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2.2 Newton’s laws

2.2.1 Interactions between objects

Forces and interactions

Definition

Force

A push or a pull that one object exerts on another object as a result of an interaction between them.

Definition

Interaction

The way in which two objects affect each other, producing a force on each of the two objects.

  1. A force is a push or a pull that one object exerts on another object.
  2. Forces are measured in newtons, which have the symbol N\text{N}N.
  3. A force can make an object speed up, slow down, change direction or change shape.
  4. Forces never come from a single object on its own, because every force is produced by an interaction between two objects.
  5. When you kick a football, the force on the ball comes from the interaction between your foot and the ball.
  6. When an apple falls from a tree, the force that pulls it down comes from the interaction between the apple and the Earth.
  7. Interactions are sorted into two groups depending on whether the objects need to touch.
    1. Non-contact interactions act across a gap, so the objects can be separated.
    2. Contact interactions act only while the surfaces of the objects are touching.

Non-contact forces

Definition

Non-contact force

A force that acts between two objects without them touching, such as a gravitational, electrostatic or magnetic force.

Definition

Gravitational force

The attractive force that acts between any two objects that have mass.

Definition

Electrostatic force

The force between electrically charged objects, which is attractive between opposite charges and repulsive between like charges.

Definition

Magnetic force

The force between two magnets, or between a magnet and a magnetic material, which acts without the objects touching.

  1. A non-contact force acts between objects that are not touching, so it can act across empty space.
  2. There are three kinds of non-contact interaction that you need to recall: gravitational, electrostatic and magnetic.
  3. Gravitational interactions happen between any two objects that have mass.
    1. The gravitational force is always attractive, so it pulls the two objects towards each other.
    2. The gravitational pull of the Earth on an object near its surface is the object's weight, which acts downwards towards the centre of the Earth.
    3. Gravity also holds the Moon in orbit around the Earth and the Earth in orbit around the Sun.
  4. Electrostatic interactions happen between objects that carry an electric charge.
    1. Objects with opposite charges, one positive and one negative, attract each other.
    2. Objects with like charges, both positive or both negative, repel each other.
    3. A balloon rubbed on a woollen jumper becomes charged and can then attract small pieces of paper or stick to a wall.
  5. Magnetic interactions happen between magnets, or between a magnet and a magnetic material such as iron, steel, nickel or cobalt.
    1. Unlike poles (a north pole and a south pole) attract, while like poles (two north poles or two south poles) repel.
    2. A magnet always attracts a magnetic material that is not itself a magnet, which is how a fridge magnet holds a shopping list in place.
  6. Gravitational forces can only attract, while electrostatic and magnetic forces can either attract or repel.
  7. Each non-contact force gets weaker as the distance between the objects increases.

Contact forces

Definition

Contact force

A force that acts between two objects only while their surfaces are touching.

Definition

Normal contact force

The push exerted by a surface on an object pressing against it, acting at right angles to the surface.

Definition

Friction

A contact force that opposes the sliding, or attempted sliding, of one surface over another.

Definition

Air resistance

The frictional force that acts on an object moving through air, in the opposite direction to its motion.

  1. A contact force acts only when the surfaces of two objects are touching each other.
  2. The normal contact force is the push that a surface exerts on an object pressing on it.
    1. The word normal means perpendicular, so this force always acts at right angles to the surface.
    2. A mug resting on a desk is held up by an upward normal contact force from the desk.
    3. A ladder leaning against a wall feels a horizontal normal contact force from the wall, because the wall surface is vertical.
  3. Friction is the force that opposes the movement, or the attempted movement, of one surface sliding over another.
    1. Friction acts parallel to the surfaces in contact, in the opposite direction to the sliding.
    2. Even surfaces that feel smooth have tiny bumps that catch on one another, and this is what produces friction.
    3. Friction between your shoes and the pavement lets you walk, and friction between a brake pad and a bicycle wheel slows the bike.
  4. Air resistance is a type of friction that acts on an object moving through air, and it gets larger as the object moves faster.
  5. A similar resistive force, called drag, acts on objects moving through water or any other liquid.
  6. Two further contact forces are common in force problems.
    1. Tension is the pull exerted by a stretched string, rope or cable on the object attached to it.
    2. Upthrust is the upward push of a liquid or gas on an object that is partly or fully immersed in it.
Example

Sorting the forces on a cyclist

  • A cyclist rides along a flat road on a windy day in Bristol.
  • The Earth pulls the cyclist and bicycle downwards with a gravitational force, which is a non-contact force because it acts without the Earth needing to touch them.
  • The road pushes up on the tyres with a normal contact force, which is a contact force because the tyres touch the road.
  • Friction between the tyres and the road pushes the bicycle forwards when the cyclist pedals, which is a contact force between two touching surfaces.
  • Air resistance acts backwards on the cyclist, which is a contact force because the air particles collide with the cyclist's body.
Example

Sorting the forces on a magnet near a paper clip

  • A bar magnet is held 2 cm2\ \text{cm}2 cm above a steel paper clip lying on a desk.
  • The paper clip lifts off the desk and jumps up to the magnet without being touched, so the magnetic force on it is a non-contact force.
  • While the paper clip is still resting on the desk, the desk exerts an upward normal contact force on it.
  • The Earth exerts a downward gravitational force on the paper clip at all times, whether or not it touches the desk.
Common Mistake
  • Do not say that a force needs contact to act, because gravitational, electrostatic and magnetic forces all act across a gap.
  • Do not describe air resistance as a non-contact force, because it is caused by air particles colliding with the surface of the object.
  • Do not write "gravity" when you mean the force itself, because the force of gravity on an object is called its weight.
  • Do not give a force the unit joule or watt, because force is measured in newtons.
Exam technique

Naming interactions

  • When a question asks for a type of interaction, use the exact physics name: gravitational, electrostatic, magnetic, normal contact force or friction.
  • When you give an example, name both objects involved, such as "the Earth and a falling apple", rather than only "gravity".
  • If you are asked to classify a force, state whether the objects are touching and use that as your reason.
  • Recall that only gravitational forces are always attractive, as this is a common one-mark distinction.
Self review
  • What is a force, and what is its unit?
  • Name the three non-contact interactions you need to know.
  • Which non-contact force can only attract?
  • In which direction does a normal contact force act compared with the surface?
  • Why is air resistance classed as a contact force?

2.2.2 Force pairs in interactions

Every interaction involves two objects

Definition

Interaction pair

The two forces produced by one interaction, one acting on each of the two interacting objects.

  1. An interaction always involves two objects, so it always produces two forces, one acting on each object.
  2. If object A pushes or pulls on object B, then object B also pushes or pulls on object A at the same time.
  3. These two forces together are called an interaction pair.
  4. The two forces in an interaction pair have three features in common.
    1. They are the same type of force, so a gravitational force is always paired with a gravitational force and a normal contact force with a normal contact force.
    2. They act along the same line but point in opposite directions.
    3. They start and stop at the same instant, because they are produced by the same interaction.
  5. The two forces act on different objects, which is the most important feature of an interaction pair.
  6. To describe a pair fully, use the pattern "the force on object A by object B" for one force and "the force on object B by object A" for the other.

Pairs in contact interactions

  1. In a contact interaction, the two surfaces press on each other or rub against each other, so each surface feels a force.
  2. When your hand pushes a door, your hand exerts a force on the door and the door exerts a force back on your hand, which is why you can feel the door pressing on your palm.
  3. When a mug rests on a desk, the mug pushes down on the desk and the desk pushes up on the mug.
    1. Both of these forces are normal contact forces, acting at right angles to the desk surface.
    2. One force acts on the desk and the other acts on the mug.
  4. When you walk, your shoe pushes backwards on the pavement and the pavement pushes forwards on your shoe.
    1. Both of these forces are friction forces between the touching surfaces.
    2. The forward force on your shoe is the force that moves you along.
  5. When a hammer hits a nail, the hammer pushes the nail into the wood and the nail pushes back on the hammer, which is why the hammer stops.

Pairs in non-contact interactions

  1. Non-contact interactions also produce a force on each of the two objects, even though there is a gap between them.
  2. In a gravitational interaction, the Earth pulls a falling apple downwards and the apple pulls the Earth upwards.
    1. The pull of the apple on the Earth has no noticeable effect, because the Earth has an enormous mass.
    2. The force on the Earth is still there, and it acts on the Earth, not on the apple.
  3. In a magnetic interaction, a magnet pulls a steel paper clip towards it and the paper clip pulls the magnet towards itself.
    1. If the magnet is placed on a floating cork, the magnet visibly moves towards a paper clip held near it, which shows the paper clip really does exert a force on the magnet.
  4. In an electrostatic interaction, two charged balloons hung side by side by threads push each other apart.
    1. Each balloon swings away from the other, which shows that both balloons experience a force.
  5. Gravity between the Earth and the Moon is also an interaction pair, with the Earth pulling the Moon and the Moon pulling the Earth.
    1. The Moon's pull on the Earth's oceans helps to cause the tides.

Choosing the object you are studying

  1. Because the two forces in a pair act on different objects, only one force from each pair acts on any object you choose to study.
  2. To find the forces acting on an object, list every interaction that object is involved in and keep only the force that acts on it from each one.
  3. For a mug resting on a desk, the mug takes part in two interactions.
    1. The interaction with the Earth gives the mug a downward gravitational force, its weight.
    2. The interaction with the desk gives the mug an upward normal contact force.
  4. The partner forces, the mug pulling up on the Earth and the mug pushing down on the desk, act on other objects, so they are left out when you study the mug.
Example

Describing the pair when a footballer heads a ball

  • A footballer heads a ball that is moving towards her.
  • The interaction is a contact interaction between her head and the ball.
  • Her head exerts a force on the ball, which changes the direction in which the ball moves.
  • The ball exerts a force on her head at the same instant, which she feels as a jolt on her forehead.
  • Both forces are normal contact forces, they point in opposite directions, and one acts on the ball while the other acts on her head.
Example

Describing the pair for a magnet and a fridge door

  • A magnet holds a takeaway menu on a steel fridge door.
  • The interaction between the magnet and the door is a magnetic interaction.
  • The fridge door is attracted towards the magnet and the magnet is attracted towards the fridge door.
  • The magnet also presses on the menu and the door, so there are contact interactions as well, each with its own pair of forces.
Common Mistake
  • Do not describe only one force when you explain an interaction, because every interaction produces a force on each object.
  • Do not put both forces of an interaction pair on the same object, because they always act on different objects.
  • Do not assume that the larger or heavier object exerts the force and the smaller object does not, because both objects exert a force on each other.
Exam technique

Describing interaction pairs

  • Name both objects in every answer, because a mark is usually given for identifying what interacts with what.
  • Use the words "on" and "by" to make it clear which object each force acts on.
  • State the type of force for both members of the pair, and make sure it is the same type.
  • If the question shows a diagram, check which object it asks about before you list any forces.
Self review
  • How many forces does a single interaction between two objects produce?
  • On which objects do the two forces of an interaction pair act?
  • Describe the interaction pair when a mug rests on a desk.
  • Why does the apple's pull on the Earth have no noticeable effect?
  • Explain why the mug pushing down on the desk is not included when you list the forces on the mug.

2.2.3 Forces as vectors

Force is a vector

Definition

Vector quantity

A quantity that has both magnitude and direction.

Definition

Scalar quantity

A quantity that has magnitude but no direction.

Definition

Magnitude

The size of a quantity, given as a number and a unit.

  1. Every physical quantity is either a scalar quantity or a vector quantity.
  2. A scalar has a size only, such as a mass of 5 kg5\ \text{kg}5 kg, a time of 20 s20\ \text{s}20 s or an energy of 300 J300\ \text{J}300 J.
  3. A vector has a size and a direction, such as a displacement of 40 m40\ \text{m}40 m north or a velocity of 3 m/s3\ \text{m/s}3 m/s to the left.
  4. Force is a vector, because a push or pull always acts in a particular direction.
  5. A 20 N20\ \text{N}20 N push to the right has a completely different effect from a 20 N20\ \text{N}20 N push to the left, even though the magnitude is the same.
  6. To describe a force fully, you must give its magnitude in newtons and its direction, such as "600 N600\ \text{N}600 N downwards" or "50 N50\ \text{N}50 N to the left".
  7. Direction can be given in words (up, down, left, right, forwards, backwards) or by using positive and negative signs along a chosen line.
    1. If you choose to the right as positive, a 50 N50\ \text{N}50 N force to the left is written as −50 N-50\ \text{N}−50 N.

Drawing forces as arrows

  1. A vector is shown on a diagram as an arrow.
  2. The direction of the arrow shows the direction in which the force acts.
  3. The length of the arrow shows the magnitude of the force, so a larger force is drawn with a longer arrow.
  4. Two forces of equal size are drawn as arrows of equal length, even if they point in different directions.
  5. The arrow is drawn starting from the point where the force acts on the object.
    1. The weight of an object is drawn from the middle of the object, pointing straight down.
    2. A push from a hand or a pull from a rope is drawn from the point where the hand or rope touches the object.
  6. An arrow drawn to scale uses a fixed length for each newton, so the size of a force can be read off the diagram.
    1. With a scale of 1 cm1\ \text{cm}1 cm to 10 N10\ \text{N}10 N, a 35 N35\ \text{N}35 N force is drawn as an arrow 3.5 cm3.5\ \text{cm}3.5 cm long.
    2. Every arrow on the same diagram must use the same scale, or the lengths cannot be compared.
  7. Each arrow is labelled with the name of the force, and with its value when the value is known.

Free body diagrams

Definition

Free body diagram

A diagram of a single object showing every force that acts on it as a labelled arrow.

  1. A free body diagram shows one object on its own, with an arrow for every force that acts on it.
  2. The object is drawn as a simple shape, such as a box or a dot, because its detailed appearance does not matter.
  3. The surroundings, such as the floor, the rope or the Earth, are not drawn, because only their forces on the object are needed.
  4. Forces that the object exerts on other things are not included, because they act on those other objects.
  5. To draw a free body diagram, work through the object's interactions one at a time.
    1. Start with weight, because every object near the Earth has a gravitational force pulling it straight down.
    2. Add a force for every surface, string, rope or fluid that touches the object.
    3. Check that each arrow points in the right direction and that larger forces have longer arrows.
    4. Label every arrow with the name of the force.
  6. The forces that appear most often on free body diagrams act in predictable directions.
    1. Weight acts vertically downwards.
    2. The normal contact force acts at right angles to the surface, so it acts upwards from a flat floor.
    3. Friction, air resistance and drag act in the opposite direction to the motion, or to the attempted motion.
    4. Tension acts along a string or rope, pulling away from the object.
    5. Thrust or driving force acts in the direction in which an engine, a motor or a person pushes the object.
    6. Upthrust acts vertically upwards on an object in a liquid or gas.
  7. In the free body diagram of the car, the vertical arrows for weight and normal contact force are the same length, while the forward thrust arrow is longer than the backward friction arrow.
  8. Reading the arrows tells you which forces are larger, which is the first step in working out how the object will move.

