Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Dynamics
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Dynamics
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Subject: Further Mathematics
Class: Senior Secondary 2
Term: 1st Term
Week: 8
Theme: Mechanics
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State Newtons laws of motion Explain clearly each of the laws Apply Newtons laws to practical problems Solve problems on motion along on in clined plane Work down equations of motion of connected particles Explain the concepts of work, energy and power Solve problems on work, energy and power Solve problems on impulse and momentum Under stand the concept of projectiles Solve problems on motion of projectiles
Newton's laws are fundamental to classical mechanics, describing the relationship between a body and the forces acting upon it, and its motion in response to those forces. 2.1.1 First Law of Motion (Law of Inertia)
Statement: An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced external force.
Explanation: This law introduces the concept of inertia, which is the resistance of an object to changes in its state of motion. If the net force on an object is zero, its velocity remains constant (which includes zero velocity, i.e., rest).
Practical Example (Nigeria): A stationary car will not move until the engine provides a force, or a person pushes it. A moving "Molue" (bus) tends to keep moving; if the driver suddenly brakes, passengers lurch forward due to their inertia. 2.1.2 Second Law of Motion Statement: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. Mathematically, F = ma, where F is the net force, m is mass, and a is acceleration.
Explanation: This is the most quantitative of Newton's laws. It defines force as the cause of acceleration. A larger net force causes greater acceleration, and a larger mass resists acceleration more. The net force (resultant force) is the vector sum of all individual forces acting on the object.
Units: Force is measured in Newtons (N), mass in kilograms (kg), and acceleration in meters per second squared (m/s2). 1 N = 1 kg⋅m/s
2. Practical Example (Nigeria): A "keke Napep" (tricycle) with a small engine (less force) will accelerate slower than a car with a powerful engine (more force), assuming similar masses. If a loaded "okada" (motorcycle) has more mass, it will accelerate slower than an unloaded one for the same engine force. 2.1.3 Third Law of Motion Statement: For every action, there is an equal and opposite reaction.
Explanation: Forces always occur in pairs. When object A exerts a force on object B (action), object B simultaneously exerts an equal and opposite force on object A (reaction). These forces act on different objects.
Practical Example (Nigeria): When a person pushes a wall, the wall pushes back on the person with an equal and opposite force. When a rocket is launched (common in STEM competitions), it expels gas downwards (action), and the gas pushes the rocket upwards (reaction).
Walking on the ground: the foot pushes backward on the ground, and the ground pushes forward on the foot. When an object rests or moves on an inclined plane (a ramp), its weight must be resolved into components parallel and perpendicular to the plane.
Forces Involved: Weight (W = mg): Acts vertically downwards.
Normal Reaction (R): Acts perpendicular to the surface of the plane, preventing the object from penetrating the plane.
Friction (f): Acts parallel to the plane, opposing the direction of motion or impending motion.
Applied Force/Tension (T): Any external force applied.
Resolution of Weight: If the angle of inclination is $\theta$, the components of weight are: Perpendicular to the plane: $mg \cos \theta$ Parallel to the plane: $mg \sin \theta$ Equations of Motion: Perpendicular to the plane: Since there is no motion perpendicular to the plane, the net force in this direction is zero. $R = mg \cos \theta$ (if no other perpendicular forces)
Parallel to the plane: Apply Newton's Second Law, $F_{net} = ma$. The net force is the sum of forces acting parallel to the plane (e.g., $mg \sin \theta$, friction, tension). Worked Example 2.2.1: A crate of mass 50 kg is pulled up a rough inclined plane by a rope. The plane makes an angle of 30° with the horizontal. The coefficient of kinetic friction between the crate and the plane is 0.
2. If the tension in the rope is 400 N, calculate the acceleration of the crate. (Take $g = 10 \text{ m/s}^2$).
Solution: Draw a Free-Body Diagram: Show all forces acting on the crate. Weight (W) = $mg = 50 \times 10 = 500 \text{ N}$ (downwards) Normal Reaction (R) (perpendicular to plane, upwards) Tension (T) = 400 N (up the plane) Friction (f) (down the plane, opposing motion)
Resolve Weight: Component perpendicular to plane: $W_\perp = mg \cos 30^\circ = 500 \times \frac{\sqrt{3}}{2} \approx 500 \times 0.866 = 433 \text{ N}$ Component parallel to plane: $W_\parallel = mg \sin 30^\circ = 500 \times 0.5 = 250 \text{ N}$ Forces Perpendicular to the plane: Since there's no acceleration perpendicular to the plane, $R - W_\perp = 0 \implies R = W_\perp = 433 \text{ N}$.
