In simple terms
A friendly intro before the formal notes — no formulas yet.
Equilibrium of Forces
Master equilibrium for Cambridge 9702 Paper 2: zero resultant force, zero resultant torque, principle of moments, couples, and the triangle of forces — with Senpai Corner diagrams and live animations.
- 1
Resultant force = 0 — no net push or pull (translational equilibrium).
- 2
Resultant moment = 0 — no net turning effect about any pivot (rotational equilibrium).
- 3
Both must hold at the same time for complete equilibrium.
What this topic covers
The official Cambridge syllabus points this lesson works through.
- 4.2.1
State and apply the principle of moments
- 4.2.2
Understand that, when there is no resultant force and no resultant torque, a system is in equilibrium
- 4.2.3
Use a vector triangle to represent coplanar forces in equilibrium
Explore the concept
Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.
Step-synced diagram — highlights what to look for in the simulation above.
Step 1
Resultant force = 0 — no net push or pull (translational equilibrium).
24 more simulations for this topic — run them in the Simulations section below
Simulations
Every simulation here runs the real model — try the steps on a card, then check what you see against the notes.
24 simulations · 5 to start with
Start herein this order — each one shows a different piece of the topic
- The Physics ClassroomStart here · 19702 4.2 · IB A.2
Equilibrium
Drag out a force vector to balance the given one, then level up to balancing two and then three force vectors
Why this one: The balancing force is the resultant of the others reversed; level up from one force to three.
Try this
- Balance a single given force vector.
- Level up and balance two force vectors.
- Try three force vectors.
Look for The balancing force equals the resultant of the given forces reversed.
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
- The Physics ClassroomStart here · 29702 4.2 · IB A.2
Go for the Gold!
Add gold coins to a bag hanging from ropes without breaking them; the rope angles set the tension, and levels get harder
Why this one: Flatter ropes mean bigger tensions for the same weight: resolve vertically to see why they snap.
Try this
- Add coins one at a time and watch the tension.
- Note the rope angles before adding more.
- Progress to the next level.
Look for Rope tension rises as the ropes get closer to horizontal.
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
- 3JCN PhysicsStart here · 39702 4.1 · 9702 4.2 · IB A.4
Torque - Balancing Act
Hang masses on a beam at chosen positions to balance it about a pivot
Why this one: Zero resultant force is not enough; moments about the pivot must also cancel.
Try this
- Hang one mass on each side at equal distances from the pivot and check balance.
- Double one mass and move it to half the distance; check balance again.
- Hang two masses on one side and balance them with one mass on the other.
Look for The beam balances when the clockwise and anticlockwise moments about the pivot are equal.
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
- The Physics ClassroomStart here · 49702 4.2 · 9702 4.1 · IB A.4
Static Equilibrium
A rod leans against a wall; control the lengths, masses and angles and see how each affects the forces
Why this one: A leaning rod needs both force and moment balance; change the angle and watch the wall force.
Try this
- Change the angle and read the force from the wall.
- Change the mass and compare the forces.
- Change the length and compare.
Look for In equilibrium the forces and the moments about any point both sum to zero.
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
- The Physics ClassroomStart here · 59702 4.2 · 9702 1.4 · IB A.2
Balance It!
Balance an angled force with forces that lie along the x- and y-axes; the Help page walks through resolving it into components
Why this one: Resolve an angled force into components and cancel each with an axis force.
Try this
- Resolve the angled force into its x- and y-components.
- Add an x-axis force equal and opposite to the x-component.
- Add a y-axis force to cancel the y-component and check the balance.
Look for An angled force is balanced when the axis forces equal and oppose its x- and y-components.
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
More simulations19 more on this topic — core ones first
- PhETCoreJava · best on a laptop9702 5.1 · 3.1 · 4.2 · IB A.3 · A.2
The Ramp
Push objects up a ramp; vary the angle, friction and mass and read the work done, energy bar charts and force graphs.
Try this
- Push a crate to the top at 15° with friction off — read the work done and the GPE gained.
- Turn friction on and repeat — the extra work shows up as thermal energy.
- Steepen the ramp — more force per metre, but the same GPE at the top.
Look for Work done = gain in GPE + energy lost to friction; the ramp trades force for distance.
Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)
- oPhysicsCore9702 3.1 · 9702 4.2 · IB A.2
Friction: Pulling a Box on a Horizontal Surface
Pull a box with a rope at adjustable tension/angle; explore static vs kinetic friction and normal force
Try this
- Increase the rope tension slowly from zero until the box starts to move.
- Raise the rope angle and note the normal force.
- Compare friction just before and just after the box starts moving.
Look for Static friction grows to match the pull up to a limit, then drops to a smaller kinetic value; an upward rope angle reduces the normal force.
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
- oPhysicsCore9702 3.1 · 9702 4.2 · IB A.2
Static and Kinetic Friction on an Inclined Plane
Vary incline angle (0-90) and friction coefficients; see weight, normal and friction vectors and the motion
Try this
- Raise the incline angle slowly until the block begins to slide.
- Change the static coefficient and find the new slipping angle.
- Set the incline near 90° and look at the normal force vector.
Look for The block slips when tan θ reaches the static coefficient, and the normal force shrinks to zero as the incline approaches vertical.
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
- oPhysicsCore9702 4.1 · 9702 4.2 · IB A.4
Equilibrium Problem: Bar with Axis Supported by a Cable
Uniform bar hinged at one end, held by a cable, with a movable mass; find cable tension and hinge force
Try this
- Slide the mass along the bar and read the cable tension.
- Move the mass to the far end.
- Compare the hinge force at both positions.
Look for Taking moments about the hinge, the cable tension rises as the mass moves outward.