A car driving along a road shown beside its free body diagram, with arrows for the normal contact force, the weight, the thrust and the friction, where the longer thrust arrow shows that thrust is larger than friction.

Example

Free body diagram of a hanging lamp

  • A lamp hangs from the ceiling of a classroom on a single cable and does not move.
  • Draw the lamp as a small box.
  • Draw an arrow straight down from the centre of the box and label it weight.
  • Draw an arrow straight up from the top of the box, where the cable is attached, and label it tension.
  • Make the two arrows the same length, because the lamp is not moving and the two forces are equal in size.
  • Do not draw the ceiling or the force of the lamp on the cable, because neither of these is a force on the lamp.
Example

Free body diagram of a box being dragged

  • A removal worker drags a heavy box across a carpet to the right by pulling on a horizontal rope, and the box speeds up.
  • The box has four forces acting on it.
    • Weight acts downwards from the centre of the box.
    • The normal contact force from the carpet acts upwards.
    • Tension in the rope acts to the right.
    • Friction from the carpet acts to the left, opposing the sliding.
  • The tension arrow is drawn longer than the friction arrow, because the box is speeding up to the right.
Example

Converting a force into an arrow length

  • A force of 240 N240\ \text{N}240 N is to be drawn with a scale of 1 cm1\ \text{cm}1 cm to 40 N40\ \text{N}40 N.
arrow length=240 N40 N per cm=6.0 cm \text{arrow length}=\dfrac{240\ \text{N}}{40\ \text{N per cm}}=6.0\ \text{cm} arrow length=40 N per cm240 N​=6.0 cm
  • The force is drawn as an arrow 6.0 cm6.0\ \text{cm}6.0 cm long in the direction of the force.
Common Mistake
  • Do not draw a force on a free body diagram only because the object is moving in that direction, because a ball rolling to the right and slowing down has only a backward horizontal force on it.
  • Do not include forces that the object exerts on other objects, such as a book pushing down on a table, on the book's free body diagram.
  • Do not draw equal forces with different arrow lengths, because examiners use arrow length to check your understanding of magnitude.
Exam technique

Drawing force diagrams for marks

  • Use a ruler, and put a clear arrowhead on every arrow.
  • Start each arrow on the object, at the point where the force acts.
  • Label every arrow with the correct physics name, and include the value with its unit when it is given.
  • Make the lengths of the arrows match the sizes of the forces, as a mark is often awarded for equal or unequal lengths drawn correctly.
Self review
  • Why is force a vector quantity?
  • What two features of a force does an arrow show?
  • What is a free body diagram?
  • In which direction does the normal contact force on a box on a flat floor act?
  • How long is the arrow for a 90 N90\ \text{N}90 N force on a scale of 1 cm1\ \text{cm}1 cm to 20 N20\ \text{N}20 N?

2.2.4 Newton's first law

Resultant force

Definition

Resultant force

The single force that has the same effect on an object as all of the forces acting on it combined.

Definition

Balanced forces

Forces acting on an object that cancel each other out so that the resultant force is zero.

  1. Most objects have several forces acting on them at once, and their combined effect decides how the object moves.
  2. The resultant force is the single force that would have the same effect on the object as all of the forces acting together.
  3. The resultant force is also called the net force.
  4. Forces acting along the same line are combined by taking their directions into account.
    1. Forces in the same direction add together.
    2. Forces in opposite directions partly or completely cancel, so the smaller force is taken away from the larger one.
    3. The resultant acts in the direction of the larger total force.
  5. When the forces on an object cancel completely, they are balanced forces and the resultant force is zero.
  6. When the forces do not cancel, they are unbalanced and there is a non-zero resultant force in one direction.
  7. A zero resultant force does not mean that no forces are acting, because several forces can act and still cancel out.

Newton's first law

Definition

Newton's first law

An object remains at rest, or continues to move at a constant velocity, unless a resultant force acts on it.

  1. Newton's first law describes what an object does when the resultant force on it is zero.
  2. If the resultant force on a stationary object is zero, the object stays at rest.
  3. If the resultant force on a moving object is zero, the object keeps moving at the same speed in the same direction, which means at a constant velocity.
  4. An object does not need a forward resultant force to keep moving, it only needs a resultant force to change the way it is moving.
  5. On Earth, moving objects usually slow down because of friction and air resistance, which is why it can seem as if motion needs a force to keep it going.
  6. In deep space, far from any planet or star, a space probe that switches off its engines keeps travelling in a straight line at a constant speed for years.

Uniform velocity

Definition

Uniform velocity

Motion at a constant speed in a constant direction.

  1. An object moving with uniform velocity covers equal distances in equal times without changing direction.
  2. An object at uniform velocity has balanced forces acting on it, so the resultant force on it is zero.
  3. For a car cruising along a straight motorway at a steady 30 m/s30\ \text{m/s}30 m/s, the forward driving force from the engine is exactly balanced by the backward resistive forces of air resistance and friction.
    1. If the driving force is 2500 N2500\ \text{N}2500 N forwards, the resistive forces add up to 2500 N2500\ \text{N}2500 N backwards.
    2. The car has the same velocity at the end of each second as it had at the start, because nothing is unbalanced.
  4. Vertically, the weight of the car is balanced by the normal contact force from the road, so the car neither rises nor sinks.

Changing speed or direction

  1. Velocity is a vector, so it changes whenever the speed changes, the direction changes, or both change.
  2. A change in velocity can only happen when there is a non-zero resultant force on the object.
  3. The way the velocity changes depends on the direction of the resultant force.
    1. A resultant force in the same direction as the motion makes the object speed up.
    2. A resultant force in the opposite direction to the motion makes the object slow down.
    3. A resultant force at an angle to the motion makes the object change direction.
    4. A resultant force on a stationary object makes it start to move in the direction of the resultant.
  4. When a bus driver presses the accelerator, the driving force becomes larger than the resistive forces, so the bus speeds up.
  5. When the driver brakes, the backward forces become larger than any forward force, so the bus slows down.
  6. When a football is kicked sideways as it rolls past a player, the sideways force changes the direction of the ball.
Example

Explaining the motion of a train

  • A train is pulled out of a station by a forward driving force of 90 kN90\ \text{kN}90 kN while resistive forces of 30 kN30\ \text{kN}30 kN act backwards.
  • The forces are unbalanced, because 90 kN90\ \text{kN}90 kN forwards is larger than 30 kN30\ \text{kN}30 kN backwards.
resultant force=90 kN−30 kN=60 kN forwards \text{resultant force}=90\ \text{kN}-30\ \text{kN}=60\ \text{kN}\ \text{forwards} resultant force=90 kN−30 kN=60 kN forwards
  • The resultant force acts in the direction of motion, so the train speeds up.
  • Later the resistive forces rise to 90 kN90\ \text{kN}90 kN, the forces become balanced and the resultant force is zero.
  • By the first law, the train then keeps moving at a constant velocity.
Example

Explaining the motion of a sliding puck

  • An ice hockey puck slides across an ice rink after being struck.
  • The friction between the puck and the ice is very small, so the resultant force on the puck is almost zero.
  • The puck therefore travels a long way in a nearly straight line at a nearly constant speed.
  • When the puck hits the barrier, the barrier exerts a large force on it, so the puck changes direction and bounces back.
Practical

Gliders on a linear air track

  • Aim: to observe how gliders move when the resultant force on them is almost zero, and how their velocities change only when they collide.
  • Apparatus: a linear air track with an air blower, two gliders, two interrupt cards, two light gates connected to a data logger, a pair of magnetic or spring buffers that fit on the gliders, and a spirit level.
  • Method:
    • Switch on the blower so that air streams out of the small holes along the track and lifts each glider on a thin cushion of air, which almost removes friction.
    • Level the track by adjusting its feet until a glider placed in the middle stays still and does not drift towards either end.
    • Fix an interrupt card upright on each glider and set up the two light gates about 50 cm50\ \text{cm}50 cm apart along the track.
    • Give one glider a gentle push and let it pass through both light gates without touching it again.
    • Record the speed shown at each light gate and compare the two values.
    • Place the second glider at rest between the two gates, push the first glider towards it and watch the collision.
    • Record the speeds of both gliders through the gates before and after the collision.
    • Repeat with the buffers swapped so that the gliders either bounce apart or stick together.
  • Results: a single glider records almost the same speed at both light gates, and each glider changes speed or direction only during the short time the gliders are in contact.
  • Conclusion: with a resultant force of almost zero the gliders move at a nearly uniform velocity, and their velocities change only when the collision gives each glider a resultant force.
  • Watch out: a track that is not level gives the glider a small resultant force down the slope, so it speeds up or slows down between the gates and the result looks like evidence against the first law.
  • Safety: keep fingers clear of the gliders during collisions and do not lean on the track, as it is easily bent.
Practical

Balloon glider on a string

  • Aim: to see how a resultant force changes the velocity of an object, and how the object moves when that force stops.
  • Apparatus: a balloon, a drinking straw, sticky tape, about 5 m5\ \text{m}5 m of smooth nylon string, two clamp stands, a metre rule or tape measure, a balloon pump and a phone to film the motion.
  • Method:
    • Thread the string through the straw, then tie the ends to the two clamp stands so that the string is tight and level across the bench.
    • Inflate the balloon with the pump, pinch the neck closed and tape the balloon to the straw with the neck pointing along the string.
    • Mark the string every 50 cm50\ \text{cm}50 cm with small pieces of tape and set the phone to film the whole length of the string.
    • Hold the balloon at the starting mark, then release the neck without pushing the balloon.
    • Use the video, played frame by frame, to note the time at which the balloon passes each mark.
    • Repeat three times with the balloon inflated to the same size each time, measured by wrapping a string around its widest part.
  • Results: while air escapes, the balloon covers more distance in each equal time interval, and once the balloon is empty it covers less distance in each interval until it stops.
  • Conclusion: escaping air gives the balloon a forward resultant force that makes it speed up, and once the air runs out, friction from the string gives a backward resultant force that slows it down.
  • Watch out: a sagging string makes the balloon climb or fall, so part of the change in speed is caused by gravity rather than by the escaping air.
  • Safety: wear eye protection, because a balloon can burst when it is over-inflated.
Common Mistake
  • Do not say that a moving object needs a resultant force to keep it moving, because a resultant force is only needed to change its velocity.
  • Do not say there are "no forces" on an object moving at constant velocity, because the forces are present but balanced.
  • Do not forget that a change in direction is a change in velocity, so it also needs a resultant force.
Exam technique

Explaining motion with the first law

  • Start by stating whether the forces are balanced or unbalanced, and say which force is larger.
  • Then state the size of the resultant force, using the phrase "resultant force is zero" for balanced forces.
  • Finish with the effect on the motion: stays at rest, constant velocity, speeds up, slows down or changes direction.
  • In a practical question about an air track, link the cushion of air to reduced friction and so to a resultant force close to zero.
Self review
  • What is meant by the resultant force on an object?
  • State Newton's first law.
  • What can you say about the forces on a car moving at a steady velocity on a straight road?
  • Give three ways in which the velocity of an object can change.
  • Why does a glider on a linear air track move at a nearly constant velocity?

2.2.5 Resolving forces with vector diagrams

Scale vector diagrams

Definition

Scale drawing

An accurate diagram in which every length represents a quantity according to a fixed scale.

  1. A vector diagram shows forces as arrows so that you can find their combined effect by drawing and measuring.
  2. Every arrow is drawn to scale, so its length represents the size of the force and its direction matches the direction of the force.
  3. Choose a scale that makes the diagram as large as the space allows, such as 1 cm1\ \text{cm}1 cm to 10 N10\ \text{N}10 N, because a large diagram gives more accurate measurements.
  4. Write the scale beside the diagram so that every length can be converted back into a force.
  5. Use a sharp pencil and a ruler for lengths, and a protractor for any angle.
Note

At GCSE, vector diagrams only involve forces that are parallel or perpendicular to each other.

Combining forces

Definition

Resultant force

The single force that has the same effect on an object as all of the forces acting on it combined.

  1. The resultant force found from a vector diagram is the single force with the same effect as all the forces acting together.
  2. To add forces on a vector diagram, use the tip-to-tail method.
    1. Draw the first force arrow to scale.
    2. Draw the second force arrow to scale, starting from the tip of the first arrow.
    3. Draw the resultant as a straight arrow from the tail of the first arrow to the tip of the last arrow.
    4. Measure the length of the resultant and convert it into newtons using the scale, then measure its angle if a direction is needed.
  3. For parallel forces, the arrows lie along one line.
    1. Two forces in the same direction give a resultant equal to their sum.
    2. Two forces in opposite directions give a resultant equal to their difference, in the direction of the larger force.
  4. For perpendicular forces, the two arrows form two sides of a right-angled triangle and the resultant is the third side.
  5. Because the triangle is right-angled, you can check a measured resultant with Pythagoras' theorem, FR=F1 2+F2 2F_{\text{R}}=\sqrt{F_1^{\,2}+F_2^{\,2}}FR​=F12​+F22​​.
  6. You can also draw both arrows from the same point, complete a rectangle and draw the diagonal, which gives the same resultant.
Example

Resultant of two perpendicular forces

  • A boat is pushed east by its engine with a force of 300 N300\ \text{N}300 N while the wind pushes it north with a force of 400 N400\ \text{N}400 N.
  • With a scale of 1 cm1\ \text{cm}1 cm to 100 N100\ \text{N}100 N, draw a 3.0 cm3.0\ \text{cm}3.0 cm arrow to the east, then a 4.0 cm4.0\ \text{cm}4.0 cm arrow to the north starting from its tip.
  • Join the tail of the first arrow to the tip of the second arrow and measure the resultant as 5.0 cm5.0\ \text{cm}5.0 cm.
FR=5.0 cm×100 N per cm=500 N F_{\text{R}}=5.0\ \text{cm}\times100\ \text{N per cm}=500\ \text{N} FR​=5.0 cm×100 N per cm=500 N
  • The protractor shows the resultant at about 53∘53^\circ53∘ to the east direction, so the boat is pushed with 500 N500\ \text{N}500 N in a direction between east and north.
check: 3002+4002=250 000=500 N \text{check: }\sqrt{300^2+400^2}=\sqrt{250\,000}=500\ \text{N} check: 3002+4002​=250000​=500 N

Resolving a force

Definition

Resolving a force

Replacing a single force with two perpendicular components that together have the same effect as the original force.

