Statics Learning Map
LESSON 01
The Particle Model
Mechanics questions begin with a mathematical model of a physical situation. In the simplest statics problems, the object is modelled as a particle.
A particle is an object whose size and shape are ignored. Its entire mass is treated as being concentrated at a single point.
This model is useful when the dimensions of the object do not affect the motion or equilibrium being considered.
What does the particle model ignore?
- The physical size of the object.
- The shape of the object.
- Any rotational effects caused by its dimensions.
- The exact position at which its mass is distributed.
If the question asks only whether forces balance on a small object, the particle model is usually appropriate. If the object's size, shape or turning effect matters, a particle model is no longer sufficient.
LESSON 02
What Does Equilibrium Mean?
A particle is in equilibrium when the resultant force acting on it is zero.
By Newton's second law,
Therefore, if the resultant force is zero,
So an object in equilibrium has zero acceleration.
At rest
An object can be stationary and remain stationary if its resultant force is zero.
Constant velocity
An object can also move with constant velocity while the resultant force is zero.
Equilibrium does not necessarily mean “not moving”. It means zero acceleration.
LESSON 03
Forces Acting on a Particle
Before solving an equilibrium problem, identify every force acting on the particle.
Weight
Acts vertically downwards and has magnitude mg.
Tension
A string or cable can pull an object along the direction of the string.
Reaction
A surface can exert a contact force on an object.
Applied force
A push or pull supplied directly by another object or external agent.
Free-body diagrams
A free-body diagram isolates the particle and shows the forces acting on it. It is one of the most useful tools in Mechanics.
Draw the forces first. Then resolve them. Do not immediately start writing equations before identifying what is acting on the particle.
LESSON 04
Resolving Forces
A force acting at an angle can be replaced mathematically by two perpendicular components.
These formulae apply when θ is measured from the horizontal.
Do not automatically use F cos θ for horizontal and F sin θ for vertical. Check where the angle is measured. If θ is measured from the vertical, the roles reverse.
Why resolve forces?
A two-dimensional equilibrium problem can be converted into two one-dimensional equations:
The horizontal and vertical directions are independent. You may choose either positive direction, but your signs must remain consistent.
LESSON 05
Weight
Weight is the gravitational force acting on a mass.
Weight always acts vertically downwards.
Mass versus weight
Mass
- Measured in kg.
- Amount of matter.
- Does not depend on location.
Weight
- Measured in N.
- Gravitational force.
- W = mg.
Worked Example
A particle has mass 6 kg. Find its weight, taking g = 9.8 m s⁻².
Use:
W = 6 × 9.8
W = 58.8 N
LESSON 06
Tension
A light, inextensible string or cable can exert a tension force on a particle.
Tension acts along the string and away from the particle.
If a particle is attached to two strings, there may be two different tensions unless the problem gives a reason for them to be equal.
Do not assume two tensions are equal simply because two strings look similar. Their directions and magnitudes must come from the equations or the conditions given.
LESSON 07
Inclined Forces
Consider a force T making an angle θ above the horizontal. Its components are:
If the force points upwards and to the right, both components are positive when right and upward are chosen as positive.
If the force points upwards and to the left, its horizontal component is negative under the same convention.
Choosing signs
- Choose a positive horizontal direction.
- Choose a positive vertical direction.
- Resolve every angled force.
- Give each component its correct sign.
- Apply ΣFx = 0 and ΣFy = 0.
LESSON 08
Three-Force Equilibrium
A particle may be acted upon by three forces and still be in equilibrium.
For example, a particle may have:
- weight acting downwards,
- tension in one string acting up and right,
- tension in another string acting up and left.
The upward components of the tensions must balance the weight, while their horizontal components must balance one another.
These questions usually test whether you can correctly interpret the geometry before applying trigonometry. The equations are often straightforward once the force directions are correct.
LESSON 09
Connected Particles
Two particles may be connected by a light inextensible string. In a suitable model, the tension is the same throughout the string.
The mass of the string is neglected.
The string does not stretch, so connected particles have linked motion.
For a particle system, analyse each particle separately. Draw a separate force diagram for each particle and write Newton's second-law equation for each one.
Never write one equation for two particles without first identifying the forces on each individual particle.
LESSON 10
The Complete Exam Method
A reliable method prevents most errors in statics questions.
- Identify the object. Decide exactly which particle you are analysing.
- Draw all forces. Include weight, tension, reactions and applied forces.
- Mark the angles carefully. Check whether each angle is measured from the horizontal or vertical.
- Resolve angled forces. Replace each angled force by horizontal and vertical components.
- Choose signs. Usually right/up positive is convenient.
- Apply horizontal equilibrium. Use ΣFx = 0 when the particle is in equilibrium.
- Apply vertical equilibrium. Use ΣFy = 0.
- Solve algebraically. Keep exact values where useful before rounding.
- Check the answer. Check units, direction and physical plausibility.
WORKED EXAMPLES
Exam-Level Worked Examples
Example 1 — Opposing Horizontal Forces
A particle is acted on by a 27 N force to the right and a force F to the left. It is in equilibrium. Find F.
