Unbalanced Force - Newton's Second Law of Motion

Key idea: Learn what unbalanced forces mean and how to use F = ma (Newton’s second law) to solve motion problems (O Level Physics).

  • SEC G3 Physics 2027
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Learning objectives

  • Distinguish contact forces from non-contact forces
  • State that mass measures the amount of matter in a body
  • Describe a gravitational field as a region where a mass experiences gravitational force
  • Define gravitational field strength as gravitational force per unit mass
  • Apply weight = mass × gravitational field strength
  • Distinguish mass from weight
  • Describe the effect of balanced and unbalanced forces on a body
  • Describe ways a force may change motion
  • Identify action–reaction pairs on interacting bodies
  • Draw free-body diagrams for force systems in at most two dimensions
  • Solve three-force static equilibrium graphically
  • Apply resultant force = mass × acceleration
  • Relate mass to resistance to change in motion
  • Explain the effects of friction on motion
  • Describe falling with and without air resistance, including terminal velocity
  • Describe a moment as a force's turning effect in everyday examples
  • Apply moment = force × perpendicular distance from the pivot
  • State the principle of moments for a body in equilibrium
  • apply the principle of moments to new situations or to solve related problems
  • show an understanding that the weight of a body may be taken as acting at a single point known as its centre of gravity
  • Explain qualitatively how centre-of-gravity position affects stability

1. Definition

A. Unbalanced forces

Forces on a body are unbalanced when the resultant force is not zero:

Fᵣₑₛᵤₗₜₐₙₜ ≠ 0

B. Newton’s second law (O Level form)

The resultant force on a body causes it to accelerate in the direction of the resultant force.

For constant mass:

Fᵣₑₛᵤₗₜₐₙₜ = ma

  • Fᵣₑₛᵤₗₜₐₙₜ = resultant force (N)
  • m = mass (kg)
  • a = acceleration (m s⁻²)

2. Key Ideas

  • Always use the resultant force, not just one force.
  • If Fᵣₑₛᵤₗₜₐₙₜ = 0, forces are balanced and a = 0 (see Balanced Forces and Newton’s First Law).
  • If Fᵣₑₛᵤₗₜₐₙₜ ≠ 0, the object accelerates:
    • same direction as velocity → speeds up
    • opposite direction to velocity → slows down (decelerates)
  • Units check: N = kg × m s⁻².
  • 1 N is the force needed to give a mass of 1 kg an acceleration of 1 m s⁻².

Newton’s 2nd law (constant mass): force vs acceleration

Force–acceleration graph for a constant mass, showing direct proportionality (straight line through origin).

Scroll across the graph to read all labels.

Force–acceleration graph for a constant mass, showing direct proportionality (straight line through origin).Force–acceleration graph for a constant mass, showing direct proportionality (straight line through origin).
For a fixed mass, acceleration is directly proportional to the resultant force: a = F/m.
Open full-size graph
View figure data
Values for Newton’s 2nd law (constant mass): force vs acceleration
Resultant force (N)m = 2 kg
00
21
42
63
84
105

3. Detailed Explanations

A. Resultant force decides acceleration

Resultant force is the vector sum of all forces on the object:

  • if forces cancel → Fᵣₑₛᵤₗₜₐₙₜ = 0 → no acceleration
  • if one direction “wins” → Fᵣₑₛᵤₗₜₐₙₜ ≠ 0 → acceleration

B. Exam workflow (how to use F = ma)

  1. Draw a free body diagram (FBD) and label all forces.
  2. Choose a positive direction (often the direction of motion).
  3. Add the forces using signs; with forward positive, subtract the backward-force magnitudes.
  4. Use Fᵣₑₛᵤₗₜₐₙₜ = ma.
  5. Check the unit and use the sign to state the direction.

See: Free Body Diagrams (FBD) and How To Add Forces.

C. Including friction and air resistance

In many questions, the applied force is not the same as the resultant force because resistive forces act in the opposite direction.

4. Common Mistakes

  • Using the applied force instead of the resultant force.
  • Forgetting friction/air resistance when the question does not say “neglect”.
  • Mixing up mass and weight (weight is a force; mass is not).
  • Unit errors (g instead of kg, km/h instead of m/s, minutes instead of seconds).
  • Sign mistakes (not stating a positive direction, or giving an acceleration direction that contradicts the forces).

5. Exam Tips

  • Start with an FBD, even for a 1-line calculation.
  • Write “resultant force = (forward forces) − (backward forces)” before substituting.
  • If the object slows down, its acceleration is opposite to its velocity.
  • Always show units and give your final answer in N or m s⁻² as needed.

6. Worked Examples

Hidden assumptions to watch for
  • “smooth” / “friction negligible”: do not include friction.
  • “constant velocity” / “constant speed in a straight line”: Fᵣₑₛᵤₗₜₐₙₜ = 0.
  • “ignore air resistance”: do not include drag.
  • Units: use kg, m s⁻¹, s before applying F = ma.

