Classical Dynamics: Newton's Laws

The Fundamental Principles Governing Motion

Learning Goals

  • Define force, mass, and acceleration with correct units.
  • Understand and apply the Law of Inertia (First Law).
  • Master the dynamical relationship $\Sigma \vec{F} = m \vec{a}$ (Second Law).
  • Accurately identify action-reaction pairs (Third Law).
  • Resolve forces into 2D components and analyze Inclined Planes.
  • NEW: Define and use **Tension** in connected systems.
  • NEW: Analyze two-body systems (Atwood and Horizontal Pulleys).

Key Vocabulary

  • Force ($\vec{F}$): Vector quantity, measured in Newtons ($N$).
  • Inertia: Resistance to change in motion, proportional to mass ($m$).
  • Net Force ($\Sigma \vec{F}$): The vector sum of all forces.
  • Equilibrium: State where $\Sigma \vec{F} = 0$.
  • Normal Force ($F_{\text{N}}$) & Friction Force ($F_{\text{f}}$).
  • Tension ($F_{\text{T}}$): Force transmitted through a rope or cable.
  • Connected System: Two or more masses accelerating together.

Do Now: Inertia and Mass

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Time Remaining

The Challenge:

A small $1000~kg$ car and a large $40,000~kg$ train are traveling on parallel tracks at the exact same velocity. If both engines suddenly fail, which object will require a larger retarding force to stop it in the same amount of time? Explain your reasoning using the concept of **inertia**.

Newton's First Law: The Law of Inertia

Formal 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."

The Equilibrium Condition

This law describes the state of **equilibrium**, where the net force is zero. Acceleration is zero.

$$ \Sigma \vec{F} = 0 $$

Inertia: Resistance to Change

Inertia is what causes passengers to lurch forward when a bus suddenly brakes. The passenger's body resists the change in velocity. The amount of inertia is directly proportional to the **mass** ($m$).

Newton's Second Law: Dynamics Quantified

The Law of Acceleration:

The net force acting on an object is equal to the mass of the object multiplied by its acceleration.

$$ \Sigma \vec{F} = m \vec{a} $$

Force ($\Sigma F$)

Causes the acceleration. Measured in $\text{N}$.

Mass ($m$)

The object's inertia. Measured in $\text{kg}$.

Acceleration ($a$)

The rate of velocity change. Measured in $\text{m/s}^2$.

Second Law Application: 1D Calculation

Applying $\Sigma F = m a$ to a simple horizontal force system.

Problem Statement:

A construction worker pushes a $40~\text{kg}$ block with an applied force ($F_{\text{A}}$) of $150~\text{N}$. The rough concrete floor exerts a friction force ($F_{\text{f}}$) of $50~\text{N}$ opposite the direction of motion. Determine the block's acceleration ($a$).

Step-by-Step Solution:

1. Calculate Net Force ($\Sigma F$). (Forward forces are positive):

$$ \Sigma F = F_{\text{A}} - F_{\text{f}} = 150 \text{ N} - 50 \text{ N} = 100 \text{ N} $$

2. Calculate Acceleration ($a$) using $a = \Sigma F / m$:

$$ a = \frac{100 \text{ N}}{40 \text{ kg}} = 2.5 \frac{\text{m}}{\text{s}^2} $$

Result: The block accelerates at $2.5~\text{m/s}^2$ in the direction of the push.

Newton's Third Law: The Interaction Pair

The Statement (Action-Reaction):

"For every action force, there is an equal and opposite reaction force."

The Core Equation

The forces are always equal in magnitude and opposite in direction.

$$ \vec{F}_{\text{A on B}} = - \vec{F}_{\text{B on A}} $$

Crucial Distinction

  • The forces always act on **different** objects.
  • This is why they **never cancel out** when analyzing the motion of a single object.

Analyzing the Reaction Pair

When Forces Act on Different Objects

The acceleration of any object is determined only by the **net force acting on that single object** ($\Sigma \vec{F} = m \vec{a}$).

Action

Hammer hits Nail (Forward)

Reaction

Nail hits Hammer (Backward)

Result

The hammer and nail experience the same force magnitude, but their accelerations depend on their mass.

2D Forces: Resolving Components

To apply $\Sigma \vec{F} = m \vec{a}$ in two dimensions, we must break every force into $x$ and $y$ components.

Trigonometric Resolution:

X-Component (Adjacent)

The force component parallel to the x-axis, typically found using **cosine**.

$$ F_x = F \cos \theta $$

Y-Component (Opposite)

The force component parallel to the y-axis, typically found using **sine**.

$$ F_y = F \sin \theta $$

($\theta$ is the angle measured from the positive x-axis, or the closest axis).

Strategy: Analyzing Inclined Planes

The most efficient way to analyze motion on a ramp is to **tilt your coordinate system.**

Tilted Axes Rules:

  1. **Draw FBD:** Include $W$ (straight down), $F_{\text{N}}$ (perpendicular to surface), and $F_{\text{f}}$ (parallel to surface).
  2. **Tilt Axes:** Set the $+x$ axis parallel to the incline (direction of motion) and the $+y$ axis perpendicular to the incline.
  3. **Resolve Weight ($W$):** The only force that requires resolution is gravity (Weight, $W$).
  4. Weight Components: The angle of the incline ($\theta$) is equal to the angle between $W$ and the new $y$-axis.
$$ W_x = W \sin \theta \quad (\text{Down the ramp}) \\ W_y = W \cos \theta \quad (\text{Perpendicular into the ramp}) $$

2D Example: Static Equilibrium on a Ramp

A $2~\text{kg}$ block is held stationary on a frictionless ramp inclined at $\theta = 30^\circ$. Find the Normal Force ($F_{\text{N}}$) exerted by the ramp. (Assume $g = 9.8~\text{m/s}^2$).

