Translational and rotational equilibrium
Key idea: A complete H2 Physics lesson on field and contact forces, Hooke’s law, moments, couples, centre of gravity and equilibrium.
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The core idea
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Big question: What must be true for a body to have neither translational nor rotational acceleration?
Complete equilibrium requires both zero resultant force and zero resultant torque. A free-body diagram supplies the forces; resolving components gives translational equations, while moments about a convenient point give rotational balance. For three non-parallel forces, a closed vector triangle represents zero resultant.
Force balance and torque balance answer different questions
ΣF = 0 rules out linear acceleration of the centre of mass. Στ = 0 rules out angular acceleration. A pair of equal opposite forces on different lines satisfies the first condition but forms a couple, so the second condition is essential.
Check your understanding: Can a body in equilibrium be moving?
Yes. It may move with constant velocity and constant angular velocity because equilibrium means no acceleration, not necessarily no motion.
A closed force triangle is a vector proof
For exactly three non-parallel forces in equilibrium, draw the force vectors head to tail in their true directions. They must close. The side lengths give force magnitudes and the angles come from the force directions, not automatically from the object’s shape.
The triangle proves translational balance. For an extended body you must still consider whether the three lines of action are concurrent or whether a remaining torque exists.
Check your understanding: What does a gap in a head-to-tail force polygon represent?
The gap vector is the non-zero resultant force.
Key ideas to keep
- Zero resultant force alone does not prevent rotation.
- Choose a pivot that removes unknown forces from the moment equation where possible.
- After moments, use force balance to recover any remaining reaction.
See the reasoning
Worked example
Beam held by an angled cable
Question: A 3.0 m uniform horizontal beam of weight 120 N is hinged at the left. A cable at the right end acts 30° above the beam towards the hinge. Find the cable tension and the hinge-force components.
Step 1: Use torque balance first
Why: Taking moments about the hinge removes both unknown hinge-force components.
Working: (T sin30°)(3.0) = 120(1.5), so T = 120 N.
Step 2: Balance horizontal forces
Why: The cable pulls left with component T cos30°, so the hinge must supply an equal rightward component.
Working: Hₓ = 120 cos30° = 104 N right.
Step 3: Balance vertical forces
Why: The cable supplies 60 N upward, leaving the hinge to support the rest of the beam’s weight.
Working: Hᵧ + 120 sin30° − 120 = 0, so Hᵧ = 60 N up.
Answer: Cable tension = 120 N; hinge force components are 104 N right and 60 N upward.
Check: The vertical support components add to 120 N and the cable’s vertical component produces exactly the beam-weight moment.
Another worked model
Question
A 100 N uniform horizontal board is supported at both ends, 4.0 m apart. A 300 N load is 1.0 m from the left end. Find both reactions.
Check the worked solution
Weight of the board acts at its centre. About the left end: Rright(4.0) = 300(1.0) + 100(2.0), so Rright = 125 N. Vertical equilibrium gives Rleft = 400 − 125 = 275 N.
Use a hint if needed
Practise with support
Try this
A block is fully submerged and held stationary by a vertical string. Draw and label its forces, then write the vertical equilibrium equation.
Hint: Include weight, upthrust and string tension; choose upward positive.
Check your answer
Weight acts downward; upthrust and string tension act upward if the string supports the block. Vertical equilibrium gives upthrust + tension − weight = 0.
Now work without the hint
Practise independently
Your turn
A 500 N sign hangs from two cables making 30° and 50° above the horizontal. State the vector-equilibrium equations needed to find both tensions.
Check your answer
With tensions T30 and T50: horizontal equilibrium gives T30 cos 30° = T50 cos 50°. Vertical equilibrium gives T30 sin 30° + T50 sin 50° = 500 N. These equations correspond to a closed vector triangle.
Avoid these traps
Common mistakes
Common mistake
Zero resultant force alone proves equilibrium.
What is wrong with this reasoning?
Show better thinking
Complete equilibrium requires both zero resultant force and zero resultant torque. A free-body diagram and a consistent moment equation must satisfy both.
Write for the examiner
Exam guidance
Show the force diagram, moment sign convention and both equilibrium conditions explicitly.
Exam-style practice [6 marks]
A uniform 4.0 m beam of weight 200 N is supported vertically at both ends. A 300 N load is 1.0 m from the left support. Find both support forces and state the two equilibrium conditions used.
Plan before you answer
- Take moments about one support.
- Use vertical force balance.
- State both conditions in words or symbols.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
Taking moments about the left support gives 4.0Rright = 200(2.0) + 300(1.0), so Rright = 175 N. From vertical balance, Rleft + 175 = 500, so Rleft = 325 N. The calculation uses ΣF = 0 and Στ = 0.
Come back in three days
Check what stayed with you
Recall question 1
State the two conditions for complete equilibrium.
Check the answer
Resultant force is zero and resultant torque about any point is zero.
Recall question 2
Why is a pivot through an unknown support often useful?
Check the answer
That support has zero moment about the pivot and drops out of the torque equation.
Recall question 3
When do three equilibrium forces form a closed triangle?
Check the answer
When the three forces are non-parallel and their vector sum is zero.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. For friction and viscous force, qualitative treatment is required; coefficients of friction and viscosity are excluded.
- GCE A-Level H2 PhysicsTopic 2(h) / Topic 2(i) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027