27 − F = 0
F = 27 N
Example 2 — Vertical Equilibrium
A 12 kg particle is supported by an upward force R. Find R when g = 9.8 m s⁻².
Weight:
W = 12 × 9.8 = 117.6 N
Since vertical forces balance:
R = 117.6 N
Example 3 — Resolving a Force
A force of 50 N acts at 30° above the horizontal. Find its horizontal and vertical components.
Horizontal:
50 cos 30° = 43.30 N
Vertical:
50 sin 30° = 25 N
Horizontal = 43.3 N, Vertical = 25.0 N
Example 4 — Finding Tension
A particle is in equilibrium under a horizontal force of 30 N and a tension T acting at 60° above the horizontal. Find T.
Horizontal equilibrium gives:
Since cos 60° = 0.5:
0.5T = 30
T = 60 N
Example 5 — Tension Supporting Weight
A 5 kg particle is held in equilibrium by a string making 40° with the horizontal. A horizontal force acts in the opposite direction. Find the vertical component of the tension.
Weight:
W = 5 × 9.8 = 49 N
Vertical equilibrium requires:
Therefore:
T = 49 / sin 40°
T ≈ 76.1 N
Example 6 — Two Tensions
A 10 kg particle is held in equilibrium by two strings. The tensions are T₁ and T₂. T₁ acts at 30° above the horizontal to the left and T₂ acts at 60° above the horizontal to the right. Find T₁ and T₂.
Vertical equilibrium:
Horizontal equilibrium:
Since cos60° = 0.5 and cos30° = √3/2:
T₂ = √3 T₁
Substitute into the vertical equation:
0.5T₁ + 0.866(√3T₁) = 98
0.5T₁ + 1.5T₁ = 98
2T₁ = 98
T₁ = 49 N
Therefore:
T₂ = 49√3 N ≈ 84.9 N
Example 7 — Force at an Angle from the Vertical
A force of 80 N acts at 25° to the vertical. Find its horizontal component.
Because the angle is measured from the vertical, the horizontal component uses sine:
Horizontal ≈ 33.8 N
Example 8 — Equilibrium with an Unknown Angle
A 100 N tension acts at angle θ above the horizontal. Its horizontal component is 60 N. Find θ.
cosθ = 0.6
θ = cos⁻¹(0.6)
θ ≈ 53.1°
Example 9 — Resultant of Perpendicular Forces
Two perpendicular forces of 6 N and 8 N act on a particle. Find the magnitude of their resultant.
By Pythagoras:
R = √100
R = 10 N
Example 10 — Three Forces
A particle is acted on by a 20 N horizontal force to the right and a 40 N force at 60° above the horizontal to the left. Find the vertical force required for equilibrium.
Vertical component of the 40 N force:
40 sin60° = 34.64 N
Therefore the required vertical force is:
34.64 N downward
Notice that horizontal equilibrium would not be satisfied by these forces alone. A further horizontal force would be needed unless another force is present.
Example 11 — Unknown Force Components
A force F acts at 37° above the horizontal. Its horizontal component is 24 N. Find F.
F = 24 / cos37°
F ≈ 30.0 N
Example 12 — Finding an Angle from Components
A force has horizontal component 12 N and vertical component 5 N. Find its magnitude and angle above the horizontal.
Magnitude:
Angle:
θ ≈ 22.6°
Example 13 — Sign Convention
Take right as positive. A particle experiences 15 N to the right, 9 N to the left and another force F to the left. Find F for equilibrium.
15 − 9 − F = 0
6 − F = 0
F = 6 N
Example 14 — Particle on an Inclined Plane
A particle of weight 98 N rests on a smooth plane inclined at 30° to the horizontal. Find the component of its weight down the plane.
The component of weight parallel to the plane is:
49 N
Example 15 — Reaction on a Smooth Plane
A 6 kg particle rests in equilibrium on a smooth plane inclined at 30°. Find the normal reaction.
Weight:
W = 6 × 9.8 = 58.8 N
Perpendicular component:
R ≈ 50.9 N
EXAM TRAPS
Common Mistakes That Lose Marks
Always inspect the triangle formed by the components. The component adjacent to the angle uses cosine and the component opposite the angle uses sine.
If mass is given, immediately consider whether W = mg needs to appear in your force diagram.
Tension pulls away from the particle along the string. It does not push toward the particle.
If right is positive, every leftward component must be negative. If upward is positive, every downward component must be negative.
Equilibrium means zero acceleration. Constant velocity is also possible.
Equal tension is a modelling consequence of a suitable light-string model, not something you should assume from the picture alone.
Keep calculator values during intermediate steps and round only at the final answer unless the question specifies otherwise.
Diagram → components → equations → solve → check.
FORMULA ATLAS
Statics Formula Sheet
These equations are not separate tricks. They all follow from Newton's laws and the geometry of resolving vectors.
FINAL ASSESSMENT
Statics Mastery Exam — 100 Questions
Test the complete topic. Every question has four options. Questions cover definitions, equilibrium, components, weight, tension, inclined planes, connected particles and exam traps.