Solution (Y-Direction):

1. Calculate Weight ($W$):

$$ W = m g = (2 \text{ kg})(9.8 \text{ m/s}^2) = 19.6 \text{ N} $$

2. Apply Second Law in the y-direction ($\Sigma F_y = 0$ since $a_y=0$):

$$ \Sigma F_y = F_{\text{N}} - W_y = 0 \\ F_{\text{N}} = W_y = W \cos \theta $$

3. Solve for $F_{\text{N}}$:

$$ F_{\text{N}} = (19.6 \text{ N}) \cos(30^\circ) \approx 16.97 \text{ N} $$

Result: The ramp pushes back with a Normal Force of approximately $17.0~\text{N}$.

2D Example: Dynamic Motion on a Ramp

If the $2~\text{kg}$ block *slides* down the $30^\circ$ incline, and the friction force is $F_{\text{f}} = 4.0~\text{N}$, what is its acceleration ($a$)?

Solution (X-Direction):

1. Identify $W_x$ (the driving force) and $F_{\text{f}}$ (the opposing force):

$$ W_x = W \sin \theta = (19.6 \text{ N}) \sin(30^\circ) = 9.8 \text{ N} $$

2. Apply Second Law in the x-direction ($\Sigma F_x = m a$):

$$ \Sigma F_x = W_x - F_{\text{f}} = m a \\ a = \frac{W_x - F_{\text{f}}}{m} $$

3. Solve for acceleration ($a$):

$$ a = \frac{9.8 \text{ N} - 4.0 \text{ N}}{2.0 \text{ kg}} = \frac{5.8 \text{ N}}{2.0 \text{ kg}} = 2.9 \frac{\text{m}}{\text{s}^2} $$

Result: The block accelerates down the ramp at $2.9~\text{m/s}^2$.

Conclusion: Mastering 2D Dynamics

By resolving forces into $x$ and $y$ components, we can apply Newton's Second Law to any complex scenario.

What's Next? (Connected Systems)

Our final step in foundational dynamics is analyzing systems where forces are transmitted through cables and ropes.

Next Class Topic: **Tension, Atwood Machines, and Pulley Systems.** (Let's start now!)

Tension: The Force in the Cable

Tension ($F_{\text{T}}$) is the force exerted by a flexible medium (like a rope, string, or cable) when it is pulled taut.

Key Properties of Tension:

  • **Direction:** Tension always pulls **away** from the object to which the rope is attached.
  • **Ideal Rope:** In introductory problems, we assume ropes are **massless**, meaning tension ($F_{\text{T}}$) is uniform throughout the entire rope.
  • **Ideal Pulley:** We assume pulleys are **massless and frictionless**. They only change the **direction** of the tension force, not its magnitude.
  • **Connected Systems:** Objects connected by a single rope or cable **must** have the same magnitude of acceleration ($a$).

System Dynamics: The Atwood Machine

A classic vertical system used to find the acceleration of two connected masses.

The System Approach:

Treat the two blocks and the rope as a single system. The **Tension forces are internal** and cancel out! Only external forces (Weight/Gravity) contribute to the system's net force ($\Sigma F_{\text{net}}$).

$$ \text{Driving Force: } W_2 = m_2 g \\ \text{Opposing Force: } W_1 = m_1 g \\ \Sigma F_{\text{net}} = W_2 - W_1 = (m_2 + m_1) a $$

Solving for Acceleration ($a$):

$$ a = \frac{m_2 g - m_1 g}{m_1 + m_2} = g \left( \frac{m_2 - m_1}{m_1 + m_2} \right) $$

System Dynamics: Horizontal Pulley

A mass $m_1$ on a table connected to a hanging mass $m_2$ via a rope and pulley.

The System Approach (with Friction):

The motion is determined by the forces acting *along the line of motion* (the $x$-axis for $m_1$, and the $y$-axis for $m_2$).

Forces Driving/Opposing:

  • **Driving:** $W_2 = m_2 g$ (Gravity on the hanging mass).
  • **Opposing:** $F_{\text{f}1}$ (Friction on the table mass $m_1$).

System Net Force ($\Sigma F_{\text{net}}$):

Tension is internal and cancels. The total mass accelerating is $m_1 + m_2$.

$$ \Sigma F_{\text{net}} = W_2 - F_{\text{f}1} = (m_1 + m_2) a $$

This is the most powerful method for finding system acceleration!

Unit Mastery Check: Classical Dynamics

You have mastered the foundational laws and techniques for analyzing forces in all standard scenarios.

Summary of Core Concepts

  • **1D Motion:** $\Sigma F = m a$ (Basic FBD and Net Force calculation).
  • **2D Motion:** **Force Resolution** ($F_x, F_y$) and **Tilted Axes** (Inclined Planes).
  • **Connected Systems:** Treating multiple masses as a single unit to find the system acceleration ($a$).
  • **Universal Tools:** Free Body Diagrams (FBDs) and Newton's Three Laws.

The Unit is Complete! Great work!