Newton's First and Second Laws

Inertia, net force, and F = ma

Newton's First and Second Laws

Newton's First Law (Law of Inertia)

Statement: An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted upon by a net external force.

If F=0, then v=constant\text{If } \sum \vec{F} = 0, \text{ then } \vec{v} = \text{constant}

Key Concepts

Inertia is the tendency of an object to resist changes in its motion.

  • More massive objects have more inertia
  • Inertia is measured by mass (kg)

Equilibrium occurs when the net force is zero.

  • Static equilibrium: object at rest (v=0v = 0)
  • Dynamic equilibrium: object moving at constant velocity (v=constantv = \text{constant}, a=0a = 0)

Common Misconceptions

Wrong: "A force is needed to keep an object moving" ✓ Correct: A force is only needed to change motion (accelerate)

Wrong: "If velocity is zero, force must be zero" ✓ Correct: Multiple forces can balance to give zero net force

Newton's Second Law

Statement: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

F=ma\sum \vec{F} = m\vec{a}

Or in component form: Fx=max\sum F_x = ma_x Fy=may\sum F_y = ma_y

Understanding the Equation

Net force (F\sum \vec{F}): Vector sum of all forces

  • Measured in Newtons (N)
  • 1 N=1 kg\cdotpm/s21 \text{ N} = 1 \text{ kg·m/s}^2

Mass (mm): Measure of inertia

  • Measured in kilograms (kg)
  • Scalar quantity (no direction)
  • Not the same as weight!

Acceleration (a\vec{a}): Rate of change of velocity

  • Measured in m/s²
  • Vector quantity (same direction as net force)
  • If F=0\sum \vec{F} = 0, then a=0a = 0

Key Relationships

  1. Direct relationship with force: If you double the force, you double the acceleration (same mass) aF (when m is constant)a \propto F \text{ (when } m \text{ is constant)}

  2. Inverse relationship with mass: If you double the mass, you halve the acceleration (same force) a1m (when F is constant)a \propto \frac{1}{m} \text{ (when } F \text{ is constant)}

  3. Direction: Acceleration is always in the same direction as the net force

Mass vs. Weight

Mass

  • Mass is the amount of matter in an object
  • Scalar quantity
  • Measured in kg
  • Same everywhere in the universe
  • Measure of inertia

Weight

  • Weight is the gravitational force on an object
  • Vector quantity (points toward Earth's center)
  • Measured in Newtons (N)
  • Depends on location (gravity varies)
  • Formula: W=mgW = mg

On Earth's surface: W=mgwhere g=9.8 m/s2W = mg \quad \text{where } g = 9.8 \text{ m/s}^2

On the Moon: gmoon1.6g_{moon} \approx 1.6 m/s² (about 1/6 of Earth)

  • Same mass, but weight is 1/6 as much!

Free Body Diagrams (FBDs)

A free body diagram shows all forces acting on a single object.

Steps to Draw an FBD

  1. Isolate the object (draw a dot or box)
  2. Draw all forces as arrows starting from the object
    • Length represents magnitude
    • Direction shows force direction
  3. Label each force (e.g., FNF_N, FgF_g, FTF_T)
  4. Do NOT include:
    • Motion (velocity, acceleration)
    • Forces the object exerts on other things

Common Forces

  • Weight (Fg\vec{F}_g or W\vec{W}): Always points downward, magnitude mgmg
  • Normal force (FN\vec{F}_N or N\vec{N}): Perpendicular to surface, pushes away from surface
  • Tension (FT\vec{F}_T or T\vec{T}): Along rope/string, pulls toward rope
  • Friction (f\vec{f}): Parallel to surface, opposes motion/attempted motion
  • Applied force (Fapp\vec{F}_{app}): Push or pull from external agent

Problem-Solving Strategy

  1. Draw a free body diagram
  2. Choose coordinate system (align with motion when possible)
  3. Write Fx=max\sum F_x = ma_x and Fy=may\sum F_y = ma_y
  4. List all forces in each direction (use + and - signs)
  5. Solve for unknowns
  6. Check units and reasonableness

