Chapter 5 – Newton’s Laws of Motion (Comprehensive Study Notes)

Introductory Context

Newton's three laws of motion form the backbone of classical mechanics. While historically they emerged in a particular order, here they are presented in the way most convenient for problem–solving. The chapter begins by accepting that we already know how to:
• assign mass to an object,
• measure a force’s magnitude & direction, and
• determine acceleration relative to a chosen reference frame.

Newton’s First Law (Law of Inertia)

If and only if the vector sum of all external forces on a particle is zero, the particle has zero acceleration; it will either remain at rest or continue with constant velocity.
$\vec a = 0 \;\Leftrightarrow\; \sum \vec F = 0$

➔ Frame–dependence: All statements about rest, motion, or acceleration are meaningful only after specifying a reference frame. Consequently, the first law does not hold indiscriminately in every frame.

Elevator-Cabin Thought Experiment

• Situation: An elevator cable snaps, causing cabin + contents to fall with $a \approx 9.8\;\text{m\,s}^{-2}$ downward.
• Observer A (inside cabin – a falling frame) sees the lamp at rest ⇒ $\vec a<em>{\text{lamp}} = 0$, concludes $W=T$. • Observer B (on Earth) measures $\vec a</em>{\text{lamp}} = 9.8\,\text{m\,s}^{-2}$ downward, concludes $W \neq T$.

At least one frame must be “bad.” Frames where the first-law statement works are called inertial; those where it fails are non-inertial.

Practical Note on Earth as a Frame

Earth is not perfectly inertial (because of its rotation & orbital motion). However, for most everyday measurements the net non-inertial effects are negligible, so Earth can be treated as an inertial frame to a good approximation.

Example 5.1 – Static Hanging Mass

Data: $m = 0.50\,\text{kg},\; g = 9.8\,\text{m\,s}^{-2}$.
Forces on the mass (Earth frame):
• Weight $W = mg = 4.9\,\text{N}$ downward.
• Tension $T$ upward.
Since $\vec a = 0$, $T = 4.9\,\text{N}$.

Inertial Frames Beyond Earth

If frame $S$ is inertial and frame $S'$ moves at constant velocity relative to $S$, then $S'$ is also inertial. Mathematically, with
$\vec a<em>{P,S} = \vec a</em>{P,S'}$ whenever $\vec a_{S',S}=0,$
we preserve “$\vec a=0 \Leftrightarrow \sum \vec F=0$.” Everyday examples: a smoothly cruising train, an airplane in level flight, or a ship gliding on calm sea.

Newton’s Second Law (Quantitative Law of Dynamics)

In any inertial frame, acceleration of a particle equals the total external force divided by its mass:
$\vec a = \frac{\sum \vec F}{m} \qquad\text{or}\qquad \sum \vec F = m\vec a.$
The force and acceleration in Eq. (5.2) are evaluated at the same instant.

Working Systematically with 1st & 2nd Laws

A four-step algorithm is advocated:

1. Decide the system – choose an object or a set of objects that share one acceleration vector.

2. Identify external forces – list only forces exerted by bodies outside the chosen system.

3. Draw a Free-Body Diagram (FBD) – represent the system as a dot and attach all force vectors.

4. Choose axes & write component equations – preferably align one axis with expected acceleration.

Subtlety with Composite Objects (Fig. 5.4)

Blocks A & B move together, block C hangs by a string, a disc D slides on the string, and point G is on another block. Although C and G cover equal horizontal distances in equal times, E moves vertically, not horizontally ⇒ (E+G) is not an acceptable single-acceleration system.

Force-Identification Example (Boy Holding a Load)

Depending on whether you choose “boy” or “load” as the system, different forces must be listed. Only forces on the system count; action–reaction counterparts acting on some other body must be omitted from that particular FBD.

Example 5.2 – Block Pulled by an Angled String

Given: Smooth horizontal table, tension makes angle $\theta$ above horizontal, acceleration $a$.
FBD yields components:
• $\sum F<em>x = T\cos\theta = Ma \Rightarrow T = \frac{Ma}{\cos\theta}.$ • $\sum F</em>y = N + T\sin\theta - Mg = 0 \Rightarrow N = Mg - Ma\tan\theta.$

Newton’s Third Law (Action–Reaction)

If body A exerts $\vec F$ on body B, then B exerts $-\vec F$ on A. The two forces:
• act along the same line,
• are equal in magnitude and opposite in direction,
• act on different bodies, so they never appear together on the same FBD.