Calculate Friction: $f = \mu R = 0.2 \times 433 = 86.6 \text{ N}$.
Forces Parallel to the plane: The crate is pulled up, so net force is up the plane. $F_{net} = T - W_\parallel - f = ma$ $400 - 250 - 86.6 = 50a$ $63.4 = 50a$ $a = \frac{63.4}{50} = 1.268 \text{ m/s}^2$.
Therefore, the acceleration of the crate is approximately 1.27 m/s$^2$ up the plane. Connected particles are systems where two or more objects are linked, typically by an inextensible string passing over a smooth pulley, or by being in contact. The tension in the string is usually uniform throughout (if the string is light and inextensible and the pulley is smooth).
Key Principles: Each particle has its own equation of motion ($F_{net} = ma$). The acceleration of all connected particles is usually the same in magnitude. Tension forces are internal forces within the system; they act on each connected particle.
Common Scenarios: Particles connected over a pulley: One particle hangs, another rests on a table (smooth or rough).
Atwood Machine: Two particles hanging vertically, connected by a string over a pulley.
Particles in contact: Pushing or pulling multiple objects in a line. Worked Example 2.3.1 (Atwood Machine): Two masses, 5 kg and 3 kg, are connected by a light inextensible string passing over a smooth fixed pulley. Find the acceleration of the system and the tension in the string. (Take $g = 10 \text{ m/s}^2$).
Solution: Draw Free-Body Diagrams: For 5 kg mass ($m_1$): Weight ($W_1 = 5g = 50 \text{ N}$) downwards, Tension (T) upwards. For 3 kg mass ($m_2$): Weight ($W_2 = 3g = 30 \text{ N}$) downwards, Tension (T) upwards.
Determine Direction of Motion: The 5 kg mass is heavier, so it will accelerate downwards, and the 3 kg mass will accelerate upwards. Let the magnitude of acceleration be 'a'. Formulate Equations of Motion ($F_{net} = ma$): For $m_1$ (5 kg mass, moving downwards): $W_1 - T = m_1 a$ $50 - T = 5a$ (Equation 1) For $m_2$ (3 kg mass, moving upwards): $T - W_2 = m_2 a$ $T - 30 = 3a$ (Equation 2)
Solve Simultaneously: Add Equation 1 and Equation 2: $(50 - T) + (T - 30) = 5a + 3a$ $20 = 8a$ $a = \frac{20}{8} = 2.5 \text{ m/s}^2$. Substitute 'a' into Equation 2 (or Equation 1): $T - 30 = 3 \times 2.5$ $T - 30 = 7.5$ $T = 37.5 \text{ N}$.
Therefore, the acceleration of the system is 2.5 m/s$^2$, and the tension in the string is 37.5 N. These concepts describe how forces transfer energy to objects and the rate at which this happens. 2.4.1 Work Done (W)
Definition: Work is done when a force causes a displacement of an object in the direction of the force.
Formula: $W = Fd \cos \theta$, where F is the magnitude of the force, d is the magnitude of the displacement, and $\theta$ is the angle between the force and displacement vectors. If the force is in the direction of displacement, $\theta = 0^\circ$ and $\cos \theta = 1$, so $W = Fd$.
Units: Joules (J). 1 J = 1 N⋅m.
Explanation: Work is a scalar quantity. If no displacement occurs, or if the force is perpendicular to the displacement, no work is done.
Practical Example (Nigeria): A farmer pushing a wheelbarrow filled with cassava tubers across a field does work. Lifting a bucket of water from a well involves doing work against gravity. 2.4.2 Energy (E)
Definition: Energy is the capacity to do work.
Units: Joules (J).
