K323 / 2027
Dynamics overview

Full chapter

Dynamics

All 7 topics and the revision summary on one page.

01

Forces, mass and weight

A force is a push or pull arising from an interaction between objects. To understand a force, identify what it acts on and what exerts it.

A string pulls on a trolley. A table pushes on a book. Earth attracts both objects. Each description names two interacting bodies, even when only one of them is moving.

Contact and non-contact forces

A contact force involves objects or materials touching. Its direction depends on the interaction.

Normal contact force
A surface pushes on an object in a direction perpendicular to the surface. A table supports a book in this way.
Tension
A stretched string or rope pulls on the object attached to it, along the string and away from the object.
Friction
Surfaces in contact exert a force that opposes their relative sliding, or their tendency to slide.
Air resistance
Air exerts a force on an object moving through it. Air is matter, so this is a contact interaction even though we cannot see the air.

Non-contact forces act without the objects touching. Gravitational attraction acts between masses; electrostatic forces act between electric charges; magnetic forces act between magnets or between a magnet and a suitable magnetic material. A magnet can attract a steel paper clip across a gap.

Forces are vectors, so give both magnitude and direction. For example, 6 N upwards describes a force of magnitude 6 newtons acting upwards. The quantity force may be written F or f; its unit is the newton, N.

Mass describes the object; weight is a force

Mass
A measure of the amount of matter in a body. Mass may be written m or M. It is a scalar, measured in kilograms (kg).
Weight
The gravitational force acting on a body. Weight is a vector, measured in newtons (N). Near Earth it acts downwards, towards Earth's centre.

A gravitational field is a region in which a mass experiences a force due to gravitational attraction. The gravitational field strength, g, is the gravitational force per unit mass at that point.

g = weight / mass   and   W = mgW = weight in N; m = mass in kg; g = gravitational field strength in N/kg. The weight acts in the direction of the gravitational field.

Near Earth's surface, g is approximately 10 N/kg: each kilogram experiences about 10 N of gravitational force. This corresponds to the approximately 10 m/s2 downward acceleration of free fall. Field strength is not 10 N/kg everywhere; use the value supplied in a question.

Worked example

The same object in two gravitational fields

An unchanged object has a mass of 1.5 kg. Find its weight where g = 10 N/kg, then where g = 1.6 N/kg.

  1. First location: W = 1.5 kg x 10 N/kg = 15 N.
  2. Second location: W = 1.5 kg x 1.6 N/kg = 2.4 N.

The kg units cancel against the kg in N/kg, leaving newtons. The mass remains 1.5 kg in both places. The weaker field produces a smaller weight; it does not remove matter from the object.

Measuring mass and weight

Use an electronic balance for a mass reading and a spring balance or newton-meter for a force reading. Choose a suitable range and check the zero. When measuring weight with a spring balance, hang the object freely and wait until it is stationary. The upward tension then has the same magnitude as the downward weight.

For example, an object gives a balance reading of 0.150 kg and a steady spring-balance reading of 1.5 N. These are different quantities. Their ratio gives g = 1.5 / 0.150 = 10 N/kg. Record the quantity and unit for each reading; do not call the force reading a mass.

For a small mass, the unit mg means milligram: 1 mg = 0.001 g = 0.000001 kg. In the equation W = mg, the letters instead mean the product m x g, mass multiplied by gravitational field strength. A unit after a mass reading and two quantities in an equation have different roles.

Convert grams to kilograms before using W = mg with g in N/kg. For example, 150 g = 0.150 kg. The prefix conversion changes the number and unit, not the amount of matter.

Optional check An unchanged object of mass 0.80 kg is taken to a place where the gravitational field strength is 3.0 N/kg. Which description is correct?
An unchanged object of mass 0.80 kg is taken to a place where the gravitational field strength is 3.0 N/kg. Which description is correct?

02

Forces on one body

Draw the forces on the chosen body, then combine them to find the resultant force.

A force comes from an interaction, such as a string pulling or a surface pushing. Review force types and weight if you need help naming the forces.

A free-body diagram shows the forces acting on one selected body. Represent the body with a dot or a simple outline. Draw labelled arrows from it: the arrowhead gives direction, and the length can represent magnitude using a chosen scale.

