Newton's Laws and Free Body Diagrams

Three laws of motion; free body diagrams; equilibrium; resolving forces

# Newton's Laws and Free Body Diagrams — IB Physics

Newton's three laws form the foundation of classical mechanics. In IB Physics, you must be able to draw free body diagrams and apply Newton's laws to solve problems.


1. Newton's Laws

First Law (Inertia)

An object remains at rest or moves at constant velocity unless acted upon by a resultant (net) force.

Second Law

Fnet=ma\boxed{F_{\text{net}} = ma}

The net force on an object equals its mass times its acceleration. Direction of FnetF_{\text{net}} = direction of aa.

Third Law

When body A exerts a force on body B, body B exerts an equal and opposite force on body A.

Third law pairs:

  • Same type of force
  • Act on different bodies
  • Equal magnitude, opposite direction

2. Free Body Diagrams

Rules:

  1. Draw the object as a point or simple shape
  2. Draw all forces acting on that object (not forces the object exerts on others)
  3. Label each force with name and magnitude if known
  4. Arrows show direction; length represents magnitude

Common Forces

  • Weight (W=mgW = mg, downward)
  • Normal reaction (NN, perpendicular to surface)
  • Friction (ff, opposing motion/tendency to move)
  • Tension (TT, along string/rope)
  • Air resistance / drag (opposing motion)
  • Applied / driving force

3. Equilibrium

An object is in translational equilibrium when Fnet=0F_{\text{net}} = 0.

  • Static equilibrium: at rest
  • Dynamic equilibrium: moving at constant velocity

For equilibrium: sum of forces = 0 in every direction.

4. Resolving Forces on an Incline

For an object on a slope at angle θ\theta:

  • Component parallel to slope: mgsinθmg\sin\theta
  • Component perpendicular to slope: mgcosθmg\cos\theta
  • Normal reaction: N=mgcosθN = mg\cos\theta

Worked Example: Lift Problem

Problem

A 70 kg person stands on scales in a lift accelerating upward at 2 m/s².

Nmg=maN - mg = ma N=m(g+a)=70(9.81+2)=827N = m(g + a) = 70(9.81 + 2) = 827 N

The scales read 827 N (apparent weight increases).

Solution

Worked Example: Two Connected Masses

Problem

A 5 kg block on a frictionless table is connected by a string over a pulley to a 3 kg hanging mass.

System: Fnet=3g=29.4F_{\text{net}} = 3g = 29.4 N, total mass = 8 kg a=29.4/8=3.68a = 29.4/8 = 3.68 m/s² T=5a=5(3.68)=18.4T = 5a = 5(3.68) = 18.4 N (or T=3g3a=29.411.0=18.4T = 3g - 3a = 29.4 - 11.0 = 18.4 N ✓)

Solution

Worked Example: Inclined Plane

Problem

A 10 kg box is on a 30° incline. μ=0.3\mu = 0.3. Find the acceleration.

F=mgsin30°=49.1F_{\parallel} = mg\sin30° = 49.1 N N=mgcos30°=84.9N = mg\cos30° = 84.9 N f=μN=0.3(84.9)=25.5f = \mu N = 0.3(84.9) = 25.5 N Fnet=49.125.5=23.6F_{\text{net}} = 49.1 - 25.5 = 23.6 N a=23.6/10=2.36a = 23.6/10 = 2.36 m/s²

Solution

6. Practice Questions

    1. State Newton's three laws of motion. (3 marks)
    1. A 1200 kg car accelerates at 2.5 m/s² against a resistive force of 400 N. Find the driving force. (3 marks)
    1. Draw a free body diagram for a ball in free fall (with air resistance). (2 marks)
    1. Give an example of a Newton's third law pair and explain why they don't cancel out. (3 marks)

    Answers

    1. N1: Object remains at rest/constant velocity unless net force acts. N2: F=maF = ma. N3: Equal and opposite force pairs between interacting bodies.

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Summary

  • N1: no net force → constant velocity; N2: F=maF = ma; N3: equal and opposite on different bodies
  • Free body diagrams show all forces on ONE object
  • Equilibrium: Fnet=0F_{\text{net}} = 0
  • On inclines: resolve into parallel (mgsinθmg\sin\theta) and perpendicular (mgcosθmg\cos\theta)

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