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\thetaF– 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 formM = 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 (
W1 d1 = W2 d2).
- Turning a bolt with a wrench – a longer wrench (larger
- 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)
- Magnitude of the force (
F). - Perpendicular distance (lever arm) from the pivot (
d). - Angle (
\theta) between the force direction and the lever arm; the effective moment isF d \sin\theta. In this experiment\theta = 90°, so\sin\theta = 1.
- Magnitude of the force (
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.
d₁ and d₂ from the pivot.
Method
- Measure and record the perpendicular distances
d₁andd₂from the centre of the pivot to the attachment points of the left and right strings. - Hang a known mass
m₁on the left string. Release the board and observe the direction of rotation (if any). - Gradually add mass to the right string until the board remains perfectly horizontal and shows no tendency to rotate. Record the balancing mass
m₂. - 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). - For each trial calculate the clockwise and anticlockwise moments and verify that
\sum M = 0within 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:
\[
M{\text{left}} = m1 g d_1,\qquad
M{\text{right}} = m2 g d_2
\]
Taking clockwise as positive, the condition \sum M = 0 gives
\[
m1 g d1 - m2 g d2 = 0 \;\Longrightarrow\; m1 d1 = m2 d2
\]
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
d₁andd₂. - 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.