Free Pearson Edexcel International A Level Physics lesson

Moments, balance and centre of gravity

Where a force acts matters as much as how large it is.

What you’ll learn

  • Calculate moments using perpendicular distance to a force’s line of action.
  • Apply both force and moment equilibrium to extended rigid bodies.
  • Locate and use the centre of gravity in balance and stability problems.

Measure the perpendicular lever arm

A force’s moment describes its turning effect about a specified axis or pivot. Its magnitude is force multiplied by the shortest perpendicular distance from the axis to the force’s line of action. Extend the force arrow into a line if needed, then draw that perpendicular. The distance from the pivot to the point where you hold a tool is only the lever arm when the force is perpendicular to that radius.

Choose clockwise or anticlockwise as positive, and assign a sign to each moment from the rotation it would tend to cause. A force whose line of action passes through the pivot has zero moment about that pivot even if the force is large. An inclined force can be handled by using its perpendicular component F sin θ with the pivot-to-application distance r, giving rF sin θ. Do not combine that component with an already shortened perpendicular distance, which would count the geometry twice.

Moment has units N m. Although a joule also has dimensions N m, moment describes rotation tendency rather than energy transferred; write N m here to preserve the meaning. A longer handle can reduce the required force for the same moment, but the material and pivot still need to withstand the load. For a static analysis the rigid-body approximation assumes deformation is small enough not to change the geometry significantly.

Moment magnitude = Fd⊥ = rF sin θ θ is between r and F

Symbols and units

  • F is force magnitude in N; r is pivot-to-force-point distance in m; θ is their included angle.
  • d⊥ = r sin θ is the perpendicular distance to the line of action in m; moment is measured in N m.

How the equation explains the idea

  1. Only the force component perpendicular to r produces turning, so moment magnitude is r(F sin θ).
  2. Equivalently move sin θ into the lever arm: moment = Fd⊥. A force directed through the pivot has d⊥ = 0 and zero moment even if its magnitude is large.
A longer perpendicular lever arm needs less force

Conditions. The required balancing moment is fixed at 10 N m. Force acts perpendicular to the lever, with d⊥ between 0.2 and 1.0 m. Model F = 10/d⊥ excludes d⊥ = 0.

Perpendicular lever arm / m; Balancing force / N.

At a 0.2 m lever arm, 50 N is needed; at 0.4 m only 25 N is needed. Force varies inversely with lever arm. The curve is not drawn to zero because no finite force through the pivot produces the required moment.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

A spanner at an angle

  1. A force of 30.0 N is applied 0.200 m from a nut at 60.0° to the spanner’s radius.
  2. Perpendicular distance to the line of action is 0.200 sin60.0° = 0.173 m.
  3. Moment = 30.0 × 0.173 = 5.20 N m. Equivalently use the perpendicular force component 26.0 N times 0.200 m.

The moment is 5.20 N m, with its sense set by the diagram.

Keep this idea. Name the pivot and rotational sense, then measure the perpendicular distance to the force’s line of action.

Balance forces and moments on an extended body

For a rigid body at rest, there must be no tendency to accelerate translationally or rotationally. Write both a force balance and a moment balance. In a vertical-load beam problem, upward support forces sum to the downward loads, and clockwise moments sum to anticlockwise moments about any chosen point. Choosing a pivot at one unknown support removes that force from the moment equation because its lever arm is zero.

Draw the full beam and locate all forces before calculating. The weight of a uniform beam acts at its geometric midpoint; for a nonuniform beam use its stated or measured centre of gravity instead. Distances in the moment equation are measured from the chosen pivot, not simply copied from coordinate labels. Include the beam’s own weight unless the problem explicitly neglects it. Once one reaction is found by moments, find the other by force balance and check the result about another pivot.

A calculated negative support reaction requires physical interpretation. An ordinary roller or resting support can push but cannot pull; a negative value may mean the beam would lose that support and tip. A bolted connection can sometimes provide a downward force, but that is a different model and must be stated. Do not claim equilibrium from a moment equation alone: a nonzero total force could still accelerate the entire beam without an initial turning tendency about your selected point.