  1. Choose the body: for example, the book, rather than the book and table together.
  2. Identify its interactions: Earth attracts the book, and the table pushes on it.
  3. Draw only forces on that body: the book's force on the table belongs on a different diagram.
  4. Check the directions and labels: weight is downwards; the normal contact force is perpendicular to the table.

Forces on a stationary book

Selected body: book

Forces on a stationary bookThe selected body is a book at rest on a horizontal table. The table pushes the book upwards and Earth pulls it downwards. The two forces have equal magnitudes and act on the same book. They are not an interaction pair.Supportby tableWeightby Earth

These forces balance on the same body.

Forces on a stationary book on a horizontal table. The upward support and downward weight balance. Both arrows are forces on the book.

The resultant is the single force equal to the vector sum of all the forces on the body. Along one line, forces in the same direction add; forces in opposite directions subtract. Keep their directions in the answer.

Balanced forces do not require the object to be at rest

Forces are balanced when their resultant is zero. The body's acceleration is then zero, so its velocity stays constant. A body already at rest remains at rest; a moving body continues at the same speed in the same direction.

Balanced forces on a moving trolley

Selected body: 2.0 kg trolley

The trolley is moving right along a horizontal track.

Balanced forces on a moving trolleyThe selected body is a 2.0 kg trolley moving right. Support by the track is 20 N upwards and weight by Earth is 20 N downwards. The string pulls with 3 N to the right and the track exerts 3 N resistance to the left. All force arrows use the same scale. The resultant is zero, so the velocity stays constant.20 NSupportby track20 NWeightby Earth3 NResistanceby track3 NPullby string

Zero resultant: constant velocity. The trolley need not be at rest.

A 2.0 kg trolley moves along a straight level track at a constant 2 m/s. The string pulls forwards with 3 N and the track resists with 3 N. Its 20 N weight and 20 N support also balance.

The trolley has forces acting on it, but their resultant is zero. The string's forward pull balances the resistance. If resistance were absent, a forward pull would no longer be needed to maintain the same velocity.

An unbalanced force changes velocity

A non-zero resultant force produces acceleration in its direction. Since velocity includes direction, several changes are possible:

  • An object at rest can start moving.
  • A resultant in the direction of motion can increase its speed.
  • A resultant opposite to its motion can reduce its speed.
  • A resultant can turn its motion. An object following a curved path has changing velocity even if its speed stays constant.

For example, increasing the trolley's string pull to 7 N while resistance remains 3 N gives a resultant of 4 N forwards. Its velocity now changes. We calculate how quickly it changes in resultant force and acceleration.

Do not add a "force of motion". Motion is what the body does; each force must come from an interaction. If a separate velocity arrow helps, label it as velocity and keep it distinct from the force diagram.

The trolley's support equals its weight here because the pull is horizontal and there is no vertical acceleration. Do not assume support always equals weight when other vertical forces or vertical acceleration are present.

Optional check A trolley travels along a straight level track at a constant 2 m/s. What must be true about the forces on it?
A trolley travels along a straight level track at a constant 2 m/s. What must be true about the forces on it?

03

Pairs of forces

When two bodies interact, each exerts a force on the other. The two forces have equal magnitudes and opposite directions.

A free-body diagram contains forces on one body. An interaction pair spans two bodies, so its members belong on different body diagrams.

These are often called action-reaction pairs. The forces occur together; "reaction" does not mean a later response. Identify the pair by naming both bodies and then reversing their roles.

One contact interaction, two different bodies. Only the paired contact forces are shown.

The table pushes the book upwards

Selected body: book

The table pushes the book upwardsOne member of the contact interaction pair: the force on the book by the table acts upwards. Only this contact force is shown; the other forces on the book are omitted.On bookby table

The book pushes the table downwards

Selected body: table

The book pushes the table downwardsThe other member of the same contact interaction pair: the force on the table by the book acts downwards. Its magnitude equals the upward force on the book. Only this contact force is shown; the other forces on the table are omitted.On tableby book
The table pushes upwards on the book; the book pushes downwards on the table. These panels show only that contact pair, with other forces omitted. They are not complete free-body diagrams of either body.