Definition

Component

One of the two perpendicular parts into which a force is split when it is resolved.

  1. Resolving a force means splitting it into two components that are perpendicular to each other.
  2. The two components are usually chosen to be horizontal and vertical, or along and at right angles to a slope.
  3. The two components together have exactly the same effect as the original force, so they replace it in a calculation rather than being added to it.
  4. Resolving is useful because forces along each perpendicular direction can then be combined separately.
  5. To resolve a force by scale drawing, follow a fixed set of steps.
    1. Draw the force to scale at the correct angle, starting from one corner of your diagram.
    2. Draw a horizontal line and a vertical line from the start of the arrow.
    3. From the tip of the arrow, draw lines at right angles back to the horizontal and vertical lines to complete a rectangle.
    4. Measure the horizontal and vertical sides of the rectangle and convert each length into newtons with the scale.
  6. The horizontal component is the part of the force that moves an object along the ground, and the vertical component is the part that lifts it or presses it down.

A force drawn at an angle to the horizontal and resolved into a horizontal component and a vertical component, which form the two sides of a rectangle with the original force as its diagonal.

Example

Resolving the pull on a sledge

  • A child pulls a sledge across snow with a rope that makes an angle of 40∘40^\circ40∘ above the horizontal, and the tension in the rope is 60 N60\ \text{N}60 N.
  • With a scale of 1 cm1\ \text{cm}1 cm to 10 N10\ \text{N}10 N, draw a 6.0 cm6.0\ \text{cm}6.0 cm arrow at 40∘40^\circ40∘ above the horizontal.
  • Complete the rectangle and measure the horizontal side as 4.6 cm4.6\ \text{cm}4.6 cm and the vertical side as 3.9 cm3.9\ \text{cm}3.9 cm.
horizontal component=4.6 cm×10 N per cm=46 N \text{horizontal component}=4.6\ \text{cm}\times10\ \text{N per cm}=46\ \text{N} horizontal component=4.6 cm×10 N per cm=46 N vertical component=3.9 cm×10 N per cm=39 N \text{vertical component}=3.9\ \text{cm}\times10\ \text{N per cm}=39\ \text{N} vertical component=3.9 cm×10 N per cm=39 N
  • Only the 46 N46\ \text{N}46 N horizontal component pulls the sledge forwards, while the 39 N39\ \text{N}39 N vertical component lifts the front of the sledge slightly.

Equilibrium

Definition

Equilibrium

The state of an object when the resultant force acting on it is zero.

  1. An object is in equilibrium when the resultant force on it is zero.
  2. An object in equilibrium either stays at rest or moves at a constant velocity.
  3. When the forces on an object in equilibrium are drawn tip to tail, the last arrow finishes exactly where the first arrow started, so the arrows form a closed shape.
    1. Two forces in equilibrium form a closed shape only if they are equal in size and opposite in direction.
    2. Three forces in equilibrium form a closed triangle.
  4. If a vector diagram does not close, the gap between the start and the finish is the resultant force, so the object is not in equilibrium.
  5. To find the force needed for equilibrium, find the resultant of the other forces and add a force of the same size in the opposite direction.
  6. Resolving gives a second test for equilibrium, because the horizontal components must cancel and the vertical components must cancel.
Example

Finding the force that keeps a crate in equilibrium

  • Two ropes pull a crate on a smooth floor, one with 80 N80\ \text{N}80 N to the east and one with 60 N60\ \text{N}60 N to the north.
  • Drawn tip to tail with a scale of 1 cm1\ \text{cm}1 cm to 20 N20\ \text{N}20 N, the arrows are 4.0 cm4.0\ \text{cm}4.0 cm and 3.0 cm3.0\ \text{cm}3.0 cm long and the gap that closes the triangle is 5.0 cm5.0\ \text{cm}5.0 cm.
F=5.0 cm×20 N per cm=100 N F=5.0\ \text{cm}\times20\ \text{N per cm}=100\ \text{N} F=5.0 cm×20 N per cm=100 N
  • A third rope must pull with 100 N100\ \text{N}100 N in the direction that closes the triangle, which points between west and south.
  • With this third force, the three arrows form a closed triangle and the crate is in equilibrium.
Common Mistake
  • Do not draw the components and the original force on the same diagram as if they were three separate forces, because the components replace the original force.
  • Do not start the second arrow from the tail of the first arrow when you use the tip-to-tail method, because this gives the wrong resultant.
  • Do not assume that an object in equilibrium must be stationary, because it may be moving at a constant velocity.
Exam technique

Scale diagram questions

  • State your scale and use the whole of the space provided, because a small diagram loses accuracy.
  • Draw the resultant with a double arrowhead or label it clearly so that it is not confused with the original forces.
  • Give the resultant with its unit and its direction, as both are usually needed for full marks.
  • Expect your measured answer to be accepted within a small range, so measure carefully rather than rounding early.
Self review
  • Describe the tip-to-tail method for adding two forces.
  • Find the resultant of a 5 N5\ \text{N}5 N force to the east and a 12 N12\ \text{N}12 N force to the north.
  • What does it mean to resolve a force?
  • Why should the components not be added to the original force?
  • What shape do three forces in equilibrium make when drawn tip to tail?

2.2.6 Forces acting on an isolated object

Isolated objects and systems

Definition

Isolated system

An object, or a group of objects, that is considered separately from its surroundings so that the external forces acting on it can be identified.

Definition

External force

A force exerted on a system by an object outside the system.

  1. To analyse the forces on something, you first draw an imaginary boundary around it and treat it as an isolated object or system.
  2. A system can be a single object, such as a ball bearing, or a group of objects that move together, such as a skydiver and her parachute.
  3. Only external forces, which are exerted on the system by objects outside the boundary, affect how the whole system moves.
  4. Forces between parts of the system are internal forces, and they cancel out in pairs, so they do not change the motion of the system as a whole.
  5. The choice of boundary decides which forces count as external.
    1. If the system is the skydiver alone, the tension in the parachute harness is an external force pulling her upwards.
    2. If the system is the skydiver and parachute together, the harness tension is internal and the external forces are the total weight and the air resistance on the whole system.
  6. For each external force, you should be able to name the object that exerts it, such as the Earth for weight or the air for air resistance.

Forces on a falling object

Definition

Drag

The resistive force that acts on an object moving through a fluid, in the opposite direction to its motion.

Definition

Terminal velocity

The constant velocity reached by a falling object when the resistive forces acting on it balance its weight.

  1. An object falling through air or a liquid has two main forces acting on it.
    1. Its weight, from the Earth, acts downwards and stays the same throughout the fall.
    2. Drag from the fluid acts upwards, opposite to the motion, and increases as the speed increases.
  2. An object in a liquid also feels a small upthrust, which acts upwards alongside the drag.
  3. The motion of the falling object follows a fixed sequence of stages.
    1. At the moment of release the speed is zero, so the drag is zero and the resultant force is the weight, which gives the largest acceleration.
    2. As the object speeds up, the drag increases, so the resultant downward force gets smaller and the acceleration decreases.
    3. Eventually the drag grows until it equals the weight, so the forces are balanced and the resultant force is zero.
    4. With a resultant force of zero, the acceleration is zero and the object falls at a constant speed called its terminal velocity.
  4. Terminal velocity depends on the size and shape of the object as well as its weight.
    1. A larger surface area facing the direction of motion produces more drag at a given speed, so terminal velocity is reached at a lower speed.
    2. A heavier object needs more drag to balance its weight, so it has a higher terminal velocity than a lighter object of the same shape.
    3. A more viscous fluid, such as cold syrup or glycerol, produces more drag at a given speed than a runny fluid, so the terminal velocity is lower.

Three free body diagrams of a falling object: at release with only weight acting, while it speeds up with weight larger than air resistance, and at terminal velocity with weight and air resistance equal and opposite.

The skydiver

  1. A skydiver jumping from a plane shows every stage of falling motion.
  2. Just after jumping, her weight is much larger than the air resistance, so she accelerates downwards.
  3. As she falls faster, the air resistance increases until it equals her weight, and she reaches a terminal velocity of roughly 55 m/s55\ \text{m/s}55 m/s.
  4. When she opens her parachute, the surface area increases greatly, so the air resistance suddenly becomes much larger than her weight.
  5. The resultant force now acts upwards, so she decelerates while still moving downwards.
  6. As she slows down, the air resistance decreases until it equals her weight again.
  7. She then falls at a new, much lower terminal velocity of about 5 m/s5\ \text{m/s}5 m/s, which is slow enough for a safe landing.

Vehicles and top speed

  1. The same ideas apply to a vehicle moving horizontally, where the driving force takes the place of weight.
  2. When a car pulls away, the driving force is larger than the resistive forces of air resistance and friction, so the car accelerates.
  3. As the car speeds up, air resistance increases, so the resultant force and the acceleration both decrease.
  4. When the resistive forces equal the maximum driving force, the resultant force is zero and the car has reached its top speed.
  5. A car with a more streamlined shape has less air resistance at each speed, so it can reach a higher top speed with the same driving force.
  6. Adding a roof box increases the air resistance, so the top speed falls and more fuel is used at motorway speeds.
Example

Stages in the fall of a ball bearing through glycerol

  • A steel ball bearing is released from rest at the surface of glycerol in a tall tube.
  • At first the drag is zero, so the resultant force is the weight minus the small upthrust, and the ball accelerates downwards.
  • As the speed rises, the drag increases, so the resultant force and the acceleration decrease.
  • When the drag plus the upthrust equals the weight, the resultant force is zero.
weight=drag+upthrust \text{weight}=\text{drag}+\text{upthrust} weight=drag+upthrust
  • From this point the ball falls at its terminal velocity, covering equal distances in equal times.
Example

Calculating a terminal velocity from measurements

  • A model parachute falls the last 1.50 m1.50\ \text{m}1.50 m of its drop at a steady speed in a mean time of 2.50 s2.50\ \text{s}2.50 s.
v=st v=\dfrac{s}{t} v=ts​ v=1.50 m2.50 s=0.60 m/s v=\dfrac{1.50\ \text{m}}{2.50\ \text{s}}=0.60\ \text{m/s} v=2.50 s1.50 m​=0.60 m/s
  • The terminal velocity of the model parachute is 0.60 m/s0.60\ \text{m/s}0.60 m/s.
Practical

Terminal velocity of a model parachute

  • Aim: to design and build a parachute for a fixed mass, measure its terminal velocity and investigate how the canopy area affects it.
  • Apparatus: squares of thin plastic bin bag with sides of 20 cm20\ \text{cm}20 cm, 30 cm30\ \text{cm}30 cm, 40 cm40\ \text{cm}40 cm, 50 cm50\ \text{cm}50 cm and 60 cm60\ \text{cm}60 cm, cotton thread, sticky tape, a 20 g20\ \text{g}20 g lump of modelling clay, scissors, a metre rule, a tape measure, masking tape, a stopwatch and a phone to film the fall.
  • Method:
    • Tape four equal 30 cm30\ \text{cm}30 cm lengths of thread to the corners of a canopy and tie their free ends to the modelling clay so that the mass hangs centrally.
    • Choose a safe drop point at least 3 m3\ \text{m}3 m high, such as a stairwell balcony supervised by your teacher, and measure the drop height with a tape measure.
    • Mark a measured section of 1.50 m1.50\ \text{m}1.50 m at the bottom of the drop with masking tape on the wall behind the parachute.
    • Hold the parachute open by the top of the canopy and release it without pushing it.
    • Film the fall against the wall, then use the video to find the time taken to fall through the marked section.
    • Check that the parachute takes the same time to fall through the top half and the bottom half of the section, as this shows it has reached terminal velocity.
    • Repeat the drop three times, discard any anomalous time and calculate a mean.
    • Calculate the terminal velocity by dividing the length of the section by the mean time.
    • Repeat the whole process for each canopy size, keeping the same mass, thread and drop height.
  • Variables: the canopy area is the independent variable, the terminal velocity is the dependent variable, and the mass, the thread length, the canopy material and the drop height are control variables.
  • Results: larger canopies take longer to fall through the marked section, so they give a lower terminal velocity.
  • Conclusion: increasing the canopy area increases the air resistance at a given speed, so the air resistance balances the weight at a lower speed.
  • Watch out: draughts from open windows or doors push the parachute sideways and change the fall time, so close them and avoid drops where the canopy swings.
  • Safety: never stand on chairs or desks to release the parachute, and keep the landing area clear of other people.
Practical

Ball bearings falling through glycerol

  • Aim: to measure the terminal velocity of ball bearings falling through glycerol and investigate how it depends on the size of the ball bearing or the temperature of the glycerol.
  • Apparatus: a 1 dm31\ \text{dm}^31 dm3 measuring cylinder filled with glycerol, steel ball bearings of different diameters, a micrometer or vernier callipers, three rubber bands, a metre rule, a stopwatch, a strong magnet, paper towels, a thermometer and a large beaker of hot water for a water bath.
  • Method:
    • Measure the diameter of each ball bearing with the micrometer.
    • Place one rubber band about 10 cm10\ \text{cm}10 cm below the surface of the glycerol, and two more bands 10 cm10\ \text{cm}10 cm and 20 cm20\ \text{cm}20 cm below the first.
    • Hold a ball bearing just above the centre of the glycerol surface and let it go so that it falls down the middle of the cylinder, away from the walls.
    • Start the stopwatch as the ball passes the first band and record the times at which it passes the second and third bands.
    • Check that the ball takes the same time to fall between each pair of bands, which shows it is travelling at terminal velocity.
    • Collect the ball by sliding the magnet up the outside of the cylinder.
    • Repeat three times for each size of ball bearing and calculate the mean time for the 20 cm20\ \text{cm}20 cm section.
    • To test temperature instead, stand the cylinder in the water bath, record the glycerol temperature with the thermometer and repeat with one size of ball bearing at several temperatures.
  • Variables: the ball bearing diameter or the glycerol temperature is the independent variable, the terminal velocity is the dependent variable, and the distance between the bands, the release point and the material of the balls are control variables.
  • Results: larger ball bearings fall faster through the measured section, and every ball falls faster through warmer glycerol.
  • Conclusion: a larger ball bearing has a weight that increases more than its drag, so it reaches a higher terminal velocity, while warmer glycerol is less viscous, so it produces less drag and gives a higher terminal velocity.
  • Watch out: a ball released near the wall of the cylinder is slowed by the wall, and parallax error when reading the bands makes the times less accurate, so view each band at eye level.
  • Safety: wipe up any spilled glycerol straight away because it makes the floor slippery, and handle the hot water bath with care.
Common Mistake
  • Do not say that there is "no force" at terminal velocity, because the weight and the drag are both still acting but are balanced.
  • Do not say that a skydiver moves upwards when the parachute opens, because she is still moving downwards but is slowing down.
  • Do not say that drag stays constant during a fall, because drag increases as the speed increases.
Exam technique

Explaining terminal velocity

  • Name both forces and say which is larger at each stage, using "weight" and "air resistance" or "drag".
  • Include the key link that air resistance increases with speed, as this is usually a separate mark.
  • At terminal velocity, write "air resistance equals weight, so the resultant force is zero and there is no acceleration".
  • For the parachute opening, explain that the larger area increases air resistance, giving an upward resultant force and a deceleration.
Self review
  • What is meant by an external force on a system?
  • Why does the acceleration of a falling object decrease as it falls?
  • State the condition for an object to reach terminal velocity.
  • Explain why a skydiver slows down when she opens her parachute.
  • Why do ball bearings fall faster through warm glycerol than through cold glycerol?

