Forces and Newton's Laws

A-Level Maths · Mechanics

Forces and Newton's Laws

Newton's laws of motion form the foundation of classical mechanics. At A-Level, you must be able to model physical situations, resolve forces, and apply F = ma to find accelerations, tensions, and reactions.

Newton's Three Laws

First Law (Inertia): A body remains at rest or continues to move at constant velocity unless acted upon by a resultant (net) force.

Second Law: The resultant force on a body equals its mass times its acceleration:

F = ma (force in newtons, mass in kg, acceleration in m/s²)

This is a vector equation — it applies independently in each direction.

Third Law: When two bodies interact, the forces they exert on each other are equal in magnitude and opposite in direction (action-reaction pairs). These forces act on different bodies.

Types of Force

ForceSymbolDescription
WeightW = mgGravitational force, always acts downward
Normal reactionR or NContact force perpendicular to surface
TensionTForce in a taut string or rope
FrictionFOpposes relative motion along the surface
Thrust/pushPApplied force
Air resistance/dragDOpposes motion through air

Resolving Forces

When forces act at angles, resolve into perpendicular components (usually horizontal and vertical):

  • Horizontal component = Fcosθ (where θ is measured from the horizontal)
  • Vertical component = Fsinθ

Then apply F = ma separately in each direction.

Worked Example: A 5 kg block is pulled along a smooth horizontal surface by a force of 30 N at 40° above the horizontal. Find the acceleration and the normal reaction.

Horizontally (→ positive): 30cos40° = 5a

a = 30cos40°/5 = 22.98/5 = 4.60 m/s²

Vertically (↑ positive): R + 30sin40° - 5g = 0

R = 5(9.8) - 30sin40° = 49 - 19.28 = 29.7 N

Connected Particles

When particles are connected by a string (possibly over a pulley), they have the same magnitude of acceleration (assuming inextensible string). Apply F = ma to each particle separately, then solve simultaneously.

Worked Example: Two particles of mass 3 kg and 5 kg are connected by a light inextensible string over a smooth pulley (Atwood's machine). Find the acceleration and the tension.

For the 5 kg mass (taking downward as positive):

5g - T = 5a ... (i)

For the 3 kg mass (taking upward as positive, since it accelerates upward):

T - 3g = 3a ... (ii)

Adding (i) and (ii): 5g - 3g = 8a

2g = 8a, so a = g/4 = 2.45 m/s²

From (ii): T = 3(2.45) + 3(9.8) = 7.35 + 29.4 = 36.75 N

Inclined Planes

On a plane inclined at angle θ to the horizontal, resolve forces parallel and perpendicular to the slope:

  • Component of weight down the slope = mgsinθ
  • Component of weight into the slope = mgcosθ
  • Normal reaction R = mgcosθ (if no other forces act perpendicular to the slope)

Worked Example: A 4 kg block slides down a smooth slope inclined at 30°. Find the acceleration.

Parallel to slope (down positive): mgsin30° = ma

a = gsin30° = 9.8 × 0.5 = 4.9 m/s²

Particles on Slopes Connected by Strings

Worked Example: A 2 kg block on a smooth slope at 30° is connected by a string over a pulley at the top of the slope to a 3 kg block hanging vertically. Find the acceleration.

For the 3 kg block (downward positive): 3g - T = 3a ... (i)

For the 2 kg block (up the slope positive): T - 2gsin30° = 2a ... (ii)

Adding: 3g - 2gsin30° = 5a

3(9.8) - 2(9.8)(0.5) = 5a

29.4 - 9.8 = 5a

a = 19.6/5 = 3.92 m/s²

Equilibrium

A body is in equilibrium if the resultant force is zero in every direction (acceleration = 0). For a body in equilibrium:

  • Sum of forces horizontally = 0
  • Sum of forces vertically = 0

Modelling Assumptions

Questions often use phrases that indicate simplifications:

PhraseMeaning
"Light string"String has zero mass (tension is the same throughout)
"Inextensible string"String does not stretch (particles have same acceleration)
"Smooth surface"No friction
"Particle"No size, no rotation — all forces act at one point
"Smooth pulley"No friction at the pulley (tension same on both sides)

Exam Tips

  • Always draw a force diagram for each body. Label every force clearly.
  • Apply F = ma to each body separately — never mix forces on different bodies in one equation.
  • On inclined planes, resolve parallel and perpendicular to the slope, not horizontally and vertically.
  • In connected particle problems, define a consistent positive direction for all bodies (e.g., the direction of acceleration).
  • Newton's Third Law pairs act on different bodies. If you include the weight of a book on a table (Earth pulls book down), the reaction pair is the book pulling the Earth up — not the normal reaction from the table.
Test yourself on Mechanics

5 quick A-Level Maths questions, marked instantly, so you know what stuck. Free account, no card needed.

Start the free quiz →
Don't understand a part?

Sign in and ask our AI tutor to explain any passage in plain English.

Try AI explanations →

More on Mechanics

Kinematics and the SUVAT Equations Moments and Equilibrium Projectiles Friction and Inclined Planes

← All A-Level Maths notes