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Solving equilibrium problems

Draw the forces, combine force and moment balance, and check whether hinges, cables and supports can provide the forces you calculate.

  • GCE A-Level H2 Physics 2027

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Two checks for a body held at rest

A rigid body in static equilibrium has zero resultant force and zero resultant moment. In a plane, write

∑ Fₓ = 0, ∑ Fy = 0, ∑ M = 0.

The force equations prevent acceleration of its centre of mass. The moment equation prevents angular acceleration. A couple shows why force balance alone is insufficient.

The principle of moments states that, for equilibrium, the total clockwise moment equals the total anticlockwise moment about any point. Choose one sign convention, such as anticlockwise positive, and keep it throughout the calculation.

Start with the physical supports

Draw only external forces acting on the chosen body. Put weight through its centre of gravity; for a uniform beam in a uniform gravitational field, this is the midpoint.

SupportWhat it can provide in the model
Taut cableA tension pulling along the cable; it cannot push.
Unattached support beneath a beamAn upward contact force; it cannot pull the beam downward.
Frictionless hingeHorizontal and vertical force components, but no reaction moment.
Fixed clampA support force and a reaction moment.

A negative answer for an assumed upward reaction means a downward force is needed. Check whether the stated support can supply it. Changing the arrow on paper cannot make an unattached support pull.

Choose a point that removes an unknown

Take moments about a point on an unknown force’s line of action. Its moment is zero there. After finding another force from moments, return to force balance to find the remaining support force.

Changing the reference point changes individual moments. For a body with zero resultant force, the total moment is the same about every point, so a correct equilibrium calculation gives the same physical result whichever point you choose.

A beam held by an angled cable

A uniform 3.00 m horizontal beam has weight 2.40 × 10² N. A frictionless hinge is at its left end and a cable at its right end pulls upward and left at 30.0° to the horizontal.

Forces and force lines on a beam held by a cableA uniform 3 metre horizontal beam is hinged at its left end and has a cable at its right end pulling up and left at 30 degrees to the horizontal. Weight W acts downward at the midpoint. The hinge force H acts up and right; its dashed components are Hx right and Hy up. Extended lines of H, T and W meet at C.A horizontal beam in equilibriumDashed lines extend force directionsHingeHTCWHₓHᵧ30°1.50 m1.50 m
Scroll across the figure to read all labels.
The hinge supplies one force H, shown with its horizontal and vertical components. T, H and W are the three external forces. Their lines meet at C in this equilibrium model; taking moments about the hinge removes H. Force arrows are schematic.

First take moments about the hinge. The horizontal tension component acts along the beam and has zero moment there; the vertical component supplies the anticlockwise moment:

T sin 30.0°(3.00) = 240(1.50), T = 240 N.

Then take right and up as positive:

Hₓ-T cos 30.0° = 0, Hy + T sin 30.0°-240 = 0.

Thus Hₓ ≈ 208 N right and Hy = 120 N up. The force vector H has these two components; they are not two separate interactions. The vertical support components add to the weight, and their moment balance holds independently.

Three-force equilibrium

For exactly three non-parallel forces on a rigid body in equilibrium, their vectors form a closed head-to-tail triangle and their lines of action meet at one point. These are different checks: the triangle checks magnitudes and directions; concurrency checks where the forces act.

To see why the lines must meet, take moments about the intersection of two force lines. Both have zero moment there, so the third must pass through that same point. In the beam figure, the three external forces are vector H, vector T and vector W; their lines meet at C.

Worked examples and independent practice

Guided practice 1

A fixed support supplies a force and a moment

About 4 min

Problem

A uniform 2.0 m beam of weight 40 N is held horizontally by a fixed support at its left end. A 60 N load hangs from the right end. Find the upward support force and the reaction moment needed for equilibrium.

Try this before viewing the solution

Unit: N
Unit: N m

Hints

Hint 1: balance vertical forces
The two downward forces total 40 + 60 N.
Show solution step by step

  1. List the vertical forces

    Method

    The fixed support reaction acts upward; beam weight and load act downward.

    Reason

    The beam is in translational equilibrium.

    Working

    ∑ Fy = R-40-60

  2. Apply vertical equilibrium

    Method

    R = 100 N upward.

    Reason

    Zero vertical resultant requires the support force to balance both downward loads.

    Working

    R-40-60 = 0 ⇒ R = 100 N

  3. Keep rotational equilibrium distinct

    Method

    The fixed support must also provide a reaction moment.

    Reason

    A single upward force at the left end cannot by itself balance the clockwise moments of the two loads.

    Working

    Mₛᵤₚₚₒᵣₜ = 40(1.0) + 60(2.0) = 160 N m anticlockwise.

Exam-style question 1

Beam supported at both ends (find reactions)

5 marks

Examination question

A uniform horizontal 3.0 m beam of weight 30 N is supported at both ends. A 60 N load hangs 2.0 m from the left end. Find the upward reactions RA and RB. [5 marks]

Try this before viewing the solution

Unit: N
Unit: N

Show solution step by step

  1. Write vertical force equilibrium

    1 mark

    Method

    RA + RB = 90 N.

    Reason

    The reactions balance the beam and added-load weights.

    Working

    RA + RB = 30 + 60

  2. Take moments about the left support

    2 marks

    Method

    3.0RB = 165 N m.

    Reason

    The left reaction has zero lever arm; the uniform beam’s weight acts at 1.5 m.

    Working

    RB(3.0) = 30(1.5) + 60(2.0) = 165

  3. Calculate the right reaction

    1 mark

    Method

    RB = 55 N.

    Reason

    Divide the balancing moment by the 3.0 m support separation.

    Working

    RB = 165/3.0 = 55 N

  4. Calculate the left reaction

    1 mark

    Method

    RA = 35 N.

    Reason

    The two reactions must sum to 90 N.

    Working

    RA = 90-55 = 35 N

A support that cannot pull

A uniform 4.00 m horizontal beam of weight 2.00 × 10² N rests on unattached supports at its left end and 3.00 m from that end. A 5.00 × 10² N downward load acts at the right end. Can it stay at rest? Calculate the support forces before deciding.

Show answer

Moments about the left support give 3.00R_right = 200(2.00) + 500(4.00), so R_right = 800 N upward. Vertical balance would then require R_left = 700-800 = -100 N with upward positive. The left support cannot supply the required downward force. The beam tends to lift away from it and turn about the right support.

Check your understanding 1: Choosing another point for moments

In the two-support worked example, take moments about the right support instead. Explain why the right reaction drops out and check the left reaction.

Show answer

The right reaction passes through the chosen point, so it has zero moment. The beam’s weight is 1.5 m left of this point and the load is 1.0 m left. They turn anticlockwise; the left reaction turns clockwise:

3.0RA = 30(1.5) + 60(1.0) = 105, RA = 35 N.

This agrees with the original calculation. Vertical balance then gives RB = 55 N.

Before you move on

Revisit Moments and couples if choosing the perpendicular distance is still difficult.

Syllabus and review details