Newton's three laws, weight, normal reaction, tension, force diagrams, F=ma.
Answer at least 3 of 3 correctly to complete this section.
Isaac Newton published his three laws of motion in 1687. Over 300 years later, they are still the foundation of mechanics at A-Level — and they are powerful enough to send rockets to the Moon.
Every mechanics problem you meet at A-Level comes down to drawing a force diagram and applying F=ma. Get this right and the rest of mechanics follows.
Newton’s First Law: A body remains at rest or moves with constant velocity unless acted on by a resultant force.
This means:
Two forces appear in almost every problem:
Mass vs weight: Mass (m) is measured in kilograms and is a property of the object. Weight (W=mg) is a force measured in newtons. A 5 kg mass has weight 5×9.8=49 N.
A book of mass 2 kg rests on a horizontal table. Find the normal reaction force.
The book is in equilibrium (not accelerating), so the resultant force is zero.
Forces acting on the book:
Resolving vertically:
R−19.6=0
R=19.6 N
This seems obvious, but setting up the force balance correctly is essential practice for harder problems.
Answer at least 3 of 3 correctly to complete this section.
Newton’s Second Law: The resultant force on a body is equal to its mass times its acceleration.
Fresultant=ma
This is the most-used equation in mechanics. It connects force, mass, and acceleration in one direction.
A person stands in a lift that accelerates upwards. Will the normal reaction force from the floor be greater than, less than, or equal to the person’s weight?
Think about it before reading on. The lift accelerates upwards, so the resultant force must be upwards. The normal reaction (up) must exceed the weight (down) to produce a net upward force. So R>W — the person feels heavier. This is why you feel heavy when a lift starts going up.
A box of mass 10 kg is pushed along a smooth horizontal surface by a horizontal force of 30 N. Find the acceleration.
Forces in the direction of motion:
There is no friction (the surface is smooth).
Applying F=ma horizontally:
30=10a
a=3 m s−2
A person of mass 60 kg stands in a lift accelerating upwards at 1.5 m s−2. Find the normal reaction between the person and the lift floor.
Forces on the person:
Taking upwards as positive (the direction of acceleration):
R−588=60×1.5
R−588=90
R=678 N
The person feels heavier because the floor pushes up harder than their weight. This is why you feel heavy when a lift accelerates upwards.
What if the lift decelerates? If the lift moves upwards but decelerates, the acceleration is downwards. The equation becomes R−588=60×(−1.5), giving R=498 N. The person feels lighter.
A force of 40 N acts at 30° above the horizontal on a 5 kg box on a smooth horizontal surface. Find the acceleration.
The force has horizontal and vertical components:
Applying F=ma horizontally (the direction of motion):
203=5a
a=43≈6.93 m s−2
Answer at least 3 of 3 correctly to complete this section.
When an object is on a slope inclined at angle θ to the horizontal, the key is to resolve forces parallel and perpendicular to the slope — not horizontally and vertically.
The weight mg resolves into two components:
The normal reaction R acts perpendicular to the slope, balancing the perpendicular component of weight:
R=mgcosθ
Why resolve along the slope? Because the object moves along the slope, so that is the direction in which we apply F=ma. The perpendicular direction has no acceleration (the object does not fly off the surface or sink into it), so the forces balance there.
A box of mass 4 kg slides down a smooth plane inclined at 30° to the horizontal. Find the acceleration.
Forces parallel to the slope (taking down the slope as positive):
There is no friction (smooth plane).
Applying F=ma parallel to the slope:
mgsinθ=ma
a=gsin30°=9.8×0.5=4.9 m s−2
Notice: On a smooth slope, the acceleration does not depend on the mass — it cancels out, just like in free fall.
A box of mass 5 kg is pushed up a smooth plane inclined at 20° to the horizontal by a force P N acting parallel to the slope. The box accelerates at 2 m s−2 up the slope. Find P.
Forces parallel to the slope (taking up the slope as positive):
P−mgsin20°=ma
P−5×9.8×sin20°=5×2
P−16.76=10
P=26.76 N (3 s.f.: 26.8 N)
Perpendicular to the slope, there is no acceleration:
R=mgcosθ
For the 5 kg box on a 20° slope:
R=5×9.8×cos20°=46.1 N (3 s.f.)
Common error: Students write R=mg on a slope. This is only true on a horizontal surface. On a slope, R=mgcosθ, which is always less than mg.
Answer at least 3 of 3 correctly to complete this section.
Newton’s Third Law: When body A exerts a force on body B, body B exerts an equal and opposite force on body A.
Key points about N3 pairs:
For example: a book on a table. The book pushes down on the table (contact force), and the table pushes up on the book (normal reaction). These are a Newton’s third law pair — same magnitude, opposite direction, acting on different objects.
Common trap: Weight and normal reaction on the same object are NOT a Newton’s third law pair. They act on the same body and are different types of force. The N3 partner of the book’s weight is the book’s gravitational pull on the Earth.
For each pair of forces, decide whether they form a Newton’s third law pair.
The weight of a ball and the normal reaction of the table on the ball. Not a N3 pair. Both act on the ball (same object), and they are different types of force (gravitational vs contact).
The Earth pulling the ball downwards and the ball pulling the Earth upwards. Yes — a N3 pair. Same type (gravitational), equal magnitude, opposite direction, different objects.
A foot pushing on the ground and the ground pushing back on the foot. Yes — a N3 pair. Same type (contact), equal magnitude, opposite direction, different objects.
Friction on a sliding box and the driving force pushing the box. Not a N3 pair. Both act on the box. The N3 partner of friction on the box is the friction the box exerts on the surface.
Tension pulling a trailer forward and tension pulling the car backward. Yes — a N3 pair. The string pulls the trailer forward and the trailer pulls the string backward (and by extension, the car). Same type (tension), different objects.
The test: A N3 pair must be (i) the same type of force, (ii) equal in magnitude, (iii) opposite in direction, and (iv) acting on different objects. If both forces act on the same object, they are not a N3 pair.
A particle of mass 3 kg lies on a smooth plane inclined at angle α to the horizontal, where sinα=53. A light inextensible string is attached to the particle and runs parallel to the slope to a pulley at the top, then hangs vertically with a mass of 2 kg attached. The system is released from rest. Find the acceleration of the system.
This is a connected particles problem (covered in detail in the next lesson), but the force diagram skills are what matter here.
For the 2 kg mass (taking downward as positive):
2g−T=2a
For the 3 kg mass on the slope (taking up the slope as positive):
T−3gsinα=3a
T−3(9.8)(53)=3a
T−17.64=3a
Adding the two equations to eliminate T:
2(9.8)−17.64=5a
19.6−17.64=5a
1.96=5a
a=0.392 m s−2
Answer at least 3 of 4 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.
Constant acceleration equations (suvat), choosing and applying the correct equation.
Systems of connected particles, pulleys, applying Newton's laws to each particle.
Friction force, coefficient of friction, limiting equilibrium, motion on rough surfaces.
Moment of a force, principle of moments, equilibrium of rigid bodies, tilting.
Resolving forces into components, inclined planes, equilibrium with angled forces.