A2-1. Newton's Laws of Motion

Vector Nature of Forces

Force is a vector quantity, meaning it has both magnitude and direction. Forces can be categorized into contact forces (e.g., friction, tension, applied push/pull) and action-at-a-distance (field) forces (e.g., gravity, magnetic force).

Resolving Forces: Any force $\vec{F}$ acting at an angle $\theta$ to the horizontal can be broken down into perpendicular components:

Horizontal Component: $F_x = F \cos\theta$
Vertical Component: $F_y = F \sin\theta$

Resultant Force: When multiple forces act on a single point, their combined effect is the resultant force. By summing the components ($F_{net,x}$ and $F_{net,y}$), the magnitude and angle of the resultant are:

$$F = \sqrt{{F_{x}}^2 + {F_{y}}^2} \quad \text{and} \quad \tan\theta = \dfrac{F_y}{F_x}$$

Newton's First Law (Law of Inertia)

Originally proposed by Galileo as the principle of inertia and refined by Descartes (who added that direction is also maintained), Newton formalized this into his First Law of Motion.

Newton's First Law: An object will remain at rest or continue to move with a constant velocity (constant speed in a straight line) unless acted upon by a resultant external force.

Inertia: The inherent property of matter to resist any change to its state of motion. Mass is the quantitative measure of inertia.
Equilibrium: If the net force $\Sigma F = 0$, acceleration $a = 0$.

Common Phenomena of Inertia

Car suddenly starts: Passengers feel pushed backward because their bodies try to maintain their initial state of rest (static inertia).
Airplane taking off: Passengers feel pressed into their seats due to static inertia.
Pulling a tablecloth quickly: The cups and plates remain on the table because they resist the sudden change in motion.
Car suddenly brakes: Passengers lurch forward because their bodies try to maintain their forward velocity (dynamic inertia).

Newton's Second Law

This law dictates exactly how an object's motion changes when a net force is applied.

Newton's Second Law: The acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. The acceleration is always in the same direction as the net force.

$$\vec{F}_{net} = m\vec{a}$$

The Newton (N): 1 N is defined as the force required to give a 1 kg mass an acceleration of $1\text{ m/s}^2$.

4 Steps to Solving Newton's Law Problems

Select the Object: Treat the object or system of interest as a point mass.
Draw the Free-Body Diagram (FBD): Identify all external forces acting on the object (e.g., gravity, normal force, applied force, friction).
Set up Coordinate Axes: Choose perpendicular x and y axes. Usually, align one axis with the direction of acceleration to simplify calculations. Resolve forces into these components.
Apply Newton's Laws: Use $\Sigma F_x = ma_x$ and $\Sigma F_y = ma_y$ to solve for the unknowns.

Common Forces: Gravity and Normal Force

Before analyzing complex motion, it is essential to understand two omnipresent forces:

Gravity (Weight, $W$ or $mg$): The universal force of attraction acting on an object due to Earth's mass. It always points straight down toward the center of the Earth. The magnitude is calculated as $W = mg$.
Normal Force ($N$ or $F_N$): The contact force exerted by a surface on an object. It always acts perpendicular (at a 90° angle) to the surface. It is the force that prevents the object from falling through the surface.

Forces on an Inclined Plane: When an object rests on a sloped surface, gravity no longer aligns perfectly with the normal force. We must resolve the gravitational force ($mg$) into components parallel and perpendicular to the incline:

Parallel Component ($mg \sin\theta$): Pulls the object down the slope.
Perpendicular Component ($mg \cos\theta$): Pushes the object into the surface (exactly balancing the Normal force, $N$, if there is no vertical acceleration).

Example 1: Block on an Incline

Problem: A 2.0 kg metal block is placed on a smooth, frictionless incline angled at 37° to the horizontal. A pushing force of 20 N is applied to the block straight up along the incline. Determine the block's acceleration. ($g = 10\text{ m/s}^2$)

Solution:

a) Resolve Gravity:
The weight of the block is $W = mg = 2.0 \times 10 = 20\text{ N}$.
The component of weight pulling it down the incline is $W_x = mg \sin(37^\circ) = 20 \times 0.6 = 12\text{ N}$.
b) Apply Newton's Second Law along the incline:
Taking 'up the incline' as the positive direction:
$$\Sigma F_x = F_{push} - W_x = 20 - 12 = 8\text{ N}$$
$$a = \dfrac{\Sigma F_x}{m} = \dfrac{8}{2.0} = \mathbf{4.0\text{ m/s}^2}$$ (directed up the incline)

Newton's Third Law (Action and Reaction)

Forces never exist in isolation; they are always mutual interactions between two objects.

Newton's Third Law: If object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A.

$$\vec{F}_{AB} = -\vec{F}_{BA}$$

Characteristics of Action-Reaction Pairs:

Equal and Opposite: The magnitude is exactly the same, but the direction is exactly opposite.
Different Objects: They act on two different bodies, so they do not cancel each other out when analyzing a single body's free-body diagram.
Same Nature: If the action is a gravitational pull, the reaction is a gravitational pull. If the action is a contact push, the reaction is a contact push.
Simultaneous: Action and reaction forces are generated and disappear at the exact same time.

Example 2: Atwood Machine

Problem: Two objects, Mass A ($m_A = 1.0\text{ kg}$) and Mass B ($m_B = 4.0\text{ kg}$), are connected by a massless string over a frictionless pulley. If released from rest, how far does Mass B move after 1.0 s? (Assume $g = 10\text{ m/s}^2$)