Balanced forces and interaction pairs answer different questions

A stationary book on a horizontal table
Two forces that balance on the bookA contact interaction pair
Table on book, upwards.
Earth on book, downwards.
Table on book, upwards.
Book on table, downwards.
Both act on the same body. Their resultant on that body is zero.They act on different bodies. They belong to the same book-table interaction.

The table's support and the book's weight can be equal and opposite, but that alone does not make them an interaction pair. One comes from contact with the table; the other comes from gravitational attraction by Earth.

The partner of Earth's gravitational force on the book is the book's gravitational force on Earth. These also have equal magnitudes and opposite directions. Earth's very large mass means its resulting acceleration from this interaction is extremely small.

Worked explanation

A hand pushing a wall

A hand pushes horizontally on a wall with a force of 12 N. The wall exerts a 12 N force on the hand in the opposite direction.

On a diagram of the hand, include the wall's force on the hand. The hand's force on the wall acts on the other body, so it does not belong on that diagram. The pair does not cancel within the hand's force diagram.

Equal forces do not require equal accelerations of the two bodies. Their masses and the other forces acting on each body matter.

Name the interaction, not just the arrow directions. "Force on the book by the table" pairs with "force on the table by the book". Swapping the bodies keeps you from confusing this pair with the book's weight.

Optional check A table pushes upwards on a book resting on it. Which force is the other member of this interaction pair?
A table pushes upwards on a book resting on it. Which force is the other member of this interaction pair?

04

Resultant force and acceleration

The resultant force determines the acceleration. Calculate it from all the forces on the body before using F = ma.

Acceleration is the change in velocity per unit time. Use a free-body diagram to find the resultant and choose a positive direction.

Resultant force = mass x acceleration
Fresultant = ma
For a body of constant mass: force in N, mass in kg and acceleration in m/s2. Acceleration is in the direction of the resultant force.

A resultant force of 1 N gives a 1 kg mass an acceleration of 1 m/s2. The equation concerns acceleration, not the velocity at that instant. An object may be moving quickly while its resultant force and acceleration are both zero.

Unbalanced forces on a moving trolley

Selected body: 2.0 kg trolley

The trolley is moving right along a horizontal track.

Unbalanced forces on a moving trolleyThe selected body is a 2.0 kg trolley moving right. Support by the track is 20 N upwards and weight by Earth is 20 N downwards. The string pulls with 7 N to the right and the track exerts 3 N resistance to the left. All force arrows use the same scale. The resultant is 4 N to the right, so the trolley accelerates to the right.20 NSupportby track20 NWeightby Earth3 NResistanceby track7 NPullby string

Horizontal resultant: 7 - 3 = 4 N right. The vertical forces balance.

Forces on a 2.0 kg trolley: a 7 N horizontal string pull forwards, 3 N resistance from the track backwards, and a balanced 20 N vertical pair. The horizontal resultant is 4 N forwards.

Worked example

Subtract the opposing force first

The trolley shown is moving forwards. Find its acceleration. Take forwards as positive.

  1. Find the horizontal resultant: Fresultant = 7 - 3 = +4 N. The forces oppose each other, so subtract their magnitudes.
  2. Rearrange: a = Fresultant / m.
  3. Substitute: a = 4 / 2.0 = 2.0 m/s2 forwards.

The balanced vertical forces give no vertical acceleration. Using 7 / 2.0 would treat the pull as the entire resultant and ignore the track's resistance.

A forward-moving object can accelerate backwards

In a different interval, suppose the trolley is still moving forwards but its string pull is 3 N and track resistance is 7 N. Keeping forwards positive gives:

Fresultant = 3 - 7 = -4 N
a = -4 / 2.0 = -2.0 m/s2
The acceleration is backwards while the velocity is forwards, so the trolley slows during this interval.

The negative sign specifies direction. It does not mean the trolley is already moving backwards. If the trolley reaches a stop, reassess the forces before predicting what happens next; the same resistance direction cannot simply be assumed throughout a reversal.

Mass and inertia

Inertia is resistance to a change in motion. Mass is the property of a body associated with this resistance. A larger mass has greater inertia.

With the same resultant force, a larger mass has a smaller acceleration. A 4 N resultant gives a 2 kg trolley an acceleration of 2 m/s2, but a 4 kg trolley an acceleration of 1 m/s2. Doubling mass halves acceleration when the resultant is unchanged.

