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Physics

Statics, Forces, and Moments

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Matthew Williams
|May 20, 2026|7 min read
Centre of GravityCSEC PhysicsForcesHooke's LawMomentsPaper 01Paper 02Section A

Types of forces, weight, turning effects and moments, the principle of moments, levers, centre of gravity, stability, and Hooke's Law.

Statics deals with objects that are in equilibrium, either at rest or moving at constant velocity. Understanding statics means understanding how forces balance and how they produce turning effects.

Types of Forces

Forces come from different physical interactions:

TypeDescriptionExample
GravitationalAttraction between massesWeight of an object
ElectrostaticAttraction/repulsion between chargesForce between charged rods
MagneticForce on moving charges or magnetic materialsForce between poles
NuclearShort-range force holding nucleus togetherProton-proton binding
Normal reactionPerpendicular contact force from a surfaceFloor pushing up on a block
TensionForce in a stretched string or springWeight on a spring
FrictionResistive force opposing relative motionBrakes on a wheel
UpthrustUpward force exerted by a fluidBuoyancy on a floating boat

Weight and Gravitational Field Strength

Weight is the gravitational force acting on a mass. It acts downward toward the centre of the Earth and is measured in newtons.

W=mgW = mgW=mg

where mmm is mass (kg) and ggg is gravitational field strength = 10 N kg⁻¹ at the Earth's surface (or gravitational acceleration = 10 m s⁻²).

Mass and weight are not the same. Mass is the amount of matter in an object and does not change with location. Weight depends on the gravitational field strength at that location, an object on the Moon has the same mass but weighs less.

Moments (Turning Effects)

A force applied at a distance from a pivot produces a moment (also called a torque). The moment measures the turning effect of the force.

T=F×dT = F \times dT=F×d

where TTT is the moment in newton-metres (N m), FFF is the applied force in newtons, and ddd is the perpendicular distance from the pivot to the line of action of the force.

The perpendicular distance matters. If a force is applied parallel to the moment arm, it produces no turning effect.

Clockwise moments and anticlockwise moments are distinguished by direction. The principle of moments states that for an object in equilibrium:

sum of clockwise moments=sum of anticlockwise moments\text{sum of clockwise moments} = \text{sum of anticlockwise moments}sum of clockwise moments=sum of anticlockwise moments

PivotEffort ELoad Ld₁d₂Principle of moments: E × d₁ = L × d₂ (for equilibrium)
Moment of a force: T = F × d (perpendicular distance from pivot)
Example/Principle of moments (2024 Paper 02, Q2)

A uniform metre rule has its centre at the 50 cm mark and is balanced on a fulcrum at the 60 cm mark. A mass of 0.24 kg hangs from the 100 cm end. Find the weight of the metre rule.

The weight of the rule acts at its centre of mass (50 cm mark). The applied mass hangs at 100 cm.

Distances from the fulcrum (at 60 cm):

  • 0.24 kg mass: 100 − 60 = 40 cm = 0.40 m (clockwise moment)
  • Rule's weight: 60 − 50 = 10 cm = 0.10 m (anticlockwise moment)

Weight of 0.24 kg mass: W=0.24×10=2.4W = 0.24 \times 10 = 2.4W=0.24×10=2.4 N

Applying principle of moments:

2.4×0.40=Wrule×0.102.4 \times 0.40 = W_{\text{rule}} \times 0.102.4×0.40=Wrule​×0.10

0.96=0.10 Wrule0.96 = 0.10 \, W_{\text{rule}}0.96=0.10Wrule​

Wrule=0.960.10=9.6NW_{\text{rule}} = \frac{0.96}{0.10} = 9.6 \text{N}Wrule​=0.100.96​=9.6N

Levers and Mechanical Advantage

A lever is a rigid bar that can rotate about a fixed pivot (fulcrum). Levers allow a small effort force to overcome a larger load force by applying the effort at a greater distance from the pivot.

Mechanical advantage (MA) is the ratio of load to effort:

MA=loadeffort\text{MA} = \frac{\text{load}}{{\text{effort}}}MA=effortload​

A lever with MA > 1 amplifies force, but the effort moves through a greater distance than the load. Energy is conserved.

Centre of Gravity

The centre of gravity of an object is the point through which its total weight appears to act. For a uniform, symmetrical object (cube, sphere, cylinder), the centre of gravity is at the geometric centre.

For an irregular object, the centre of gravity can be found experimentally by suspending it from two or more points in turn and drawing the vertical plumb line through each suspension point. The centre of gravity is where the lines intersect.

Stability

Whether an object topples when tilted depends on the position of its centre of gravity and the width of its base:

TypeConditionBehaviour when tilted
Stable equilibriumCentre of gravity is low; base is wideReturns to original position
Unstable equilibriumCentre of gravity is high; base is narrowTopples over
Neutral equilibriumCentre of gravity at same height throughoutStays in new position

A tall, narrow object is unstable. A wide, low object is stable. Racing cars are designed with low centres of gravity and wide wheelbases.

Hooke's Law

When a spring (or elastic material) is stretched, the extension is proportional to the applied force, provided the elastic limit is not exceeded. This is Hooke's Law:

F=keF = keF=ke

where FFF is the applied force (N), kkk is the spring constant (N m⁻¹), and eee is the extension (m).

The elastic limit is the maximum force beyond which the spring does not return to its original length when the force is removed. Below the elastic limit, deformation is elastic (reversible). Above it, deformation is plastic (permanent).

On a force-extension graph, the relationship is linear up to the elastic limit. Beyond it, the graph curves and the spring becomes permanently deformed.

Hooke's Law: extension is proportional to force up to the elastic limit
Example/Spring constant (2021 Paper 02, Q1)

A student attaches masses to a spring and records the extension. At a force of 4.2 N, the extension is 18.0 cm.

Step 1: Convert units.

e=18.0cm=0.180me = 18.0 \text{cm} = 0.180 \text{m}e=18.0cm=0.180m, F=4.2NF = 4.2 \text{N}F=4.2N

Step 2: Apply Hooke's Law.

k=Fe=4.20.180=23.3N m−1k = \frac{F}{e} = \frac{4.2}{0.180} = 23.3 \text{N m}^{-1}k=eF​=0.1804.2​=23.3N m−1

Step 3: If the graph is linear and passes through the origin up to this point, the spring obeys Hooke's Law.

The spring constant is 23.3 N m⁻¹.

Exam Tip

In the Hooke's Law graph question, the gradient of the force-extension graph equals the spring constant kkk. Calculate it using two well-separated points on the straight-line portion only. Do not use points beyond the elastic limit in your gradient calculation.

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Scalars and Vectors
Next in syllabus order
Motion and Dynamics