Newton's Laws and Application of Newton's Laws
Grade 12
1. Vectors, Resultants and Components
Revision from Grades 10 and 11
Every force problem in Grade 12 adds vectors or resolves them into components. This section revises vectors in one dimension (Grade 10) and in two dimensions (Grade 11) in full.
Vectors and scalars
A physical quantity is anything in physics that can be measured, such as time, mass, weight, force and charge. Some quantities are fully described by a number and a unit; others also need a direction.
Definition: scalar
A scalar is a physical quantity that has magnitude only.
Definition: vector
A vector is a physical quantity that has both magnitude and direction.
Scalars
- mass (kg)
- time (s)
- distance (m)
- speed (m·s⁻¹)
- charge (C)
- energy (J)
Vectors
- force (N)
- weight (N)
- displacement (m)
- velocity (m·s⁻¹)
- acceleration (m·s⁻²)
A mass of 5 kg is complete as it stands, but a force of 5 N is not: it matters whether you push the box to the left or to the right. Weight is a force, so it is a vector that always points downwards, while mass is a scalar.
Writing vectors. A symbol with an arrow above it, F→, represents the force vector: its magnitude and its direction. The same letter without the arrow, F, represents only the magnitude of the force, for example F = 20 N.
Drawing vectors. A vector is drawn as an arrow. The length of the arrow, drawn to a chosen scale, shows the magnitude, and the arrowhead shows the direction. The start of the arrow is its tail and the end with the arrowhead is its head.
Two force vectors drawn to scale: 40 N to the right and 20 N to the left. The first arrow is twice as long as the second because its force is twice as large.
Directions in one dimension. Along a straight line there are only two directions. Choose one as positive (for example to the right or east); the opposite direction is then negative. A force of −20 N, with right as positive, is 20 N to the left.
Adding vectors in one dimension
Equal vectors have the same magnitude and the same direction. The negative of a vector has the same magnitude but the opposite direction: if A is 30 N to the right, −A is 30 N to the left.
A and B are equal vectors (30 N to the right). C is the negative of A (30 N to the left). D points the same way as A but is larger, so it is not equal to A.
Definition: resultant vector
The resultant vector is the single vector that has the same effect as all the original vectors acting together.
The resultant of force vectors is also called the net force. It can be found by drawing or by calculation.
The tail-to-head method
- Choose a scale. Draw a start line.
- On the top line, draw the vectors in the direction with the larger total one after the other, tail to head, starting at the start line.
- On the line underneath, draw the vectors in the opposite direction tail to head, starting at the head of the last vector on the top line and pointing back.
- Draw the resultant on the bottom line from the start line to the head of the last vector there, so the two arrowheads meet. Measure its length and use the scale to find its magnitude; it points the same way as the top line.
Forces of 15 N east, 40 N west and 10 N east drawn tail to head, with east to the right. F₂ points the way with the larger total, so it goes on the top line. On the line underneath, F₁ and F₃ follow tail to head, starting at the head of F₂. R runs along the bottom line from the start line to the head of F₃. R is 3 squares long, which is 15 N west.
Finding the resultant by calculation
- Choose a positive direction.
- Give each vector a sign: + in the positive direction and − in the opposite direction.
- Add the signed values.
- Give the magnitude of the answer and turn its sign back into a direction.
Forces in the same direction add up to a larger resultant; forces in opposite directions partly or completely cancel. For two forces along a line, the largest possible resultant is their sum and the smallest is their difference.
Subtracting vectors. To subtract a vector, add its negative: A − B = A + (−B). This works for up to four force vectors in the same way, one at a time.
Three horizontal forces on a crate. With east as positive, R = (+30 N) + (+12 N) + (−50 N) = −8 N, so the resultant is 8 N west.
The resultant of perpendicular vectors
In Grade 10 all the vectors acted along one straight line. In two dimensions they can act in any direction in a flat plane, so we work on a Cartesian plane: a horizontal x-axis and a vertical y-axis that cross at right angles. East (or to the right) is usually the positive x-direction and north (or upwards) the positive y-direction.
The method works for force vectors and for displacement vectors, and for up to four vectors at a time.
Step 1: add the co-linear vectors on each axis
Vectors along the same line are co-linear. Add the co-linear horizontal vectors, with signs, to get the net horizontal vector Rₓ. Add the co-linear vertical vectors to get the net vertical vector Rᵧ. This is the Grade 10 method, used once for each axis.
Four forces act on one point. Horizontally: Rₓ = (+20 N) + (−8 N) = +12 N, so Rₓ = 12 N east. Vertically: Rᵧ = (+14 N) + (−5 N) = +9 N, so Rᵧ = 9 N north.
Step 2: sketch Rₓ and Rᵧ and the resultant
Sketch Rₓ and Rᵧ on the Cartesian plane, then sketch the resultant R in one of two ways:
- Tail-to-head method: draw Rₓ, then draw Rᵧ with its tail at the head of Rₓ. R runs from the tail of Rₓ to the head of Rᵧ.
- Tail-to-tail (parallelogram) method: draw Rₓ and Rᵧ from the same point (tail to tail). Complete the parallelogram with dashed lines; because Rₓ and Rᵧ are perpendicular, it is a rectangle. R is the diagonal that starts at the common tail.
Tail to head: Rᵧ starts at the head of Rₓ, and R closes the right-angled triangle. θ is the angle between R and the x-axis.
Tail to tail: Rₓ and Rᵧ start at the same point. The rectangle is completed with dashed lines, and R is the diagonal from the common tail.
Step 3: the magnitude, with the theorem of Pythagoras
Rₓ, Rᵧ and R form a right-angled triangle with R as the hypotenuse, so the theorem of Pythagoras gives the magnitude of the resultant:
R² = Rₓ² + Rᵧ²
Step 4: the direction, with a trigonometric ratio
The angle θ between R and the x-axis follows from the tangent ratio. Use the magnitudes of the components, then describe the direction in words from the sketch.
tan θ = RᵧRₓ
Give the direction as an angle from a named direction, for example "36,87° north of east" or "36,87° above the positive x-axis". A direction can also be given as a compass bearing, measured clockwise from north: 36,87° north of east is a bearing of 53,13°.
Common mistakes
- Adding the magnitudes: 12 N east and 9 N north do not give 21 N, because they do not act along one line.
- Forgetting the square root: R² = 225 N², so R = 15 N, not 225 N.
- Turning the ratio upside down: tan θ = RₓRᵧ gives the angle from the y-axis, not from the x-axis.
- Giving the size without the direction: a resultant vector needs both.
Finding the resultant by drawing. Choose a scale, draw the vectors tail to head with a ruler and protractor, and draw R from the tail of the first vector to the head of the last. Measure the length of R and convert it with the scale; measure θ with the protractor. This graphical tail-to-head method works for up to four vectors in any directions. The calculation (component method) gives the same answer without a scale drawing.
Definition: closed vector diagram
A closed vector diagram is a vector diagram in which the vectors are drawn tail to head and the head of the last vector ends at the tail of the first vector, so that the resultant of the vectors is zero.
If the vectors drawn tail to head end exactly where the first one started, there is no gap for a resultant to fill: the resultant is zero. For forces, a closed vector diagram means the net force on the object is zero. Three forces that keep an object in equilibrium always form a closed triangle, and any one of them is equal in size and opposite in direction to the resultant of the other two.
F₁ = 12 N east, F₂ = 9 N north and F₃ = 15 N at 36,87° south of west, drawn tail to head. The head of F₃ ends at the tail of F₁: a closed vector diagram, so the resultant is zero.
Practical: forces on a force board
Hang masses from three strings tied at a knot, with two strings running over pulleys at the edges of a force board. When the knot stays still, draw the three forces (weights) to scale in their directions. Drawn tail to head, they form a closed triangle, which shows that the resultant of the three non-linear forces is zero.
How to plan, record and write up a practical: see the scientific investigation skills guide.
Resolving a vector into components
Resolving works in the opposite direction to finding a resultant. Any vector can be replaced by two perpendicular vectors, a horizontal x-component and a vertical y-component, that together have exactly the same effect as the original vector. The vector is the resultant of its components.
How to resolve a vector
- Draw a sketch of the vector on the Cartesian plane, starting at the origin. Show its magnitude and the angle θ between the vector and the x-axis.
- Drop dashed lines from the head of the vector to each axis to show the components.
- Calculate the components with the formulas below.
- Give each component a sign or a direction from the sketch: a component pointing left (west) or down (south) is negative.
Rₓ = R cos θ
Rᵧ = R sin θ
A force of 80 N at 30° to the positive x-axis. Its components are Fₓ = 80 N × cos 30° = 69,28 N and Fᵧ = 80 N × sin 30° = 40 N.
Check where the angle is measured from
R cos θ gives the x-component only when θ is measured from the x-axis: the x-component is the side adjacent to θ. If the angle is given from the vertical (the y-axis), the x-component is opposite that angle, so it is R sin of that angle, and the y-component is R cos of that angle. Always sketch the vector first.
Force and displacement. The same method resolves any vector. A suitcase pulled with a force along a handle that slopes upwards moves because of the horizontal component of the pull, while the vertical component lifts some of its weight. A displacement of 15 km at 40° north of west has a westward component of 15 cos 40° = 11,49 km and a northward component of 15 sin 40° = 9,64 km.
A displacement of 15 km at 40° north of west. The angle is measured from the west side of the x-axis, so the x-component is −11,49 km (west) and the y-component is +9,64 km (north).
The component method for up to four vectors
- Resolve every vector that is not along an axis into its x- and y-components.
- Add all the x-components, with signs, to get Rₓ. Add all the y-components to get Rᵧ.
- Sketch Rₓ and Rᵧ and find R with R² = Rₓ² + Rᵧ².
- Find the direction with tan θ = RᵧRₓ and describe it from the sketch.
- If Rₓ = 0 and Rᵧ = 0, the vectors form a closed vector diagram and the resultant is zero.
Worked Examples
- 1Take east as positive: R = (+15 N) + (−40 N) + (+10 N) = −15 N, so R = 15 N west.
- 1Take right as positive: A = +25 N and B = −10 N.
- 2A − B = A + (−B) = (+25 N) + (+10 N) = +35 N
- 1Take east and north as positive. Horizontally: Rₓ = (+20 N) + (−8 N) = +12 N, so Rₓ = 12 N east.
- 2Vertically: Rᵧ = (+14 N) + (−5 N) = +9 N, so Rᵧ = 9 N north.
- 3R² = Rₓ² + Rᵧ² = (12)² + (9)² = 225
- 4R = 15 N
- 5tan θ = 912, so θ = 36,87°
- 1The two displacements are perpendicular: Δy = 1,2 km north and Δx = 0,9 km east.
- 2Δx² + Δy² = R², so R² = (0,9)² + (1,2)² = 2,25
- 3R = 1,5 km
- 4tan θ = 1,20,9, so θ = 53,13° (from east, towards north)
- 1Sketch the force from the origin at 30° above the positive x-axis.
- 2Fₓ = F cos θ = 80 × cos 30° = 69,28 N
- 3Fᵧ = F sin θ = 80 × sin 30° = 40 N
- 1Resolve the 50 N force: x-component = 50 cos 60° = 25 N east; y-component = 50 sin 60° = 43,3 N north.
- 2Rₓ = (+30) + (+25) + (−20) = 35 N, so Rₓ = 35 N east
- 3Rᵧ = (+43,3) + (−25) = 18,3 N, so Rᵧ = 18,3 N north
- 4R² = (35)² + (18,3)², so R = 39,5 N
- 5tan θ = 18,335, so θ = 27,6°
2. Kinds of Forces
Revision from Grade 11
Weight, the normal force, friction, applied forces and tension, and the difference between static and kinetic friction. This work is examined directly in Paper 1 and is used again in momentum, vertical projectile motion and work, energy and power.
A force is a push or a pull, measured in newtons (N). In mechanics you will meet five kinds of force again and again: weight, the normal force, friction, applied forces and tension.
Weight, w
Weight is the gravitational force of the Earth on an object. It always acts vertically downwards, from the centre of the object, and its size is w = mg, with g = 9,8 m·s⁻² near the Earth's surface. (Weight is defined in section 7 of this topic.)
Definition: normal force
The normal force, N, is the force or the component of a force which a surface exerts on an object with which it is in contact, and which is perpendicular to the surface.
The normal force acts perpendicular to the surface, whether the plane is horizontal or inclined. On a horizontal floor it points straight up; on an inclined plane it points away from the slope at right angles to it. The normal force is not always equal to the weight: on an incline it is smaller than the weight, and when a rope pulls a box upwards at an angle, part of the pull lifts the box and the normal force becomes smaller too.
Definition: frictional force
The frictional force, f, is the force that opposes the motion of an object and which acts parallel to the surface.
Friction acts parallel to the surface and opposes the motion (or the tendency to move) of the object relative to the surface. It is caused by the tiny irregularities of the two surfaces interlocking, which impedes motion. People have long used friction: the first people to make fire did so by rubbing sticks together.
A force diagram of a crate pushed to the right on a rough floor: the applied force F, friction f along the floor opposing the motion, the weight w downwards and the normal force N perpendicular to the floor.
Static and kinetic friction
Definition: static frictional force
The static frictional force, fₛ, is the force that opposes the tendency of motion of a stationary object relative to a surface.
Definition: kinetic frictional force
The kinetic frictional force, fₖ, is the force that opposes the motion of a moving object relative to a surface.
Static friction adjusts itself. When you push a heavy crate gently and it does not move, the static frictional force is exactly as large as your push. Push harder and static friction grows to match, but only up to a maximum value, the maximum static frictional force fsmax, reached just before the crate starts to move:
fsmax = μsN
If the applied force exceeds fsmax, the object starts to move and a net force accelerates it. Once it slides, kinetic friction acts. For a given pair of surfaces the kinetic frictional force is constant:
fk = μkN
μs and μk are the coefficients of static and kinetic friction. They have no unit and depend only on the two surfaces. For most surfaces μk is smaller than μs, which is why it is easier to keep a crate sliding than to get it started. Use these formulas for objects at rest or sliding on horizontal and on inclined planes: in each case, N is the normal force on that surface.
What friction depends on
- A frictional force is proportional to the normal force: press the surfaces together twice as hard and the friction doubles.
- It is independent of the area of contact: a brick slides just as easily on its large face as on its small face.
- It is independent of the velocity of motion: kinetic friction is the same at 1 m·s⁻¹ and at 3 m·s⁻¹.
Investigation: normal force and maximum static friction
Pull a block with a spring balance and read the force just before it starts to move: that reading is the maximum static friction. Add masses on top of the block to increase the normal force and repeat; then keep the block the same and change the surface (rough, smooth, cloth). The maximum static friction increases in proportion to the normal force and depends on the surfaces.
How to plan, record and write up a practical: see the scientific investigation skills guide.
Applied forces and tension
An applied force is a push or a pull by a person or a machine, labelled F (or FA). Tension, T, is the pulling force in a string, rope or cable. A string always pulls: it pulls on the objects at both of its ends, along the string. In these problems the strings are light (their mass is negligible) and inextensible (they do not stretch), so the tension is the same all along the string and objects joined by it have the same acceleration.
Worked Examples
- 1On a horizontal floor with only vertical forces besides the push: N = w = mg = (20)(9,8) = 196 N
- 2(a) fsmax = μsN = (0,45)(196) = 88,2 N
- 3(b) The push of 60 N is less than 88,2 N, so the crate stays at rest and static friction balances the push.
- 1N = mg = 196 N
- 2fk = μkN = (0,3)(196) = 58,8 N
3. Force Diagrams and Free-Body Diagrams
Revision from Grade 11
Force diagrams, free-body diagrams and resolving forces on an inclined plane. This work is examined directly in Paper 1 and is used again in momentum, vertical projectile motion and work, energy and power.
Before any calculation with forces, draw the forces. Two kinds of drawing are used.
Force diagram
A force diagram is a picture of the object or objects of interest with all the forces acting on it (them) drawn in as arrows, each starting where the force acts.
Free-body diagram
In a free-body diagram the object of interest is drawn as a dot, and all the forces acting on it are drawn as arrows pointing away from the dot.
Definition: free-body diagram
A free-body diagram is a diagram that shows the relative magnitudes and directions of forces acting on a body or particle that has been isolated from its surroundings.
The free-body diagram of the pushed crate from the previous section. Only forces that act ON the crate are drawn; the arrow lengths show the relative magnitudes (here N = w and F is larger than f, so the crate accelerates).
Drawing a free-body diagram
- Choose one object and draw it as a dot.
- Draw every force that acts on that object as an arrow away from the dot: weight, normal force, friction, tension, applied forces.
- Make the arrow lengths show the relative sizes of the forces.
- Label every arrow with a name or symbol (w or Fg, N or FN, f, T, F). Never draw the net force as an extra force.
Resolving forces on an inclined plane
On an inclined plane, choose the x-axis parallel to the slope and the y-axis perpendicular to it. The normal force and friction then lie along the axes, and only the weight has to be resolved. The angle between the weight and the perpendicular to the slope equals the angle of the incline, θ, so:
w∥ = mg sin θ (parallel to the slope, pointing down it)
w⊥ = mg cos θ (perpendicular to the slope, into it)
A block on an incline: its weight w and the components w∥ down the slope and w⊥ into the slope. The normal force N balances w⊥, so on a plain incline N = mg cos θ.
The net force in the x-direction is the vector sum of all the force components in the x-direction, and the net force in the y-direction is the vector sum of all the components in the y-direction. On an incline, nothing accelerates the block into or out of the slope, so the perpendicular components always add up to zero.
A force applied at an angle is resolved in the same way: a pull F at θ above the horizontal has a horizontal component F cos θ and a vertical component F sin θ that lifts part of the weight, so the normal force becomes N = mg − F sin θ.
A block pulled by a force F at an angle above the horizontal. The vertical component of F lifts part of the weight, so N is smaller than w.
Worked Examples
- 1w = mg = (5)(9,8) = 49 N
- 2w∥ = mg sin θ = (49) sin 30° = 24,5 N down the slope
- 3w⊥ = mg cos θ = (49) cos 30° = 42,44 N into the slope
- 4Perpendicular to the slope the forces add up to zero: N − w⊥ = 0, so N = 42,44 N
- 1F cos θ = 15: cos θ = 1518, so θ = 33,56°
- 2Vertical component: Fy = F sin θ = (18) sin 33,56° = 9,95 N upwards
- 3Vertically Fnet = 0: N + Fy − w = 0, so N = (5)(9,8) − 9,95 = 39,05 N
- 4The block is at rest, so horizontally fsmax = Fx = 15 N
- 5fsmax = μsN: 15 = μs(39,05), so μs = 0,38
4. Newton's First and Second Laws
Revision from Grade 11
Newton's first law and inertia, and Newton's second law. This work is examined directly in Paper 1 and is used again in momentum, vertical projectile motion and work, energy and power.
Newton's first law and inertia
Definition: Newton's first law of motion
Newton's first law of motion states that a body will remain in its state of rest or motion at constant velocity unless a non-zero resultant (net) force acts on it.
An object at rest stays at rest, and an object that moves keeps moving in a straight line at a constant speed, unless a net force acts on it. Both situations are called equilibrium: the net force is zero. Objects do not need a force to keep moving; they need a net force to change their motion (to speed up, slow down or change direction).
Definition: inertia
Inertia is the property of an object that makes it resist any change in its state of rest or uniform motion.
Mass is the measure of an object's inertia. The larger the mass, the larger the inertia and the harder it is to change the object's motion: a loaded truck is much harder to start, stop or turn than a bicycle. Newton's first law is therefore also called the law of inertia.
Examples of inertia
- When a bus pulls away suddenly, standing passengers seem to fall backwards: their bodies tend to stay at rest while the bus moves forward under them.
- When the bus brakes suddenly, the passengers lurch forward: their bodies tend to keep moving at the original velocity.
- A tablecloth pulled away quickly leaves the dishes in place, because their inertia keeps them at rest.
Why wearing seatbelts matters
In a collision a car stops in a very short time. According to Newton's first law, a passenger who is not wearing a seatbelt continues to move forward at the car's original velocity, because no force acts on them to stop them with the car. They keep moving until something stops them: the steering wheel, the dashboard, the windscreen or the road. A seatbelt supplies the unbalanced (net) force that slows the passenger down together with the car, spread over a longer time and over the strong parts of the body. Headrests work the same way in a collision from behind, where the body is pushed forward and the head tends to stay behind.
Newton's second law
Definition: Newton's second law of motion
Newton's second law of motion states that when a net force acts on an object, the object will accelerate in the direction of the force and the acceleration is directly proportional to the force and inversely proportional to the mass of the object.
Fnet = ma
Fnet is the net (resultant) force in newtons, m the mass in kilograms and a the acceleration in m·s⁻². One newton is the net force that gives a mass of 1 kg an acceleration of 1 m·s⁻²: 1 N = 1 kg·m·s⁻². The acceleration is always in the direction of the net force.
What the law says
- For a constant mass, the acceleration is directly proportional to the net force: double the net force and the acceleration doubles.
- For a constant net force, the acceleration is inversely proportional to the mass: double the mass and the acceleration halves.
- If Fnet = 0, then a = 0: the object is at rest or moves at constant velocity (Newton's first law).
Objects in equilibrium and accelerating objects
Draw force diagrams and free-body diagrams for objects that are in equilibrium (at rest or moving with constant velocity) and for objects that are accelerating (non-equilibrium). For an object in equilibrium, all the forces along the plane of the motion add up to zero, and so do all the forces perpendicular to it. For an accelerating object, the arrows in the direction of motion do not balance: the longer arrow shows the direction of the net force.
Applying Fnet = ma
- Draw a free-body diagram of each object.
- Choose a positive direction, usually the direction of the acceleration.
- Apply Fnet = ma separately in the x-direction and in the y-direction. Perpendicular to the plane of the motion the forces always add up to zero.
- If there is more than one object, draw a free-body diagram for each object and apply Fnet = ma to each object separately.
- Substitute with signs and units, and solve.
Common mistakes
- Putting the net force into the free-body diagram as if it were an extra force: Fnet is the sum of the forces already drawn.
- Using the weight instead of the normal force in the friction formula when a force pulls or pushes at an angle, or on an incline.
- Using the applied force instead of the net force in F = ma.
Prescribed experiment: verifying Newton's second law
Pull a trolley along a runway with one, two, three and four identical stretched rubber bands (so the force increases in equal steps) and record its motion with a ticker timer. Calculate the acceleration for each force from the tape. A graph of acceleration against net force is a straight line through the origin: a is directly proportional to Fnet for a constant mass. Repeating with the same force and extra masses on the trolley shows that a is inversely proportional to the mass (a against 1 ÷ m is a straight line).
How to plan, record and write up a practical: see the scientific investigation skills guide.
Worked Examples
- 1Newton's first law: a body remains in its state of motion at constant velocity unless a non-zero net force acts on it.
- 2The wall exerts a force on the taxi and stops it, but that force does not act on the passenger.
- 3The passenger therefore continues to move forward at 60 km·h⁻¹ until the dashboard or windscreen exerts a large force on them over a very short time.
- 1Inertia is the property of an object that makes it resist any change in its state of rest or uniform motion.
- 2Mass is the measure of inertia, and the 10 kg brick has the larger mass.
- 1Take the direction of motion as positive.
- 2Fnet = ma: F + (−f) = ma
- 350 − 20 = 10a
- 4a = 3 m·s⁻²
- 1Resolve the pull: Fx = 60 cos 30° = 51,96 N; Fy = 60 sin 30° = 30 N
- 2y-direction, Fnet = 0: N + Fy − mg = 0, so N = (8)(9,8) − 30 = 48,4 N
- 3fk = μkN = (0,2)(48,4) = 9,68 N
- 4x-direction, Fnet = ma: 51,96 − 9,68 = 8a
- 5a = 5,29 m·s⁻²
5. Applying Newton's Second Law
Revision from Grade 11
A single object on horizontal and inclined planes, lifts and rockets, and the four two-body systems. This work is examined directly in Paper 1 and is used again in momentum, vertical projectile motion and work, energy and power.
Every problem in this section is solved the same way: a free-body diagram for each object, a positive direction, and Fnet = ma applied to each object in each direction.
A single object on a horizontal or inclined plane
On a horizontal plane the normal force balances the weight (unless another force has a vertical component), and Fnet along the plane is the applied force minus friction. On an inclined plane, take the direction of motion along the slope as positive. For a block sliding down a frictionless incline, the only force along the slope is w∥:
mg sin θ = ma so a = g sin θ
On a rough incline, kinetic friction acts up the slope, against the motion, and N = mg cos θ, so fk = μkmg cos θ. If the block is pushed or pulled up the slope, both w∥ and friction act down the slope. A block at rest on an incline has static friction up the slope equal to w∥.
A block sliding down a rough incline: friction f acts up the slope, opposite to the motion.
Vertical motion: lifts, rockets and apparent weight
For vertical motion, take the direction of the acceleration as positive. A lift cable pulls up with tension T; a rocket engine pushes up with a thrust. For a person standing on a scale in a lift, the scale reads the normal force it exerts on the person. This reading is the apparent weight, which differs from the true weight mg whenever the lift accelerates:
- Accelerating upwards (or slowing down while moving down): N − mg = ma, so N = m(g + a) and the person feels heavier.
- At rest or moving at constant velocity: N = mg; the apparent weight equals the weight.
- Accelerating downwards (or slowing down while moving up): mg − N = ma, so N = m(g − a) and the person feels lighter.
- In free fall (a = g) the normal force is zero and the person feels weightless.
Two-body systems
When two objects are joined by a light inextensible string, they move together with the same magnitude of acceleration, and the tension is the same at both ends of the string. Draw a free-body diagram for EACH body and apply Newton's second law to EACH body separately. This gives two equations, which are solved together for the acceleration and the tension. The four arrangements in CAPS are:
- Both on a flat horizontal plane, with or without friction.
- One on a horizontal plane and one hanging vertically from a string over a frictionless pulley.
- Both on an inclined plane, with or without friction.
- Both hanging vertically from a string over a frictionless pulley.
Both on a horizontal plane: F pulls the front block, and the string pulls the rear block along.
One on a horizontal table, one hanging over a frictionless pulley: the weight of the hanging block drives the system.
Both on an inclined plane: F pulls the upper block up the slope, and the string pulls the lower block.
Both hanging over a frictionless pulley: the heavier block accelerates down and the lighter block up.
Combining Newton's second law with the equations of motion
Many exam questions first ask for the acceleration from Fnet = ma and then use it in an equation of motion, or work the other way: find the acceleration from the motion, then the force. The acceleration is the link between the two. The equations of motion hold only while the acceleration is constant, so the net force must be constant too.
vf = vi + aΔt
Δx = viΔt + ½aΔt²
vf² = vi² + 2aΔx
Δx = (vi + vf2)Δt
Force and motion in one problem
- Draw a free-body diagram and choose a positive direction.
- Use Fnet = ma to find the acceleration (or use the motion to find it).
- Use the same acceleration, with its sign, in an equation of motion.
Worked Examples
- 1Take down the slope as positive. Perpendicular to the slope N = mg cos θ.
- 2(a) mg sin θ = ma, so a = g sin θ = (9,8) sin 25° = 4,14 m·s⁻²
- 3(b) mg sin θ − μkmg cos θ = ma, so a = g(sin θ − μk cos θ)
- 4a = (9,8)(sin 25° − (0,15) cos 25°) = 2,81 m·s⁻²
- 1(a) Up positive: N − mg = ma, so N = m(g + a) = (70)(9,8 + 1,5) = 791 N
- 2(b) Down positive: mg − N = ma, so N = m(g − a) = (70)(9,8 − 1,5) = 581 N
- 14 kg block: T = 4a
- 26 kg block: 30 − T = 6a
- 3Add the equations: 30 = 10a, so a = 3 m·s⁻²
- 4T = (4)(3) = 12 N
- 1Table block: N = mg = 39,2 N; fk = (0,2)(39,2) = 7,84 N. T − 7,84 = 4a
- 2Hanging block (down positive): 19,6 − T = 2a
- 3Add: 19,6 − 7,84 = 6a, so a = 1,96 m·s⁻²
- 4T = 19,6 − (2)(1,96) = 15,68 N
- 1Up the slope positive. Lower block: T − 2(9,8) sin 25° = 2a
- 2Upper block: 40 − T − 3(9,8) sin 25° = 3a
- 3Add: 40 − 20,71 = 5a, so a = 3,86 m·s⁻²
- 4T = 2(3,86) + 8,28 = 16 N
- 1The 5 kg block accelerates down and the 3 kg block up.
- 23 kg block (up positive): T − 29,4 = 3a
- 35 kg block (down positive): 49 − T = 5a
- 4Add: 19,6 = 8a, so a = 2,45 m·s⁻²
- 5T = 29,4 + (3)(2,45) = 36,75 N
- 1N = mg = (4)(9,8) = 39,2 N; fk = μkN = (0,25)(39,2) = 9,8 N
- 2Fnet = ma: 30 − 9,8 = 4a, so a = 5,05 m·s⁻²
- 3vf = vi + aΔt = 0 + (5,05)(3) = 15,15 m·s⁻¹
- 4Δx = viΔt + ½aΔt² = 0 + ½(5,05)(3)² = 22,73 m
6. Newton's Third Law
Revision from Grade 11
Newton's third law and action-reaction pairs. This work is examined directly in Paper 1 and is used again in momentum, vertical projectile motion and work, energy and power.
Definition: Newton's third law of motion
Newton's third law of motion states that when object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of equal magnitude on object A.
Forces always come in pairs, called action-reaction pairs. If you push on a wall, the wall pushes back on you just as hard. Neither force comes first: they act at the same time.
Properties of action-reaction pairs
- The two forces are equal in magnitude.
- They act in opposite directions.
- They act on different objects: one on A and one on B.
- They act simultaneously: they appear and disappear together.
- They are the same type of force (both gravitational, both contact forces, and so on).
- Because they act on different objects, they never cancel each other out.
Identifying action-reaction pairs
Describe each force in the form "A exerts a force on B". Its pair is then "B exerts a force on A".
A book on a table
- The Earth pulls down on the book (its weight). Pair: the book pulls up on the Earth with an equal gravitational force.
- The table pushes up on the book (the normal force). Pair: the book pushes down on the table with an equal force.
- The weight of the book and the normal force on the book are equal and opposite, but they are NOT an action-reaction pair: both act on the same object (the book), and they are different types of force. They balance because the book is in equilibrium (Newton's first law).
A donkey pulling a cart
- The donkey pulls the cart forward; the cart pulls the donkey backward with an equal force.
- The donkey pushes backward on the ground with its hooves; the ground pushes the donkey forward with an equal force.
- The cart accelerates because the forward pull of the donkey on the cart is larger than the friction on the cart. The backward pull of the cart acts on the donkey, not on the cart, so it does not cancel the forward pull.
Other examples: a swimmer pushes the water backwards and the water pushes the swimmer forwards; a rocket pushes exhaust gases down and the gases push the rocket up.
Worked Examples
- 1(a) Action: the Earth pulls the learner down. Reaction: the learner pulls the Earth up.
- 2(b) Action: the floor pushes the learner up. Reaction: the learner pushes the floor down.
- 1The two forces of the pair act on different objects, so they cannot cancel.
- 2To find whether the cart accelerates, add only the forces that act on the cart: the donkey's pull forward and friction backward.
- 3The ground pushes the donkey forward; if this is larger than the cart's pull on the donkey, the donkey accelerates too.
7. Newton's Law of Universal Gravitation
Revision from Grade 11
Newton's law of universal gravitation, weight, mass and weightlessness. This work is examined directly in Paper 1 and is used again in momentum, vertical projectile motion and work, energy and power.
Definition: Newton's law of universal gravitation
Newton's law of universal gravitation states that each body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
F = Gm1m2r²
F is the gravitational force (in N) that each mass exerts on the other, m1 and m2 are the masses (in kg), r is the distance between their centres (in m) and G is the universal gravitational constant, G = 6,67 × 10⁻¹¹ N·m²·kg⁻². The forces on the two masses are equal in size and opposite in direction: they are an action-reaction pair.
r is measured from centre to centre, not from surface to surface.
An inverse-square law
- Double one of the masses and the force doubles.
- Double the distance between the centres and the force becomes ¼ as large; triple it and the force becomes ⅑ as large.
- Halve the distance and the force becomes four times as large.
Because G is so small, the force between everyday objects is tiny and cannot be felt. It becomes large only when at least one of the masses is enormous, such as a planet.
Weight and gravitational acceleration
Definition: weight
Weight is the gravitational force the Earth exerts on any object on or near its surface.
w = mg
At the Earth's surface, the gravitational force on an object of mass m is its weight. Setting mg = GMmR² gives the gravitational acceleration at the surface of a planet of mass M and radius R:
g = GMR²
With the Earth's mass M = 5,98 × 10²⁴ kg and radius R = 6,38 × 10⁶ m, g = 9,8 m·s⁻²: the value used near the Earth. The same formula gives g on any planet from its own mass and radius, and w = mg then gives an object's weight on that planet.
Mass and weight
- Mass is the amount of matter in an object and a measure of its inertia. It is a scalar, measured in kilograms (kg), and it is the same everywhere.
- Weight is a force: the gravitational pull of the Earth (or another planet) on the object. It is a vector, measured in newtons (N), and it changes from planet to planet because g changes.
- A 60 kg astronaut has a mass of 60 kg on the Earth, on the Moon and in space, but her weight is much smaller on the Moon.
Definition: weightlessness
Weightlessness is the sensation of having no weight, experienced when an object is in free fall and no contact force such as a normal force acts on it.
Astronauts in an orbiting spacecraft are not beyond the reach of gravity: at that height g is still almost 90% of its value at the surface. They feel weightless because they and the spacecraft are falling around the Earth together, so the floor exerts no normal force on them. You feel the same sensation for a moment in a lift whose cable snaps, or at the top of a fast roller-coaster hill.
Experiment: verifying g
Let a mass fall freely while it pulls a ticker tape through a ticker timer. The distances between dots increase uniformly; from them calculate the acceleration, which should be close to 9,8 m·s⁻². Data loggers can be used instead of the ticker timer.
How to plan, record and write up a practical: see the scientific investigation skills guide.
Worked Examples
- 1F = Gm1m2r² = (6,67 × 10⁻¹¹)(5,98 × 10²⁴)(70)(6,38 × 10⁶)²
- 2F = 685,94 N
- 3Check with w = mg: (70)(9,8) = 686 N, the same to within rounding.
- 1g = GMR² = (6,67 × 10⁻¹¹)(6,42 × 10²³)(3,4 × 10⁶)² = 3,7 m·s⁻²
- 2w = mg = (60)(3,7) = 222 N