Newton’s Laws of Motion & Applications – Comprehensive Lecture 3 Notes

MECHANICS OVERVIEW

• Mechanics: Branch of physics focused on relationships among force, matter, and motion.- Sub-branches:

  • KINEMATICS – purely descriptive; deals with displacement, velocity, acceleration for scenarios such as

    • Motion along a straight line

    • Motion with constant acceleration

    • Freely-falling bodies

    • Projectile motion

  • DYNAMICS – explanatory; studies how forces cause motion.- Modern foundation laid by Sir Isaac Newton

    • > Newton’s three laws of motion.

FORCES: DEFINITION AND TYPES

• Force: A push or pull; an interaction between two objects or between an object and its environment.- Vector quantity (has magnitude & direction).

CONTACT FORCES

• Require physical touch between bodies.- Normal force $n$: Surface pushes perpendicular to itself whenever an object rests or pushes on it.

  • Friction force $f$: Surface exerts a parallel force that opposes impending or actual motion.

  • Tension force $T$: Pulling force transmitted by ropes, cords, cables, etc.

LONG-RANGE FORCES

• Act even through empty space.- Weight $w$: Gravitational pull on an object; direction toward Earth’s center.

• SI unit of any force: $1\;\text{N} = 1\;\text{kg}\cdot\text{m}/\text{s}^2$

SUPERPOSITION OF FORCES & MASS

• Net force: Vector sum of all forces on a body.- $\vec F<em>{net} = \sum \vec F = \vec F</em>1 + \vec F<em>2 + \vec F</em>3 + \cdots$

• Mass (scalar):- Qualitative: "Amount of matter" in an object.

  • Quantitative: Measure of inertia (resistance to acceleration).

  • Larger mass (\to) greater inertia (\to) smaller acceleration for the same net force.

NEWTON’S LAWS OF MOTION

FIRST LAW – LAW OF INERTIA

• Statement: "A body at rest remains at rest and a body in motion continues motion with constant velocity in a straight line unless acted on by an external unbalanced force."

• Equilibrium criterion: $\sum \vec F = 0$ (\to) body either at rest or moves at constant velocity.

SECOND LAW – LAW OF ACCELERATION

• Unbalanced/net force causes acceleration.

• Quantitative form: $\sum \vec F = m\vec a$- Component form: $\sum F<em>x = ma</em>x,\; \sum F<em>y = ma</em>y,\; \sum F<em>z = ma</em>z$

• Acceleration direction matches net force direction; magnitude proportional to $\Vert\vec F_{net}\Vert$, inversely proportional to $m$.

THIRD LAW – ACTION–REACTION

• If object A exerts force on B, B exerts equal-magnitude, opposite-direction force on A.- $\vec F<em>{A\,on\,B} = -\vec F</em>{B\,on\,A}$

• Forces always occur in pairs; isolated single force cannot exist.

PROBLEM-SOLVING STRATEGY (NEWTON’S 2ND LAW)

1. Draw a simple diagram of the physical setup.

2. Isolate the object of interest; create a free-body diagram (FBD) showing all external forces on that object only.- For multi-object systems, draw separate FBD for each.

3. Choose convenient coordinate axes; resolve each force into components.

4. Apply $\sum F = ma$ along each axis; solve simultaneous equations for unknowns (accelerations, tensions, masses, etc.). Must have as many independent equations as unknowns.

FRICTION

• General property: opposes relative motion (or attempted motion) between surfaces in contact; direction parallel to surface.

STATIC FRICTION (BEFORE SLIDING)

• Exists when object does not move despite applied force.

• Magnitude range: $0 \le f<em>s \le f</em>{s\,max} = \mu_s n$

• $\mu_s$ = coefficient of static friction; depends on surface pair.

KINETIC FRICTION (DURING SLIDING)

• Object slides steadily; opposing force magnitude:- $f<em>k = \mu</em>k n$

• \muk < \mus for same materials (less force needed to keep sliding than to start it).

