107 N = 21.4 N

Calculate the net force in the horizontal direction:

Fnet(x) = Fax - Fk = 69.28 N - 21.4 N = 47.88 N

Calculate the acceleration:

a = Fnet(x) / m = 47.88 N / 15 kg = 3.19 m/s²

Example 2: A 5 kg block is placed on an inclined plane that makes an angle of 25° with the horizontal. The coefficient of static friction between the block and the plane is 0.

4. Will the block slide down the plane? If it does, calculate its acceleration.

Solution:

Draw a free-body diagram: Include Fg, FN, and Fs (static friction). Resolve Fg into Fg|| and Fg⊥.

Calculate the components of the weight force:

Fg|| = mg sin(25°) = 5 kg 9.8 m/s² sin(25°) = 20.71 N

Fg⊥ = mg cos(25°) = 5 kg 9.8 m/s² cos(25°) = 44.41 N

Calculate the maximum static friction:

Fs(max) = μs FN = μs Fg⊥ = 0.4 44.41 N = 17.76 N

Compare Fg|| and Fs(max):

Fg|| (20.71 N) > Fs(max) (17.76 N).

Therefore, the block will slide down the plane.

Since the block slides, we need to calculate the kinetic friction: Assume μk = 0.3 (given or look up a reasonable value).

Fk = μk FN = 0.3 * 44.41 N = 13.32 N

Calculate the net force along the plane:

Fnet = Fg|| - Fk = 20.71 N - 13.32 N = 7.39 N

Calculate the acceleration:

a = Fnet / m = 7.39 N / 5 kg = 1.48 m/s²