Lecture 8 Newton's Laws and Applications: Equilibrium, Dynamics, Mass, Weight, and Normal Forces

Newton's Laws of Motion

Newton's Second Law

Newton's Second Law describes the relationship between the net force acting on an object, its mass, and its resulting acceleration. This law is fundamental to understanding how forces affect motion.

• Definition: The acceleration a of an object with mass m is proportional to the net force Fnet acting on it and inversely proportional to its mass.

• Equation: $\vec{a} = \frac{\vec{F}_{\text{net}}}{m}$

• Net Force: The net force is the vector sum of all individual forces acting on the object: $\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \ldots$

• Direction: The acceleration points in the direction of the net force.

• Example: If multiple forces act on a block, the block accelerates in the direction of the vector sum of those forces.

Newton's Third Law

Newton's Third Law states that forces always occur in pairs, known as action-reaction pairs. These pairs act on different objects and are equal in magnitude but opposite in direction.

• Definition: For every action force, there is an equal and opposite reaction force.

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

• Action-Reaction Pair: If object A exerts a force on object B, then object B exerts an equal and opposite force on object A.

• Example: When you push against a wall, the wall pushes back against you with equal force in the opposite direction.

Equilibrium

Static and Dynamic Equilibrium

Equilibrium occurs when the net force acting on an object is zero. There are two types: static equilibrium (object at rest) and dynamic equilibrium (object moving at constant velocity).

• Static Equilibrium: Object is at rest; $\vec{a} = 0$.

• Dynamic Equilibrium: Object moves in a straight line at constant speed; $\vec{a} = 0$.

• Equilibrium Condition: The sum of the forces in each direction is zero: $\sum F_x = ma_x = 0$ $\sum F_y = ma_y = 0$

• Example: A book resting on a table is in static equilibrium; a car moving at constant speed on a straight road is in dynamic equilibrium.

Problem-Solving Strategies

Equilibrium Problems

To solve equilibrium problems, follow these steps:

• Prepare: Make simplifying assumptions and check that $a = 0$.

• Identify Forces: Draw a free-body diagram showing all forces.

• Apply Newton's Second Law: Write equations for each component: $\sum F_x = 0$ $\sum F_y = 0$

• Example: Finding the tension in a rope towing a car at constant speed.

Dynamics Problems

Dynamics problems involve objects that are accelerating. The approach is similar but includes kinematics.

• Prepare: Identify known quantities and what is to be found.

• Draw Diagrams: Sketch a pictorial representation and free-body diagram.

• Apply Newton's Second Law: Write equations for each component: $\sum F_x = ma_x$ $\sum F_y = ma_y$

• Use Kinematics: If needed, use kinematic equations to relate acceleration, velocity, and position.

• Example: Calculating the tension in a rope towing a car that accelerates from rest.

Mass and Weight

Definitions and Differences

Mass and weight are related but distinct physical quantities.

• Mass: A measure of an object's inertia; its resistance to acceleration. Mass is constant regardless of location.

• Weight: The gravitational force exerted on an object. Weight depends on the local gravitational acceleration.

• Equation: $w = mg$

• Example: An object has the same mass on Earth and Jupiter, but its weight is greater on Jupiter due to stronger gravity.

Apparent Weight

Apparent weight is the magnitude of the contact force supporting an object, such as the reading on a scale. It can differ from true weight if the object is accelerating vertically.

• Definition: $w_{\text{app}} = w + ma_y$

• Example: In an elevator accelerating upward, the scale reads more than your true weight; accelerating downward, it reads less.

• Weightlessness: In free fall, apparent weight is zero, but true weight (gravitational force) still acts.

Normal Forces

Normal Force on a Flat Surface

The normal force is the perpendicular contact force exerted by a surface to support an object resting on it.

• Definition: The normal force adjusts to balance the object's weight and any additional downward force.

• Equation (static equilibrium): $n = w + F$ where $F$ is any extra downward force.

• Example: A book pressed down on a table with extra force has a normal force greater than its weight.

Normal Force on an Incline

On an inclined surface, the normal force is perpendicular to the surface, not necessarily equal to the object's weight.

• Decomposition: The weight can be decomposed into components parallel and perpendicular to the incline.

• Equations: Perpendicular component: $w_y = mg \cos \theta$ Parallel component: $w_x = mg \sin \theta$

• Normal Force: $n = mg \cos \theta$

• Common Mistakes: The normal force is always perpendicular to the surface; the weight always points straight down.

• Example: A skier on a slope experiences a normal force less than their weight.

Applications and Examples

Example: Tension in a Rope (Equilibrium)

A car of mass 1500 kg is towed at constant speed by a rope at a 20° angle. The friction force is 320 N. Find the tension in the rope.

• Horizontal component: $T \cos 20° = 320\,\text{N}$

• Solution: $T = \frac{320\,\text{N}}{\cos 20°}$

Example: Tension in a Rope (Acceleration)

If the car accelerates from rest to 12 m/s in 10 s, use kinematics to find acceleration, then apply Newton's second law:

• Acceleration: $a = \frac{12\,\text{m/s}}{10\,\text{s}} = 1.2\,\text{m/s}^2$

• Equation: $T \cos 20° = ma + f$

• Solution: $T = \frac{ma + f}{\cos 20°}$

Example: Apparent Weight in an Elevator

A person of mass 70 kg stands on a scale in an elevator moving at 5.0 m/s. As the elevator stops, the scale reads 750 N. Was the elevator moving up or down?

• True weight: $w = (70\,\text{kg})(9.8\,\text{m/s}^2) = 686\,\text{N}$

• Apparent weight: $w_{\text{app}} = 750\,\text{N}$

• Acceleration: $a = \frac{w_{\text{app}} - w}{m} = \frac{750 - 686}{70} = 0.91\,\text{m/s}^2$

• Interpretation: Apparent weight greater than true weight means elevator is slowing down while moving downward.

Example: Acceleration Down an Incline

A skier slides down a 27° slope. Neglecting friction, find the acceleration.

• Equation: $a = g \sin \theta$

• Calculation: $a = (9.8\,\text{m/s}^2) \sin 27° = 4.4\,\text{m/s}^2$

• Interpretation: For $\theta = 0°$, $a = 0$; for $\theta = 90°$, $a = g$ (free fall).

Summary Table: Key Equations and Concepts

Concept	Equation	Notes
Newton's Second Law	$\vec{a} = \frac{\vec{F}_{\text{net}}}{m}$	Acceleration is proportional to net force, inversely to mass
Equilibrium (x, y components)	$\sum F_x = 0$, $\sum F_y = 0$	Applies when $a = 0$
Weight	$w = mg$	Depends on local gravity
Apparent Weight	$w_{\text{app}} = w + ma_y$	Scale reading; differs if accelerating vertically
Normal Force (flat surface)	$n = w + F$	Includes extra downward force $F$
Normal Force (incline)	$n = mg \cos \theta$	Perpendicular to surface
Acceleration down incline	$a = g \sin \theta$	Neglecting friction

Additional info:

• All equations are vector equations; always resolve forces into components for analysis.

• Free-body diagrams are essential for visualizing forces and solving problems.

• Apparent weight is a key concept in elevator and free-fall scenarios.