1. Title and aim
Principle of moments and rotational equilibrium of parallel forces. Verify the principle of moments for parallel coplanar forces acting on a rigid beam, demonstrate that rotational equilibrium requires the algebraic sum of clockwise and counter-clockwise torques to vanish about an off-center pivot axis, and determine the unknown mass and center of gravity of a uniform meter rule using ordinary least-squares regression.
2. Abstract
Five observations were generated from the ideal teaching model over the settings listed below. A straight-line fit gave an estimate of meter rule mass of 0.1200 kg, differing by 0.000% from the model value. The reported scatter comes from scale rounding and any stated model approximation, not measured laboratory noise.
3. Apparatus and fixed settings
- Hardwood meter rule: A seasoned, uniform hardwood meter rule of total length 100.0 cm, having a rectangular cross-section, sharp millimeter graduations along both top and bottom edges, and a verified center of gravity at x_cg = 50.00 cm.
- Wedge fulcrum and heavy retort stand: A hardened tool-steel knife-edge prism clamped horizontally atop a heavy cast-iron laboratory stand fitted with leveling foot-screws, providing a frictionless line contact at the fixed graduation mark x_0 = 35.0 cm.
- Slotted brass masses and suspension loop: A calibrated set of precision brass masses (60.0 g to 150.0 g) suspended from a lightweight wire stirrup with a fine nylon loop that slides smoothly along the meter rule arm to achieve exact rotational balance.
- Miniature spirit level and analytical balance: A tubular glass spirit level placed along the center axis of the meter rule to indicate true horizontal alignment, and an electronic laboratory balance (0.01 g precision) used to determine the independent reference comparison mass.
- Hardwood meter rule length L = 100.0 cm; fulcrum pivot fixed at knife-edge position x_0 = 35.0 cm.
- Rule center of gravity x_cg = 50.0 cm; fixed moment arm d_rule = x_cg - x_0 = 15.0 cm (0.150 m).
- Reference rule mass M_rule = 0.1200 kg (120.0 g); local gravity g = 9.80665 m s⁻².
4. Theory and graph choice
When a rigid body is subjected to a system of coplanar parallel forces, static mechanical equilibrium requires two independent and simultaneous criteria: translational equilibrium, where the vector sum of all acting forces vanishes (Σ F_y = 0), and rotational equilibrium, where the algebraic sum of the moments of all forces about any arbitrary pivot axis perpendicular to the plane of action equals zero (Σ τ_O = 0). In this classical mechanics investigation, a uniform hardwood meter rule of unknown mass M_rule is supported on a low-friction steel knife-edge fulcrum positioned deliberately away from its center of gravity (fixed at x_0 = 35.0 cm, while its center of gravity lies at x_cg = 50.0 cm). This asymmetric mounting creates an unbalanced clockwise gravitational torque due to the distributed self-weight of the ruler, equivalent to a concentrated force M_rule · g acting at an effective moment arm d_rule = x_cg - x_0 = 15.0 cm (0.150 m). To restore perfect horizontal equilibrium without external clamping, a standardized brass mass m is suspended from the shorter arm of the rule at an adjustable distance d_m on the opposite side of the fulcrum. Under balanced horizontal conditions, the counter-clockwise torque produced by the suspended load exactly neutralizes the clockwise torque of the ruler's self-weight: m · g · d_m = M_rule · g · d_rule. Eliminating the local acceleration of gravity g leads directly to the hyperbolic governing relation d_m = (M_rule · d_rule) · (1/m). By measuring the precise balancing distance d_m for five distinct standardized masses m, and plotting d_m against the reciprocal mass 1/m in SI units, the resulting straight line possesses a slope S = M_rule · d_rule. From this regression slope, the total mass of the meter rule is deduced with high statistical precision as M_rule = S / d_rule, demonstrating how rotational equilibrium can be transformed into an analytical weighing instrument without employing a spring scale or chemical balance.
The calculation fits y = mx + b in SI units. The graph may display scaled axes for readability; those display factors are removed before calculating the physical result. The gradient has units m kg.
5. Method
- Place the unweighted hardwood meter rule horizontally across the steel knife-edge fulcrum and adjust its position until it rocks gently and balances in a steady horizontal attitude without tipping. Note the graduation mark directly above the knife edge as the center of gravity x_cg (nominally 50.0 cm for a homogeneous rule).
- Shift the knife-edge fulcrum stand and position the meter rule so that graduation mark x_0 = 35.0 cm rests securely upon the prism edge. Confirm that the rule tilts downward on the longer arm due to the unbalanced clockwise gravitational torque exerted by the rule's self-weight.
- Load a total mass m = 60.0 g onto the suspension stirrup and slide its fine thread loop along the shorter arm of the rule (between 0.0 cm and 35.0 cm). Slowly shift the loop until the meter rule levels out horizontally, confirmed by the central spirit level bubble resting centered between its index lines.