Modelled example 1

Direct use of F = ma

Core

Problem

A 20 kg box has a resultant force of 50 N. Find its acceleration.
Study the worked solution
  1. Use the resultant force

    Method

    Rearrange Fᵣₑₛᵤₗₜₐₙₜ = ma for acceleration.

    Reason

    The question already gives the vector sum of forces, so no further force combination is needed.

    Working

    a = Fᵣₑₛᵤₗₜₐₙₜ/m = 50/20 = 2.5 m s⁻²

Guided practice 2

Find force using acceleration from velocity change

About 5 min

Problem

A 1000 kg car accelerates from rest to 20 m s⁻¹ in 5.0 s. Neglect resistance and find the driving force.

Find acceleration before force

Unit: m s^-2
Unit: N

Hints

Hint 1: find the rate of velocity change
Use a = (v-u)/t.
Hint 2: apply Newton's second law
Multiply the 1000 kg mass by the acceleration.
View solution step by step
  1. Calculate acceleration

    Method

    Find the velocity change per unit time.

    Reason

    Force is related to acceleration, not directly to velocity.

    Working

    a = (20-0)/5.0 = 4.0 m s⁻²
  2. Calculate the resultant force

    Reason

    With resistive forces neglected, the driving force is the resultant.

    Working

    F = ma = (1000)(4.0) = 4.0 × 10³ N

Common misconception 3

What happens when the push is removed?

Find and correct the mistake

Learner response

A box moves across rough ground at constant speed. A student says it will continue at constant speed when the push is removed because moving objects keep moving. Locate the missing force and predict the motion.

Rebuild the resultant after removal

Motion after the push is removed

View solution step by step
  1. Compare forces before and after removal

    Method

    Remove only the applied push from the FBD.

    Reason

    Friction still acts opposite the motion on the rough ground.

    Working

    Before removal, push balances friction; after removal, friction is the resultant force.
  2. Predict the acceleration

    Method

    Place acceleration opposite the velocity.

    Reason

    The backward resultant makes the box slow rather than stop instantaneously.

    Working

    The box slows to rest.

Examiner practice 4

Including friction (resultant force first)

3 marks

Examination question

A 10 kg crate is pulled right with 60 N while friction is 20 N left. Find its acceleration. [3 marks]

Show resultant before applying F = ma

View solution step by step
  1. Find the resultant

    1 mark

    Method

    Subtract the opposing force.

    Reason

    Newton’s second law uses the vector resultant, not the 60 N pull alone.

    Working

    Fᵣₑₛᵤₗₜₐₙₜ = 60-20 = 40 N right
  2. Apply Newton's second law

    1 mark

    Method

    Divide resultant force by mass.

    Reason

    Fᵣₑₛᵤₗₜₐₙₜ = ma.

    Working

    a = 40/10
  3. State magnitude and direction

    1 mark

    Reason

    Acceleration points with the rightward resultant.

    Working

    a = 4.0 m s⁻² right

Challenge 5

Deceleration (resultant opposite to motion)

Minimal support

Signed-force transfer

An 800 kg car moves right while a 2400 N resultant braking force acts left. Take right as positive and find the acceleration.

Carry force direction into the sign

Unit: m s^-2

Hints

Hint 1: assign the force sign
Right is positive, so the braking resultant is -2400 N.
Hint 2: divide by mass
Use a = Fᵣₑₛᵤₗₜₐₙₜ/m.
View solution step by step
  1. Apply the sign convention

    Method

    Write the leftward resultant as negative.

    Reason

    The chosen positive direction is right.

    Working

    Fᵣₑₛᵤₗₜₐₙₜ = -2400 N
  2. Calculate and interpret

    Reason

    Acceleration has the same direction as the resultant force.

    Working

    a = (-2400)/800 = -3.0 m s⁻² = 3.0 m s⁻² left

7. Mind Stretchers

Mind stretcher 1: How does acceleration change as speed increases?Extension

A car starts from rest. The driving force is roughly constant, but the air resistance increases as the car speeds up.

Describe what happens to the car’s acceleration as time goes on.

Show Answer

As speed increases, air resistance increases, so the resultant force decreases.

Since a = Fᵣₑₛᵤₗₜₐₙₜ/m, the acceleration decreases. Eventually the driving force balances the resistive forces, so the resultant force becomes zero and the car reaches constant velocity (acceleration zero).

A trolley of mass 2.0 kg has a straight-line velocity–time graph with gradient 3.0 m s⁻².

Find the resultant force on the trolley.

Show Answer

Gradient of a velocity–time graph is acceleration, so a = 3.0 m s⁻².

F = ma = (2.0)(3.0) = 6.0 N

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027