When a car brakes, a passenger tends to continue with the existing forward velocity. The seat belt exerts a force that slows the passenger with the car. There is no new forward "inertia force" to add to the passenger's force diagram; inertia describes the body's resistance to changing its motion.

Investigating the relationship

The following idealised data describe a trolley of fixed mass 0.50 kg. The forces listed are resultant forces, after accounting for resistance.

Model data for a fixed 0.50 kg mass
Resultant force / NAcceleration / (m/s2)
0.200.40
0.400.80
0.601.20

Doubling the resultant from 0.20 to 0.40 N doubles acceleration from 0.40 to 0.80 m/s2. The ratio Fresultant / a is 0.50 kg in every row, consistent with the fixed mass. An acceleration-against-resultant-force graph would be a straight line through the origin for this model.

For a real investigation, keep the trolley's total mass unchanged and use a level track. Change the horizontal pull, estimate acceleration from motion readings, and account for resistance when finding the resultant. A motion sensor or timed velocity readings can provide the changes needed for a = (v - u) / t.

Repeat readings to judge variation. A slightly sloping track adds a component of weight along the track, while changing resistance can make a measured pull different from the assumed resultant. Address those causes instead of calling every disagreement "human error". Real data need not fall exactly on the model values.

Optional check A 4.0 kg trolley has a horizontal pull of 14 N to the right and a resistance of 6 N to the left. Its vertical forces balance. What is its acceleration?
A 4.0 kg trolley has a horizontal pull of 14 N to the right and a resistance of 6 N to the left. Its vertical forces balance. What is its acceleration?

05

Friction and motion

Friction acts between surfaces in contact. It opposes their relative sliding, or their tendency to slide.

Always identify the body whose force you are describing. The direction of friction must follow the particular contact, just like any other force on a free-body diagram.

Friction can slow a sliding object

A box slides to the right across a stationary floor. Its lower surface slides right relative to the floor, so friction from the floor on the box acts left. If this is the only horizontal force, the resultant is leftwards and the box slows.

Friction on a box sliding right

Selected body: box

The box slides right over a stationary horizontal floor.

Friction on a box sliding rightThe selected body is the sliding box. Friction from the floor acts left, opposing the box sliding right relative to the floor. The floor supports the box upwards and Earth pulls it downwards. The vertical forces balance. Arrow lengths are schematic.Supportby floorWeightby EarthFrictionby floor

Friction opposes the relative sliding at this contact. Arrow lengths are schematic.

The box moves right relative to the floor. Friction on the box acts left. The motion is stated separately from the force arrows.

If you pull the box steadily across the same floor, the pull can balance friction. The box then moves at constant velocity even though both horizontal forces are present.

Friction can prevent sliding

A gentle horizontal push does not always make a box move. Static friction can balance the push while the surfaces remain at rest relative to each other. For example, if a 2 N push is balanced by 2 N of friction, the horizontal resultant is zero.

Static friction adjusts to the situation up to a limit. Increasing the push may increase the balancing friction while the box remains still; it does not mean friction always has one fixed magnitude. If the available friction cannot balance the push, the box starts sliding.

Friction can help you move forwards

As you push off to start walking forwards, your planted foot pushes backwards on the ground. The foot would tend to slip backwards without enough grip. Friction from the ground on the foot acts forwards, helping accelerate you forwards.

Friction helps a person start walking

Selected body: planted foot

When pushing off, the foot tends to slip backwards (left) over the ground.

Friction helps a person start walkingThe selected body is the planted foot of a person starting to walk forwards, to the right. The foot tends to slip backwards over the ground. Friction on the foot by the ground acts forwards, to the right. Only this friction force is shown; other forces on the foot are omitted.Friction on footby ground

Ground-on-foot friction acts forwards. Only this contact force is shown.

During this push-off, the ground exerts forward friction on the foot. The shoe grips without sliding; friction opposes its tendency to slip backwards.

This is why "friction always opposes the object's motion" is too broad. In the sliding-box case it acts backwards; in this walking case it acts forwards. In both, the direction follows the relative sliding or tendency to slide at the contact.

Grip, braking and resistance

Friction provides grip between a tyre and the road. Friction between brake pads and a rotating wheel or brake disc helps slow the rotation. It can also cause unwanted wear and resistance in moving machine parts. Reducing friction can make some movements easier, while reducing grip where it is needed can make slipping more likely.