⚠ Caveat: At very large separations or when electromagnetic retardation is significant, the simple equal-and-opposite statement needs refinement.

Tension in Strings & Massless Approximation

For a light (≈ massless) string, tension is uniform along its length unless an intermediate massive particle is attached.

Example 5.3 – String of Mass m Below Point A

Lower-end tension (string on block): $T = Mg.$
Tension at A (upper part pulling lower part): $T' = T + mg = (M+m)g.$

Example 5.4 – String with Moving Block (Acceleration a)

Block M descends with $a$, string mass below A is $m$. Treating “block + lower string” as one system:
$T = (M+m)(g-a).$
If the string were massless, setting $m\to 0$ gives the lower-end tension $T'' = M(g-a).$

In compound setups (Fig. 5.12), each span separated by a pulley may possess its own constant tension $T<em>1,\,T</em>2,\,T<em>3,\ldots$. Light, frictionless pulleys equalize tensions on their two sides (e.g. $T</em>1=T_2$ if pulley B has negligible mass).

Pseudo (Inertial) Forces in Non-Inertial Frames

Suppose non-inertial frame $S'$ accelerates by $\vec a<em>0$ relative to inertial frame $S$. For a mass $m$, Newton’s 2nd law in $S$ is $m\vec a</em>{S} = \sum \vec F.$
Acceleration seen from $S'$ is $\vec a<em>{S'} = \vec a</em>{S} - \vec a<em>0.$ Rewriting: $m\vec a</em>{S'} = \sum \vec F - m\vec a<em>0.$ To keep the familiar “$\sum F = m a$” inside the accelerating frame, introduce a pseudo force $\boxed{\vec F</em>{\text{pseudo}} = - m\vec a_0}.$
It acts on every mass m, opposite to the frame’s acceleration.

Example 5.5 – Car-Pendulum

A car accelerates horizontally with magnitude $a<em>0$. Pendulum bob (mass $m$) is stationary relative to car; real forces: tension $T$, weight $mg$. Add pseudo force $m a</em>0$ opposite to car’s acceleration. Balance forces in components:
$T\sin\theta = m a<em>0, \qquad T\cos\theta = m g \;\Rightarrow\; \tan\theta = \frac{a</em>0}{g}.$
Hence the string deflects by $\theta = \tan^{-1}(a_0/g)$ from the vertical.

Horse-and-Cart Paradox Clarified

FBD for cart alone: forward pull $F<em>1$ by the shaft, backward friction (road on wheels) $f'$ ⇒ acceleration $a = (F</em>1-f')/M<em>c$. FBD for horse alone: backward pull $F</em>2$ from cart, forward ground-reaction (friction) $f$ ⇒ acceleration $a = (f-F<em>2)/M</em>h$.
Action–reaction pair $F<em>1=F</em>2$ cancel between different bodies, not within one equation. Properly accounting for ground friction resolves the apparent contradiction: net external force on the coupled horse–cart system comes from the ground on the horse/wheels, allowing forward acceleration.

Inertia Revisited

A body’s resistance to changes in its state of motion is called inertia. Quantitatively, if equal forces are applied to two particles, the one with larger mass experiences the smaller acceleration, reflecting greater inertia.

Mathematically: for equal $\sum \vec F$,
$a<em>1 : a</em>2 = \frac{1}{m<em>1} : \frac{1}{m</em>2}.$
Hence mass is a direct measure of inertia.

Ethical & Practical Implications

• Engineers exploit approximate inertial nature of Earth for everyday design (bridges, vehicles, buildings).
• Understanding pseudo forces is crucial for designing accelerometers, aircraft instruments, and amusement-park rides.
• Clarifying internal vs external forces (horse–cart) prevents fallacious energy or momentum arguments in popular debates.

Recurring Problem-Solving Themes

1. Always specify the system first.

2. Draw an accurate FBD free of extraneous action–reaction forces.

3. Identify whether the working frame is inertial; if not, add pseudo forces.

4. Align axes with expected accelerations to decouple equations.

5. When multiple objects share constraints (ropes, pulleys), link their accelerations algebraically before applying Newton’s laws.