CCEA GCE Mathematics (2210) · Forces and Newton's laws
Mini-Lesson
Forces and Newton's laws
This mini-lesson covers Newton's three laws, F = ma, weight and normal reaction, resolving forces, friction (F ≤ μR) and limiting equilibrium, inclined planes, connected particles and pulleys, and lift problems. Take g = 9.8 m s⁻².
Where this sits:Unit AS 2: Applied Mathematics — forces and Newton's laws: force diagrams, equilibrium, F = ma, connected particles. Unit A2 2 — resolving in two dimensions, friction and motion on an inclined plane.
Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Press Start when you are ready.
The three laws
Newton's laws of motion
First law: a body stays at rest or moves with constant velocity unless a resultant force acts. So constant velocity ⇔ zero resultant force (equilibrium).
Second law:F = ma — the resultant force equals mass × acceleration, and acts in the same direction as the acceleration.
Third law: if A pushes B, then B pushes A with an equal and opposite force. The two forces of the pair act on different bodies, so they never cancel each other out.
F = maALWAYS start with a force diagram, then resolve in the direction of the acceleration
Calculate
Newton's second law
1A car of mass 1500 kg experiences a resultant force of 4500 N. Find its acceleration.
m/s²
Hint: a = F ÷ m = 4500 ÷ 1500.
Vertical forces
Weight, normal reaction and lifts
A body on a horizontal surface has weight mg down and normal reaction R up. If it is not accelerating vertically, R = mg. But in a lift, R changes:
Worked example — a lift accelerating upwards
Person of mass 60 kg, lift accelerating upwards at 2 m s⁻². Take up as positive.
R − mg = ma ⇒ R = m(g + a) = 60(9.8 + 2) = 60 × 11.8 = 708 N
Accelerating down instead would give R = m(g − a) = 60 × 7.8 = 468 N — you feel lighter.
The reaction is not always mg. It equals mg only when there is no vertical acceleration and no other vertical force.
Calculate
Lift problem
2A person of mass 60 kg stands in a lift accelerating upwards at 2 m s⁻². Taking g = 9.8, find the normal reaction from the floor.
N
Hint: Resolve upwards: R − mg = ma, so R = m(g + a) = 60 × (9.8 + 2) = 60 × 11.8.
Friction
Friction and limiting equilibrium
F ≤ μRfriction opposes RELATIVE motion · at the point of slipping F = μR (limiting friction) · μ is the coefficient of friction
Friction is self-adjusting: it only pushes back as hard as it needs to, up to its maximum μR. Below that maximum, the body stays still.
Worked example
A 10 kg block rests on a rough horizontal floor, μ = 0.3.
R = mg = 10 × 9.8 = 98 N ⇒ maximum friction = μR = 0.3 × 98 = 29.4 N
A 20 N push does not move it (friction rises to 20 N and balances it). A 35 N push does move it, with resultant 35 − 29.4 = 5.6 N.
Calculate
Limiting friction
3A 10 kg block sits on a rough horizontal surface with μ = 0.3. Taking g = 9.8, find the maximum friction force.
N
Hint: R = mg = 98 N, so F(max) = μR = 0.3 × 98.
Inclines
Resolving on an inclined plane
Resolve along and perpendicular to the slope — never horizontally and vertically. For a plane at angle θ:
along the slope: mg sin θ · perpendicular: R = mg cos θthe steeper the slope, the bigger the sin θ component pulling the body down it
Worked example — smooth slope, 30°
A 4 kg particle on a smooth 30° slope. Along the slope: mg sin 30° = ma
a = g sin 30° = 9.8 × 0.5 = 4.9 m s⁻² down the slope (independent of the mass!)
If the slope were rough with μ = 0.2: friction = μmg cos 30° = 0.2 × 4 × 9.8 × 0.866 = 6.79 N, and ma = mg sin30 − friction = 19.6 − 6.79 = 12.81 N ⇒ a = 3.20 m s⁻².
Calculate
Smooth incline
4A 4 kg particle slides down a smooth plane inclined at 30° to the horizontal. Taking g = 9.8, find its acceleration.
m/s²
Hint: Along the slope: mg sin θ = ma, so a = g sin 30° = 9.8 × 0.5. The mass cancels.
Connected particles
Pulleys and strings
For particles joined by a light inextensible string over a smooth pulley:
the tension is the same throughout the string;
both particles have the same magnitude of acceleration;
write F = ma for each particle separately, taking its own direction of motion as positive, then solve the pair of equations.
Worked example — 5 kg and 3 kg over a smooth pulley
5 kg (descending): 5g − T = 5a · 3 kg (ascending): T − 3g = 3a
Add: 2g = 8a ⇒ a = 2(9.8)/8 = 2.45 m s⁻²
Then T = 3(g + a) = 3(9.8 + 2.45) = 36.75 N (check in the other equation: 5(9.8) − 36.75 = 12.25 = 5 × 2.45 ✓)
Calculate
Pulley acceleration
5Masses of 5 kg and 3 kg hang either side of a smooth light pulley on a light inextensible string. Taking g = 9.8, find the acceleration of the system.