Air resistance is a resistive force from the surrounding air. Its direction opposes motion relative to the air. An object falling downwards through still air experiences upward air resistance; see falling with air resistance for its changing effect on motion.

Optional explanation: a box stays still under a 3 N horizontal push

Assume the floor's friction is the only other horizontal force. What is the friction on the box, and why?

The box remains at rest, so horizontal acceleration and resultant force are zero. Friction is therefore 3 N opposite to the push. There is no horizontal motion, but there is still a force of friction.

06

Three-force equilibrium

For a stationary point mass, the forces have zero resultant. Three force vectors drawn head to tail therefore form a closed triangle.

Use graphical vector addition: preserve each arrow's direction, choose a scale, then measure lengths. If needed, review W = mg to find the known weight first.

A point-mass model represents a small body as a single point. Here we use it to find the tensions in two supporting strings. Each string pulls away from the mass along its own direction.

Draw the forces before drawing the triangle

A 0.50 kg mass hangs stationary from two strings. The left string is 30° above the horizontal to the left; the right string is 45° above the horizontal to the right. Take g = 10 N/kg.

Three forces on a stationary point mass

Selected body: 0.50 kg point mass

Forces on a mass held by two stringsA stationary 0.50 kg point mass experiences three forces. Its 5.0 N weight acts vertically downwards. Left-string tension TL acts upwards and left, 30 degrees above the left horizontal. Right-string tension TR acts upwards and right, 45 degrees above the right horizontal. Both tensions pull away from the mass. The dashed horizontal line is an angle reference, not a force.TLLeft stringTRRight string30°45°5.0 NWeightby Earth

The dashed horizontal line is an angle reference. TL and TR act along the strings, away from the mass.

Forces on the point mass: a known 5.0 N weight downwards, an unknown left tension TL, and an unknown right tension TR. The strings supply the two tension directions.

The weight is W = mg = 0.50 x 10 = 5.0 N downwards. There are three forces on the mass. Their directions are known, but the two tensions' magnitudes are not.

Worked construction

Find both tensions with a scale drawing

  1. Choose a paper scale: let 1 cm represent 1 N.
  2. Draw the known force: draw AB 5.0 cm vertically downwards to represent the weight.
  3. Draw the left-tension direction: from B, draw a ray upwards and left, at 30° above the left horizontal.
  4. Draw a line parallel to the right tension: through A, draw a line at 45° to the horizontal. Extend it backwards, downwards and left, until it meets the first ray at C.
  5. Check the arrow directions: the closed sequence is A to B, B to C, then C to A. BC points along the left tension; CA points along the right tension.
  6. Measure and convert: BC is about 3.7 cm and CA about 4.5 cm. At this scale, TL is about 3.7 N and TR about 4.5 N.

A closed triangle of forces

Follow the arrowheads: A to B to C to A.

Head-to-tail construction for three-force equilibriumThe closed force triangle follows A to B to C to A. AB is 5.0 N vertically downwards. BC is TL, about 3.7 N, directed upwards and left at 30 degrees above the left horizontal. CA is TR, about 4.5 N, directed upwards and right at 45 degrees above the right horizontal. The three arrows return to A, so their resultant is zero. All three sides use the same force scale. On paper, a scale of 1 centimetre per newton gives AB 5.0 centimetres, BC about 3.7 centimetres and CA about 4.5 centimetres.ABC5.0 NTRabout 4.5 NTLabout 3.7 N

For your paper drawing, use 1 cm per N. Use the labelled values on screen; its size changes with your device.

The force triangle closes: AB is the weight, BC is TL and CA is TR. On paper at 1 cm per N, the supplied approximate lengths are 5.0 cm, 3.7 cm and 4.5 cm respectively.

The free-body diagram shows the forces acting at the mass. The separate triangle moves those arrows head to tail to add them; it does not show the physical positions of the strings. Returning to A means that the resultant vector is zero.

Use a paper construction for ruler measurements. A screen diagram changes size with the display and need not preserve physical centimetres. Use the stated scale and angles, draw carefully, then measure. Small differences in graphical answers are expected.

