Dynamics
Key idea: Balanced and unbalanced forces, action–reaction pairs, one-dimensional free-body diagrams, F = ma and friction.
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The core idea
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Learning objectives
- 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 one-dimensional force systems
- Apply resultant force = mass × acceleration
- Explain the effects of friction on motion
Syllabus and review details
Free-body diagrams are limited to forces acting in one dimension. Two-dimensional diagrams, graphical equilibrium, inertia and terminal velocity are not part of this topic.
- K223 / K224 Science Physics componentK223 / K224 · 2027Checked against the syllabus · complete topic coverageK223/K224 2027 syllabus, Dynamics topic 4
Resultant force determines whether velocity changes
The resultant force is the single force with the same overall effect as all the forces acting on one body. Forces are balanced when their vector sum is zero and unbalanced when it is not zero.
Resultant force = 0
The body has zero acceleration. It remains at rest or continues at constant velocity.
Resultant force ≠ 0
The body accelerates in the direction of the resultant force.
A force can change speed or direction
An unbalanced force changes velocity. It can start or stop a body, increase or decrease its speed, or change its direction. A force that changes direction still causes acceleration even if the speed stays constant.
- A forward resultant increases forward speed.
- A resultant opposite to the velocity decreases speed.
- A sideways resultant changes direction.
- Zero resultant preserves the current velocity; it does not necessarily stop the body.
Action–reaction forces act on different bodies
When body A exerts a force on body B, body B exerts an equal-magnitude, opposite-direction force on body A. The pair is the same type of interaction and acts at the same time.
Earth pulls the student downward with a gravitational force.
The student pulls Earth upward with an equal gravitational force.
Draw a one-dimensional free-body diagram
A free-body diagram isolates one body and shows only the external forces acting on it. For this course, the examined cases keep all forces along one line.
- Name the body being analysed and replace it with a dot or simple box.
- Identify each external interaction with that body.
- Draw one labelled arrow per force from the body in the force direction.
- Choose a positive direction before finding the signed resultant.
| Force | Direction | Interaction |
|---|---|---|
| Tension, T | Upward | Cable pulls the lift |
| Weight, W | Downward | Earth attracts the lift |
Apply resultant force = mass × acceleration
Apply the relationship to the resultant force, not automatically to one named force. Mass is in kilograms, acceleration in metres per second squared, and force in newtons.
- Draw or inspect the free-body diagram.
- Choose a positive direction.
- Add forces with signs to obtain the resultant force.
- Use Fresultant = ma and interpret the sign of the answer.
Friction affects the resultant force and energy transfer
Friction is a contact force that opposes relative motion or attempted relative motion between surfaces. It can reduce acceleration, cause deceleration, prevent slipping, and transfer energy to internal energy stores.
Common mistakes
- “A moving body must have a forward resultant force.”
- Constant velocity needs zero resultant force. A forward resultant is needed for forward acceleration.
- “Balanced forces are an action–reaction pair.”
- Balanced forces act on one body. Action–reaction forces act on two interacting bodies.
- “The largest force is the resultant force.”
- The resultant is the vector sum of all forces, including their directions.
- “F = ma can be applied to any force in the diagram.”
- The F in this relationship is the resultant force in the chosen direction.
Worked applications
Resultant force and acceleration in one dimension
A 4.0 kg trolley is pushed right with 18 N while friction is 6.0 N left. Find its acceleration.
Take right as positive.
Fresultant = 18 − 6.0 = 12 N to the right.
a = F/m = 12/4.0 = 3.0 m/s² to the right.
Constant velocity
A cyclist travels at constant velocity while the resistive force is 35 N. Find the forward driving force.
Constant velocity means zero acceleration, so the resultant force is zero.
The forward driving force therefore balances the resistance: 35 N.
Connect the complete Dynamics method
A 5.0 kg lift has an upward cable tension of 65 N and a downward weight of 50 N. The free-body diagram has only these two vertical forces. Taking upward as positive gives Fresultant = 65 − 50 = 15 N, so a = F/m = 15/5.0 = 3.0 m/s² upward. The equal force that the lift exerts on the cable is the action–reaction partner; it acts on the cable and is not part of the lift's diagram.
Guided practice
A 4.0 kg hanging object has 19 N upward and 7.0 N downward. Draw the force diagram, find the acceleration, and explain what changes if the downward force rises to 19 N.