Newton's Laws Summary

| Law | Statement | Equation | |-----|-----------|----------| | First | Object maintains velocity unless net force acts | F=0v=const\sum \vec{F} = 0 \Rightarrow \vec{v} = \text{const} | | Second | Net force causes acceleration | F=ma\sum \vec{F} = m\vec{a} | | Third | Forces come in equal/opposite pairs | FAB=FBA\vec{F}_{AB} = -\vec{F}_{BA} |

Special Cases

Zero Acceleration

If a=0a = 0, then F=0\sum F = 0 (equilibrium)

  • Object at rest OR moving at constant velocity
  • All forces balance

Constant Acceleration

If a=constanta = \text{constant}, then F=constant\sum F = \text{constant}

  • Can use kinematic equations
  • Common in free fall, inclined planes

Variable Acceleration

If aa changes, then F\sum F changes

  • More complex analysis required
  • Need calculus or numerical methods

📚 Practice Problems

1Problem 1medium

Question:

A 5.0 kg box is pushed across a frictionless floor with a force of 20 N. (a) What is the acceleration of the box? (b) If the box starts from rest, what is its velocity after 3.0 seconds? (c) How far has it traveled in this time?

💡 Show Solution

Solution:

Given: m = 5.0 kg, F = 20 N, friction = 0

(a) Acceleration: Using Newton's 2nd Law: F_net = ma 20 = 5.0a a = 4.0 m/s²

(b) Velocity after 3.0 s: Using v = v₀ + at v = 0 + 4.0(3.0) v = 12 m/s

(c) Distance traveled: Using Δx = v₀t + ½at² Δx = 0 + ½(4.0)(3.0)² Δx = 2.0(9.0) Δx = 18 m

2Problem 2easy

Question:

A 55 kg box experiences a net force of 2020 N to the right. What is its acceleration?

💡 Show Solution

Given:

  • Mass: m=5m = 5 kg
  • Net force: F=20\sum F = 20 N (to the right)

Find: Acceleration aa

Use Newton's Second Law: F=ma\sum F = ma

Solve for acceleration: a=Fm=205=4 m/s2a = \frac{\sum F}{m} = \frac{20}{5} = 4 \text{ m/s}^2

Direction: To the right (same as net force)

Answer: The acceleration is 4 m/s² to the right.

Check:

  • Units: N/kg = (kg·m/s²)/kg = m/s² ✓
  • Larger force → larger acceleration ✓
  • Direction matches force ✓

3Problem 3medium

Question:

A 5.0 kg box is pushed across a frictionless floor with a force of 20 N. (a) What is the acceleration of the box? (b) If the box starts from rest, what is its velocity after 3.0 seconds? (c) How far has it traveled in this time?

💡 Show Solution

Solution:

Given: m = 5.0 kg, F = 20 N, friction = 0

(a) Acceleration: Using Newton's 2nd Law: F_net = ma 20 = 5.0a a = 4.0 m/s²

(b) Velocity after 3.0 s: Using v = v₀ + at v = 0 + 4.0(3.0) v = 12 m/s

(c) Distance traveled: Using Δx = v₀t + ½at² Δx = 0 + ½(4.0)(3.0)² Δx = 2.0(9.0) Δx = 18 m

4Problem 4hard

Question:

Three forces act on a 2.0 kg object: F₁ = 10 N east, F₂ = 8.0 N north, and F₃ = 6.0 N west. (a) Find the net force (magnitude and direction). (b) Find the acceleration of the object. (c) If the object starts from rest, what is its velocity after 5.0 seconds?

💡 Show Solution

Solution:

(a) Net force: Choose coordinate system: +x = east, +y = north

F_net,x = F₁ - F₃ = 10 - 6.0 = 4.0 N (east) F_net,y = F₂ = 8.0 N (north)

Magnitude: F_net = √(4.0² + 8.0²) = √(16 + 64) = √80 = 8.9 N

Direction: θ = tan⁻¹(F_y/F_x) = tan⁻¹(8.0/4.0) = tan⁻¹(2) = 63° north of east

(b) Acceleration: F_net = ma 8.9 = 2.0a a = 4.5 m/s² (in direction of F_net)

(c) Velocity after 5.0 s: v = v₀ + at = 0 + 4.5(5.0) = 22.5 m/s at 63° north of east

Components: vₓ = 4.0/2.0 × 5.0 = 10 m/s, vᵧ = 8.0/2.0 × 5.0 = 20 m/s

5Problem 5hard

Question:

Three forces act on a 2.0 kg object: F₁ = 10 N east, F₂ = 8.0 N north, and F₃ = 6.0 N west. (a) Find the net force (magnitude and direction). (b) Find the acceleration of the object. (c) If the object starts from rest, what is its velocity after 5.0 seconds?