Do not add the two tension magnitudes and set that sum equal to the weight. Their directions differ, so they must be added as vectors. Drawing C to B instead of B to C would also reverse the left tension and break the intended head-to-tail sequence.

Try a different arrangement

This optional paper task uses a 6.0 N downward weight, supported by two strings that each make 60° with the horizontal on their respective sides. Draw the free-body diagram, choose a scale and construct the closed force triangle. There is no answer to submit.

Show the construction result

At 1 cm per N, draw the 6.0 cm downward weight. Construct the left-tension direction from its head, then the line parallel to the right tension through its tail, just as in the worked method.

Each tension side measures about 3.5 cm, so both tensions are approximately 3.5 N. The two sides have equal lengths because the arrangement is symmetric. Their vector sum balances the weight; the tension is not 6.0 / 2 = 3.0 N because both strings pull at an angle.

07

Falling with air resistance

A falling object can speed up while its acceleration gets smaller. Increasing air resistance reduces the downward resultant force.

Weight is mg, and acceleration comes from resultant force / mass. Take downwards as positive in the examples on this page.

Without significant air resistance

For a body of constant mass in a uniform gravitational field, weight is constant. If air resistance is negligible, weight is the only force, so:

a = Fresultant / m = mg / m = gNear Earth, this is approximately 10 m/s2 downwards. A released body gains about 10 m/s of downward velocity each second while this model applies.

The mass cancels, which explains why objects of different masses have the same free-fall acceleration in this model. Gravity has not disappeared: it is the force causing the acceleration. With no air resistance, this model does not produce a terminal speed.

As a body falls through still air

Consider the same body released from rest, with its shape and orientation unchanged. At release its speed relative to the air is zero, so air resistance is zero. As its downward speed increases, upward air resistance grows, while its weight remains the same.

Same object and same weight throughout. Each snapshot uses the same force scale.

1. Released from rest

Selected body: 0.50 kg falling object

1. Released from restThe selected body is the same 0.50 kg object. Its weight is 5.0 N downwards. Air resistance is 0.0 N upwards. There is no upward force arrow because air resistance is zero at release in still air. The resultant is 5.0 N downwards and acceleration is 10.0 metres per second squared downwards.5.0 NWeight by Earth

Air resistance: 0 N. Resultant: 5.0 N downwards.

2. A later instant

Selected body: 0.50 kg falling object

2. A later instantThe selected body is the same 0.50 kg object. Its weight is 5.0 N downwards. Air resistance is 2.0 N upwards. The upward arrow uses the same force scale as the weight. The resultant is 3.0 N downwards and acceleration is 6.0 metres per second squared downwards.5.0 NWeight by Earth2.0 NDrag by air

Resultant: 3.0 N downwards. The object is still speeding up.

3. Terminal motion

Selected body: 0.50 kg falling object

3. Terminal motionThe selected body is the same 0.50 kg object. Its weight is 5.0 N downwards. Air resistance is 5.0 N upwards. The upward arrow uses the same force scale as the weight. The resultant is 0.0 N and acceleration is zero. The object continues downwards at a non-zero constant velocity.5.0 NWeight by Earth5.0 NDrag by air

Equal forces. Zero acceleration; non-zero downward velocity.

A constant-mass body at three stages: release, a later downward-moving instant, and terminal motion. Weight stays at 5.0 N; the supplied upward air resistances are 0 N, 2.0 N and 5.0 N.

Worked example

Follow the resultant, not just the weight

The object has mass 0.50 kg and g = 10 N/kg. Its weight is 5.0 N throughout. Subtract the upward air resistance from the downward weight.

Downwards is positive; mass is 0.50 kg
Air resistance / NDownward resultant / NAcceleration / (m/s2)
05.0 - 0 = 5.05.0 / 0.50 = 10
2.05.0 - 2.0 = 3.03.0 / 0.50 = 6.0
5.05.0 - 5.0 = 00 / 0.50 = 0

At the middle stage, acceleration is still downwards, so the downward speed is still increasing. It increases less rapidly than at release because the resultant has fallen from 5.0 N to 3.0 N.

Terminal velocity

When air resistance equals weight, the resultant and acceleration are zero. The object continues downwards at a constant, non-zero velocity called its terminal velocity, while these conditions remain unchanged.

Velocity increases while acceleration decreases

Downwards is positive. This is a schematic graph.

A falling object approaches terminal velocityA schematic velocity-time graph takes downwards as positive. Starting at zero velocity, the curve rises steeply and then becomes progressively flatter, approaching a positive, non-zero terminal velocity. Its gradient, the downward acceleration, decreases towards zero. No numerical times or terminal speed are specified by this graph.Downward velocityTime0Terminal velocity(non-zero)

The curve approaches a level above zero. The force snapshots do not supply numerical times or speeds.

For release from rest, the downward velocity rises while the curve becomes less steep. Its gradient approaches zero as the velocity approaches a constant terminal value. This is a schematic graph: no numerical times or speeds are specified.

Read the two features separately: increasing graph height means increasing downward velocity; decreasing gradient means decreasing acceleration. A horizontal line at a positive velocity represents steady downward motion, not rest.

The supplied force values let us calculate acceleration at the three stages. They do not tell us the times of those stages or the numerical terminal velocity. A real falling object may reach the ground before it approaches terminal motion.

What if the air resistance changes?

If an already downward-moving object suddenly experiences air resistance greater than its weight, the resultant is upwards. Its acceleration is then upwards, so it slows while still travelling downwards. Zero acceleration occurs only when the forces balance.

Terminal velocity does not mean zero weight or zero velocity. Both weight and air resistance still act. Their balance gives zero acceleration and a constant downward velocity.

Optional check An object is falling downwards at terminal velocity through still air. Which explanation is correct?
An object is falling downwards at terminal velocity through still air. Which explanation is correct?

Revision summary

Choose the body, identify its interactions and find the resultant before predicting its motion.

Weight and gravitational field
W = mg. Weight W in N; mass m in kg; gravitational field strength g in N/kg. Near Earth, g is approximately 10 N/kg. Use the value given.
Resultant force and acceleration
Fresultant = ma, so a = Fresultant / m. Acceleration is in the resultant's direction. Use all the relevant forces, not just the applied pull.

From forces to motion

  • Zero resultant: zero acceleration. The body stays at rest or continues at constant velocity.
  • Non-zero resultant: velocity changes. The body can start, speed up, slow down or change direction.
  • A fast-moving object need not have a large resultant force. A slowing object still has acceleration.
  • Draw only forces on the chosen body. Identify the interaction behind each arrow; motion and inertia are not extra forces.

Balanced forces or an interaction pair?

Keep the bodies and interactions explicit
Balanced forcesAction-reaction pair
Act on the same body and sum to zero.Act on different bodies in the same interaction.
Example: table on book upwards and Earth on book downwards.Example: table on book upwards and book on table downwards.

Mass, weight and force types

Mass is a measure of matter and is measured in kg; weight is gravitational force and is measured in N. An unchanged object's mass stays the same when its gravitational field changes. Greater mass also means greater inertia: more resistance to a change in motion.

Contact examples: normal force, tension, friction and air resistance. Non-contact examples: gravitational, electrostatic and magnetic forces. Normal force is perpendicular to the contact surface; tension pulls along a string.

Friction direction

Identify the contact and the relative sliding or tendency to slide. Friction can slow a sliding box, prevent a stationary box from sliding, or push a walking person forwards during push-off.

Three-force equilibrium

  1. Draw the three forces on the stationary point mass.
  2. Choose a paper scale and draw the known force.
  3. Use the other force directions to construct a closed head-to-tail triangle.
  4. Check arrow directions, measure lengths and convert them to forces.

The triangle closes because the resultant is zero. Use the stated scale on paper; a resized screen does not preserve centimetres.

Falling bodies

  • Negligible air resistance: weight is the only force and a = g in a uniform field.
  • Air resistance below weight during downward motion: the body speeds up downwards. As resistance grows, the downward acceleration decreases.
  • Air resistance equal to weight: zero resultant and zero acceleration; terminal motion has a constant, non-zero downward velocity.
  • Air resistance above weight: upward acceleration slows a downward-moving body.

Check a practical explanation

State which mass or force you measured and its unit. For a force-motion comparison, keep total mass controlled, account for resistance and check that a slope has not introduced an extra force along the track. Name a limitation and explain its effect.

Back to forces, mass and weight