GRAPHICAL REPRESENTATION (FIG. 5.19 INSIGHTS)

• Stage 0: No pull (\to) $f_s = 0$

• Stage 1: Small pull $T$ < $f<em>{s\,max}$ (\to) box still at rest, $f</em>s = T$.

• Stage 2: Pull increases to $f_{s\,max}$ (\to) imminent motion.

• Stage 3: Box starts sliding (\to) abrupt drop to kinetic region where $f_k$ nearly constant though small variations occur as molecular bonds form/break.

COEFFICIENTS OF FRICTION (SELECTED VALUES – TABLE 5.1)

• Steel/steel: $\mu<em>s \approx 0.74,\; \mu</em>k \approx 0.57$

• Aluminum/steel: $\mu<em>s 0.61,\; \mu</em>k 0.47$

• Glass/glass: $\mu<em>s 0.94,\; \mu</em>k 0.40$

• Teflon/steel or Teflon/Teflon: both $\mu<em>s = \mu</em>k = 0.04$ (extremely low friction; engineering relevance for non-stick surfaces).

• Rubber on dry concrete: $\mu<em>s 1.0,\; \mu</em>k 0.8$ (reason car tires grip roads well).

PRACTICAL IMPLICATIONS & CONNECTIONS

• Engineering: Determining tensions and friction crucial for bridge cables, elevators, conveyors, automobile braking.

• Safety: Understanding maximum static friction prevents slippage in ladders, tires, footwear.

• Philosophical/Scientific Impact: Newton’s laws unified celestial and terrestrial mechanics, shifting worldview from Aristotelian impetus theory to universal law-based predictability.

• Ethical Usage: Responsible application of mechanics principles essential in construction standards, transport systems, amusement-park rides.

KEY FORMULA SUMMARY (QUICK REFERENCE)

WORKED EXAMPLES

Example 1 – Unbalanced Horizontal Force

Problem: Calculate the acceleration of an object given a net force and its weight.

• Given: Net force $F = 50\;\text{N}$ on object weighing $W = 100\;\text{N}$.

• Mass: $m = W/g = 100/9.81 \approx 10.2\;\text{kg}$ (approx $10\,kg$ if $g=9.8$).

• Acceleration: $a = F/m \approx 50 / 10.2 \approx 4.9\,\text{m/s}^2$

Example 2 – Body on Smooth Horizontal Surface

Problem: (a) Determine the mass of an object, and (b) calculate how much further it travels after the force is removed.
Given:

• Constant force $F = 40\,\text{N}$, displacement $s = 100\,\text{m}$ in $t = 5\,\text{s}$ , starts from rest $(u=0)$.

  (a) Mass

• Kinematics: $s = ut + \tfrac12 a t^2 \Rightarrow 100 = 0 + 0.5 a (5)^2 \Rightarrow a = 8\,\text{m/s}^2$.

• $m = F/a = 40/8 = 5\,\text{kg}$.

  (b) Further travel after force removed

• At $t=5\,\text{s}$, velocity $v = u + at = 0 + 8(5) = 40\,\text{m/s}$.

• With no force (neglecting friction) (\to) no acceleration; hence constant velocity.

• Distance in next $5\,\text{s}$: $s = vt = 40(5) = 200\,\text{m}$

Example 3 – Light Pulley, Masses on Table & Hanging

Problem: (a) Find the acceleration of the system, and (b) determine the tension in the cord.

• Mass on table: $m_1 = 100\,\text{g} = 0.10\,\text{kg}$ (frictionless)

• Hanging mass: $m<em>2 = 10\,\text{g} = 0.010\,\text{kg}$ (a) Acceleration:- Net external force = weight of hanging mass $m</em>2 g$.

  • Total mass accelerated = $m<em>1 + m</em>2$.

  • $a = \dfrac{m<em>2 g}{m</em>1 + m_2} = \dfrac{0.010 g}{0.110} \approx 0.89\,\text{m/s}^2$.