2.2.7 Free body diagrams and resultant force

Reading a free body diagram

Definition

Free body diagram

A diagram of a single object showing every force that acts on it as a labelled arrow.

Definition

Unbalanced forces

Forces acting on an object that do not cancel each other out, so that there is a non-zero resultant force.

  1. A free body diagram shows one object with an arrow for each force acting on it, so it lets you see at a glance whether the forces are balanced or unbalanced.
  2. Forces are unbalanced when they do not cancel out, so there is a resultant force in one direction.
  3. A resultant force makes the object accelerate in the direction of the resultant force.
  4. To find the resultant force from a free body diagram, work in two perpendicular directions, usually horizontal and vertical.
    1. Add up the forces in one direction and subtract the forces in the opposite direction.
    2. Do the same for the other direction.
    3. If one direction gives zero, the resultant force lies along the other direction only.
  5. The direction of the resultant is the direction of the larger total force, so it is shown by the longer arrows on the diagram.
  6. The resultant force is not an extra force on the object, so it is not drawn as another arrow alongside the real forces.

Resultant force along one line

Definition

Resultant force

The single force that has the same effect on an object as all of the forces acting on it combined.

  1. When all of the forces act along one line, choose one direction as positive and treat forces in the other direction as negative.
  2. The resultant force is the sum of the forces with their signs.
  3. A positive answer means the resultant acts in the positive direction, and a negative answer means it acts in the opposite direction.
  4. More than two forces can act along the same line, such as a car with a driving force forwards and both air resistance and friction backwards.
Example

Resultant force on an accelerating car

  • A car pulling onto a dual carriageway has a driving force of 3200 N3200\ \text{N}3200 N forwards, air resistance of 700 N700\ \text{N}700 N backwards and friction of 500 N500\ \text{N}500 N backwards.
  • Take forwards as positive.
FR=3200 N−700 N−500 N=2000 N F_{\text{R}}=3200\ \text{N}-700\ \text{N}-500\ \text{N}=2000\ \text{N} FR​=3200 N−700 N−500 N=2000 N
  • The vertical forces, weight and normal contact force, are equal and opposite, so they make no contribution.
  • The resultant force is 2000 N2000\ \text{N}2000 N forwards, so the car speeds up.

Examples of unbalanced forces

  1. Free body diagrams can describe many everyday situations where two or more forces give a resultant force.
  2. A car speeding up on a level road has a forward driving force that is larger than the total backward resistive forces.
  3. A cyclist braking has a backward friction force from the brakes and road that is larger than any forward force, so the resultant force acts backwards and the cyclist slows down.
  4. A skydiver straight after jumping has a weight that is much larger than the upward air resistance, so the resultant force acts downwards and she speeds up.
  5. A lift starting to rise has an upward tension in its cable that is larger than its weight, so the resultant force acts upwards.
  6. A rocket at lift-off has an upward thrust from its engines that is larger than its weight, so it accelerates upwards.
  7. A ball thrown upwards has its weight and air resistance both acting downwards while it rises, so the resultant force is downwards and the ball slows down.
Example

Resultant force on a lift

  • A lift with a weight of 8000 N8000\ \text{N}8000 N is pulled upwards by a cable tension of 9200 N9200\ \text{N}9200 N.
  • The free body diagram shows a longer upward arrow for tension and a shorter downward arrow for weight.
FR=9200 N−8000 N=1200 N F_{\text{R}}=9200\ \text{N}-8000\ \text{N}=1200\ \text{N} FR​=9200 N−8000 N=1200 N
  • The resultant force is 1200 N1200\ \text{N}1200 N upwards, so the lift accelerates upwards.
Example

Resultant force on a ball thrown upwards

  • A tennis ball with a weight of 0.6 N0.6\ \text{N}0.6 N is moving upwards, and the air resistance on it is 0.1 N0.1\ \text{N}0.1 N.
  • The ball is moving upwards, so air resistance acts downwards in the same direction as the weight.
FR=0.6 N+0.1 N=0.7 N F_{\text{R}}=0.6\ \text{N}+0.1\ \text{N}=0.7\ \text{N} FR​=0.6 N+0.1 N=0.7 N
  • The resultant force is 0.7 N0.7\ \text{N}0.7 N downwards, so the ball slows down as it rises.
Common Mistake
  • Do not choose the largest single force as the resultant, because the resultant depends on every force and its direction.
  • Do not draw a forward force on an object only because it is moving forwards, because a ball rising upwards has no upward force on it after it leaves your hand.
  • Do not add an arrow labelled "resultant force" to a free body diagram unless the question asks for it, because it is not a separate force.
Exam technique

Describing a resultant from a diagram

  • Name each force and state which forces are larger, using evidence from the arrow lengths or the values given.
  • Show your working for the resultant as a single line, such as 3200−700−500=2000 N3200-700-500=2000\ \text{N}3200−700−500=2000 N, so a method mark can be given even if you slip.
  • Always give the resultant with a unit and a direction, such as "2000 N2000\ \text{N}2000 N forwards".
  • Finish by linking the resultant force to the effect on the motion: speeding up, slowing down or changing direction.
Self review
  • What are unbalanced forces?
  • How do you find the resultant of several forces acting along one line?
  • Find the resultant force on a car with a driving force of 1500 N1500\ \text{N}1500 N and resistive forces totalling 900 N900\ \text{N}900 N.
  • Why does a ball thrown upwards slow down as it rises?
  • Why should the resultant force not be drawn as an extra arrow on a free body diagram?

2.2.8 Free body diagrams for balanced forces

Balanced forces

Definition

Balanced forces

Forces acting on an object that cancel each other out so that the resultant force is zero.

Definition

Equilibrium

The state of an object when the resultant force acting on it is zero.

  1. Forces on an object are balanced when they cancel each other out completely, so the resultant force is zero.
  2. An object with balanced forces acting on it is in equilibrium.
  3. Balancing is a special case, because the forces must match in both size and direction.
    1. For two forces to balance, they must be equal in size, act in opposite directions and act along the same line.
    2. For several forces to balance, the total force in each direction must equal the total force in the opposite direction.
  4. A zero resultant force means the object does not accelerate, so its velocity stays the same.
    1. A stationary object with balanced forces stays stationary.
    2. A moving object with balanced forces keeps moving at a constant speed in a straight line.
  5. Balanced forces therefore tell you that the velocity is not changing, not that the object is at rest.

Drawing balanced free body diagrams

Definition

Free body diagram

A diagram of a single object showing every force that acts on it as a labelled arrow.

  1. A free body diagram for balanced forces shows each force on the object as a labelled arrow, with balancing forces drawn as arrows of equal length pointing in opposite directions.
  2. Draw the object as a box or a dot, and draw only the forces acting on that object.
  3. Consider the vertical forces and the horizontal forces separately.
    1. The upward arrows must have the same total length as the downward arrows.
    2. The arrows pointing one way along the horizontal must have the same total length as the arrows pointing the other way.
  4. If a situation has no horizontal forces, such as an object hanging still, the diagram shows only vertical arrows.

Balanced forces on stationary objects

  1. A book on a shelf has its weight acting downwards and an equal normal contact force from the shelf acting upwards.
  2. A hanging basket outside a café has its weight acting downwards and an equal tension from its chain acting upwards.
  3. A boat floating still in a harbour has its weight acting downwards and an equal upthrust from the water acting upwards.
  4. Two teams in a tug of war that is not moving pull on the rope with equal forces in opposite directions, so the rope stays still.
  5. In each case the arrows on the free body diagram are equal and opposite, so the object stays at rest.

Balanced forces on moving objects

  1. A car at a steady speed on a straight, level road has a forward driving force balanced by backward air resistance and friction, and its weight balanced by the normal contact force from the road.
  2. A skydiver at terminal velocity has her weight balanced by an equal upward air resistance, so she falls at a constant speed.
  3. A lift rising at a steady speed has its weight balanced by an equal upward tension in its cable.
  4. A plane cruising at a constant height and speed has its weight balanced by an upward lift force and its thrust balanced by air resistance.
  5. In each case the object keeps moving, but its velocity does not change, because the resultant force is zero.
Example

Free body diagram of a lift rising at a steady speed

  • A lift moves upwards between floors at a constant speed.
  • Draw the lift as a box with one arrow pointing up, labelled tension, and one arrow pointing down, labelled weight.
  • Make the two arrows the same length, because the lift is not speeding up or slowing down.
  • The forces are equal in size and opposite in direction, so the resultant force is zero and the lift keeps rising at the same speed.
Example

Free body diagram of a car at a steady speed

  • A car drives along a flat stretch of the M6 at a constant speed.
  • Draw four arrows from the box representing the car: weight down, normal contact force up, driving force forwards and air resistance with friction backwards.
  • Make the up and down arrows equal in length, and make the forward arrow equal in length to the backward arrow.
  • Both pairs of forces balance, so the resultant force is zero and the car continues at a constant velocity.
Common Mistake
  • Do not say that balanced forces mean an object must be stationary, because it may be moving at a constant velocity.
  • Do not write "there is no force" when the forces are balanced, because the forces are still acting and only the resultant is zero.
  • Do not draw balancing arrows with different lengths, because unequal arrows show a resultant force.
Exam technique

Describing balanced situations

  • Name each force and the force that balances it, for example "the weight is balanced by the upthrust".
  • Use the phrase "equal in size and opposite in direction" for each balanced pair.
  • State that the resultant force is zero, then give the consequence: the object stays at rest or keeps moving at a constant velocity.
  • On a diagram, use a ruler so that equal forces are clearly the same length.
Self review
  • What is the resultant force on an object in equilibrium?
  • What three conditions must two forces meet to balance each other?
  • Describe the free body diagram of a boat floating at rest.
  • Which forces balance on a car moving at a steady speed on a level road?
  • Can an object with balanced forces be moving? Explain your answer.

2.2.9 Newton's second law

Newton's second law

Definition

Newton's second law

The acceleration of an object is directly proportional to the resultant force acting on it and inversely proportional to its mass.

Definition

Mass

The amount of matter in an object, measured in kilograms.

  1. Newton's second law tells you how much an object accelerates when a resultant force acts on it.
  2. The acceleration is directly proportional to the resultant force when the mass stays the same.
    1. Doubling the resultant force on a trolley doubles its acceleration.
  3. The acceleration is inversely proportional to the mass when the resultant force stays the same.
    1. Doubling the mass of a trolley halves its acceleration for the same resultant force.
  4. The acceleration always acts in the same direction as the resultant force.
  5. These ideas combine into one equation, force=mass×acceleration\text{force}=\text{mass}\times\text{acceleration}force=mass×acceleration, which you must recall.
  6. In symbols the equation is F=maF=maF=ma.
    1. FFF is the resultant force in newtons, N\text{N}N.
    2. mmm is the mass in kilograms, kg\text{kg}kg.
    3. aaa is the acceleration in metres per second squared, m/s2\text{m/s}^2m/s2.

The newton

Definition

Newton

The unit of force, equal to the resultant force that gives a mass of one kilogram an acceleration of one metre per second squared.

  1. The equation F=maF=maF=ma is used to define the unit of force.
  2. One newton is the resultant force that gives a mass of 1 kg1\ \text{kg}1 kg an acceleration of 1 m/s21\ \text{m/s}^21 m/s2.
  3. This means that 1 N=1 kg m/s21\ \text{N}=1\ \text{kg m/s}^21 N=1 kg m/s2.
  4. Large forces are often given in kilonewtons, where 1 kN=1000 N1\ \text{kN}=1000\ \text{N}1 kN=1000 N.

