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Topic 2/3
15 Flashcards in this deck.
Sir Isaac Newton's First Law of Motion, often referred to as the Law of Inertia, states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. This principle highlights the inherent resistance of objects to changes in their state of motion.
Mathematically, the First Law can be expressed as: $$ F_{net} = 0 \implies v = \text{constant} $$ where \( F_{net} \) is the net external force acting on the object, and \( v \) is its velocity.
**Examples:**
Newton's Second Law quantifies the relationship between force, mass, and acceleration. It states that the acceleration of an object is directly proportional to the net external force acting upon it and inversely proportional to its mass.
The Second Law is mathematically represented as: $$ F = m \cdot a $$ where:
**Examples:**
The Third Law of Motion states that for every action, there is an equal and opposite reaction. This means that forces always occur in pairs; if object A exerts a force on object B, object B simultaneously exerts a force of equal magnitude but opposite direction on object A.
Mathematically, this can be expressed as: $$ F_{A \rightarrow B} = -F_{B \rightarrow A} $$ where \( F_{A \rightarrow B} \) is the force exerted by object A on object B, and \( F_{B \rightarrow A} \) is the force exerted by object B on object A.
**Examples:**
Newton's Laws are not only theoretical constructs but also have practical applications across various fields of physics and engineering.
**Engineering:**
**Everyday Life:**
Delving deeper into the mathematical foundations, Newton's Second Law can be derived from basic principles of motion.
Starting with the definition of acceleration: $$ a = \frac{dv}{dt} $$ Where \( v \) is velocity and \( t \) is time. Multiplying both sides by mass \( m \): $$ F = m \cdot \frac{dv}{dt} = m \cdot a $$ This derivation solidifies the direct proportionality between force and acceleration, and the inverse relationship with mass.
**Example Problem:** Calculate the acceleration of a 10 kg mass subjected to a net force of 50 N.
**Solution:** $$ a = \frac{F}{m} = \frac{50 \, \text{N}}{10 \, \text{kg}} = 5 \, \text{m/s}^2 $$
Friction is a critical factor that influences the application of Newton's Laws. It acts as an external force that can impede motion.
The force of friction can be categorized into two types:
Mathematically, friction is expressed as: $$ f = \mu \cdot N $$ where:
Understanding friction is essential for accurately applying Newton's Second Law in real-world scenarios.
Newton's Laws hold true in inertial frames of reference, where objects either remain at rest or move at a constant velocity unless acted upon by external forces. However, in non-inertial (accelerating) frames, fictitious forces must be introduced to apply Newtonian mechanics accurately.
**Example:** In a rotating carousel, a rider feels an outward force, which is not a real force but a result of being in an accelerating (non-inertial) frame.
Expanding upon Newton's Laws, the equations of motion provide a more detailed mathematical framework for predicting an object's future position and velocity.
Starting with Newton's Second Law: $$ F = m \cdot a = m \cdot \frac{d^2x}{dt^2} $$ Integrating with respect to time, assuming constant acceleration: $$ v = v_0 + a \cdot t $$ $$ x = x_0 + v_0 \cdot t + \frac{1}{2} a \cdot t^2 $$ where \( v_0 \) and \( x_0 \) are the initial velocity and position, respectively.
These equations are fundamental in solving complex motion problems, including projectile motion and harmonic oscillators.
Momentum (\( p \)) is defined as the product of an object's mass and its velocity: $$ p = m \cdot v $$ Impulse (\( J \)) is the change in momentum resulting from a force applied over a time interval: $$ J = F \cdot \Delta t = \Delta p $$
**Conservation of Momentum:** In a closed system with no external forces, the total momentum remains constant. This principle is crucial in collision analysis.
**Example Problem:** A 1500 kg car moving at 20 m/s collides with a 1000 kg truck moving at 15 m/s in the opposite direction. Calculate the final velocity of the system post-collision (assuming a perfectly inelastic collision).
**Solution:** $$ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f $$ $$ 1500 \cdot 20 + 1000 \cdot (-15) = (1500 + 1000) v_f $$ $$ 30000 - 15000 = 2500 v_f $$ $$ v_f = \frac{15000}{2500} = 6 \, \text{m/s} $$
While Newton's Laws primarily address linear motion, their principles extend to rotational motion through torque and angular momentum.
**Torque (\( \tau \))** is the rotational equivalent of force: $$ \tau = r \cdot F \cdot \sin(\theta) $$ where \( r \) is the distance from the pivot point, \( F \) is the force applied, and \( \theta \) is the angle between the force and lever arm.
**Angular Momentum (\( L \))** is given by: $$ L = I \cdot \omega $$ where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
Extending Newton's Second Law to rotation: $$ \tau = I \cdot \alpha $$ where \( \alpha \) is the angular acceleration.
**Example:** Calculating the torque required to accelerate a wheel requires knowledge of its moment of inertia and desired angular acceleration.
Newton's Laws bridge various scientific disciplines, underpinning concepts in engineering, astronomy, and even economics.
**Engineering:**
**Astronomy:**
**Economics:**
These connections illustrate the universal applicability of Newtonian mechanics across various fields.
Tackling complex physics problems involving Newton's Laws often requires a systematic approach:
**Example Problem:** A block of mass 5 kg is pulled up an incline at 30° with a constant velocity. If the coefficient of kinetic friction is 0.2, calculate the applied force.
**Solution:** Since the block moves at constant velocity, \( F_{net} = 0 \).
Forces along the incline:
Normal force: $$ N = m \cdot g \cdot \cos(\theta) = 5 \cdot 9.81 \cdot \cos(30°) \approx 5 \cdot 9.81 \cdot 0.866 = 42.5 \, \text{N} $$ Friction: $$ f_k = \mu \cdot N = 0.2 \cdot 42.5 = 8.5 \, \text{N} $$ Gravitational component: $$ m \cdot g \cdot \sin(\theta) = 5 \cdot 9.81 \cdot 0.5 = 24.525 \, \text{N} $$ Applying \( F - 24.525 - 8.5 = 0 \): $$ F = 24.525 + 8.5 = 33.025 \, \text{N} $$
While Newton's Laws focus on forces and motion, energy concepts like kinetic and potential energy complement the analysis.
**Kinetic Energy (\( KE \)):** $$ KE = \frac{1}{2} m v^2 $$ **Potential Energy (\( PE \)):** $$ PE = m \cdot g \cdot h $$
Understanding the interplay between force, work, and energy provides a more holistic view of physical systems.
Aspect | Newton's First Law | Newton's Second Law | Newton's Third Law |
---|---|---|---|
Definition | Law of Inertia: Objects remain at rest or in uniform motion unless acted upon by external forces. | F=ma: The acceleration of an object is directly proportional to the net force and inversely proportional to its mass. | Action-Reaction: For every action, there is an equal and opposite reaction. |
Key Equation | $F_{net} = 0 \implies v = \text{constant}$ | $F = m \cdot a$ | $F_{A \rightarrow B} = -F_{B \rightarrow A}$ |
Applications | Predicting object behavior in the absence of forces. | Calculating acceleration and designing motion systems. | Understanding interactions between objects, such as propulsion. |
To excel in exams, remember the acronym F.A.M.: