1.2 Forces and Newton's laws
⤓ Presentation⤓ PowerPointPresentation: opens offline in any browser, simulations live. PowerPoint: editable.
AS content in Unit 1 (Mechanics and Materials): specification points 8, 9, 10 and 12. Friction is always given to you as a force in this course.
What this page covers
8 points
- Free-body diagrams for a particle and for an extended body, with the weight acting at the centre of gravity
- Weight, mass and gravitational field strength: \(g = F/m\) and \(W = mg\)
- Newton's first law and equilibrium, including resolving forces
- Forces on slopes: components of the weight along and perpendicular to the slope
- Newton's second law, \(\Sigma F = ma\), for a body of constant mass
- Connected bodies (towing) and the forces on a person in a lift
- Terminal velocity, explained with forces
- Newton's third law and the properties of the pair of forces in an interaction
In one minute
7 points
- One diagram — the free-body diagram — and one equation — \(\Sigma F = ma\) — link force to motion, from a sign hanging still to a skydiver opening a parachute.
- A free-body diagram shows every force acting on one body, drawn as arrows starting on that body. The weight of an extended body acts at its centre of gravity.
- Weight \(W = mg\). Mass (kg) is the same everywhere; weight (N) depends on \(g\), which is 9.81 N kg⁻¹ at the Earth's surface.
- First law: zero resultant force ⇔ at rest or constant velocity. Resolve in two perpendicular directions and set each sum to zero.
- Second law: \(\Sigma F = ma\). The acceleration is in the direction of the resultant force.
- Terminal velocity: drag grows with speed until it equals the weight; the resultant force, and so the acceleration, become zero.
- Third law: if A pushes or pulls B, B pushes or pulls A with a force of equal size, opposite direction and the same type — acting on the other body.
Contents
14 sections
- What is a free-body diagram?
- What is the difference between mass and weight?
- What does Newton's first law tell you?
- How do you resolve forces on a slope?
- How do you use Newton's second law?
- How do you analyse connected bodies and lifts?
- What is terminal velocity?
- What does Newton's third law say?
- Exam tips for 1.2
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is a free-body diagram?
Almost every mistake in a forces question comes from a force that is missing, doubled or acting on the wrong body. A free-body diagram prevents this by isolating one body and asking a single question: what is pushing or pulling this object?
| Force | Caused by | Direction |
|---|---|---|
| Weight \(W\) | the gravitational pull of the Earth | vertically down, from the centre of gravity |
| Normal contact force \(R\) | a surface pushing on the body | perpendicular to the surface, away from it |
| Friction \(F\) | a surface resisting sliding | along the surface, opposing the (tendency to) slide |
| Tension \(T\) | a stretched string, rope, cable or tow bar | along the string, pulling away from the body |
| Drag / air resistance \(D\) | a fluid the body moves through | opposite to the velocity |
| Upthrust \(U\) | a fluid the body is immersed in | vertically up |
| Driving force | the road pushing forward on driven wheels (or the water on a propeller) | forwards |
Drawing one well
- Draw only the body you are interested in — a dot for a particle, an outline for an extended body.
- Every arrow starts on the body and points the way the force acts; label it with its name or symbol and value.
- Make arrow lengths roughly proportional to the sizes of the forces — balanced forces get equal arrows.
- For an extended body, draw the weight from the centre of gravity: the single point at which the whole weight can be taken to act. For a uniform body it is at the geometric centre.
- Never include forces that the body exerts on other things.
Adding a "force of motion" or a "force of the throw" to a moving body. Once a ball leaves the hand, the hand exerts no force on it: the only forces are its weight and air resistance.
What is the difference between mass and weight?
In everyday speech "weight" is often measured in kilograms, but in physics the two are different quantities. Mass tells you how much matter there is and how hard the body is to accelerate. Weight is a force: the pull of a gravitational field on that mass.
At the Earth's surface \(g = 9.81\ \text{N kg}^{-1}\). Since \(1\ \text{N} = 1\ \text{kg m s}^{-2}\), this is the same as a free-fall acceleration of 9.81 m s⁻².
| Mass | Weight | |
|---|---|---|
| What it is | the amount of matter; how hard the body is to accelerate | the gravitational force on the body |
| Type | scalar | vector (down, towards the centre of the planet) |
| Unit | kg | N |
| Changes from place to place? | no | yes, with \(g\) |
| Measured with | a balance (compares masses) | a newton-meter or force sensor |
A planetary rover has a mass of 1025 kg. The gravitational field strength at the surface of Mars is 3.71 N kg⁻¹. Calculate the weight of the rover on Earth and on Mars, and state its mass on Mars.
Step 1 Weight on Earth: \(W = mg = 1025 \times 9.81 = 10\,055\ \text{N} = \) 1.01 × 10⁴ N
Step 2 Weight on Mars: \(W = 1025 \times 3.71 = 3803\ \text{N} = \) 3.80 × 10³ N
Step 3 Mass does not depend on the field: mass on Mars = 1025 kg
What does Newton's first law tell you?
The law says that motion does not need a cause; changes in motion do. Zero resultant force means zero acceleration: the body is at rest or moving in a straight line at constant speed. It is the special case \(a = 0\) of the second law. A body with zero resultant force is in equilibrium (translational equilibrium; turning effects are on the moments page). Constant speed round a bend is not equilibrium: the direction of the velocity changes, so there is a resultant force.
Solving an equilibrium problem
- Draw the free-body diagram.
- Choose two perpendicular directions (horizontal and vertical, or along and perpendicular to a slope).
- Resolve every force and write: sum of components = 0 in each direction.
- Or draw the forces tip to tail — for three forces in equilibrium they form a closed triangle.
A café sign of mass 15.0 kg hangs at rest from two cables. Each cable makes an angle of 40° with the horizontal, and the arrangement is symmetrical. Calculate the tension in each cable.
Step 1 The sign is in equilibrium. Horizontally, the two horizontal components (\(T\cos 40^\circ\)) cancel, so the tensions are equal.
Step 2 Vertically, the upward components balance the weight: \(2T\sin 40^\circ = 15.0 \times 9.81 = 147.2\ \text{N}\)
Step 3 Rearrange: \(T = \dfrac{147.2}{2\sin 40^\circ} = 114.5\ \text{N} \approx \) 114 N
Each tension is bigger than half the weight (73.6 N), because only its vertical component, 73.6 N, holds the sign up.
The flatter the cables, the smaller \(\sin\theta\) and the larger the tension. A cable pulled perfectly horizontal could never support a weight — a useful check that your answer gets bigger as the angle gets smaller.
How do you resolve forces on a slope?
On a slope, horizontal and vertical are the wrong axes. Resolve along the slope and perpendicular to it: the normal contact force and friction then each lie along one direction, and only the weight needs resolving. For a slope at angle \(\theta\) to the horizontal, the weight \(mg\) has components
Perpendicular to the slope there is no acceleration, so \(R = mg\cos\theta\) (when no other force has a component in that direction). A quick check on which component gets sin: if the slope were flat (\(\theta = 0\)) there would be no pull down the slope, and \(\sin 0 = 0\).
A crate of mass 8.00 kg rests without sliding on a loading ramp inclined at 20° to the horizontal. Calculate (a) the frictional force on the crate and (b) the normal contact force on the crate.
Step 1 Weight: \(W = 8.00 \times 9.81 = 78.5\ \text{N}\)
(a) Along the slope, friction balances the component of the weight down the slope: \(F = 78.5\sin 20^\circ = \) 26.8 N up the slope
(b) Perpendicular to the slope, \(R\) balances the other component: \(R = 78.5\cos 20^\circ = \) 73.7 N perpendicular to the slope
Check: \(\sqrt{26.8^2 + 73.7^2} = 78.5\ \text{N}\) — friction and the normal contact force together exactly cancel the weight.
Writing \(R = mg\) for a body on a slope. The normal contact force balances only the component of the weight perpendicular to the slope, \(mg\cos\theta\), which is always less than \(mg\).
How do you use Newton's second law?
When the forces do not balance, the body accelerates — and the second law tells you by how much.
A reliable method
- Draw the free-body diagram and choose the positive direction — the direction of the acceleration is usually best.
- Find the resultant force along that direction: forces that way count positive, forces the other way negative.
- Write \(\Sigma F = ma\) and solve; then use the equations of motion if you need a speed, time or distance.
An electric car of mass 1450 kg has a forward driving force of 3200 N. At one instant, the total resistive force on it is 850 N. (a) Calculate the acceleration of the car. (b) State and explain the resistive force when the car reaches a steady top speed with the same driving force.
(a) Resultant force, forwards positive: \(\Sigma F = 3200 - 850 = 2350\ \text{N}\). Then \(a = 2350 \div 1450 = 1.62\ \text{m s}^{-2}\) \(\approx\) 1.6 m s⁻² forwards
(b) Steady speed means constant velocity, so the resultant force is zero (first law): resistive force = 3200 N, equal and opposite to the driving force.
A child on a toboggan, total mass 42 kg, slides from rest down a straight snow slope at 18° to the horizontal. A constant frictional force of 60 N acts up the slope; air resistance is negligible. Calculate (a) the acceleration and (b) the speed after 4.0 s.
Step 1 Component of the weight down the slope: \(42 \times 9.81 \times \sin 18^\circ = 127.3\ \text{N}\)
(a) Resultant force down the slope, then \(\Sigma F = ma\): \(a = \dfrac{127.3 - 60}{42} = 1.60\ \text{m s}^{-2}\) \(\approx\) 1.6 m s⁻² down the slope
(b) Constant force, so constant acceleration: \(v = u + at = 0 + 1.603 \times 4.0 = \) 6.4 m s⁻¹
The normal contact force does not appear: it is perpendicular to the slope, so it has no component along the motion.
\(\Sigma F\) means the resultant force. Substituting just the driving force, or just the weight component, into \(F = ma\) gives the wrong acceleration — write the subtraction out in full so that every force can be seen.
How do you analyse connected bodies and lifts?
Connected bodies
Bodies joined by a taut rope or rigid tow bar have the same acceleration. Treat the whole system first: the tension is internal (it pulls the two bodies equally in opposite directions), so it cancels and \(\Sigma F_{\text{external}} = (\text{total mass}) \times a\). Then take one body on its own to find the tension: its free-body diagram now includes the tension. Pick the body with fewer forces for the second step, and use the other one as a check.
A car of mass 1250 kg tows a trailer of mass 350 kg along a straight, level road. The driving force on the car is 2800 N. The resistive forces are 520 N on the car and 280 N on the trailer. Calculate (a) the acceleration and (b) the tension in the tow bar.
(a) Whole system — the tension cancels: \(a = \dfrac{2800 - 520 - 280}{1250 + 350} = \dfrac{2000}{1600} = \) 1.25 m s⁻²
(b) Trailer alone — forward tension, backward resistance: \(T - 280 = 350 \times 1.25\), so \(T = 717.5\ \text{N} \approx \) 718 N
Check with the car alone: \(2800 - 520 - 717.5 = 1562.5\ \text{N} = 1250 \times 1.25\) ✓
A person in a lift
Two forces act on the person: weight \(mg\) down and the normal contact force \(R\) from the floor (or scales) up. Scales read \(R\), the apparent weight — what you feel through your feet. Taking upwards as positive, \(R - mg = ma\), so \(R = m(g + a)\).
- Accelerating upwards (speeding up going up, or slowing down going down): \(R > mg\).
- Constant velocity, in either direction: \(R = mg\).
- Accelerating downwards (speeding up going down, or slowing down going up): \(R < mg\).
It is the direction of the acceleration, not of the velocity, that decides.
A student of mass 60.0 kg stands on bathroom scales in a lift. The lift accelerates upwards at 1.50 m s⁻², then moves at constant velocity, then accelerates downwards at 1.50 m s⁻² as it slows to a stop on a higher floor. (a) Calculate the force the scales exert on the student in each stage. (b) The scales are marked in kilograms (they divide \(R\) by 9.81). Calculate their reading in the first stage.
(a) \(R = m(g + a)\), upwards positive: \(60.0 \times (9.81 + 1.50) = 679\ \text{N}\); \(60.0 \times 9.81 = 589\ \text{N}\); \(60.0 \times (9.81 - 1.50) = 499\ \text{N}\). 679 N, 589 N, 499 N
(b) Reading \(= R \div 9.81 = 678.6 \div 9.81 = \) 69.2 kg. The student's mass is still 60.0 kg — the scales only measure the contact force.
Using \(g + a\) whenever the lift is "going up". A lift slowing down on its way up is accelerating downwards, so \(R = m(g - a)\).
What is terminal velocity?
The key fact is that drag (air resistance) increases with speed, while the weight stays the same. Follow a skydiver from the moment of jumping:
- At release the speed is zero, so there is no drag: the resultant force is \(mg\) and the acceleration is \(g\).
- As the speed rises the drag rises, the resultant force \(mg - D\) falls, so the acceleration falls — the \(v\)–\(t\) graph gets less steep.
- When \(D = mg\) the resultant force is zero, so the acceleration is zero (first law): the body falls at terminal velocity.
- Opening a parachute greatly increases the drag at that speed: \(D > mg\), the resultant force is upwards and the skydiver decelerates. As the speed falls the drag falls, until \(D = mg\) again at a new, lower terminal velocity.
The same idea applies horizontally: a vehicle reaches top speed when the resistive forces equal the driving force.
An "explain" answer about terminal velocity needs the full chain: speed increases → drag increases → resultant force decreases → acceleration decreases → when drag = weight, resultant force = 0, so acceleration = 0 and velocity is constant. Each arrow is a separate idea; leaving one out leaves a gap.
What does Newton's third law say?
Forces always come in pairs, because a force is an interaction between two bodies. The two forces of a pair have four properties:
- They act on different bodies — so they never cancel each other.
- They are equal in magnitude and opposite in direction, along the same line.
- They are the same type of force (both gravitational, both contact, both electrostatic…).
- They act at the same time, for the same duration.
The third law explains how anything starts moving. Walking: your foot pushes backwards on the ground; the ground pushes forwards on you with an equal friction force — that is the force that accelerates you. A rocket engine pushes exhaust gas backwards; the gas pushes the rocket forwards. A swimmer's hands push water backwards; the water pushes the swimmer forwards.
Naming the weight and the normal contact force on a resting body as a third-law pair. Test any pair with three questions: are they on different bodies? the same type? described by swapping the words "A on B" to "B on A"? All three must be yes.
Exam tips for 1.2
Sketch a free-body diagram for every forces question, even when it is not asked for. A missing resistive force, a tension in the wrong direction or a force on the wrong body is easy to spot on a diagram and hard to spot in an equation.
When asked to label or identify forces, write who exerts the force on whom, or use the standard name: "normal contact force of the floor on the box", not "reaction"; "weight", not "gravity".
On a slope, resolve along and perpendicular to the slope. For a hanging object, resolve horizontally and vertically. The right choice leaves only one force to resolve.
\(\Sigma F = ma\) needs the mass. If a question gives a weight, divide by 9.81 first; if it gives grams, convert to kilograms.
State needs a short answer; calculate needs working and a unit; explain needs the physics reason linked step by step; draw a free-body diagram needs labelled arrows starting on the body.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A moving body needs a resultant force to keep moving. | The first law: with no resultant force, velocity stays constant. | A resultant force is needed to change the velocity. |
| Weight and the normal contact force are a third-law pair. | They act on the same body and are different types of force. | The pair of the weight is the pull of the body on the Earth. |
| Third-law forces cancel, so nothing can accelerate. | The two forces act on different bodies. | Only forces on the same body are added to find its resultant. |
| Mass and weight are the same thing. | Mass is a scalar in kg; weight is a force in N that depends on \(g\). | \(W = mg\); mass is the same on the Moon, weight is not. |
| At terminal velocity there are no forces on the skydiver. | The weight still acts. | The forces are balanced: drag = weight, resultant = 0. |
| Opening a parachute makes the skydiver move upwards. | An upward resultant force gives an upward acceleration, which only slows the downward motion. | The skydiver still moves down, but slows down. |
How this topic connects
1.1 Motion, vectors and projectiles
A constant resultant force gives the constant acceleration the equations of motion need, and resolving vectors is the tool for every slope and equilibrium problem here.
1.3 Momentum
Newton's second law can also be written as force = rate of change of momentum; together with the third law it leads to conservation of momentum in collisions.
1.4 Moments, Materials and Unit 4
Extended bodies in equilibrium also need balanced turning effects, using the centre of gravity met here. Drag and upthrust on a sphere in a fluid are calculated with Stokes' law in the Materials topic. In Unit 4, a body moving in a circle at constant speed needs a resultant force towards the centre — the second law again.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(\Sigma F = ma\) | resultant force and acceleration, constant mass | Yes |
| \(g = F/m\) | gravitational field strength | Yes |
| \(W = mg\) | weight from mass | Yes |
| \(mg\sin\theta,\ mg\cos\theta\) | components of weight along and perpendicular to a slope | No — derive it |
| \(R = m(g + a)\) | normal contact force in a lift, upwards positive | No — derive it from \(\Sigma F = ma\) |
| \(\Sigma F_x = 0,\ \Sigma F_y = 0\) | equilibrium (first law) | No — learn it |
Check your understanding
1. An astronaut has a mass of 72 kg. State her mass on the Moon and calculate her weight there (g = 1.62 N kg⁻¹).
Mass is unchanged: 72 kg. Weight \(= 72 \times 1.62 = 116.6\ \text{N} \approx\) 117 N.
2. A swimmer's hands push backwards on the water. State the Newton's third law partner of this force.
The water pushes forwards on the swimmer's hands, with a force of equal size — a contact force, acting on the swimmer.
3. A 900 kg lift car accelerates downwards at 0.60 m s⁻². Calculate the tension in its cable.
Upwards positive: \(T - 900 \times 9.81 = 900 \times (-0.60)\), so \(T = 900 \times 9.21 = \) 8.29 × 10³ N.
4. Explain why a falling raindrop reaches a terminal velocity.
As its speed increases, the air resistance on it increases while its weight stays constant, so the resultant force and the acceleration decrease. When the air resistance equals the weight the resultant force is zero, so the acceleration is zero and the velocity stays constant.
5. A 2.0 kg block slides down a smooth (frictionless) slope at 30° to the horizontal. Calculate its acceleration and the normal contact force on it.
\(a = g\sin 30^\circ = \) 4.9 m s⁻² down the slope (the mass cancels). \(R = 2.0 \times 9.81 \times \cos 30^\circ = \) 17 N.
Specification checklist
- (8) I can draw and interpret free-body diagrams for a particle and for an extended body, drawing the weight at the centre of gravity.
- (9) I can use resultant force = mass × acceleration for a body of constant mass, including bodies on slopes, connected bodies and people in lifts.
- (9) I know that zero resultant force means the body is at rest or moving at constant velocity (Newton's first law), and I can resolve forces to solve equilibrium problems.
- (9) I can explain, using forces, how and why a body reaches terminal velocity.
- (10) I can use \(g = F/m\) and \(W = mg\), and I can explain the difference between mass and weight.
- (12) I can state Newton's third law, identify the pair of forces in an interaction and list their properties.