💡 Show Solution

Solution:

(a) Net force: Choose coordinate system: +x = east, +y = north

F_net,x = F₁ - F₃ = 10 - 6.0 = 4.0 N (east) F_net,y = F₂ = 8.0 N (north)

Magnitude: F_net = √(4.0² + 8.0²) = √(16 + 64) = √80 = 8.9 N

Direction: θ = tan⁻¹(F_y/F_x) = tan⁻¹(8.0/4.0) = tan⁻¹(2) = 63° north of east

(b) Acceleration: F_net = ma 8.9 = 2.0a a = 4.5 m/s² (in direction of F_net)

(c) Velocity after 5.0 s: v = v₀ + at = 0 + 4.5(5.0) = 22.5 m/s at 63° north of east

Components: vₓ = 4.0/2.0 × 5.0 = 10 m/s, vᵧ = 8.0/2.0 × 5.0 = 20 m/s

6Problem 6medium

Question:

A 12001200 kg car accelerates from rest to 2020 m/s in 55 seconds on a straight road. What is the net force on the car?

💡 Show Solution

Given:

  • Mass: m=1200m = 1200 kg
  • Initial velocity: v0=0v_0 = 0 m/s
  • Final velocity: v=20v = 20 m/s
  • Time: t=5t = 5 s

Find: Net force F\sum F

Step 1: Find acceleration a=vv0t=2005=4 m/s2a = \frac{v - v_0}{t} = \frac{20 - 0}{5} = 4 \text{ m/s}^2

Step 2: Apply Newton's Second Law F=ma\sum F = ma F=(1200)(4)\sum F = (1200)(4) F=4800 N\sum F = 4800 \text{ N}

Answer: The net force on the car is 4800 N (in the direction of motion).

Alternative: 48004800 N =4.8= 4.8 kN (kilonewtons)

Physical interpretation: This is a significant force—equivalent to the weight of about 490 kg! The engine must provide this force (minus friction) to accelerate the car.

7Problem 7hard

Question:

A 22 kg book rests on a table. Draw a free body diagram and find the normal force. Then, if someone pushes down on the book with a force of 1010 N, what is the new normal force?

💡 Show Solution

Part 1: Book at rest

Free Body Diagram:

  • Weight W=mgW = mg pointing down
  • Normal force NN pointing up

Given:

  • Mass: m=2m = 2 kg
  • g=10g = 10 m/s² (approximate)
  • Acceleration: a=0a = 0 (at rest)

Apply Newton's Second Law (vertical direction):

Choose up as positive.

Fy=may\sum F_y = ma_y NW=0N - W = 0 Nmg=0N - mg = 0 N=mg=(2)(10)=20 NN = mg = (2)(10) = 20 \text{ N}

Part 2: Someone pushes down with 10 N

Updated Free Body Diagram:

  • Weight W=20W = 20 N pointing down
  • Normal force NN pointing up
  • Applied force Fapp=10F_{app} = 10 N pointing down

Apply Newton's Second Law: Fy=may\sum F_y = ma_y NWFapp=0(still at rest, so a=0)N - W - F_{app} = 0 \quad \text{(still at rest, so } a = 0\text{)} N2010=0N - 20 - 10 = 0 N=30 NN = 30 \text{ N}

Answers:

  • Part 1: Normal force = 20 N (equals weight)
  • Part 2: Normal force = 30 N (equals weight + push)

Key insight: Normal force is NOT always equal to weight! It adjusts to prevent the object from accelerating through the surface. Here, the table "pushes back harder" to support both the weight and the downward push.