- Read the exact position of the suspension thread on the rule's millimeter scale, using an eye position perpendicular to the rule to eliminate optical parallax error. Calculate the balancing moment arm distance d_m = x_0 - x_m. Repeat the leveling check three times to verify mechanical repeatability.
- Systematically increase the suspended load to m = 80.0 g, 100.0 g, 120.0 g, and 150.0 g. At each mass increment, slide the suspension stirrup to find the new horizontal balance position, verify with the spirit level, and record the corresponding balancing distance d_m.
- Convert all masses to kilograms and compute reciprocal mass 1/m (in kg⁻¹). Convert balancing distances to meters. Plot d_m against 1/m, fit an ordinary least-squares regression line, compute the slope S, and divide by d_rule = 0.150 m to deduce the meter rule mass M_rule.
6. Observations and calculation table
All observations use fixed model settings. Values are rounded only at the instrument reading stage; further calculations use full precision. Repeated physical trials are not claimed.
| No. | Balancing mass m (g) | Balancing distance d_m (cm) | x (SI) | y (SI) | Fitted y (SI) | Residual (SI) |
|---|---|---|---|---|---|---|
| 1 | 60.0 | 30.0 | 16.66667 | 0.30000 | 0.30000 | 0.00000 |
| 2 | 80.0 | 22.5 | 12.50000 | 0.22500 | 0.22500 | 2.7756e-17 |
| 3 | 100.0 | 18.0 | 10.00000 | 0.18000 | 0.18000 | 0.00000 |
| 4 | 120.0 | 15.0 | 8.33333 | 0.15000 | 0.15000 | 0.00000 |
| 5 | 150.0 | 12.0 | 6.66667 | 0.12000 | 0.12000 | 0.00000 |
Download the sample observations (CSV) ↓
7. Worked calculation for one observation
8. Graph, gradient and final calculation
Dots show the sample observations. The line is the ordinary least-squares fit.
Σ(x − x̄)² = 61.11111
Σ(x − x̄)(y − ȳ) = 1.10000
m = Σ(x − x̄)(y − ȳ) / Σ(x − x̄)² = 0.01800000
b = ȳ − mx̄ = 0.00000000
Slope S = 0.018000 m kg; M_rule = S / d_rule = 0.018000 / 0.150 m = 0.12000 kg (120.00 g)
9. Uncertainty and comparison
Percentage difference = |estimate − model value| / |model value| × 100 = 0.0000%. Percentage difference is not experimental uncertainty.
For n = 5 observations, Σresidual² = 7.7037e-34; s² = Σresidual²/(n − 2). Gradient standard error = √[s²/Σ(x − x̄)²] = 2.0499e-18. Intercept standard error = √[s²(1/n + x̄²/Σ(x − x̄)²)] = 2.3335e-17.
Propagating only the relevant fit coefficient gives a fit-only standard error of 1.3666e-17 kg. This is a one-standard-error quantity under the regression assumptions. Deterministic rounded model data do not supply a complete physical uncertainty budget.
Reading resolution of the meter scale (±0.5 mm), knife-edge fulcrum width and bearing friction, alignment of the center of gravity relative to the geometric center (wood grain non-uniformity), and hanger string mass contribute uncertainty. The linear regression standard error quantifies random placement scatter, but does not capture zero tare offsets in the slotted weights or ambient air currents acting on the extended meter rule arm.
10. Result and discussion
Model comparison value: 0.1200 kg
The observations follow the stated model relationship. Check whether the intercept and residual pattern suggest a systematic offset or a missing approximation. Agreement with the model does not independently verify its input constants.
11. Precautions and limitations
- Align the knife-edge fulcrum perfectly perpendicular to the rule's longitudinal axis to prevent lateral twisting or bevel friction.
- Ensure the suspension stirrup thread is light, thin, and hangs strictly vertically beneath the rule without brushing against the support stand.
- Verify true horizontal equilibrium using the central spirit level or by observing symmetric damping of small angular oscillations.
- Keep the slotted masses centered symmetrically on the hanger rod so the center of gravity of the load remains directly below the suspension index thread.
The model assumes a perfectly rigid, homogeneous beam, zero knife-edge contact bearing width, frictionless knife-edge pivot, and negligible air buoyancy on the extended arms. Simulated readings reflect a seasoned hardwood meter rule (reference mass M_rule = 0.1200 kg = 120.0 g) with millimeter scale reading resolution of 0.5 mm.
12. Conclusion
The worked example illustrates how to obtain meter rule mass from measurements, graph variables and a fitted coefficient. The estimate is 0.1200 kg. A real experiment must additionally evaluate the apparatus-specific uncertainties and corrections described above.