    (b) Tension $T$ in cord:

  • Apply $\sum F = m a$ to hanging mass: $m<em>2 g - T = m</em>2 a \Rightarrow T = m_2 (g - a) \approx 0.010(9.81-0.89) \approx 0.089\,\text{N}$.

Example 4 – Two Inclined Planes, Masses Connected

Problem: Analyze the motion of two connected masses on inclined planes to determine acceleration and tension.

• Weights: $W<em>1 = 8\,\text{N},\; W</em>2 = 10\,\text{N}$.

• Angles shown (not numerically specified in transcript) — analysis typical:- Resolve each weight parallel to its plane; set directions; write coupled equations; solve for acceleration $a$ and tension $T$.

• Key takeaway: even different slopes, the single tension equilibrates along string.

Example 5 – Traffic-Light Three-Cable System

Problem: Determine the tensions in the cables supporting a traffic light in static equilibrium.

• Weight of light: $W = 100\,\text{N}$.

• Geometry (figure): central vertical cable supports weight; two side cables at angles to support absorb part of load.

• For static equilibrium, sum forces zero.- Vertical components of side tensions + central tension = $100\,\text{N}$.

• Horizontal components cancel each other (symmetry if angles equal).

• Solve simultaneous equations to find each tension; shows how city engineers design cable strengths.

Example 6 – Finding Friction Coefficients (Horizontal Pull)

Problem: Calculate the coefficients of static and kinetic friction for a box on a surface.
Given:

• Box weight $W = 100\,\text{N}$.

• Start of motion: $F<em>{start} = 25\,\text{N}$ (\Rightarrow) $f</em>{s\,max}=25\,\text{N}$.

  • $\mu<em>s = f</em>{s\,max}/n = 25/100 = 0.25$

• Uniform motion: $F<em>{uniform} = 20\,\text{N} = f</em>k$.- $\mu<em>k = f</em>k/n = 20/100 = 0.20$

(b) Pull at $30^\circ$ above horizontal

• Normal force reduced by vertical component of applied force.

• Recompute $n$, observe new limiting static and kinetic forces, then reverse solve for $\mu<em>s$ and $\mu</em>k$ (exercise left for practice).

Example 7 – Up-Slope Pull with Oblique Force & Friction

Problem: Determine the force $F$ required to pull an object up an inclined plane at a constant velocity, considering an oblique force and kinetic friction.

• Data: $W = 50\,\text{N}$; plane dims 4 ft (\times) 3 ft (\to) $\theta = \tan^{-1}(3/4) \approx 37^\circ$.

• Pull angle: $30^\circ$ above plane; $\mu_k = 0.25$.

• Equilibrium (uniform motion) equation resolved along plane:

  $\; F \cos 30^\circ + \mu<em>k F \sin 30^\circ = W \sin 37^\circ + \mu</em>k W \cos 37^\circ$

• Solving yields $F \approx 40.4\,\text{N}$ (shown algebraically in slide).

Seatwork #2 – Problem 5.74 (Challenge)

Problem: Given an inclined-plane block and a hanging mass connected by a cord, determine the hanging mass required for the system to descend a specific distance within a given time, accounting for friction.

• Inclined-plane block $m<em>1 = 20.0\,\text{kg}$ on $\theta = 53.1^\circ$ slope; coefficient $\mu</em>k = 0.40$.

• Hanging mass $m_2$ must descend $12.0\,\text{m}$ in first $3.00\,\text{s}$ after release from rest.- Required acceleration from kinematics: $s = \tfrac12 a t^2 \Rightarrow a = 2s/t^2 = 2(12)/9 = 2.67\,\text{m/s}^2$.

  • Write force equations for each mass (include friction on $m<em>1$) and solve for unknown $m</em>2$:

    $m<em>2 g - T = m</em>2 a$

    $T - m<em>1 g \sin\theta - \mu</em>k m<em>1 g \cos\theta = m</em>1 a$

  • Eliminate $T$, solve (\to) students expected to compute $m_2$ (\approx) $??\,\text{kg}$ (numerical result when solved gives (\approx) 10 kg; verify independently).