Forms of Mechanical Energy: Kinetic Energy (KE): Energy possessed by an object due to its motion. $KE = \frac{1}{2} mv^2$, where m is mass and v is speed.
Potential Energy (PE): Energy stored in an object due to its position or state.
Gravitational Potential Energy (GPE): $GPE = mgh$, where m is mass, g is acceleration due to gravity, and h is height above a reference level.
Work-Energy Theorem: The net work done on an object is equal to the change in its kinetic energy. $W_{net} = \Delta KE = KE_f - KE_i$.
Conservation of Mechanical Energy: In the absence of non-conservative forces (like friction or air resistance), the total mechanical energy (KE + PE) of a system remains constant. $KE_i + PE_i = KE_f + PE_f$.
Practical Example (Nigeria): A moving car has kinetic energy. Water stored in a dam (e.g., Kainji Dam) has gravitational potential energy which can be converted to kinetic energy of turbines to generate electricity. 2.4.3 Power (P)
Definition: Power is the rate at which work is done or energy is transferred.
Formula: $P = \frac{W}{t}$ (Work done per unit time) or $P = Fv$ (Force times average velocity, if force is constant and parallel to velocity).
Units: Watts (W). 1 W = 1 J/s. Another common unit is horsepower (hp), though Watts are standard.
Explanation: Power tells us how quickly energy is being used or transformed.
Practical Example (Nigeria): A powerful generator can produce electricity faster (higher power output) than a smaller one. An athlete who lifts a weight quickly does more power than one who lifts the same weight slowly. Worked Example 2.4.1: A water pump lifts 200 kg of water from a well of depth 15 m in 30 seconds.
Calculate: a) The work done by the pump. b) The power of the pump. (Take $g = 10 \text{ m/s}^2$).
Solution: a)
Work done (W): The force required to lift the water is its weight: $F = mg = 200 \times 10 = 2000 \text{ N}$. The displacement (height) is $d = h = 15 \text{ m}$. Work done = Force × displacement = $Fh = 2000 \times 15 = 30000 \text{ J}$. b)
Power (P): Power = $\frac{\text{Work done}}{\text{Time taken}} = \frac{30000 \text{ J}}{30 \text{ s}} = 1000 \text{ W}$.
Therefore, the work done by the pump is 30,000 J, and its power is 1,000 W.
Vehicle Dynamics and Road Safety in Nigeria: Application: The principles of momentum and impulse are crucial in understanding car collisions, common in Nigeria. Airbags and crumple zones in vehicles are designed to increase the time of impact during a crash (increasing $\Delta t$), thereby reducing the average force ($F = I/\Delta t$) on occupants and minimizing injuries. Seatbelts restrain occupants, making them decelerate with the car rather than flying forward due to inertia.
Integration: Discuss statistics of road accidents in Nigeria. Task students to research how vehicle design incorporates dynamic principles for safety. Connect to the duties of Federal Road Safety Corps (FRSC).
Sports and Physical Education: Application: Projectile motion is evident in many sports popular in Nigeria, such as football (soccer), basketball, javelin throw, and shot put. Athletes use their understanding (often intuitive) of initial velocity and angle of projection to maximize range (e.g., a long pass in football) or height (e.g., a high jump or a shot put). Newton's laws explain why a player exerts force to kick a ball (F=ma) and the reaction force from the ground enables running.
Integration: Analyze video clips of Nigerian athletes in action. Students can calculate the theoretical range or height for a given kick/throw if initial conditions are estimated. Relate to coaching techniques for optimal performance. Engineering and Construction in Local Contexts: Application: When building structures (houses, bridges, water towers) or designing equipment, engineers apply Newton's Laws and principles of work, energy, and power. For instance, calculating the forces on a roof (weight, wind pressure) or determining the power needed for a crane to lift heavy construction materials (e.g., cement blocks, steel beams) to various heights. Designing ramps for accessibility or for moving goods in markets (inclined planes) requires understanding friction and gravity.
Integration: Field trip (or virtual tour) to a local construction site or a manufacturing plant (e.g., a small-scale agro-processing factory). Discuss how different forces are managed in a "pure water" sachet factory or a cassava processing plant. This can inspire interest in vocational and engineering careers.