Translational and rotational equilibrium

Key idea: Complete equilibrium requires no linear acceleration and no angular acceleration. That means both the resultant force and resultant torque are zero.

GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 3 of 3

Check your understanding

By the end of this lesson, you should be able to

Apply the principle of moments to unfamiliar arrangements.
Use both equilibrium conditions rather than force balance alone.
Represent equilibrium with a free-body diagram and, for three forces, a closed vector triangle.

Learn the idea

Big question: What evidence is needed to show that an extended body will neither accelerate nor start rotating?

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.

Scroll diagram horizontally to read all labels.

For exactly three forces on a static point mass, equilibrium means the scaled head-to-tail vector triangle closes.

Key ideas

A body may be moving at constant velocity while in equilibrium.
Any pivot gives the same physical answer, but a helpful pivot reduces algebra.
Angles in a force triangle follow the force directions, not necessarily the shape of the object.

Relationships to know

ΣFₓ = 0
ΣFᵧ = 0
Στ about any point = 0

Follow 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.

Now try it with support

Practise with support

A 3.0 m uniform beam of weight 120 N is hinged at the left and held horizontal by a vertical cable at the right. Find the cable tension.

Hints

The beam’s weight acts 1.5 m from the hinge.
Take moments about the hinge.

View the guided answer

3.0T = 120(1.5), so T = 60 N. Vertical balance then gives an upward hinge force of 60 N.

Your turn

Practise independently

A uniform 4.0 m beam of weight 200 N is supported at both ends. A 300 N load is 1.0 m from the left support. Find both support forces.

Check your answer

Taking moments about the left support: 4.0Rright = 200(2.0) + 300(1.0), so Rright = 175 N. Vertical balance gives Rleft + 175 = 500, hence Rleft = 325 N.

Common mistakes and exam guidance

Watch out for

Stopping after ΣF = 0 even though a couple can still cause angular acceleration.
Drawing forces that the beam exerts on its supports instead of forces acting on the beam.

In an exam

Write the pivot beside your moment equation.
After solving, check the answer with the unused force-balance condition.

Put the ideas together

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.

View the marking points and model answer

Marking points

Uses moment equation about either support.
Includes beam weight at midpoint.
Obtains right support 175 N.
Uses vertical force balance.
Obtains left support 325 N.
States zero resultant force and zero resultant torque.

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.

Finish from memory

Three-question recap

State the two conditions for complete equilibrium.
Check
Resultant force is zero and resultant torque about any point is zero.
Why is a pivot through an unknown support often useful?
Check
That support has zero moment about the pivot and drops out of the torque equation.
When do three equilibrium forces form a closed triangle?
Check
When the three forces are non-parallel and their vector sum is zero.

Try this next

Try an equilibrium question with an angled cable and resolve its tension before taking moments.

Check the whole topic Return to Forces and Moments Review Moments, couples and centre of gravity

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H1 Physics
Edition: GCE A-Level H1 Physics 2027