Second law calculations

  1. The equation can be rearranged to find any one of the three quantities.
    1. To find the acceleration, divide both sides by mmm to give a=Fma=\dfrac{F}{m}a=mF​.
    2. To find the mass, divide both sides by aaa to give m=Fam=\dfrac{F}{a}m=aF​.
  2. Always use the resultant force, so work it out first when more than one force acts along the line of motion.
  3. Convert every value into SI units before you substitute.
    1. A mass in grams is divided by 100010001000 to give kilograms, so 250 g=0.25 kg250\ \text{g}=0.25\ \text{kg}250 g=0.25 kg.
    2. A force in kilonewtons is multiplied by 100010001000 to give newtons, so 4.5 kN=4500 N4.5\ \text{kN}=4500\ \text{N}4.5 kN=4500 N.
  4. A negative acceleration from your calculation means the object is slowing down, which happens when the resultant force acts against the motion.
Example

Acceleration of a delivery van

  • A delivery van with a mass of 2500 kg2500\ \text{kg}2500 kg has a resultant forward force of 5000 N5000\ \text{N}5000 N.
a=Fm a=\dfrac{F}{m} a=mF​ a=5000 N2500 kg=2.0 m/s2 a=\dfrac{5000\ \text{N}}{2500\ \text{kg}}=2.0\ \text{m/s}^2 a=2500 kg5000 N​=2.0 m/s2
  • The van accelerates at 2.0 m/s22.0\ \text{m/s}^22.0 m/s2 in the direction of the resultant force.
Example

Acceleration when two forces act

  • A cyclist and bicycle with a total mass of 80 kg80\ \text{kg}80 kg have a forward driving force of 200 N200\ \text{N}200 N and air resistance of 40 N40\ \text{N}40 N.
  • Find the resultant force first.
F=200 N−40 N=160 N F=200\ \text{N}-40\ \text{N}=160\ \text{N} F=200 N−40 N=160 N a=Fm=160 N80 kg=2.0 m/s2 a=\dfrac{F}{m}=\dfrac{160\ \text{N}}{80\ \text{kg}}=2.0\ \text{m/s}^2 a=mF​=80 kg160 N​=2.0 m/s2
  • The cyclist accelerates forwards at 2.0 m/s22.0\ \text{m/s}^22.0 m/s2.
Example

Force needed to stop a football

  • A goalkeeper stops a 450 g450\ \text{g}450 g football, giving it a deceleration of 60 m/s260\ \text{m/s}^260 m/s2.
  • Convert the mass into kilograms, so m=0.45 kgm=0.45\ \text{kg}m=0.45 kg.
F=ma=0.45 kg×60 m/s2=27 N F=ma=0.45\ \text{kg}\times60\ \text{m/s}^2=27\ \text{N} F=ma=0.45 kg×60 m/s2=27 N
  • The goalkeeper's gloves exert a force of 27 N27\ \text{N}27 N on the ball, opposite to its motion.
Practical

Investigating force and acceleration

  • Aim: to investigate how the acceleration of a trolley depends on the resultant force acting on it, with the total mass kept constant.
  • Apparatus: a dynamics trolley, a 2 m2\ \text{m}2 m runway, wooden blocks, a bench pulley, strong thread, a 100 g100\ \text{g}100 g mass hanger with five 100 g100\ \text{g}100 g slotted masses, a double interrupt card, a light gate and a data logger, a metre rule, a top-pan balance and a box of sand or a cushion.
  • Method:
    • Clamp the pulley to the end of the bench and lay the runway on the bench so that the trolley can roll towards the pulley.
    • Raise the starting end of the runway with blocks until a trolley given a gentle push rolls down at a steady speed, which shows that friction has been balanced by a small component of the weight.
    • Fix the double interrupt card upright on the trolley and position the light gate near the pulley end of the runway.
    • Place four 100 g100\ \text{g}100 g masses on the trolley and tie the thread from the trolley over the pulley to the empty hanger, so that the hanger hangs just below the pulley with the box of sand beneath it.
    • Each 100 g100\ \text{g}100 g that hangs from the thread has a weight of 1 N1\ \text{N}1 N, so the hanger alone gives an accelerating force of 1 N1\ \text{N}1 N.
    • Hold the trolley at a marked start line, then release it without pushing.
    • Record the acceleration calculated by the data logger from the speeds of the two halves of the card and the time between them.
    • Repeat twice more and calculate the mean acceleration.
    • Move one 100 g100\ \text{g}100 g mass from the trolley to the hanger, increasing the force by 1 N1\ \text{N}1 N, and repeat the measurements.
    • Continue until all of the masses are on the hanger, which gives five values of force from 1 N1\ \text{N}1 N to 5 N5\ \text{N}5 N.
    • Plot a graph of mean acceleration on the vertical axis against force on the horizontal axis.
  • Variables: the accelerating force is the independent variable, the acceleration is the dependent variable, and the total mass of the trolley, hanger and masses, the runway slope and the start line are control variables.
  • Results: the acceleration increases in equal steps as the force increases in equal steps, giving a straight line that passes through the origin.
  • Conclusion: the acceleration is directly proportional to the resultant force when the mass is constant.
  • Watch out: moving masses from the trolley to the hanger keeps the total moving mass constant, but adding new masses to the hanger would change the mass as well as the force and spoil the test.
  • Safety: keep feet away from the falling hanger and let it land in the box of sand, and stop the trolley before it hits the pulley.
Common Mistake
  • Do not use one force from a diagram in F=maF=maF=ma when several forces act, because the equation needs the resultant force.
  • Do not substitute a mass in grams, because the equation only works with mass in kilograms.
  • Do not confuse mass and weight, because mass is measured in kg\text{kg}kg and weight is a force measured in N\text{N}N.
Exam technique

Newton's second law calculations

  • Write F=maF=maF=ma from memory, because it is not usually given to you.
  • Rearrange the equation before substituting when the unknown is aaa or mmm, and show the rearranged form on its own line.
  • Look out for masses in grams and forces in kilonewtons, as a conversion is often where a mark is lost.
  • In practical questions, explain that moving masses from the trolley to the hanger keeps the total mass constant.
Self review
  • State Newton's second law in words.
  • Write the equation that links resultant force, mass and acceleration, with the unit of each quantity.
  • Define one newton.
  • Calculate the acceleration of a 1200 kg1200\ \text{kg}1200 kg car with a resultant force of 3000 N3000\ \text{N}3000 N.
  • Why are masses moved from the trolley to the hanger, rather than added to the hanger, in the force and acceleration practical?

2.2.10 Inertia and inertial mass

Inertia

Definition

Inertia

A measure of how difficult it is to change the velocity of an object.

  1. Inertia is a measure of how difficult it is to change the velocity of an object.
  2. A change in velocity includes starting to move, speeding up, slowing down, stopping and changing direction.
  3. Every object has inertia, whether it is moving or at rest.
    1. A stationary object resists being set in motion.
    2. A moving object resists being slowed down, stopped or turned.
  4. The greater the mass of an object, the greater its inertia, so the harder it is to change its velocity.
  5. An empty supermarket trolley is easy to start, stop and steer, while a fully loaded trolley needs a much bigger push to do the same things.
  6. A loaded lorry needs a much larger braking force than a small car to stop in the same distance, because it has far more inertia.
  7. Inertia explains why passengers on a bus lurch forwards when it brakes suddenly, because their bodies keep moving forwards until a force from the seat belt or the seat in front changes their velocity.
  8. Inertia is a property of the object, not a force, so it is never drawn on a free body diagram.

Inertial mass

Definition

Inertial mass

The ratio of the resultant force acting on an object to the acceleration that the force produces.

  1. The inertial mass of an object puts a number on its inertia.
  2. Inertial mass is defined as the ratio of the resultant force to the acceleration it produces.
  3. In words, inertial mass=resultant forceacceleration\text{inertial mass}=\dfrac{\text{resultant force}}{\text{acceleration}}inertial mass=accelerationresultant force​.
  4. In symbols, m=Fam=\dfrac{F}{a}m=aF​, where mmm is in kg\text{kg}kg, FFF is in N\text{N}N and aaa is in m/s2\text{m/s}^2m/s2.
  5. An object with a large inertial mass gains only a small acceleration from a given resultant force.
    1. A resultant force of 600 N600\ \text{N}600 N gives a 60 kg60\ \text{kg}60 kg sprinter an acceleration of 10 m/s210\ \text{m/s}^210 m/s2.
    2. The same 600 N600\ \text{N}600 N gives a 1200 kg1200\ \text{kg}1200 kg car an acceleration of only 0.5 m/s20.5\ \text{m/s}^20.5 m/s2.
  6. Measuring the force on an object and its acceleration is one way to find its mass without using a balance.
  7. This method works in places where a balance would be useless, such as on board the International Space Station, where objects appear weightless but still have inertia.
    1. Astronauts find their mass by measuring how quickly a known force accelerates them.
  8. The force in the equation must be the resultant force, so opposing forces are combined before you divide by the acceleration.
Example

Inertial mass of a car

  • A resultant force of 3600 N3600\ \text{N}3600 N gives a car an acceleration of 2.4 m/s22.4\ \text{m/s}^22.4 m/s2.
m=Fa m=\dfrac{F}{a} m=aF​ m=3600 N2.4 m/s2=1500 kg m=\dfrac{3600\ \text{N}}{2.4\ \text{m/s}^2}=1500\ \text{kg} m=2.4 m/s23600 N​=1500 kg
  • The inertial mass of the car is 1500 kg1500\ \text{kg}1500 kg.
Example

Comparing the inertia of two trolleys

  • The same resultant force of 12 N12\ \text{N}12 N acts in turn on two trolleys, A and B.
  • Trolley A accelerates at 3.0 m/s23.0\ \text{m/s}^23.0 m/s2 and trolley B accelerates at 1.5 m/s21.5\ \text{m/s}^21.5 m/s2.
mA=12 N3.0 m/s2=4.0 kg m_{\text{A}}=\dfrac{12\ \text{N}}{3.0\ \text{m/s}^2}=4.0\ \text{kg} mA​=3.0 m/s212 N​=4.0 kg mB=12 N1.5 m/s2=8.0 kg m_{\text{B}}=\dfrac{12\ \text{N}}{1.5\ \text{m/s}^2}=8.0\ \text{kg} mB​=1.5 m/s212 N​=8.0 kg
  • Trolley B has twice the inertial mass of trolley A, so it has the greater inertia and its velocity is harder to change.
Common Mistake
  • Do not call inertia a force, because it is a property that describes how hard it is to change an object's velocity.
  • Do not say that inertia only affects moving objects, because a stationary object also resists being made to move.
  • Do not use a single force in m=Fam=\dfrac{F}{a}m=aF​ when other forces act, because inertial mass is defined using the resultant force.
Exam technique

Defining inertia and inertial mass

  • For inertia, write "a measure of how difficult it is to change the velocity of an object", using the word velocity rather than "movement".
  • For inertial mass, write that it is "the ratio of resultant force to acceleration", and give the equation m=Fam=\dfrac{F}{a}m=aF​.
  • In a comparison, link a larger mass to a smaller acceleration for the same resultant force.
Self review
  • What is inertia?
  • How is inertial mass defined?
  • Explain why a loaded lorry is harder to stop than a small car.
  • Calculate the inertial mass of an object that accelerates at 4.0 m/s24.0\ \text{m/s}^24.0 m/s2 when the resultant force on it is 20 N20\ \text{N}20 N.
  • Why can a balance not be used to find the mass of an astronaut on the International Space Station?

2.2.11 Momentum in collisions

Momentum

Definition

Momentum

The product of the mass and the velocity of an object.

  1. Every moving object has momentum, which depends on both its mass and its velocity.
  2. Momentum is calculated with momentum=mass×velocity\text{momentum}=\text{mass}\times\text{velocity}momentum=mass×velocity, an equation you must recall.
  3. In symbols the equation is p=mvp=mvp=mv.
    1. ppp is the momentum in kilogram metres per second, kg m/s\text{kg m/s}kg m/s.
    2. mmm is the mass in kilograms, kg\text{kg}kg.
    3. vvv is the velocity in metres per second, m/s\text{m/s}m/s.
  4. A heavy lorry moving slowly can have the same momentum as a light car moving quickly.
  5. A stationary object has zero momentum, because its velocity is zero.
  6. Momentum is a vector, because velocity is a vector, so momentum always has a direction.
  7. For motion along a line, choose one direction as positive, so an object moving the other way has a negative velocity and a negative momentum.
Example

Momentum of a rugby player

  • A rugby player with a mass of 95 kg95\ \text{kg}95 kg runs at 8.0 m/s8.0\ \text{m/s}8.0 m/s towards the try line.
p=mv p=mv p=mv p=95 kg×8.0 m/s=760 kg m/s p=95\ \text{kg}\times8.0\ \text{m/s}=760\ \text{kg m/s} p=95 kg×8.0 m/s=760 kg m/s
  • The player has a momentum of 760 kg m/s760\ \text{kg m/s}760 kg m/s towards the try line.

Momentum in collisions

Definition

Conservation of momentum

The principle that the total momentum of colliding objects before a collision equals their total momentum after it, provided no external forces act.

  1. When two objects collide, they push on each other, so the velocity of each object changes and the momentum of each object changes.
  2. One object loses momentum and the other gains exactly the same amount, because the forces on the two objects are equal in size and opposite in direction and act for the same time.
  3. The total momentum of the two objects therefore stays the same, as long as no external forces such as friction act on them.
  4. This is the conservation of momentum: the total momentum before a collision equals the total momentum after the collision.
  5. Momentum is conserved whether the objects bounce apart or stick together.
  6. When you calculate total momentum, add the momenta with their signs, because momenta in opposite directions partly cancel.
  7. Collisions in everyday life show the same pattern.
    1. When a moving snooker ball hits a stationary ball head-on, the first ball can stop dead while the second ball moves off with almost the same velocity.
    2. When a car runs into the back of a stationary car and the two lock together, they move off together more slowly than the first car was moving.
    3. When two dodgem cars collide head-on, each one bounces back, but the total momentum of the pair is unchanged.
Example

Two trolleys that stick together

  • A 2.0 kg2.0\ \text{kg}2.0 kg trolley moving at 3.0 m/s3.0\ \text{m/s}3.0 m/s collides with a stationary 1.0 kg1.0\ \text{kg}1.0 kg trolley, and the two trolleys stick together.
total momentum before=2.0×3.0+1.0×0=6.0 kg m/s \text{total momentum before}=2.0\times3.0+1.0\times0=6.0\ \text{kg m/s} total momentum before=2.0×3.0+1.0×0=6.0 kg m/s
  • The total momentum after the collision is also 6.0 kg m/s6.0\ \text{kg m/s}6.0 kg m/s, carried by a combined mass of 3.0 kg3.0\ \text{kg}3.0 kg.
v=pm=6.0 kg m/s3.0 kg=2.0 m/s v=\dfrac{p}{m}=\dfrac{6.0\ \text{kg m/s}}{3.0\ \text{kg}}=2.0\ \text{m/s} v=mp​=3.0 kg6.0 kg m/s​=2.0 m/s
  • The joined trolleys move off at 2.0 m/s2.0\ \text{m/s}2.0 m/s in the original direction.
Example