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
- oPhysicsCore9702 4.1 · 9702 4.2 · IB A.2
Stability, Equilibrium, and Center of Mass
Tilt objects to explore stability, equilibrium and where the centre of mass must sit over the base
Try this
- Tilt an object until it topples.
- Try an object with a wider base.
- Lower the centre of mass and tilt again.
Look for An object topples once its centre of mass passes outside its base; a wider base or lower centre of mass needs a larger tilt.
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
- The Physics ClassroomCore9702 3.1 · 9702 4.2 · 9702 5.1
Inclined Plane Simulation
Set the angle, mass, initial velocity and coefficients of static and kinetic friction; observe the forces, motion and energy changes
Try this
- Set both friction coefficients to zero and vary the angle.
- Add kinetic friction and compare the acceleration.
- Increase the mass and compare the acceleration.
Look for Acceleration down a slope is g(sin θ − μ cos θ), independent of mass.
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
Key formulas
Tap any symbol to reveal exactly what it means and its units.
Tap a symbol — great for exam definitions
Tap a symbol — great for exam definitions
Tap a symbol — great for exam definitions
Full topic notes
Formal explanation with the rigour you need for the exam.
Two conditions for equilibrium
Resultant force = 0 — no net push or pull (translational equilibrium).
Resultant moment = 0 — no net turning effect about any pivot (rotational equilibrium).
Both must hold at the same time for complete equilibrium.
Moments — the turning effect of a force
A moment is the turning effect of a force about a pivot. Only the perpendicular distance from the pivot to the line of action counts. On a spanner, if the force is at angle θ to the handle, use the perpendicular component: .
Reference diagram (Senpai Corner): perpendicular force on a beam gives . Angled force on a spanner gives . See the moments reference image in resources — the live animation above shows a balanced beam.
Principle of moments
For a body in rotational equilibrium, the total clockwise moment about any chosen pivot equals the total anticlockwise moment about that same pivot.
Choose a pivot through an unknown force to eliminate it from the moment equation.
Include the weight of uniform beams at their centre (half-length from either end).
Units: moment in N m; never confuse with energy (N m = J only when work is done).
Couples and torque
A couple is two equal, parallel, opposite forces acting along different lines. They produce rotation only — no resultant force. Examples: turning a steering wheel, using a screwdriver.
Triangle of forces (coplanar forces)
When three coplanar forces act on an object in translational equilibrium, draw them tip-to-tail as vectors. If they form a closed triangle, the resultant is zero. This is the graphical test for equilibrium in one plane.
Use the live diagram (Triangle of forces tab) to see vectors a, b and c closing tip-to-tail. The Senpai Corner reference below matches this layout.
Centre of gravity
The centre of gravity is the single point where the entire weight of an object may be considered to act. For a uniform object in a uniform gravitational field, it coincides with the geometric centre — place the weight there when modelling beams and rods.
Worked examples
See the formulas applied — reveal one step at a time, like the exam.
A uniform beam of length 4.0 m and weight 80 N rests on supports at each end. A 120 N load is placed 1.5 m from the left support. Find the support forces.
- 1
Forces: Let left support = , right support = . Up = down: N.
A uniform ladder of length 5.0 m and weight 200 N leans against a smooth vertical wall at an angle of 60° to the rough horizontal ground. Find the reaction forces from the wall and ground, and the frictional force.
- 1
Diagram & Forces: Draw the ladder. Forces are: Weight (W=200 N) acting down from the centre (2.5 m), ground normal reaction () up, ground friction () horizontally inwards, wall normal reaction () horizontally outwards.
How it all connects
The big idea sits in the middle — tap a linked idea to explore the link.
Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.
Glossary
Key terms for this topic — skim now; the Check step will test them.
- moment
A moment is the turning effect of a force about a pivot. Only the perpendicular distance from the pivot to the line of action counts.
- rotational equilibrium
For a body in rotational equilibrium, the total clockwise moment about any chosen pivot equals the total anticlockwise moment about that same pivot.
- couple
A couple is two equal, parallel, opposite forces acting along different lines.
- different lines
A couple is two equal, parallel, opposite forces acting along different lines. They produce rotation only — no resultant force.
- rotation only
No resultant force. Examples: turning a steering wheel, using a screwdriver.
- closed triangle
If they form a closed triangle, the resultant is zero. This is the graphical test for equilibrium in one plane.
- live diagram
Use the live diagram (Triangle of forces tab) to see vectors a, b and c closing tip-to-tail.
- centre of gravity
The centre of gravity is the single point where the entire weight of an object may be considered to act.
Quick check
Write your answer first, then compare it with the model one — the gap is what you would have lost.
Teach it back
If you can explain it simply, you own it — gaps here are marks you’d lose.
Teach it back
Explain this topic as if teaching a friend. We name the gaps an examiner would still dock.
Revision flashcards
Guess first, then flip — retrieval beats re-reading.
Key takeaways
Review these before you close the topic — retrieval beats re-reading.
Resultant force = 0 — no net push or pull (translational equilibrium).
Resultant moment = 0 — no net turning effect about any pivot (rotational equilibrium).
Both must hold at the same time for complete equilibrium.
Practice — then mark it
The whole point: a real Cambridge question, marked mark-by-mark.
State the conditions necessary for an object to be in equilibrium.
The trapdoor is in equilibrium when F is 1.7N.
Calculate d.
d = .............................................................. m
Extra simulations & links
PhET, GeoGebra and other curated tools — open in a new tab.
Frequently asked
Checkpoint
One marked question is worth ten re-reads — close the loop before you move on.
Reading it isn’t knowing it — prove it.
Before you move on: do 9702/22 · Q3(b) on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.
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