Describe an experiment to demonstrate that there is no resultant moment on an object in equilibrium

1.5.2 Turning Effect of Forces

Objective

Design and carry out an experiment that demonstrates that an object in static equilibrium experiences no resultant moment (net torque) about any axis.

Key Concepts

• Moment (torque) – a measure of the turning effect of a force

  The moment of a force about a chosen pivot quantifies how strongly the force tends to rotate the body about that point.

  General equation:
  M = F\,d\sin\theta

  • F – magnitude of the force (N)
  • d – perpendicular distance from the line of action of the force to the pivot (m)
  • \theta – angle between the force direction and the lever arm (°)
  • When the force is perpendicular to the lever arm, \theta = 90° and \sin\theta = 1, giving the familiar form M = Fd.
  • Unit: newton‑metre (N·m).
  • Moment is a vector; its direction is given by the right‑hand rule (out of the page for anticlockwise, into the page for clockwise).
• Everyday examples
  • Turning a bolt with a wrench – a longer wrench (larger d) produces a larger turning effect for the same applied force.
  • A seesaw balances when the product of each child's weight and its distance from the fulcrum is equal on both sides (W_1 d_1 = W_2 d_2).
• Sign convention for moments

  Choose one convention and apply it consistently throughout the experiment and calculations:

  • Clockwise = positive, anticlockwise = negative, or
  • Anticlockwise = positive, clockwise = negative.
• Conditions for static equilibrium (syllabus requirement) \[ \sum \mathbf{F}=0 \qquad\text{and}\qquad \sum M=0 \]
  • The vector sum of all external forces must be zero.
  • The algebraic sum of all moments about any chosen point must be zero.
• Factors that affect the magnitude of a moment (required by the syllabus)
  1. Magnitude of the force (F).
  2. Perpendicular distance (lever arm) from the pivot (d).
  3. Angle (\theta) between the force direction and the lever arm; the effective moment is F d \sin\theta. In this experiment \theta = 90°, so \sin\theta = 1.

Apparatus

Rigid wooden board (≈ 0.5 m long)	Low‑friction central pivot (fulcrum)
Clamp stand (optional, to hold the board horizontally when not rotating)	Two identical strings with hooks
Set of calibrated masses (e.g., 50 g, 100 g, 200 g)	Meter rule or measuring tape
Safety glasses

Safety Notes

• Check that the pivot turns smoothly; a sticky pivot can produce a spurious resisting torque.
• Secure all masses to the strings before releasing the board to avoid sudden drops.
• Wear safety glasses in case a mass detaches.
• Do not stand directly above the board when it is released.

Experimental Arrangement

The board rests on a central, low‑friction pivot so it can rotate freely about its centre. Two strings are attached to the board at known distances d₁ (left side) and d₂ (right side) from the pivot. Hanging masses create vertical forces at those points.

Method

1. Measure and record the perpendicular distances d₁ and d₂ from the centre of the pivot to the attachment points of the left and right strings.
2. Hang a known mass m₁ on the left string. Release the board and observe the direction of rotation (if any).
3. Gradually add mass to the right string until the board remains perfectly horizontal and shows no tendency to rotate. Record the balancing mass m₂.
4. Repeat steps 1–3 for at least three different sets of distances (e.g., d₁=0.10 m, d₂=0.20 m; d₁=0.15 m, d₂=0.25 m; d₁=0.12 m, d₂=0.30 m).
5. For each trial calculate the clockwise and anticlockwise moments and verify that \sum M = 0 within experimental uncertainty.

Data Table

Trial	d₁ (m)	d₂ (m)	m₁ (kg)	m₂ (kg)	Moment left M₁ = m₁ g d₁ (N·m)	Moment right M₂ = m₂ g d₂ (N·m)	\sum M (N·m)
1	0.10	0.20	0.200	0.100	0.200 × 9.8 × 0.10 = 0.196	0.100 × 9.8 × 0.20 = 0.196	≈ 0
2	0.15	0.25	0.150	0.090	0.150 × 9.8 × 0.15 = 0.221	0.090 × 9.8 × 0.25 = 0.221	≈ 0
3	0.12	0.30	0.250	0.100	0.250 × 9.8 × 0.12 = 0.294	0.100 × 9.8 × 0.30 = 0.294	≈ 0

Theoretical Explanation

When the board is in equilibrium the clockwise and anticlockwise moments balance:

\[ M_{\text{left}} = m_1 g d_1,\qquad M_{\text{right}} = m_2 g d_2 \]

Taking clockwise as positive, the condition \sum M = 0 gives

\[ m_1 g d_1 - m_2 g d_2 = 0 \;\Longrightarrow\; m_1 d_1 = m_2 d_2 \]

Thus, for any pair of vertical forces the product of the force magnitude and its perpendicular lever arm must be equal for the net moment to vanish. The pivot supplies a vertical reaction force, so the separate condition \sum \mathbf{F}=0 is also satisfied.

Sources of Error

• Friction at the pivot introduces a small resisting torque.
• Inaccurate measurement of the distances d₁ and d₂.
• Masses not centred on the strings, altering the effective lever arm.
• Air currents, vibrations, or elasticity of the strings.
• Parallax error when reading the scale on the masses.

Conclusion

The experiment confirms the Cambridge IGCSE (0625) principle that an object in static equilibrium experiences no resultant moment. By adjusting the masses until m₁ d₁ = m₂ d₂, the clockwise and anticlockwise torques cancel, giving \sum M = 0. Simultaneously, the support at the pivot provides an equal and opposite reaction, ensuring \sum \mathbf{F}=0. This demonstrates that both the net force and the net moment must vanish for true static equilibrium.