A head-on collision

  • Trolley A, of mass 1.5 kg1.5\ \text{kg}1.5 kg, moves right at 2.0 m/s2.0\ \text{m/s}2.0 m/s, and trolley B, of mass 1.0 kg1.0\ \text{kg}1.0 kg, moves left at 1.0 m/s1.0\ \text{m/s}1.0 m/s.
  • Take right as positive, so the velocity of trolley B is −1.0 m/s-1.0\ \text{m/s}−1.0 m/s.
total momentum before=1.5×2.0+1.0×(−1.0)=3.0−1.0=2.0 kg m/s \text{total momentum before}=1.5\times2.0+1.0\times(-1.0)=3.0-1.0=2.0\ \text{kg m/s} total momentum before=1.5×2.0+1.0×(−1.0)=3.0−1.0=2.0 kg m/s
  • After the collision trolley A moves right at 0.40 m/s0.40\ \text{m/s}0.40 m/s, so the momentum of trolley B is found from the conserved total.
pB=2.0−1.5×0.40=1.4 kg m/s p_{\text{B}}=2.0-1.5\times0.40=1.4\ \text{kg m/s} pB​=2.0−1.5×0.40=1.4 kg m/s
  • Trolley B moves off to the right with a momentum of 1.4 kg m/s1.4\ \text{kg m/s}1.4 kg m/s, which is a velocity of 1.4 m/s1.4\ \text{m/s}1.4 m/s.
Practical

Momentum of colliding trolleys

  • Aim: to measure the momentum of two trolleys before and after a collision and test whether the total momentum is conserved.
  • Apparatus: two dynamics trolleys, one fitted with a pin and the other with a cork so they stick together, a 2 m2\ \text{m}2 m runway, wooden blocks, two interrupt cards of measured length, two light gates connected to a data logger, a top-pan balance, slotted masses and a box of foam at the end of the runway.
  • Method:
    • Measure the mass of each trolley, including its interrupt card and any added masses, with the top-pan balance.
    • Raise one end of the runway until a trolley given a gentle push rolls at a steady speed, so that friction is balanced.
    • Set up light gate 1 near the start of the runway and light gate 2 further down, and place trolley B at rest just beyond light gate 1.
    • Give trolley A a gentle push so that its card passes through light gate 1 before it hits trolley B.
    • Let the joined trolleys pass through light gate 2, and catch them in the foam.
    • Use the blocking times and the card lengths to find the velocity of trolley A before the collision and of the joined trolleys after it.
    • Calculate the momentum before the collision as mAvAm_{\text{A}}v_{\text{A}}mA​vA​ and the momentum after as (mA+mB)v(m_{\text{A}}+m_{\text{B}})v(mA​+mB​)v.
    • Repeat three times, then add slotted masses to trolley B and repeat for several different masses.
  • Variables: the mass of trolley B is the independent variable, the momentum after the collision is the dependent variable, and the runway slope, the trolleys and the card lengths are control variables.
  • Results: the total momentum after each collision is very close to the momentum before it, and the joined trolleys move more slowly when trolley B is heavier.
  • Conclusion: the total momentum is conserved in the collision, within the uncertainty of the measurements.
  • Watch out: if the runway is not friction-compensated, friction removes momentum between the gates, so the momentum after the collision comes out noticeably smaller than before.
  • Safety: catch the trolleys in the foam at the end of the runway so that they do not fall off the bench.
Common Mistake
  • Do not confuse momentum, measured in kg m/s\text{kg m/s}kg m/s, with a moment, which is a turning effect measured in N m\text{N m}N m.
  • Do not forget the sign of each velocity, because adding two momenta in opposite directions as if they were in the same direction gives the wrong total.
  • Do not say that each object keeps its own momentum in a collision, because it is the total momentum that is conserved.
Exam technique

Momentum questions

  • Write p=mvp=mvp=mv from memory and give the unit kg m/s\text{kg m/s}kg m/s with every answer.
  • For a collision, write a line for the total momentum before and a line for the total momentum after, then set them equal.
  • State the direction you have chosen as positive, and keep the signs in your working.
  • When you describe conservation of momentum, say that the total momentum before equals the total momentum after, provided no external forces act.
Self review
  • Write the equation for momentum and give its unit.
  • Why is momentum a vector quantity?
  • What is the momentum of a stationary object?
  • State the law of conservation of momentum.
  • Calculate the momentum of a 0.16 kg0.16\ \text{kg}0.16 kg cricket ball moving at 30 m/s30\ \text{m/s}30 m/s.

2.2.12 Relating force, mass, velocity and acceleration

Linking force and acceleration

Definition

Directly proportional

A relationship between two quantities in which multiplying one by a factor multiplies the other by the same factor.

Definition

Inversely proportional

A relationship between two quantities in which multiplying one by a factor divides the other by the same factor.

  1. Force, mass, velocity and acceleration are tied together by two equations that you must recall.
    1. F=maF=maF=ma links the resultant force to the mass and the acceleration.
    2. a=v−uta=\dfrac{v-u}{t}a=tv−u​ links the acceleration to the change in velocity and the time taken.
  2. For a fixed mass, the acceleration is directly proportional to the resultant force.
    1. If the resultant force on a shopping trolley is tripled, its acceleration is tripled.
  3. For a fixed resultant force, the acceleration is inversely proportional to the mass.
    1. If the same push is given to a trolley with three times the mass, its acceleration is one third as large.
  4. Proportional reasoning lets you predict changes without working out every value from the beginning.
    1. If the force is doubled and the mass is also doubled, the acceleration does not change, because 2F2m=Fm\dfrac{2F}{2m}=\dfrac{F}{m}2m2F​=mF​.
    2. If the force is halved and the mass is doubled, the acceleration falls to one quarter, because F/22m=14×Fm\dfrac{F/2}{2m}=\dfrac{1}{4}\times\dfrac{F}{m}2mF/2​=41​×mF​.

Force and change in velocity

Definition

Initial velocity

The velocity of an object at the start of the time interval being considered.

Definition

Final velocity

The velocity of an object at the end of the time interval being considered.

  1. Substituting a=v−uta=\dfrac{v-u}{t}a=tv−u​ into F=maF=maF=ma gives F=m(v−u)tF=\dfrac{m(v-u)}{t}F=tm(v−u)​.
  2. This equation shows how the resultant force depends on the mass, the change in velocity and the time over which the change happens.
    1. A larger mass needs a larger force to produce the same change in velocity in the same time.
    2. A larger change in velocity in the same time needs a larger force.
    3. The same change in velocity spread over a longer time needs a smaller force.
  3. The last of these ideas explains why stopping something gently needs a smaller force than stopping it suddenly.
    1. A fielder who pulls their hands back while catching a cricket ball makes the stopping time longer, so the force on their hands is smaller.
    2. Bending your knees when you land from a jump makes the time to stop longer, so the force on your legs is smaller.
  4. If the time is not known, the acceleration can be found from the velocities and the distance with v2−u2=2asv^2-u^2=2asv2−u2=2as, which you need to select and apply rather than recall.
  5. Combining v2−u2=2asv^2-u^2=2asv2−u2=2as with F=maF=maF=ma shows that, for a given mass and change of speed, a larger force changes the velocity over a shorter distance.

Solving multi-step problems

  1. Many questions need two equations, with the answer from the first used in the second.
  2. Work out which quantity links the two equations, which is usually the acceleration.
  3. Find the acceleration from the motion data first, then use it in F=maF=maF=ma, or find the acceleration from F=maF=maF=ma first, then use it to find a velocity, a time or a distance.
  4. Keep unrounded values in your calculator between steps and round only the final answer.
  5. Check the sign of the acceleration, because a negative value means the resultant force acts against the motion and the object is slowing down.
Example

Force to accelerate a car

  • A car of mass 1200 kg1200\ \text{kg}1200 kg speeds up from 8.0 m/s8.0\ \text{m/s}8.0 m/s to 20 m/s20\ \text{m/s}20 m/s in 6.0 s6.0\ \text{s}6.0 s.
  • Find the acceleration first.
a=v−ut=20 m/s−8.0 m/s6.0 s=2.0 m/s2 a=\dfrac{v-u}{t}=\dfrac{20\ \text{m/s}-8.0\ \text{m/s}}{6.0\ \text{s}}=2.0\ \text{m/s}^2 a=tv−u​=6.0 s20 m/s−8.0 m/s​=2.0 m/s2
  • Then use Newton's second law.
F=ma=1200 kg×2.0 m/s2=2400 N F=ma=1200\ \text{kg}\times2.0\ \text{m/s}^2=2400\ \text{N} F=ma=1200 kg×2.0 m/s2=2400 N
  • The resultant force on the car is 2400 N2400\ \text{N}2400 N forwards.
Example

Time for a force to change a velocity

  • A resultant force of 150 N150\ \text{N}150 N acts on a 60 kg60\ \text{kg}60 kg cyclist who starts from rest.
a=Fm=150 N60 kg=2.5 m/s2 a=\dfrac{F}{m}=\dfrac{150\ \text{N}}{60\ \text{kg}}=2.5\ \text{m/s}^2 a=mF​=60 kg150 N​=2.5 m/s2
  • Rearrange a=v−uta=\dfrac{v-u}{t}a=tv−u​ to find the time to reach 10 m/s10\ \text{m/s}10 m/s.
t=v−ua=10 m/s−02.5 m/s2=4.0 s t=\dfrac{v-u}{a}=\dfrac{10\ \text{m/s}-0}{2.5\ \text{m/s}^2}=4.0\ \text{s} t=av−u​=2.5 m/s210 m/s−0​=4.0 s
  • The cyclist reaches 10 m/s10\ \text{m/s}10 m/s after 4.0 s4.0\ \text{s}4.0 s.
Example

Braking force from a stopping distance

  • A 900 kg900\ \text{kg}900 kg car travelling at 20 m/s20\ \text{m/s}20 m/s brakes and stops in a distance of 25 m25\ \text{m}25 m.
  • No time is given, so select the equation that links uuu, vvv, aaa and sss.
a=v2−u22s=02−2022×25=−40050=−8.0 m/s2 a=\dfrac{v^2-u^2}{2s}=\dfrac{0^2-20^2}{2\times25}=\dfrac{-400}{50}=-8.0\ \text{m/s}^2 a=2sv2−u2​=2×2502−202​=50−400​=−8.0 m/s2 F=ma=900 kg×(−8.0 m/s2)=−7200 N F=ma=900\ \text{kg}\times(-8.0\ \text{m/s}^2)=-7200\ \text{N} F=ma=900 kg×(−8.0 m/s2)=−7200 N
  • The braking force is 7200 N7200\ \text{N}7200 N, and the negative sign shows that it acts opposite to the motion.
Example

Predicting a change with proportion

  • A loaded van has twice the mass of the same van when empty, and the same resultant force acts on it in both cases.
  • The acceleration is inversely proportional to the mass, so the loaded van has half the acceleration of the empty van.
  • With half the acceleration, the loaded van takes twice as long to reach the same speed from rest.
Common Mistake
  • Do not use the time in F=maF=maF=ma directly, because you must find the acceleration first.
  • Do not drop the minus sign from a deceleration partway through a calculation, because the sign tells you the direction of the force.
  • Do not assume that a larger force always gives a larger acceleration, because a larger mass can reduce the acceleration more than the force increases it.
Exam technique

Multi-step force and motion problems

  • Write the two equations you plan to use at the start, so the examiner can follow your method.
  • Label each intermediate answer, such as "acceleration =2.0 m/s2=2.0\ \text{m/s}^2=2.0 m/s2", as this often carries its own mark.
  • For a "how would the acceleration change" question, use proportion and give a factor, such as "halves" or "doubles", rather than only "decreases".
  • When you explain why a longer stopping time helps, link it to a smaller deceleration and so to a smaller resultant force.
Self review
  • How does the acceleration of an object change if the resultant force doubles and the mass stays the same?
  • Write the equation that combines F=maF=maF=ma and a=v−uta=\dfrac{v-u}{t}a=tv−u​.
  • Explain why bending your knees on landing reduces the force on your legs.
  • Calculate the resultant force on a 50 kg50\ \text{kg}50 kg skater who speeds up from 2.0 m/s2.0\ \text{m/s}2.0 m/s to 5.0 m/s5.0\ \text{m/s}5.0 m/s in 1.5 s1.5\ \text{s}1.5 s.
  • What happens to the acceleration if both the force and the mass are tripled?

2.2.13 Work done by a force

Work done by a force

Definition

Work done

The energy transferred when a force moves an object through a distance in the direction of the force.

Definition

Line of action

The straight line along which a force acts.

Definition

Joule

The unit of energy, equal to the work done when a force of one newton moves an object one metre along the line of action of the force.

  1. Work is done whenever a force makes an object move.
  2. The work done depends on the size of the force and how far the object moves along the line of action of the force.
  3. The work done is calculated with work done=force×distance\text{work done}=\text{force}\times\text{distance}work done=force×distance, an equation you must recall.
  4. In symbols the equation is W=FsW=FsW=Fs.
    1. WWW is the work done in joules, J\text{J}J.
    2. FFF is the force in newtons, N\text{N}N.
    3. sss is the distance moved along the line of action of the force, in metres, m\text{m}m.
  5. One joule of work is done when a force of 1 N1\ \text{N}1 N moves an object through 1 m1\ \text{m}1 m along the line of action of the force.
  6. Rearranging the equation gives F=WsF=\dfrac{W}{s}F=sW​ to find a force and s=WFs=\dfrac{W}{F}s=FW​ to find a distance.

Distance along the line of action

  1. Only movement in the direction of the force counts towards the work done by that force.
  2. When you lift a bag of shopping onto a table, the lifting force acts upwards, so the distance to use is the vertical height the bag is raised.
  3. When you push a wheelie bin along a path, the push acts horizontally, so the distance to use is the horizontal distance moved.
  4. If an object does not move, no work is done on it, however large the force.
    1. Pushing hard against a solid wall does no work on the wall, because the wall does not move.
  5. A force at right angles to the motion does no work.
    1. When you carry a box along a level corridor, your upward supporting force does no work on the box, because the box moves horizontally, not upwards.
  6. When a force opposes the motion, such as friction or air resistance, work is still done, but it is work done against the motion.

Energy transferred by work

Definition

Energy store

A way in which energy is held by an object or system, such as a kinetic, gravitational, elastic, thermal or chemical store.

  1. Doing work is a way of transferring energy, so the work done by a force equals the energy it transfers.
  2. A transfer by a force is called a mechanical transfer, which is one of the main ways energy moves between stores.
  3. Every time work is done, energy leaves one energy store and enters another.
    1. Lifting a box transfers energy from the chemical store of your muscles to the gravitational store of the box.
    2. A car engine accelerating a car transfers energy from the chemical store of the fuel to the kinetic store of the car.
    3. Stretching a spring transfers energy to the elastic store of the spring.
  4. Work done against friction transfers energy to the thermal stores of the surfaces, which is why your hands warm up when you rub them together.
  5. When a car brakes, the brakes do work against the motion, so energy is transferred from the kinetic store of the car to the thermal store of the brake discs and pads.
  6. To describe a transfer fully, name the force doing the work, the store that loses energy and the store that gains energy.
Example

Work done pushing a wheelie bin

  • A resident pushes a wheelie bin 15 m15\ \text{m}15 m down a path with a steady horizontal force of 40 N40\ \text{N}40 N.
W=Fs W=Fs W=Fs W=40 N×15 m=600 J W=40\ \text{N}\times15\ \text{m}=600\ \text{J} W=40 N×15 m=600 J
  • The push does 600 J600\ \text{J}600 J of work, transferring 600 J600\ \text{J}600 J of energy, which is dissipated to the thermal stores of the wheels, the path and the surroundings by friction.
Example

Work done lifting a suitcase

  • A suitcase with a weight of 180 N180\ \text{N}180 N is lifted at a steady speed onto a luggage rack 1.6 m1.6\ \text{m}1.6 m above the floor.
  • Lifting at a steady speed needs a force equal to the weight, so F=180 NF=180\ \text{N}F=180 N, and the distance along the line of action is the vertical height.
W=Fs=180 N×1.6 m=288 J W=Fs=180\ \text{N}\times1.6\ \text{m}=288\ \text{J} W=Fs=180 N×1.6 m=288 J
  • The work done is 288 J288\ \text{J}288 J, so 288 J288\ \text{J}288 J is transferred to the gravitational store of the suitcase.
Example

Distance from the work done

  • A tow truck does 36 000 J36\,000\ \text{J}36000 J of work pulling a car with a force of 1800 N1800\ \text{N}1800 N.
s=WF=36 000 J1800 N=20 m s=\dfrac{W}{F}=\dfrac{36\,000\ \text{J}}{1800\ \text{N}}=20\ \text{m} s=FW​=1800 N36000 J​=20 m
  • The car is pulled 20 m20\ \text{m}20 m along the line of the tow rope.
Practical

Work done lifting weights and climbing stairs

  • Aim: to measure the work done in lifting weights and in walking up a flight of stairs.
  • Apparatus: a set of 1 kg1\ \text{kg}1 kg masses, a top-pan balance, bathroom scales, a metre rule, a tape measure and a flight of stairs.
  • Method, lifting weights:
    • Measure the mass of the load on the top-pan balance and convert it to kilograms.
    • Find the weight of the load in newtons by multiplying its mass in kilograms by the gravitational field strength, 10 N/kg10\ \text{N/kg}10 N/kg.
    • Stand a metre rule upright on the bench and lift the load steadily from the bench top to a marked height, such as 0.80 m0.80\ \text{m}0.80 m.
    • Calculate the work done by multiplying the weight by the vertical height.
    • Repeat with different loads and heights.
  • Method, climbing stairs:
    • Measure your mass on the bathroom scales and convert it into a weight in newtons.
    • Measure the height of one step with the metre rule, holding it vertically against the riser.
    • Count the number of steps in the flight and multiply by the height of one step to find the total vertical height.
    • Walk up the stairs at a steady pace.
    • Calculate the work done against your weight by multiplying your weight by the total vertical height.
  • Results: a pupil with a weight of 550 N550\ \text{N}550 N who climbs 151515 steps, each 0.18 m0.18\ \text{m}0.18 m high, rises 2.7 m2.7\ \text{m}2.7 m and does about 1500 J1500\ \text{J}1500 J of work against their weight.
  • Conclusion: the work done depends only on the weight lifted and the vertical height gained, not on the length of the staircase or the path taken.
  • Watch out: the distance along the slope of the stairs is longer than the vertical height, so using it overestimates the work done against weight.
  • Safety: walk rather than run on the stairs, keep one hand on the rail, and lift loads with a straight back.
Common Mistake
  • Do not use the distance travelled along a slope or a staircase when the force is the weight, because the distance must be along the line of action of the force.
  • Do not give work done in newtons, because work done is an energy and is measured in joules.
  • Do not say that holding a heavy object still does work on it, because the object does not move.
Exam technique

Work done answers

  • Write W=FsW=FsW=Fs from memory, substitute the values and give the answer in joules.
  • Say explicitly which distance you used, such as "vertical height", when the question involves lifting.
  • When asked to describe the energy transfer, name the store that loses energy and the store that gains energy, not only "energy is transferred".
  • Use "thermal store" rather than "heat energy" when describing work done against friction.
Self review
  • Write the equation that links work done, force and distance.
  • What is meant by the line of action of a force?
  • Why is no work done when you push against a wall that does not move?
  • Calculate the work done when a 250 N250\ \text{N}250 N force pushes a crate 3.0 m3.0\ \text{m}3.0 m along the floor.
  • Describe the energy transfer when a person lifts a box onto a shelf.

2.2.14 Work, stored energy and units of energy

Work and stored energy

Definition

Energy store

A way in which energy is held by an object or system, such as a kinetic, gravitational, elastic, thermal or chemical store.

  1. The work done by a force is equal to the energy transferred by that force.
  2. This means you can use W=FsW=FsW=Fs to calculate how much energy goes into or out of an energy store.
  3. When a force does work on an object, the energy in one of the object's stores increases by the amount of work done.
    1. A crane that does 40 kJ40\ \text{kJ}40 kJ of work lifting a steel beam at a steady speed increases the gravitational store of the beam by 40 kJ40\ \text{kJ}40 kJ.
    2. A bow that is drawn back with 60 J60\ \text{J}60 J of work stores 60 J60\ \text{J}60 J in its elastic store, ready to transfer to the arrow.
  4. When an object does work against a force, the energy in one of its stores decreases by the amount of work done.
    1. A bicycle that freewheels to a stop on a level road loses all of the energy in its kinetic store as it does work against friction and air resistance.
  5. Energy is never created or destroyed in these transfers, so the energy lost from one store always equals the total energy gained by the other stores.
  6. Some of the energy transferred usually ends up in the thermal stores of the surroundings, where it spreads out and is no longer useful.

Calculating energy transfers

  1. Energy transfer calculations follow the same steps whichever stores are involved.
    1. Identify the force doing the work and the distance moved along its line of action.
    2. Calculate the work done with W=FsW=FsW=Fs.
    3. Use the work done as the energy transferred, then add it to or take it away from the store involved.
  2. A transfer can happen in several stages, and the energy can be followed from one store to the next.
    1. When a ball is lifted and then dropped, the work done lifting it fills its gravitational store, and as it falls this energy moves to its kinetic store.
    2. When the ball hits the ground, the energy moves to the thermal stores of the ball and ground, with some carried away by sound.
  3. If a store is given a value, such as the energy in the kinetic store of a moving car, you can use W=FsW=FsW=Fs in reverse to find a force or a distance.
Example

Energy in the gravitational store of a lifted crate

  • A forklift truck lifts a crate with a steady upward force of 2400 N2400\ \text{N}2400 N through a height of 3.5 m3.5\ \text{m}3.5 m.
W=Fs=2400 N×3.5 m=8400 J W=Fs=2400\ \text{N}\times3.5\ \text{m}=8400\ \text{J} W=Fs=2400 N×3.5 m=8400 J
  • The energy in the gravitational store of the crate increases by 8400 J8400\ \text{J}8400 J, which is 8.4 kJ8.4\ \text{kJ}8.4 kJ.
Example

Braking force from the kinetic store

  • A car has 300 kJ300\ \text{kJ}300 kJ in its kinetic store and the driver brakes to a stop in a distance of 50 m50\ \text{m}50 m.
  • All of the energy in the kinetic store is transferred by the work done against the braking force.
F=Ws=300 000 J50 m=6000 N F=\dfrac{W}{s}=\dfrac{300\,000\ \text{J}}{50\ \text{m}}=6000\ \text{N} F=sW​=50 m300000 J​=6000 N
  • The average braking force is 6000 N6000\ \text{N}6000 N, and the energy ends up mainly in the thermal stores of the brakes.
Example

Energy remaining after work against friction

  • A sledge has 900 J900\ \text{J}900 J in its kinetic store at the bottom of a hill and slides 20 m20\ \text{m}20 m across flat snow against a friction force of 25 N25\ \text{N}25 N.
W=Fs=25 N×20 m=500 J W=Fs=25\ \text{N}\times20\ \text{m}=500\ \text{J} W=Fs=25 N×20 m=500 J energy left in the kinetic store=900 J−500 J=400 J \text{energy left in the kinetic store}=900\ \text{J}-500\ \text{J}=400\ \text{J} energy left in the kinetic store=900 J−500 J=400 J
  • The sledge is still moving after 20 m20\ \text{m}20 m, with 400 J400\ \text{J}400 J in its kinetic store, while 500 J500\ \text{J}500 J has been dissipated to the thermal stores of the sledge and snow.

Newton-metres and joules

Definition

Newton-metre

The unit obtained by multiplying a force in newtons by a distance in metres, which is equal in size to one joule.

  1. Multiplying a force in newtons by a distance in metres gives an answer in newton-metres, N m\text{N m}N m.
  2. Work done is an energy, so its answer can also be given in joules, J\text{J}J.
  3. The two units are identical in size, so 1 N m=1 J1\ \text{N m}=1\ \text{J}1 N m=1 J.
  4. Converting between them changes only the unit, not the number, so 750 N m=750 J750\ \text{N m}=750\ \text{J}750 N m=750 J.
  5. Large amounts of energy are given with prefixes, which do change the number.
    1. 1 kJ=1000 J1\ \text{kJ}=1000\ \text{J}1 kJ=1000 J, so 4.2 kJ=4200 J=4200 N m4.2\ \text{kJ}=4200\ \text{J}=4200\ \text{N m}4.2 kJ=4200 J=4200 N m.
    2. 1 MJ=1 000 000 J1\ \text{MJ}=1\,000\,000\ \text{J}1 MJ=1000000 J, so 0.35 MJ=350 000 J0.35\ \text{MJ}=350\,000\ \text{J}0.35 MJ=350000 J.
  6. Convert prefixed values into joules before substituting them into W=FsW=FsW=Fs, so that the force comes out in newtons and the distance in metres.
Example

Converting units in a work calculation

  • A winch does 1.8 kJ1.8\ \text{kJ}1.8 kJ of work raising a load through 12 m12\ \text{m}12 m.
  • Convert the energy into joules, so W=1800 JW=1800\ \text{J}W=1800 J, which is also 1800 N m1800\ \text{N m}1800 N m.
F=Ws=1800 N m12 m=150 N F=\dfrac{W}{s}=\dfrac{1800\ \text{N m}}{12\ \text{m}}=150\ \text{N} F=sW​=12 m1800 N m​=150 N
  • The winch pulls with a force of 150 N150\ \text{N}150 N.
Common Mistake
  • Do not multiply or divide by 100010001000 when converting between newton-metres and joules, because 1 N m1\ \text{N m}1 N m is exactly 1 J1\ \text{J}1 J.
  • Do not forget the 100010001000 when converting kilojoules into joules, because the prefix "kilo" means a thousand.
  • Do not write "heat energy" or "movement energy", because the correct names are the thermal store and the kinetic store.
Exam technique

Energy transfer calculations

  • State which store gains energy and which store loses energy before you calculate, as this often earns the first mark.
  • Show the conversion from kJ\text{kJ}kJ or MJ\text{MJ}MJ into J\text{J}J as a separate line of working.
  • If a question asks for an answer in N m\text{N m}N m, give the same number you found in joules with the new unit.
  • In a multi-stage transfer, describe each stage in order and name the store at each point.
Self review
  • How is the work done by a force related to the energy transferred?
  • What happens to the energy in the kinetic store of a bicycle that freewheels to a stop?
  • Convert 2.5 kJ2.5\ \text{kJ}2.5 kJ into newton-metres.
  • A 40 N40\ \text{N}40 N force does work over 6.0 m6.0\ \text{m}6.0 m, so how much energy is transferred?
  • A car with 200 kJ200\ \text{kJ}200 kJ in its kinetic store brakes to a stop in 40 m40\ \text{m}40 m, so what is the average braking force?

2.2.15 Power as the rate of energy transfer

Power

Definition

Power

The rate at which energy is transferred.

  1. Power is the rate at which energy is transferred, which means how much energy is transferred each second.
  2. When a force does work, the energy transferred equals the work done, so power is also the rate at which work is done.
  3. Power is calculated with power=work donetime\text{power}=\dfrac{\text{work done}}{\text{time}}power=timework done​, an equation you must recall.
  4. In symbols the equation is P=WtP=\dfrac{W}{t}P=tW​.
    1. PPP is the power in watts, W\text{W}W.
    2. WWW is the work done, or energy transferred, in joules, J\text{J}J.
    3. ttt is the time taken in seconds, s\text{s}s.
  5. The same equation can be written as P=EtP=\dfrac{E}{t}P=tE​, where EEE is the energy transferred in joules.
  6. Rearranging gives W=PtW=PtW=Pt to find the work done and t=WPt=\dfrac{W}{P}t=PW​ to find the time.

The watt

Definition

Watt

The unit of power, equal to an energy transfer of one joule per second.

  1. Power is measured in watts, with the symbol W\text{W}W.
  2. One watt is a rate of energy transfer of one joule per second, so 1 W=1 J/s1\ \text{W}=1\ \text{J/s}1 W=1 J/s.
  3. Larger powers are given in kilowatts and megawatts.
    1. 1 kW=1000 W1\ \text{kW}=1000\ \text{W}1 kW=1000 W, which is roughly the power of a strong cyclist sprinting for a few seconds.
    2. 1 MW=1 000 000 W1\ \text{MW}=1\,000\,000\ \text{W}1 MW=1000000 W, which is the scale of the power of a train engine or a large wind turbine.
  4. Times must be in seconds before you substitute, so a time of 222 minutes becomes 120 s120\ \text{s}120 s.

Comparing power

  1. A more powerful machine transfers energy more quickly than a less powerful one.
  2. Two things decide the power: the energy transferred and the time taken.
    1. If two machines transfer the same energy, the one that takes less time has the greater power.
    2. If two machines run for the same time, the one that transfers more energy has the greater power.
  3. Two pupils of equal weight who climb the same staircase do the same work against their weight, but the pupil who runs up does it in less time and so develops a greater power.
  4. A crane lifting a load to the top of a building does the same work whether it lifts slowly or quickly, but the faster crane needs a more powerful motor.
  5. Power describes how quickly energy is transferred, not how much, so a low-power machine running for a long time can transfer more energy than a high-power machine running briefly.
Example

Power of a lift motor

  • A lift motor does 54 000 J54\,000\ \text{J}54000 J of work raising a lift between floors in 18 s18\ \text{s}18 s.
P=Wt P=\dfrac{W}{t} P=tW​ P=54 000 J18 s=3000 W P=\dfrac{54\,000\ \text{J}}{18\ \text{s}}=3000\ \text{W} P=18 s54000 J​=3000 W
  • The power of the lift motor is 3000 W3000\ \text{W}3000 W, which is 3.0 kW3.0\ \text{kW}3.0 kW.
Example

Power of a pupil running upstairs

  • A pupil with a weight of 600 N600\ \text{N}600 N runs up a staircase with a vertical height of 3.0 m3.0\ \text{m}3.0 m in 4.0 s4.0\ \text{s}4.0 s.
  • Find the work done against the pupil's weight first.
W=Fs=600 N×3.0 m=1800 J W=Fs=600\ \text{N}\times3.0\ \text{m}=1800\ \text{J} W=Fs=600 N×3.0 m=1800 J P=Wt=1800 J4.0 s=450 W P=\dfrac{W}{t}=\dfrac{1800\ \text{J}}{4.0\ \text{s}}=450\ \text{W} P=tW​=4.0 s1800 J​=450 W
  • The pupil develops a power of 450 W450\ \text{W}450 W while running up the stairs.
Example

Time for a crane to do work

  • A 2.5 kW2.5\ \text{kW}2.5 kW crane motor does 75 kJ75\ \text{kJ}75 kJ of work lifting a load.
  • Convert to base units, so P=2500 WP=2500\ \text{W}P=2500 W and W=75 000 JW=75\,000\ \text{J}W=75000 J.
t=WP=75 000 J2500 W=30 s t=\dfrac{W}{P}=\dfrac{75\,000\ \text{J}}{2500\ \text{W}}=30\ \text{s} t=PW​=2500 W75000 J​=30 s
  • The crane takes 30 s30\ \text{s}30 s to lift the load.
Common Mistake
  • Do not define power as "the amount of energy", because power is the rate of energy transfer.
  • Do not confuse the unit watt, W\text{W}W, with the symbol for work done, WWW, which is measured in joules.
  • Do not substitute a time in minutes or hours, because the equation needs the time in seconds.
Exam technique

Power questions

  • Write P=WtP=\dfrac{W}{t}P=tW​ from memory, because you are expected to recall it.
  • Use the word rate in every definition of power, for example "the rate at which energy is transferred".
  • When you compare two machines, mention both the energy transferred and the time taken before you conclude which is more powerful.
  • In a two-step question, find the work done with W=FsW=FsW=Fs first, then divide by the time.
Self review
  • Define power.
  • Write the equation that links power, work done and time.
  • How many joules per second is 1 kW1\ \text{kW}1 kW?
  • Calculate the power of a motor that does 1200 J1200\ \text{J}1200 J of work in 8.0 s8.0\ \text{s}8.0 s.
  • Two cranes lift identical loads to the same height, so how can you tell which is more powerful?

2.2.16 Newton's third law

Newton's third law

Definition

Newton's third law

When two objects interact, they exert forces on each other that are equal in size and opposite in direction.

Definition

Interaction pair

The two forces produced by one interaction, one acting on each of the two interacting objects.

  1. Newton's third law states that when two objects interact, they exert forces on each other that are equal in size and opposite in direction.
  2. The two forces form an interaction pair, often called a third law pair.
  3. Every third law pair has the same set of features.
    1. The forces are equal in size.
    2. The forces act in opposite directions along the same line.
    3. The forces act on different objects.
    4. The forces are the same type of force, such as two gravitational forces or two normal contact forces.
    5. The forces start and stop at the same time.
  4. The law applies to every interaction, whether the objects are touching or not, and whether they are moving or still.
  5. A short way to state a pair is "if A exerts a force on B, then B exerts an equal and opposite force on A".

Third law pairs and balanced forces

Definition

Balanced forces

Forces acting on an object that cancel each other out so that the resultant force is zero.

  1. Third law pairs never cancel each other out, because the two forces act on different objects.
  2. Forces can only cancel when they act on the same object, and then they are balanced forces.
  3. A mug resting on a desk shows the difference clearly.
    1. The weight of the mug (the Earth pulling the mug down) and the normal contact force (the desk pushing the mug up) both act on the mug, so they are a pair of balanced forces, not a third law pair.
    2. The third law partner of the mug's weight is the mug pulling the Earth upwards.
    3. The third law partner of the desk pushing the mug up is the mug pushing the desk down.
  4. Two forces that are equal and opposite are therefore not automatically a third law pair, so you must check which object each force acts on.

Situations in equilibrium

  1. In equilibrium situations, the object you are studying has balanced forces on it, but third law pairs are still present between it and every object it touches or attracts.
  2. When you lean on a wall without moving, your hand pushes the wall and the wall pushes your hand back with an equal force, which you feel in your palm.
  3. When a picture hangs on a hook, the picture pulls down on the hook and the hook pulls up on the picture with an equal force.
  4. When two teams in a tug of war are not moving, each team pulls the rope and the rope pulls back on each team with an equal force.

Situations that are not in equilibrium

  1. Third law pairs also act when objects are speeding up, slowing down or changing direction.
  2. Both objects feel the same size of force, but the effects of the forces can be very different because the objects can have very different masses.
  3. When you walk, your foot pushes backwards on the ground and the ground pushes forwards on your foot with an equal force, which accelerates you forwards.
  4. When a swimmer pushes water backwards with her arms, the water pushes her forwards with an equal force.
  5. When a rocket fires its engines, it pushes hot exhaust gases downwards and the gases push the rocket upwards with an equal force.
  6. When a gun is fired, it pushes the bullet forwards and the bullet pushes the gun backwards with an equal force, so the gun recoils.
    1. The bullet has a tiny mass, so it gets a huge acceleration.
    2. The gun has a much larger mass, so it gets a much smaller acceleration.
  7. When a car hits a fly, the force on the fly is exactly equal to the force on the car, but the fly is squashed because its mass is so small.
Example

Identifying the pairs for a rowing boat

  • A rower pulls the oars so that the blades push water backwards.
  • The oars exert a backward force on the water.
  • The water exerts an equal forward force on the oars, and so on the boat.
  • These two forces are the same type, a contact force between the blades and the water, and they act on different objects, the water and the oars.
  • The forward force from the water accelerates the boat because it is larger than the drag on the hull.
Example

Comparing the effects on an astronaut and a satellite

  • An astronaut floating in space pushes a satellite with a force of 40 N40\ \text{N}40 N.
  • By the third law, the satellite pushes the astronaut with a force of 40 N40\ \text{N}40 N in the opposite direction.
  • The astronaut has a mass of 100 kg100\ \text{kg}100 kg and the satellite has a mass of 400 kg400\ \text{kg}400 kg.
aastronaut=40 N100 kg=0.40 m/s2 a_{\text{astronaut}}=\dfrac{40\ \text{N}}{100\ \text{kg}}=0.40\ \text{m/s}^2 aastronaut​=100 kg40 N​=0.40 m/s2 asatellite=40 N400 kg=0.10 m/s2 a_{\text{satellite}}=\dfrac{40\ \text{N}}{400\ \text{kg}}=0.10\ \text{m/s}^2 asatellite​=400 kg40 N​=0.10 m/s2
  • The forces are equal, but the astronaut moves away four times as fast because he has a quarter of the mass.
Common Mistake
  • Do not say that the forces in a third law pair cancel out, because they act on different objects.
  • Do not call an object's weight and the normal contact force on it a third law pair, because both forces act on the same object.
  • Do not say that the larger or faster object exerts the larger force, because the two forces in a pair are always equal in size.
Exam technique

Third law answers

  • Name both objects and describe both forces, using "on" and "by" to show which object each force acts on.
  • State that the forces are equal in size and opposite in direction, as each of these can earn a separate mark.
  • If a question asks why the forces do not cancel, say that they act on different objects.
  • When the effects on two objects differ, explain the difference using their masses, not a difference in force.
Self review
  • State Newton's third law.
  • Give four features shared by the two forces in a third law pair.
  • What is the third law partner of the weight of a mug?
  • Why do third law pairs never cancel out?
  • Explain why a gun recoils with a much smaller acceleration than the bullet.

2.2.17 Circular motion and changing velocity

Speed and velocity

Definition

Speed

The distance travelled by an object per unit time.

Definition

Velocity

The speed of an object in a stated direction.

  1. Speed is a scalar, so it tells you only how fast an object is moving.
  2. Velocity is a vector, so it tells you how fast an object is moving and in which direction.
  3. Two cars travelling at 13 m/s13\ \text{m/s}13 m/s, one heading north and one heading east, have the same speed but different velocities.
  4. The velocity of an object changes if its speed changes, if its direction changes, or if both change.
  5. This means an object can keep a constant speed while its velocity changes, as long as its direction is changing.

Moving in a circle

Definition

Tangent

A straight line that touches a curve at a single point and has the same gradient as the curve at that point.

  1. An object moving in a circle is always changing the direction in which it travels.
  2. At any instant, the object moves along the tangent to the circle, which is at right angles to the radius at that point.
  3. As the object moves round, the tangent points in a different direction at every point on the circle.
  4. If the object moves at a constant speed, the velocity arrows all have the same length, but each one points in a different direction.
  5. The velocity is therefore continuously changing, even though the speed is not.
  6. A car driving round a roundabout at a steady 10 m/s10\ \text{m/s}10 m/s shows this clearly.
    1. At one point the car is heading east at 10 m/s10\ \text{m/s}10 m/s.
    2. A quarter of the way round, the car is heading north at 10 m/s10\ \text{m/s}10 m/s.
    3. Halfway round, the car is heading west at 10 m/s10\ \text{m/s}10 m/s, which is the opposite velocity to the one it started with.

Acceleration in a circle

Definition

Acceleration

The rate of change of velocity.

  1. Acceleration is the rate of change of velocity, not only the rate of change of speed.
  2. An object moving in a circle at a constant speed has a changing velocity, so it is accelerating all the time.
  3. By Newton's first law, a changing velocity needs a resultant force, so there must be a resultant force on any object moving in a circle.
  4. This resultant force acts towards the centre of the circle, which is why the direction of motion keeps turning inwards.
  5. Different forces provide this inward resultant force in different situations.
    1. For a conker whirled on a string, the tension in the string pulls the conker towards the centre.
    2. For a car on a roundabout, friction between the tyres and the road pushes the car towards the centre.
    3. For a rider on a fairground ride, the normal contact force from the seat or the wall pushes the rider towards the centre.
  6. If the inward force suddenly disappears, the object stops turning and carries on in a straight line along the tangent, as Newton's first law predicts.
    1. If the string of a whirling conker snaps, the conker flies off along the tangent, not outwards along the radius.

An object moving in a circle at a steady speed, with its velocity along the tangent and its acceleration pointing towards the centre of the circle along the radius.

Example

Explaining the changing velocity of a cyclist on a velodrome

  • A cyclist rides round a circular velodrome track at a steady speed of 15 m/s15\ \text{m/s}15 m/s.
  • Velocity is a vector, so it depends on direction as well as speed.
  • As the cyclist moves round the track, the direction of motion changes continuously.
  • So the velocity of the cyclist changes continuously, even though the speed stays at 15 m/s15\ \text{m/s}15 m/s.
  • A changing velocity means the cyclist is accelerating, and the resultant force causing this acts towards the centre of the track.
Example

Following a hammer thrower's hammer

  • A hammer thrower swings a heavy ball on a wire in a horizontal circle at a steady speed.
  • The tension in the wire pulls the ball towards the centre of the circle, so its direction keeps changing.
  • When the thrower lets go, the tension disappears and there is no longer an inward resultant force.
  • The ball then flies off along the tangent to the circle at the point of release.
Common Mistake
  • Do not say that an object moving in a circle at a constant speed has a constant velocity, because its direction is changing.
  • Do not say that an object moving in a circle at a constant speed is not accelerating, because any change in velocity is an acceleration.
  • Do not say that an object flies outwards along the radius when the inward force is removed, because it moves off along the tangent.
Exam technique

Explaining circular motion

  • Start by stating that velocity is a vector, so it has a direction.
  • Then say that the direction of the object changes continuously as it moves round the circle.
  • Conclude that the velocity changes even though the speed is constant, and that the object is therefore accelerating.
  • Avoid writing only "because it is moving in a circle", as this repeats the question without explaining it.
Self review
  • What is the difference between speed and velocity?
  • In which direction does an object moving in a circle travel at any instant?
  • Explain why an object moving in a circle at a constant speed has a changing velocity.
  • In which direction does the resultant force on an object moving in a circle act?
  • Describe what happens to a whirling conker if its string snaps.

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Free-body diagrams of a book on a table, a box with a 12 N push and 5 N friction, and a falling ball at terminal velocity with forces labelled

A force is a push or pull caused by an interaction between objects, and it is measured in newtons, N. Force is a vector, so size and direction both matter.

A free-body diagram shows only the object you are analysing and the forces acting on it. The resultant force is the single overall force that has the same effect as all the forces acting together.

When forces balance, the resultant force is zero and the object is in equilibrium. In the middle diagram, 12 N12 \, \text{N}12N right and 5 N5 \, \text{N}5N left give a resultant of 7 N7 \, \text{N}7N to the right.

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Which of these is a non-contact force?

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A force is a [     ] caused by an [     ] between objects.

2.2 Newton’s laws Revision Guide

  1. GCSE
  2. /Physics
  3. /2.2 Newton’s laws

Revision notes for OCR GCSE Physics 2.2 Newton’s laws: explanations and worked examples on 2.2.1 Interactions between objects, 2.2.2 Force pairs in interactions, 2.2.3 Forces as vectors, 2.2.4 Newton's first law, 2.2.5 Resolving forces with vector diagrams, 2.2.6 Forces acting on an isolated object, 2.2.7 Free body diagrams and resultant force, 2.2.8 Free body diagrams for balanced forces, 2.2.9 Newton's second law, 2.2.10 Inertia and inertial mass, 2.2.11 Momentum in collisions, 2.2.12 Relating force, mass, velocity and acceleration, 2.2.13 Work done by a force, 2.2.14 Work, stored energy and units of energy, 2.2.15 Power as the rate of energy transfer, 2.2.16 Newton's third law, and 2.2.17 Circular motion and changing velocity.

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