Classical Mechanics · Chapter 3

Newton's Laws

Three laws, written in 1687The PrincipiaNewton's Philosophiæ Naturalis Principia Mathematica unified terrestrial motion and celestial motion with the same mechanics. That was the move: apples and planets became one subject., that governed all of physics for 230 years — and still govern most of engineering today.

PrerequisitesKinematics⋅Vectors⋅BasicalgebraKinematics \cdot Vectors \cdot Basic algebra
Learning Goals
  • State Newton's three laws and identify the object each force acts on.
  • Draw free-body diagrams that include every external force on one isolated object.
  • Resolveforcesintocomponentsandapply\sumF=maalongeachaxisResolve forces into components and apply \sumF = ma along each axis.
  • Distinguish static friction from kinetic friction and use the correct inequality or equation.
  • Solve connected-body and incline problems with consistent sign conventions.

3.1 The Three Laws

Theorem 3.1 — Newton's First Law — Inertia
An object at rest remains at rest, and an object in motion remains in uniform motion (constant velocity), unless acted upon by a net external force.

This is a statement about inertial reference frames: in the absence of forces, objects maintain their state of motion. Inertia is the resistance to change — and it scales with mass.
Theorem 3.2 — Newton's Second Law — Force and Acceleration
The net force on an object equals its mass times acceleration:Fnet=maor\sumF=maF_{net} = ma \qquad or \qquad \sumF = maMoreprecisely,Fnet=dp/dt(therateofchangeofmomentum).Forceisavector—directMore precisely, F_{net} = dp/dt (the rate of change of momentum). Force is a vector — direction matters. When multiple forces act, they add vectorially to give the net force.
Theorem 3.3 — Newton's Third Law — Action-Reaction
For every force exerted by object A on object B, there is an equal and opposite force exerted by B on A:F_{A on B} = -F_{B on A}Critical: these forces act on differentobjects. A horse pulling a cart exerts a force on the cart; the cart exerts an equal force back on the horse. They don't cancel (they're on different objects) — which is why the horse-cart system can still accelerate.

3.2 Free Body Diagrams

A free body diagram (FBD) is a sketch that shows a single object with all forces acting on it as labeled arrows. Drawing a correct FBD is the essential first step in any force problem.

Rules for drawing FBDs:

  1. Isolate the object — draw a dot or simple shape representing it.
  2. Identify every force acting on that object (not forces it exerts on other things).
  3. Draw each force as an arrow from the object, with length proportional to magnitude.
  4. Label each force with its type and magnitude if known.
  5. Choose a coordinate system and resolve forces into components.

3.3 Friction

Definition 3.1 — Static and Kinetic Friction
Friction is the contact force that opposes relative sliding between surfaces.
  • Static friction fs≤μsN:preventssliding.Adjuststomatchappliedforceuptoamaximumf_{s} \le \mu_s N: prevents sliding. Adjusts to match applied force up to a maximum
  • Kinetic friction fk=μkN:constantonceslidingbegins.Alwaysμk<μsf_{k} = \mu_k N: constant once sliding begins. Always \mu_k < \mu_s
fk=μkN=μkmg(onflatsurface)direction:opposesmotionf_{k} = \mu_k N = \mu_k mg (on flat surface) \qquad direction: opposes motion
a=Fapplied−μkmgm=Fappliedm−μkga = \frac{F_\mathrm{applied} - \mu_k mg}{m} = \frac{F_\mathrm{applied}}{m} - \mu_k g(3.1)

The block won't move at all if F_applied < μ_s N. Once moving, use μ_k. In the simulation, we model the transition: if |v| < 0.01 m/s and |F| < μN, acceleration is zero.

Mass5 kg
Applied Force15 N
Friction coefficient μ0.30
Force breakdown
Applied (F)15 N
Weight (W = mg)49.1 N ↓
Normal (N = mg)49.1 N ↑
Max static friction14.7 N
Net acceleration0.06 m/s²
Figure 3.1. Force diagram simulation. Forces are drawn proportionally to magnitude. The velocity bar (top right) shows speed and direction. Negative applied force reverses the block.
Example 3.1 — Block on an Incline

A5kgblocksitsona30°inclinewithμk=0.2.FindtheaccelerationdowntheslopeA 5 kg block sits on a 30° incline with \mu_k = 0.2. Find the acceleration down the slope.

