Newton's Laws of Motion

Misconceptions

Key Concepts

Newton's First Law (Law of Inertia)

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.

Key insight: Inertia is the tendency to resist changes in motion. Mass is a measure of inertia.

Newton's Second Law

The acceleration of an object is directly proportional to the net force and inversely proportional to its mass.

ΣF = ma    or    a = ΣF/m

• •ΣF = net force (vector sum of all forces)
• •m = mass (scalar, in kg)
• •a = acceleration (vector, in m/s²)

Newton's Third Law

For every action, there is an equal and opposite reaction.

Key insight: Forces always come in pairs acting on different objects.

• •If A pushes B with force F, then B pushes A with force -F
• •These forces are equal in magnitude, opposite in direction

Free Body Diagrams (FBD)

A diagram showing ALL forces acting on a SINGLE object:

1. •Draw the object as a point or simple shape
2. •Draw arrows for each force, starting from the object
3. •Label each force (W, N, T, f, F_applied, etc.)
4. •Arrow length should indicate relative magnitude

Common Forces

• •Weight (W or Fg): W = mg, always points down
• •Normal force (N): Perpendicular to surface, pushes away from surface
• •Tension (T): Along rope/string, pulls away from object
• •Friction (f): Parallel to surface, opposes motion or tendency to move
• •Applied force (F): Any external push or pull

Equations

[1] ΣF = ma (Newton's Second Law)
[2] W = mg (Weight)
[3] ΣFx = max (x-component)
[4] ΣFy = may (y-component)
[5] For equilibrium: ΣF = 0 (or ΣFx = 0 and ΣFy = 0)

Worked Examples

Example 1: Horizontal Push

Problem: A 10 kg box is pushed with a force of 50 N on a frictionless surface. What is its acceleration?

Solution:

1. •Draw FBD: Weight (W) down, Normal (N) up, Applied force (F) horizontal
2. •Apply Newton's 2nd Law in x-direction:
   • •ΣFx = ma
   • •50 = 10 × a
   • •a = 5.0 m/s²

Example 2: Elevator Problem

Problem: A 70 kg person stands on a scale in an elevator accelerating upward at 2 m/s². What does the scale read?

Solution:

1. •Draw FBD: Weight (W = mg) down, Normal force (N = scale reading) up
2. •Choose up as positive
3. •Apply Newton's 2nd Law:
   • •ΣF = ma
   • •N - mg = ma
   • •N = m(g + a) = 70(9.8 + 2) = 70 × 11.8 = 826 N

Example 3: Two Blocks

Problem: Two blocks (3 kg and 5 kg) are pushed across a frictionless surface by a 24 N force applied to the 3 kg block. Find the force between the blocks.

Solution:

1. •Find total acceleration: a = F/(m₁ + m₂) = 24/8 = 3 m/s²
2. •For the 5 kg block alone, the only horizontal force is from the 3 kg block:
   • •F_contact = m₂ × a = 5 × 3 = 15 N

Explanation Patterns

1. •Always draw a Free Body Diagram first - identify all forces on the object
2. •Choose a coordinate system - typically x horizontal, y vertical
3. •Break forces into components if they're at angles
4. •Write ΣF = ma for each direction (x and y separately)
5. •Solve the resulting equations - often simultaneous equations
6. •Check your answer: units should be N or m/s², sign should make physical sense

Common Problem Types

1. •Single object, horizontal forces: Direct application of F = ma
2. •Inclined plane (without friction): Break weight into components
3. •Connected objects: Use same acceleration, different forces
4. •Elevator/vertical acceleration: Scale reads N, not mg
5. •Equilibrium: ΣF = 0, solve for unknown force
6. •Action-reaction identification: Forces on different objects