Check the guided reasoning
Initially, F = 19 − 7.0 = 12 N upward, so a = 12/4.0 = 3.0 m/s² upward. If the opposing force rises to 19 N, the resultant becomes zero: acceleration is zero, so the body remains at rest or continues at constant velocity.
Independent practice
Repeat the method for a 6.0 kg body with 31 N upward and 13 N downward. State the acceleration and identify one genuine action–reaction pair involving the applied force.
Check your answer
The resultant force is 31 − 13 = 18 N upward, so a = F/m = 18/6.0 = 3.0 m/s² upward. A genuine action–reaction pair is the 31 N force exerted on the body and the equal 31 N force that the body exerts back on the object applying it. The two forces act on different bodies, so they do not cancel on one free-body diagram.
Challenge yourself
Two identical trolleys are pushed apart by a compressed spring. One trolley is then loaded and the test is repeated. Compare the forces on the trolleys and predict which trolley has the larger acceleration.
Check your thinking
The spring exerts equal-sized forces in opposite directions on the two trolleys. The loaded trolley has the smaller acceleration because a = F/m. Equal force does not mean equal acceleration when the masses differ.
Independent self-check
- State the motion of a body when its resultant force is zero.
- A 3.0 kg trolley has 14 N right and 5.0 N left. Calculate its acceleration.
- Name the two forces on a hanging lamp supported by one vertical cable.
- Identify the reaction force when a foot pushes the ground backward.
- Explain why friction can reduce acceleration.
Check your answers
- It stays at rest or continues with constant velocity.
- F = 14 − 5.0 = 9.0 N; a = 9.0/3.0 = 3.0 m/s² right.
- Tension upward and weight downward.
- The ground pushes the foot forward.
- Friction opposes the driving force, reducing the resultant force and therefore the acceleration.
Try this next
A 2.5 kg cart experiences 11 N forward and 3.5 N backward. Calculate its acceleration.
Show answer
F = 11 − 3.5 = 7.5 N, so a = 7.5/2.5 = 3.0 m/s² forward.
Practise this topic
The Dynamics check matches this course. Use the feedback to revisit the right explanation, then come back later for a fresh check.
Practise
Practise: Dynamics
A text-first Dynamics assessment with labelled controls and explicit bodies, force directions, motion, quantities and units.
About 10 minutes
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Questions are selected when you start. Use the feedback to decide what to practise next; this does not prove mastery.
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Beyond the syllabus: optional enrichment that does not count towards your progress.
Practise
Practise after feedback: Dynamics
A text-first Dynamics assessment with labelled controls and explicit bodies, force directions, motion, quantities and units.
About 10 minutes
Practise
Questions are selected when you start. Use the feedback to decide what to practise next; this does not prove mastery.
Recent attempts
History is stored only in this browser.
No completed attempts are saved yet.
Beyond the syllabus: optional enrichment that does not count towards your progress.
Check what I know
Check what I know: Dynamics
A text-first Dynamics assessment with labelled controls and explicit bodies, force directions, motion, quantities and units.
About 8 minutes
Check what I know
Answer 6 short questions. This starting check helps choose what to work on; it does not prove mastery.
Recent attempts
History is stored only in this browser.
No completed attempts are saved yet.
Beyond the syllabus: optional enrichment that does not count towards your progress.
Check my progress
Check my progress: Dynamics
A text-first Dynamics assessment with labelled controls and explicit bodies, force directions, motion, quantities and units.
About 10 minutes
Check my progress
Answer 6 questions. If accepted, this result can contribute to your course progress.
Recent attempts
History is stored only in this browser.
No completed attempts are saved yet.
Beyond the syllabus: optional enrichment that does not count towards your progress.
Check again
Check again: Dynamics
A text-first Dynamics assessment with labelled controls and explicit bodies, force directions, motion, quantities and units.
About 10 minutes
Check again
Answer 6 questions. If accepted, this result can contribute to your course progress.
Recent attempts
History is stored only in this browser.
No completed attempts are saved yet.
Beyond the syllabus: optional enrichment that does not count towards your progress.
Review
Review: Dynamics
A text-first Dynamics assessment with labelled controls and explicit bodies, force directions, motion, quantities and units.
About 10 minutes
Review
Answer 6 questions. A scheduled review can contribute to your course progress only when it is due and the result is accepted.
Recent attempts
History is stored only in this browser.
No completed attempts are saved yet.
Beyond the syllabus: optional enrichment that does not count towards your progress.
Course and syllabus information
- Course
- SEC G2 Science Physics component
- Edition
- SEC G2 Science Physics component 2027