Weight components:Alongslope:mgsin30°=5×9.81×0.5=24.5N(down).Normal:N=mgcos30°=5×9.81×0.866Along slope: mg sin30° = 5\times9.81\times0.5 = 24.5 N (down). Normal: N = mg cos30° = 5\times9.81\times0.866 = 42.5 N
Friction:fk=\muN=0.2×42.5=8.5N(uptheslope,opposingmotionf_{k} = \muN = 0.2\times42.5 = 8.5 N (up the slope, opposing motion
Net force:Fnet=24.5−8.5=16N(downslopeF_{net} = 24.5 - 8.5 = 16 N (down slope
Acceleration:a=Fnet/m=16/5=a = F_{net}/m = 16/5 = 3.2m/s23.2 m/s^{2} (down the incline)
Example 3.2 — Atwood Machine

Twomassesm1=3kgandm2=5kghangoverafrictionlesspulley.FindtheaccelerationTwo masses m_{1} = 3 kg and m_{2} = 5 kg hang over a frictionless pulley. Find the acceleration and tension.

System equation:Netforce=(m2−m1)g=(5−3)×9.81=19.62N.Totalmass=8kgNet force = (m_{2} - m_{1})g = (5-3)\times9.81 = 19.62 N. Total mass = 8 kg.
Acceleration:a=(m2−m1)g/(m1+m2)=19.62/8=a = (m_{2}-m_{1})g/(m_{1}+m_{2}) = 19.62/8 = 2.45m/s22.45 m/s^{2}
Tension:Fromm1:T−m1g=m1a→T=m1(g+a)=3×12.26=From m_{1}: T - m_{1}g = m_{1}a \to T = m_{1}(g+a) = 3\times12.26 = 36.8 N
Verify from m₂:m2g−T=m2a→T=m2(g−a)=5×7.36=36.8N✓m_{2}g - T = m_{2}a \to T = m_{2}(g-a) = 5\times7.36 = 36.8 N ✓
Definition 3.2 — Common Traps
  • Third-law pairs do not cancel: they act on different objects.
  • Normal force is not always mg: inclines, elevators, and added vertical forces change it.
  • Static friction is an inequality: usefs≤μsNuntilslidingbeginsuse f_{s} \le \mu_sN until sliding begins
  • Choose axes intelligently: for inclines, align one axis along the slope.
  • Tension is shared only under assumptions: massless rope and frictionless pulley make the tension uniform.
Exercises — 3.1–3.3 Newton's Laws
1.
A1,200kgcaracceleratesat3m/s2.Rollingfrictioncoefficientis0.02.WhatforcemusA 1,200 kg car accelerates at 3 m/s^{2}. Rolling friction coefficient is 0.02. What force must the engine provide?
N
Straightforward
2.
An80kgblocksitsonasurfacewithμs=0.4.A200Nforceisappliedhorizontally.DoAn 80 kg block sits on a surface with \mu_s = 0.4. A 200 N force is applied horizontally. Does the block move? If not, what is the friction force?
N
Straightforward
3.Abox(μs=0.35)juststartstoslidewhenaninclineistiltedtoangleθ.Findθ.InteA box (\mu_s = 0.35) just starts to slide when an incline is tilted to angle \theta. Find \theta. Interestingly, the answer is independent of mass — explain why.
Intermediate
4.A 5 kg block sits on a frictionless table, connected by a rope over a frictionless pulley to a hanging 3 kg mass. Find the acceleration and rope tension.
Intermediate
5.A block slides at constant velocity down a 25° incline. Derive the kinetic friction coefficient \mu_ fromtheangle.(Hint:drawtheFBDandapplyNewton′s2ndlawwitha=0from the angle. (Hint: draw the FBD and apply Newton's 2nd law with a = 0
Challenging
Key Takeaways
  • Newton's 1st: Objects resist changes in motion (inertia). Force is needed to change velocity.
  • Newton′s2nd:Fnet=maisthequantitativelinkbetweenforceandmotionNewton's 2nd: F_{net} = ma is the quantitative link between force and motion.
  • Newton's 3rd: Forces come in pairs — but they act on different objects, so they don't cancel.
  • Free body diagrams: isolate one object, draw all forces acting ON it, resolve components.
  • Staticfriction≤μsNpreventsmotion;kineticfriction=μkN(always<μs)resistssStatic friction \le \mu_s N prevents motion; kinetic friction = \mu_k N (always < \mu_s) resists sliding.