4.1 Forces and equilibrium
- Syllabus
- 9709–2028–2029
- Topic
- 4.1
- Level
- A2
Learning objectives
4.1.1Forces• identify the forces acting in a given situation e.g. by drawing a force diagram.4.1.2The vector nature of force• understand the vector nature of force, and find and use components and resultants Calculations are always required, not approximate solutions by scale drawing.4.1.3Forces• use the principle that, when a particle is in equilibrium, the vector sum of the forces acting is zero, or equivalently, that the sum of the components in any direction is zero Solutions by resolving are usually expected, but equivalent methods (e.g. triangle of forces, Lami's Theorem, where suitable) are also acceptable; these other methods are not required knowledge, and will not be referred to in questions.4.1.4Friction• understand that a contact force between two surfaces can be represented by two components, the normal component and the frictional component4.1.5The model of a 'smooth' contact• use the model of a 'smooth' contact, and understand the limitations of this model4.1.6Equilibrium• understand the concepts of limiting friction and limiting equilibrium, recall the definition of coefficient of friction, and use the relationship FR n= or FRG n, as appropriate Terminology such as 'about to slip' may be used to mean 'in limiting equilibrium' in questions.4.1.7Newton's laws• use Newton's third law. e.g. the force exerted by a particle on the ground is equal and opposite to the force exerted by the ground on the particle.
Isolate one particle and label every external force
| Interaction | Force on chosen particle |
|---|---|
| Earth | weight mg vertically downward |
| taut light string | tension along string, pulling away |
| surface contact | normal reaction perpendicular; friction if rough |
| applied push/pull | along stated line of action |
Choose the particle/body, replace it by a point or simple outline, inspect every external interaction, draw one labelled arrow per force from the body, and include angles needed for later resolution.
A particle resting on a rough inclined plane may have weight mg downward, normal reaction perpendicular to the plane and friction along the plane. Motion or acceleration is not an extra force.
Draw only forces acting on the selected body. A force that this body exerts on something else belongs on the other body’s diagram.
Velocity, acceleration and “resultant force” are not additional interaction arrows. Do not cancel third-law partners across different bodies.
Resolve forces and calculate the resultant exactly
For force $F$ at angle $\theta$ from the positive $x$-axis:F_x=F\cos\theta,\qquad F_y=F\sin\theta,withsignssetbydirection.
Resolve every force along the same perpendicular axes, add signed components to (Rx,Ry), then calculate R=Rx2+Ry2 and determine its quadrant-correct direction.
Forces $(3,4)$ N and $(-1,2)$ N give\mathbf R=(2,6)\text{ N},\quad |\mathbf R|=\sqrt{40}=2\sqrt{10}\text{ N}.
Choose axes along an incline or along a force when that reduces unknown components. A negative component means opposite to the chosen positive direction.
Calculations are required; a scale drawing is not an accepted substitute. Adding magnitudes ignores the included angle.
Equilibrium requires the resultant force to vanish in every independent direction
A particle is in equilibrium when ΣF=0. Resolve horizontally and vertically (or along chosen axes), giving one scalar equation per independent direction.
Include all external forces, use geometry to express angles, and solve the component equations together. A zero horizontal resultant alone does not ensure equilibrium.
A weight supported by two symmetric strings has equal tensions; horizontal components cancel and vertical components sum to the weight.
Equilibrium does not mean no forces act; it means their vector sum is zero.
Split one rough contact into normal and friction components
The force exerted by a rough surface on a particle is represented by two perpendicular components: normal reaction R perpendicular to the surface and friction F parallel to the surface.
The normal component pushes away from the surface. Friction opposes actual relative motion or the tendency/impending relative motion between the contacting surfaces.
On an incline, draw R perpendicular to the plane and F along it. Weight remains vertical; it is not one of the contact components.
Ifneeded,thesingleresultantcontactforcehasmagnitudeC=\sqrt{R^2+F^2},butequilibriumisusuallysolvedusingtheseparatecomponents.
Do not assume F=μR here. That equality belongs only to limiting friction in objective 6.
Use a smooth contact as a zero-friction idealisation
A smooth surface exerts only a normal reaction perpendicular to the surface: the tangential/friction component is modelled as zero.
Remove friction from the free-body diagram, keep weight and all other forces, choose axes along/perpendicular to the surface, and solve with the remaining reaction.
For a particle on a smooth incline, R is perpendicular to the plane and the component mgsinθ acts down the plane; the model predicts no contact resistance along the plane.
Real surfaces generally have friction. The smooth model is unsuitable when tangential resistance, sticking, impending slip or energy loss materially affects the result.
Smooth means frictionless contact, not force-free contact: the normal reaction remains.
Decide whether friction adjusts or has reached its limit
The coefficient of friction is $\mu=F_{\max}/R$. Static friction satisfies0\le F\le\mu R.Inlimitingequilibrium(“abouttoslip”),F=\mu R.
Identify the impending motion, draw friction opposite that tendency, resolve equilibrium to find F and R, then use equality only if the wording indicates limiting equilibrium; otherwise verify F≤μR.
For a particle about to slide down a rough plane of angle $\theta$:R=mg\cos\theta,\quad F=mg\sin\theta=\mu R,so $\mu=\tan\theta$.
“About to slip”, “on the point of moving” and “limiting equilibrium” all signal maximum static friction. If the tendency reverses, friction direction reverses.
In ordinary static equilibrium friction takes the value needed up to the limit; it is not always μR.
Pair Newton-third-law forces across two bodies
If body A exerts a force on body B, then B simultaneously exerts an equal-magnitude, opposite-direction force of the same interaction type on A.
A valid third-law pair has:
| Test | Requirement |
|---|---|
| bodies | forces act on different bodies |
| interaction | same pair of interacting bodies/type |
| size/direction | equal magnitude, opposite direction |
| timing | simultaneous |
The ground pushes upward on a particle with normal reaction R; the particle pushes downward on the ground with force R. These arrows belong on different free-body diagrams.
Weight is Earth pulling the particle; its partner is the particle pulling Earth. The normal reaction is therefore not the third-law partner of weight.
Third-law partners cannot cancel in one particle’s equilibrium equation because they do not act on the same particle. Newton’s first/second laws are outside this exact objective.