Rigid-body equilibrium: ΣF = 0 and ΣM = 0 Clockwise moments = anticlockwise moments

Symbols and units

  • ΣF is resultant force in N; ΣM is signed resultant moment about a chosen point in N m.
  • Choose one rotational sense as positive; every lever arm is perpendicular to its force.

How the equation explains the idea

  1. Zero resultant force prevents translational acceleration. Zero resultant moment independently prevents angular acceleration.
  2. For two opposing moments F₁d₁ and F₂d₂, equilibrium gives F₁d₁ − F₂d₂ = 0, hence F₂ = F₁d₁/d₂.
  3. Choosing a pivot through an unknown reaction removes that force’s moment but does not remove it from the force balance.

Worked example

Two supports share an uneven load

  1. A uniform 4.0 m beam weighs 200 N and is supported at both ends. An extra 300 N load sits 1.0 m from the left end.
  2. Taking moments about the left support: Rright × 4.0 = 200 × 2.0 + 300 × 1.0 = 700 N m.
  3. Rright = 175 N. Vertical balance gives Rleft = 200 + 300 − 175 = 325 N.

The reactions are unequal because the loading is not symmetric.

Keep this idea. Locate the forces, choose a useful pivot, solve moments, then check the force balance and whether contacts can supply the result.

Find centre of gravity and judge stability

The centre of gravity is the point through which an extended body’s resultant weight acts in a given gravitational field. In an approximately uniform field it coincides with the centre of mass. Symmetry locates it for a uniform regular object, but it need not lie inside the material: a ring’s centre is in empty space. A nonuniform rod’s weight should not automatically be drawn at its midpoint.

For a rod, move a narrow support until the rod balances horizontally and mark the support position; repeat from both directions and assess the width of the balance region. For a thin lamina, suspend it freely from one point and use a plumb line to mark the vertical below that suspension; repeat at another point. Their intersection locates the centre of gravity, with further suspensions checking consistency. Use secured apparatus and safe edges; for irregular or heavy objects plan handling with supervision. These procedures need actual measurements before a result can be claimed.

A standing object remains stable against slow tipping while the vertical line through its centre of gravity lies inside its support base. At the tipping threshold the line passes through an edge; beyond it, weight has an overturning moment about that edge. Lowering the centre of gravity or widening the base usually increases the tilt needed to reach this threshold. This is a quasi-static tipping model: sliding, deformation, acceleration or external pulls may produce failure earlier and must be considered separately.

For point masses in uniform g: xCG = Σ(mᵢxᵢ)/Σmᵢ At tipping: weight’s line of action passes through the support edge

Symbols and units

  • mᵢ is each point mass in kg; xᵢ is its position along the same axis in m.
  • xCG is centre-of-gravity position in m, for an approximately uniform gravitational field.

How the equation explains the idea

  1. Match the moment of total weight to the sum of individual moments: (Σmᵢ)g xCG = Σ(mᵢgxᵢ).
  2. Cancel g and divide by total mass to obtain xCG = Σ(mᵢxᵢ)/Σmᵢ. A heavier mass pulls the weighted mean closer to its own position.
  3. At the tipping threshold the weight line passes through the support edge; beyond it, weight provides an overturning moment.

Worked example

A rod with a heavy end fitting

  1. Treat a uniform 1.0 m rod of mass 0.60 kg as acting at x = 0.50 m. Add a 0.40 kg fitting at x = 1.00 m.
  2. Balance position from the left end: xCG = (0.60 × 0.50 + 0.40 × 1.00)/(0.60 + 0.40).
  3. xCG = 0.70 m. A support there makes the net weight moment zero.

The combined centre of gravity shifts toward the heavier end fitting.

Keep this idea. Use mass distribution or a balance/suspension measurement to locate weight, and use its line of action to test slow tipping.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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