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).
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The core idea
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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.
View figure data
| Resultant force (N) | m = 2 kg |
|---|---|
| 0 | 0 |
| 2 | 1 |
| 4 | 2 |
| 6 | 3 |
| 8 | 4 |
| 10 | 5 |
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)
- Draw a free body diagram (FBD) and label all forces.
- Choose a positive direction (often the direction of motion).
- Add the forces using signs; with forward positive, subtract the backward-force magnitudes.
- Use Fᵣₑₛᵤₗₜₐₙₜ = ma.
- 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.
- friction: Friction
- air resistance and terminal velocity: Terminal Velocity
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
- “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,sbefore applying F = ma.
Modelled example 1
Direct use of F = ma
Problem
Study the worked solution
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
Problem
Find acceleration before force
Hints
Hint 1: find the rate of velocity change
Hint 2: apply Newton's second law
View solution step by step
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⁻²Calculate 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?
Learner response
Rebuild the resultant after removal
View solution step by step
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.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)
Examination question
Show resultant before applying F = ma
View solution step by step
Find the resultant
1 markMethod
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 rightApply Newton's second law
1 markMethod
Divide resultant force by mass.Reason
Fᵣₑₛᵤₗₜₐₙₜ = ma.Working
a = 40/10State magnitude and direction
1 markReason
Acceleration points with the rightward resultant.Working
a = 4.0 m s⁻² right
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark resultant force, Newton's second law and the final directed acceleration.
Challenge 5
Deceleration (resultant opposite to motion)
Signed-force transfer
Carry force direction into the sign
Hints
Hint 1: assign the force sign
Hint 2: divide by mass
View solution step by step
Apply the sign convention
Method
Write the leftward resultant as negative.Reason
The chosen positive direction is right.Working
Fᵣₑₛᵤₗₜₐₙₜ = -2400 NCalculate 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).
Mind stretcher 2: Link to kinematics graphs (find force from a gradient)Extension
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